Instruction: What is the monthly installment?

Response: The Basics of Annuity Theory A series of payments made at equal intervals of time is called an annuity. Common examples are house rents, mortgage payments on homes, and installments payments on automobiles. An annuity where payments are guaranteed to occur for a ﬁxed period of time is called anannuity−certain. For example, mortgage payments on a home. The ﬁxed period of time for which payments are made is called term of the annuity. For example, in the case of a home mortgage a term can be either a 15−year loan or a 30 −year loan. Annuities that are not certain are called contingent annuities. For example, a pension is paid so long as the person survives. That is, regular payments are made as long as the person is alive. Pension is an example of contingent annuity also called life annuity. Unless otherwise indicated, the annuity−certain is the type of annuity we will assume in this book, and the “certain” will be dropped from the name. The interval between annuity payments is called a payment period, often just called a period. In Sections 15 through 21, we consider annuities for which the payment period and the interest conversion period are equal. Also, payments are of level amount, i.e. have the same ﬁxed monetary value for each period. In Sections 22 −29, we will discuss annuities for which payments are made more or less frequently than interest is converted and annuities with varying payments. With level annuities, we will most of the time assume payments of 1 since any other level annuity can be obtained from this by a simple multiplication. 143 144 THE BASICS OF ANNUITY THEORY 15 Present and Accumulated Values of an Annuity-Immediate An annuity under which payments of 1 are made at the end of each period for n periods is called an annuity−immediate or ordinary annuity. The cash stream represented by the annuity can be visualized on a time diagram as shown in Figure 15.1 . Figure 15.1 The ﬁrst arrow shows the beginning of the ﬁrst period, at the end of which the ﬁrst payment is due under the annuity. The second arrow indicates the last payment date −just after the payment has been made. Let idenote the interest rate per period. The present value of the annuity at time 0 will be denoted by ani or simply an. See Figure 15.2. Figure 15.2 Using the equation of value with comparison date at time t= 0, we can write an = ν+ ν2 + ··· + νn. That is, the present value of the annutiy is the sum of the present values of each of the npayments. We recognize the expression on the right −hand side as a geometric progression. Thus, multiplying both sides by ν to obtain νan = ν2 + ν3 + ··· + νn + νn+1. Subtracting this from the previous equation we ﬁnd (1 −ν)an = ν(1 −νn). Hence, an = ν·1 −νn 1 −ν = ν·1 −νn iν = 1 −(1 + i)−n i . (15.1) 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 145 Example 15.1 Calculate the present value of an annuity −immediate of amount $100 paid annually for 5 years at the rate of interest of 9%. Solution. The answer is 100a5 = 1001−(1.09)−5 0.09 ≈388.97 Formula (15.1) can be rewritten as 1 = νn + ian. This last equation is the equation of value at time t= 0 of an investment of $1 for nperiods during which an interest of iis received at the end of each period and is reinvested at the same rate i, and at the end of the nperiods the original investment of $1 is returned. Figure 15.3 describes the time diagram of this transaction. Figure 15.3 Next, we will determine the accumulated value of an annuity−immediate right after thenth payment is made. It is denoted by sn. See Figure 15.4. Figure 15.4 Writing the equation of value at the comparison date t= n we ﬁnd sn = 1 + (1 +i) + (1 +i)2 + ··· + (1 +i)n−1. That is, sn is the sum of the accumulated value of each of the n payments. Using the deﬁnition of sn and the reasoning used to establish the formula for an we can write sn =1 + (1 +i) + (1 +i)2 + ··· + (1 +i)n−1 =(1 + i)n −1 (1 + i) −1 = (1 + i)n −1 i . 146 THE BASICS OF ANNUITY THEORY This last equation is equivalent to 1 + isn = (1 + i)n. This last equation is the equation of value at time t= nof an investment of $1 for nperiods during which an interest of iis received at the end of each period and is reinvested at the same rate i, and at the end of the nperiods the original investment of $1 is returned. Figure 15.5 describes the time diagram of this transaction. Figure 15.5 Example 15.2 Calculate the future value of an annuity −immediate of amount $100 paid annually for 5 years at the rate of interest of 9%. Solution. The answer is 100s5 = 100 ×(1.09)5−1 0.09 ≈$598.47 Example 15.3 Show that am+n = am + νman = an + νnam. Interpret this result verbally. Solution. We have am + νman = 1−νm i + νm ·1−νn i = 1−νm+νm−νm+n i = 1−νm+n i = am+n. A verbal interpretation is as follows: The present value of the ﬁrst mpayments of an (m+ n)−year annuity−immediate of 1 is am.The remaining npayments have valuean at time t= m; discounted to the present, these are worth νman at time t= 0 Example 15.4 At an eﬀective annual interest rate i, you are given (1) the present value of an annuity immediate with annual payments of 1 for n years is 40 (2) the present value of an annuity immediate with annual payments of 1 for 3 n years is 70. Calculate the accumulated value of an annuity immediate with annual payments of 1 for 2 n years. Solution. Using Example 15.3 we can write a3n = a2n + ν2nan = an + νnan + ν2nan = an(1 + νn + ν2n). Hence, we obtain the quadratic equation ν2n + νn + 1 = 7 4 . Solving this equation we ﬁnd νn = 1 2 . 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 147 Again, using Example 15.3 we can write a3n = an + νna2n which implies that νna2n = 70 −40 = 30 and therefore a2n = 60. Finally, s2n = ν−2na2n = 4(60) = 240 We next establish a couple of relationships between an and sn. Theorem 15.1 With an and sn as deﬁned above we have (i) sn = (1 + i)nan. That is, the accumulated value of a principal of an after n periods is just sn. (ii) 1 an = 1 sn + i. Proof. (i) We have sn = (1+i)n−1 i = (1 + i)n ·1−(1+i)−n i = (1 + i)nan. (ii) We have 1 sn + i= i (1 + i)n −1 + i= i+ i(1 + i)n −i (1 + i)n −1 = i 1 −νn = 1 an A verbal interpretation of (ii) will be introduced when we discuss the concepts of amortization and sinking funds. Example 15.5 For a given interest rate i, an = 8.3064 and sn = 14.2068. (a) Calculate i. (b) Calculate n. Solution. (a) Using part (ii) of the previous theorem we ﬁnd i= 1 8.3064 − 1 14.2068 = 5%. (b) Using part (i) of the previous theorem we ﬁnd n= 1 ln (1 +i) ln (sn an ) = 11 The type of annuity discussed in this section involves compound interest rate. It is possible to deﬁne annuity−immediate not involving compound interest such as simple interest rate, simple discount rate, and force of interest. For example, we wish to ﬁnd the present value of an n−period annuity immediate in which each payment is invested at simple interest rate i. The present value is equal to the sum of the present value of the individual payments. Thus, we obtain an = 1 1 + i + 1 1 + 2i + ··· + 1 1 + ni. 148 THE BASICS OF ANNUITY THEORY The accumulated value of such an annuity is equal to the accumulated value of the individual payments. That is, sn = 1 + (1 +i) + (1 + 2i) + ··· + [1 + (n−1)i]. Example 15.6 Find an expression for an assuming each payment of 1 is valued at simple discount rate d. Solution. The present value is the sum of the present value of individual payments. That is, an = (1 −d) + (1−2d) + ··· + (1 −nd) = n−d(1 + 2 +··· + n) = n−n(n+ 1) 2 d 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 149 Practice Problems Problem 15.1 Consider an investment of $5,000 at 6% convertible semiannually. How much can be withdrawn each half−year to use up the fund exactly at the end of 20 years? Problem 15.2 The annual payment on a house is $18,000. If payments are made for 40 years, how much is the house worth assuming annual interest rate of 6%? Problem 15.3 If d= 0.05, calculate a12 . Problem 15.4 Calculate the present value of 300 paid at the end of each year for 20 years using an annual eﬀective interest rate of 8%. Problem 15.5 If an = x and a2n = y, express d as a function of x and y. Problem 15.6 (a) Given: a7 = 5.153, a11 = 7.036, a18 = 9.180. Find i. (b) You are given that an = 10.00 and a3n = 24.40. Determine a4n. Problem 15.7 Show that sm+n = sm + (1 +i)msn = sn + (1 +i)nsm. Interpret this result verbally. Problem 15.8 A grandmother has a granddaughter entering university next year. Her granddaughter expects to remain in school for ten years and receive a PhD. This grandmother wishes to provide $1,000 a year to her granddaughter for entertainment expenses. Assuming a 3.5% eﬀective annual interest rate, how much does the grandmother have to deposit today to provide ten annual payments starting one year from now and continuing for ten years? Problem 15.9 In the previous problem, if the PhD student saves the income from her grandmother in an account also paying 3.5% eﬀective annual interest, how much will she have when she receives the ﬁnal $1,000 payment? 150 THE BASICS OF ANNUITY THEORY Problem 15.10 Find the present value of an annuity which pays $200 at the end of each quarter −year for 12 years if the rate of interest is 6% convertible quarterly. Problem 15.11 Compare the total amount of interest that would be paid on a $3,000 loan over a 6 −year period with an eﬀective rate of interest of 7.5% per annum, under each of the following repayment plans: (a) The entire loan plus accumulated interest is paid in one lump sum at the end of 6 years. (b) Interest is paid each year as accrued, and the principal is repaid at the end of 6 years. (c) The loan is repaid with level payments at the end of each year over the 6-year period. Problem 15.12 A loan of $20,000 to purchase a car at annual rate of interest of 6% will be paid back through monthly installments over 5 years, with 1st installment to be made 1 month after the release of the loan. Problem 15.13 Over the next 20 years, you deposit money into a retirement account at the end of each year according to the following schedule: Time Amount invested each year 1 - 5 $2000 6 - 10 $3000 11 - 20 $5000 The eﬀective annual rate of interest is 9%. Find the accumulated value of your account at time 20. Problem 15.14 A family wishes to accumulate 50,000 in a college education fund by the end of 20 years. If they deposit 1,000 into the fund at the end of each of the ﬁrst 10 years, and 1 ,000 + X at the end of each of the second 10 years, ﬁnd X to the nearest unit if the fund earns 7% eﬀective. Problem 15.15 An annuity provides a payment of n at the end of each year for n years. The annual eﬀective interest rate is 1 n. What is the present value of the annuity? Problem 15.16 The cash price of a new automobile is 10,000. The purchaser is willing to ﬁnance the car at 18% convertible monthly and to make payments of 250 at the end of each month for 4 years. Find the down payment which will be necessary. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 151 Problem 15.17 You have $10,000 down payment on a $20,000 car. The dealer oﬀers you the following two options: (a) paying the balance with end −of−month payments over the next three years at i(12) = 0.12; (b) a reduction of $500 in the price of the car, the same down payment of $10,000, and bank ﬁnancing of the balance after down payment, over 3 years with end −of−month payments at i(12) = 0.18. Which option is better? Problem 15.18 Nancy has 10,000 in a bank account earning 6% compounded monthly. Calculate the amount that she can withdraw at the end of each month from the account if she wants to have zero in the account after 12 months. Problem 15.19 Megan purchased a new car for 18,000. She ﬁnances the entire purchase over 60 months at a nominal rate of 12% compounded monthly. Calculate Megan’s monthly payment. Problem 15.20 ‡ Seth, Janice, and Lori each borrow 5,000 for ﬁve years at a nominal interest rate of 12%, compounded semiannually. Seth has interest accumulated over the ﬁve years and pays all the interest and principal in a lump sum at the end of ﬁve years. Janice pays interest at the end of every six−month period as it accrues and the principal at the end of ﬁve years. Lori repays her loan with 10 level payments at the end of every six −month period. Calculate the total amount of interest paid on all three loans. Problem 15.21 If d(12) = 12%,calculate the accumulated value of 100 paid at the end of each month for 12 months. Problem 15.22 The accumulated value of an nyear annuity−immediate is four times the present value of the same annuity. Calculate the accumulated value of 100 in 2 n years. Problem 15.23 You are given the following information: (i) The present value of a 6 n−year annuity−immediate of 1 at the end of every year is 9.7578. (ii) The present value of a 6 n−year annuity−immediate of 1 at the end of every second year is 4.760. (iii) The present value of a 6 n−year annuity−immediate of 1 at the end of every third year is K. Determine K assuming an annual eﬀective interest rate of i. 152 THE BASICS OF ANNUITY THEORY Problem 15.24 ‡ Which of the following does not represent a deﬁnition of an? (a) νn [ (1+i)n−1 i ] (b) 1−νn i (c) ν+ ν2 + ··· + νn (d) ν [1−νn 1−ν ] (e) sn (1+i)n−1 Problem 15.25 ‡ To accumulate 8000 at the end of 3 n years, deposits of 98 are made at the end of each of the ﬁrst n years and 196 at the end of each of the next 2 n years. The annual eﬀective rate of interest is i. You are given (1 + i)n = 2.0. Determine i. Problem 15.26 ‡ For 10,000, Kelly purchases an annuity −immediate that pays 400 quarterly for the next 10 years. Calculate the annual nominal interest rate convertible monthly earned by Kelly’s investment. Hint: Use linear interpolation. Problem 15.27 ‡ Susan and Jeﬀ each make deposits of 100 at the end of each year for 40 years. Starting at the end of the 41st year, Susan makes annual withdrawals of X for 15 years and Jeﬀ makes annual withdrawals of Y for 15 years. Both funds have a balance of 0 after the last withdrawal. Susan’s fund earns an annual eﬀective interest rate of 8%. Jeﬀ’s fund earns an annual eﬀective interest rate of 10%. Calculate Y −X. Problem 15.28 A loan of 10,000 is being repaid by 10 semiannual payments, with the ﬁrst payment made one−half year after the loan. The ﬁrst 5 payments are K each, and the ﬁnal 5 are K+ 200 each. What is K if i(2) = 0.06? Problem 15.29 Smith makes deposits of 1,000 on the last day of each month in an account earning interest at rate i(12) = 0.12. The ﬁrst deposit is January 31, 2005 and the ﬁnal deposit is December 31, 2029. The accumulated account is used to make monthly payments of Y starting January 31, 2030 with the ﬁnal one on December 31, 2054. Find Y. Problem 15.30 A loan of $1,000 is to be repaid with annual payments at the end of each year for the next 20 years. For the ﬁrst ﬁve years the payments are k per year; the second 5 years, 2 k per year; the third 5 years, 3k per year; and the fourth 5 years, 4 k per year. Find an expression for k. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 153 Problem 15.31 ‡ Happy and ﬁnancially astute parents decide at the birth of their daughter that they will need to provide 50,000 at each of their daughter’s 18th, 19th, 20th and 21st birthdays to fund her college education. They plan to contribute X at each of their daughter’s 1st through 17th birthdays to fund the four 50,000 withdrawals. If they anticipate earning a constant 5% annual eﬀective rate on their contributions, which of the following equations of value can be used to determine X,assuming compound interest? (A) X(ν+ ν2 + ··· + ν17) = 50,000(ν+ ν2 + ν3 + ν4) (B) X[(1.05)16 + (1.05)15 + ··· + (1.05)] = 50,000(1 + ν+ ν2 + ν3) (C) X[(1.05)17 + (1.05)16 + ··· + (1.05) + 1] = 50,000(1 + ν+ ν2 + ν3) (D) X[(1.05)17 + (1.05)16 + ··· + (1.05)] = 50,000(1 + ν+ ν2 + ν3) (E) X(1 + ν+ ν2 + ··· + ν17) = 50,000(ν18 + ν19 + ν20 + ν21 + ν22) Problem 15.32 For time t> 0, the discount function is deﬁned by [a(t)]−1 = 1 1 + 0.01t. Consider a ﬁve-year annuity with payments of 1 at times t= 1,2,3,4,5. Consider the following: A calculates a5 as the sum of the present value of the individual payments. However, B accumulates the payments according to the accumulation function a(t) = 1 + 0.01t and then multiplies the result by 1 a(5) . By how much do the answers of A and B diﬀer? Remark 15.1 It can be shown that the above two processes do not produce the same answers in general for any pattern of interest other than compound interest. That’s why, it is always recommended to avoid dealing with annuities not involving compound interest, if possible. Problem 15.33 Simplify the sum ∑40 n=15 sn. Problem 15.34 A 20 year annuity pays 100 every other year beginning at the end of the second year, with additional payments of 300 each at the ends of years 3, 9, and 15. The eﬀective annual interest rate is 4%. Calculate the present value of the annuity. 154 THE BASICS OF ANNUITY THEORY Problem 15.35 An annuity pays 1 at the end of each 4 −year period for 40 years. Given a8 i = k, ﬁnd the present value of the annuity. Problem 15.36 Smith is negotiating a price for a new car. He is willing to pay $250 at the end of each month for 60 months using the 4.9% compounded monthly interest rate that he qualiﬁes for. Smith estimates that tax, title, and license for the new car will increase the negotiated price by 10%, and he estimates that he will receive $500 trade −in value for his current car. Calculate the highest negotiated price that Smith is willing to pay for the car. Problem 15.37 An account is credited interest using 6% simple interest rate from the date of each deposit into the account. Annual payments of 100 are deposited into this account. Calculate the accumulated value of the account immediately after the 20 th deposit. Problem 15.38 A homeowner signs a 30 year mortgage that requires payments of $971.27 at the end of each month. The interest rate on the mortgage is 6% compounded monthly. If the purchase price of the house is $180,000 then what percentage down payment was required? Problem 15.39 A 25 −year−old worker begins saving for retirement, making level annual deposits at the end of each year. The savings are invested at an annual eﬀective interest rate of 8%, and are of an amount that is projected to equal 1,000,000 when the worker is 65 (after the 40th deposit is made on that date). After making ﬁve annual deposits, the worker becomes unemployed for a period of time and, as a result, skips the next three annual deposits. Assuming that the account earns an 8% annual eﬀective rate in all 40 years, what amount will the worker need to deposit in each of the remaining 32 years in order to achieve the original goal of a 1,000,000 balance at age 65? Problem 15.40 Paul lends 8000 to Peter. Peter agrees to pay it back in 10 annual installments at 7% with the ﬁrst payment due in one year. After making 4 payments, Peter renegotiates to pay oﬀ the debt with 4 additional annual payments. The new payments are calculated so that Paul will get a 6.5% annual yield over the entire 8−year period. Determine how much money Peter saved by renegotiating. Problem 15.41 Mario deposits 100 into a fund at the end of each 2 year period for 20 years. The fund pays interest at an annual eﬀective rate of i. The total amount of interest earned by the fund during the 19th and 20th years is 250. Calculate the accumulated amount in Marios account at the end of year 20. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 155 Problem 15.42 Show that 1 1−ν10 = 1 s10 ( s10 + 1 i ) . Problem 15.43 If a4 = 3.2397 and s4 = 4.5731 what is the value of a8 ? Problem 15.44 Smith borrows $5,000 on January 1, 2007. He repays the loan with 20 annual payments, starting January 1, 2008. The payments in even-number year are 2 X each; the payments in odd-number years are X each. If d= 0.08, ﬁnd the total amount of all 20 payments. Problem 15.45 (a) Show that am−n = am −νmsn where 0 <n<m. (b) Show that sm−n = sm −(1 + i)man where 0 <n<m. Problem 15.46 Show that sn ≥n≥an. Thus far, we have only considered annuities where the payments are made at the end of the period, but it is possible that the circumstances may be such that the annuity is to run for a given number of periods and a portion of a period. The purpose of the remaining problems is to deﬁne terms such as an+k and sn+k where n is a positive integer and 0 <k <1. Problem 15.47 Let kbe a positive real number. Find an expression for the error involved in approximating (1+i)k−1 i by the number k. Hint: Use the Taylor series expansion of (1 + i)k around i= 0. Problem 15.48 (a) Find the interest accrued of a principal of 1 at t = n to time t = n+ k, assuming compound interest. (b) Show that i(ν+ ν2 + ν3 + ··· + νn) + [(1 +i)k −1]νn+k + νn+k = 1. (c) Let an+ 1 m denote the present value of an annuity consisting of n payments of 1 at the end of each period and a ﬁnal payment of (1+i) 1 m−1 i at the end of the 1 mth fraction of the ( n+ 1) period. Deﬁne an+ 1 m = 1−νn+ 1 m i . Show that an+ 1 m = an + νn+ 1 m [ (1 + i) 1 m −1 i ] . 156 THE BASICS OF ANNUITY THEORY Thus, an+k is the sum of the present value of an n−period annuity−immediate of 1 per period, plus a ﬁnal payment at time n+ k of (1+i)k−1 i . Problem 15.49 Show that an+k ≈an + kνn+ 1 m. Hint: Use Problems 15.47 and 15.48. Problem 15.50 Compute a5.25 if i= 5% using the following deﬁnitions: (a) The formula established in Problem 15.48. (b) A payment of 0.25 at time 5.25. (c) A payment of 0.25 at time 6. Problem 15.51 (a) Deﬁne sn+k = (1+i)n+k−1 k .Show that sn+k = (1+ i)ksn + (1+i)k−1 i .Thus, sn+k can be interprested as the accumulated value of an n−period annuity−immediate at time t= n+ k with an additional payment of (1+i)k−1 i at time t= n+ k. (b) Show that sn+k ≈(1 + i)ksn + k. 16 ANNUITY IN ADVANCE: ANNUITY DUE 157 16 Annuity in Advance: Annuity Due An annuity−due is an annuity for which the payments are made at the beginning of the payment periods. The cash stream represented by the annuity can be visualized on a time diagram as shown in Figure 16.1 . Figure 16.1 The ﬁrst arrow shows the beginning of the ﬁrst period at which the ﬁrst payment is made under the annuity. The second arrow appears n periods after arrow 1, one period after the last payment is made. Let idenote the interest rate per period. The present value of the annuity at time 0 will be denoted by ¨an. To determine ¨an, we can proceed as in the case of determining an. In this case, we use the time diagram shown in Figure 16.2 Figure 16.2 Considering the equation of value at time t= 0 we can write ¨an = 1 + ν+ ν2 + ··· + νn−1. That ¨an is is equal to the sum of the present values of each of the n payments. We recognize the expression on the right −hand side as a geometric progression. Thus, multiplying both sides by ν to obtain ν¨an = ν+ ν2 + ν3 + ··· + νn−1 + νn. Subtracting this from the previous equation we ﬁnd (1 −ν)¨an = (1 −νn). 158 THE BASICS OF ANNUITY THEORY Hence, ¨an = 1 −νn 1 −ν . Since 1 −ν = d, we have ¨an = 1 −νn d = 1 −(1 + i)−n d = 1 −(1 −d)n d . Example 16.1 Find ¨a8 if the eﬀective rate of discount is 10%. Solution. Since d= 0.10, we have ν = 1 −d= 0.9. Hence, ¨a8 = 1−(0.9)8 0.1 = 5.6953279 Example 16.2 What amount must you invest today at 6% interest rate compounded annually so that you can withdraw $5,000 at the beginning of each year for the next 5 years? Solution. The answer is 5000¨a5 = 5000 ·1−(1.06)−5 0.06(1.06)−1 = $22,325.53 Now, let ¨sn denote the accumulated value of an annuity −due at the end of n periods. To de- termine ¨sn we proceed in a way analogous to determining sn.We consider the time diagram shown in Figure 16.3. Figure 16.3 Writing the equation of value at time t= n we ﬁnd ¨sn =(1 + i) + (1 +i)2 + ··· + (1 +i)n−1 + (1 +i)n =(1 + i)(1 + i)n −1 (1 + i) −1 =(1 + i)n −1 iν = (1 + i)n −1 d 16 ANNUITY IN ADVANCE: ANNUITY DUE 159 Example 16.3 What amount will accumulate if we deposit $5,000 at the beginning of each year for the next 5 years? Assume an interest of 6% compounded annually. Solution. The answer is 5000¨s5 = 5000 · (1.06)5−1 0.06(1.06)−1 = $29,876.59 The following theorem provides a couple of relationships between ¨an and ¨sn Theorem 16.1 (a) ¨sn = (1 + i)n¨an (b) 1 ¨an = 1 ¨sn + d Proof. (a) We have ¨sn = (1 + i)n −1 d = (1 + i)n ·1 −(1 + i)−n d = (1 + i)n¨an (b) 1 ¨sn + d= d (1 + i)n −1 + d(1 + i)n −1 (1 + i)n −1 =d+ d[(1 + i)n −1] (1 + i)n −1 = d(1 + i)n (1 + i)n −1 = d 1 −(1 + i)−n = 1 ¨an The result in (a) says that if the present value at time 0, ¨ an, is accumulated forward to time n, then you will have its future value, ¨sn. Example 16.4 For a given interest rate i, ¨an = 8.3064 and ¨sn = 14.2068. (a) Calculate d. (b) Calculate n. Solution. (a) Using part (b) of the previous theorem we ﬁnd d= 1 8.3064 − 1 14.2068 = 5%. (b) Using part (a) of the previous theorem we ﬁnd n= 1 ln (1 +i) ln (¨sn ¨an ) = 10.463 It is possible to relate the annuity-immediate and the annuity-due as shown in the following theorem. 160 THE BASICS OF ANNUITY THEORY Theorem 16.2 (a) ¨an = (1 + i)an. (b) ¨sn = (1 + i)sn. Proof. The two results can be obtained directly by using the time diagram shown in Figure 16.4 Figure 16.4 An algebraic justiﬁcation is given next. (a) Since d= i i+1 , we have ¨an = 1−(1+i)−n d = (1 + i) ·1−(1+i)−n i = (1 + i)an. (b) Since d= i i+1 , we have: ¨sn = (1+i)n−1 d = (1 + i) ·(1+i)n−1 i = (1 + i)sn An annuity−due starts one period earlier than an annuity −immediate and as a result, earns one more period of interest, hence more proﬁt is made. Example 16.5 Over the next 30 years, you deposit money into a retirement account at the beginning of each year. The ﬁrst 10 payments are 200 each. The remaining 20 payments are 300 each. The eﬀective annual rate of interest is 9%. Find the present value of these payments. Solution. The time diagram of this situation is shown in Figure 16.5 Figure 16.5 The answer is 300¨a30 −100¨a10 = 1.09(300a30 −100a10 ) = $2659.96 More relationships between annuity-immediate and annuity-due are given below. 16 ANNUITY IN ADVANCE: ANNUITY DUE 161 Theorem 16.3 (a) ¨an = 1 + an−1 . (b) sn = ¨sn−1 + 1. Proof. (a) We have ¨an = 1−(1+i)−n d = i+1 i [1 −(1 + i)−n] = 1−(1+i)−n+1+i i = 1 + an−1 . The result has the following verbal interpretation: An additional payment of 1 at time 0 results in an−1 becoming n payments that now commence at the beginning of each year whose present value is ¨an. (b) We have ¨sn−1 = (1+i)n−1−1 d = i+1 i [(1 + i)n−1 −1] = (1+i)n−1−i i = sn −1 An interpretation of (b) is as follows: A withdrawal of 1 at time n results in sn becoming n−1 payments that commence at the beginning of each year (starting at t= 1) whose accumulated value at time t= n is ¨sn−1 Remark 16.1 Most compound interest tables do not include values of annuities −due. Thus, the formulas of Theorems 16.2 - 16.3 must be used in ﬁnding numerical values for annuities −due. Example 16.6 An investor wishes to accumulate $3000 at the end of 15 years in a fund which earns 8% eﬀective. To accomplish this, the investor plans to make deposits at the beginning of each year, with the ﬁnal payment to be made one year prior to the end of the investment period. How large should each deposit be? Solution. Let R be the payment at the beginning of each year. The accumulated value of the investment at the end of the investment period is to be $3000, so the equation of value at time t = 15 is 3000 = R¨s14 . Solving for R we ﬁnd R= 3000 ¨s14 = 3000 s15 −1 = $114.71 Example 16.7 Show that ¨an = an + 1 −νn. Interpret the result verbally. Solution. We have ¨an = (1 + i)an = an + ian = an + i·1−νn i = an + 1 −νn. An additional payment of 1 at time 0 results in an becoming n+ 1 payments that now commence at the beginning of the year and whose present value is ¨an + νn 162 THE BASICS OF ANNUITY THEORY Example 16.8 Show that ¨sn = sn −1 + (1 +i)n. Interpret the result verbally. Solution. We have ¨sn = (1 + i)sn = sn + i·(1+i)n−1 i = sn −1 + (1 +i)n. An additional payment of 1 at time n results in ¨sn becoming n+ 1 payments that now commence at the beginning of the year and whose accumulated value at time t= n is sn + (1 +i)n Remark 16.2 The names annuity−immediate and annuity-due are used traditionally, although they do not seem to be logical. The ﬁrst payment of an annuity−immediate is not made immediately at the beginning of the ﬁrst payment period, it is due at the end of it. 16 ANNUITY IN ADVANCE: ANNUITY DUE 163 Practice Problems Problem 16.1 An 8-year annuity due has a present value of $1,000. If the eﬀective annual interest rate is 5%, then what is the value of the periodic payment? Problem 16.2 An 8-year annuity due has a future value of $1,000. Find the periodic payment of this annuity if the eﬀective annual interest rate is 5%. Problem 16.3 A 5-year annuity due has periodic cash ﬂows of $100 each year. Find the accumulated value of this annuity if the eﬀective annual interest rate is 8%. Problem 16.4 A 5-year annuity due has periodic cash ﬂows of $100 each year. Find the present value of this annuity if the eﬀective annual interest rate is 8%. Problem 16.5 Calculate the accumulated value immediately after the last payment of a 20 year annuity due of annual payments of 500 per year. The annual eﬀective interest rate is 7%. Problem 16.6 Megan wants to buy a car in 4 years for 18,000. She deposits X at the beginning of each month for four years into an account earning 6% compounded monthly. Calculate X. Problem 16.7 Kathy wants to accumulate a sum of money at the end of 10 years to buy a house. In order to accomplish this goal, she can deposit 80 per month at the beginning of the month for the next ten years or 81 per month at the end of the month for the next ten years. Calculate the annual eﬀective rate of interest earned by Kathy. Problem 16.8 An annuity pays $500 every six months for 5 years. The annual rate of interest is 8% convertible semiannually. Find each of the following: (a) The PV of the annuity six months (one period) before the ﬁrst payment, (b) the PV of the annuity on the day of the ﬁrst payment, (c) the FV of the annuity on the day of the last payment, (d) and the FV of the annuity six months after the last payment. 164 THE BASICS OF ANNUITY THEORY Problem 16.9 You will retire in 30 years. At the beginning of each month until you retire, you will invest X earning interest at 9% convertible monthly. Starting at year 30, you will withdraw $4,000 at the beginning of each month for the next 15 years. Also, starting at year 30, your fund will only earn interest at 6% convertible monthly. Find X such that your account will be empty after the last withdrawal. Problem 16.10 ‡ Chuck needs to purchase an item in 10 years. The item costs 200 today, but its price increases by 4% per year. To ﬁnance the purchase, Chuck deposits 20 into an account at the beginning of each year for 6 years. He deposits an additional X at the beginning of years 4, 5, and 6 to meet his goal. The annual eﬀective interest rate is 10%. Calculate X. Problem 16.11 Which of the following are true: (i) ¨s9 + 1 = s10 (ii) a10 −a3 = v2a7 (iii) a5 (1 + i) −1 = a4 ? Problem 16.12 Find the present value of payments of $200 every six months starting immediately and continuing through four years from the present, and $100 every six months thereafter through ten years from the present, if i(2) = 0.06. Problem 16.13 A worker aged 40 wishes to accumulate a fund for retirement by depositing $1,000 at the beginning of each year for 25 years. Starting at age 65 the worker plans to make 15 annual withdrawals at the beginning of each year. Assuming all payments are certain to be made, ﬁnd the amount of each withdrawal starting at age 65 to the nearest dollar, if the eﬀective rate of interest is 8% during the ﬁrst 25 years but only 7% thereafter. Problem 16.14 Which one is greater sn −an or ¨sn −¨an? Problem 16.15 A (2n−1)−payment annuity-immediate has payments 1,2,··· ,n −1,n,n −1,··· ,2,1.Show that the present value of the annuity one payment period before the ﬁrst payment is an ·¨an. 16 ANNUITY IN ADVANCE: ANNUITY DUE 165 Problem 16.16 Jeﬀ makes payments at the end of each year into an account for 10 years. The present value of Jeﬀ’s payments is 5,000. Ryan makes a payment equal to Jeﬀ’s payment at the beginning of each year for 11 years into the same account. The present value of Ryan’s payments is 5,900. Calculate the amount of Jeﬀ’s payment. Problem 16.17 A person deposits 100 at the beginning of each year for 20 years. Simple interest at an annual rate of i is credited to each deposit from the date of deposit to the end of the twenty year period. The total amount thus accumulated is 2,840. If instead, compound interest had been credited at an eﬀective annual rate of i,what would the accumulated value of these deposits have been at the end of twenty years? Problem 16.18 You plan to accumulate 100,000 at the end of 42 years by making the following deposits: X at the beginning of years 1-14 No deposits at the beginning of years 15-32; and Y at the beginning of years 33-42. The annual eﬀective interest rate is 7%. Suppose X−Y = 100. Calculate Y. Problem 16.19 ‡ Jim began saving money for his retirement by making monthly deposits of 200 into a fund earning 6% interest compounded monthly. The ﬁrst deposit occurred on January 1, 1985. Jim became unemployed and missed making deposits 60 through 72. He then continued making monthly deposits of 200. How much did Jim accumulate in his fund on December 31, 1999 ? Problem 16.20 ‡ An investor accumulates a fund by making payments at the beginning of each month for 6 years. Her monthly payment is 50 for the ﬁrst 2 years, 100 for the next 2 years, and 150 for the last 2 years. At the end of the 7th year the fund is worth 10,000. The annual eﬀective interest rate is i, and the monthly eﬀective interest rate is j. Which of the following formulas represents the equation of value for this fund accumulation? (a) ¨s24 i(1 + i)[(1 + i)4 + 2(1 +i)2 + 3] = 200 (b) ¨s24 i(1 + j)[(1 + j)4 + 2(1 +j)2 + 3] = 200 (c) ¨s24 j(1 + i)[(1 + i)4 + 2(1 +i)2 + 3] = 200 (d) s24 j(1 + i)[(1 + i)4 + 2(1 +i)2 + 3] = 200 (e) s24 i(1 + j)[(1 + j)4 + 2(1 +j)2 + 3] = 200 166 THE BASICS OF ANNUITY THEORY Problem 16.21 ‡ Carol and John shared equally in an inheritance. Using his inheritance, John immediately bought a 10-year annuity-due with an annual payment of 2,500 each. Carol put her inheritance in an investment fund earning an annual eﬀective interest rate of 9%. Two years later, Carol bought a 15-year annuity-immediate with annual payment of Z. The present value of both annuities was determined using an annual eﬀective interest rate of 8%. Calculate Z. Problem 16.22 ‡ Jerry will make deposits of 450 at the end of each quarter for 10 years. At the end of 15 years, Jerry will use the fund to make annual payments of Y at the beginning of each year for 4 years, after which the fund is exhausted. The annual eﬀective rate of interest is 7% . Determine Y. Problem 16.23 If a(t) = 1 log2 (t+2)−log2 (t+1) , ﬁnd an expression for ¨ an by directly taking the present value of the payments. Problem 16.24 The accumulated value of an nyear annuity-due is four times the present value of the same annuity. Calculate the accumulated value of 100 in 2 n years. Problem 16.25 Show that sn ·¨an >n2 for i> 0 and n> 1. Problem 16.26 ‡ A man turns 40 today and wishes to provide supplemental retirement income of 3000 at the begin- ning of each month starting on his 65th birthday. Starting today, he makes monthly contributions of X to a fund for 25 years. The fund earns a nominal rate of 8% compounded monthly. Each 1000 will provide for 9.65 of income at the beginning of each month starting on his 65th birthday until the end of his life. Calculate X. Problem 16.27 An annuity pays 3 at the beginning of each 3 year period for 30 years. Find the accumulated value of the annuity just after the ﬁnal payment, using i(2) = 0.06. Problem 16.28 A worker aged 30 wishes to accumulate a fund for retirement by depositing $3,000 at the beginning 16 ANNUITY IN ADVANCE: ANNUITY DUE 167 of each year for 35 years. Starting at age 65 the worker plans to make 20 equal annual withdrawals at the beginning of each year. Assuming that all payments are certain to be made, ﬁnd the amount of each withdrawal starting at age 65, if the annual eﬀective rate of interest is 9% during the ﬁrst 35 years but only 6% thereafter. Problem 16.29 Irene deposits 100 at the beginning of each year for 20 years into an account in which each deposit earns simple interest at a rate of 10% from the time of the deposit. Other than these deposits Irene makes no other deposits or withdrawals from the account until exactly 25 years after the ﬁrst deposit was made, at which time she withdraws the full amount in the account. Determine the amount of Irene’s withdrawal. Problem 16.30 Show that ¨s2n ¨sn + ¨sn ¨s2n −¨s3n ¨s2n = 1. Problem 16.31 If ¨ap = x and sq = y, show that ap+q = νx+y 1+iy. 168 THE BASICS OF ANNUITY THEORY 17 Annuity Values on Any Date: Deferred Annuity Evaluating annuities thus far has always been done at the beginning of the term( either on the date of, or one period before the ﬁrst payment) or at the end of the term (either on the date of, or one period after the last payment). In this section, we shall now consider evaluating the (1) present value of an annuity more than one period before the ﬁrst payment date, (2) accumulated value of an annuity more than one period after the last payment date, (3) current value of an annuity between the ﬁrst and last payment dates. We will assume that the evaluation date is always an integral number of periods from each payment date. (1) Present values more than one period before the ﬁrst payment date Consider the question of ﬁnding the present value of an annuity −immediate with periodic interest rate iand m+ 1 periods before the ﬁrst payment date. Figure 17.1 shows the time diagram for this case where “?” indicates the present value to be found. Figure 17.1 The present value of an n−period annuity−immediate m+ 1 periods before the ﬁrst payment date (called a deferred annuity since payments do not begin until some later period) is the present value at time m discounted for m time periods, that is, vman. It is possible to express this answer strictly in terms of annuity values. Indeed, am+n −am = 1 −νm+n i −1 −νm i = νm −νm+n i = νm1 −νn i = νman. Such an expression is convenient for calculation, if interest tables are being used. Example 17.1 Exactly 3 years from now is the ﬁrst of four $200 yearly payments for an annuity-immediate, with an eﬀective 8% rate of interest. Find the present value of the annuity. Solution. The answer is 200ν2a4 = 200(a6 −a2 ) = 200(4.6229 −1.7833) = $567.92 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 169 The deferred −annuity introduced above uses annuity −immediate. It is possible to work with a deferred annuity−due. In this case, one can easily see that the present value is given by νm¨an = ¨am+n −¨am. Example 17.2 Calculate the present value of an annuity −due paying annual payments of 1200 for 12 years with the ﬁrst payment two years from now. The annual eﬀective interest rate is 6%. Solution. The answer is 1200(1.06)−2¨a12 = 1200(¨a14 −¨a2 ) = 1200(9.8527 −1.9434) ≈9,491.16 (2) Accumulated values more than 1 period after the last payment date Consider the question of ﬁnding the accumulated value of an annuity −immediate with periodic interest rate i and m periods after the last payment date. Figure 17.2 shows the time diagram for this case where “?” indicates the sought accumulated value. Figure 17.2 The accumulated value of an n−period annuity−immediate m periods after the last payment date is the accumulated value at time naccumulated for mtime periods, that is, (1 + i)msn.Notice that sm+n −sm =(1 + i)m+n −1 i −(1 + i)m −1 i =(1 + i)m+n −(1 + i)m i = (1 + i)m(1 + i)n −1 i = (1 + i)msn Example 17.3 For four years, an annuity pays $200 at the end of each year with an eﬀective 8% rate of interest. Find the accumulated value of the annuity 3 years after the last payment. Solution. The answer is 200(1 + 0.08)3s4 = 200(s7 −s3 ) = 200(8.9228 −3.2464) = $1135.28 It is also possible to work with annuities −due instead of annuities −immediate. The reader should verify that (1 + i)m¨sn = ¨sm+n −¨sm. 170 THE BASICS OF ANNUITY THEORY Example 17.4 A monthly annuity−due pays 100 per month for 12 months. Calculate the accumulated value 24 months after the ﬁrst payment using a nominal rate of 4% compounded monthly. Solution. The answer is 100 ( 1 + 0.04 12 )12 ¨s12 0.04 12 = 1,276.28 (3) Current value between the ﬁrst and last payment date Next, we consider the question of ﬁnding the present value of an n−period annuity −immediate after the payment at the end of mth period where 1 ≤m≤n.Figure 17.3 shows the time diagram for this case. Figure 17.3 The current value of an n−period annuity−immediate immediately upon the mth payment date is the present value at time 0 accumulated for m time periods which is equal to the accumulated value at time n discounted for n−m time periods, that is, (1 + i)man = νn−msn. One has the following formula, (1 + i)man = νn−msn = sm + an−m. To see this, (1 + i)man =(1 + i)m ·1 −(1 + i)−n i =(1 + i)m −(1 + i)m−n i =(1 + i)m −1 i + 1 −(1 + i)m−n i =sm + an−m. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 171 Example 17.5 For four years, an annuity pays $200 at the end of each half −year with an 8% rate of interest convertible semiannually. Find the current value of the annuity immediately upon the 5th payment (i.e., middle of year 3). Solution. The answer is 200(1.04)5a8 0.04 = 200(s5 0.04 + a3 0.04) = 200(5.4163 + 2.7751) = $1,638.28 For annuity−due we have a similar formula for the current value (1 + i)m¨an = νn−m¨sn = ¨sm + ¨an−m. Example 17.6 Calculate the current value at the end of 2 years of a 10 year annuity due of $100 per year using a discount rate of 6%. Solution. We have (1 + i)−1 = 1 −d= 1 −0.06 = 0.94 and i= 0.06 0.94 . Thus, 100(.94)−2¨a10 = $870.27 Up to this point, we have assumed that the date is an integral number of periods. In the case the date is not an integral number of periods from each payment date, the value of an annuity is found by ﬁnding the value on a date which is an integral number of periods from each payment date and then the value on this date is either accumulated or discounted for the fractional period to the actual evaluation date. We illustrate this situation in the next example. Example 17.7 An annuity−immediate pays $1000 every six months for three years. Calculate the present value of this annuity two months before the ﬁrst payment using a nominal interest rate of 12% compounded semiannually. Solution. The present value at time t= 0 is 1000a6 0.06 = 10001 −(1.06)−6 0.06 = $4917.32. Let j be the interest rate per 2-month. Then 1 + j = (1 + 0.06) 1 3 . The present value two months before the ﬁrst payment is made is 4917.32(1.06) 2 3 = $5112.10 172 THE BASICS OF ANNUITY THEORY Practice Problems Problem 17.1 For four years, an annuity pays $200 at the end of each half-year with an 8% rate of interest convertible semiannually. Find the current value of the annuity three months after the 5th payment (i.e., nine months into year 3). Problem 17.2 Which of the following is equal to the current value shown on the time diagram? (i) s3 + a4 (ii) ¨s3 + a4 (iii) ¨a6 (1 + i)3 (iv) v4s6 (v) ¨s2 + (1 +i)a4 Problem 17.3 Calculate the present value of an annuity immediate with 20 annual payments of 500 if the ﬁrst payment of the annuity immediate starts at the end of the ﬁfth year. The annual eﬀective interest rate is 8%. Problem 17.4 Calculate the current value at the end of 5 years of an annuity due paying annual payments of 1200 for 12 years. The annual eﬀective interest rate is 6%. Problem 17.5 A monthly annuity due pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. Problem 17.6 A monthly annuity immediate pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 173 Problem 17.7 Annuities X and Y provide the following payments: End of Year Annuity X Annuity Y 1 - 10 1 K 11 - 20 2 0 21 - 30 1 K Annuities X and Y have equal present values and an annual eﬀective interest rate of i such that v10 = 0.5. Determine K. Problem 17.8 Payments of $100 per quarter are made from June 7, Z through December 7, Z+ 11, inclusive. If the nominal interest convertible quarterly is 6%: (a) Find the present value on September 7, Z−1. (b) Find the current value on March 7, Z+ 8. (c) Find the accumulated value on June 7, Z+ 12. Problem 17.9 Find the current value to the nearest dollar on January 1 of an annuity which pays $2,000 every six months for ﬁve years. The ﬁrst payment is due on the next April 1 and the rate of interest is 9% convertible semiannually. Problem 17.10 John buys a series of payments. The ﬁrst payment of 50 is in six years. Annual payments of 50 are made thereafter until 14 total payments have been made. Calculate the price John should pay now to realize an annual eﬀective return of 7%. Problem 17.11 Which of the following are true? (i) ¨a10 −¨a3 = a9 −a2 (ii) ν3¨a3 = ν2a3 (iii) ν8s12 = ¨a3 + ¨s9 Problem 17.12 A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the ﬁfth year and to continue thereafter for as long as necessary. The eﬀective rate of discount is 5%. Find the amount of the ﬁnal payment if it is to be larger than the regular payments. 174 THE BASICS OF ANNUITY THEORY Problem 17.13 Using an annual eﬀective interest rate i> 0, you are given: (i) The present value of 2 at the end of each year for 2 n years, plus an additional 1 at the end of each of the ﬁrst n years, is 36. (ii) The present value of an n−year deferred annuity-immediate paying 2 per year for n years is 6. Calculate i. Problem 17.14 Show that ∑15 t=10(¨st −st) = s16 −s10 −6. Problem 17.15 It is known that a7 a11 = a3 +sx ay+sz . Find x,y, and z. Problem 17.16 Simplify a15 (1 + ν15 + ν30) to one symbol. Problem 17.17 The present value of an annuity-immediate which pays $200 every 6 months during the next 10 years and $100 every 6 months during the following 10 years is $4,000. The present value of a 10-year deferred annuity-immediate which pays $250 every 6 months for 10 years is $2,500. Find the present value of an annuity-immediate which pays $200 every 6 months during the next 10 years and $300 every 6 months during the following 10 years. Problem 17.18 ‡ At an annual eﬀective interest rate of i,i> 0, both of the following annuities have a present value of X : (i) a 20-year annuity-immediate with annual payments of 55; (ii) a 30-year annuity-immediate with annual payments that pays 30 per year for the ﬁrst 10 years, 60 per year for the second 10 years, and 90 per year for the ﬁnal 10 years. Calculate X. Problem 17.19 Tom borrows 100 at an annual eﬀective interest rate of 4% and agrees to repay it with 30 annual installments. The amount of each payment in the last 20 years is set at twice that in the ﬁrst 10 years. At the end of 10 years, Tom has the option to repay the entire loan with a ﬁnal payment X, in addition to the regular payment. This will yield the lender an annual eﬀective rate of 4.5% over the 10 year period. Calculate X. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 175 Problem 17.20 You are given an annuity-immediate with 11 annual payments of 100 and a ﬁnal larger payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all payments is 1000. Using an annual eﬀective interest rate of 1%, calculate the present value at the beginning of the ninth year of all remaining payments. Round your answer to the nearest integer. Problem 17.21 The proceeds of a 10,000 death beneﬁt are left on deposit with an insurance company for seven years at an eﬀective annual interest rate of 5%. The balance at the end of seven years is paid to the beneﬁciary in 120 equal monthly payments of X, with the ﬁrst payment made immediately. During the payout period, interest is credited at an annual eﬀective rate of 3%. Problem 17.22 The present value today of a 20 year annuity-immediate paying 500 every 6 months but with the ﬁrst payment deferred tyears has a present value of 5,805.74. If i(12) = .09 is the interest rate used to calculate the present value, ﬁnd t. Problem 17.23 John buys an enterprise that sells dental equipment in the amount of 60 million dollars. He decides to ﬁnance the purchase by making 20 semiannual payments with the ﬁrst payment in two years. Find the value of each payment if the nominal interest rate is 12% compounded semiannually. Problem 17.24 Peter wants to accumulate $20,000 in ﬁve years by making monthly payment at the end of each month for the ﬁrst three years into a savings account that pays annual interest rate of 9% com- pounded monthly and then leave the accumulated amount in the account for the remaining two years. What is the value of his regular monthly payment? 176 THE BASICS OF ANNUITY THEORY 18 Annuities with Inﬁnite Payments: Perpetuities A perpetuity is an annuity whose term is inﬁnite, i.e., an annuity whose payments continue forever with the ﬁrst payment occurs either immediately (perpetuity −due) or one period from now (perpetuity−immediate). Thus, accumulated values of perpetuities do not exist. Let us determine the present value of a perpetuity −immediate at the time one period before the ﬁrst payment, where a payment of 1 is made at the end of each period. The present value will be denoted by a∞. A time diagram of this case is given in Figure 18.1 Figure 18.1 Using the equation of value at time t= 0 we ﬁnd a∞ =ν+ ν2 + ··· =inﬁnite geometric progression with ν <1 = ν 1 −ν = ν iν = 1 i. The verbal interpretation of this formula is as follows: If the periodic eﬀective rate of interest is i then one can invest a principal of 1 i for one period and obtain a balance of 1 + 1 i at the end of the ﬁrst period $1. A payment of $1 is made and the remaining balance of 1 i is reinvested for the next period. This process continues forever. Now, since an = 1−νn i and limn→∞νn = 0 for 0 <ν <1 we have a∞ = lim n→∞ an = 1 i. Example 18.1 Suppose a company issues a stock that pays a dividend at the end of each year of $10 indeﬁnitely, and the the companies cost of capital is 6%. What is the the value of the stock at the beginning of the year? Solution. The answer is 10 ·a∞ = 10 · 1 0.06 = $166.67 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 177 Analogously to perpetuity−immediate, we may deﬁne a perpetuity −due to be an inﬁnite sequence of equal payments where each payment is made at the beginning of the period. Let ¨ a∞ denote the present value of a perpetuity−due at the time of ﬁrst payment is made. A time diagram describing this case is given in Figure 18.2. Figure 18.2 The equation of value at time t= 0 is ¨a∞ = 1 + ν+ ν2 + ··· = 1 1 −ν = 1 d = lim n→∞ ¨an. This formula can be obtained from ﬁnding the present value of a payment of $1 at the beginning of the ﬁrst period and an annutiy −immediate. Indeed, 1 + a∞ = 1 + 1 i = 1 + i i = 1 d = ¨a∞. Example 18.2 What would you be willing to pay for an inﬁnite stream of $37 annual payments (cash inﬂows) beginning now if the interest rate is 8% per annum? Solution. The answer is 37¨a∞ = 37 0.08(1.08)−1 = $499.50 Remark 18.1 We pointed out at the beginning of the section that accumulated values for perpetuities do not exist since payments continue forever. We can argue mathematically as follows: If a perpetuity−immediate has an accumulated value, denoted by s∞, then we expect to have s∞ = limn→∞sn. But lim n→∞ sn = lim n→∞ (1 + i)n −1 i does not exist since 1 + i> 1 and limn→∞(1 + i)n = ∞. 178 THE BASICS OF ANNUITY THEORY Perpetuities are useful in providing verbal explanations of identities. For example, the formula an = 1 −νn i = 1 i −νn i = a∞ −νna∞. can be interpreted as the diﬀerence between payments for two perpetuities each paying 1 at the end of each period; the ﬁrst payment of the ﬁrst perpetuity is one period from now and the ﬁrst payment of the second perpetuity is n+ 1 periods from now. The present value of the ﬁrst perpetuity is 1 i and that of the second perpetuity is νn i . An identical relationship holds for perpetuity due. That is ¨an = 1 −νn d = 1 d −νn d = ¨a∞ −νn¨a∞. Example 18.3 You can receive one of the following two sets of cash ﬂows. Under Option A,you will receive $5,000 at the end of each of the next 10 years. Under Option B, you will receive X at the beginning of each year, forever. The annual eﬀective rate of interest is 10%. Find the value of X such that you are indiﬀerent between these two options. Solution. The equation of value at time t= 0 is 5000a10 =X d = X (1 i + 1 ) 30722.84 =11X X =$2,792.99 Similar to deferred annuities, one can discuss deferred perpetuities. The present value P0 of a deferred perpetuity −immediate with periodic payment of 1 that starts in n periods time, with a ﬁrst cash ﬂow at the beginning of period n+ 1, is given by the equation of value at time t= n (1 + i)nP0 = a∞. Example 18.4 Fifty dollars is paid at the end of each year forever starting six years from now. Assume the annual eﬀective rate of interest is 10%, ﬁnd the present value of the investment. Solution. The deferred period is t= 5. The answer is 50(1 + i)−5a∞ = 50(1.1)−5 · 1 0.1 = $310.46 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 179 Example 18.5 A deferred perpetuity-immediate paying $2000 monthly is bought for a sum of $229,433.67. Find the deferred period if the interest rate compounded monthly is 6%. Solution. The monthly interest rate is 0.06 12 = 0 .005. Let n be the deferred period, i.e. the time it takes $229,433.67 to accumulate to 2000 a∞. Thus, n satisﬁes the equation 2000a∞ = 229,433.67(1.005)n ⇒2000 · 1 0.005 = 229,433.67(1.005)n. Solving this equation for n we ﬁnd n= 111.45 months 180 THE BASICS OF ANNUITY THEORY Practice Problems Problem 18.1 What would you be willing to pay for an inﬁnite stream of $37 annual payments (cash inﬂows) beginning one year from today if the interest rate is 8%? Problem 18.2 Which of the following are equal to a3 . (i) ν+ ν2 + ν3 (ii) ¨a4 −1 (iii) a∞(1 −ν3) Problem 18.3 An annuity−due pays 100 at the beginning of each year forever. The eﬀective annual rate of interest is 25%. What is the present value of the annuity? Problem 18.4 A special perpetuity pays $200 at the end of each year for the ﬁrst 10 years and $100 at the end of each year thereafter. If the eﬀective annual rate of interest is 10%, ﬁnd the present value of the perpetuity. Problem 18.5 ‡ A perpetuity −immediate pays X per year. Brian receives the ﬁrst n payments, Colleen receives the next n payments, and Jeﬀ receives the remaining payments. Brian’s share of the present value of the original perpetuity is 40%, and Jeﬀ’s share is K. Calculate K. Problem 18.6 A sum P is used to buy a deferred perpetuity −due of 1 payable annually. The annual eﬀective interest rate is i> 0. Find an expression for the deferred period. Problem 18.7 Deposits of $1,000 are placed in a fund at the beginning of each year for the next 20 years. After 30 years annual payments commence and continue forever, with the ﬁrst payment at the end of the 30th year. Find an expression for the amount of each payment. Problem 18.8 A benefactor leaves an inheritance to four charities, A,B,C, and D.The total inheritance is a series of level payments at the end of each year forever. During the ﬁrst n years A,B, and C share each payment equally. All payments after n years revert to D. If the present values of the shares of A,B,C, and D are all equal, ﬁnd (1 + i)n. 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 181 Problem 18.9 A level perpetuity−immediate is to be shared by A,B,C, and D. Areceives the ﬁrst n payments. B the second n payments, C the third n payments, and D all payments thereafter. It is known that the ratio of the present value of C′s share to A′s share is 0.49. Find the ratio of the present value of B′s share to D′s share. Problem 18.10 Adam buys a perpetuity due of 1,000 per month for 100,000. Calculate the annual eﬀective rate of interest used to calculate the price of this perpetuity. Problem 18.11 The present value of a perpetuity immediate where the payment is P is 1,000 less than the present value of a perpetuity due where the payment is P. Calculate P. Problem 18.12 A trust has been established such that RJ will receive a perpetuity of 1000 a year with the ﬁrst payment at the end of 5 years. Calculate the present value of the perpetuity at a discount rate of d= 8%. Problem 18.13 Julie, Chris, and Allen will share an annual perpetuity immediate of 1200. Julie will receive the ﬁrst 9 payments. Chris will receive the next 16 payments. Allen will receive all remaining payments. At an annual eﬀective interest rate of 5%, order the value of each person’s share of the perpetuity. Problem 18.14 John is receiving annual payments from a perpetuity immediate of 12,000. Krista is receiving annual payments from a perpetuity due of 10,000. If the present value of each perpetuity is equal, calculate i. Problem 18.15 The present value of an annual perpetuity immediate of 150 is equal to the present value of an annual perpetuity immediate that pays 100 at the end of the ﬁrst 20 years and 200 at the end of year 21 and each year thereafter. Calculate i. Problem 18.16 ‡ An estate provides a perpetuity with payments of X at the end of each year. Seth, Susan, and Lori share the perpetuity such that Seth receives the payments of X for the ﬁrst n years and Susan receives the payments of X for the next m years, after which Lori receives all the remaining payments of X.Which of the following represents the diﬀerence between the present value of Seth’s 182 THE BASICS OF ANNUITY THEORY and Susan’s payments using a constant rate of interest? (a) X[an −νnam] (b) X[¨an −νn¨am] (c) X[an −νn+1am] (d) X[an −νn−1am] (e) X[van −νn+1am] Problem 18.17 ‡ Which of the following are characteristics of all perpetuities?(including non −level payments) I. The present value is equal to the ﬁrst payment divided by the annual eﬀective interest rate. II. Payments continue forever. III. Each payment is equal to the interest earned on the principal. Problem 18.18 ‡ A perpetuity paying 1 at the beginning of each 6-month period has a present value of 20 . A second perpetuity pays X at the beginning of every 2 years. Assuming the same annual eﬀective interest rate, the two present values are equal. Determine X. Problem 18.19 What is the present value of receiving $30 at the end of each year forever, starting 9 years from now? Assume an annual rate of interest of 7% Problem 18.20 Jim buys a perpetuity of 100 per year, with the ﬁrst payment 1 year from now. The price for the perpetuity is 975.61, based on a nominal yield of i compounded semiannually. Immediately after the second payment is received, the perpetuity is sold for 1642.04, to earn for the buyer a nominal yield of j compounded semiannually. Calculate i−j. Problem 18.21 Victor wants to purchase a perpetuity paying 100 per year with the ﬁrst payment due at the end of year 11. He can purchase it by either (1) paying 90 per year at the end of each year for 10 years, or (2) paying K per year at the end of each year for the ﬁrst 5 years and nothing for the next 5 years. Calculate K. Problem 18.22 A perpetuity pays 1 at the end of every year plus an additional 1 at the end of every second year. The present value of the perpetuity is K for i≥0. Determine K. 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 183 Problem 18.23 The accumulated value just after the last payment under a 12-year annuity-immediate of 1000 per year, paying interest at the rate of 5% per annum eﬀective, is to be used to purchase a perpetuity at an interest rate of 6%, with ﬁrst payment to be made 1 year after the last payment under the annuity. Determine the size of the payments under the perpetuity. 184 THE BASICS OF ANNUITY THEORY 19 Solving for the Unknown Number of Payments of an An- nuity In this section we consider the question of ﬁnding the number of payments n given the regular payment R, the interest per period i and either the present value or the accumulated value of an annuity. We will assume an annuity−immediate. A similar calculation applies for annuity −due. Let P be the present value of an annuity −immediate. A time diagram is given in Figure 19.1 Figure 19.1 The equation of value at time t= 0 is P = Rani, where P,R, and i are known quantities and n is the unknown. Solving this equation for n we ﬁnd P = Rani = R (1 −νn i ) ⇒νn = 1 −i (P R ) ⇒n= ln [ 1 −i (P R )] ln ν . This last expression is not necessary a positive integer. Thus, the equation P = Ran is replaced by P = Ran+k where n is a positive integer and 0 <k <1. In this case, using the results of Problems 15.47-15.48 we can write P = Ran+ki = R (1 −νn+k i ) ⇒νn+k = 1 −i (P R ) which implies n+ k= ln [ 1 −i (P R )] ln ν . It follows that, for the annuity to have the present value P, nregular payments of Rmust be made and an additional one payment in the amount of R [(1 + i)k −1 i ] to be made at time t= n+ k, that is, at the fractional period k of the (n+ 1)th period. In practice, the last smaller payment is made either at the same time as the last regular payment making the last payment larger than the regular payment (such a payment is called a balloon 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 185 payment) or at the end of the period following the last regular payment. In this case the smaller payment is called drop payment. The following examples illustrate how to deal with the question of ﬁnding the unknown number of payments. Example 19.1 An investment of $80,000 is to be used to make payments of $5000at the end of every six months for as long as possible. If the fund earns a nominal interest rate of 12% compounded semiannually, ﬁnd how many regular payments can be made and ﬁnd the amount of the smaller payment: (a) to be paid along the last regular payment, (b) to be paid six months after the last regular payment, (c) to be paid during the six months following the last regular payment. Solution. We ﬁrst solve the equation 8000 = 5000 an+k0.06 as follows 80000 = 5000an+k0.06 = 5000 [1 −(1.06)−(n+k) i ] ⇒1 −(1.06)−(n+k) = 80000 5000 (0.06) = 0.96 from which we ﬁnd 1 −0.96 = 0.04 = (1.06)−(n+k) ⇒n+ k= −ln 0.04 ln 1.06 = 55.242. Thus, n= 55 and k= 0.242. (a) Let X be the amount of the smaller payment to be be made at the end of the 55th period. A time diagram of this situation is given in Figure 19.2. Figure 19.2 Note that every period on the time diagram consists of six months. The equation of value at time t= 0 is 80000 = 5000a55 0.06 + X(1.06)−55 ⇒80000 = 79952.715 + (0.0405674)X ⇒X = $1165.59. Thus, in this case, we have 54 payments of $5000 each, and a last payment of 5000 + 1165 .59 = 6165.59 186 THE BASICS OF ANNUITY THEORY (b) This situation is illustrated in Figure 19.3. Figure 19.3 The equation of value at time t= 0 is 80000 = 5000a55 0.06 + Y(1.06)−56 ⇒80000 = 79952.715 + (0.0382712)Y ⇒Y = $1235.53. In this case, we have 55 payments of $5000 each and a last payment of $1235.53 six months after the last regular payment. (c) In this case we have 55 payments of $5000 each and one last payment of 5000 [(1 + 0.06)0.242 −1 0.06 ] = $1182.13 to be made 44 days (0 .242 ×365 2 = 44.165) after the last regular payment A similar type of calculation can be done with known accumulated value instead of present value as illustrated in the next example. Example 19.2 In order to accumulate $5,000, you will deposit $100 at the end of each month for as long as necessary. Interest is 15% compounded monthly. (a) Find the number of regular payments and the fractional period that are required to accumulate the $5,000? (b) If a ﬁnal fractional payment will be added to the last regular payment, what must this fractional payment be? (c) If a ﬁnal fractional payment will be made one month after the last regular payment, what must this fractional payment be? (d) If the fractional payment to be made during the month following the last regular payment. Solution. (a) The monthly interest rate is 0.15 12 = 0.0125. The equation of value at time t= n+ k is 5000 = 100sn+k0.0125 = 100 ·(1 + i)n+k −1 i = 100 ·(1.0125)n+k −1 0.0125 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 187 Solving for n+ k we ﬁnd n+ k= ln ( 1 + 5000(0.0125) 100 ) ln 1.0125 = 39.08 Thus, there will be 39 regular payments and a payment at the fractional time 0 .08 = 0.08 ×365 12 = 2.433, i.e., two days after the last regular payment. (b) Let X be the fractional payment added to the last regular payment. The time diagram of this case is shown in Figure 19.4. Figure 19.4 The equation of value at time t= 39 is 5000 = 100s39 0.0125 + X. Solving for x we ﬁnd X = $13.38. (c) Let Y be such a payment. The time diagram of this case is shown in Figure 19.5. Figure 19.5 The equation of value at time t= 40 is 5000 = 100 s39 (1.0125) + Y. Solving for Y we ﬁnd Y = −48.96. What does this negative answer mean? If 39 regular payments are made at the end of each month, then at time 39 (measured in months) the account will have accumulated 100 s39 0.0125 = 4986.62. This is extremely close to the 5000 that we wanted in the ﬁrst place. If no additional payment is made from time 39 to time 40, then interest alone will make the account have 4986 .62(1.0125) = 5048.96. This means that there will be no additional payment required at time 40. In fact, $48.96 would have to be withdrawn from the account to make the account have only $5000 at time 40. (d) In tis case, 39 regular payments of $100 each to be made and one additional payment in the amount of 100 [(1 + 0.0125)0.08 −1 0.0125 ] = $7.95 to be made 2 days after the last regular payment 188 THE BASICS OF ANNUITY THEORY Example 19.3 An investment of $10,000 is to be used to make payments of $700 at the end of every year for as long as possible. If the eﬀective annual rate if interest is 6%, ﬁnd: (a) the number of regular payments the fund will make, (b) the size of the smaller ﬁnal payment if the payment will be made in addition to the last regular payment, (c) the size of the smaller ﬁnal payment if the payment will be made one year after the last regular payment. (d) the size of the smaller ﬁnal payment if the payment is to be made during the year following the last regular payment. Solution. (a) We have 10000 = 700an or 10000 = 700 ·1−(1+0.06)−n 0.06 . Solving for n we ﬁnd n= ln ( 1 −10000(0.06) 700 ) −ln 1.06 = 33.40 Thus, there will be 33 regular payments. (b) Let x be the size of the payment. The equation of value at t= 0 is 10000 = 700a33 + v33x or 10000(1.06)33 = 700s33 + x. Solving for x we ﬁnd x= 265.68. (c) Let x be the size of the payment. The equation of value at time t= 0 is 10000 = 700a33 + v34x. Solving for x we ﬁnd x= 281.62. (d) The equation of value at time t= 0 is 700a33+k = 10000 or 0.07a33+k = 1 where 0 < k <1. (See Problem 15.29). Thus, we have 0 .071−v33+k i = 1 ⇒v33+k = 1 −6 7 = 1 7 ⇒ 33 + k= ln 7 ln 1.06 ≈33.395 ⇒k= 0.395. In this case, the equation of value at time t= 0 is 700a33 + (1.06)−33.395x= 10000. Solving for x we ﬁnd x = $271.87. That is, the ﬁnal irregular payment of $271.87 is paid at time 33.395 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 189 Practice Problems Problem 19.1 Allison pays $27,506.28 for an annuity due today. This annuity will pay her $1,750 at the beginning of each month for n months with the ﬁrst payment coming today. If i(12) = 12%, ﬁnd n. Round your answer to the nearest integer. Problem 19.2 In order to accumulate $2,000, you will deposit $100 at the end of each month for as long as necessary. Interest is i(12) = 4%. (a) How many regular payments are required to accumulate the $2,000? (b) If a ﬁnal fractional payment will be added to the last regular payment, what must this fractional payment be? Problem 19.3 ‡ The present values of the following three annuities are equal: 1. perpetuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of 7.25%; 2. 50-year annuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of j%; 3. n−year annuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of (j−1)%. Calculate n. Problem 19.4 Jenna is the beneﬁciary of a fund of 20,000 that pays her 1,000 at the end of each month. The fund earns 6% compounded monthly. The ﬁnal payment to exhaust the fund will be a balloon payment. Calculate the amount of the balloon payment. Problem 19.5 Jordan inherits 50,000. This inheritance is invested in a fund earning an annual rate of interest of 6%. He withdraws 5,000 per year beginning immediately. Once Jordan can no longer withdraw a full 5,000, he will withdraw a ﬁnal payment one year after the prior regular payment. Calculate the ﬁnal payment. Problem 19.6 A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the 5th year, and to continue thereafter for as long as necessary. Find the time and amount of the ﬁnal payment if the ﬁnal payment is to be larger than the regular payments. Assume i= 4.5%. 190 THE BASICS OF ANNUITY THEORY Problem 19.7 A fund of 2,000 is to be accumulated by nannual payments of 50 by the end of each year, followed by n annual payments of 100 by the end of each year, plus a smaller ﬁnal payment made 1 year after the last regular payment. If the eﬀective rate of interest is 4.5%, ﬁnd nand the amount of the ﬁnal irregular payment. Problem 19.8 One annuity pays 4 at the end of each year for 36 years. Another annuity pays 5 at the end of each year for 18 years. The present values of both annuities are equal at eﬀective rate of interest i.If an amount of money invested at the same rate i will double in n years, ﬁnd n. Problem 19.9 A fund earning 8% eﬀective is being accumulated with payments of 500 at the beginning of each year for 20 years. Find the maximum number of withdrawals of 1,000 which can be made at the end of each year under the condition that once withdrawals start they must continue through the end of the 20-year period. Problem 19.10 A borrower has the following two options for repaying a loan: (i) Sixty monthly payments of 100 at the end of each month. (ii) A single payment of 6,000 at the end of K months. Interest is at the nominal annual rate of 12% convertible monthly. The two options have the same present value. Find K to the nearest integer. Problem 19.11 ‡ An annuity pays 1 at the end of each year fornyears. Using an annual eﬀective interest rate of i,the accumulated value of the annuity at time ( n+ 1) is 13.776 . It is also known that (1 + i)n = 2.476. Calculate n. Problem 19.12 Ten annual deposits of 1,000 each are made to account A, starting on January 1, 1986. Annual deposits of 500 each are made to account B indeﬁnitely, also starting on January 1, 1986. Interest on both accounts is i= 0.05, with interest credited on December 31. On what date will the balance in account B be larger than the balance in account A? Assume the only transactions are deposits and interest credited to the accounts on December 31. Problem 19.13 ‡ You are given an annuity-immediate with 11 annual payments of 100 and a ﬁnal payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all payments is 1,000. Calculate the ﬁnal payment. 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 191 Problem 19.14 A loan of 10,000 is being repaid with payment of 300 at the end of each month for as long as necessary plus an additional payment at the time of the last regular payment. What is the amount of the additional payment using an interest rate of 9% compounded monthly. Problem 19.15 A loan of 10,000 is being repaid with payments of 500 starting one month after the loan is made and lasting as long as necessary. A ﬁnal smaller payment is made one month after the last regular payment of 500. What is the amount of the additional smaller payment using an interest rate of 12% compounded monthly? Problem 19.16 You are given an annuity immediate with 11 annual payments of 100 and a ﬁnal balloon payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all the payments is 1,000. Using an annual eﬀective interest rate of 1%, calculate the present value at the beginning of the ninth year of all remaining payments. Problem 19.17 On the ﬁrst day of every January, April, July and October Smith deposits 100 in an account earning i(4) = 0.16.He continues the deposits until he accumulates a suﬃcient balance to begin withdrawals of 200 every 3 months, starting three months after the ﬁnal deposit such that he can make twice as many withdrawals as he made deposits. How many deposits are needed? Problem 19.18 On the ﬁrst day of the month, starting January 1, 1995, Smith deposits 100 in an account earning i(12) = 0.09,with interest credited the last day of each month. In addition, Smith deposits 1,000 in the account every December 31. Around what date does the account ﬁrst exceed 100,000? 192 THE BASICS OF ANNUITY THEORY 20 Solving for the Unknown Rate of Interest of an Annuity In this section we consider the problem of ﬁnding the interest rate given the number of payments and either the present value or the accumulated value of the payments. That is, we want to solve an equation like an = k or sn = k for i. We consider three methods in determining an unknown rate of interest. The ﬁrst is to solve for i by algebraic techniques. For example, the expression k= an = ν+ ν2 + ··· + νn is a polynomial in ν of degree n. If the roots of this polynomial can be determined algebraically, then iis immediately determined. That is, i= ν−1 −1.This method is generally practical only for small values of n. Example 20.1 If a2 = 1.75, ﬁnd an exact expression for i> 0. Solution. We have a2 = 1.75 ⇔ν+ ν2 = 1.75 ⇔ν2 + ν−1.75 = 0 ⇔ν = −1 + √ 8 2 . Since (1 + i)−1 = −1+ √ 8 2 , solving this for i we ﬁnd i= 2√ 8−1 −1 ≈0.09384 Example 20.2 Solve s3 i = 3.31 for i. Solution. We have (1 + i)3 −1 i = 3.31. This reduces to i3 + 3i2 −0.31i= 0 and can be factored to i(i+ 3.1)(i−0.1) = 0. i= 0 is clearly an extraneous solution and is not the correct answer. Hence, since i >0 the correct answer is i= 0.1 = 10% 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 193 Example 20.3 The Honors Society decides to set up a scholarship for university students. They deposit $1,000 in an account for 11 years with the ﬁrst payment a year from now, and then transfer the accumulated value to a perpetuity-immediate paying $500 each year to a deserving student. Assuming the account and the perpetuity are at the same eﬀective annual interest rate i, ﬁnd i. Solution. The accumulated value in the account after the last deposit is 100 s11 . The present value of the perpetuity at that time is 500 a∞; the accumulated value and the value of the perpetuity must be equal. 1000s11 =500a∞ 2 ·(1 + i)11 −1 i =1 i (1 + i)11 =1.5 i=1.5 1 11 −1 ≈3.75% The second method for determining the unknown rate is to use linear interpolation (see Section 13). Example 20.4 An annuity−immediate pays you $100 in each of the next eight years. Its present value is $680. Using linear interpolation of the interest table, ﬁnd an estimate of the annual eﬀective rate i. i s8 a8 2.5% 8.7361 7.1701 3% 8.8923 7.0197 3.5% 9.0517 6.8740 4% 9.2142 6.7327 4.5% 9.3800 6.5959 5% 9.5491 6.4632 6% 9.8975 6.2098 7% 10.2598 5.9713 Solution. From the given table we see that a8 0.035 = 6.8740 and a8 0.04 = 6.7327. Using linear interpolation (See Section 13), we ﬁnd i≈0.035 + (0.04 −0.035) · 6.80 −6.8740 6.7327 −6.8740 = 3.762% 194 THE BASICS OF ANNUITY THEORY Example 20.5 At what interest rate, convertible quarterly, is $16,000 the present value of $1,000 paid at the end of every quarter for ﬁve years? Solution. Let j = i(4) 4 so that the equation of value at time t= 0 becomes 1000a20 j = 16,000 or a20 j = 16. From the tables of interest we see that a20 0.02 = 16.3514 and a20 0.0250 = 15.5892.Thus, using linear interpolation we ﬁnd j = 0.02 + (0.0250 −0.0200) · 16 −16.3514 15.5892 −16.3514 = 0.0223 which gives i(4) = 4(0.0223) = 0.0892 = 8.92% i a20 1.0% 18.0456 1.5% 17.1686 2.0% 16.3514 2.5% 15.5892 The third method is to use the Newton-Raphson iteration method. This method is used to approxi- mate the zeros of the equation f(x) = 0 where f is a diﬀerentiable function. The idea of the method is to start with an initial guess x0 and then ﬁnd the equation of the tangent line at x0.We next ﬁnd the x−intercept of this line say x1, which is closer to the real solution than x0. See Figure 20.1. One can easily see that x1 = x0 −f(x0) f′(x0). Next, we ﬁnd the equation of the tangent line at x= x1 and ﬁnd the x−intercept of this line. Say, x2 = x1 −f(x1) f′(x1). The iteration continues so that we generate a sequence x0,x1,··· ,xn,··· with xn+1 = xn −f(xn) f′(xn) 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 195 xn converging to the solution of f(x) = 0. Figure 20.1 In our context, we want to solve an = k for i using the above mentioned method. Equivalently, we want to solve the equation f(i) = 1 −(1 + i)−n i −k= 0. The iteration formula in this case is given by is+1 =is − [ [1 −(1 + is)−n]i−1 s −k n(1 + is)−n−1i−1 s + [1 −(1 + is)−n]i−2 s ] =is [ 1 + 1 −(1 + is)−n −kis 1 −(1 + is)−n−1{1 + is(n+ 1)} ] The next question regarding this third method is the selection of the initial guessi0.From the power series of (1 + i)−n we ﬁnd an =1 i(1 −(1 + i)−n) =1 i ( 1 − [ 1 −ni+ (−n)(−n−1) 2! i2 + ··· ]) =n [ 1 −n+ 1 2! i+ (n+ 1)(n+ 2) 3! i2 −··· ] 196 THE BASICS OF ANNUITY THEORY Thus, 1 an = 1 n [ 1 − (n+ 1 2! i−(n+ 1)(n+ 2) 3! i2 + ··· )]−1 = 1 n [ 1 + (n+ 1 2! i−(n+ 1)(n+ 2) 3! i2 + ··· ) + (n+ 1 2! i−(n+ 1)(n+ 2) 3! i2 + ··· )2 + ··· ] = 1 n [ 1 + n+ 1 2! i+ n2 −1 12 i2 −n(n2 −1) 24 i3 + ··· ] . The rate of convergence of this series is much faster than the one for the series expansion of an. Now, for the initial guess we use the approximation 1 k ≈1 n [ 1 + n+ 1 2 i0 ] . Solving for i0 we ﬁnd i0 = 2(n−k) k(n+ 1). In practice, the iterations will be carried out until is+1 = is to the required degree of accuracy. Example 20.6 Rework Example 20.5 using Newton-Raphson iterations Solution. Our starting value is i0 = 2(n−k) k(n+1) = 2(20−16) 16(20+1) = 0.0238. Finding several iterations we obtain i1 =0.0238 [ 1 + 1 −(1.0238)−20 −16(0.0238) 1 −(1.0238)−21{1 + 0.0238 ×21} ] = 0.022246 i2 =0.022246 [ 1 + 1 −(1.022246)−20 −16(0.022246) 1 −(1.022246)−21{1 + 0.022246 ×21} ] = 0.0222623 i3 =0.0222623 [ 1 + 1 −(1.0222623)−20 −16(0.0222623) 1 −(1.0222623)−21{1 + 0.0222623 ×21} ] = 0.0222623 Thus, a more accurate interest rate to Example 20.5 is i(4) = 4(0.0222623) = 8.9049% Next, consider the problem of ﬁnding i solution to the equation sni = k. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 197 For this, we let f(i) = (1 + i)n −1 i −k. The Newton-Raphson iterations are given by is+1 = is [ 1 + (1 + is)n −1 −kis (1 + is)n−1{1 −is(n−1)}−1 ] and the initial guess is i0 ≈2(k−n) k(n−1). See Problem 20.20 198 THE BASICS OF ANNUITY THEORY Practice Problems Problem 20.1 (a) The accumulated value of a ﬁve-year annuity-immediate with semiannual payments of $2,000 is $25,000. Use linear interpolation of the interest table to ﬁnd an estimate for i. (b) The present value of a 11-year annuity-due with annual payments of $2,000 is $18,000. Use linear interpolation of the interest table to ﬁnd an estimate for i. Problem 20.2 ‡ To accumulate 8,000 at the end of 3 n years, deposits of 98 are made at the end of each of the ﬁrst n years and 196 at the end of each of the next 2 n years. The annual eﬀective rate of interest is i. You are given (1 + i)n = 2.0. Determine i. Problem 20.3 ‡ For 10,000, Kelly purchases an annuity-immediate that pays 400 quarterly for the next 10 years. Calculate the annual nominal interest rate convertible monthly earned by Kelly’s investment. Problem 20.4 ‡ A 10-year loan of $20,000 is to be repaid with payments at the end of each year. It can be repaid under the following two options: (X) Equal annual payments at the annual eﬀective rate of interest of 8%; (Y) Installments of $2,000 each year plus interest on the unpaid balance at an annual eﬀective rate of i. The sum of the payments under option (X) equals the sum of the payments under option (Y). Determine i. Problem 20.5 Given a∞ = X and ¨a∞ = 1.25X. Calculate the interest rate used. Problem 20.6 A perpetuity pays 1 at the beginning of every year. The present value is 10. Calculate the annual eﬀective rate of interest earned by the perpetuity. Problem 20.7 If ¨a2 ·s2 = 4.05, calculate i. Problem 20.8 A deferred perpetuity pays 500 annually beginning at the end of year 5. The present value of the deferred perpetuity is 4,992. Calculate the annual eﬀective interest rate used to calculate the present value. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 199 Problem 20.9 An annuity due which pays 100 per month for 12 years has a present value of 7,908. Calculate the annual eﬀective interest rate used to determine the present value. Problem 20.10 An annuity which pays 200 at the end of each quarter for 5 years has a present value of 3,600. Calculate the nominal rate of interest compounded quarterly. Problem 20.11 An annuity immediate pays 750 per year for 15 years. The accumulated value of the annuity after 15 years is 15,000. Calculate the annual eﬀective rate of interest used to calculate the accumulated value. Problem 20.12 Adam buys a perpetuity due of 1000 per month for 100,000. Calculate the annual eﬀective rate of interest used to calculate the price of this perpetuity. Problem 20.13 A beneﬁciary receives a 10,000 life insurance beneﬁt. If the beneﬁciary uses the proceeds to buy a 10-year annuity-immediate, the annual payout will be 1,538. If a 20-year annuity-immediate is purchased, the annual payout will be 1,072. Both calculations are based on an annual eﬀective interest rate of i. Find i. Problem 20.14 If ¨sn = 11 and sn+1 = 2an+1 , ﬁnd i. Problem 20.15 If sn = 10 and s2n = 24, ﬁnd νn. Problem 20.16 ‡ Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual eﬀective interest rate of i. The accumulated amount in the account at the end of 40 years is X,which is 5 times the accumulated amount in the account at the end of 20 years. Calculate X. Problem 20.17 The following two investment plans result in the same accumulated amount at the end of 10 years. 1. You lend $10,000, to be repaid in 10 equal installments at the end of each year, at the annual rate of interest of 6%. The installments are deposited to a savings account paying 3% eﬀective annual rate of interest. 2. You deposit $10,000 to an account paying the annual eﬀective rate of interest i. Find i. 200 THE BASICS OF ANNUITY THEORY Problem 20.18 ‡ At an annual eﬀective interest rate of i,i> 0%, the present value of a perpetuity paying 10 at the end of each 3-year period, with the ﬁrst payment at the end of year 3, is 32. At the same annual eﬀective rate of i,the present value of a perpetuity paying 1 at the end of each 4-month period, with ﬁrst payment at the end of 4 months, is X. Calculate X. Problem 20.19 A fund of $17,000 is to be accumulated at the end of ﬁve years with payments at the end of each half-year. The ﬁrst ﬁve payments are $1000 each, while the second ﬁve are $2000 each. Find the nominal rate of interest convertible semiannually earned on the fund. Problem 20.20 Consider solving the equation sn = k. (a) Show that sn = n ( 1 + n−1 2! i+ (n−1)(n−2) 3! i2 + ··· ) . (b) Show that [sn]−1 = 1 n ( 1 −n−1 2 i+ ··· ) . (c) Show that the Newton-Raphson iterations are given by is+1 = is [ 1 + (1 + is)n −1 −kis (1 + is)n−1{1 −(n−1)is}−1 ] . (d) Show that an initial guess for Newton-Raphson method is given by i0 ≈2(n−k) k(n−1). Problem 20.21 Use the Newton-Raphson iteration method with three iterations to estimate the rate of interest if s20 i = 25. Problem 20.22 Eric receives 12000 from a life insurance policy. He uses the fund to purchase two diﬀerent annuities, each costing 6000. The ﬁrst annuity is a 24 year annuity immediate paying K per year to himself. The second annuity is an 8 year annuity immediate paying 2 K per year to his son. Both annuities are based on an annual eﬀective interest rate of i> 0. Determine i. Problem 20.23 Jeﬀ deposits 100 at the end of each year for 13 years into Fund X. Antoinette deposits 100 at the end of each year for 13 years into Fund Y. Fund X earns an annual eﬀective rate of 15% for the ﬁrst 5 years and an annual eﬀective rate of 6% thereafter. Fund Y earns an annual eﬀective rate of i. At the end of 13 years, the accumulated value of Fund X equals the accumulated value of Fund Y. Calculate i. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 201 Problem 20.24 Dottie receives payments of X at the end of each year for nyears. The present value of her annuity is 493. Sam receives payments of 3 X at the end of each year for 2 n years. The present value of his annuity is 2748. Both present values are calculated at the same annual eﬀective interest rate. Determine νn. Problem 20.25 ‡ A discount electronics store advertises the following ﬁnancing arrangement: “We don’t oﬀer you confusing interest rates. We’ll just divide your total cost by 10 and you can pay us that amount each month for a year.” The ﬁrst payment is due on the date of sale and the remaining eleven payments at monthly intervals thereafter. Calculate the eﬀective annual interest rate the store’s customers are paying on their loans. 202 THE BASICS OF ANNUITY THEORY 21 Varying Interest of an Annuity In this section we consider situations in which interest can vary each period, but compound interest is still in eﬀect. Let ik denote the rate of interest applicable from time k−1 to time k. We consider ﬁrst the present value of an n−period annuity-immediate. We consider the following two variations. The ﬁrst is when ik is applicable only for period k regardless of when the payment is made. That is, the rate ik is used only in period kfor discounting all payments. In this case the present value is given by an = (1 + i1)−1 + (1 +i1)−1(1 + i2)−1 + ··· + (1 +i1)−1(1 + i2)−1 ···(1 + in)−1. The second variation is when the rate ik is used as the eﬀective rate for each period i≤k when a payment is made at time k. In this case, the present value is an = (1 + i1)−1 + (1 +i2)−2 + ··· + (1 +in)−n. Present value of annuity-due can be obtained from present value of annuity-immediate by using the formula ¨an = 1 + an−1 . We now turn to accumulated values. We will consider an annuity-due. Again we consider two diﬀerent situations. If ik is applicable only for period k regardless of when the payment is made, then the accumulated value is given by ¨sn = (1 + i1)(1 + i2) ···(1 + in) + ··· + (1 +in−1)(1 + in) + (1 +in). For ik applicable for all periods i≥k, the accumulated value is ¨sn = (1 + i1)n + (1 +i2)n−1 + ··· + (1 +in). Accumulated values of annuity-immediate can be obtained from the accumulated values of annuity- due using the formula sn+1 = ¨sn + 1. Example 21.1 Find the accumulated value of a 12-year annuity-immediate of $500 per year, if the eﬀective rate of interest (for all money) is 8% for the ﬁrst 3 years, 6% for the following 5 years, and 4% for the last 4 years. 21 VARYING INTEREST OF AN ANNUITY 203 Solution. The accumulated value of the ﬁrst 3 payments to the end of year 3 is 500s3 0.08 = 500(3.2464) = $1623.20. The accumulated value of the ﬁrst 3 payments to the end of year 8 at 6% and then to the end of year 12 at 4% is 1623.20(1.06)5(1.04)4 = $2541.18. The accumulated value of payments 4, 5, 6, 7, and 8 at 6% to the end of year 8 is 500s5 0.06 = 500(5.6371) = $2818.55. The accumulated value of payments 4, 5, 6, 7, and 8 to the end of year 12 at 4% is 2818.55(1.04)4 = $3297.30. The accumulated value of payments 9, 10, 11, and 12 to the end of year 12 at 4% is 500s4 0.04 = 500(4.2465) = $2123.23. The accumulated value of the 12-year annuity immediate is 2541.18 + 3297.30 + 2123.23 = $7961.71 Example 21.2 How much must a person deposit now into a special account in order to withdraw $1,000 at the end of each year for the next ﬁfteen years, if the eﬀective rate of interest is equal to 7% for the ﬁrst ﬁve years, and equal to 9% for the last ten years? Solution. The answer is PV =1000(a5 0.07 + a10 0.09(1.07)−5) =1000(4.1002 + 4.5757) = $8675.90 Example 21.3 Find s5 , if δt = 0.02t for 0 ≤t≤5. Solution. s5 is equal to the sum of the accumulated value of the individual payments. That is, s5 = e ∫5 1 δtdt + e ∫5 2 δtdt + ··· + 1 Since e ∫5 r δtdt = e ∫5 r 0.02tdt = e.25−0.01r2 we ﬁnd s5 = e0.24 + e0.21 + e0.16 + e0.09 + 1 ≈5.7726 204 THE BASICS OF ANNUITY THEORY Practice Problem Problem 21.1 James deposits 1,000 into an account at the end of each year for the next 6 years. The account earns 5% interest. James also deposits 1,000 at the end of each of years 7 through 10 into another account earning 4%. Calculate the total amount James will have in both accounts at the end of ten years. Problem 21.2 A fund earns 5% during the next six years and 4% during years 7 through 10. James deposits 1,000 into the account now. Calculate his accumulated value after 10 years. Problem 21.3 A fund earns 5% during the next six years and 4% during years 7 through 10. James deposits 1,000 into the account at the end of each year for the next ten years. Calculate his accumulated value after 10 years. Problem 21.4 A perpetuity pays $1,200 at the beginning of each year with the ﬁrst payment being made imme- diately. The trust funding the perpetuity will earn an annual eﬀective interest rate of 10% for the ﬁrst 10 years, 8% for the second 10 years and 5% thereafter. Calculate the amount needed to fund the perpetuity immediately before the ﬁrst payment. Problem 21.5 You are given: (i) X is the current value at time 2 of a 20-year annuity-due of 1 per annum. (ii) The annual eﬀective interest rate for year t is 1 8+t. Find X. Problem 21.6 Fund Apays interest of 4% on all money deposited in the ﬁrst 5 years and 5% on all money deposited thereafter. Fund B pays interest of 4% during the ﬁrst 5 years and 5% thereafter without regard to the date the deposit was made. (a) If Kevin deposits 500 into each fund now, how much will he have after 10 years in each fund? (b) Heather deposits 100 at the start of each year for 10 years into Fund B. Lisa deposits 100 at the end of each year into Fund A.Who will have more after 10 years and how much more will that person have? 21 VARYING INTEREST OF AN ANNUITY 205 Problem 21.7 A loan P is to be repaid by 10 annual payments beginning 6 months from the date of the loan. The ﬁrst payment is to be half as large as the others. For the ﬁrst 4 1 2 years interest is ieﬀective; for the remainder of the term interest is j eﬀective. Find an expression for the ﬁrst payment. Problem 21.8 Find the present value of an annuity-immediate which pays 1 at the end of each half-year for ﬁve years, if the rate of interest is 8% convertible semiannually for the ﬁrst three years and 7% convertible semiannually for the last two years. Problem 21.9 Find the present value of an annuity-immediate which pays 1 at the end of each half-year for ﬁve years, if the payments for the ﬁrst three years are discounted at 8% convertible semiannually and the payments for the last two years are discounted at 7% convertible semiannually. Problem 21.10 Given that δt = 1 20−t, t≥0, ﬁnd s10 . Problem 21.11 What is the cost of an annuity of $300 per year for 15 years where the future interest rate for the ﬁrst 5 years is 4%, the rate for the next 5 years is 5%, and the rate for the last 5 years is 6%. Problem 21.12 A perpetuity pays $200 each year, with the ﬁrst payment scheduled one year from now. Assume that the eﬀective annual interest rate for the next ﬁve years is 8%, and thereafter it is 5%. Find the present value of the perpetuity. 206 THE BASICS OF ANNUITY THEORY 22 Annuities Payable at a Diﬀerent Frequency than Interest is Convertible In this section we address annuities for which payment period and the interest conversion period diﬀer and for which the payments are level amount. For example, an annuity-due with monthly payment of $2,000 and with interest rate say 12% convertible quarterly. The approach we use in this section for solving this type of problem consists of the following two steps: (i) Find the rate of interest convertible at the same frequency as payments are made, which is equivalent to the given rate of interest. In the example above, we need to ﬁnd the interest rate convertible monthly which is equivalent to the rate of 12% convertible quarterly. In this case, we ﬁnd i(12) = 12 [( 1 + 0.12 4 )4 12 −1 ] = 11.9%. (ii) Using this new interest rate, ﬁnd the value of the annuity using the techniques discussed in the previous sections. We illustrate this approach in the following examples. Example 22.1 You want to accumulate $250,000. You intend to do this by making deposits of $2,000 into an investment account at the beginning of each month. The account earns 12% interest, convertible quarterly. How many months will it take to reach your goal? (Round to the nearest month.) Solution. We are given an interest rate of 12% per quarter. Let j be the equivalent rate of interest per month, which is the payment period. We have (1 + j)12 = ( 1 + 0.12 4 )4 . Solving for j we ﬁnd j = (1.03) 1 3 −1 = 0.9901634%.Thus, 250,000 = 2000¨snj →(1+j)n−1 j j+1 = 125 →(1 +j)n = 2.22557 → n= ln 2.22557 ln 1.009901634 = 81.195 ≈81 months Example 22.2 A loan of $3,000 is to be repaid with quarterly installments at the end of each quarter for 5 years. If the rate of interest charged on the loan is 10% convertible semi-annually, ﬁnd the amount of each quarterly payment. Solution. We are given an interest rate of 5% per half-year. Letjbe the equivalent rate of interest per quarter, which is the payment period. We have (1 + j)4 = (1 + 0.05)2 →j = (1.05) 1 2 −1 = 0.024695. If R denote the quarterly payment then the equation of value at time t= 0 is Ra20 j = 3000.Solving for R we ﬁnd R= 3000 15.6342 = $191.89 22 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE207 Example 22.3 The nominal rate of interest, convertible semiannually, is 6%. An annuity-immediate pays $50 each month for ﬁve years. Find the accumulated value of this annuity at the time of the last payment. Solution. We are given an interest rate of 3% per half-year. Let j be the equivalent rate of interest per month, which is the payment period. We have (1 + j)12 = (1 + 0.03)2 →j = (1.03) 1 6 −1 = 0.49386%. The accumulated value is 50s60 j = 50 ·(1+j)60−1 j = $3,481.91 Example 22.4 At what annual eﬀective rate of interest will payments of $100 at the end of every quarter accumulate to $2500 at the end of ﬁve years? Solution. Let j be the eﬀective rate per quarter. We are given that 100 s20 j = 2500 or (1 +j)20 −1 = 25j.Let f(j) = (1 +j)20 −25j−1.By trial and error we ﬁnd f(0.02) = −0.014053 and f(0.024) = 0.006938. Using linear interpolation we ﬁnd j ≈0.02 + 0.014053 × 0.024 −0.02 0.006938 + 0.014053 = 0.02268. Let i be the annual eﬀective rate of interest. Then i= (1 + 0.02268)4 −1 = 9.39%. Alternatively, we can use the Newton-Raphson iterations to estimate j. Letting j0 = 2(k−n) k(n−1) = 2(25−20) 25(20−1) ≈0.021053 be the starting value of the interations, the ﬁrst three iterations of Newton- Raphson method are j1 =0.021053 [ 1 + 1.02105320 −1 −25(0.021053) (1.021053)19(1 −0.021053 ×19) −1 ] = 0.02288 j2 =0.02288 [ 1 + 1.0228820 −1 −25(0.02288) (1.02288)19(1 −0.02288 ×19) −1 ] = 0.02277 j3 =0.02277 [ 1 + 1.0227720 −1 −25(0.02277) (1.02277)19(1 −0.02277 ×19) −1 ] = 0.02285 j4 =0.02285 [ 1 + 1.0228520 −1 −25(0.02285) (1.02285)19(1 −0.02285 ×19) −1 ] = 0.02285 Thus, j = 2.285% and consequently i= (1.02285)4 −1 = 0.0946 = 9.46% 208 THE BASICS OF ANNUITY THEORY Practice Problems Problem 22.1 Calculate the present value of an annuity due that pays 500 per month for 10 years. The annual eﬀective interest rate is 6%. Problem 22.2 Calculate the present value of an annuity immediate of 100 per quarter for 6 years using a nominal interest rate of 9% compounded monthly. Problem 22.3 Calculate the accumulated value of an annuity which pays 1,000 at the beginning of each year for 10 years. Use an interest rate of i(12) = 0.08. Problem 22.4 A perpetuity pays 1,000 at the end of each quarter. Calculate the present value using an annual eﬀective interest rate of 10%. Problem 22.5 A perpetuity pays 100 at the beginning of each quarter. Calculate the present value using a force of interest of 0.06. Problem 22.6 A perpetuity due pays 6,000 at the start of each year. Calculate the present value using i(4) = 0.06. Problem 22.7 Find the accumulated value at the end of four years of an investment fund in which $100 is deposited at the beginning of each quarter for the ﬁrst two years and $200 is deposited at the beginning of each quarter for the second two years, if the fund earns 12% convertible monthly. Problem 22.8 A 20-year annuity-due pays $1,000 each year. If i(4) = 8%, ﬁnd the present value of this annuity three years before the ﬁrst payment. Problem 22.9 Find the accumulated value 18 years after the ﬁrst payment is made of an annuity-immediate on which there are 8 payments of $2,000 each made at two-year intervals. The nominal interest rate is 7% convertible semiannually. Round your answer to the nearest dollar. 22 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE209 Problem 22.10 Find the present value of a ten-year annuity which pays $400 at the beginning of each quarter for the ﬁrst 5 years, increasing to $600 per quarter thereafter. The annual eﬀective rate of interest is 12%. Round to the nearest dollar. Problem 22.11 A sum of $100 is placed into a fund at the beginning of every other year for eight years. If the fund balance at the end of eight years is $520, ﬁnd the rate of simple interest earned by the fund. Problem 22.12 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using an annual eﬀective interest rate of 12%. Problem 22.13 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using a nominal interest rate of 12% compounded monthly. Problem 22.14 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using a nominal interest rate of 12% compounded six times per year. Problem 22.15 A 30−year annuity pays $1,000 every quarter. Ifi(12) = 6%,what is the present value of this annuity six months prior to the ﬁrst payment? Problem 22.16 You deposit $2,500 into an account at the beginning of every month for 15 years. The interest rate on the account is i(2) = 8%. Find the accumulated value in the account nine months after the last deposit. Problem 22.17 A perpetuity will make annual payments of $3,000, with the ﬁrst payment occurring 5 months from now. The interest rate is 12% convertible monthly. Find the present value of this annuity. Problem 22.18 You want to accumulate $2,000,000 over the next 30 years. You intend to do this by making deposits of X into an investment account at the end of each month, for 30 years. The account earns 12% convertible semi-annually. Find X. 210 THE BASICS OF ANNUITY THEORY 23 Analysis of Annuities Payable Less Frequently than In- terest is Convertible In this section we analyze annuities where the compounding frequency is larger than the payment frequency. We start with the case of annuity −immediate. Let kbe the number of interest conversion periods in one payment period. Consider an annuity−immediate that pays 1 at the end of each payment period. Let i be the rate per conversion period and n the total number of conversion periods for the term of the annuity. We will assume that each payment period contains an integral number of interest conversion periods so that n and k are positive in- tegers and also we assume that n is divisible by k. The total number of annuity payments made is then n k. Let Ldenote the present value of an annuity−immediate which pays 1 at the end of each k interest conversion periods for a total of n interest conversion periods. A time diagram of this sitatuation is shown in Figure 32.1. Figure 23.1 Using the equation of value at time t= 0 we ﬁnd L=νk + ν2k + ·+ ν n k·k =νk(1 + νk + ν2k + ·+ ν(n k−1)k) =νk ×1 −(νk) n k 1 −νk = νk1 −νn 1 −νk = 1 −νn (1 + i)k −1 = 1−νn i (1+i)k−1 i = an sk . The accumulated value of this annuity immediately after the last payment is made is (1 + i)nan sk = sn sk . 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE211 Remark 23.1 The annuity of making a payment of $1 at the end of each k interest conversion periods for a total of n k payments is equivalent to the annuity that consists of a periodic payment of 1 sk for n periods. A time diagram illustrating this annuity is given in Figure 23.2. Figure 23.2 Example 23.1 Find the present value and the accumulated value of an annuity −immediate with payments of $1,500 every 4 years from years 4 to 40. The interest earned is 8% converted annually. Solution. With i= 0.08,the present value is 1500 a40 s4 = 1500 ·11.9246 4.5061 = $3,969.48 and the accumulated value is (1.08)40(3969.48) = $86,235.05 Next, we consider the case of an annuity−due. Let ¨Lbe the present value of an annuity which pays 1 at the beginning of each k interest conversion periods for a total of nconversion interest periods. Then we have the following time diagram. Figure 23.3 Using the equation of value at time t= 0 we ﬁnd 212 THE BASICS OF ANNUITY THEORY ¨L=1 + νk + ν2k + ·+ νn−k =1 + (νk) + (νk)2 + ·+ (νk) n−k k =1 −(νk) n−k k +1 1 −νk = 1 −νn 1 −νk = 1−νn i 1−νk i = an ak . The accumulated value of this annuity k interest conversion periods after the last payment is (1 + i)nan ak = sn ak . Remark 23.2 It is easy to see that the annuity described above is equivalent to the annuity that consists of n payments of 1 ak at the end of each interest conversion period. Example 23.2 An annuity pays 1,000 at time 0, time 3, time 6, etc until 20 payments have been made. Calculate the present value and the accumulated value using an annual eﬀective interest rate of 6%. Solution. With i= 0.06,the present value is 1000 a60 a3 = 1000 ·16.16142771 2.67301194 = 6046.15 and the accumulated value is (1.06)60(6046.15) = 199,448.48 In the case of a perpetuity −immediate, the present value is νk + ν2k + ··· = νk 1 −νk = 1 (1 + i)k −1 = 1 i·sk = lim n→∞ an sk . Example 23.3 The present value of a perpetuity paying 1 at the end of every 3 years is 125 91 . Find i. 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE213 Solution. We have 1 i·s3 = 125 91 .Thus, 1 (1+i)3−1 = 125 91 →(1 +i)3 = (6 5 )3 →1 +i= 6 5 →i= 6 5 −1 = 0.2 = 20% In the case of a perpetuity-due, the present value is 1 + νk + ν2k + ··· = 1 1 −νk =1 i · i 1 −(1 + i)−k = 1 i·ak = lim n→∞ an ak . Example 23.4 A perpetuity paying 1 at the beginning of each year has a present value of 20. If this perpetuity is exchanged for another perpetuity paying Rat the beginning of every 2 years with the same eﬀective annual rate as the ﬁrst perpetuity, ﬁnd R so that the present values of the two perpetuities are equal. Solution. An equation of value now for the ﬁrst perpetuity is 1 d = 20 implying i = 1 19 . Also, given that 20 = R ia2 = R 1−ν2 = R (1−ν)(1+ν) = R (0.05)(1.95) →R= (20)(0.0975) = 1.95 Another problem that comes under the category of annuities payable less frequently than inter- est is convertible is the problem of ﬁnding the value of a series of payments at a given force of interest δ.We illustrate this in the next example. Example 23.5 Find an expression for the present value of an annuity on which payments are 100 per quarter for 5 years, just before the ﬁrst payment is made, if δ= 0.08. Solution. Let j be the rate per quarter equivalent to δ. Then (1 + j)4 = e0.08 or 1 + j = e0.02. Thus, PV = 100¨a20 j = 100 [ 1−(1+j)−20 1−(1+j)−1 ] = 100 [ 1−e−0.40 1−e−0.02 ] In some cases, the number of conversion periods in a payment period is not an integral, i.e. k >1 but k is not an integer. In this case, we handle this problem using the basic principles, i.e. to ﬁnd the present value or the accumulated value we write a sum of present values or accumulated values of the individual payments. An illustration of this type is shown next. 214 THE BASICS OF ANNUITY THEORY Example 23.6 Find an expression for the present value of an annuity on which payments are 1 at the beginning of each 4−month period for 12 years, assuming a rate of interest per 3 −month period i. Solution. Let i be the eﬀective rate of interest per 3-month period and j the eﬀective interest rate per 4−month. Then (1 + j)3 = (1 + i)4 ⇒1 + j = (1 + i) 4 3 . The present value is given by ¨a36 j = 1−(1+j)−36 1−(1+j)−1 = 1−ν48 1−ν 4 3 We conclude this section by pointing out that the approach discussed in this section can be general- ized to ﬁnding annuities payable less frequently than interest is convertible on any date, as discussed in Section 17. Example 23.7 Find the present value of an annuity−immediate in which there are a total of 5 payments of 1, the ﬁrst to be made at the end of seven years, and the remaining four payments at three-year intervals, at an annual eﬀective rate of 6%. Solution. The time diagram for this example is given in Figure 23.4. Figure 23.4 The present value is given by ν4 a15 s3 = (1.06)−4 1 −(1.06)−15 (1.06)3 −1 = $2.416 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE215 Practice Problems Problem 23.1 An annuity pays 1,000 at time 0, time 3, time 6, etc until 20 payments have been made. Calculate the accumulated value immediately after the last payment using an annual eﬀective interest rate of 6%. Problem 23.2 Give an expression of the the present value, 3 years before the ﬁrst payment is made, of an annuity on which there are payments of 200 every 4 months for 12 years: (a) expressed as an annuity-immediate; (b) expressed as an annuity-due. Assume monthly rate of interest. Problem 23.3 Find an expression for the present value of an annuity-due of 600 per annum payable semiannually for 10 years, if d(12) = 0.09. Problem 23.4 An annuity pays 500 at the start of each year for 15 years. Calculate the accumulated value of the annuity after 15 years assuming a constant force of interest of 5%. Problem 23.5 A perpetuity pays 100 at the end of every third year. The ﬁrst payment is in three years. Calculate the present value of the perpetuity using a constant force of interest of 10%. Problem 23.6 Find an expression for the present value of an annuity in which there are a total of rpayments of 1, the ﬁrst to be made at the end of seven years, and the remaining payments at three-year intervals, at an annual eﬀective rate i, expressed as (a) annuity-immediate (b) annuity-due. Problem 23.7 A 30-year annuity-immediate with 2,000 payable at the end of every 6 months and a 30-year annuity- immediate with 10,000 payable at the end of every 6 years are to be replaced by a perpetuity paying R every 3 months. You are given that i(4) = .08. Find R. 216 THE BASICS OF ANNUITY THEORY Problem 23.8 The payments you received from a 20 year annuity −immediate paying 500 every 6 months have been left to accumulate in a fund and are now worth 40,000. If i(12) = 0.06 is the rate earned by your fund, calculate how long it is since the last annuity payment was made. Problem 23.9 A perpetuity of $1,000 payable at the end of every 6 months and a perpetuity of $10,000 payable at the end of every 6 years are to be replaced by a 30 year annuity paying R every 3 months. You are given that i(4) = 0.08. Find R. Problem 23.10 An investment of $1,000 is used to make payments of $100 at the end of each year for as along as possible with a smaller payment to be made at the time of the last regular payment. If interest is 7% convertible semiannually, ﬁnd the number of payments and the amount of the total ﬁnal payment. Problem 23.11 Show that the present value at time 0 of 1 payable at times 7, 11, 15, 19, 23, and 27, where the eﬀective rate per annum is i, is given by a28 −a4 s3 + a1 . Problem 23.12 ‡ Fence posts set in soil last 9 years and cost $Y each while fence posts set in concrete last 15 years and cost $(Y + X). The posts will be needed for 35 years. Find an expression of X such that a fence builder would be indiﬀerent between the two types of posts? Problem 23.13 A perpetuity of $750 payable at the end of every year and a perpetuity of $750 payable at the end of every 20 years are to be replaced by an annuity of R payable at the end of every year for 30 years. If i(2) = 0.04, ﬁnd an expression for R. Problem 23.14 Given that δt = 2 10+t, t≥0, ﬁnd a4 . Problem 23.15 At a nominal rate of interest i, convertible semiannually, the present value of a series of payments of 1 at the end of every 2 years forever, is 5.89. Calculate i. 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE217 Problem 23.16 Gus deposits 25 into a bank account at the beginning of each 3-year period for 18 years (i.e. there is no deposit in year 18). The account credits interest at an annual eﬀective rate of interest of i. The accumulated amount in the account after 18 years is X, which is four times as large as the accumulated amount in the account after 9 years (excluding the deposit made at time t = 9). Calculate X. Problem 23.17 Find the present value of an annuity−due in which there are a total of 5 payments of 1, the ﬁrst to be made at the end of seven years, and the remaining four payments at three-year intervals, at an annual eﬀective rate of 6%. Problem 23.18 The present value today of a 20 year annuity −immediate paying 500 every 6 months but with the ﬁrst payment deferred tyears has a present value of 5,805.74. If i(12) = 0.09 is the interest rate used to calculate the present value, ﬁnd t. Problem 23.19 A 20−year annuity−due pays $1,000 each year. If i(4) = 8%, ﬁnd the present value of this annuity three years before the ﬁrst payment. 218 THE BASICS OF ANNUITY THEORY 24 Analysis of Annuities Payable More Frequently than In- terest is Convertible In a discussion parallel to the one in section 23, we discuss annuities payable more frequently than interest is convertible. We ﬁrst consider the case of an annuity-immediate. Let m denote the number of payments per one interest conversion period. Let n be the total number of conversion periods in the term of the annuity. Then the total number of payments for the term of the annuity is mn.Let ibe the interest rate per conversion period. We will assume that the number of payments per conversion period is an integral number. Payments of 1 are being made per interest conversion period with 1 m being made at the end of each mth of an interest conversion period. The present value of such an annuity will be denoted by a(m) n . We have the following time diagram. Figure 24.1 The formula for the present value is derived as follows a(m) n = 1 m [ ν 1 m + ν 2 m + ··· + νn−1 m + νn ] = ν 1 m m [ 1 −(ν 1 m)mn 1 −ν 1 m ] = 1 −νn m [ (1 + i) 1 m −1 ]= 1 −νn i(m) . The accumulated value of this annuity immediately after the last payment is made is given by s(m) n = (1 + i)na(m) n = (1 + i)n −1 i(m) . Formulas of a(m) n and s(m) n in terms of an and sn are a(m) n = i i(m) an = s(m) 1 an and s(m) n = i i(m) sn = s(m) 1 sn. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE219 One consideration which is important in practice involves the proper coeﬃcients for annuity payable mthly. Each payment made is of amount 1 m,while the coeﬃcient of the symbol a(m) n is 1. In general, the proper coeﬃcient is the total amount paid during one interest conversion period, and not the amount of each actual payment. The total amount paid during one interest conversion period is called periodic rent of the annuity. Example 24.1 A loan of $3,000 is to be repaid with quarterly installments at the end of each quarter for ﬁve years. If the rate of interest charged is 10% converted semiannually, ﬁnd the amount of each quarterly payment. Solution. Let Rbe the amount of quarterly payment. In one interest conversion period there are two payments. Then 2Ra10 (2) 0.05 = 3000 →R= 1500 1−(1.05)−10 2[(1.05)0.5−1] = 191.89 Example 24.2 What is the accumulated value following the last payment of $100 deposited at the end of each month for 30 years into an account that earns an eﬀective annual rate of 10%? Solution. Each conversion period contains 12 payment periods. The term of the loan is 30 interest conversion periods. Thus, 12 ·100 · (1.1)30−1 12[(1.1) 1 12 −1] = 206,284.33 Example 24.3 The accumulated amount of an annuity-immediate of $R per year payable quarterly for seven years is $3317.25. Find R if i(1) = 0.05. Solution. Each R is paid quarterly at R 4 . The equation of value at time t= 7 years is Rs(4) 7 = 3317.25. But s(4) 7 = i i(4) s7 . On the other hand, ( 1 + i(4) 4 )4 = 1.05 ⇒i(4) = 4[(1.05) 1 4 −1] = 0.0491. 220 THE BASICS OF ANNUITY THEORY Hence, we obtain 3317.25 = R 0.05 0.0491 (8.1420). Solving for R; we ﬁnd R= $400 In the case of annuity-due, the amount 1 m is payable at the beginning of the mth period of an interest conversion period for a total of n interest conversion period. The present value is denoted by ¨a(m) n . A time diagram of this case is shown in Figure 24.2. Figure 24.2 The present value is determined as follows: ¨a(m) n = 1 m[1 + ν 1 m + ν 2 m + ··· + νn−1 m] = 1 m [ 1 −(ν 1 m)mn 1 −ν 1 m ] = 1 −νn m[1 −(1 + i)−1 m] = 1 −νn d(m) . The accumulated value one mth of an interest conversion period after the last payment is made is given by ¨s(m) n = (1 + i)n¨a(m) n = (1 + i)n −1 d(m) . It follows from the formulas of an,sn,¨a(m) n , and ¨s(m) n that ¨a(m) n = i d(m) an = ¨s(m) 1 an and ¨s(m) n = i d(m) sn = ¨s(m) 1 sn. Example 24.4 What is the present value of the payment of $150 per quarter for 10 years if the eﬀective annual interest rate is 6% and the ﬁrst payment is due today. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE221 Solution. The answer is 600·¨a10 (4) 0.06 = 600·a10 0.06 · i d(4) = 600 ·7.3601 ·1.037227 = $4,580.46 Example 24.5 Express ¨a(12) n in terms of a(2) n with an adjustment factor. Solution. We have ¨a(12) n = i d(12) an and a(2) n = i i(2) an. Thus, ¨a(12) n = i(2) d(12) a(2) n The following identities involving an,sn,¨a(m) n , and ¨s(m) n are analogous to the identity involving an,sn,¨an, and ¨sn discussed in earlier sections. Theorem 24.1 (a) 1 a(m) n = 1 s(m) n + i(m) (b) 1 ¨a(m) n = 1 ¨s(m) n + d(m) (c) ¨a(m) n = (1 + i) 1 ma(m) n = [ i i(m) + i m ] an (d) ¨s(m) n = (1 + i) 1 ms(m) n = [ i i(m) + i m ] sn (e) ¨a(m) n = 1 m + a(m) n−1 m (f) ¨s(m) n = s(m) n+ 1 m −1 m Proof. See Problem 24.10 Next, consider an inﬁnite payment of 1 m at the end of mth of an interest conversion period. Let a(m) ∞ denote the present value. A time diagram describing this case is shown in Figure 24.3. 222 THE BASICS OF ANNUITY THEORY The present value is determined as follows. a(m) ∞ = 1 m[ν 1 m + ν 2 m + ν 3 m + ···] = 1 m ∞∑ p=1 ν p m = 1 m ν 1 m 1 −ν 1 m = 1 m(1 + i) 1 m(1 −ν 1 m) = 1 m((1 + i) 1 m −1) = 1 i(m) Figure 24.3 Example 24.6 An annuity pays 100 at the end of each month forever. Calculate the present value of the annuity using an annual eﬀective interest rate of 8%. Solution. The present value is 12 ·100a(12) ∞ = 1200 12[(1.08) 1 12 −1] = 15,542.36 Now, consider an inﬁnite payment of 1 m at the beginning of mth of an interest conversion pe- riod. Let the present value be denoted by ¨a(m) ∞ . Figure 24.4 describes this situation. Figure 24.4 The present value is determined as follows. ¨a(m) ∞ = 1 m[1 + ν 1 m + ν 2 m + ν 3 m + ···] = 1 m ∞∑ p=0 ν p m = 1 m[1 −ν 1 m] = 1 d(m) , ν 1 m <1 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE223 Example 24.7 At what annual eﬀective rate of interest is the present value of a series of payments of $1 for every six months forever, with the ﬁrst payment made now, equal to $10? Solution. We have 10 = 2¨a(2) ∞ = 2 d(2) →d(2) = 0.2 →i= (1 −0.1)−2 −1 = 23.46% In some cases, each conversion period does not contain an integral number of payments, i.e. m> 1 but mis not an integer. In this case, we handle this problem using the basic principles, i.e. to ﬁnd the present value or the accumulated value we write a sum of present values or accumulated values of the individual payments. An illustration of this type is shown next. Example 24.8 Find an expression for the present value of an annuity which pays 1 at the beginning of each 3-month period for 12 years, assuming a rate of interest per 4-month period. Solution. Working in years the annual payments are 4 ×1 = $4. Thus, the present value is 4¨a(4) 12 = 4 ·1 −(1 + i)−12 d(4) . But 1 + i= ( 1 −d(4) 4 )−4 ⇒d(4) = 4[1 −(1 + i)−1 4 ]. Thus 4¨a(4) 10 =4 1 −(1 + i)−12 4[1 −(1 + i)−1 4 ] =1 −ν12 1 −ν 1 4 =1 −ν48 j 1 −νj =1 −ν36 k 1 −ν 3 4 k where j is the eﬀective interest rate per 3-month and k is the eﬀective interest rate per 4-month Finally, it is possible to generalize the approach to ﬁnding annuity values on any date, as discussed in Section 17, to annuities payable more frequently than interest is convertible. 224 THE BASICS OF ANNUITY THEORY Example 24.9 Payments of $400 at the end of each month are made over a ten-year period. Find expressions for (a) the present value of these payments two years prior to the ﬁrst payment; (b) the accumulated value three years after the ﬁnal payment. Use symbols based on an eﬀective rate of interest i. Solution. (a) The present value is given by 12 ·400νa(12) 10 . (b) The accumulated value is given by 12 ·400s(12) 10 (1 + i)3 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE225 Practice Problems Problem 24.1 The present value of an annuity is denoted by 120a(12) n .What is the amount of the monthly payment of this annuity? Problem 24.2 The present value of an annuity that pays 100 at the end of each year for n years using an annual eﬀective interest rate of 10.25% is 1,000. Calculate the present value of an annuity that pays 100 at the end of every six months for n years using the same interest rate. (Note: You should work this problem without ﬁnding n.) Problem 24.3 Suppose you deposit $200 per month, at the end of each month, in a savings account paying the eﬀective annual rate of interest 6%. How much will be in the account after ten years? (i.e. just after your 120th deposit?) Problem 24.4 A loan of $150,000 is repaid by equal installments at the end of each month, for 25 years. Given that the nominal annual rate of interest, convertible semiannually, is 8%, compute the amount of monthly installment. Problem 24.5 Suppose you deposit $300 at the end of every three months for 10 years into a bank account paying the annual eﬀective rate of interest 4%. What is the accumulated amount after 10 years? Problem 24.6 An annuity pays 100 at the end of each month for 20 years. Using a nominal rate of interest of 4% compounded quarterly, calculate the current value of the annuity at the end of the 3rd year. Problem 24.7 Which of the following are true? (i) an = i(m) i a(m) n (ii) s(m) n = (1 + i)na(m) n (iii) a(∞) ∞ = ¨a(∞) ∞ Problem 24.8 If 3a(2) n = 2a(2) 2n = 45s(2) 1 , ﬁnd i. 226 THE BASICS OF ANNUITY THEORY Problem 24.9 A sum of 10,000 is used to buy a deferred perpetuity-due paying $500 every six months forever. Find an expression for the deferred period expressed as a function of d, the annual discount rate. Problem 24.10 Prove the following identities: (a) 1 a(m) n = 1 s(m) n + i(m) (b) 1 ¨a(m) n = 1 ¨s(m) n + d(m) (c) ¨a(m) n = (1 + i) 1 ma(m) n = [ i i(m) + i m ] an (d) ¨s(m) n = (1 + i) 1 ms(m) n = [ i i(m) + i m ] sn (e) ¨a(m) n = 1 m + a(m) n−1 m (f) ¨s(m) n = s(m) n+ 1 m −1 m Problem 24.11 Find the present value of a ten-year annuity which pays $400 at the beginning of each quarter for the ﬁrst 5 years, increasing to $600 per quarter thereafter. The annual eﬀective rate of interest is 12%. Round to the nearest dollar. Use the approach developed in this section. Problem 24.12 A family wishes to provide an annuity of $100 at the end of each month to their daughter now entering college. The annuity will be paid for only nine months each year for four years. Prove that the present value one month before the ﬁrst payment is 1200¨a4 a(12) 9/12 . Problem 24.13 The nominal rate of interest, convertible semiannually, is 6%. An annuity−immediate pays $50 each month for ﬁve years. Find the accumulated value of this annuity at the time of the last payment. Problem 24.14 You deposit $2,500 into an account at the beginning of every month for 15 years. The interest rate on the account is i(2) = 8%. Find the accumulated value in the account nine months after the last deposit. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE227 Problem 24.15 You want to accumulate $1,000,000. You intend to do this by making deposits of $5,000 into an investment account at the end of each month, until your account balance equals $1,000,000. The account earns 8% convertible quarterly. Determine the number of monthly deposits you will need to make to achieve your goal. 228 THE BASICS OF ANNUITY THEORY 25 Continuous Annuities In this section we consider annuities with a ﬁnite term and an inﬁnte frequency of payments . Formulas corresponding to such annuities are useful as approximations corresponding to annuities payable with great frequency such as daily. Consider an annuity in which a very small payment dtis made at time tand these small payments are payable continuously for n interest conversion periods. Let i denote the periodic interest rate. Then the total amount paid during each period is ∫ k k−1 dt= [t]k k−1 = $1. Let an denote the present value of an annuity payable continuously forninterest conversion periods so that 1 is the total amount paid during each interest conversion period. Then the present value can be found as follows: an = ∫ n 0 νtdt= νt ln ν ⏐⏐⏐⏐ n 0 =νn −1 ln ν = νn −1 −ln (1 +i) =1 −νn δ . It is easy to see the following an = lim m→∞ a(m) n = lim m→∞ 1 −νn i(m) = 1 −νn δ since lim m→∞ i(m) = δ. Similarly, an = lim m→∞ ¨a(m) n = lim m→∞ 1 −νn d(m) = 1 −νn δ since lim m→∞ d(m) = δ. Moreover, an = i δan = d δ¨an = 1 −e−nδ δ . 25 CONTINUOUS ANNUITIES 229 Example 25.1 Starting four years from today, you will receive payment at the rate of $1,000 per annum, payable continuously, with the payment terminating twelve years from today. Find the present value of this continuous annuity if δ= 5%. Solution. The present value is PV = 1000v4 ·a8 = 1000e−0.20 ·1−e−0.40 0.05 = $5,398.38 The next example exhibit the situation when the constant force above is replaced by a variable force of interest. Example 25.2 Find an expression for an if δt = 1 1+t. Solution. The present value of the paymentdtat the exact time twith variable force of interest ise− ∫t 0 δrdrdt= e−ln (1+t)dt= (1 + t)−1dt. Hence, an = ∫ n 0 dt 1 + t = ln (1 + t)|n 0 = ln (1 +n) Next, let sn denote the accumulated value at the end of the term of an annuity payable continuously for ninterest conversion periods so that 1 is the total amound paid during each interest conversion period. Then sn =(1 + i)nan = ∫ n 0 (1 + i)n−tdt = −(1 + i)n−t ln (1 +i) ⏐⏐⏐⏐ n 0 =(1 + i)n −1 δ . It is easy to see that sn = lim m→∞ s(m) n = lim m→∞ ¨s(m) n = enδ −1 δ = i δsn = d δ¨sn. Example 25.3 Find the force of interest at which the accumulated value of a continuous payment of 1 every year for 8 years will be equal to four times the accumulated value of a continuous payment of 1 every year for four years. 230 THE BASICS OF ANNUITY THEORY Solution. We have s8 =4s4 e8δ −1 δ =4 ·e4δ −1 δ e8δ −4e4δ + 3 =0 (e4δ −3)(e4δ −1) =0 If e4δ = 3 then δ= ln 3 4 ≈0.275 = 27.5%. If e4δ = 1 then δ= 0, an extraneous solution Example 25.4 The annual eﬀective rate is i = 5% . Find the accumulated value after six years of an annuity that oﬀers payments of $1,000 per annum, convertible continuously for ﬁve years. After ﬁve years, payments terminate, but the balance still earns interest during the sixth year. Solution. The accumulated value is AV = 1000 s5 (1.05). But δ = ln (1 + i) = ln 1 .05 = 0 .0487902. Thus, AV = 1000 ·(1.05)5−1 0.0487902 ·(1.05) = $5,945.78 The present value of a perpetuity payable continuously with total of 1 per period is given by a∞ = lim n→∞ an = 1 δ. Example 25.5 A perpetuity paid continuously at a rate of 100 per year has a present value of 800. Calculate the annual eﬀective interest rate used to calculate the present value. Solution. The equation of value at time t= 0 is 800 = 100 δ = 100 ln (1 +i). Thus, i= e 1 8 −1 = 13.3% 25 CONTINUOUS ANNUITIES 231 Practice Problems Problem 25.1 Calculate the present value of a continuous annuity of 1,000 per annum for 8 years at: (a) An annual eﬀective interest rate of 4%; (b) A constant force of interest of 4%. Problem 25.2 Find the force of interest at which s20 = 3s10 . Problem 25.3 If an = 4 and sn = 12, ﬁnd δ. Problem 25.4 There is $40,000 in a fund which is accumulating at 4% per annum convertible continuously. If money is withdrawn continuously at the rate of $2,400 per annum, how long will the fund last? Problem 25.5 Annuity A oﬀers to pay you $100 per annum, convertible continuously, for the next ﬁve years. Annuity B oﬀers you $ X at the end of each year for ten years. The annual eﬀective interest rate i is 8%. Find X such that you are indiﬀerent between the two annuities. Problem 25.6 Given δ= 0.1. Evaluate a(12) 10 a10 . Problem 25.7 You are given d dtst = (1.02)2t. Calculate δ. Problem 25.8 Given an = n−4 and δ= 10%, ﬁnd ∫n 0 atdt. Problem 25.9 Payments will be made to you at a continuous rate of $100 per year for the next ﬁve years. The eﬀective rate of interest over this period is i = 0.06. Calculate the present value of this payment stream. Problem 25.10 Show that d dtat = νt s1 . 232 THE BASICS OF ANNUITY THEORY Problem 25.11 Show that an <a(m) n <an <¨a(m) n <¨an. Hint: See Example 10.15. Problem 25.12 Find an expression for t, 0 <t< 1, such that 1 paid at time t is equivalent to 1 paid continuously between 0 and 1. Problem 25.13 A bank makes payments continuously at a rate of $400 a year. The payments are made between 5 and 7 years. Find the current value of these payments at time 2 years using an annual rate of discount of 4%. Problem 25.14 A company makes payments continuously at a rate of $200 per year. The payments are made between 2 and 7 years. Find the accumulated value of these payments at time 10 years using an annual rate of interest of 6.5%. Problem 25.15 Lauren is being paid a continuous perpetuity payable at a rate of 1000 per year. Calculate the present value of the perpetuity assuming d(12) = 0.12. Problem 25.16 If i = 0.04, calculate the accumulated value of a continuous annuity payable at a rate of 100 per year for 10 years. Problem 25.17 If δ= 0.06, calculate the present value of a continuous annuity of 1 payable for 20 years. Problem 25.18 You are given ∫n 0 atdt= 100. Calculate an. Problem 25.19 An n−year continuous annuity that pays at a rate of $748 per year has a present value of $10,000 when using an interest rate of 6%, compounded continuously. Determine n. Problem 25.20 Which of the following are true? (I) ( an −d δ ) (1 + i) = an−1 . 25 CONTINUOUS ANNUITIES 233 (II) The present value of a 10 year annuity immediate paying 10 per month for the ﬁrst eight months of each year is 120 a10 a(12) 8/12 . (III) The present value of a perpetuity paying one at the end of each year, except paying nothing every fourth year, is s3 is4 . Problem 25.21 Payments of $3650 per year are made continuously over a ﬁve-year period. Find the present value of this continuous annuity two years prior to the ﬁrst payment. Given the nominal rate i(365) = 8%. 234 THE BASICS OF ANNUITY THEORY 26 Varying Annuity-Immediate Thus far in this book, all annuities that we have considered had level series of payments, that is, payments are all equal in values. In this and the next three sections we consider annuities with a varying series of payments. In this section, we assume that the payment period and interest conversion period coincide. Annuities with varying payments will be called varying annuities. Any type of annuities can be evaluated by taking the present value or the accumulated value of each payment seperately and adding the results. There are, however, several types of varying annuities for which relatively simple compact expressions are possible. The only general types we will study vary in either arithmetic progression or geometric progression. Payments Varying in an Arithmetic Progression First, let us assume that payments vary in arithmetic progression. That is, the ﬁrst payment is P and then the payments increase by Q thereafter, continuing for n years as shown in the time diagram of Figure 26.1. Figure 26.1 If PV is the present value for this annuity-immediate, then PV = Pν + (P + Q)ν2 + (P + 2Q)ν3 + ··· + [P + (n−1)Q]νn. (26.1) Multiplying Equation (26.1) by (1 + i), we obtain (1 + i)PV = P + (P + Q)ν+ (P + 2Q)ν2 + ··· + [P + (n−1)Q]νn−1. (26.2) Subtracting Equation(26.1) from Equation (26.2) we obtain iPV = P(1 −νn) + (ν+ ν2 + ··· + νn)Q−nνnQ or PV = Pan + Q[an −nνn] i . (26.3) The accumulated value of these payments at time n is AV = (1 + i)nPV = Psn + Q[sn −n] i . (26.4) 26 VARYING ANNUITY-IMMEDIATE 235 Two special cases of the above varying annuity often occur in practice. The ﬁrst of these is the increasing annuity where P = Q= 1 as shown in Figure 26.2. Figure 26.2 The present value of such an annuity is (Ia)n = an + an −nνn i = (1 + i)an −nνn i = ¨an −nνn i . (26.5) The accumulated value at time n is given by (Is)n = (1 + i)n(Ia)n = ¨sn −n i = sn+1 −(n+ 1) i . Example 26.1 The following payments are to be received: $500 at the end of the ﬁrst year, $520 at the end of the second year, $540 at the end of the third year and so on, until the ﬁnal payment is $800. Using an annual eﬀective interest rate of 2% (a) determine the present value of these payments at time 0; (b) determine the accumulated value of these payments at the time of the last payment. Solution. In n years the payment is 500 + 20( n−1). So the total number of payments is 16. The given payments can be regarded as the sum of a level annuity immediate of $480 and an increasing annuity-immediate $20,$40,··· ,$320. (a) The present value at time t= 0 is 480a16 + 20(Ia)16 = 480(13.5777) + 20(109.7065) = $8,711.43. (b) The accumulated value at time t= 16 is 480s16 + 20(Is)16 = 480(18.6393) + 20(150.6035) = $11,958.93 Example 26.2 Show that (Ia)n = ∑n−1 t=0 νtan−t. 236 THE BASICS OF ANNUITY THEORY Solution. We have (Ia)n = n−1∑ t=0 νtan−t = n−1∑ t=0 νt1 −νn−t i =1 i n−1∑ t=0 νt −νn i n−1∑ t=0 1 =¨an i −nνn i = ¨an −nνn i . The second special case is the decreasing annuity-immediate where P = n and Q = −1 as shown in Figure 26.3. Figure 26.3 In this case, the present value one year before the ﬁrst payment (i.e., at time t= 0) is given by (Da)n = nan −an −nνn i = n−nνn −an + nνn i = n−an i . (26.6) The accumulated value at time n is given by (Ds)n = (1 + i)n(Da)n = n(1 + i)n −sn i = (n+ 1)an −(Ia)n. Example 26.3 John receives $400 at the end of the ﬁrst year, $350 at the end of the second year, $300 at the end of the third year and so on, until the ﬁnal payment of $50. Using an annual eﬀective rate of 3.5%, calculate the present value of these payments at time 0. Solution. In year nthe payment is 400−50(n−1).Since the ﬁnal payment is 50 we must have 400−50(n−1) = 50. Solving for n we ﬁnd n= 8. Thus, the present value is 50(Da)8 = 50 ·8 −a8 0.035 = $1,608.63 26 VARYING ANNUITY-IMMEDIATE 237 Example 26.4 Calculate the accumulated value in Example 26.3. Solution. The answer is 50(Ds)8 = 50 · 8(1.035)8−s8 0.035 = $2,118.27 Example 26.5 Show that (Ia)n + (Da)n = (n+ 1)an. Solution. We have (Ia)n + (Da)n =an + an −nνn i + n−an i =an + nan = (n+ 1)an Besides varying annuities immediate, it is also possible to have varying perpetuity −immediate. Consider a perpetuity−immediate with payments that form an arithmetic progression (and of course P >0 and Q >0). The present value for such a perpetuity with the ﬁrst payment at the end of the ﬁrst period is PV = lim n→∞ [ Pan + Q[an −nνn] i ] == P lim n→∞ an + Q lim n→∞ [an −nνn] i =P lim n→∞ an + Q[limn→∞an −limn→∞nνn] i =Pa∞ + Qa∞ i = P i + Q i2 since a∞ = 1 i and limn→∞nνn = 0 (by L’Hopital’s rule). For the special case P = Q= 1 we ﬁnd (Ia)∞ = 1 i + 1 i2 . Example 26.6 Find the present value of a perpetuity-immediate whose successive payments are 1, 2, 3, 4, ··· at an eﬀective rate of 6%. Solution. We have (Ia)∞ = 1 i + 1 i2 = 1 0.06 + 1 0.062 = $294.44 238 THE BASICS OF ANNUITY THEORY Payments Varying in a Geometric Progression Next, we consider payments varying in a geometric progression. Consider an annuity-immediate with a term of n periods where the interest rate is i per period, and where the ﬁrst payment is 1 and successive payments increase in geometric progression with common ratio 1 + k. The present value of this annuity is ν+ ν2(1 + k) + ν3(1 + k)2 + ··· + νn(1 + k)n−1 =ν·1 −[(1 + k)ν]n 1 −(1 + k)ν = 1 1 + i · 1 − (1+k 1+i )n 1 − (1+k 1+i ) = 1 − (1+k 1+i )n i−k provided that k ̸= i. If k = i then the original sum consists of a sum of n terms of ν which equals to nν. Example 26.7 The ﬁrst of 30 payments of an annuity occurs in exactly one year and is equal to $500. The payments increase so that each payment is 5% greater than the preceding payment. Find the present value of this annuity with an annual eﬀective rate of interest of 8%. Solution. The present value is given by PV = 500 ·1 − (1.05 1.08 )30 0.08 −0.05 = $9,508.28 For an annuity-immediate with a term ofnperiods where the interest rate is iper period, and where the ﬁrst payment is 1 and successive payments decrease in geometric progression with common ratio 1 −k. The present value of this annuity is ν+ ν2(1 −k) + ν3(1 −k)2 + ··· + νn(1 −k)n−1 =ν·1 −[(1 −k)ν]n 1 −(1 −k)ν = 1 1 + i · 1 − (1−k 1+i )n 1 − (1−k 1+i ) = 1 − (1−k 1+i )n i+ k 26 VARYING ANNUITY-IMMEDIATE 239 provided that k̸= i. If k= i then the original sum becomes ν+ ν2(1 −i) + ν3(1 −i)2 + ··· + νn(1 −i)n−1 = 1 2i [ 1 − (1 −i 1 + i )n] Finally, we consider a perpetuity with payments that form a geometric progression where 0 < 1 +k <1 +i.The present value for such a perpetuity with the ﬁrst payment at the end of the ﬁrst period is ν+ ν2(1 + k) + ν3(1 + k)2 + ··· = ν 1 −(1 + k)ν = 1 i−k. Observe that the value for these perpetuities cannot exist if 1 + k≥1 + i. Example 26.8 What is the present value of a stream of annual dividends, which starts at 1 at the end of the ﬁrst year, and grows at the annual rate of 2%, given that the rate of interest is 6% ? Solution. The present value is 1 i−k = 1 0.06−0.02 = 25 240 THE BASICS OF ANNUITY THEORY Practice Problems Problem 26.1 Smith receives $400 in 1 year, $800 in 2 years, $1,200 in 3 years and so on until the ﬁnal payment of $4,000. Using an eﬀective interest rate of 6%, determine the present value of these payments at time 0. Problem 26.2 Find an expression for the present value at time 0 of payments of $55 at time 1 year, $60 at time 2 years, $65 at time 3 years and so on, up to the last payment at time 20 years. Problem 26.3 Find an expression for the accumulated value at time 20 years of payments of $5 at time 1 year, $10 at time 2 years, $15 at time 3 years, and so on, up to $100 at time 20 years. Problem 26.4 Find the present value of a perpetuity-immediate whose successive payments are $5 ,$10,$15,··· assuming an annual eﬀective rate of interest of 5%. Problem 26.5 An annuity pays $100 at the end of one month. It pays $110 at the end of the second month. It pays $120 at the end of the third month. The payments continue to increase by $10 each month until the last payment is made at the end of the 36th month. Find the present value of the annuity at 9% compounded monthly. Problem 26.6 An annual annuity-immediate pays $100 at the end of the ﬁrst year. Each subsequent payment is 5% greater than the preceding payment. The last payment is at the end of the 20th year. Calculate the accumulated value at: (a) an annual eﬀective interest rate of 4%; (b) an annual eﬀective interest rate of 5%. Problem 26.7 A perpetuity pays $100 at the end of the ﬁrst year. Each subsequent annual payment increases by $50. Calculate the present value at an annual eﬀective interest rate of 10%. Problem 26.8 A small business pays you an annual proﬁt at the end of each year for 20 years. The payments grow at an annual rate of 2.5%, the ﬁrst payment is $10,000. What is the present value of this stream of payments at the annual eﬀective rate of interest 6%? 26 VARYING ANNUITY-IMMEDIATE 241 Problem 26.9 What is the present value of a 30 year immediate annuity at 10% interest where the ﬁrst payment is $100 and each payment thereafter is increased by 5%? Problem 26.10 A perpetuity-immediate has annual payments of 1, 3, 5, 7, ··· .If the present values of the 6th and 7th payments are equal, ﬁnd the present value of the perpetuity. Problem 26.11 If X is the present value of a perpetuity of 1 per year with the ﬁrst payment at the end of the 2nd year and 20X is the present value of a series of annual payments 1, 2, 3, ··· with the ﬁrst payment at the end of the 3rd year, ﬁnd d. Problem 26.12 There are two perpetuities-immediate. The ﬁrst has level payments of p at the end of each year. The second is increasing such that the payments are q,2q,3q,··· .Find the rate of interest that will make the diﬀerence in present value between these perpetuities a) zero; b) a maximum. Problem 26.13 An annuity pays 10 at the end of the ﬁrst year, 20 at the end of the second year, 30 at the end of the third year, etc. The last payment is made at the end of the 12th year. Calculate the accumulated value of this annuity immediately after the last payment using an annual eﬀective rate of 4%. Problem 26.14 A 20 year annuity-immediate pays 500 + 50t at the end of year t. Calculate the present value of this annuity using an annual eﬀective rate of 6%. Problem 26.15 An annuity pays 10 at the end of year 2, and 9 at the end of year 4. The payments continue decreasing by 1 each two year period until 1 is paid at the end of year 20. Calculate the present value of the annuity at an annual eﬀective interest rate of 5%. Problem 26.16 Find an expression for the present value of a perpetuity under which a payment of 1 is made at the end of the 1st year, 2 at the end of the 2nd year, increasing until a payment of n is made at the end of the nth year, and thereafter payments are level at n per year forever. 242 THE BASICS OF ANNUITY THEORY Problem 26.17 Find an expression for the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of n, and then decrease by 1 each period up to a ﬁnal payment of 1. Problem 26.18 An 11-year annuity has a series of payments 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, with the ﬁrst payment made at the end of the second year. The present value of this annuity is 25 at interest rate i. A 12-year annuity has a series of payments 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, with the ﬁrst payment made at the end of the ﬁrst year. Calculate the present value of the 12-year annuity at interest rate i. Problem 26.19 Olga buys a 5-year increasing annuity for X. Olga will receive 2 at the end of the ﬁrst month, 4 at the end of the second month, and for each month thereafter the payment increases by 2 . The nominal interest rate is 9% convertible quarterly. Calculate X. Problem 26.20 An annuity-immediate has semiannual payments of 800, 750, 700, ··· , 350, at i(2) = 0 .16. If a10 0.08 = A, ﬁnd the present value of the annuity in terms of A. Problem 26.21 ‡ A perpetuity costs 77.1 and makes annual payments at the end of the year. The perpetuity pays 1 at the end of year 2, 2 at the end of year 3, ··· ,n at the end of year ( n+ 1).After year (n+ 1),the payments remain constant at n. The annual eﬀective interest rate is 10.5%. Calculate n. Problem 26.22 ‡ A perpetuity-immediate pays 100 per year. Immediately after the ﬁfth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the ﬁrst year. Each subsequent annual payment will be 8% greater than the preceding payment. The annual eﬀective rate of interest is 8%. Calculate X. Problem 26.23 ‡ A perpetuity-immediate pays 100 per year. Immediately after the ﬁfth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the ﬁrst year. Each subsequent annual payment will be 8% greater than the preceding payment. Immediately after the 10th payment of the 25-year annuity, the annuity will be exchanged for a perpetuity-immediate paying Y per year. The annual eﬀective rate of interest is 8%. Calculate Y. 26 VARYING ANNUITY-IMMEDIATE 243 Problem 26.24 ‡ Mike buys a perpetuity-immediate with varying annual payments. During the ﬁrst 5 years, the payment is constant and equal to 10. Beginning in year 6, the payments start to increase. For year 6 and all future years, the current year’s payment is K% larger than the previous year’s payment. At an annual eﬀective interest rate of 9.2%, the perpetuity has a present value of 167.50. Calculate K, given K <9.2. Problem 26.25 ‡ A company deposits 1000 at the beginning of the ﬁrst year and 150 at the beginning of each subsequent year into perpetuity. In return the company receives payments at the end of each year forever. The ﬁrst payment is 100. Each subsequent payment increases by 5%. Calculate the company’s yield rate for this transaction. Problem 26.26 ‡ Megan purchases a perpetuity-immediate for 3250 with annual payments of 130. At the same price and interest rate, Chris purchases an annuity-immediate with 20 annual payments that begin at amount P and increase by 15 each year thereafter. Calculate P. Problem 26.27 ‡ The present value of a 25-year annuity-immediate with a ﬁrst payment of 2500 and decreasing by 100 each year thereafter is X. Assuming an annual eﬀective interest rate of 10%, calculate X. Problem 26.28 ‡ The present value of a series of 50 payments starting at 100 at the end of the ﬁrst year and increasing by 1 each year thereafter is equal to X. The annual eﬀective rate of interest is 9%. Calculate X. Problem 26.29 ‡ An annuity-immediate pays 20 per year for 10 years, then decreases by 1 per year for 19 years. At an annual eﬀective interest rate of 6%, the present value is equal to X. Calculate X. Problem 26.30 ‡ At an annual eﬀective interest rate of i, the present value of a perpetuity-immediate starting with a payment of 200 in the ﬁrst year and increasing by 50 each year thereafter is 46,530. Calculate i. Problem 26.31 ‡ An insurance company has an obligation to pay the medical costs for a claimant. Average annual claims costs today are $5,000, and medical inﬂation is expected to be 7% per year. The claimant is expected to live an additional 20 years. Claim payments are made at yearly intervals, with the ﬁrst claim payment to be made one year from today. Find the present value of the obligation if the annual interest rate is 5%. 244 THE BASICS OF ANNUITY THEORY Problem 26.32 ‡ Sandy purchases a perpetuity-immediate that makes annual payments. The ﬁrst payment is 100, and each payment thereafter increases by 10. Danny purchases a perpetuity-due which makes annual payments of 180. Using the same annual eﬀective interest rate, i> 0,the present value of both perpetuities are equal. Calculate i. Problem 26.33 ‡ Joe can purchase one of two annuities: Annuity 1: A 10-year decreasing annuity-immediate, with annual payments of 10 ,9,8,··· ,1. Annuity 2: A perpetuity-immediate with annual payments. The perpetuity pays 1 in year 1, 2 in year 2, 3 in year 3, ··· , and 11 in year 11 . After year 11, the payments remain constant at 11. At an annual eﬀective interest rate of i,the present value of Annuity 2 is twice the present value of Annuity 1 . Calculate the value of Annuity 1. Problem 26.34 ‡ Mary purchases an increasing annuity-immediate for 50,000 that makes twenty annual payments as follows: i) P,2P,··· ,10P in years 1 through 10; and ii) 10(1.05)P,10(1.05)2P,··· ,10(1.05)10P in years 11 through 20. The annual eﬀective interest rate is 7% for the ﬁrst 10 years and 5% thereafter. Calculate P. Problem 26.35 Find the present value of an annuity-immediate such that payments start at 1, each payment thereafter increases by 1 until reaching 10, and then remain at that level until 25 payments in total are made. Problem 26.36 Find an expression of the present value of an annuity that pays 10 at the end of the ﬁfth year and decreases by 1 thereafter until reaching 0. Problem 26.37 Annual deposits are made into a fund at the beginning of each year for 10 years. The ﬁrst ﬁve deposits are $1000 each and deposits increase by 5% per year thereafter. If the fund earns 8% eﬀective, ﬁnd the accumulated value at the end of 10 years. Problem 26.38 Find the present value of a 20-year annuity-immediate with ﬁrst payment of $600 and each subse- quent payment is 5% greater than the preceding payment. The annual eﬀective rate of interest is 10.25%. 26 VARYING ANNUITY-IMMEDIATE 245 Problem 26.39 A perpetuity pays 200 at the end of each of the ﬁrst two years, 300 at the end of years 3 and 4, 400 at the end of years 5 and 6, etc. Calculate the present value of this perpetuity if the annual eﬀective rate of interest is 10%. Problem 26.40 Janice is receiving a perpetuity of 100 at the end of each year. Megan is receiving a perpetuity that pays 10 at the end of the ﬁrst year, 20 at the end of the second year, 30 at the end of the third year, etc. The present values of the two perpetuities are equal. Calculate the annual eﬀective interest rate used to determine the present values. Problem 26.41 Gloria borrows 100,000 to be repaid over 30 years. You are given (1) Her ﬁrst payment is X at the end of year 1 (2) Her payments increase at the rate of 100 per year for the next 19 years and remain level for the following 10 years (3) The eﬀective rate of interest is 5% per annum Calculate X. Problem 26.42 You are given a perpetual annuity immediate with annual payments increasing in geometric pro- gression, with a common ratio of 1.07. The annual eﬀective interest rate is 12%. The ﬁrst payment is 1. Calculate the present value of this annuity. Problem 26.43 An annuity immediate pays 10 at the ends of years 1 and 2, 9 at the ends of years 3 and 4, etc., with payments decreasing by 1 every second year until nothing is paid. The eﬀective annual rate of interest is 5%. Calculate the present value of this annuity immediate. Problem 26.44 Barbara purchases an increasing perpetuity with payments occurring at the end of every 2 years. The ﬁrst payment is 1, the second one is 2, the third one is 3, etc. The price of the perpetuity is 110. Calculate the annual eﬀective interest rate. Problem 26.45 You are given (Ia)n−1 = K d , where d is the annual eﬀective discount rate. Calculate K. Problem 26.46 Francois purchases a 10 year annuity immediate with annual payments of 10 X. Jacques purchases 246 THE BASICS OF ANNUITY THEORY a 10 year decreasing annuity immediate which also makes annual payments. The payment at the end of year 1 is equal to 50. At the end of year 2, and at the end of each year through year 10, each subsequent payment is reduced over what was paid in the previous year by an amount equal to X. At an annual eﬀective interest rate of 7.072%, both annuities have the same present value. Calculate X, where X <5. Problem 26.47 ‡ 1000 is deposited into Fund X,which earns an annual eﬀective rate of 6%. At the end of each year, the interest earned plus an additional 100 is withdrawn from the fund. At the end of the tenth year the fund is depleted. The annual withdrawals of interest and principal are deposited into Fund Y, which earns an annual eﬀective rate of 9%. Determine the accumulated value of Fund Y at the end of year 10. Problem 26.48 You are given two series of payments. Series A is a perpetuity with payments of 1 at the end of each of the ﬁrst 2 years, 2 at the end of each of the next 2 years, 3 at the end of each of the next 2 years, and so on. Series B is a perpetuity with payments of K at the end of each of the ﬁrst 3 years, 2K at the end of the next 3 years, 3 K at the end of each of the next 3 years, and so on. The present values of the two series of payments are equal. Calculate K. Problem 26.49 John deposits 100 at the end of each year for 20 years into a fund earning an annual eﬀective interest rate of 7%. Mary makes 20 deposits into a fund at the end of each year for 20 years. The ﬁrst 10 deposits are 100 each, while the last 10 deposits are 100 + X each. The fund earns an annual eﬀective interest rate of 8% during the ﬁrst 10 years and 6% annual eﬀective interest thereafter. At the end of 20 years, the amount in John’s fund equals the amount in Mary’s fund. Calculate X. Problem 26.50 An annuity-immediate pays an initial beneﬁt of 1 per year, increasing by 10.25% every four years. The annuity is payable for 40 years. Using an annual eﬀective interest rate of 5%, determine an expression for the present value of this annuity. Problem 26.51 An increasing perpetuity with annual payments has a present value of 860 at an annual eﬀective discount rate, d.The initial payment at the end of the ﬁrst year is 3 and each subsequent payment is 2 more than its preceding payment. Calculate d. 26 VARYING ANNUITY-IMMEDIATE 247 Problem 26.52 A 10-year increasing annuity-immediate paying 5 in the ﬁrst year and increasing by 5 each year thereafter has a present value of G. A 10-year decreasing annuity-immediate paying y in the ﬁrst year and decreasing by y 10 each year thereafter has a present value of G. Both present values are calculated using an annual eﬀective interest rate of 4%. Calculate y to the nearest integer. Problem 26.53 A loan of 10,000 is being repaid by 10 semiannual payments, with the ﬁrst payment made one-half year after the loan. The ﬁrst 5 payments are K each, and the ﬁnal 5 are K + 200 each. Given i(2) = 0.6, determine K. Problem 26.54 An annuity immediate has the following payment pattern: 1 ,2,··· ,n −1,n,n −1,··· ,2,1. The present value of this annuity using an annual eﬀective interest rate of 9% is 129.51. Determine n. Problem 26.55 An increasing 25-year annuity immediate has an initial payment in year one of 4 and each subsequent annual payment is 3 more than the preceding one. Find the present value of this annuity using an annual eﬀective interest rate of 7%. Problem 26.56 A 10-year annuity-immediate has a ﬁrst payment of 2,000. Each subsequent annual payment is 100 less than the preceding payment. (The payments are 2,000; 1,900; 1,800; etc.) At an annual eﬀective interest rate of 5%, what is the accumulated value of this stream of payments on the date of the ﬁnal payment? Problem 26.57 A payment of 100 purchases an increasing perpetuity with a ﬁrst payment of 10 at the end of the ﬁrst year. Each annual payment thereafter is 5% greater than the prior year’s payment. What is the annual interest rate for this stream of payments? Problem 26.58 You are given a perpetuity, with annual payments as follows: (1) Payments of 1 at the end of the ﬁrst year and every three years thereafter. (2) Payments of 2 at the end of the second year and every three years thereafter. (3) Payments of 3 at the end of the third year and every three years thereafter. (4) The interest rate is 5% convertible semi-annually. Calculate the present value of this perpetuity. 248 THE BASICS OF ANNUITY THEORY Problem 26.59 (a) You just paid $50 for a share of stock which pays annual dividends. You expect to receive the ﬁrst dividend payment of $5 one year from now. After that, you expect dividends to increase by $X each year, forever. The price you paid implies that you would be happy with an annual return of 12%. What is X? (b) What value $Y could you sell your share for if the new buyer has the same desired annual return but expects the annual dividend increase to be X+ 1? Problem 26.60 A 30-year annuity-immediate pays 20,000 in the ﬁrst year. Thereafter, the payments decrease 500 each year. If d(4) is 6%, what is the present value of this annuity? Problem 26.61 You are about to buy shares in a company. The ﬁrst annual dividend, of $3 per share, is expected to be paid 3 years from now. Thereafter, you expect that the dividend will grow at the rate of 10% per year for ever. Assuming your desired return is 12%, what are you willing to pay per share for this stock? Problem 26.62 A perpetuity with annual payments is payable beginning 10 years from now. The ﬁrst payment is 50. Each annual payment thereafter is increased by 10 until a payment of 150 is reached. Subsequent payments remain level at 150. This perpetuity is purchased by means of 10 annual premiums, with the ﬁrst premium of P due immediately. Each premium after the ﬁrst is 105% of the preceding one. The annual eﬀective interest rates are 5% during the ﬁrst 9 years and 3% thereafter. Calculate P. Problem 26.63 Show that (Da)n = ∑n t=1 at. Problem 26.64 Simplify ∑10 t=1(t+ 1)νt. 27 VARYING ANNUITY-DUE 249 27 Varying Annuity-Due In this section, we examine the case of an increasing annuity-due. Consider an annuity with the ﬁrst payment is P at the beginning of year 1 and then the payments increase by Qthereafter, continuing for n years. A time diagram of this situation is given in Figure 27.1 Figure 27.1 The present value for this annuity-due is PV = P + (P + Q)ν+ (P + 2Q)ν2 + ··· + [P + (n−1)Q]νn−1. (27.1) Multiplying Equation (28.1) by ν, we obtain νPV = Pν + (P + Q)ν2 + (P + 2Q)ν3 + ··· + [P + (n−1)Q]νn. (27.2) Subtracting Equation(28.2) from Equation (28.1) we obtain (1 −ν)PV = P(1 −νn) + (ν+ ν2 + ··· + νn)Q−nνnQ or PV = P¨an + Q[an −nνn] d . The accumulated value of these payments at time n is AV = (1 + i)nPV = P¨sn + Q[sn −n] d . In the special case when P = Q= 1 we ﬁnd (I¨a)n = ¨an −nνn d and (I¨s)n = ¨sn −n d = sn+1 −(n+ 1) d . 250 THE BASICS OF ANNUITY THEORY Example 27.1 Determine the present value and future value of payments of $75 at time 0, $80 at time 1 year, $85 at time 2 years, and so on up to $175 at time 20 years. The annual eﬀective rate is 4%. Solution. The present value is 70¨ a21 + 5(I¨a)21 = $1 ,720.05 and the future value is (1 .04)21(1,720.05) = $3919.60 In the case of a decreasing annuity-due where P = n and Q = −1, the present value at time 0 is (D¨a)n = n−an d and the accumulated value at time n is (D¨s)n = (1 + i)n(D¨a)n = n(1 + i)n −sn d . Example 27.2 Calculate the present value and the accumulated value of a series of payments of $100 now, $90 in 1 year, $80 in 2 years, and so on, down to $10 at time 9 years using an annual eﬀective interest rate of 3%. Solution. The present value is 10(D¨a)10 = 10·10−8.530203 0.03/1.03 = $504.63 and the accumulated value is (1.03)10(504.63) = $678.18 Next, we consider a perpetuity-due with payments that form an arithmetic progression (and of course P >0 and Q >0). The present value for such a perpetuity with the ﬁrst payment at time 0 is PV = lim n→∞ [ P¨an + Q[an −nνn] d ] =P lim n→∞ ¨an + Q lim n→∞ [an −nνn] d =P lim n→∞ ¨an + Q[limn→∞an −limn→∞nνn] d =P¨a∞ + Qa∞ d = P d + Q(1 + i) i2 27 VARYING ANNUITY-DUE 251 since a∞ = 1 i, ¨a∞ = 1 d and limn→∞nνn = 0 (by L’Hopital’s rule). In the special case P = Q= 1 we ﬁnd (I¨a)∞ = 1 d2 . Example 27.3 Determine the present value at time 0 of payments of $10 paid at time 0, $20 paid at time 1 year, $30 paid at time 2 years, and so on, assuming an annual eﬀective rate of 5%. Solution. The answer is 10(I¨a)∞ = 10 d2 = 10 (1.05 0.05 )2 = $4,410.00 Next, we consider payments varying in a geometric progression. Consider an annuity-due with a term of nperiods where the interest rate is iper period, and where the ﬁrst payment is 1 at time 0 and successive payments increase in geometric progression with common ratio 1 + k. The present value of this annuity is 1 + ν(1 + k) + ν2(1 + k)2 + ··· + νn−1(1 + k)n−1 =1 −[(1 + k)ν]n 1 −(1 + k)ν = 1 − (1+k 1+i )n 1 − (1+k 1+i ) =(1 + i) 1 − (1+k 1+i )n i−k provided that k ̸= i. If k = i then the original sum consists of a sum of n terms of 1 which equals to n. Example 27.4 An annual annuity due pays $1 at the beginning of the ﬁrst year. Each subsequent payment is 5% greater than the preceding payment. The last payment is at the beginning of the 10th year. Calculate the present value at: (a) an annual eﬀective interest rate of 4%; (b) an annual eﬀective interest rate of 5%. Solution. (a) PV = (1.04) [ 1−(1.05 1.04 ) 10] (0.04−0.05) = $10.44. (b) Since i= k, PV = n= $10.00 252 THE BASICS OF ANNUITY THEORY For an annuity-due withnpayments where the ﬁrst payment is 1 at time 0 and successive payments decrease in geometric progression with common ratio 1 −k. The present value of this annuity is 1 + ν(1 −k) + ν2(1 −k)2 + ··· + νn−1(1 −k)n−1 =1 −[(1 −k)ν]n 1 −(1 −k)ν = 1 − (1−k 1+i )n 1 − (1−k 1+i ) =(1 + i) 1 − (1−k 1+i )n i+ k provided that k̸= i. If k= i then the original sum is 1 + ν(1 −i) + ν2(1 −i)2 + ··· + νn−1(1 −i)n−1 = 1 2d [ 1 − (1 −i 1 + i )n] . Example 27.5 ‡ Matthew makes a series of payments at the beginning of each year for 20 years. The ﬁrst payment is 100. Each subsequent payment through the tenth year increases by 5% from the previous payment. After the tenth payment, each payment decreases by 5% from the previous payment. Calculate the present value of these payments at the time the ﬁrst payment is made using an annual eﬀective rate of 7%. Solution. The present value at time 0 of the ﬁrst 10 payments is 100 [ 1 − (1.05 1.07 )10 0.07 −0.05 ] ·(1.07) = 919.95. The value of the 11th payment is 100(1 .05)9(0.95) = 147 .38. The present value of the last ten payments is 147.38 [ 1 − (0.95 1.07 )10 0.07 + 0.05 ] ·(1.07)(1.07)−10 = 464.71. The total present value of the 20 payments is 919 .95 + 464.71 = 1384.66 Finally, the present value of a perpetuity with ﬁrst payment of 1 at time 0 and successive pay- ments increase in geometric progression with common ration 1 + k is 1 + ν(1 + k) + ν2(1 + k)2 + ··· = 1 1 −(1 + k)ν = 1 + i i−k. Observe that the value for these perpetuities cannot exist if 1 + k≥1 + i. 27 VARYING ANNUITY-DUE 253 Example 27.6 Perpetuity A has the following sequence of annual payments beginning on January 1, 2005: 1,3,5,7,··· Perpetuity B is a level perpetuity of 1 per year, also beginning on January 1, 2005. Perpetuity C has the following sequence of annual payments beginning on January 1, 2005: 1,1 + r,(1 + r)2,··· On January 1, 2005, the present value of Perpetuity A is 25 times as large as the present value of Perpetuity B, and the present value of Perpetuity A is equal to the present value of Perpetuity C. Based on this information, ﬁnd r. Solution. The present value of Perpetuity A is 1 d + 2(1+i) i2 . The present value of Perpetuity B is 1 d. The present value of Perpetuity C is 1+i i−r. We are told that 1 + i i + 2(1 + i) i2 = 25(1 + i) i . This is equivalent to 12i2 + 11i−1 = 0. Solving for i we ﬁnd i= 1 12 . Also, we are told that 1 + i i + 2(1 + i) i2 = 1 + i i−r or 25(12)(1 + 1 12) = 1 + 1 12 1 12 −r. Solving for r we ﬁnd r= 0.08 = 8% 254 THE BASICS OF ANNUITY THEORY Practice Problems Problem 27.1 A 20 year increasing annuity due pays 100 at the start of year 1, 105 at the start of year 2, 110 at the start of year 3, etc. In other words, each payment is 5% greater than the prior payment. Calculate the present value of this annuity at an annual eﬀective rate of 5%. Problem 27.2 A perpetuity pays 1000 at the beginning of the ﬁrst year. Each subsequent payment is increased by inﬂation. If inﬂation is assumed to be 4% per year, calculate the present value of this perpetuity using an annual eﬀective interest rate of 10%. Problem 27.3 A perpetuity pays 1000 at the beginning of the ﬁrst year. Each subsequent payment is increased by inﬂation. If inﬂation is assumed to be 4% per year, calculate the present value of this perpetuity using an annual eﬀective interest rate of 4%. Problem 27.4 A 10-year decreasing annuity-due makes a payment of 100 at the beginning of the ﬁrst year, and each subsequent payment is 5 less then the previous. What is the accumulated value of this annuity at time 10 (one year after the ﬁnal payment)? The eﬀective annual rate of interest is 5%. Problem 27.5 A 10 year annuity due pays 1,000 as the ﬁrst payment. Each subsequent payment is 50 less than the prior payment. Calculate the current value of the annuity at the end of 5 years using an annual eﬀective rate of 8%. Problem 27.6 You buy an increasing perpetuity-due with annual payments starting at 5 and increasing by 5 each year until the payment reaches 100. The payments remain at 100 thereafter. The annual eﬀective interest rate is 7.5%. Determine the present value of this perpetuity. Problem 27.7 Debbie receives her ﬁrst annual payment of 5 today. Each subsequent payment decreases by 1 per year until time 4 years. After year 4, each payment increases by 1 until time 8 years. The annual interest rate is 6%. Determine the present value. 27 VARYING ANNUITY-DUE 255 Problem 27.8 Determine the accumulated value at time 20 years of payments of $10 at time 0, $20 at time 1 year, $30 at time 2 years, and so on, up to $200 at time 19 years. The annual eﬀective rate of interest is 4%. Problem 27.9 Perpetuity X has payments of 1,2,3,··· at the beginning of each year. Perpetuity Y has payments of q,q, 2q,2q,3q,3q,··· at the beginning of each year. The present value of X is equal to the present value of Y at an eﬀective annual interest rate of 10%. Calculate q. Problem 27.10 Kendra receives $900 now, $970 in 1 year, $1040 in 2 years, $1,110 in 3 years, and so on, until the ﬁnal payment of $1600. Using an annual eﬀective rate of interest of 9%, ﬁnd (a) the present value of these payments at time 0; (b) the accumulated value at time 11 years. Problem 27.11 Mary is saving money for her retirement. She needs $750,000 in 10 years to purchase a retirement apartment in Florida. She invests X, now, X−5,000 in 1 year, X−10,000 in 2 years, and so on, down to X−45,000 in 9 years. Using an annual eﬀective rate of interest of 5%, ﬁnd X. Problem 27.12 Chris makes annual deposits into a bank account at the beginning of each year for 20 years. Chris’ initial deposit is equal to 100, with each subsequent deposit k% greater than the previous year’s deposit. The bank credits interest at an annual eﬀective rate of 5%. At the end of 20 years, the accumulated amount in Chris’ account is equal to 7276.35. Given k >5, calculate k. Problem 27.13 You are given: (i) The present value of an annuity-due that pays 300 every 6 months during the ﬁrst 15 years and 200 every 6 months during the second 15 years is 6000. (ii) The present value of a 15-year deferred annuity-due that pays 350 every 6 months for 15 years is 1580. (iii) The present value of an annuity-due that pays 100 every 6 months during the ﬁrst 15 years and 200 every 6 months during the next 15 years is X. The same interest rate is used in all calculations. Determine X. Problem 27.14 A 20 year annuity-due with annual payments has a ﬁrst payment of 100 and each subsequent annual payment is 5% more than its preceding payment. Calculate the present value of this annuity at a nominal interest rate of 10% compounded semiannually. 256 THE BASICS OF ANNUITY THEORY Problem 27.15 The payments of a perpetuity-due are three years apart. The initial payment is 7 and each sub- sequent payment is 5 more than its previous payment. Find the price (present value at time 0) of this perpetuity-due, using an annual eﬀective discount rate of 6%. Problem 27.16 You are considering the purchase of a share ofXYZ stock. It pays a quarterly dividend. You expect to receive the ﬁrst dividend, of $2.50, 3 months from now. You expect the dividend to increase by 6% per year, forever, starting with the 5th dividend. If the desired return on your investment is 15%, what is the amount your are willing to pay for a share of XYZ stock? Problem 27.17 You plan to make annual deposits at the start of each year for 25 years into your retirement fund. The ﬁrst 5 deposits are to be 2,000 each. Thereafter, you plan to increase the deposits by X per year. Assuming the fund can earn an eﬀective annual interest rate of 6%, what must X be for you to have accumulated 250,000 at the end of the 25 years? 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE257 28 Varying Annuities with Payments at a Diﬀerent Fre- quency than Interest is Convertible In Sections 26 and 27, we discussed varying annuities where the payment period coincide with the interest conversion period. In this section we consider varying annuities with payments made more or less frequently than interest is convertible. We will limit our discussion to increasing annuities. Decreasing annuities can be handled in a similar fashion. Varying Annuities Payable Less Frequently Than Interest is Convertible Consider annuities payable less frequently than interest is convertible. We let k be the number of interest conversion periods in one payment period, n the term of the annuity measured in interest conversion periods, and i the rate of interest per conversion period. It follows that the number of payments over the term of the annuity is given by n k which we assume is a positive integer. Let PV be the present value of a generalized increasing annuity-immediate with payments 1, 2, 3, etc. occurring at the end of each interval of k interest conversion periods. Then, PV = νk + 2ν2k + 3ν3k + ··· + (n k −1 ) νn−k + n kνn. (28.1) Multiply Equation (28.1) by (1 + i)k to obtain (1 + i)kPV = 1 + 2νk + 3ν2k + ··· + (n k −1 ) νn−2k + n kνn−k. (28.2) Now subtracting Equation (28.1) from Equation (28.2) we ﬁnd PV [(1 + i)k −1] = 1 + νk + ν2k + ··· + νn−k −n knun = an ak −n kνn. Hence, PV = an ak −n kνn isk . The accumulated value at time t= n is AV = (1 + i)nPV = (1 + i)n an ak −n kνn isk . Example 28.1 A 10-year annuity-immediate pays 1 in two years, 2 in four years, 3 in six years, 4 in eight years, and 5 in ten years. Using an eﬀective annual interest rate 5%, ﬁnd the present value and the accumulated value of this annuity. 258 THE BASICS OF ANNUITY THEORY Solution. The present value of this annuity at time 0 year is PV = a10 a2 −5(1.05)−10 0.05s2 = 7.721735 1.85941 −5(1.05)−10 (1.05)2 −1 = 10.568. The accumulated value is 10.568(1.05)10 = $17.21 Now, consider an increasing annuity-due with payments 1, 2, 3, etc. occurring at the start of each interval of k interest conversion periods. Then PV = 1 + 2νk + 3ν2k + ··· + (n k −1 ) νn−2k + n kνn−k. (28.3) Multiply Equation (28.3) by νk to obtain νkPV = νk + 2ν2k + 3ν3k + ··· + (n k −1 ) νn−k + n kνn. (28.4) Now subtracting Equation (28.4) from Equation (28.3) we ﬁnd PV [1 −νk] = 1 + νk + ν2k + ··· + νn−k −n kνn = an ak −n kνn. Hence, PV = an ak −n kνn iak . The accumulated value at time t= n is AV = (1 + i)n an ak −n kνn iak . Example 28.2 Find the present value and the accumulated value of a 5-year annuity in which payments are made at the beginning of each half-year, with the ﬁrst payment of $50, the second payment of $100, the third payment of $150, and so on. Interest is 10% convertible quarterly. Solution. The present value is PV = 50 a20 a2 −10ν20 0.025a2 = 15.58916 1.92742 −10(1.025)−20 1 −(1.025)−2 = $2060.15. 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE259 The accumulated value at time t= 20 is (2060.15)(1.025)20 = $3375.80 Varying Annuities Payable More Frequently Than Interest is Convertible We next consider increasing annuities where payments are made more frequently than interest is convertible. We examine two diﬀerent types. The ﬁrst is the increasing mthly annuity where the increase occurs once per conversion period. The second is the mthly increasing mthly annuity where the increase takes place with each mthly payment. Annuities Payable mthly Consider ﬁrst an increasing annuity-immediate with constant payments during each interest con- version period with increases occurring only once per interest conversion period. Let the annuity be payable mthly with each payment in the ﬁrst period equal to 1 m, each payment in the second period equal to 2 m,··· , each payment in the nth period equal to n m as shown in Figure 28.1. Figure 28.1 Let i be the rate per one conversion period and i(m) be the nominal interest rate payable m times per interest conversion period. Making the mthly payments of k m in period k is the same as making one payment equal to the accumulated value of the mthly payments at the end of the period, that is, one payment of k msmj = k m ·(1 + j)m −1 j = k i i(m) where j = i(m) m . The time diagram of this is given in Figure 28.2. Figure 28.2 260 THE BASICS OF ANNUITY THEORY Consequently, the increasing annuity payable mthly is the same as an increasing annuity with P = Q= i i(m) which implies the present value (Ia)(m) n = i i(m) (Ia)n = ¨an −nνn i(m) . The accumulated value is (Is)(m) n = (1 + i)n(Ia)(m) n = ¨sn −n i(m) . Example 28.3 Determine the present value and the accumulated value of 1 at the end of each quarter in the ﬁrst year, 2 at the end of each quarter in the second year, 3 at the end of each quarter in the third year, 4 at the end of each quarter in the fourth year, and 5 at the end of each quarter in the ﬁfth year. The annual eﬀective interest rate is 4% Solution. The factor ( Ia)(4) 5 values a payment of 1/4 at the end of each quarter during the ﬁrst year, 2/4 at the end of each quarter during the second year, and so on. So, the present value is 4(Ia)(4) 5 = 4 · ¨a5 −5ν5 i(4) . But i(4) =4[(1.04) 1 4 −1] = 3.9414% ¨a5 =1 −(1.04)−5 0.04/1.04 = 4.629895 (Ia)(4) 5 =4.629895 −5(1.04)−5 0.039414 = 13.199996 4(Ia)(4) 5 =52.80. The accumulated value is (Is)(4) 5 = (1.04)5(Ia)(4) 5 = 1.045 ×52.80 = 64.24 In the case of an annuity-due, one can easily show that (I¨a)(m) n = ¨an −nνn d(m) and (I¨s)(m) n = (1 + i)n(I¨a)(m) n = ¨sn −n d(m) . 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE261 Example 28.4 Determine the present value and accumulated value of 10 at the start of each quarter in the ﬁrst year, 20 at the start of each quarter in the second year, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The present value is 40(I¨a)(4) 5 =40 ×¨a5 −5ν5 d(4) =40 ×4.629895 −5(1.04)−5 0.039029 =533.20 where d(4) = 4[1 −(1.04)−1 4 ] = 3.9029%. The accumulated value is (1.04)5 ×533.20 = 648.72 mthly Increasing mthly Payable Annuities Consider next a situation in which payments vary within each interest conversion period as shown in Figure 28.3. Such an annuity is referred to as mthly increasing mthly annuity. Figure 28.3 Denoting the present value of such an annuity by ( I(m)a)(m) n we ﬁnd (I(m)a)(m) n = 1 m2 [ν 1 m + 2ν 2 m + ··· + mnν mn m ]. (28.5) Multiply both sided by (1 + i) 1 m to obtain (1 + i) 1 m(I(m)a)(m) n = 1 m2 [1 + 2ν 1 m + ··· + mnνn−1 m]. (28.6) 262 THE BASICS OF ANNUITY THEORY Subtract Equation (28.5) from Equation (28.6) to obtain (I(m)a)(m) n [(1 + i) 1 m −1] = 1 m2 [1 + ν 1 m + ··· + νn−1 m −nmνn] = 1 m[¨a(m) n −nνn] (I(m)a)(m) n = ¨a(m) n −nνn m[(1 + i) 1 m −1] =¨a(m) n −nνn i(m) where ¨a(m) n = 1 −νn d(m) = i d(m) an. The accumulated value is (I(m)s)(m) n = (1 + i)n(I(m)a)(m) n = ¨s(m) n −n i(m) where ¨s(m) n = (1 + i)m −1 d(m) . Example 28.5 Determine the present value and the accumulated value of 1 at the end of the ﬁrst quarter, 2 at the end of the second quarter, 3 at the end of the third quarter, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The factor ( I(4)a)(4) 5 values a payment of 1 16 at the end of the ﬁrst quarter, 2 16 at the end of the second quarter, and so on. So the present value is 16(I(4)a)(4) 5 = 16 ×¨a(4) 5 −5ν5 i(4) = 16 ×4.56257 −5(1.04)−5 0.039414 = 183.87 and the accumulated value is (1.04)5 ×183.87 = 223.71 In the case of an annuity-due, it can be easily shown that (I(m)¨a)(m) n = ¨a(m) n −nνn d(m) 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE263 and (I(m)¨s)(m) n = (1 + i)n(I(m)¨a)(m) n = ¨s(m) n −n d(m) . Example 28.6 Determine the present value and the accumulated value of 1 at the start of the ﬁrst quarter, 2 at the start of the second quarter, 3 at the start of the third quarter, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The present value is 16(I(5)¨a)(4) 5 = 16 ×¨a(4) 5 −5ν5 d(4) = 4.56257 −5(1.04)−5 0.03903 = 185.68 and the accumulated value is (1.04)5 ×185.68 = 225.91 Other Considerations Types of annuities like the one discussed above but that vary in geometric progression can be handled by expressing the annuity value as a summation of the present value or accumulated value of each payment. This summation is a geometric progression which can be directly evaluated. We illutrate this in the next example. Example 28.7 Find the accumulated value at the end of eight years of an annuity in which payments are made at the beginning of each quarter for four years. The ﬁrst payment is $3,000, and each of the other payments is 95% of the previous payment. Interest is credited at 6% convertible semiannually. Solution. We count periods in quarters of a year. If j is the interest rate per quarter then 1 + j = (1.03) 1 2 . The accumulated value at t= 8 ×4 = 32 is 3000[(1.03)16 + (0.95)(1.03)15.5 + (0.95)2(1.03)15 + ··· + (0.95)15(1.03)8.5] =3000(1.03)16 [ 1 + ( 0.95√ 1.03 ) + ( 0.95√ 1.03 )2 + ··· + ( 0.95√ 1.03 )15] =3000(1.03)16 · 1 − ( 0.95√ 1.03 )16 1 − ( 0.95√ 1.03 ) = $49,134.18 264 THE BASICS OF ANNUITY THEORY Example 28.8 Find an expression for the present value of a perpetuity which pays 1 at the end of the third year, 2 at the end of the sixth year, 3 at the end of the ninth year and so on. Solution. The present value satisﬁes the equation PV = ν3 + 2ν6 + 3ν9 + ··· Thus, PV (1 −ν3) = ν3 + ν6 + ν9 + ··· = ν3 1 −ν3 or PV = ν3 (1 −ν3)2 Some of the problems can be solved using the basic principles. Example 28.9 Scott deposits 1 at the beginning of each quarter in year 1, 2 at the beginning of each quarter in year 2, ··· , 8 at the beginning of each quarter in year 8. One quarter after the last deposit Scott withdraws the accumulated value of the fund and uses it to buy a perpetuity immediate with level payments of X at the end of each year. All calculations assume a nominal interest rate of 10% per annum compounded quarterly. Calculate X. Solution. Regard the annuity as an annuity-due with 8 annual payments: k¨a4 0.025 where 1 ≤k ≤8. The annual eﬀective rate of interest is 1 + i= (1.025)4. The accumulated value at time t= 32 is ¨a4 0.025(I¨s)8 i = ¨a4 0.025 [s9 i−9](1+i) i = 196.77. On the other hand, we have X (1.025)4 −1 = 196.77. Solving this equation for X we ﬁnd X = 20.43 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE265 Practice Problems Problem 28.1 (a) Find the total sum of all the payments in ( Ia)(12) 2 . (b) Find the total sum of all the payments in ( I(12)a)(12) 2 . Problem 28.2 Find an expression for ( I(m)a)(m) ∞ . Problem 28.3 An annuity pays $100 at the end of each month in the ﬁrst year, $200 at the end of each month in the second year, and continues to increase until it pays 1000 at the end of each month during the 10th year. Calculate the present value of the annuity at an annual eﬀective interest rate of 6%. Problem 28.4 A monthly annuity due pays $1 at the beginning of the ﬁrst month. Each subsequent payment increases by $1. The last payment is made at the beginning of the 240th month. Calculate the accumulated value of the annuity at the end of the 240th month using: (a) an interest rate of 6% compounded monthly; (b) an annual eﬀective interest rate of 6%. Problem 28.5 A 20 year annuity pays 10 at the beginning of each quarter during the ﬁrst year, 20 at the beginning of each quarter during the second year, etc with 200 being paid at the beginning of each quarter during the last year. Calculate the present value of the annuity assuming an annual eﬀective interest rate of 12%. Problem 28.6 A 20 year annuity pays 10 at the beginning of each quarter during the ﬁrst year, 20 at the beginning of each quarter during the second year, etc with 200 being paid at the beginning of each quarter during the last year. Calculate the accumulated value of the annuity assuming a nominal interest rate of 6% compounded monthly. Problem 28.7 Show that the present value of a perpetuity on which payments are 1 at the end of the 5th and 6th years, 2 at the end of the 7th and 8th years, 3 at the end of the 9th and 10th years, etc, is ν4 i−νd. 266 THE BASICS OF ANNUITY THEORY Problem 28.8 A perpetuity has payments at the end of each four-year period. The ﬁrst payment at the end of four years is 1. Each subsequent paymet is 5 more than the previous payment. It is known that v4 = 0.75. Calculate the present value of this perpetuity. Problem 28.9 A perpetuity provides payments every six months starting today. The ﬁrst payment is 1 and each payment is 3% greater than the immediately preceding payment. Find the present value of the perpetuity if the eﬀective rate of interest is 8% per annum. Problem 28.10 Find the accumulated value at the end of ten years of an annuity in which payments are made at the beginning of each half-year for ﬁve years. The ﬁrst payment is $2,000, and each of the other payments is 98% of the previous payment. Interest is credited at 10% convertible quarterly. Problem 28.11 Find the present value of payments of 5 now, 10 in 6 months, 15 in one year, 20 in 18 months, and so on for 6 years. The nominal discount rate is 12% convertible semiannually. Problem 28.12 Marlen invests $ X now in order to receive $5 in 2 months, $10 in 4 months, $15 in 6 months, and so on. The payments continue for 10 years. The annual eﬀective rate of interest is 8%. Determine X. Problem 28.13 A 10-year annuity has the following schedule of payments: $100 on each January 1, $200 on each April 1, $300 on each July 1, and $400 on each October 1. Show that the present value of this annuity on January 1 just before the ﬁrst payment is made is 1600¨a10 (I(4)¨a)(4) 1 . Problem 28.14 Calculate the accumulated value at time 15 years of payments of 35 at the start of every quarter during the ﬁrst year, 70 at the start of every quarter during the second year, 105 at the start of every quarter during the third year, and so on for 15 years. The nominal interest is 12% convertible quarterly. Problem 28.15 A perpetuity pays 1000 immediately. The second payment is 97% of the ﬁrst payment and is made at the end of the fourth year. Each subsequent payment is 97% of the previous payment and is paid four years after the previous payment. Calculate the present value of this annuity at an annual eﬀective rate of 8%. 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE267 Problem 28.16 A 20 year annuity pays 10 at the end of the ﬁrst quarter, 20 at the end of the second quarter, etc with each payment increasing by 10 until the last quarterly payment is made at the end of the 20th year. Calculate the present value of the annuity assuming a nominal interest rate of 6% compounded monthly. Problem 28.17 A 20 year annuity pays 10 at the end of the ﬁrst quarter, 20 at the end of the second quarter, etc with each payment increasing by 10 until the last quarterly payment is made at the end of the 20th year. Calculate the present value of the annuity assuming an annual eﬀective interest rate of 12%. Problem 28.18 Joan has won a lottery that pays 1000 per month in the ﬁrst year, 1100 per month in the second year, 1200 per month in the third year, and so on. Payments are made at the end of each month for 10 years. Using an eﬀective interest rate of 3% per annum, calculate the present value of this prize. Round your answer to the nearest integer. Problem 28.19 On his 65th birthday Smith would like to purchase a ten-year annuity-immediate that pays 6000 per month for the ﬁrst year, 5500 per month for the second year, 5000 per month for the third year, and so on. Using an annual eﬀective interest rate of 5%, how much will Smith pay for such an annuity? Problem 28.20 On Susan’s 65 th birthday, she elects to receive her retirement beneﬁt over 20 years at the rate of 2,000 at the end of each month. The monthly beneﬁt increases by 4% each year. Assuming an annual eﬀective interest rate of 8.16%, and letting j denote the equivalent monthly eﬀective interest rate, ﬁnd the present value on Susan’s 65 th birthday of her retirement beneﬁt. 268 THE BASICS OF ANNUITY THEORY 29 Continuous Varying Annuities In this section we look at annuities in which payments are being made continuously at a varying rate. Consider an annuity for n interest conversion periods in which payments are being made contin- uously at the rate f(t) at exact moment t and the interest rate is variable with variable force of interest δt. Then f(t)e− ∫t 0 δrdrdt is the present value of the payment f(t)dt made at exact moment t. Hence, the present value of this n−period continuous varying annuity is PV = ∫ n 0 f(t)e− ∫t 0 δrdrdt. (29.1) Example 29.1 Find an expression for the present value of a continuously increasing annuity with a term of nyears if the force of interest is δ and if the rate of payment at time t is t2 per annum. Solution. Using integration by parts process, we ﬁnd ∫ n 0 t2e−δtdt= −t2 δe−δt ]n 0 + 2 δ ∫ n 0 te−δtdt = −n2 δ e−δn − [2t δ2 e−δt ]n 0 + 2 δ2 ∫ n 0 e−δtdt = −n2 δ e−δn −2n δ2 e−δn − [2 δ3 e−δt ]n 0 = −n2 δ e−δn −2n δ2 e−δn − 2 δ3 e−δn + 2 δ3 = 2 δ3 −e−δn [n2 δ + 2n δ2 + 2 δ3 ] Under compound interest, i.e., δt = ln (1 +i), formula (29.1) becomes PV = ∫ n 0 f(t)νtdt. 29 CONTINUOUS VARYING ANNUITIES 269 Under compound interest and with f(t) = t (an increasing annuity), the present value is (Ia)n = ∫ n 0 tνtdt = tνt ln ν ⏐⏐⏐⏐ n 0 − ∫ n 0 νt ln νdt = tνt ln ν ⏐⏐⏐⏐ n 0 − νt (ln ν)2 ⏐⏐⏐⏐ n 0 = −nνn δ −νn δ2 + 1 δ2 =1 −νn δ2 −nνn δ =an −nνn δ and the accumulated value at time n years is (Is)n = (1 + i)n(Ia)n = sn −n δ . The above two formulas can be derived from the formulas of ( I(m)a)(m) n and (I(m)s)(m) n . Indeed, (Ia)n = lim m→∞ (I(m)a)(m) n = lim m→∞ ¨a(m) n −nνn i(m) = an −nνn δ and (Is)n = lim m→∞ (I(m)s)(m) n = lim m→∞ ¨s(m) n −n i(m) = sn −n δ . Example 29.2 Sam receives continuous payments at an annual rate of 8 t + 5 from time 0 to 10 years. The continuously compounded interest rate is 9%. (a) Determine the present value at time 0. (b) Determine the accumulated value at time 10 years. Solution. (a) The payment stream can be split into two parts so that the present value is 8(Ia)10 + 5a10 . 270 THE BASICS OF ANNUITY THEORY Since i=e0.09 −1 = 9.4174% a10 =1 −(1.094174)−10 0.09 = 6.59370 (Ia)10 =6.59370 −10(1.094174)−10 0.09 = 28.088592 we obtain 8(Ia)10 + 5a10 = 8 ×28.088592 + 5×6.59370 = 257.68. (b) The accumulated value at time 10 years is 257.68 ×(1.094174)10 = 633.78 Example 29.3 If δ= 0.06, calculate the present value of 10 year continuous annuity payable at a rate of t at time t. Solution. The present value is (Ia)10 = 1 −ν10 δ2 −10ν10 δ = 1 −e−10×0.06 0.062 −10e−10×0.06 0.06 = 33.86 For a continuous payable continuously increasing perpetuity (where f(t) = t), the present value at time 0 is (Ia)∞ = lim n→∞ an −nνn δ = lim n→∞ 1−(1+i)−n δ −n(1 + i)−n δ = 1 δ2 . Example 29.4 Determine the present value of a payment stream that pays a rate of 5 t at time t. The payments start at time 0 and they continue indeﬁnitely. The annual eﬀective interest rate is 7%. Solution. The present value is 5(Ia)∞ = 5 [ln (1.07)]2 = 1,092.25 We conclude this section by considering the case of a continuously decreasing continuously payable stream in which a continuous payment is received from time 0 to time nyears. The rate of payment 29 CONTINUOUS VARYING ANNUITIES 271 at time t is f(t) = n−t, and the force of interest is constant at δ. The present value is (Da)n =nan −(Ia)n =n1 −νn δ −an −nνn δ =n−an δ Example 29.5 Otto receives a payment at an annual rate of 10 −t from time 0 to time 10 years. The force of interest is 6%. Determine the present value of these payments at time 0. Solution. Since i= e0.06 −1 = 6.184% a10 = 1 −(1.06184)−10 0.06 = 7.5201 the present value is then (Da)10 = 10 −7.5201 0.06 = 41.33 Example 29.6 Using the information from the previous example, determine the accumulated value at time 10 years of the payments received by Otto. Solution. The accumulated value at time 10 years is (1.06184)10 ×(Da)10 = (1.06184)10 ×41.33 = 75.31 272 THE BASICS OF ANNUITY THEORY Practice Problems Problem 29.1 Calculate the present value at an annual eﬀective interest rate of 6% of a 10 year continuous annuity which pays at the rate of t2 per period at exact moment t. Problem 29.2 Calculate the accumulated value at a constant force of interest of 5% of a 20 year continuous annuity which pays at the rate of t+1 2 per period at exact moment t. Problem 29.3 Evaluate (Ia)∞ if δ= 0.08. Problem 29.4 Payments under a continuous perpetuity are made at the periodic rate of (1 + k)t at time t. The annual eﬀective rate of interest is i, where 0 < k < i.Find an expression for the present value of the perpetuity. Problem 29.5 A perpetuity is payable continuously at the annual rate of 1 + t2 at time t. If δ = 0.05, ﬁnd the present value of the perpetuity. Problem 29.6 A one-year deferred continuous varying annuity is payable for 13 years. The rate of payment at time tis t2 −1 per annum, and the force of interest at time tis (1 +t)−1.Find the present value of the annuity. Problem 29.7 If an = a and a2n = b, express (Ia)n in terms of a and b. Problem 29.8 A 10 year continuous annuity pays at a rate of t2 + 2t+ 1 at time t. Calculate the present value of this annuity if δt = (1 + t)−1. Problem 29.9 A 30 year continuous annuity pays at a rate of t at time t. If δ = 0.10, calculate the current value of the annuity after ten years. Problem 29.10 A 10 year continuous annuity pays at a rate of t at time t. If νt = .94t, calculate the accumulated value of the annuity after ten years. 29 CONTINUOUS VARYING ANNUITIES 273 Problem 29.11 Nancy receives a payment stream from time 2 to time 7 years that pays an annual rate of 2 t−3 at time t. The force of interest is constant at 6% over the period. Calculate the accumulated value at time 7 years. Problem 29.12 Find the ratio of the total payments made under ( Ia)10 during the second half of the term of the annuity to those made during the ﬁrst half. Problem 29.13 ‡ Payments are made to an account at a continuous rate of (8 k+ tk), where 0 ≤t ≤10. Interest is credited at a force of interest δt = 1 8+t. After 10 years, the account is worth 20,000. Calculate k. Problem 29.14 Show that (Ia)n <(I(m)a)(m) n <(Ia)(m) n whenever nδ <1. Hint: Use Problem 25.11. Problem 29.15 An investment fund is started with an initial deposit of 1 at time 0. New deposits are made continuously at the annual rate 1 + t at time t over the next n years. The force of interest at time t is given by δt = (1 + t)−1. Find the accumulated value in the fund at the end of n years. Problem 29.16 Money was deposited continuously at the rate of 5,000 per annum into a fund for 7 years. If the force of interest is 0.05 from t= 0 to t= 10, what will the fund balance be after 10 years? Problem 29.17 A 5−year annuity makes payments continuously at a ratef(t) = e ( t2 24 +t ) .The force of interest varies continuously at a rate of δt = t 12 . Find the present value of this annuity. 274 THE BASICS OF ANNUITY THEORY Rate of Return of an Investment In this chapter, we present a number of important results for using the theory of interest in more complex “real-world” contexts than considered in the earlier chapters. In this chapter, the concept of yield rates is introduced: A yield rate (also known as the internal rate of return) is that interest rate at which the present value of all returns from the investment is equal to the present value of all contributions into the investment. Thus, yield rates are solutions to the equation NPV (i) = 0 where NPV stands for the net present value and is the diﬀerence between the present value of all returns and that of all contributions. Techniques for measuring the internal rate of return will be discussed. Yield rates are used as an index to see how favorable or unfavorable a particular transaction is. When lower, yield rates favor the borrower and when higher they favor the lender. We point out here, as always the case thus far, that the eﬀect of taxes will be ignored throughout the discussions. 275 276 RATE OF RETURN OF AN INVESTMENT 30 Discounted Cash Flow Technique In this section we consider the question of proﬁtability of investment projects that involve cash ﬂows. More speciﬁcally, we look at the present values of projects consisting of cash ﬂows. Such projects can be considered as a generalization of annuities with any pattern of payments and/or withdrawals compared to annuities previously discussed where the annuities consisted of regular series of payments. The process of ﬁnding the present value of the type of projects consisting of a stream of cash ﬂows will be referred to as the discounted cash ﬂow technique, abbreviated by DCF. We examine two DCF measures for assessing a project: net present value and internal rate of return. Consider an investment project with contributions (outﬂow) C0,C1,··· ,Cn at time t0 <t1 <t2 < ··· <tn and returns (inﬂow) R0,R1,··· ,Rn at the same point of time. Let ct = Rt−Ct be the net change ( or net cash ﬂow) in the account at time t which can be positive or negative. From the vantage point of the lender, if ct >0, then there is a net cash deposit into the investment at time t. If ct < 0 then there is a net cash withdrawal from the investment at time t. Thus, if at time 1 a deposit of $1,000 is made and at the same time a withdrawal of $2,000 is made then c1 = 1000 −2000 = −1,000 so there is a net cash withdrawal of $1,000 from the investment at that time. We have adopted the vantage point of the lender. From the vantage point of the borrower the signs in the above discussion are switched. The following example illustrates these deﬁnitions. Example 30.1 In order to develop a new product and place it in the market for sale, a company has to invest $80,000 at the beginning of the year and then $10,000 for each of the next three years. The product is made available for sale in the fourth year. For that, a contribution of $20,000 must be made in the fourth year. The company incurs maintenance expenses of $2000 in each of the next ﬁve years. The project is expected to provide an investment return of $12,000 at the end of the fourth year, $30,000 at the end of the ﬁfth year, $40,000 at the end of the sixth year, $35,000 at the end of the seventh year, $25,000 at the end of the eigth year, $15,000 at the end of the ninth year and $8,000 at the end of the tenth year. After the tenth year, the product is withdrawn from the market. Create a chart to describe the cash ﬂows of this project. Solution. Table 30.1 describes the cash ﬂows of this project 30 DISCOUNTED CASH FLOW TECHNIQUE 277 Year Contributions Returns ct 0 80,000 0 −80,000 1 10,000 0 −10,000 2 10,000 0 −10,000 3 10,000 0 −10,000 4 20,000 12,000 −8,000 5 2,000 30,000 28,000 6 2,000 40,000 38,000 7 2,000 35,000 33,000 8 2,000 25,000 23,000 9 2,000 15,000 13,000 10 0 8,000 8,000 Table 30.1 Now, suppose that the rate of interest per period is i.This interest rate is sometimes referred to as the required return of the investment or the cost of capital. Using the discounted cash ﬂow technique, the net present value at rate i of the investment is deﬁned by NPV (i) = n∑ k=0 νtkctk, ν = (1 + i)−1. Thus, the net present value of a series of cash ﬂows is the present value of cash inﬂows minus the present value of cash outﬂows. Expressed in another way, NPV is the sum of the present values of the net cash ﬂows over n periods. The value of NPV (i) can be positive, negative, or zero depending on how large or small the value of i is. Example 30.2 Find the net present value of the investment discussed in the previous example. Solution. The net present value is NPV (i) = −80,000 −10,000ν−10,000ν2 −10,000ν3 −8,000ν4 + 28,000ν5 + 38,000ν6 + 33,000ν7 + 23,000ν8 + 13,000ν9 + 8,000ν10 Notice that NPV (0.03) = 1488.04 >0, NPV(0.032180) = 0, and NPV (0.04) = −5122.13 <0 The value of i for which NPV (i) = 0 is called the yield rate (or internal rate of return ); 278 RATE OF RETURN OF AN INVESTMENT that is, the yield rate is the rate of interest at which the present value of contributions from the investment is equal to the present value of returns into the investment . Stated diﬀerently, the yield rate is the rate that causes investment to break even. According to the eqution NPV (i) = 0 , yield rates are the same from either the borrower’s or lender’s perspective. They are totally determined by the cash ﬂows deﬁned in the transaction and their timing. Yield rates are often used to measure how favorable or unfavorable a transaction might be. From a lender’s perspective, a higher yield rate makes a transaction more favorable. From a borrower’s perspective, a lower yield rate makes a transaction more favorable. Based on ﬁnancial factors alone, projects with positive NPV are considered acceptable. Projects with NPV (i) < 0 should be rejected. When NPV (i) = 0, the investment would neither gain nor lose value; We should be indiﬀerent in the decision whether to accept or reject the project. Example 30.3 Determine the net present value of the investment project with the following cash ﬂow, at a cost of capital of 4.8%. Time 0 1 2 3 4 5 Flow −1000 70 70 70 70 1070 Solution. The answer is NPV (i) = −1000 + 70a5 + 1000ν5 = 95.78 Example 30.4 Assume that cash ﬂows from the construction and sale of an oﬃce building is as follows. Year 0 1 2 Amount −150000 −100000 300000 Given a 7% cost of capital, create a present value worksheet and show the net present value, NPV. Solution. t vt ct PV 0 1.0 −150000 −150000 1 1 1.07 = 0.935 −100000 −93500 2 1 1.072 = 0.873 300000 261900 NPV= 18400 30 DISCOUNTED CASH FLOW TECHNIQUE 279 Finding the yield rate may require the use of various approximation methods, since the equations that have to be solved may be polynomials of high degree. Example 30.5 An investment project has the following cash ﬂows: Year Contributions Returns 0 100 0 1 200 0 2 10 60 3 10 80 4 10 100 5 5 120 6 0 60 (a) Calculate the net present value at 15%. (b) Calculate the internal rate of return (i.e. the yield rate) on this investment. Solution. (a) NPV (0.15) = −100 −200(1.15)−1 + 50(1.15)−2 + 70(1.15)−3 + 90(1.15)−4 + 115(1.15)−5 + 60(1.15)−6 = −$55.51. (b) We must solve the equation −100 −200ν+ 50ν2 + 70ν3 + 90ν4 + 115ν5 + 60ν6 = 0.Using linear interpolation with the points (7% ,4.53) and (8%,−4.52) one ﬁnds i≈0.07 −4.53 × 0.08 −0.07 −4.52 −4.53 = 7.50% Yield rate needs not be unique as illustrated in the following example. Example 30.6 In exchange for receiving $230 at the end of one year, an investor pays $100 immediately and pays $132 at the end of two years. Find the yield rate. Solution. The net present value is NPV (i) = −100 −132ν2 + 230ν. To ﬁnd the IRR we must solve 132 ν2 − 230ν+ 100 = 0. Solving this equation by the quadratic formula, we ﬁnd i= 10% or i= 20% Consider the situation where a bank makes a loan to an individual. In this case the bank acts as a lender or investor, and the individual acts as the borrower. Payments that are received by the investor (cash inﬂows) are taken to be positive cash ﬂows. Payments that are made by the investor (cash outﬂows) are considered to be negative. For the borrower, the cash ﬂows will be of the opposite sign. 280 RATE OF RETURN OF AN INVESTMENT Example 30.7 A bank lends Diane $8,000 now. She repays $600 at the end of each quarter for 5 years. Find the annual eﬀective rate of return. Solution. Let j be the eﬀective IRR per quarter. Then j satisﬁes the equation −8000 + 600a20 j = 0 or 3(1 −(1 + j)−20) −40j = 0. Using linear interpolation with the points (4% ,0.030839) and (4.5%,−0.04392) we ﬁnd j ≈0.04 −0.030839 × 0.045 −0.04 −0.04392 −0.030839 = 0.042. Thus, the annual rate of return is i= 1.0424 −1 = 17.89% Yield rates need not be positive. If a yield rate is 0, then the investor(lender) received no return on investment. If a yield rate is negative, then the investor(lender) lost money on the investment. We will assume that such a negative yield rate satisﬁes −1 <i< 0.A yield rate i< −1 implies full loss of the investment. Example 30.8 Find the yield rates of Example 30.1 Solution. Using a scientiﬁc calculator one ﬁnds the two real solutions i= −1.714964 and i= 0.032180.Since, i> −1, the only yield rate is i= 3.218% 30 DISCOUNTED CASH FLOW TECHNIQUE 281 Practice Problems Problem 30.1 Assuming a nominal rate of interest, convertible quarterly, of 12%, ﬁnd the net present value of a project that requires an investment of $100,000 now, and returns $16,000 at the end of each of years 4 through 10. Problem 30.2 The internal rate of return for an investment with contributions of $3,000 at time 0 and $1,000 at time 1 and returns of $2,000 at time 1 and $4,000 at time 2 can be expressed as 1 n. Find n. Problem 30.3 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. Find NPV (0.10). Problem 30.4 An investment project has the following cash ﬂows: Year Contributions Returns 0 100,000 0 1 5,000 0 2 4,000 10,000 3 2,000 10,000 4 0 20,000 5 0 40,000 6 0 60,000 7 0 80,000 (a) Calculate the net present value at 15%. (b) Calculate the internal rate of return on this investment. Problem 30.5 You are the President of XYZ Manufacturing Company. You are considering building a new factory. The factory will require an investment of 100,000 immediately. It will also require an additional investment of 15,000 at the beginning of year 2 to initiate production. Finally, maintenance costs for the factory will be 5,000 per year at the beginning of years 3 through 6. The factory is expected to generate proﬁts of 10,000 at the end of year one, 15,000 at the end of year two, 20,000 at the end of year 3, and 30,000 at the end of years 4 through 6. Calculate the internal rate of return on the potential factory. 282 RATE OF RETURN OF AN INVESTMENT Problem 30.6 An investment project has the following cash ﬂows: Year Contributions Returns 0 100,000 0 1 0 10,000 2 0 20,000 3 0 30,000 4 0 20,000 5 0 10,000 Calculate the internal rate of return on this investment. Problem 30.7 What is the internal rate of return of a project that requires an investment of $1,000,000 now, and returns $150,000 at the end of each of years 1 through 15? Problem 30.8 Assuming an annual eﬀective discount rate of 6%, what is the net present value of a project that requires an investment of $75,000 now, and returns 2 ,000(13 −t) at times t = 1 ,2,··· ,12 (i.e, $24,000 at time 1, $22,000 at time 2, $20,000 at time 3, etc.)? Problem 30.9 What is the internal rate of return on a project that requires a $10,000 investment now, and returns $4,000 four years from now and $12,000 eight years from now? Problem 30.10 A project has the following out-ﬂows and in-ﬂows: out-ﬂows: 12,000 at t= 0, 6,000 at t= 3, 9,000 at t= 6 and 12,000 at t= 9 in-ﬂows: 3,000 at t= 2 through t= 14 plus 12,000 at t= 15. If your desired internal rate of return was 8%, would you accept this project? Problem 30.11 A company pays $120,000 to purchase a property. The company pays $3,000 at the end of each of the next 6 months to renovate the property. At the end of the eigth month, the company sells the property for $150,000. Find the net present value of this project for the company at an annual eﬀective rate of 8%. Problem 30.12 What is the project’s monthly internal rate of return of the previous problem? 30 DISCOUNTED CASH FLOW TECHNIQUE 283 Problem 30.13 A project requires an initial investment of $50,000. The project will generate net cash ﬂows of $15,000 at the end of the ﬁrst year, $40,000 at the end of the second year, and $10,000 at the end of the third year. The project’s cost of capital is 13%. According to the DCF technique, should you invest in the project? Problem 30.14 A ten-year investment project requires an initial investment of $100,000 at inception and mainte- nance expenses at the beginning of each year. The maintenance expense for the ﬁrst year is $3,000, and is anticipated to increase 6% each year thereafter. Projected annual returns from the project are $30,000 at the end of the ﬁrst year decreasing 4% per year thereafter. Find c6 = R6 −C6. Problem 30.15 Determine the net present value for a project that costs $104,000 and would yield after −tax cash ﬂows of $16,000 the ﬁrst year, $18,000 the second year, $21,000 the third year, $23,000 the fourth year, $27,000 the ﬁfth year, and $33,000 the sixth year. Your ﬁrm’s cost of capital is 12%. Problem 30.16 Determine the net present value for a project that costs $253,494.00 and is expected to yield after−tax cash ﬂows of $29,000 per year for the ﬁrst ten years, $37,000 per year for the next ten years, and $50,000 per year for the following ten years. Your ﬁrm’s cost of capital is 12%. Problem 30.17 An investment project has the following cash ﬂows: Year Contributions Returns 0 10,000 0 1 5,000 0 2 1,000 0 3 1,000 0 4 1,000 0 5 1,000 0 6 1,000 8,000 7 1,000 9,000 8 1,000 10,000 9 1,000 11,000 10 0 12,000 Find the yield rate of this investment project. 284 RATE OF RETURN OF AN INVESTMENT Problem 30.18 Suppose a sum of money could be invested in project A which pays a 10% eﬀective rate for six years, or be invested in project B which pays an 8% eﬀective rate for 12 years. (a) Which option we choose as an investor? (b) Find the eﬀective rate of interest necessary for the six years after projectAends so that investing in project B would be equivalent. Problem 30.19 ‡ Project P requires an investment of 4000 at time 0. The investment pays 2000 at time 1 and 4000 at time 2. Project Qrequires an investment of X at time 2. The investment pays 2000 at time 0 and 4000 at time 1. The net present values of the two projects are equal at an interest rate of 10%. Calculate X. Problem 30.20 X corporation must decide whether to introduce a new product line. The new product will have startup costs, operational costs, and incoming cash ﬂows over six years. This project will have an immediate (t= 0) cash outﬂow of $100,000 (which might include machinery, and employee training costs). Other cash outﬂows for years 1-6 are expected to be $5,000 per year. Cash inﬂows are expected to be $30,000 per year for years 1-6. All cash ﬂows are after-tax, and there are no cash ﬂows expected after year 6. The required rate of return is 10%. Determine the net present value at this rate. Problem 30.21 An investment account is established on which it is estimated that 8% can be earned over the next 20 years. If the interest each year is subject to income tax at a 25% tax rate, ﬁnd the percentage reduction in the accumulated interest at the end of 20 years. 31 UNIQUENESS OF IRR 285 31 Uniqueness of IRR As pointed out in Example 30.4, an internal rate of return may not be unique. In this section, we look at conditions under which the equation ∑n i=0 ctiνti = 0 has exactly one solution. For the ease of the analysis we choose time to be equally spaced so that NPV (i) = ∑n t=0 ctνt = 0, that is, the net present value function is a polynomial of degree n in ν. By the Fundamental Theorem of Algebra, NPV (i) = 0 has n roots, counting repeated roots and complex roots. Example 31.1 Consider an investment with C0 = 1,C1 = 0,C2 = 1.36,R0 = 0,R1 = 2.3,R2 = 0.Find the internal rate of return. Solution. We have c0 = −1,c1 = 2.3,c2 = −1.36. The internal rate of return is a solution to the equation −1 + 2.3ν−1.36ν2 = 0. Solving this equation using the quadratic formula we ﬁnd ν = 2.3±i √ 0.0111 2.72 so no real root for i Example 31.2 Consider an investment with C0 = 1,C1 = 0,C2 = 1.32,R0 = 0,R1 = 2.3,R2 = 0.Find the internal rate of return. Solution. We have c0 = −1,c1 = 2.3,c2 = −1.32. Thus, the equation of value at time t = 0 is −1 + 2.3ν− 1.32ν2 = 0.Solving this equation using the quadratic formula we ﬁndν = −2.3±0.1 −2.64 so either ν = 0.833 or ν = 0.91. Hence, either i= 20% or i= 10% so two internal rates of return Example 31.3 Consider an investment with C0 = 1 ,C1 = 0 ,C2 = 1 .2825,R0 = 0 ,R1 = 2 .3,R2 = 0 . Find the internal rate of return. Solution. We have c0 = −1,c1 = 2.3,c2 = −1.2825. Thus, the equation of value at time t= 0 is −1 + 2.3ν− 1.2825ν2 = 0. Solving this equation we ﬁnd ν = −2.3±0.4 −2.565 so either ν = 0.74 or ν = 1.053. Hence, either i= 35% or i= −5% What we would like to do next is to formulate conditions on the ct that guarantee a unique yield rate i> −1. A ﬁrst set of conditions is given in the following theorem. 286 RATE OF RETURN OF AN INVESTMENT Theorem 31.1 Let k be an integer such that 0 <k <n. Suppose that either (i) ct ≤0 for 0 ≤t≤k and ct ≥0 for k+ 1 ≤t≤n; or (ii) ct ≥0 for 0 ≤t≤k and ct ≤0 for k+ 1 ≤t≤n. Then there is a unique interest rate i> −1 such that NPV (i) = 0. This theorem states that a unique internal rate of return exists for a transaction in which the net payments are all of one sign for the ﬁrst portion of the transaction and then have the opposite sign for the remainder of the transaction. To prove the theorem, we require the concept of Descartes’ Rule of signs which is a technique for determining the number of positive or negative roots of a polynomial: If the terms of a single-variable polynomial with real coeﬃcients and a nonzero constant term are ordered by descending variable exponent, then the number of positive roots counting multiplicity of a polynomial f(x) is either equal to the number of sign changes between adjacent nonzero coeﬃcients, or less than it by a multiple of 2. The number of negative zeros of a polynomial f(x) counting multiplicity is equal to the number of positive zeros of f(−x), so it is the number of changes of signs of f(−x) or that number decreased by an even integer. Complex roots always come in pairs. That’s why the number of positive or number of negative roots must decrease by two. Example 31.4 Use Descartes’ rule of sign to ﬁnd the possible number of positive roots and negative roots of the polynomial f(x) = x2 + x+ 1. Solution. The polynomial f(x) has no sign change between adjacent coeﬃcients. Thus, f(x) has no positive roots. The polynomial f(−x) = x2 −x+ 1 has two sign changes between adjacent coeﬃcients so that f(x) has either two negative roots or zero negative roots. Thus, the two possibilities are either two negative roots or two complex roots Example 31.5 Use Descartes’ rule of sign to ﬁnd the possible number of positive roots and negative roots of the polynomial f(x) = x3 + x2 −x−1. Solution. The polynomial f(x) has one sign change between the second and third terms. Therefore it has at most one positive root. On the other hand, f(−x) = −x3 + x2 + x−1 has two sign changes, so f(x) has 2 or 0 negative roots. Since complex roots occur in conjugate, the two possibilities are either one positive root and two negative roots or one positive root and two complex roots. Thus, f(x) 31 UNIQUENESS OF IRR 287 has exactly one positive root Proof of Theorem 31.1. According to either (i) and (ii), the coeﬃcients of NPV (i) changes signs once. So by Descartes’ rule of signs, there will be at most one positive real root. If such a root exists, then ν >0 implies i> −1. For such values of i we deﬁne the function f(i) = (1 + i)kNPV (i) = k∑ t=0 ct(1 + i)k−t + n∑ t=k+1 ct(1 + i)k−t. Taking the ﬁrst derivative we ﬁnd f′(i) = k−1∑ t=0 ct(k−t)(1 + i)k−t−1 + n∑ t=k+1 ct(k−t)(1 + i)k−t−1. If (i) is satisﬁed then ct(k−t) ≤0 for 0 ≤t≤n.Hence, f′(i) ≤0 and the function f(i) is decreasing. If (ii) is satisﬁed then f′(i) ≥0 and f(i) is increasing. In case (i), the function f(i) is decreasing. Moreover, as i→−1+ the ﬁrst sum in f(i) approaches ck whereas the second sum approaches ∞. Hence, lim i→−1+ f(i) = ∞. Similarly, as i approaches +∞ the ﬁrst sum in f(i) approaches −∞whereas the second sum approaches 0. Hence, lim i→∞f(i) = −∞. Since f is decreasing, there is a unique i >−1 such that NPV (i) = 0 . A similar argument holds for case (ii) Example 31.6 A project requires an initial investment of $10,000 and it produces net cash ﬂows of $10,000 one year from now and $2,000 two years from now. Show that there is a unique internal rate of return. Solution. The net present value is NPV (i) = −10,000+10 ,000ν+2,000ν2.The result follows from Theorem 31.1(i) with k= 1 The following theorem establishes uniqueness of yield rates under a broader set of conditions than the one given in the previous theorem. Theorem 31.2 288 RATE OF RETURN OF AN INVESTMENT Let i> −1 be a solution to NPV (i) = 0. Suppose that B0 =c0 >0 B1 =c0(1 + i) + c1 >0 B2 =c0(1 + i)2 + c1(1 + i) + c2 >0 ... Bn−1 =c0(1 + i)n−1 + c1(1 + i)n−2 + ··· + cn−1 >0 Then (i) Bn = c0(1 + i)n + c1(1 + i)n−1 + ··· + cn = 0 (ii) i is unique. Proof. (i) We are given that ∑n t=0 ctνt = 0.This is an nth degree polynomial in ν and could be written as an nth degree polynomial in i by simply multiplying both sides by (1 + i)n to obtain c0(1 + i)n + c1(1 + i)n−1 + ··· + cn = 0. Thus, the assumption of (i) is established. (ii) Suppose that j satisﬁes NPV (j) = 0 and j >i.Then we have B′ 0 =c0 = B0 >0 B′ 1 =B′ 0(1 + j) + c1 = c0(1 + j) + c1 >B1 >0 B′ 2 =B′ 1(1 + j) + c2 = c0(1 + j)2 + c1(1 + j) + c2 >B2 >0 ... B′ n−1 =c0(1 + j)n−1 + c1(1 + j)n−2 + ··· + cn−1 >Bn−1 >0 B′ n =B′ n−1(1 + j) + cn = c0(1 + j)n + c1(1 + j)n−1 + ··· + cn >Bn = 0 The last inequality shows that j is not a yield rate, a contradiction. A similar argument holds for −1 <j <i. Hence, we conclude that j = i Remark 31.1 Note that Bt is the outstanding balance at timet.According to the theorem above, if the outstanding balance is positive throughout the period of investment, then the yield rate will be unique. Also, note that c0 > 0 and cn = −Bn−1(1 + i) < 0, but that ct for t = 1 ,2,··· ,n −1 may be either positive, negative, or zero. 31 UNIQUENESS OF IRR 289 Example 31.7 Show that we cannot guarantee uniqueness of the yield rate if c0 and cn have the same sign. Solution. The conclusion of uniqueness depends on the outstanding balance being positive at all time during the investment. If c0 >0 and cn >0 then the outstanding balance must be negative prior to t= n. The negative outstanding balance implies no guarantee of uniqueness. Similarly, if c0 < 0 and cn <0, the outstanding balance at the start of the investment is negative so that no guarantee of uniqueness of yield rate The discussion of this section focused on the existence of yield rates. However, it is possible that no yield rate exists or all yield rates are imaginary (See Example 31.1). Example 31.8 An investor is able to borrow $1,000 at 8% eﬀective for one year and immediately invest the $1,000 at 12% eﬀective for the same year. Find the investor’s yield rate on this transaction. Solution. The proﬁt at the end of the year is $40, but there is no yield rate, since the net investment is zero 290 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 31.1 Use Descartes’ rule of signs to determine the maximum number of real zeroes of the polynomial f(x) = x5 −x4 + 3x3 + 9x2 −x+ 5. Problem 31.2 Consider the following transaction: Payments of $100 now and $108.15 two years from now are equivalent to a payment of $208 one year from now. Find the absolute value of the diﬀerence of the two rates that result from the equation of value at time t= 0. Problem 31.3 An investor pays $100 immediately and $X at the end of two years in exchange for $200 at the end of one year. Find the range of X such that two yield rates exist which are equal in absolute value but opposite in sign. Problem 31.4 What is the yield rate on a transaction in which an investor makes payments of $100 immediately and $101 at the end of two years, in exchange for a payment of $200 at the end of one year? Problem 31.5 ‡ A company deposits 1,000 at the beginning of the ﬁrst year and 150 at the beginning of each subsequent year into perpetuity. In return the company receives payments at the end of each year forever. The ﬁrst payment is 100. Each subsequent payment increases by 5%. Calculate the company’s yield rate for this transaction. Problem 31.6 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. What is the maximum number of possible yield rates using Descartes’ rule of signs? Problem 31.7 An investment costs $85 and it will pay $100 in 5 years. Determine the internal rate of return for this investment. Problem 31.8 A project requires an initial investment of $50,000. The project will generate net cash ﬂows of $15,000 at the end of the ﬁrst year, $40,000 at the end of the second year, and $10,000 at the end of the third year. Show that an internal rate of return exists and is unique. 31 UNIQUENESS OF IRR 291 Problem 31.9 Suppose payments of 100 now and 109.20 two years from now are to be made in return for receiving 209 one year from now. Determine the internal rate of return. Problem 31.10 What is the internal rate of return (in terms of an eﬀective annual interest rate) on a project that requires a $1,000 investment now, and provides returns to the investor of $500 one year from now and $800 two years from now? Problem 31.11 Suppose a project requires you to invest $10,000 now and $20,000 one year from now. The project returns $29,000 six months from now. Find all the yield rates (internal rates of return) of this project. Express these yield rates as annual eﬀective rates. Problem 31.12 For the previous problem, ﬁnd the range of annual interest rates which will produce a net present value greater than zero. Problem 31.13 Two growing perpetuities, each with annual payments, have the same yield rate (internal rate of return). The ﬁrst perpetuity has an initial payment of $50 one year from now, and each subsequent annual payment increases by $5. The present value of this ﬁrst perpetuity is $2,000. The second perpetuity (a perpetuity-due) has an initial payment of $100 now, and each subsequent annual payment increases by 3%. Find the present value, now, of the second perpetuity. 292 RATE OF RETURN OF AN INVESTMENT 32 Interest Reinvested at a Diﬀerent Rate In this section we consider transactions where interest may be reinvested at a rate which may or may not be equal to the original investment rate. We analyze two situations in which reinvestment rates are directly taken into consideration. First, consider an investment of 1 for n periods at rate i where the interest is reinvested at rate j. The time diagram of this situation is shown in Figure 32.1. Figure 32.1 The accumulated value at the end of the n periods is equal to the principal plus the accumulated value of the annuity-immediate with periodic payments of i at the end of each period and periodic rate j. That is, AV = 1 + isnj. Note that in the case i= j,the formula reduces to the familiar formula AV = (1 +i)n.That is, for compound interest, the reinvestment rate is equal to the original rate. Example 32.1 100 is invested in a Fund A now. Fund A will pay interest at 10% each year. When the interest is paid in Fund A, it will be immediately removed and invested in Fund B paying 8 %. Calculate the sum of the amounts in Fund A and Fund B after 10 years. Solution. At the end of 10 years, Fund A will just have the original principal: $100. Fund B will have the accumulated value of the interest payments at the rate 8%: 10s10 0.08 = 10(1.0810 −1) 0.08 = 144.87. So the total amount of money in both funds is about $244.87 Example 32.2 Brown deposits 5000 in a 5-year investment that pays interest quarterly at 8%, compounded quar- terly. Upon receipt of each interest payment, Brown reinvests the interest in an account that earns 6%, compounded quarterly. Determine Brown’s yield rate over the 5 year period, as a nominal interest rate compounded quar- terly. 32 INTEREST REINVESTED AT A DIFFERENT RATE 293 Solution. The equation of value after 20 quarters is 5000 ( 1 + i(4) 4 )20 = 5000 + 5000(0.08)s20 0.015. Solving this equation we ﬁnd i(4) = 4 [1 + 0.08s20 0.015] 1 20 −1 = 7.68% Second, consider an investment of 1 at the end of each period for n periods at rate i where the interest is reinvested at rate j. The time diagram of this situation is shown in Figure 32.2. Figure 32.2 The accumulated value at the end of the n periods is the sum of the annuity payments and the accumulated value of the interest,i.e. AV = n+ i(Is)n−1 j = n+ i [snj −n j ] . Note that the above formula simpliﬁes to the familiar result AV = sn when j = i. Example 32.3 Payments of $1,000 are invested at the end of each year for 10 years. The payments earn interest at 7% eﬀective, and the interest is reinvested at 5% eﬀective. Find the (a) amount in the fund at the end of 10 years, (b) purchase price an investor(a buyer) must pay for a yield rate (to the seller) of 8% eﬀective. Solution. (a) The amount in the fund at the end of 10 years is 1000 ( 10 + 0.07 [s10 0.05 −10 0.05 ]) =1000 ( 10 + 0.07 [12.5779 −10 0.05 ]) =$13,609.06. (b) The purchase price is 13 ,609.06(1.08)−10 = $6,303.63 294 RATE OF RETURN OF AN INVESTMENT Example 32.4 Esther invests 100 at the end of each year for 12 years at an annual eﬀective interest rate of i.The interest payments are reinvested at an annual eﬀective rate of 5%. The accumulated value at the end of 12 years is 1748.40. Calculate i. Solution. The equation of value at time t= 12 is 1748.40 = 1200 + 100i [s12 0.05 −12 0.05 ] . Solving this equation for i we ﬁnd i= [1748.40 −1200](0.05) 100[s12 0.05 −12] = 7% If the payments of 1 are made at the beginning of each period (instead of at the end), the accumu- lated value at the end of the n periods is AV = n+ i(Is)nj = n+ i [sn+1 j −(n+ 1) j ] . Example 32.5 David pays 1,000 at the beginning of each year into a fund which earns 6%. Any interest earned is reinvested at 8%. Calculate the total that David will have at the end of 7 years. Solution. Note that payment are made at the beginning of the year (annuity-due). The answer is AV = 7,000 + 1000(0.06) [ 1.088−1 0.08 −8 0.08 ] = 8,977.47 An important consideration to a lender (investor) is the rate of repayment by the borrower. A faster rate of repayment by the borrower, results in a higher yield rate for the investor. This is illustrated in the following example. Example 32.6 Three loan repayment plans are described for a $3,000 loan over a 6-year period with an eﬀective rate of interest of 7.5%. If the repayments to the lender can be reinvested at an eﬀective rate of 32 INTEREST REINVESTED AT A DIFFERENT RATE 295 6%, ﬁnd the yield rates (for the lender) (a) if the entire loan plus accumulated interest is paid in one lump sum at the end of 6 years; (b) if interest is paid each year as accrued and the principal is repaid at the end of 6 years; (c) if the loan is repaid by level payments over the 6-year period. Solution. (a) The lump sum is 3000(1.075)6 = $4629.90.Since there is no repayment to reinvest during the 6- year period, the yield rate is obviously 7.5%. This answer can be also found by solving the equation 3000(1 + i)6 = 4629.90. (b) At the end of each year during the 5-year period, the payment is 3000(0.075) = $225. The accumulated value of all payments at the end of the 6-year period is 3000 + 225 s6 0.06 = 3000+225(6.9753) = $4569.44.To ﬁnd the yield rate, we solve the equation 3000(1+i)6 = 4569.44 → i= 0.07265 = 7.265% (c) Each year during the 6-year period, the payment isRwhere 3000 = Ra6 0.075 →R= 639.13.The accumulated value of all payments at the end of the 6-year period is 639.13s6 0.06 = 639.13(6.9753) = $4458.12. To ﬁnd the yield rate, we solve the equation 3000(1 + i)6 = 4458.12 →i = 0.06825 = 6.825% 296 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 32.1 Vanessa invests $5,000 at the end of each of the next 15 years and her investment earns interest at an annual eﬀective rate of 10%. The interest that she receives at the end of each year is reinvested and earns interest at an annual eﬀective rate of 5%. Calculate the accumulated value at time 15 years. Problem 32.2 Lauren has 100,000 invested in a fund earning 5%. Each year the interest is paid to Lauren who is able to invest the interest at 8%. Calculate the amount that Lauren will have at the end of 10 years. Problem 32.3 Julie has a sum of money, S, invested in a fund which earns 10%. Each year the fund pays the interest earned to Julie. Julie can only reinvest the interest at an annual eﬀective interest rate of 6%. After 20 years, Julie has 100,000 total including the amount in the fund plus the reinvested interest. Calculate S. Problem 32.4 Megan invests 500 at the end of each year. The investment earns 8% per year which is paid to Megan who reinvests the interest at 6%. Calculate how much Megan will have after 5 years. Problem 32.5 Thomas invests X into Fund 1 at the beginning of each year for 10 years. Fund 1 pays interest annually into Fund 2. Fund 1 earns 7% annually while Fund 2 earns 6% annually. After 10 years, Thomas has a total of 50,000. Calculate X. Problem 32.6 Chris is investing 1,000 at the beginning of each year into Fund A. Fund A earns interest at a nominal interest rate of 12% compounded monthly. Fund A pays Chris interest monthly. Chris reinvests that interest in Fund B earning an annual eﬀective rate of 8%. Calculate the total amount in Fund A and Fund B after 10 years. Problem 32.7 John invests 100 at the end of year one, 200 at the end of year two, etc until he invests 1,000 at the end of year ten. The investment goes into a bank account earning 4%. At the end of each year, the interest is paid into a second bank account earning 3%. Calculate the total amount John will have after 10 years. 32 INTEREST REINVESTED AT A DIFFERENT RATE 297 Problem 32.8 Kathy pays 1,000 at the end of each year into Fund A which earns interest at an annual eﬀective interest rate of i.At the end of each year, the interest earned is transferred to Fund B earning 10% interest. After 10 years. Kathy has 15,947.52. Calculate i. Problem 32.9 It is desired to accumulate a fund of 1,000 at the end of 10 years by equal deposits at the beginning of each year. If the deposits earn interest at 8% eﬀective, but the interest can only be reinvested at only 4% eﬀective, show that the deposit necessary is 1000 2s11 0.04 −12. Problem 32.10 A loan of 10,000 is being repaid with payments of 1,000 at the end of each year for 20 years. If each payment is immediately reinvested at 5% eﬀective, ﬁnd the yield rate of this investment. Problem 32.11 An investor purchases a 5-year ﬁnancial instrument having the following features: (i) The investor receives payments of 1,000 at the end of each year for 5 years. (ii) These payments earn interest at an eﬀective rate of 4% per annum. At the end of the year, this interest is reinvested at the eﬀective rate of 3% per annum. Find the purchase price to the investor to produce a yield rate of 4%. Problem 32.12 An investor deposits 1,000 at the beginning of each year for ﬁve years in a fund earning 5% eﬀective. The interest from this fund can be reinvested at only 4% eﬀective. Show that the total accumulated value at the end of ten years is 1250(s11 0.04 −s6 0.04 −1) Problem 32.13 A invests 2,000 at an eﬀective interest rate of 17% for 10 years. Interest is payable annually and is reinvested at an eﬀective rate of 11%. At the end of 10 years the accumulated interest is 5,685.48. B invests 150 at the end of each year for 20 years at an eﬀective interest rate of 14%. Interest is payable annually and is reinvested at an eﬀective rate of 11%. Find B’s accumulated interest at the end of 20 years. Problem 32.14 ‡ Victor invests 300 into a bank account at the beginning of each year for 20 years. The account pays out interest at the end of every year at an annual eﬀective interest rate of i%. The interest is reinvested at an annual eﬀective rate of (i 2 ) %. The yield rate on the entire investment over the 20 year period is 8% annual eﬀective. Determine i. 298 RATE OF RETURN OF AN INVESTMENT Problem 32.15 ‡ Sally lends 10,000 to Tim. Tim agrees to pay back the loan over 5 years with monthly payments payable at the end of each month. Sally can reinvest the monthly payments from Tim in a savings account paying interest at 6%, compounded monthly. The yield rate earned on Sally’s investment over the ﬁve-year period turned out to be 7.45%, compounded semi-annually. What nominal rate of interest, compounded monthly, did Sally charge Tim on the loan? Problem 32.16 Paul invests 1,000 at the beginning of each year for 10 years into a Fund A earning Y% each year. At the end of the year, any interest earned is moved to Fund B which earns 5%. Combining both Funds, Paul has a total of 11,924.08 at the end of 10 years. Determine Y. Problem 32.17 Lisa invests 1200 at the beginning of each year for 8 years into an account earning 8%. The interest earned each year is transferred to an account earning 6%. At the end of 8 years, the total amount is paid out. Calculate the amount an investor would pay now for the ﬁnal payout if the investor wanted a return of 10%. Problem 32.18 Amy invests 1000 at an eﬀective annual rate of 14% for 10 years. Interest is payable annually and is reinvested at an annual eﬀective rate of i. At the end of 10 years the accumulated interest is 2341.08. Bob invests 150 at the end of each year for 20 years at an annual eﬀective rate of 15%. Interest is payable annually and is reinvested at an annual eﬀective rate of i. Find Bob’s accumulated interest at the end of 20 years. Problem 32.19 ‡ An investor pays $100,000 today for a 4-year investment that returns cash ﬂows of $60,000 at the end of each of years 3 and 4. The cash ﬂows can be reinvested at 4.0% per annum eﬀective. If the rate of interest at which the investment is to be valued is 5.0%, what is the net present value of this investment today? Problem 32.20 ‡ An investor wishes to accumulate 10,000 at the end of 10 years by making level deposits at the beginning of each year. The deposits earn a 12% annual eﬀective rate of interest paid at the end of each year. The interest is immediately reinvested at an annual eﬀective interest rate of 8%. Calculate the level deposit. 32 INTEREST REINVESTED AT A DIFFERENT RATE 299 Problem 32.21 ‡ Payments of X are made at the beginning of each year for 20 years. These payments earn interest at the end of each year at an annual eﬀective rate of 8%. The interest is immediately reinvested at an annual eﬀective rate of 6%. At the end of 20 years, the accumulated value of the 20 payments and the reinvested interest is 5600. Calculate X. Problem 32.22 A deposit of 1 is made at the end of each year for 30 years into a bank account that pays interest at the end of each year at j per annum. Each interest payment is reinvested to earn an annual eﬀective interest rate of j 2 .The accumulated value of these interest payments at the end of 30 years is 72.88. Determine j. Problem 32.23 ‡ Susan invests Z at the end of each year for seven years at an annual eﬀective interest rate of 5%. The interest credited at the end of each year is reinvested at an annual eﬀective rate of 6%. The accumulated value at the end of seven years is X. Lori invests Z at the end of each year for 14 years at an annual eﬀective interest rate of 2.5%. The interest credited at the end of each year is reinvested at an annual eﬀective rate of 3%. The accumulated value at the end of 14 years is Y. Calculate Y/X. Problem 32.24 ‡ Jason deposits 3960 into a bank account at t= 0.The bank credits interest at the end of each year at a force of interest δt = 1 8 + t. Interest can be reinvested at an annual eﬀective rate of 7%. The total accumulated amount at time t= 3 is equal to X. Calculate X. Problem 32.25 ‡ Eric deposits 12 into a fund at time 0 and an additional 12 into the same fund at time 10. The fund credits interest at an annual eﬀective rate of i. Interest is payable annually and reinvested at an annual eﬀective rate of 0 .75i. At time 20 the accumulated amount of the reinvested interest payments is equal to 64. Calculate i,i> 0. Problem 32.26 ‡ At time t = 0, Sebastian invests 2000 in a fund earning 8% convertible quarterly, but payable an- nually. He reinvests each interest payment in individual separate funds each earning 9% convertible quarterly, but payable annually. The interest payments from the separate funds are accumulated in 300 RATE OF RETURN OF AN INVESTMENT a side fund that guarantees an annual eﬀective rate of 7%. Determine the total value of all funds at t= 10. Problem 32.27 A loan of 1000 is being repaid in 10 years by semiannual installments of 50, plus interest on the unpaid balance at 4% per annum compounded semiannually. The installments and interest payments are reinvested at 5% per annum compounded semiannually. Calculate the annual eﬀective yield rate of the loan. Problem 32.28 John invests a total of 10,000. He purchases an annuity with payments of 1000 at the beginning of each year for 10 years at an eﬀective annual interest rate of 8%. As annuity payments are received, they are reinvested at an eﬀective annual interest rate of 7%. The balance of the 10,000 is invested in a 10-year certiﬁcate of deposit with a nominal annual interest rate of 9%, compounded quarterly. Calculate the annual eﬀective yield rate on the entire 10,000 investment over the 10 year period. Problem 32.29 Brown deposits $1000 at the beginning of each year for 5 years, but makes no additional deposits in years 6 through 10. The fund pays interest annually at an annual eﬀective interest rate of 5%. Interest is reinvested at only a 4% annual eﬀective interest rate. What is Brown’s total accumulated value at the end of 10 years? Problem 32.30 You have $20,000 to invest. Alternative A allows you to invest your money at 8% compounded annually for 12 years. Alternative B pays you simple interest at the end of each year of 6%. At what rate must you reinvest the interest in alternative B to have the same accumulation after 12 years? Problem 32.31 Henry invests 2000 at the beginning of the year in a fund which credits interest at an annual eﬀective rate of 9%. Henry reinvests each interest payment in a separate fund, accumulating at an annual eﬀective rate of 8%. The interest payments from this fund accumulate in a bank account that guarantees an annual eﬀective rate of 7%. Determine the sum of the principal and interest at the end of 10 years. Problem 32.32 Jim borrowed 10,000 from Bank X at an annual eﬀective rate of 8%. He agreed to repay the bank with ﬁve level annual installments at the end of each year. At the same time, he also borrowed 32 INTEREST REINVESTED AT A DIFFERENT RATE 301 15,000 from Bank Y at an annual eﬀective rate of 7.5%. He agreed to repay this loan with ﬁve level annual installments at the end of each year. He lent the 25,000 to Wayne immediately in exchange for four annual level repayments at the end of each year, at an annual eﬀective rate of 8.5%. Jim can only reinvest the proceeds at an annual eﬀective rate of 6%. Immediately after repaying the loans to the banks in full, determine how much Jim has left. 302 RATE OF RETURN OF AN INVESTMENT 33 Interest Measurement of a Fund: Dollar-Weighted Inter- est Rate An investment fund usually is a ﬁrm that invests the pooled funds of investors for a fee. In this section we examine a method of determining the interest rate earned by an investment fund. In practice, it is common for a fund to be incremented with new principal deposits, decremented with principal withdrawals, and incremented with interest earnings many times throughout a pe- riod. Since these occurrences are often at irregular intervals, we devise notation for the purpose of obtaining the eﬀective rate of interest i over one measurement period: A = the amount in the fund at the beginning of the period,i.e. t= 0. B = the amount in the fund at the end of the period, i.e. t= 1. I = the amount of interest earned during the period. ct = the net amount of principal contributed at time t (inﬂow less outﬂow at time t) where 0 ≤t≤1. C = ∑ tct = total net amount of principal contributed during the period (if negative, this is an indication of net withdrawal). The amount at the end of a period is equal to the amount at the beginning of the period plus the net principal contributions plus the interest earned: B = A+ C+ I. (33.1) To be consistent with the deﬁnition of the eﬀective rate we will assume that all the interest earned I is received at the end of the period. Assuming compound interest i throughout the period, the exact equation of value for the interest earned over the period 0 ≤t≤1 is I = iA+ ∑ 0≤t≤1 ct[(1 + i)1−t −1]. (33.2) That is, the amount of interest earned is the sum of amounts of interest earned by each individual contribution, plus the amount of interest earned on the balance at the beginning. Note that (1 + i)1−t −1 is the eﬀective rate for period from t to 1,i.e., A(1) −A(t) A(t) = A(1) A(t) −1 = (1 + i)1−t −1. Substituting equation(33.2) into equation (33.1) we ﬁnd the equation of value B = A(1 + i) + ∑ 0≤t≤1 ct(1 + i)1−t. (33.3) 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE303 The interest isatisfying equation (33.3) is called the dollar-weighted rate of interest. Note that the required return of the investment, cost of capital and dollar weighted rate of return stand for the same quantity. Example 33.1 At the beginning of a year, an investment fund was established with an initial deposit of $3,000. At the end of six months, a new deposit of $1,500 was made. Withdrawals of $500 and $800 were made at the end of four months and eight months respectively. The amount in the fund at the end of the year is $3,876. Set up the equation of value to calculate the dollar-weighted rate of interest. Solution. The equation of value is 3,000(1 + i) + 1,500(1 + i)0.5 −500(1 + i)(1−4 12 ) −800(1 + i)(1−8 12 ) = 3,876 Finding ifrom the equation (33.2) requires approximation methods and this is not an easy problem. However, in practice one uses a simple interest approximation (1 + i)1−t −1 = 1 + (1−t)i−1 = (1 −t)i obtaining I ≈iA+ ∑ 0≤t≤1 ct(1 −t)i which leads to the approximation i≈ I A+ ∑ 0≤t≤1 ct(1 −t). (33.4) It is shown that this approximation is very good as long as the ct’s are small in relation to Awhich is often the case in practice. However, if the c′ ts are not small in relation to A, then the error can become signiﬁcant. The denominator in the above approximation is commonly called the exposure associated with i. Alternatively, by using equation (33.3) one ﬁnds i by solving the equation B = A(1 + i) + ∑ 0≤t≤1 ct[1 + (1−t)i]. Example 33.2 Using Example 33.1, ﬁnd the approximate eﬀective rate of interest earned by the fund during the year using the dollar-weighted rate of return formula. 304 RATE OF RETURN OF AN INVESTMENT Solution. The interest earned I is computed to be I = 3876 + 500 + 800−(3000 + 1500) = 676. The exposure associated with i is 3000 + (−500) ( 1 −1 3 ) + (1500) ( 1 −1 2 ) + (−800) ( 1 −2 3 ) = 3150. Thus, the approximate eﬀective rate is 676 3150 = 0.2146 = 21.46%. Alternatively, i can be found by solving the equation 3,000(1 + i) + 1,500(1 + 0.5i) −500(1 + 2 3i) −800(1 + i 3) = 3,876 giving i= 21.46% Formula (33.4) is in a form which can be directly calculated. However, it is tedious to use it because of the summation term in the denominator. Therefore, estimating the summation is fa- vored. One way for doing that is to assume that the net principal contributions C occur at time k that can be approximated by the arithmetic weighted average k= 1 C ∑ 0≤t≤1 t·ct obtaining a simpler approximation i≈ I A+ (1 −k)C = I A+ (1 −k)(B−A−I) = I kA+ (1 −k)B−(1 −k)I. Example 33.3 A fund has 100,000 on January 1 and 125,000 on December 31. Interest earned by the fund during the year totaled 10,000. Calculate the net yield earned by the fund assuming that the net contributions occurred on April 1. 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE305 Solution. Using the formula, i≈ I [kA+(1−k)B−(1−k)I] , we have i≈ 10,000 [(1/4)100,000 + (3/4)125,000 −(3/4)10,000] = 10,000 111,250 = 8.98876404% One useful special case is when the principal deposits or withdrawals occur uniformly throughout the period. Thus, on average, we might assume that net principal contributions occur at timek= 1 2 , in which case we have i≈ I 0.5A+ 0.5B−0.5I = 2I A+ B−I. (33.5) This formula has been used for many years by insurance company regulators in both Canada and the US to allow them to get a feel for the investment returns a particular company is obtaining. Example 33.4 You have a mutual fund. Its balance was $10,000 on 31/12/1998. You made monthly contributions of $100 at the start of each month and the ﬁnal balance was $12,000 at 31/12/1999. What was you approximate return during the year? Solution. We are given that A = $10,000, B= $12,000, and C = $1,200. Thus, I = B−(A+ C) = $800 and i= 2(800) 10,000 + 12,000 −800 = 7.54717% Example 33.5 Find the eﬀective rate of interest earned during a calendar year by an insurance company with the following data: Assets, beginning of year 10,000,000 Premium income 1,000,000 Gross investment income 530,000 Policy beneﬁts 420,000 Investment expenses 20,000 Other expenses 180,000 306 RATE OF RETURN OF AN INVESTMENT Solution. We are given A=10,000,000 B =10,000,000 + 1,000,000 + 530,000 −(420,000 + 20,000 + 180,000) =10,910,000 I =530,000 −20,000 = 510,000 Thus, i= 2I A+ B−I = 2(510,000) 10,000,000 + 10,910,000 −510,000 = 5% 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE307 Practice Problems Problem 33.1 A fund has 10,000 at the start of the year. During the year $5,000 is added to the fund and $2,000 is removed. The interest earned during the year is $1,000. Which of the following are true? (I) The amount in the fund at the end of the year is $14,000. (II) If we assume that any deposits and withdrawals occur uniformly throughout the year, i is approximately 8.33%. (III) If the deposit was made on April 1 and the withdrawal was made on August 1, then i is approximately 7.74%. Problem 33.2 The funds of an insurance company at the beginning of the year were $500,000 and at the end of the year $680,000. Gross interest earned was $60,000, against which there were investment expenses of $5,000. Find the net eﬀective rate of interest yielded by the fund. Problem 33.3 A fund earning 4% eﬀective has a balance of $1,000 at the beginning of the year. If $200 is added to the fund at the end of three months and if $300 is withdrawn from the fund at the end of nine months, ﬁnd the ending balance under the assumption that simple interest approximation is used. Problem 33.4 At the beginning of the year an investment fund was established with an initial deposit of $1,000. A new deposit of $500 was made at the end of four months. Withdrawals of $200 and $100 were made at the end of six months and eight months, respectively. The amount in the fund at the end of the year is $1,272. Find the approximate eﬀective rate of interest earned by the fund during the year using the dollar-weighted rate of return formula. Problem 33.5 ‡ An association had a fund balance of 75 on January 1 and 60 on December 31. At the end of every month during the year, the association deposited 10 from membership fees. There were withdrawals of 5 on February 28, 25 on June 30, 80 on October 15, and 35 on October 31. Calculate the dollar-weighted rate of return for the year. Problem 33.6 ‡ An insurance company earned a simple rate of interest of 8% over the last calendar year based on the following information: 308 RATE OF RETURN OF AN INVESTMENT Assets, beginning of year 25,000,000 Sales revenue X Net investment income 2,000,000 Salaries paid 2,200,000 Other expenses paid 750,000 All cash ﬂows occur at the middle of the year. Calculate the eﬀective yield rate. Problem 33.7 ‡ At the beginning of the year, an investment fund was established with an initial deposit of 1,000. A new deposit of 1,000 was made at the end of 4 months. Withdrawals of 200 and 500 were made at the end of 6 months and 8 months, respectively. The amount in the fund at the end of the year is 1,560. Calculate the dollar-weighted yield rate earned by the fund during the year. Problem 33.8 On January 1, an investment account is worth $500,000. On April 1, the account value has increased to $530,000, and $120,000 of new principal is deposited. On August 1, the account value has decreased to $575,000, and $250,000 is withdrawn. On January 1 of the following year, the account value is $400,000. Compute the yield rate using the dollar-weighted method. Problem 33.9 A fund has 10,000 at the beginning of the year and 12,000 at the end of the year. Net contributions of 1,000 were made into the fund during year. Calculate the net yield earned by the fund assuming that the net contributions were contributed uniformly throughout the year. Problem 33.10 A fund has 100,000 on January 1 and 125,000 on December 31. Interest earned by the fund during the year totaled 10,000. The net yield earned by the fund during the year was 9.6385%. Two contributions were made to the fund during the year and there were no withdrawals. The contributions were for equal amounts made two months apart. Determine the date of the ﬁrst contribution. Problem 33.11 An investor fund has a balance on January 1 of $273,000 and a balance on December 31 of $372,000. The amount of interest earned during the year was $18,000 and the computed yield rate on the fund was 6%. What was the average date for contributions to and withdrawals from the fund? 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE309 Problem 33.12 You deposited 5,000 into a fund on January 1, 2002. On March 1,2002, you withdrew 1,000 from the fund. On August 1, 2002, you deposited 700 into the fund. On October 1, 2002, you deposited another 750 into the fund. If your annual dollar-weighted rate of return on your investment was 12.00%, what was your fund worth on December 31, 2002? Problem 33.13 At the beginning of the year a fund has deposits of 10,000. A deposit of 1000 is made after 3 months and a withdrawal of 2000 is made after 9 months. The amount in the fund at the end of the year is 9500. Find the dollar weighted rate of interest (a) using formula (33.3) (b) using formula (33.4) (c) using formula (33.5). Problem 33.14 Deposits of 10000 are made into an investment fund at time zero and 1. The fund balance at time 2 is 21000. Compute the eﬀective yield using the dollar weighted method. Problem 33.15 An investor puts $100 into a mutual fund in the ﬁrst year and $50 in the second year. At the end of the ﬁrst year the mutual pays a dividend of $10. The investor sells the holdings in the mutual fund at the end of the second year for $180. Find the dollar weighted return. Problem 33.16 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the exact dollar weighted rate of return( i.e. using (33.3)). Problem 33.17 A fund has $2,000,000 at the beginning of the year. On June 1 and September 1, two withdrawals of $100,000 each were made. On November 1, a contribution of $20,000 was made. The value in the fund at the end of the year is $1,900,000. Find (a) the exact dollar weighted rate of return; (b) the dollar weigthed rate of return using formula (33.4). Problem 33.18 The following table shows the history of account balances, deposits, and withdrawals for an account 310 RATE OF RETURN OF AN INVESTMENT during 2008: Date Value before dep/withd Dep/Withd January 1,2008 10,000 0 March 1,2008 9,500 3,000 June 1,2008 14,000 −2,000 September 1,2008 12,000 −2,000 January 1,2009 12,000 What is this account’s dollar-weighted rate of return?(Assume 30-day months.) Problem 33.19 On January 1, 2000, the balance in account is $25200. On April 1, 2000, $500 are deposited in this account and on July 1, 2001, a withdraw of $1000 is made. The balance in the account on October 1, 2001 is $25900. What is the annual rate of interest in this account according with the dollarweighted method? Problem 33.20 You invest $10,000 in a fund on January 1, 2008. On May 1, 2008, you withdraw $3,000 from the fund. On November 1, 2008, you deposit $4,000 into the fund. On December 31, 2008, your fund is worth $13,500. What was the annual dollar-weighted rate of return on your investment? Problem 33.21 You invest $5,000 in a fund on 1/1/07. On 3/1/07, you withdraw X from the fund. On 7/1/07, you deposit $2,500 into the fund. On 9/1/07, you withdraw $1,000 from the fund. On 12/31/07, your fund is worth $5,700. The annual dollar −weighted rate of return on your investment was 12.0%. Find X. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST311 34 Interest Measurement of a Fund: Time-Weighted Rate of Interest The dollar-weighted rate of interest depends on the precise timing and amount of the cash ﬂows. In practice, professional fund managers who direct investment funds have no control over the timing or amounts of the external cash ﬂows. Therefore, if we are comparing the performance of diﬀerent fund managers, the dollar weighted rate of interest doesn’t always provide a fair comparison. In this section, we consider an alternative measure of performance that does not depend on the size or the timing of the cash ﬂows, namely the time-weighted rate of interest also known as the time-weighted rate of return. Consider the following one-year investment. An amount X is invested in a fund at the beginning of the year. In six months, the fund is worth X 2 at which time the investor can decide to add to the fund, to withdraw from the fund, or to do nothing. At the end of the year, the fund balance is double the balance at six months. We next examine three situations related to the above investment: Consider three investors A,B, and C. Investor A initially invests X = $1,000. At the end of six months his investment is worth $500. He decideds not to withdraw from the fund or deposits into the fund. His fund at the end of the year worths $1,000. Since the interest earned is I = 0, the dollar-weighted rate of interest is i= 0. Next, investor B invests initially X = $1 ,000. Again, after six months the balance is $500. At the end of six months the investor withdraws half the fund balance (i.e. $250). His balance in the fund at the end of the year is $500. The equation of value at t = 1 for this transaction is 1000(1 +i) −250(1 +i) 1 2 = 500.Solving this quadratic equation we ﬁnd (1 + i) 1 2 = 0.84307 and this implies a yield rate of i= −0.2892 or −28.92%. Finally, investor C invests initially X = $1 ,000 and at six months deposits an amount equal to the fund balance(i.e. $500). Hence, his balance at the end of the year is $2,000. The equation of value at time t = 1 is 1000(1 + i) + 500(1 + i) 1 2 = 2000. Solving this quadratic equation we ﬁnd (1 + i) 1 2 = 1.18614. Hence, the yield rate in this case is i= 0.4069 or 40.69%. The yield rate for investor C is so much better than the yield rate for either investor A or investor B, because of the decisions made by each investor. The dollar-weighted rate of interest measures both the behavior of the fund and the skills of the investor. If the investor did not perform any transactions, then the dollar-weighted rate of interest would be close to 0%. Now how can we evaluate the decisions made by the fund manager? In the illustration above, we ﬁnd the yield rate for the ﬁrst six months to be j1 = −50% and for the second six months to be j2 = 100%. If i is the yield rate for the entire year then 1 + i= (1 + j1)(1 + j2) = 1 312 RATE OF RETURN OF AN INVESTMENT which implies that i = 0 and this is regardless of when cash is deposited or withdrawn. This indicates that the manager did a poor job of maintaining the fund. Yield rates computed by considering only changes in interest rate over time (which is what was previously done to evaluate the fund manager’s performance) are called time-weighted rates of interest. We can generalize the above approach as follows. Suppose m−1 deposits or withdrawals are made during a year at times t1,t2,··· ,tm−1 (so that no contributions at t0 = 0 and tm = 1). Thus, the year is divided into m subintervals. For k = 1,2,··· ,m we let jk be the yield rate over the kth subinterval. For k = 1,··· ,m −1, let Ctk be the net contribution at exact time tk and Btk the value of the fund before the contribution at time tk. Note that C0 = Cm = 0 and B0 is the initial investment and B1 is the value of the fund at the end of the year. The yield rate of the fund from time tk−1 to time tk satisﬁes the equation of value Btk = (1 + jk)(Btk−1 + Ctk−1) or 1 + jk = Btk Btk−1 + Ctk−1 , k = 1,2,··· ,m. The overall yield rate i for the entire year is given by 1 + i= (1 + j1)(1 + j2) ···(1 + jm) or i= (1 + j1)(1 + j2) ···(1 + jm) −1. We call i the time-weighted rate of return. Example 34.1 Suppose that an investor makes a series of payments and withdrawals, as follows: Date Flow Balance before Balance after 01/01/2003 0 100,000 100,000 30/06/2003 +1,000,000 74,681 1,074,681 31/12/2003 0 1,474,081 1,474,081 (a) Compute the dollar-weighted rate of interest that the investor has realized. (b) Compute the time-weigted rate of interest. Solution. (a) We assume that the contribution of $1,000,000 occurred exactly in the middle of the year. We 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST313 have A = $100 ,000, B = $1 ,474,081, C = $1 ,000,000. Hence, I = 1 ,474,081 −1,000,000 − 100,000 = 374,081 and i= 2(374,081) 100,000 + 1,474,081 −374,081 = 62.35%. (b) We have i= (74,681 100,000 )(1,474,081 1,074,681 ) −1 = 2.44% It follows from the previous example that the time-weighted method does not provide a valid measure of yield rate as does the dollar weighted method. However, the time-weighted method does provide a better indicator of underlying investment performance than the dollar-weighted method. Example 34.2 Your balance at time 0 is $2,000. At time t= 1 3 ,the balance has grown to $2,500 and a contribution of $1,000 is made. At time t= 2 3 , the balance has dropped to $3,000 and a withdrawal of $1,500 is made. At the end of the year, the fund balance is $2,000. What is the time weighted rate of return? Solution. We create the following chart k tk Btk Ctk jk 0 0.0000 $2,000 $0 0% 1 1/3 $2,500 $1,000 25.00% 2 2/3 $3,000 -$1,500 -14.29% 3 1 $2,000 $0 33.33% The time weighted rate of return is i= (1.25)(.8571)(1.3333) −1 = .428464 Example 34.3 You are given the following information about an investment account: Date June 30,01 Sep. 30,01 Dec. 31,01 March 31,02 June 30,02 Bk 12,000 10,000 14,000 13,000 X Deposit 2,000 2,000 Withdrawal 2,000 2,000 If the eﬀective annual dollar-weighted rate of return from June 30, 2001 to June 30, 2002 was exactly 10%, what was the eﬀective annual time-weighted rate of return during that same period? 314 RATE OF RETURN OF AN INVESTMENT Solution. First, we ﬁnd X : X = 10,000(1.1) + 2,000(1.1)0.75 −2,000(1.1)0.5 + 2,000(1.1)0.25 = 13,098.81 The time-weighted rate of return is then i= 10,000 12,000 −2,000 · 14,000 10,000 + 2,000 · 13,000 14,000 −2,000 · 13,098.81 13,000 −2,000 −1 = 10.37% 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST315 Practice Problems Problem 34.1 Which of the following are true? (I) Time-weighted method provides better indicator for the fund underlying performance than the dollar-weighted method. (II) Time weighted rates of interest will always be higher than dollar weighted rates of interest. (III) Dollar weighted rate of interest provide better indicators of underlying investment performance than do time weighted rates of interest. (IV) Dollar weighted rates of interest provide a valid measure of the actual investment results. Problem 34.2 A fund has 1,000 at beginning of the year. Half way through the year, the fund value has increased to 1,200 and an additional 1,000 is invested. At the end of the year, the fund has a value of 2,100. (a) Calculate the exact dollar weighted rate of return using compound interest. (b) Calculate the estimated dollar weighted rate of return using the simple interest assumptions. (c) Calculate the time weighted rate of return. Problem 34.3 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the time weighted rate of return. Problem 34.4 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the exact dollar weighted rate of return. Problem 34.5 A fund has 10,000 at the start of the year. Six months later, the fund has a value of 15,000 at which time, Stuart removes 5,000 from the fund. At the end of the year, the fund has a value of 10,000. Calculate the exact dollar weighted rate of return less the time weighted rate of return. Problem 34.6 ‡ You are given the following information about an investment account: 316 RATE OF RETURN OF AN INVESTMENT Date Value Immediately Deposit Before Deposit January 1 10 July 1 12 X December 31 X Over the year, the time-weighted return is 0%, and the dollar-weighted (money-weighted) return is Y. Determine Y. Problem 34.7 ‡ An investor deposits 50 in an investment account on January 1 . The following summarizes the activity in the account during the year: Date Value Immediately Deposit Before Deposit March 15 40 20 June 1 80 80 October 1 175 75 On June 30, the value of the account is 157.50 . On December 31, the value of the account is X. Using the time-weighted method, the equivalent annual eﬀective yield during the ﬁrst 6 months is equal to the (time-weighted) annual eﬀective yield during the entire 1-year period. Calculate X. Problem 34.8 On January 1, an investment account is worth $100,000. On May 1, the account value has increased to $112,000, and $30,000 of new principal is deposited. On November 1, the account value has decreased to $125,000, and $42,000 is withdrawn. On January 1 of the following year, the account value is $100,000. Compute the yield rate using (a) the dollar-weighted method and (b) the time-weighted method. Problem 34.9 In Problem 34.8, change May 1 to June 1 and November 1 to October 1. (a) Would the yield rate change when computed by the dollar-weighted method? (b) Would the yield rate change when computed by the time-weighted method? Problem 34.10 In Problem 34.8, assume that everything is unchanged except that an additional $5,000 is withdrawn on July 1. (a) Would the yield rate change when computed by the dollar-weighted method? (b) Explain why the yield rate cannot be computed by the time-weighted method. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST317 Problem 34.11 A mutual fund account has the balance $100 on January 1. On July 1 (exactly in the middle of the year) the investor deposits $250 into the account. The balance of the mutual fund at the end of the year is $400. (a) Calculate the exact dollar-weighted annual rate of interest for the year. (b) Suppose that the balance of the fund immediately before the deposit on July 1 is $120. Calculate the time-weighted annual rate of interest for the year. Problem 34.12 You are given the following information about two funds: Fund X: January 1, 2005: Balance 50,000 May 1, 2005: Deposit 24,000; Pre-deposit Balance 50,000 November 1, 2005: Withdrawal 36,000; Pre-withdrawal Balance 77,310 December 31, 2005: Balance 43,100 Fund Y: January 1, 2005: Balance 100,000 July 1, 2005: Withdrawal 15,000; Pre-withdrawal Balance 105,000 December 31, 2005: Balance F. Fund Y’s time-weighted rate of return in 2005 is equal to Fund X’s dollar-weighted rate of return in 2005. Calculate F. Problem 34.13 You invested 10,000 in a fund on January 1, 2000. On April 1, 2000, your fund was worth 11,000, and you added 2,000 to it. On June 1, 2000, your fund was worth 12,000 and you withdrew 2,000 from the fund. On December 31, 2000, your fund was worth 9,500. What was the time-weighted rate of return on your investment from January 1, 2000 to December 31, 2000? Problem 34.14 Deposits of $1000 are made into an investment fund at time 0 and 1. The fund balance is $1200 at time 1 and $2200 at time 2. (a) Compute the annual eﬀective yield rate computed by a dollar-weighted method. (b) Compute the annual eﬀective yield rate which is equivalent to that produced by a time-weighted method. Problem 34.15 Let A be the fund balance on January 1, B the fund balance on June 30, and C the balance on December 31. 318 RATE OF RETURN OF AN INVESTMENT (a) If there are no deposits or withdrawals, show that yield rates computed by the dollar-weighted method and the time-weighted method are both equal. (b) If there was a single deposit of D immediately after the June 30 balance was calculated, ﬁnd expressions for the dollar-weighted and time-weighted yield rates. (c) If there was a single deposit of D immediately before the June 30 balance was calculated, ﬁnd expressions for the dollar-weighted and time-weighted yield rates. (d) Compare the dollar-weighted yield rates in (b) and (c). (e) Compare the time-weighted yield rates in (b) and (c). Problem 34.16 100 is deposited into an investment account on January 1, 1998. You are given the following information on investment activity that takes place during the year: April 19,1998 October 30,1998 Value immediately prior to deposit 95 105 deposit 2X X The amount in the account on January 1, 1999 is 115. During 1998, the dollar-weighted return is 0% and the time-weighted return is Y. Calculate Y. Problem 34.17 An investment manager’s portfolio begins the year with a value of 100,000. Eleven months through the year a withdrawal of 50,000 is made and the value of the portfolio after the withdrawal is 57,000. At the end of the year the value of the portfolio is 60,000. Find the time-weighted yield rate less the dollar-weighted yield rate. Problem 34.18 On January 1, 1997 an investment account is worth 100,000. On April 1, 1997 the value has increased to 103,000 and 8,000 is withdrawn. On January 1, 1999 the account is worth 103,992. Assuming a dollar weighted method for 1997 and a time weighted method for 1998 the annual eﬀective interest rate was equal to x for both 1997 and 1998. Calculate x. Problem 34.19 On January 1, 1999, Luciano deposits 90 into an investment account. On April 1, 1999, when the amount in Luciano’s account is equal to X, a withdrawal of W is made. No further deposits or withdrawals are made to Luciano’s account for the remainder of the year. On December 31, 1999, the amount in Luciano’s account is 85. The dollar weighted return over the 1 year period is 20%. The time weighted return over the 1 year period is 16%. Calculate X. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST319 Problem 34.20 You are given the following information about the activity in two diﬀerent investment accounts. Account K Activity Date Fund value before activity Deposit Withdrawal January 1,1999 100 July 1,1999 125 X October 1,1999 110 2X December 31,1999 125 Account L Actyivity Date Fund value before activity Deposit Withdrawal January 1,1999 100 July 1,1999 125 X December 31,1999 105.8 During 1999 the dollar weighted return for investment account K equals the time weighted return for investment account L, which equals i. Calculate i. Problem 34.21 The following table shows the history of account balances, deposits, and withdrawals for an account during a particular year: Date Value before dep/withd Dep/Withd January 1,X 1,000 0 May 1,X 1,050 1,100 September 1,X 2,250 −900 January 1,X+1 16,00 What is the relationship among this account’s dollar-weighted rate of return (DW), its time-weighted rate of return (TW), and its internal rate of return (IRR)?Assume 30-day months. 320 RATE OF RETURN OF AN INVESTMENT 35 Allocating Investment Income: Portfolio and Investment Year Methods Consider an investment fund that involves many investors with seperate accounts. An example would be a pension fund for retirement in which each plan participant has an account but with commingled pool of assets. Each investor owns shares in the fund. In this section we consider the issue of allocating investment income to the various accounts. Two common methods are in use: the portfolio method and the investment year method also known as the new money method. The Portfolio Method Under the portfolio method an average rate based on the earnings of the entire fund is computed and credited to each account. All new deposits will earn this same portfolio rate. There are disadvantages of this method during time of ﬂuctuating interest rates. For example, sup- pose that the market rates have been rising signiﬁcantly during the recent years while the average portfolio rate is being lowered. In this case, there is a higher tendency for less new deposits to the fund and more withdrawals from the fund. The investment year method is an attempt to address this problem and to allocate investment income in a more equitable manner. Investment Year Method (IYM) We will describe this method on an annual basis, although typically smaller time intervals would be used. Under the investment year method, an investors’s contribution will be credited during the year with the interest rate that was in eﬀect at the time of the contribution. This interest rate is often referred to as the new money rate. For example, a portion of the IYM chart for a fund might look something like the one shown in Table 35.1. Suppose we invested $1,000 on January 1, 1994 and $500 on January 1, 1995. Then our total accumulation on Jan. 1, 2003 would be (1.064)(1.068)(1.071)(1.069)(1.073)(1.051)(1.049)(1.048)(1.047)1000 = 1688.75 from our 1994 contribution, plus (1.069)(1.070)(1.070)(1.074)(1.050)(1.049)(1.048)(1.047)500 = 794.31 from our 1995 contribution, for a total of 1688.75 + 794.31 = 2483.06. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS321 Purchase Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 1994 6.4% 6.8% 7.1% 6.9% 7.3% 5.1% 4.9% 4.8% 4.7% 1995 6.9% 7.0% 7.0% 7.4% 5.0% 4.9% 4.8% 4.7% 1996 7.1% 7.3% 7.3% 5.5% 5.4% 4.8% 4.7% 1997 7.0% 7.4% 5.4% 5.2% 4.6% 4.7% 1998 7.2% 5.7% 5.5% 4.5% 4.4% 1999 5.8% 5.1% 4.3% 4.7% 2000 5.0% 4.1% 4.6% 2001 4.0% 4.5% 2002 4.1% Table 35.1 Notice that from the chart above, investments made between 1994-1996 all earned the same rate of return both in 2001 (4.8%) and 2002 (4.7%). This illustrates the principle that under the investment year method, funds on investment longer than a certain ﬁxed period (5 years in our example) are assumed to grow at the overall yield rate for the fund (the portfolio rate), regardless of when the funds were invested. To better indicate the switch from the investment year method to the portfolio method, the data from the chart above would typically be displayed by a two-dimensional table such as the one in Table 35.2 Ifyis the calendar year of deposit, and mis the number of years for which the investment year method is applicable, then the rate of interest credited for the tth year of investment is denoted as iy t for t= 1,2,··· ,m. For t>m, the portfolio method is applicable, and the portfolio rate of interest credited for calendar year y is denoted as iy. The investment year method is more complicated than the portfolio method, but it was deemed necessary to attract new deposits and to discourage withdrawals during periods of rising interest rates. We illustrate the use of the investment year method in the following example. Example 35.1 You are given the following table of interest rates (in percentages). A person deposits 1,000 on January 1, 1997. Find the accumulated amount on January 1, 2000 (a) under the investment year method; (b) under the portfolio method; (c) when the balance is withdrawn at the end of every year and is reinvested at the new money rate. 322 RATE OF RETURN OF AN INVESTMENT y iy 1 iy 2 iy 3 iy 4 iy 5 iy+5 Portfolio Year 1992 8.25 8.25 8.4 8.5 8.5 8.35 1997 1993 8.5 8.7 8.75 8.9 9.0 8.6 1998 1994 9.0 9.0 9.1 9.1 9.2 8.85 1999 1995 9.0 9.1 9.2 9.3 9.4 9.1 2000 1996 9.25 9.35 9.5 9.55 9.6 9.35 2001 1997 9.5 9.5 9.6 9.7 9.7 1998 10.0 10.0 9.9 9.8 1999 10.0 9.8 9.7 2000 9.5 9.5 2001 9.0 Table 35.2 Solution. (a) The sequence of interest rates beginning from a given year of investment runs horizontally through the row for that year and then down the last column of rates. An investment made at the beginning of 1992 would earn the investment year rate of 8.25% in 1992, the money rate of 8.25% in 1993, 8.4% in 1994, 8.5% in 1995 and 1996. Starting in 1997, an investment made at the beginning of 1992 would earn the portfolio average rates. The 1992 investment would earn the portfolio rate of 8.35% in 1997, the portfolio rate of 8.6% in 1998, and so on. An investment made at the beginning of 2001 would earn the investment year rate of 9% in 2001- 2005. Starting the year 2006, the investment made in 2001 would earn the portfolio rate of 8.35% in 2006, the portfolio rate of 8.6% in 2007, and so on. (a) The accumulated value is 1000 ·(1.095)(1.095)(1.096) = 1,314.13. (b) The accumulated value is Q= 1000 ·(1.0835)(1.086)(1.0885) = 1,280.82. (c) The accumulated value is 1000 ·(1.095)(1.1)(1.1) = 1,324.95 Example 35.2 An investment of 1000 is made on January 1996 in an investment fund crediting interest according to Table 35.2. How much interest is credited from January 1, 1999 to January 1, 2001? Solution. The accumulated value on January 1, 1999 is 1000(1.0925)(1.0935)(1.095) = 1308.14. The accumulated value on January 1, 2001 is 1000(1.0925)(1.0935)(1.095)(1.0955)(1.096) = 1570.64. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS323 Thus, the total amount of interest credited is 1570.64 −1308.14 = 262.50 Example 35.3 Using Table 35.2, ﬁnd (a) the interest rates credited in calendar year 2000 for deposits made in 1994 - 2000; (b) the new money rates credited in the ﬁrst year of investment for deposits made in 1992 - 2001. Solution. (a) 9.5% credited for new deposits made in 2000, 9.8% for deposits made in 1999, 9.9% for deposits made in 1998, 9.7% for deposits made in 1997, 9.6% for deposits made in 1996, 9.1% for deposits made in 1994 - 1995. Thus, the interest credited in the calendar year 2000 appear on an upwardly diagonal within the table. (b) The new money rates credited in the ﬁrst year of investment appear in the ﬁrst column of the table 324 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 35.1 Using Table 35.2, what were the interest rates credited in calendar year 2001 for deposits made in 2001 and before 2001? Problem 35.2 If 100 is invested in the fund at the beginning of 2002, ﬁnd the accumulated amount at the end of 2003 using: (a) The investment year method (b) The portfolio method (c) The investment year method if the amount is withdrawn and then reinvested at the end of 2002. y iy 1 iy 2 iy 3 iy 4 iy+4 Portfolio Year 1998 7.0 6.5 6.0 5.8 5.9 2002 1999 6.4 6.1 5.8 5.9 6.0 2003 2000 6.2 6.0 5.9 6.0 6.2 2004 2001 6.1 5.9 6.1 6.4 6.6 2005 2002 6.0 6.1 6.3 6.6 2003 6.4 6.6 6.7 2004 6.8 7.0 2005 7.5 Problem 35.3 The following table lists the interest rate credited under an investment year method of crediting interest. y iy 1 iy 2 iy 3 iy 4 iy 5 iy+5 1999 7.0 6.75 6.5 6.25 6.0 5.5 2000 6.0 5.5 5.25 5.1 5.0 2001 5.0 4.8 4.6 4.3 2002 4.0 3.75 3.5 2003 3.0 3.2 2004 4.0 Becky invests $1,000 on January 1, 1999 and an additional $500 on January 1, 2003. How much money does Becky have on December 31, 2004? Use the following chart of interest rates to answer Problems 35.4 through 35.6. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS325 y iy 1 iy 2 iy 3 iy 4 iy+4 Calendar year of portfolio rate y+ 4 1997 0.0650 0.0625 0.0600 0.0575 0.0550 2001 1998 0.0600 0.0575 0.0550 0.0525 0.0500 2002 1999 0.0500 0.0475 0.0460 0.0450 0.0450 2003 2000 0.0450 0.0440 0.0430 0.0420 0.0410 2004 2001 0.0400 0.0390 0.0380 0.0370 2002 0.0300 0.0300 0.0325 2003 0.0300 0.0325 2004 0.0300 Problem 35.4 Jordan invests 1,000 in a fund on January 1, 1998. The fund uses the investment year method of determining interest rates. Calculate the amount that Jordan will have at the end of 2004. Problem 35.5 Jenna invests 1,000 in a fund on January 1, 2002. The fund uses the portfolio method of determining interest rates. Calculate the amount that Jenna will have at the end of 2004. Problem 35.6 James invests 1,000 into a fund at the beginning of each year from 2002 to 2004. The fund pays interest using the investment year interest rate. Calculate the amount of money that James will have at the end of 2004. Use the following chart of interest rates to answer Questions 35.7 through 35.10. y iy 1 iy 2 iy 3 iy+3 1995 3.7% 3.6% 3.5% 6.0% 1996 3.2% 3.1% 3.0% 5.5% 1997 2.7% 2.6% 2.5% 5.0% 1998 2.2% 2.1% 2.0% 4.5% 1999 1.7% 1.6% 1.5% 4.0% Problem 35.7 A deposit of $100 is made at the beginning of 1997. How much interest was credited during 1998? What is the accumulated value at the end of 2002? 326 RATE OF RETURN OF AN INVESTMENT Problem 35.8 How much interest is credited in the calendar years 1997 through 1999 inclusive to a deposit of $100 made at the beginning of 1995? Problem 35.9 What were the interest rates credited in calendar year 1999 for deposits made in 1999, 1998, 1997, 1996, and 1995? Problem 35.10 What were the new money rates credited in the ﬁrst year of investment for deposits made in 1995, 1996, 1997, and so on? Problem 35.11 The following table shows the annual eﬀective interest rates credited to an investment fund by calendar year of investment. The investment year method applies for the ﬁrst two years, after which a portfolio method is used. y iy 1 iy 2 iy+2 1995 t 5.5% 4.5% 1996 6.0% 6.1% 5.0% 1997 7.0% t+ 2.5% 5.5% An investment of $100 is made at the beginning of 1995 and 1997. The total amount of interest credited by the fund during the year 1998 is $13.81. Find t. Problem 35.12 It is known that 1 + iy t = (1.08 + 0.005t)1+0.01y for integral t, 0 ≤t≤5,and integral y, 0 ≤y≤10. If $1,000 is invested for three years beginning in year y = 5, ﬁnd the equivalent eﬀective rate of interest. Problem 35.13 For a certain portfolio, iy t = (1.06 + 0.004t)1+0.01y for 1 ≤t≤5 and 0 ≤y≤10. (a) What is the accumulated value at the end of year 4 of a single investment of 1,000 at the beginning of year 2? (b) What is the accumulated value at the end of year 4 of a deposit of 1,000 at the beginning of each of years 2 through 4? Problem 35.14 ‡ The following table shows the annual eﬀective interest rates being credited by an investment account, by calendar year of investment. The investment year method is applicable for the ﬁrst 3 years, after which a portfolio rate is used. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS327 y iy 1 iy 2 iy 3 iy+3 1990 10% 10% t% 8% 1991 12% 5% 10% (t−1)% 1992 8% ( t−2)% 12% 6% 1993 9% 11% 6% 9% 1994 7% 7% 10% 10% An investment of 100 is made at the beginning of years 1990, 1991, and 1992. The total amount of interest credited by the fund during the year 1993 is equal to 28.40. Calculate t. Problem 35.15 Find ¨s5 from the data in Table 35.2 assuming the ﬁrst payment is made in calendar year 1995. Problem 35.16 You are given the following table of interest rates: y iy 1 iy 2 iy+5 Portfolio Year 2004 9.00% 10.00% 11.00% 2006 2005 7.00% 8.00% 2006 5.00% 1000 is invested at the beginning of calendar years 2004, 2005, and 2006. What is the total amount of interest credited for calendar year 2006? Problem 35.17 You are given: y iy 1 iy 2 iy 3 iy+5 2001 5.0 5.5 6.0 3.0 2002 4.5 5.0 5.5 2.5 2003 3.0 3.5 4.0 2.0 2004 4.0 X 5.0 2005 5.0 5.5 6.0 2006 4.5 5.0 5.5 The accumulated value on 1/1/2007 of a deposit on 1/1/2001 is 9.24% greater than the accumulated value on 1/1/2007 of the same amount deposited on 1/1/2004. Determine X. 328 RATE OF RETURN OF AN INVESTMENT 36 Yield Rates in Capital Budgeting One of the use of yield rates is in capital budgeting. An investor always is faced with the need to allocate an amount of capital among various alternative investments in order to achieve the highest possible level of return on the investments. This process of making such ﬁnancial decisions is referred to as capital budgeting. In this section we discuss brieﬂy this concept. Moreover, we will assume that risk is non-existent in the alternative investments being compared. The two major approaches to capital budgeting that are encountered in practice are the yield rate method and the net present value method. In the yield rate method the investor computes the yield rate(s) for each alternative investment by solving the equation NPV (i) = 0.Then these rates are compared to an interest preference rate set by the investor. This is usually the minimum rate of return acceptable by the investor. All yield rates that are lower than the interest preference rate are rejected. The yield rates that are higher than the interest preference rateare the only one considered. They are ranked from highest to lowest and are selected in descending order until the amount of capital available for investment is exhausted. In the net present value method or the NPV method, the investor computes NPV (i) for each alternative investment, where i is the interest preference rate. Negative values of NPV (i) are rejected and only the positive values are considered since the present value of returns is larger than that of the contributions. Capital is then allocated among those investments with positive NPV (i) in such a manner that the total present value of returns from the investment (computed at the interest preference rate) minus the contributions to the investment is maximized. It has been shown in Finance Theory that the NPV method usually provides better decisions than other methods when making capital investments. Consequently, it is the more popular evaluation method of capital budgeting projects. Example 36.1 Consider the investment project given in the table below. (a) Find the yield rate of this project. (b) Assuming an interest preference rate of 3%, would you accept this project when using the yield rate method? the net present value method? Solution. (a) Solving the equation NPV (i) = −80,000 −10,000ν−10,000ν2 −10,000ν3 −8,000ν4 + 28,000ν5 + 38,000ν6 + 33,000ν7 + 23,000ν8 + 13,000ν9 + 8,000ν10 = 0 we ﬁnd the two solutions i= −1.714964 <−1 and i= 0.032180. Thus, the yield rate is 3.218%. (b) Using the yield rate method, since 3 .218% > 3%, the investor would accept this project 36 YIELD RATES IN CAPITAL BUDGETING 329 for consideration. Using the net present value method the investor would also accept it, since NPV (0.03) = 1488.04 >0 Year Contributions Returns ct 0 80,000 0 −80,000 1 10,000 0 −10,000 2 10,000 0 −10,000 3 10,000 0 −10,000 4 20,000 12,000 −8,000 5 2,000 30,000 28,000 6 2,000 40,000 38,000 7 2,000 35,000 33,000 8 2,000 25,000 23,000 9 2,000 15,000 13,000 10 0 8,000 8,000 Example 36.2 Repeat the same problem as above but with an interest preference rate of 4%. Solution. Using the yield rate method, since 4% >3.218%, the investor would reject this project. Using the net present value method the investor would also reject it, since NPV (0.04) = −5122.13 <0 330 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 36.1 An investment project has the following cash ﬂows: Year Contributions Returns 0 100 0 1 200 0 2 10 60 3 10 80 4 10 100 5 5 120 6 0 60 (a) Using the net present value method with an interest preference rate of 15%, should the invest- ment be accepted or rejected? (b) Answer the same question when using the yield rate method. Problem 36.2 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. (a) Calculate NPV (0.09) and NPV (0.10). (b) If an investor’s interest preference rate is 12%, should the investment be accepted or rejected? Problem 36.3 A used car can be purchased for $5000 cash or for $2400 down and $1500 at the end of each of the next two years. Should a purchaser with an interest preference rate of 10% pay cash or ﬁnance the car? Problem 36.4 Consider an investment in which a person makes payments of $100 immediately and $132 at the end of one year. What method would be better to use, the yield rate method or the net present value method if the preference interest rate is 15%? Problem 36.5 Consider an investment in which a person makes payments of $100 immediately and $101 at the end of two years in exchange for a payment of $200 at the end of one year. Explain why the yield rate method is not applicable in this case. 36 YIELD RATES IN CAPITAL BUDGETING 331 Problem 36.6 A borrower needs $800. The funds can be obtained in two ways: (i) By promising to pay $900 at the end of the period. (ii) By borrowing $1000 and repaying $1120 at the end of the period. If the interest preference rate for the period is 10%, which option should be chosen? 332 RATE OF RETURN OF AN INVESTMENT Loan Repayment Methods Loan is an arrangement in which a lender gives money or property to a borrower, and the borrower agrees to return the property or repay the money, usually along with interest, at some future point(s) in time. Various methods of repaying a loan are possible. We will consider two of them: The amortization method and the sinking fund method. The amortization method: In this method the borrower makes installment payments to the lender. Usually these payments are at a regularly spaced periodic intervals; the progressive reduction of the amount owed is described as the amortization of the loan. Examples include car loan, home mortgage repayment. The sinking fund method: In this method the loan will be repaid by a single lump sum payment at the end of the term of the loan. The borrower pays interest on the loan in installments over this period. However, the borrower may prepare himself for the repayment by making deposits to a fund called a sinking fund to accumulate the repayment amount. 333 334 LOAN REPAYMENT METHODS 37 Finding the Loan Balance Using Prospective and Retro- spective Methods. When using the amortization method, the payments form an annuity whose present value is equal to the original amount of the loan. In this section, we want to determine the unpaid balance, also referred to as the outstanding loan balance or unpaid principal at any time after the inception of the loan. There are two approaches used in ﬁnding the amount of the outstanding balance: the prospective and the retrospective method. With the prospective method, the outstanding loan balance at any point in time is equal to the present value at that date of the remaining payments. With the retrospective method, the outstanding loan balance at any point in time is equal to the original amount of the loan accumulated to that date less the accumulated value at that date of all payments previously made. In general, the two approaches are equivalent. At the time of inception of the loan we have the following equality Present Value of All Payments = Amount of Loan Accumulate each side of the equation to the date at which the outstanding loan balance is desired, obtaining Current Value of Payments = Accumulated Value of Loan Amount But payments can be divided into past and future payments giving Accumulated Value of Past Payments + Present Value of Future Payments = Accumulated Value of Loan Amount or Present Value of Future Payments = Accumulated Value of Loan Amount −Accumulated Value of Past Payments. But the left side of this equation represents the prospective approach and the right side represents the retrospective approach. We can prove that the two methods are equivalent algebraically as follows. LetBp t and Br t denote the outstanding loan balances at time t using the prospective and retrospective methods respectively. We denote the initial loan by L. 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.335 Suppose we want to repay the loan by level payments of P for nperiods at a periodic interest of i.. Then P satiﬁes the equation P = L an . For 0 <t<n, the outstanding loan balance at time t computed after making the tth payment is Bp t = Pan−t by the prospective method and Br t = L(1 + i)t −Pst by the retrospective method. We will show that Br t = Bp t. Indeed, Br t =L(1 + i)t −Pst =Pan(1 + i)t −Pst =P [(1 + i)t −(1 + i)−(n−t) −(1 + i)t + 1 i ] =P1 −νn−t i = Pan−t = Bp t. Example 37.1 ‡ A loan is being repaid with 25 annual payments of 300 each. With the 10th payment, the borrower pays an extra 1,000, and then repays the balance over 10 years with a revised annual payment. The eﬀective rate of interest is 8%. Calculate the amount of the revised annual payment. Solution. The balance after 10 years, prospectively, is Bp 10 = 300a15 = $2,567.84. If the borrower pays an additional $1,000, the balance becomes $1,567.84. An equation of value for the revised payment, denoted by R, is Ra10 = 1,567.84 or R= 1,567.84 a10 = $233.66 336 LOAN REPAYMENT METHODS Example 37.2 A loan is being repaid with 16 quarterly payments, where the ﬁrst 8 payments are each $200 and the last 8 payments are each $400. If the nominal rate of interest convertible quarterly is 10%, use both the prospective method and the retrospective method to ﬁnd the outstanding loan balance immediately after the ﬁrst six payments are made. Solution. With the prospective method, we have Bp 6 =200(ν+ ν2) + 400ν2(ν+ ν2 + ··· + ν8) =400(ν+ ν2 + ··· + ν10) −200(ν+ ν2) =400a10 −200a2 =400(8.7521) −200(1.9274) = $3,115.36. With the retrospective method, we have that the original loan amount is L=200(ν+ ν2 + ··· + ν8) + 400ν8(ν+ ν2 + ··· + ν8) =400(ν+ ν2 + ··· + ν16) −200(ν+ ν2 + ··· + ν8) =400a16 −200a8 =400(13.0550) −200(7.1701) = $3,787.98 The outstanding loan balance is Br 6 = 3787.98(1.025)6 −200s6 = 4392.90 −200(6.3877) = $3,115.36 Example 37.3 Megan is buying a car for 30,000 using a 60-month car loan with an interest rate of 9% compounded monthly. For the ﬁrst two years, Megan makes the required payment. Beginning with the ﬁrst payment in the third year, Megan begins paying twice the required payment. Megan will completely pay oﬀ her loan by making a smaller ﬁnal payment. Determine the total number of payments that Megan will make. Solution. The monthly eﬀective interest rate is 0.09 12 = 0.0075. The original required payment is P such that 30000 = Pa60 0.0075 = P(1−1.0075−60) 0.0075 →P = 30000×0.0075 (1−1.0075−60) = 622.7506568. At the end of the ﬁrst 2 years, Megan’s outstanding loan balance, prospectively, is 622.7506568a36 0.0075 = 622.7506568(1 −1.0075−36) 0.0075 = 19583.51862 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.337 From this point on, Megan pays 2P = 1245.501314 every month. Then 19583.51862 = 1245.501314an0.0075 = 1245.501314(1−1.0075−n) 0.0075 ,where nis the number of months until the loan is repaid. Thus, 0.1179255197 = (1 −1.0075−n) →1.0075−n = 0.8820744803 →n = −ln 0.8820744803 ln 1.0075 = 16.79315649. So Megan will make 16 full payments and 1 smaller payment for a total of 17 additional payments and 17 + 24 = 41 total payments over the course of repaying the loan Example 37.4 A loan of 1,000 is being repaid with quarterly payments at the end of each quarter for 5 years, at 6% convertible quarterly. Find the outstanding loan balance at the end of the 2nd year. Solution. The quarterly eﬀective interest rate is 0.06 4 = 0.015. Let P be the quarterly payment. Then 1000 = Pa20 0.015 = P(1−1.015−20) 0.015 →P = 1000×0.015 (1−1.015−20) = 58.24573587. The outstanding loan balance at the end of the 2nd year, prospectively, is 58 .24573587a12 0.015 = 58.24573587(1−1.015−12) 0.015 = $635.32 338 LOAN REPAYMENT METHODS Practice Problems Problem 37.1 A loan is being repaid with level annual payments of $1,000. Calculate the outstanding balance of the loan if there are 12 payments left. The next payment will be paid one year from now and the eﬀective annual interest rate is 5%. Problem 37.2 A loan of 10,000 is being repaid will 20 non-level annual payments. The interest rate on the loan is an annual eﬀective rate of 6%. The loan was originated 4 years ago. Payments of 500 at the end of the ﬁrst year, 750 at the end of the second year, 1,000 at the end of the third year and 1,250 at the end of the fourth year have been paid. Calculate the outstanding balance immediately after the fourth payment. Problem 37.3 Calculate the outstanding balance to the loan in the previous problem one year after the fourth payment immediately before the ﬁfth payment. Problem 37.4 Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. What is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 37.5 If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 37.6 A loan 20,000 is being repaid with annual payments of 2,000 at the end of each year. The interest rate charged on the loan is an annual eﬀective rate of 8%. Calculate the outstanding balance of the loan immediately after the 5th payment. Problem 37.7 The interest rate on a 30 year mortgage is 12% compounded monthly. The mortgage is being repaid by monthly payments of 700. Calculate the outstanding balance at the end of 10 years. Problem 37.8 The interest rate on a 30 year mortgage is 12% compounded monthly. Lauren is repaying the mortgage by paying monthly payments of 700. Additionally, to pay oﬀ the loan early, Lauren has made additional payments of 1,000 at the end of each year. Calculate the outstanding balance at the end of 10 years. 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.339 Problem 37.9 A loan is being repaid with 20 payments of 1,000. The total interest paid during the life of the loan is 5,000. Calculate the amount of the loan. Problem 37.10 A loan of 10,000 is being repaid by installments of 2,000 at the end of each year, and a smaller ﬁnal payment made one year after the last regular payment. Interest is at the eﬀective rate of 12%. Find the amount of outstanding loan balance remaining when the borrower has made payments equal to the amount of the loan. Problem 37.11 A loan is being repaid by quarterly installments of 1,500 at the end of each quarter, at 10% con- vertible quarterly. If the loan balance at the end of the ﬁrst year is 12,000, ﬁnd the original loan balance. Problem 37.12 A loan is being repaid by 15 annual payments at the end of each year. The ﬁrst 5 installments are 4,000 each, the next 5 are 3,000 each, and the ﬁnal 5 are 2,000 each. Find expressions for the outstanding loan balance immediately after the second 3,000 installment. (a) prospectively; (b) retrospectively. Problem 37.13 A loan is to be repaid with level installments payable at the end of each half-year for 3 1 2 years, at a nominal rate of interest of 8% convertible semiannually. After the fourth payment the outstanding loan balance is 5,000. Find the initial amount of the loan. Problem 37.14 A 20,000 loan is to be repaid with annual payments at the end of each year for 12 years. If (1 + i)4 = 2, ﬁnd the outstanding balance immediately after the fourth payment. Problem 37.15 ‡ A 20-year loan of 1,000 is repaid with payments at the end of each year. Each of the ﬁrst ten payments equals 150% of the amount of interest due. Each of the last ten payments is X. The lender charges interest at an annual eﬀective rate of 10%. Calculate X. Problem 37.16 ‡ A loan is amortized over ﬁve years with monthly payments at a nominal interest rate of 9% com- pounded monthly. The ﬁrst payment is 1,000 and is to be paid one month from the date of the loan. Each succeeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made. 340 LOAN REPAYMENT METHODS Problem 37.17 A mortgage loan is being repaid with level annual payments of 5,000 at the end of the year for 20 years. The interest rate on the mortgage is 10% per year. The borrower pays 10 payments and then is unable to make payments for two years. Calculate the outstanding balance at the end of the 12th year. Problem 37.18 ‡ An investor took out a loan of 150,000 at 8% compounded quarterly, to be repaid over 10 years with quarterly payments of 5483.36 at the end of each quarter. After 12 payments, the interest rate dropped to 6% compounded quarterly. The new quarterly payment dropped to 5134.62. After 20 payments in total, the interest rate on the loan increased to 7% compounded quarterly. The investor decided to make an additional payment of X at the time of his 20th payment. After the additional payment was made, the new quarterly payment was calculated to be 4265.73, payable for ﬁve more years. Determine X. Problem 37.19 ‡ A small business takes out a loan of 12,000 at a nominal rate of 12%, compounded quarterly, to help ﬁnance its start-up costs. Payments of 750 are made at the end of every 6 months for as long as is necessary to pay back the loan. Three months before the 9th payment is due, the company reﬁnances the loan at a nominal rate of 9%, compounded monthly. Under the reﬁnanced loan, payments of R are to be made monthly, with the ﬁrst monthly payment to be made at the same time that the 9th payment under the old loan was to be made. A total of 30 monthly payments will completely pay oﬀ the loan. Determine R. Problem 37.20 A loan of $1,000 is being repaid with quarterly payments at the end of each quarter for 5 years, at 6% convertible quarterly. Find the outstanding loan balance at the end of the second year. Problem 37.21 A 20,000 mortgage is being repaid with 20 annual installments at the end of each year. The borrower makes 5 payments, and then is temporarily unable to make payments for the next 2 years. Find an expression for the revised payment to start at the end of the 8th year if the loan is still to be repaid at the end of the original 20 years. Problem 37.22 A loan of 1 was originally scheduled to be repaid by 25 equal annual payments at the end of each 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.341 year. An extra payment K with each of the 6th through the 10th scheduled payments will be suﬃcient to repay the loan 5 years earlier than under the original schedule. Show that K = a20 −a15 a25 a5 . Problem 37.23 Bob takes out a loan of 1000 at an annual eﬀective interest rate of i. You are given: (a) The ﬁrst payment is made at the end of year 6. (b) Ten equal annual payments are made to repay the loan in full at the end of 15 years. (c) The outstanding principal after the payment made at the end of year 10 is 908.91. Calculate the outstanding principal at the end of year 5. Problem 37.24 The original amount of an inheritance was just suﬃcient at 3.5% eﬀective to pay $10,000 at the end of each year for 10 years. The payments were made as scheduled for the ﬁrst ﬁve years even though the fund actually earned 5% eﬀective. How much excess interest was in the fund at the end of the ﬁfth year? 342 LOAN REPAYMENT METHODS 38 Amortization Schedules When a loan is being repaid with the amortization method, each payment is partially a repayment of principal and partially a payment of interest. Determining the amount of each for a payment can be important (for income tax purposes, for example). An amortization schedule is a table which shows the division of each payment into principal and interest, together with the outstanding loan balance after each payment is made. Consider a loan of an at interest rate i per period being repaid with payments of 1 at the end of each period for n periods. At the end of period 1 (i.e. after the ﬁrst payment), the interest paid is ian = 1 −νn so that the principal repaid is νn,and the outstanding loan balance is an −νn = an−1 . Next, at the end of the second period, the interest paid is ian−1 = 1 −νn−1 so that the principal repaid is νn−1,and the outstanding loan balance is an−1 −νn−1 = an−2 .Continuing this process, we see that at the end of period k, the interest paid is ian−k+1 = 1 −νn−k+1 and the principal repaid is νn−k+1. The outstanding loan balance is an−k+1 −νn−k+1 = an−k = Bp k. The amortization table is shown below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 an 1 1 ian = 1 −νn νn an −νn = an−1 2 1 ian−1 = 1 −νn−1 νn−1 an−1 −νn−1 = an−2 ... ... ... ... ... k 1 ian−k+1 = 1 −νn−k+1 νn−k+1 an−k+1 −νn−k+1 = an−k ... ... ... ... ... n−1 1 ia2 = 1 −ν2 ν2 a2 −ν2 = a1 n 1 ia1 = 1 −ν ν a1 −ν = 0 Total n n−an an Observe each of the following from this table: First, it should be noted that the outstanding loan balance agrees with that obtained by the prospective method. Second, the sum of the principal repayments equals to the original amount of the loan. Third, the sum of interest payments is equal to the diﬀerence between the sum of the total payments and the sum of the principal repayments. Fourth, the sum of principal repayments is a geometric progression with common ration (1 + i). Example 38.1 Create an amortization schedule for a loan of $1,000 repaid over four years if the annual eﬀective rate of interest is 8%. 38 AMORTIZATION SCHEDULES 343 Solution. If R is the periodic payment then R= 1000 a4 = $301.92. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 1000.00 1 301.92 80.00 221.92 778.08 2 301.92 62.25 239.67 538.41 3 301.92 43.07 258.85 279.56 4 301.92 22.36 279.56 0 Table 38.1 In the previous example, the last line exactly balances. That is, after the last payment is made the loan is paid oﬀ. In practice, there will be rounding errors as the table is generated line by line, and the last line may not lead to a zero balance. Standard practice is to adjust the last payment so that it is exactly equal to the amount of interest for the ﬁnal period plus the outstanding loan balance at the beginning of the ﬁnal period, in order to bring the outstanding loan balance to 0. Example 38.2 Create an amortization schedule for a loan of $10,000 repaid over three years if the annual eﬀective rate of interest is 7%. Solution. If R is the periodic payment then R= 10,000 a3 = $3,810.52. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 10,000.00 1 3,810.52 700.00 3,110.52 6,889.48 2 3,810.52 482.26 3,328.26 3,561.22 3 3,810.50 249.28 3,561.22 0 It should be noted that if it is desired to ﬁnd the amount of principal and interest in one particular payment, it is not necessary to construct the entire amortization schedule. The outstanding loan balance at the beginning of the period in question can be determined by either the retrospective or the prospective methods, and then that one line of the schedule can be created. 344 LOAN REPAYMENT METHODS Example 38.3 Jones borrows $20,000 from Smith and agrees to repay the loan with equal quarterly installments of principal and interest at 10% convertible quarterly over eight years. At the end of three years, Smith sells the right to receive future payments to Collins at a price which produces a yield rate of 12% convertible quarterly for Smith. Find the total amount of interest received (a) by Collins, and (b) by Smith. Solution. (a) Each quarterly payment by Jones is 20000 a32 0.025 = $915.37. Total payments by Jones to Collins over the last ﬁve years are (20)915.37 = $18307.40. After three years, the price Collins pays to Smith is 915.37a20 0.03 = 915.37(14.8775) = $13618.42. Total amount of interest received by Collins is 18307.40 −13618.42 = $4688.98. (b) Total payments by Jones to Smith over the ﬁrst three years are (12)915.37 = $10984.44. After three years, the outstanding loan balance is 915.37a20 0.025 = 915.37(15.5892) = $14269.89. Recall from (a) that the price Collins pays to Smith after three years is $13618.42. Total amount of interest received by Smith is 13618.42 + 10984.44 −20000 = $4602.86 Let It denote the amount of interest paid in the tth installment; Pt be the amount of principal and Bt be the loan balance. 38 AMORTIZATION SCHEDULES 345 Example 38.4 Show that for n= 100 and with a periodic payment of R, we have P11 + P12 + ··· + P50 = B10 −B50 and I11 + I12 + ··· + I50 = 40R−(B10 −B50). Solution. Let P = P11 + P12 + ··· + P50. Then P =R 50∑ k=11 ν100−k+1 = Rν101 50∑ k=11 (1 + i)k =Rv101( 50∑ k=0 (1 + i)k − 10∑ k=0 (1 + i)k) =R(1 + i)−50 −(1 + i)−90 i =Ra90 −Ra50 = B10 −B50 Now, let I = I11 + I12 + ··· + I50 then I = ∑50 j=11 R−∑50 j=11 Pj = 40R−(B10 −B50) Amortization schedules of perpetuities do not exist since the entire payment represents interest and therefore the loan balance remains unchanged. Example 38.5 A $5,000 loan is being repaid by payments of $X at the end of each half year for as long as necessary until a smaller ﬁnal payment is made. The nominal rate of interest convertible semiannually is 14%. (a) If X = $400 ﬁnd the amount of principal and the interest in the sixth payment. (b) If X = $350, ﬁnd the principal in the sixth payment, and interpret this. Solution. (a) The outstanding balance at the beginning of the sixth half-year is Br 5 = 5000(1.07)5 −400s5 = 7012.76 −400(5.7507) = $4712.48. The interest in the sixth payment is (0 .07)(4712.48) = $329.87 and the principal is 400 −329.87 = $70.13. (b) The outstanding balance at the beginning of the sixth half-year is Br 5 = 5000(1.07)5 −350s5 = 7012.76 −350(5.7507) = $5000.01. 346 LOAN REPAYMENT METHODS If X ≤350, then the loan will never be paid oﬀ, because every payment will count only toward interest Thus far, in creating an amortization schedule we have assumed a constant rate of interest, the conversion period and payment period are the same, and the payments are leveled. Example 38.2 is an example where the payments are not all leveled. It is possible to create an amortization schedule with varying interest rate as illustrated in the next example. Example 38.6 An amount is invested at an annual eﬀective rate of interest i which is exactly suﬃcient to pay 1 at the end of each year for n years. In the ﬁrst year, the fund earns rate i and 1 is paid at the end of the year. However, in the second year, the fund earns rate j > i.If X is the revised payment which could be made at the end of years 2 to n, then ﬁnd X assuming that (a) the rate reverts back to i again after this one year, (b) the rate earned remains at j for the rest of the n−year period. Solution. (a) From the amortization table, the balance at the end of the ﬁrst year is an−1 i,and therefore the balance after two years must be (1 + j)an−1 i −X. However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j)an−1 i −X =Xan−2 i X(1 + an−2 i) =(1 + j)an−1 i X(1 + i)an−1 i =(1 + j)an−1 i X =1 + j 1 + i. (b) As in (a), the balance at the end of two years is (1 + j)an−1 i−X. However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j)an−1 i −X =Xan−2 j X(1 + an−2 j) =(1 + j)an−1 i X(1 + j)an−1 j =(1 + j)an−1 i X = an−1 i an−1 j Situations where payment period and interest conversion period do not coincide are best handled 38 AMORTIZATION SCHEDULES 347 from ﬁrst principles rather than from developing formulas. With an amortization schedule in which payments are made at a diﬀerent frequency than interest is convertible, the following two-step procedure can be followed: (1) Find the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. (2) Construct the amortization schedule with the rate of interest in step 1. Example 38.7 A 25-year mortgage for $100,000 bears an interest rate of 5.5% compounded semi-annually. The payments are made at the end of each month. Find the size of the payment. Solution. The monthly interest rate is j = (1.0275) 1 6 −1 = .00453168. The size of each payment is R= 100,000 a300 j = 610.39 Example 38.8 A loan of $3,000 is being amortized by 20 quarterly payments. Payments 11 and 12 are not made. At the designated time of the 12th payment, the loan is renegotiated so that the 13th payment is $N and payments 14, 16, 18, and 20 are each $40 more that the preceding payment. If the rate of interest is 8% convertible semiannually, ﬁnd the value of N which would provide that the loan be completely repaid after 20 quarters. Solution. The quarterly eﬀective rate of interest isj = (1.02) 1 2 −1.The original quareterly payment isR= 3000 a20 j . The outstanding balance at time t = 10, by the prospective method, is Ra10 j. The outstanding balance at time t= 12 is Ra10 j(1 + j)2 = 3000a10 j a20 j (1 + j)2 = 1712.46. Now, the new payments stream starting at time t= 13 are: N,N + 40,N + 40,N + 80,N + 80,N + 120,N + 120,N + 160. By the prospective method 1712.46 = Na8 j + 40(ν2 j + ν3 j + 2ν4 j + 2ν5 j + 3ν6 j + 3ν7 j + 4ν8 j). Solving this equation for N we ﬁnd N = 155.73 348 LOAN REPAYMENT METHODS Practice Problems Problem 38.1 A loan of 10,000 is being repaid with annual payments of 1500 for 11 years. Calculate the amount of principal paid over the life of the loan. Problem 38.2 A loan of 10000 is being repaid with annual payments of 1500 for 11 years. Calculate the amount of interest paid over the life of the loan. Problem 38.3 A loan is being repaid with level annual payments based on an annual eﬀective interest rate of 8%. If the amount of principal in the 10th payment is 100, calculate the amount of principal in the 5th payment. Problem 38.4 A loan is being repaid with level annual payments based on an annual eﬀective interest rate of 8%. If the outstanding balance immediately after the 10th payment is 1000, calculate the amount of interest in the 11th payment. Problem 38.5 A loan of 10,000 is being repaid with annual payments of 1500 for n years. The total principal paid in the ﬁrst payment is 685.58. Calculate the interest rate on the loan. Problem 38.6 For a loan with level annual payments, the principal repaid by the 10th payment is 10,000 while the principle repaid by the 11th payment is 11,000. Calculate the principal repaid by the 15th payment. Problem 38.7 A 60-month loan is to be repaid with level payments of 1000 at the end of each month. The interest in the last payment is 7.44. Calculate the total interest paid over the life of the loan. Problem 38.8 A 60-month loan is to be repaid with level payments of 1000 at the end of each month. The principal in the ﬁrst payment is 671.21. Calculate the eﬀective annual interest rate. Problem 38.9 Jenna is repaying a 120-month loan with interest compounded monthly at 12%. Calculate the payment in which the absolute value of the diﬀerence between the interest paid and the principal repaid is minimized. 38 AMORTIZATION SCHEDULES 349 Problem 38.10 A loan is being repaid with level payments at the end of each year for 20 years. The principal repaid in the 10th payment is 1000 and the principal repaid in the 15th payment is 1200. Calculate the amount of the loan. Problem 38.11 A loan of 10,000 is being repaid with annual payments over 10 years at 8% compound monthly. Create the amortization table for this loan. Problem 38.12 A loan is being repaid with quarterly installments of 1,000 at the end of each quarter for 5 years at 12% convertible quarterly. Find the amount of principal in the 6th installment. Problem 38.13 Consider a loan which is being repaid with installments of 1 at the end of each period for nperiods. Find an expression at issue for the present value of the interest which will be paid over the life of the loan. Problem 38.14 A loan of 10,000 is being repaid with 20 installments at the end of each year, at 10% eﬀective. Show that the amount of interest in the 11th installment is 1000 1 + ν10 . Problem 38.15 A loan is being repaid with 20 installments at the end of each year at 9% eﬀective. In what installment are the principal and interest portions most nearly equal to each other? Problem 38.16 A loan is being repaid with a series of payments at the end of each quarter, for 5 years. If the amount of principal in the 3rd payment is 100, ﬁnd the amount of principal in the last 5 payments. Interest is at the rate of 10% convertible quarterly. Problem 38.17 A loan is being repaid with installments of 1 at the end of each year for 20 years. Interest is at eﬀective rate i for the ﬁrst 10 years, and eﬀective rate j for the second 10 years. Find expressions for (a) the amount of interest paid in the 5th installment; (b) the amount of principal repaid in the 15th installment. 350 LOAN REPAYMENT METHODS Problem 38.18 A loan of 25,000 is being repaid with annual payments of 2,243 at the end of the year. The interest rate on the loan 7.5%. Calculate the interest in the 5th payment. Problem 38.19 ‡ Seth borrows X for four years at an annual eﬀective interest rate of 8%, to be repaid with equal payments at the end of each year. The outstanding loan balance at the end of the second year is 1076.82 and at the end of the third year is 559.12 . Calculate the principal repaid in the ﬁrst payment. Problem 38.20 ‡ A bank customer takes out a loan of 500 with a 16% nominal interest rate convertible quarterly. The customer makes payments of 20 at the end of each quarter. Calculate the amount of principal in the fourth payment. Problem 38.21 ‡ A loan is repaid with level annual payments based on an annual eﬀective interest rate of 7%. The 8th payment consists of 789 of interest and 211 of principal. Calculate the amount of interest paid in the 18th payment. Problem 38.22 ‡ Kevin takes out a 10-year loan of L, which he repays by the amortization method at an annual eﬀective interest rate of i.Kevin makes payments of 1000 at the end of each year . The total amount of interest repaid during the life of the loan is also equal to L. Calculate the amount of interest repaid during the ﬁrst year of the loan. Problem 38.23 ‡ Ron is repaying a loan with payments of 1 at the end of each year for n years. The amount of interest paid in period tplus the amount of principal repaid in period t+ 1 equals X. Calculate X. Problem 38.24 A 35-year loan is to be repaid with equal installments at the end of each year. The amount of interest paid in the 8th installment is 135. The amount of interest paid in the 22nd installment is 108. Calculate the amount of interest paid in the 29th installment. Problem 38.25 A 1000 loan is repaid with equal payments at the end of each year for 20 years. The principal portion of the 13th payment is 1.5 times the principal portion of the 5th payment. Calculate the total amount of interest paid on the loan. 38 AMORTIZATION SCHEDULES 351 Problem 38.26 A loan is being amortized with payments at the end of each quarter for 25 years. If the amount of principal repaid in the third payment is $100, ﬁnd the total amount of principal repaid in the forty payments consisting of payments eleven through ﬁfty. Interest is at the rate of 8% convertible quarterly. Problem 38.27 A loan is being amortized using an interest rate of 8% convertible quarterly with level payments at the end of each quarter. If the amount of principal repaid in the third payment is $100, ﬁnd the total amount of principal repaid in the ﬁve payments consisting of payments eleven through ﬁfteen. Problem 38.28 A 100,000 loan is to be repaid by 30 equal payments at the end of each year. The outstanding balance is amortized at 4%. In addition to the annual payments, the borrower must pay an origination fee at the time the loan is made. The fee is 2% of the loan but does not reduce the loan balance. When the second payment is due, the borrower pays the remaining loan balance. Determine the yield to the lender considering the origination fee and the early pay-oﬀ of the loan. Problem 38.29 A loan of 10,000 is amortized by equal annual payments for 30 years at an annual eﬀective interest rate of 5%. Determine the year in which the interest portion of the payment is most nearly equal to one-third of the payment. Problem 38.30 A loan of 1000 at a nominal rate of 12% convertible monthly is to be repaid by six monthly payments with the ﬁrst payment due at the end of 1 month. The ﬁrst three payments are X each, and the ﬁnal three payments are 3X each. Determine the sum of the principal repaid in the third payment and the interest paid in the ﬁfth payment. Problem 38.31 Carl borrows $10,000 at 12% compounded monthly, and Carl will repay the loan with 60 monthly amortization payments beginning at the end of the ﬁrst month. Find: (a) B20 (b) B30 (c) ∑30 j=21 Pj (d) ∑30 j=21 Ij Problem 38.32 ‡ A bank customer borrows X at an annual eﬀective rate of 12.5% and makes level payments at the end of each year for nyears. The interest portion of the ﬁnal payment is 153.86. The total principal repaid as of time ( n−1) is 6009.12. The principal repaid in the ﬁrst payment is Y. Calculate Y. 352 LOAN REPAYMENT METHODS Problem 38.33 A loan is to be amortized by n level annual payments of X where n> 5. You are given (1) The amount of interest in the ﬁrst payment is 604.00 (2) The amount of interest in the third payment is 593.75 (3) The amount of interest in the ﬁfth payment is 582.45 Calculate X. Problem 38.34 A loan is to be repaid by annual installments of X at the end of each year for 10 years. You are given (1) the total principal repaid in the ﬁrst 3 years is 290.35 (2) the total principal repaid in the last 3 years is 408.55 Calculate the total amount of interest paid during the life of the loan. Problem 38.35 Iggy borrows X for 10 years at an annual eﬀective rate of 6%. If he pays the principal and accumulated interest in one lump sum at the end of 10 years, he would pay 356.54 more in interest than if he repaid the loan with 10 level payments at the end of each year. Calculate X. Problem 38.36 A loan is being amortized by means of level monthly payments at an annual eﬀective interest rate of 8%. The amount of principal repaid in the 12th payment is 1000 and the amount of principal repaid in the tth payment is 3700. Calculate t. Problem 38.37 John is repaying a loan with payments of $3,000 at the end of every year over an unknown period of time. if the amount on interest in the third installment is $2,000, ﬁnd the amount of principal in the sixth installment. Assume the interest is 10% convertible quarterly. 39 SINKING FUND METHOD 353 39 Sinking Fund Method An alternative for repaying a loan in installments by the amortization method, a borrower can accumulate a fund which will exactly repay the loan in one lump sum at the end of a speciﬁed period of time. This fund is called a sinking fund. It is generally required that the borrower periodically pay interest on the loan, sometimes referred to as a service. In some cases payments into a sinking fund may vary irregularly at the discretion of the borrower. However, we will be dealing with sinking funds with regular contributions. Because the balance in the sinking fund could be applied against the loan at any point, the net amount of the loan is equal to the original amount of the loan minus the sinking fund balance. We next show that if the rate of interest paid on the loan equals the rate of interest earned on the sinking fund, then the amortization method and the sinking fund method are equivalent. To see this, Suppose that the eﬀective annual rate on the loan is i, and the sinking fund earns the same rate. Suppose that the amount of the loan is 1 , and the loan term is n periods. With the amortization method, the payment at the end of each period is 1 an .With the sinking fund method, to accumulate the amount of 1 in the sinking fund, the borrower deposits of 1 sn at the end of each year for n years. At the same moment the borrower also pays i per period to the lender. That is, payments of size i (interest) plus payments of size 1 sn are required. But we know from Section 15 that 1 sn + i= 1 an . Thus, the two methods are equivalent. As was seen in Section 38, a way to visualize the activity of an amortization method is the creation of an amortization schedule. The same idea can be used with sinking funds. The following example illustrates the creation of a sinking fund schedule. Example 39.1 Create a sinking fund schedule for a loan of $1000 repaid over four years if the annual eﬀective rate of interest is 8% Solution. If R is the sinking fund deposit then R= 1000 s4 = $221.92. 354 LOAN REPAYMENT METHODS Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1000.00 1 80.00 221.92 0 221.92 778.08 2 80.00 221.92 17.75 461.59 538.41 3 80.00 221.92 36.93 720.44 279.56 4 80.00 221.92 57.64 1000.00 0 Table 39.1 Notice from this table that the amount in the sinking fund after the tth payment is found by multiplying the sinking fund deposit by stj where j is the sinking fund rate of interest. Comparing Table 38.1 and Table 39.1 we notice the following: (1) For each period, Interest paid + Sinking Fund deposit (in Table 39.1) = Payment amount (in Table 38.1). (2) For each period, Interest paid −Interest earned on sinking fund (in Table 39.1) = Interest paid (in Table 38.1). (3) For each period, Sinking fund deposit + Interest earned on sinking fund (in Table 39.1) = principal repaid (in Table 38.1). (4) For each period, the Net amount of loan (in Table 39.1) = Outstanding loan balance in (Table 38.1) Notice that the net amount of loan concept plays the same role for the sinking fund method that the outsanding loan balance does for the amortization method. Next, we consider the situation in which the interest rate on the loan and the interest rate earned on the sinking fund diﬀers. The rate on the loan is denoted by i, and the rate on the sinking fund is denoted by j. Usually, j is less than i because the sinking fund wouldn’t normally be riskier than the loan. In this case, an amount of i will be deducted from the sinking fund deposit and the remaining amount will be invested in the sinking fund at a rate of j. Let ani&j be the present value of an annuity which pays 1 at the end of each period for n periods with i and j as previously deﬁned. If a loan of 1 is made, by the amortization method the amount of the loan will be repaid by interest payments of iani&j at the end of each year for nyears together with yearly deposits of 1 −iani&j into a sinking fund which earns interest at the eﬀective annual rate of interest j. The sinking ﬁnd should accumulate ani&j at the end of n years. That is, (1 −iani&j)snj = ani&j or ani&j = snj 1 + isnj 39 SINKING FUND METHOD 355 which is equivalent to 1 ani&j = 1 snj + i. In other words, loan of 1 can be repaid by interest payments of i to the lender and depositing 1 snj into the sinking fund at the end of each year for n years. Now since 1 anj = 1 snj + j it follows that 1 ani&j = 1 anj + (i−j) or ani&j = anj 1 + (i−j)anj . It should be noted that if i= j, then ani&j = ani. Example 39.2 ‡ John borrows 10,000 for 10 years at an annual eﬀective interest rate of 10%. He can repay this loan using the amortization method with payments of 1,627.45 at the end of each year. Instead, John repays the 10,000 using a sinking fund that pays an annual eﬀective interest rate of 14%. The deposits to the sinking fund are equal to 1,627.45 minus the interest on the loan and are made at the end of each year for 10 years. Determine the balance in the sinking fund immediately after repayment of the loan. Solution. Under the amortization method, the periodic installment is 1627.45. The periodic interest payment on the loan is 0.1(10000) = 1000.Thus, deposits into the sinking fund are 1627 .45 −1000 = 627.45. Thus, the amount in sinking fund immediately after repayment of the loan is 627.45s10 0.14−10,000 = 2130 In general, the sinking fund schedule at two rates of interest is identical to the sinking fund schedule at one rate of interest equal to the rate of interest earned on the sinking fund, except that a constant addition of (i−j) times the amount of the original loan is added to the interest paid column. We illustrate this in the next example. Example 39.3 Create a sinking fund schedule for a loan of $1,000 repaid over four years if the annual eﬀective rate of interest is 10% and the sinking fund interest of 8% 356 LOAN REPAYMENT METHODS Solution. If R is the sinking fund deposit then R= 1,000 s4 0.08 = $221.92. Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1000.00 1 100.00 221.92 0 221.92 778.08 2 100.00 221.92 17.75 461.59 538.41 3 100.00 221.92 36.93 720.44 279.56 4 100.00 221.92 57.64 1000 0 Note that this table is identical to Table 39.1 except each entry in the interest paid column in the table above is equal to each entry in the interest paid column in Table 39.1 increased by a constant equals to 20 which is (0 .10 −0.08) ×1000 The following is an important example whose results are to keep in mind. Example 39.4 A loan of 1 yields the lender rate iper period for nperiods, while the borrower replaces the capital in a sinking fund earning rate j per period. Find expressions for the following if 1 ≤t≤n: (a) Periodic interest paid to the lender. (b) Periodic sinking fund deposit. (c) Interest earned on sinking fund during the tth period. (d) Amount in sinking fund at end of the tth period. (e) Net amount of loan at the end of the tth period. (f) Net interest paid in period t. (g) Principal repaid in period t. Solution. (a) The yield rate is i, so the lender must be receiving an amount of i each period. (b) The capital is 1, so the borrower is depositing ( snj)−1 regularly into the sinking fund. (c) At the beginning of the tth period the balance in the sinking fund is ( st−1 j/snj); during the period it earns interest in the amount of j· st−1 j snj payable at the end of the period, i.e., at time t. (d) The amount in sinking fund at end of the tth period is stj snj . 39 SINKING FUND METHOD 357 (e) The net amount of loan at the end of the tth period is the excess of 1 over the balance in the sinking fund, i.e., 1 − stj snj . (f) The net interest paid in the tth period is the excess of interest paid over interest earned, i.e. i−j st−1 j snj . (g) By (e) above, the change in the amount of the loan between the ( t−1)th and the tth payment is ( 1 −stj snj ) − ( 1 − st−1 j snj ) = (1 + i)t−1 snj In creating the sinking fund schedule the payment period and the conversion interest period can be diﬀerent. These cases can be handled from basic principles as illustrated in the next two examples. Remember to use the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. Example 39.5 John borrows $5,000 for 10 years at 10% convertible quarterly. John does not pay interest currently and will pay all accrued interest at the end of 10 years together with the principal. Find the annual sinking fund deposit necessary to liquidate the loan at the end of 10 years if the sinking fund earns 7% convertible semi-annually. Solution. Let j be the annual interest rate. Then j = (1.025)4 −1. The loan balance at the end of 10 years is 5000(1 + j)10 = 5000(1.025)40 = 13425.32. Hence, the annual sinking fund deposit is R= 13425.32 s10 (1.035)2−1 = 966.08 Example 39.6 Create the sinking fund table for the following: a 3 year loan of $10,000, with interest payable semiannually at the nominal interest rate of 8.00% is to be retired by a sinking fund funded by quarterly deposits earning an eﬀective nominal interest rate of 6.00% compounded semiannually. Solution. Every six months the interest paid on the loan is 0 .04(10,000) = 400 . The quarterly rate on the sinking fund is j = 1.030.5 −1. The quarterly sinking fund deposit is D= 10,000 s12 j = 767.28. 358 LOAN REPAYMENT METHODS Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 10,000.00 0.25 0 767.28 0 767.28 9,232.72 0.50 400.00 767.28 11.42 1,545.98 8,454.02 0.75 0 767.28 23.02 2,336.27 7,663.73 1.00 400.00 767.28 34.79 3,138.33 6,861.67 1.25 0 767.28 46.73 3,952.33 6,047.67 1.50 400.00 767.28 58.85 4,778.45 5,221.55 1.75 0 767.28 71.15 5,616.88 4,383.12 2.00 400.00 767.28 83.63 6,467.78 3,532.22 2.25 0 767.28 96.30 7,331.36 2,668.64 2.50 400.00 767.28 109.16 8,207.79 1,792.21 2.75 0 767.28 122.21 9,097.27 902.73 3.00 400.00 767.28 135.45 10,000.00 0.00 39 SINKING FUND METHOD 359 Practice Problems Problem 39.1 A loan of 10,000 is being repaid with annual payments for 10 years using the sinking fund method. The loan charges 10% interest and the sinking fund earns 8%. (a) Calculate the interest payment that is paid annually to service the loan. (b) Calculate the sinking fund payment made annually. (c) Calculate the amount in the sinking fund immediately after the deposit made at the end of 5 years. (d) Create the sinking fund schedule for the loan. Problem 39.2 If the loan in Problem 39.1 was repaid using the amortization method, but the annual payment was equal to the sum of the interest payment and the sinking fund deposit, calculate the interest rate under the amortization method. Problem 39.3 A loan of 20,000 is being repaid with annual payments for 5 years using the sinking fund method. The loan charges 10% interest compounded twice a year. The sinking fund earns 8% compounded monthly. Calculate the interest payment that is paid annually to service the loan and the sinking fund deposit. Problem 39.4 A loan of 20,000 is being repaid with monthly payments for 5 years using the sinking fund method. The loan charges 10% interest compounded twice a year. The sinking fund earns an annual eﬀective interest rate of 8%. (a) Calculate the interest payment that is paid monthly to service the loan and the sinking fund deposit paid monthly. (b) Calculate the amount in the sinking fund immediately after the 30th payment. Problem 39.5 Julie agrees to repay a loan of 10,000 using the sinking fund method over 10 years. The loan charges an annual eﬀective interest rate of 7% while the sinking fund earns 6%. Calculate the amount paid into the sinking fund each year less the amount of interest paid on the loan each year. Problem 39.6 Kathy can take out a loan of 50,000 with Bank A or Bank B. With Bank A, she must repay the loan with 60 monthly payments using the amortization method with interest at 7% compounded 360 LOAN REPAYMENT METHODS monthly. With Bank B, she can repay the loan with 60 monthly payments using the sinking fund method. The sinking fund will earn 6.5% compounded monthly. What interest rate can Bank B charge on the loan so that Kathy’s payment will be the same under either option? Problem 39.7 Lauren is repaying a loan of 100,000 using the sinking fund method. At the end of each year she pays 7,000 into a sinking fund earning 8%. At the end of 5 years, Lauren pays oﬀ the loan using the sinking fund plus an additional payment of X. Calculate X. Problem 39.8 Lauren is repaying a loan of 100,000 using the sinking fund method. At the end of each year she pays 7,000 into a sinking fund earning 8%. At the end of year Y, Lauren will have suﬃcient money in the sinking fund to repay the loan. Calculate Y. Problem 39.9 Ryan takes out a loan of 100,000 and agrees to repay it over 10 years using the sinking fund method. Ryan agrees to pay interest to the lender at the end of each year. The interest rate is .01(11 −t) in year t. The sinking fund will earn 5% per year. Determine the amount in the sinking fund after 10 years if the total payment made by Ryan at the end of each year is 12,000. Problem 39.10 On a loan of 10,000, interest at 9% eﬀective must be paid at the end of each year. The borrower also deposits X at the beginning of each year into a sinking fund earning 7% eﬀective. At the end of 10 years the sinking fund is exactly suﬃcient to pay oﬀ the loan. Calculate X. Problem 39.11 A borrower is repaying a loan with 10 annual payments of 1,000. Half of the loan is repaid by the amortization method at 5% eﬀective. The other half of the loan is repaid by the sinking fund method, in which the lender receives 5% eﬀective on the investment and the sinking fund accumulates at 4% eﬀective. Find the amount of the loan. Problem 39.12 A borrows 12,000 for 10 years, and agrees to make semiannual payments of 1,000. The lender receives 12% convertible semiannually on the investment each year for the ﬁrst 5 years and 10% convertible semiannually for the second 5 years. The balance of each payment is invested in a sinking fund earning 8% convertible semiannually. Find the amount by which the sinking fund is short of repaying the loan at the end of the 10 years. 39 SINKING FUND METHOD 361 Problem 39.13 A borrower takes out a loan of 3000 for 10 years at 8% convertible semiannually. The borrower replaces one-third of the principal in a sinking fund earning 5% convertible semiannually, and the other two-thirds in a sinking fund earning 7% convertible semiannually. Find the total semiannual payment. Problem 39.14 A payment of 36,000 is made at the end of each year for 31 years to repay a loan of 400,000. If the borrower replaces the capital by means of a sinking fund earning 3% eﬀective, ﬁnd the eﬀective rate paid to the lender on the loan. Problem 39.15 A loan of 30,000 is to be repaid using the sinking fund method over 6 years. The interest on the loan is paid at the end of each year and the interest rate is 10%. The sinking fund payment is made at the beginning of each year with the sinking fund earning 6%. Calculate the amount paid into the sinking fund each year. Problem 39.16 You have two equivalent ways to repay a loan of $100,000 over 20 years. (a) Using the sinking fund method, where the rate of interest on the loan is 8% and the rate of interest earned by the sinking fund is 6%. (b) Using the amortization method where the rate of interest is k. Find k. Problem 39.17 ‡ A 12-year loan of 8000 is to be repaid with payments to the lender of 800 at the end of each year and deposits of X at the end of each year into a sinking fund. Interest on the loan is charged at an 8% annual eﬀective rate. The sinking fund annual eﬀective interest rate is 4%. Calculate X. Problem 39.18 ‡ A 20-year loan of 20,000 may be repaid under the following two methods: i) amortization method with equal annual payments at an annual eﬀective rate of 6.5% ii) sinking fund method in which the lender receives an annual eﬀective rate of 8% and the sinking fund earns an annual eﬀective rate of j Both methods require a payment of X to be made at the end of each year for 20 years. Calculate j. Problem 39.19 ‡ Lori borrows 10,000 for 10 years at an annual eﬀective interest rate of 9%. At the end of each year, she pays the interest on the loan and deposits the level amount necessary to repay the principal to a sinking fund earning an annual eﬀective interest rate of 8%. The total payments made by Lori over the 10-year period is X. Calculate X. 362 LOAN REPAYMENT METHODS Problem 39.20 John borrows 10,000 for 10 years and uses a sinking fund to repay the principal. The sinking fund deposits earn an annual eﬀective interest rate of 5%. The total required payment for both the interest and the sinking fund deposit made at the end of each year is 1445.04. Calculate the annual eﬀective interest rate charged on the loan. Problem 39.21 Joe repays a loan of 10,000 by establishing a sinking fund and making 20 equal payments at the end of each year. The sinking fund earns 7% eﬀective annually. Immediately after the ﬁfth payment, the yield on the sinking fund increases to 8% eﬀective annually. At that time, Joe adjusts his sinking fund payment to X so that the sinking fund will accumulate to 10,000 20 years after the original loan date. Determine X. Problem 39.22 A corporation borrows 10,000 for 25 years, at an eﬀective annual interest rate of 5%. A sinking fund is used to accumulate the principal by means of 25 annual deposits earning an eﬀective annual interest rate of 4%. Calculate the sum of the net amount of interest paid in the 13 th installment and the increment in the sinking fund for the ninth year. Problem 39.23 A loan of 1,000 is taken out at an annual eﬀective interest rate of 5%. Level annual interest payments are made at the end of each year for 10 years, and the principal amount is repaid at the end of 10 years. At the end of each year, the borrower makes level annual payments to a sinking fund that earns interest at an annual eﬀective rate of 4%. At the end of 10 years the sinking fund accumulates to the loan principal. Calculate the diﬀerence between the interest payment on the loan and the interest earned by the sinking fund in the ﬁfth year. Problem 39.24 John borrows 10,000 for 10 years at an annual eﬀective interest rate ofi.He accumulates the amount necessary to repay the loan by using a sinking fund. He makes 10 payments of X at the end of each year, which includes interest on the loan and the payment into the sinking fund, which earns an annual eﬀective rate of 8%. If the annual eﬀective rate of the loan had been 2 i,his total annual payment would have been 1.5X. Calculate i. Problem 39.25 Jason and Margaret each take out a 17 year loan of L.Jason repays his loan using the amortization method, at an annual eﬀective interest rate of i. He makes an annual payment of 500 at the end of each year. Margaret repays her loan using the sinking fund method. She pays interest annually, also at an annual eﬀective interest rate of i. In addition, Margaret makes level annual deposits at 39 SINKING FUND METHOD 363 the end of each year for 17 years into a sinking fund. The annual eﬀective rate on the sinking fund is 4.62% and she pays oﬀ the loan after 17 years. Margaret’s total payment each year is equal to 10% of the original loan amount. Calculate L. Problem 39.26 A 10 year loan of 10,000 is to be repaid with payments at the end of each year consisting of interest on the loan and a sinking fund deposit. Interest on the loan is charged at a 12% annual eﬀective rate. The sinking fund’s annual eﬀective interest rate is 8%. However, beginning in the sixth year the annual eﬀective interest rate on the sinking fund drops to 6%. As a result, the annual payment to the sinking fund is then increased by X. Calculate X. Problem 39.27 NTL Corp. has just taken out a 15 year, 200,000 loan on which it has to make semi-annual interest payments of 7,000. As part of the loan agreement, NTL Corp. needs to make a deposit at the end of every 6 months into a sinking fund in order to retire the loan at the end of the 15th year. NTL Corp. plans to make semi-annual deposits of 4,929.98 into the sinking fund. (a) What eﬀective annual interest rate is NTL Corp. assuming it can earn on the sinking fund? (b) What is the net interest cost in the 6 months following the 10th sinking fund deposit? Problem 39.28 E Corp has a $100,000 loan outstanding on which it has been paying semi-annual interest payments of $4,000. E Corp has also been accumulating deposits made at the end of every 6 months in a sinking fund earning 5% eﬀective annual interest so that it can retire the loan at the end of the 15th year. The lender has oﬀered to accept 103% of the sinking fund balance immediately after the 29th deposit in exchange for the outstanding loan balance and the remaining loan interest payment. At what eﬀective annual rate was the lender calculating the present value of the remaining amounts in order to make this oﬀer equivalent in value? Problem 39.29 A borrower takes out a loan of $2,000 for two years. Create a sinking fund schedule if the lender re- ceives 10% eﬀective on the loan and if the borrower replaces the amount of the loan with semiannual deposits in a sinking fund earning 8% convertible quarterly. Problem 39.30 John borrows $10,000 for ﬁve years at 12% convertible semi-annually. John replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Find the total dollar amount that John must pay over the ﬁve-year period to completely repay the loan. 364 LOAN REPAYMENT METHODS Problem 39.31 A borrower is repaying a loan of 300,000 by the sinking fund method. The sinking fund earns an annual eﬀective interest rate of 6.75%. Payments of $22,520 are made at the end of each year for 20 years to repay the loan. These payments consist of both the interest payment to the lender and also the sinking fund deposit. What is the annual eﬀective interest rate paid to the lender of the loan? 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE365 40 Loans Payable at a Diﬀerent Frequency than Interest is Convertible In Sections 37 - 39, we considered loans where the payment period coincides with the interest conversion period. In this section we examine loans where the payments are made at a diﬀerent frequency than interest is convertible. The same observations made about the amortization method and the sinking fund method apply to the loans considered in this section. We will restrict our discussion to payments made at the end of an interest conversion period. A similar argument holds for payments made at the beginning of an interest conversion period. Consider ﬁrst the amortization schedule of a loan with payments made less frequently than interest is convertible. Consider a loan of an sk at interest rate i per period being repaid with payments of 1 at the end of each k interest conversion periods for a total of n interest conversion periods. Thus, the total number of payments is n k which we assume is an integral number. At the end of k interest conversion periods a payment of 1 is made, the interest paid is [(1 + i)k − 1]an sk = 1−νn so that the principal repaid is νn,and the outstanding loan balance is an sk −νn = an−k sk . Next, at the end of 2 k interest conversion periods, the interest paid is [(1 + i)k −1] an−k sk = 1 −νn−k so that the principal repaid is νn−k, and the outstanding loan balance is an−k sk −νn−k = an−2k sk . Continuing this process, we see that at the end of period mk, the interest paid is [(1 + i)k − 1] an−(m−1)k sk = 1 −νn−(m−1)k and the principal repaid is νn−(m−1)k.The outstanding loan balance is an−(m−1)k sk −νn−(m−1)k = an−mk sk . The amortization table is shown below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 an sk k 1 [(1 + i)k −1]an sk = 1 −νn νn an sk −νn = an−k sk 2k 1 [(1 + i)k −1] an−k sk = 1 −νn−k νn−k an−k sk −νn−k = an−2k sk ... ... ... ... ... mk 1 [(1 + i)k −1] an−(m−1)k sk = 1 −νn−(m−1)k νn−(m−1)k an−(m−1)k sk −νn−(m−1)k = an−mk sk ... ... ... ... ... n−k 1 [(1 + i)k −1] a2k sk = 1 −ν2k ν2k a2k sk −ν2k = ak sk n 1 [(1 + i)k −1] ak sk = 1 −νk νk ak sk −νk = 0 Total n k n k −an sk an sk 366 LOAN REPAYMENT METHODS Note that the principal repaid column is a geometric progression with common ratio (1 + i)k. Example 40.1 A loan of $15,000 is to be repaid by means of six annual payments of $3719.18 each. The nominal interest rate is 12% compounded monthly. Create an amortization schedule for this transaction. Solution. The schedule is given below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 15,000 12 3719.18 1902.37 1816.81 13,183.19 24 3719.18 1671.96 2047.22 11,135.97 36 3719.18 1412.32 2306.86 8829.11 48 3719.18 1119.75 2599.43 6229.68 60 3719.18 790.08 2929.10 3300.58 72 3719.18 418.60 3300.58 0 Next, consider a loan of a(m) n at interest rate i per period being repaid with payments of 1 m at the end of each mth of an interest conversion period for a total of n interest conversion periods. Thus, the total number of payments is mn which we assume is an integral number. At the end of the ﬁrst mth of an interest conversion period a payment of 1 m is made, the interest paid is i i(m) a(m) n = 1 m(1 −νn) so that the principal repaid is 1 mνn, and the outstanding loan balance is a(m) n −1 mνn = a(m) n−1 m .Next, at the end of second mth of an interest conversion period, the interest paid is i i(m) a(m) n−1 m = 1 m(1 −νn−1 m) so that the principal repaid is 1 mνn−1 m, and the outstanding loan balance is a(m) n−1 m −1 mνn−1 m = a(m) n−2 m .Continuing this process, we see that at the end of period t n, the interest paid is i i(m) a(m) n−(t−1) m = 1 m(1 −νn−t−1 m ) and the principal repaid is 1 mνn−(t−1) m .The outstanding loan balance is a(m) n−t−1 m −1 mνn−t−1 m = a(m) n−t m . The amortization table is shown below. 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE367 Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 a(m) n 1 m 1 m i i(m) a(m) n = 1 m(1 −νn) 1 mνn a(m) n −1 mνn = a(m) n−1 m 2 m 1 m i i(m) a(m) n−1 m = 1 m(1 −νn−1 m) 1 mνn−1 m a(m) n−1 m −1 mνn−1 m = a(m) n−2 m ... ... ... ... ... t m 1 m i i(m) a(m) n−(t−1) m = 1 m(1 −νn−t−1 m ) 1 mνn−(t−1) m a(m) n−t−1 m −1 mνn−t−1 m = a(m) n−t m ... ... ... ... ... n−1 m 1 m i i(m) a(m) 2 m = 1 m(1 −ν 2 m) 1 mν 2 m a(m) n−2 m −1 mνn−2 m = a(m) n−1 m n 1 m i i(m) a(m) 1 m = 1 m(1 −ν 1 m) 1 mν 1 m a(m) n−1 m −1 mνn−1 m = a(m) n−1 m −1 mν 1 m = 0 Total n n−a(m) n a(m) n Note that the principal repaid column is a geometric progression with common ratio (1 + i) 1 m. Example 40.2 A debt is being amortized by means of monthly payments at an annual eﬀective rate of interest of 11%. If the amount of principal in the third payment is $1000, ﬁnd the amount of principal in the 33rd payment. Solution. Recall that the principal repaid column is a geometric progression with common ratio (1+ i) 1 m.The interval of time from the 3rd payment to the 33rd payment is 33−3 12 = 2.5 years. Thus, the principal in the 33rd payment is 1000(1.11)2.5 = $1298.10 Example 40.3 A loan of $10,000 is to be repaid by means of twelve monthly payments. The nominal interest rate is 12% compounded quarterly. Create an amortization schedule for this transaction. Solution. Let Rbe the amount of monthly payment. Then Rsatisﬁes the equation 10000 = 3Ra4 (3) 0.03.Solving this equation for R we ﬁnd R= $887.94. Moreover, we have i(3) 3 a4 (3) 0.03 = 0.02970490211 3 = 99.02. The schedule is given below. 368 LOAN REPAYMENT METHODS Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 10,000 1 3 887.94 99.02 788.92 9211.08 2 3 887.94 91.20 796.74 8414.34 1 887.94 83.32 804.62 7609.72 4 3 887.94 75.35 812.59 6797.13 5 3 887.94 67.30 820.64 5976.49 2 887.94 59.18 828.76 5147.73 7 3 887.94 50.97 836.97 4310.76 8 3 887.94 42.68 845.26 3465.50 3 887.94 34.31 853.63 2611.87 10 3 887.94 25.86 862.08 1749.79 11 3 887.94 17.33 870.61 879.18 4 887.94 8.71 879.23 0 It is not recommended that the reader depends upon the memorization of formulas in amortization and sinking fund schedules. It is rather more preferable to create these schedules using the basic principles. We illustrate the use of basic principles in creating a sinking fund schedule in the next example. Example 40.4 John borrows $2000 for two years at an annual eﬀective interest rate of 10%. He replaces the principal by means of semiannual deposits for two years in a sinking fund that earns 8% convertible quarterly. Create a sinking fund schedule for this transaction. Solution. The interest payment on the loan is 2000 ×10% = $200 at the end of each year. If R is the semiannual deposit in the sinking fund then Rs8 0.02 s2 0.02 = 2000. Solving this equation for R we ﬁnd R= 2000s2 0.02 s8 0.02 = $470.70. The sinking fund schedule is given below. 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE369 Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1/4 0 0 0 0 2000.00 1/2 0 470.70 0 470.70 1529.30 3/4 0 0 9.41 480.11 1519.89 1 200.00 470.70 9.60 960.41 1039.59 1 1/4 0 0 19.21 979.62 1020.38 1 1/2 0 470.70 19.59 1469.91 530.09 1 3/4 0 0 29.40 1499.31 500.69 2 200.00 470.70 29.99 2000.00 0 370 LOAN REPAYMENT METHODS Practice Problems Problem 40.1 A debt is being amortized by means of annual payments at a nominal interest rate of 12% com- pounded semiannually. If the amount of principal in the third payment is $1000, ﬁnd the amount of principal in the 33rd payment. Problem 40.2 A borrower is repaying a loan by making quarterly payments of $500 over 10 years. Portion of each payment is principal and the rest is interest. Assume an annual eﬀective rate of 8%, how much interest is paid by the borrower over the 10-year period? Problem 40.3 A lender receives payments of $3000 at the end of every year over an unknown number of years. If the interest portion of the third payment is $2000, ﬁnd the amount of principal in the sixth payment. Assume a nominal interest rate of 10% payable quarterly. Problem 40.4 A borrows $10,000 for ﬁve years at 12% convertible semiannually. A replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Find the total dollar amount which A must pay over the ﬁve-year period to completely repay the loan. Problem 40.5 A borrows $5000 for 10 years at 10% convertible quarterly. A does not pay interest currently and will pay all accrued interest at the end of 10 years together with the principal. Find the annual sinking fund deposit necessary to liquidate the loan at the end of 10 years if the sinking fund earns 7% convertible quarterly. Problem 40.6 A borrows $10,000 for ﬁve years at 12% convertible semiannually. A replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Create a sinking fund schedule for this loan. 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 371 41 Amortization with Varying Series of Payments In this section, we consider amortization methods with more general patterns of payments, not necessarily leveled. We will keep assuming that the interest conversion period and the payment period are equal and coincide. Consider a loan L to be repaid with n periodic payments (that include principal and interest) R1,R2,··· ,Rn. The equation of value at time t= 0 is L= n∑ t=1 νtRt. In most cases the series of payments Rt follows some regular pattern encountered with annuties so that the results of Section 26 can be used. Example 41.1 A borrower is repaying a loan with payments at the end of each year for 10 years, such that the payment the ﬁrst year is $200, the second year is $190, and so forth, until the 10th year it is $110. Find an expression of the amount of the loan. Solution. The amount of the loan is L= 100a10 + 10(Da)10 For the type of installments considered in this section, amortization schedules can be constructed from ﬁrst principles as discussed in Section 38. Furthermore, the outstanding loan balance column can be found retrospectively or prospectively as in Section 37, from which the remaining columns of interest paid and principal repaid can be found. Example 41.2 A loan is repaid with payments which start at $200 the ﬁrst year and increase by $50 per year until a payment of $1000 is made, at which time payments cease. If the interest is 4% eﬀective, ﬁnd the amount of interest and principal paid in the fourth payment. Solution. The fourth payment is R4 = $350.00.The outstanding loan balance after making the third payment is Bp 3 = 300a14 + 50(Ia)14 . 372 LOAN REPAYMENT METHODS The interest paid after the fourth payment is I4 =iBp 3 = 300(1 −ν14) + 50(¨a14 −14ν14) =300(1 −0.57747) + 50(10.98565 −8.08465) =$271.81 The principal portion of the fourth payment is P4 = R4 −I4 = 350.00 −271.81 = $78.19 One common pattern of periodic installments is when the borrower makes level payments of princi- pal. Clearly, this leads to successive total payments (consisting of interest and principal) to decrease due to the decrease in successive outstanding loan balance (which in turn results in a decrease in successive interest paid). We illustrate this in the following example. Example 41.3 A borrower is repaying a $1000 loan with 10 equal payments of principal. Interest at 6% compounded semiannually is paid on the outstanding balance each year. Find the price to yield an investor 10% convertible semiannually. Solution. The period principal payment is $100 each. The periodic interest payments are :30 ,27,24,··· ,9,3. Thus, the price at time t= 0 is just the present value of the principal and interest paid. That is, Price = 100a10 0.05 + 3(Da)10 0.05 = 908.87 It is possible that when using varying payments, the loan payment for a period is less than the interest charged over that period. This leads to an increase in the outstanding balance which in turn means a negative principal. In ﬁnance, such a case is referred to as negative amortization or deferred interest . Credit card interest rates and payment plans are examples of negative amortization. Most credit cards carry high interest rates yet require a low minimum payment. Example 41.4 A loan of 1000 is being repaid with annual payments over 10 years. The payments in the last ﬁve years are 5 times the payments in the ﬁrst 5 years. If i= 0.08,calculate the principal amortized in the ﬁfth payment. Solution. Let K be the size of the payment in the ﬁrst 5 years. Then 1000 = K(1 −1.08−5) 0.08 + 5K(1.08−5)(1 −1.08−5) 0.08 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 373 or 1000 = 17.57956685K →K = 56.88422296. The outstanding loan balance after making the 4th payment is 56.88422296 1.08 + 5(56.88422296)(1.08−1)(1 −1.08−5) 0.08 = 1104.16228. The outstanding loan balance after making the 5th payment is 5(56.88422296)(1 −1.08−5) 0.08 = 1135.61104. The principal amortized in the ﬁfth payment is 1104 .16228 −1135.61104 = −$31.45 Next, consider a varying series of payments with the sinking fund method. We will assume that the interest paid to the lender is constant each period so that only the sinking fund deposits vary. Let the varying payments be denoted by R1,R2,··· ,Rn and assume that i ̸= j. Let L denote the amount of the loan. Then the sinking fund deposit for the tth period is Rt −iL. Since the accumulated value of the sinking fund at the end of n periods must be L, we have L=(R1 −iL)(1 + j)n−1 + (R2 −iL)(1 + j)n−2 + ··· + (Rn −iL) = n∑ t=1 Rt(1 + j)n−t −iLsnj where i is rate of interest paid on the loan and j rate of interest earned on the sinking fund. Solving for L we obtain L= ∑n t=1 Rt(1 + j)n−t 1 + isnj = ∑n t=1 νt jRt 1 + (i−j)anj . Note that if i= jthen L= ∑n t=1 Rtνt.Thus, the amortization method and the sinking fund method are equivalent. Example 41.5 A borrower is repaying a loan at 5% eﬀective with payments at the end of each year for 10 years, such that the payment the ﬁrst year is $200, the second year is $190, and so forth, until the 10th year it is $110. Find the amount of the loan if the borrower pays 6% eﬀective on the loan and accumulates a sinking fund to replace the amount fo the loan at 5% eﬀective. Solution. We have L= 100a10 0.05 + 10(Da)10 0.05 1 + (0.06 −0.05)a10 0.05 = 1227.83 1 + (0.01)(7.7217) = $1139.82 374 LOAN REPAYMENT METHODS Practice Problems Problem 41.1 Aborrows $10,000 from B and agrees to repay it with a series of 10 installments at the end of each year such that each installment is 20% greater than the preceding installment. The rate of interest on the loan is 10% eﬀective. Find the amount of principal repaid in the ﬁrst three installments. Problem 41.2 John borrows 10,000 from Tony and agrees to repay the loan with 10 equal annual installments of principal plus interest on the unpaid balance at 5% eﬀective. Immediately after the loan is made, Tony sells the loan to Chris for a price which is equal to the present value of John’s future payments calculated at 4%. Calculate the price that Chris will pay for the loan. Problem 41.3 A loan is being repaid with 10 payments of 1000 t at the end of year t. (In other words, the ﬁrst payment is 1000, the second payment is 2000, etc.) The interest rate on the loan is 6%. Calculate the outstanding principal immediately after the third payment. Problem 41.4 A loan of 100,000 is being repaid with annual payments at the end of each year for 10 years. The interest rate on the loan is 10.25%. Each annual payment increases by 5% over the previous annual payment. Calculate the principal in the ﬁfth payment. Problem 41.5 ‡ A 30-year loan of 1000 is repaid with payments at the end of each year. Each of the ﬁrst ten payments equals the amount of interest due. Each of the next ten payments equals 150% of the amount of interest due. Each of the last ten payments is X.The lender charges interest at an annual eﬀective rate of 10%. Calculate X. Problem 41.6 ‡ A loan is repaid with 20 increasing annual installments of 1,2,3, ··· ,20. The payments begin one year after the loan is made. Find the principal contained in the 10th payment, if the annual interest rate is 4%. Problem 41.7 In order to repay a school loan, a payment schedule of 200 at the end of the year for the ﬁrst 5 years, 1200 at the end of the year for the next 5 years, and 2200 at the end of the year for the ﬁnal 5 years is agreed upon. If interest is at the annual eﬀective rate of i= 6%, what is the loan value? 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 375 Problem 41.8 Create an amortization schedule of the previous problem for the ﬁrst seven years. Problem 41.9 A borrows $20,000 from B and agrees to repay it with 20 equal annual installments of principal in addition to the interest on the unpaid balance at 3% eﬀective. After 10 years B sells the right to future payments to C, at a price which yields C 5% eﬀective over the next 5 years and 4% eﬀective over the ﬁnal 5 years. Find the price which C should pay to the nearest dollar. Problem 41.10 A loan of 3000 at an eﬀective quarterly interest rate ofj = .02 is amortized by means of 12 quarterly payments , beginning one quarter after the loan is made. Each payment consists of a principal repayment of 250 plus interest due on the previous quarter’s outstanding balance. Construct the amortization schedule. Problem 41.11 A loan is being repaid with 10 annual payments. The ﬁrst payment is equal to the interest due only, the second payment is twice the ﬁrst, the third payment is three time the ﬁrst, and so forth. Prove that at the rate of interest on the loan is: ( Ia)10 = a∞. Problem 41.12 A loan is being repaid with 10 payments. The ﬁrst payment is 10, the second 9, the third is 8, and so forth with the tenth payment being 1. Find an expression for the amount of interest in the sixth payment. Problem 41.13 A has money invested at eﬀective rate i. At the end of the ﬁrst year A withdraws 162.5% of the interest earned; at the end of the second year Awithdraws 325% of the interest earned, and so forth with the withdrawal factor increasing in arithmetic progression. At the end of 16 years the fund is exhausted. Find i. Problem 41.14 A loan is amortized over ﬁve years with monthly payments at a nominal interest rate of 9% com- pounded monthly. The ﬁrst payment is 1000 and is to be paid one month from the date of the loan. Each succeeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made. Problem 41.15 June borrows 20,000 from April and agrees to repay it with a series of ten installments at the end 376 LOAN REPAYMENT METHODS of each year, such that each installment is 15% greater than the preceding installment. The rate of interest on the loan is 10% annual eﬀective. Let Pj denote the amount of principal repaid in year j. Calculate P1 + P2. Problem 41.16 Smith borrows 30,000 at an annual eﬀective interest rate of 10%. He agrees to make annual payments at the end of each year for 9 years, and an additional balloon payment, X, at the end of the tenth year. Each of the ﬁrst 9 payments equals 25% more than what is owed in interest at the time of the payment. The balloon payment, X, equals the amount needed to pay oﬀ the loan. Determine X. Problem 41.17 A loan is being repaid by the amortization method using 10 semiannual payments with the ﬁrst payment due six months after the loan is made. The ﬁrst payment is 50 and each subsequent payment is 50 more than its previous payment. The interest rate on the loan is 10%, compounded semiannually. Determine the amount of principle repaid in the seventh payment. Problem 41.18 A 10 year loan with an eﬀective annual interest rate of 5% is to be repaid with the following payments (1) 100 at the end of the second year (2) 200 at the end of the fourth year (3) 300 at the end of the sixth year (4) 400 at the end of the eighth year (5) 500 at the end of the tenth year Calculate the amount of interest included in the second payment. Problem 41.19 John takes out a 10 year loan. The loan is repaid by making 10 annual repayments at the end of each year. The ﬁrst loan repayment is equal to X, with each subsequent repayment 10.16% greater than the previous year’s repayment. The annual eﬀective interest rate being charged on the loan is 8%. The amount of interest repaid during the ﬁrst year is equal to 892.20. Calculate X. Problem 41.20 Don takes out a 10 year loan of L which he repays with annual payments at the end of each year using the amortization method. Interest on the loan is charged at an annual eﬀective rate of i.Don repays the loan with a decreasing series of payments. He repays 1000 in year one, 900 in year two, 800 in year three, ··· , and 100 in year ten. The amount of principal repaid in year three is equal to 600. Calculate L. 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 377 Problem 41.21 ‡ Betty borrows 19,800 from Bank X.Betty repays the loan by making 36 equal payments of principal at the end of each month. She also pays interest on the unpaid balance each month at a nominal rate of 12%, compounded monthly. Immediately after the 16th payment is made, Bank X sells the rights to future payments to Bank Y. Bank Y wishes to yield a nominal rate of 14%, compounded semi-annually, on its investment. What price does Bank X receive? Problem 41.22 Two loans for equal amounts are repaid at an eﬀective interest rate of 4%. Loan A is repaid with 30 equal annual payments. Loan B is to be repaid by 30 annual payments, each containing equal principal amounts and an interest amount based on the unpaid balance. Payments are made at the end of each year. The annual payment for Loan A ﬁrst exceeds the annual payment for Loan B with the kth payment. Find k. Problem 41.23 A borrows $2000 at an eﬀective rate of interest of 10% per annum and agrees to repay the loan with payments at the end of each year. The ﬁrst payment is to be $400 and each payment thereafter is to be 4% greater than the preceding payment, with a smaller ﬁnal payment made one year after the last regular payment. (a) Find the outstanding loan balance at the end of three years. (b) Find the principal repaid in the third payment. 378 LOAN REPAYMENT METHODS Bonds and Related Topics This chapter is about bonds and the valuation of bonds. We will learn how to determine the price of a bond that should be paid by an investor for a desired given yield rate, determine the yield rate given the price, how a bond is amortized, and ﬁnally determine the value of a bond on a given date after it has been purchased. 379 380 BONDS AND RELATED TOPICS 42 Types of Bonds Perhaps the simplest way for a company or a government agency to raise cash is for them to sell bonds to the public. A bond is an interest bearing security which promises to pay a stated amount (or amounts) of money at some future date (or dates). The company or government branch which is issuing the bond outlines how much money it would like to borrow and speciﬁes a length of time, along with an interest rate it is willing to pay. Investors who then lend the requested money to the issuer become the issuer’s creditors through the bonds that they hold. The term of the bond is the length of time from the date of issue until the date of ﬁnal payment. The date of the ﬁnal payment is called the maturity date. Bonds with an inﬁnite term are called perpetuals. The detailed conditions of the bond may permit the bond to be called at some date prior to the maturity date, at which time the issuer (i.e., the lender) will repay the commitment, possibly with some additional amounts. Such a bond is callable. Any date prior to, or including, the maturity date on which a bond may be redeemed is termed a redemption date. The par valueor face valueof a bond is the amount that the issuer agrees to repay the bondholder by the maturity date. There are several classiﬁcation of bonds: •Accumulation bond is one in which the redemption price includes the original loan plus accu- mulated interest. Examples of such bonds are the Series E Savings bonds. •Bonds with coupons are periodic payments of interest made by the issuer of the bond prior to redemption. It is called a “coupon” because some bonds literally have coupons attached to them. Holders receive interest by stripping oﬀ the coupons and redeeming them. This is less common today as more records are kept electronically. In what follows, we we use the term “bond” we mean bonds with coupons. Zero coupon bonds are bonds that pay no periodic interest payments. It just pays a lump sum at redemption date. •Registered and Unregistered bonds. A registered bond is a bond issued with the name of the owner printed on the face of the certiﬁcate. If the owner decides to sell the bond, the change must be reported to the borrower. The coupon payments are paid by the borrower to the owners of record on each coupon payment date. An unregistered bond or bearer bond is one in which the lender is not listed in the records of the borrower. In this case, the bond belongs to whomever has legal possession of it. Again, these bonds are occasionally called coupon bonds, due to the physically attached coupons. •Fixed-rate and ﬂoated-rate bonds. A ﬁxed-rate bond is a bond that has a ﬁxed rate over the term of the bond. On the other hand, a bond that has a ﬂuctuated interest rate over the term of the bond is called a ﬂoating-rate bond. •Mortgage and debenture bonds. A mortgage bond is a more secured bond backed by a 42 TYPES OF BONDS 381 collateral such as a mortgage on a property. Mortgage bonds are backed by real estate or physical equipment that can be liquidated. A debenture bond is an unsecured bond issued by a civil or governmental corporation or agency and backed only by the credit standing of the issuer. •Income or adjustment bonds. They are a type of high risk bonds in which coupons are paid only if the borrower has earned suﬃcient income to pay them. •Junk bonds. A junk bond is a high-risk bond of default in payments. The risk that a bond issuer does not pay the coupon or principal payments is called default risk. Because of this risk, these bonds typically pay higher yields than better quality bonds in order to make them attractive to investors. •Convertible bond. A convertible bond is a bond that can be converted into the common stock of the company at the option of the bond owner. The owner of the bond is compensated with the ability to convert the bond to common stock, usually at a substantial discount to the stock’s market value. •Serial bonds. A set of bonds issued at the same time but having diﬀerent maturity dates. These are used when a borrower is in need of a large amount of money. •Treasury bonds. Issued by the US Treasury. Terms of seven or more years. •Treasury bills. Short term debt with maturities of 13, 26, or 52 weeks. T-bills yields are com- puted as rates of discount. These yields are computed on a simple discount basis. The basis for counting time periods for T-bills is actual/360. •Municipal bonds. These are bonds issued by state and local governments to ﬁnance large, long−term capital projects (e.g., hospitals, highways, schools). Example 42.1 A 10 year bond matures for its par value of 5000. The price of the bond is 4320.48 at an 8% yield convertible semi-annually. Calculate the coupon rate convertible semi-annually. Coupon payment is deﬁned to be the product of the face value and the coupon periodic rate. Also, the price of a bond is the present value of all future payments. Solution. Let k be the semiannual eﬀective coupon rate. Then 4320.48 = 5000 1.0420 + 5000k(1 −1.04−20) 0.04 or 5000k(1 −1.04−20) 0.04 = 2038.545269 Thus, k= 2038.545269 0.04 [5000(1 −1.04−20)] = 0.029999 382 BONDS AND RELATED TOPICS the coupon rate convertible semi-annually is 2 k= 0.059999 = 6.00% Example 42.2 A $1000 par value bond bearing a 6% coupon rate payable semi-annually will be redeemed at 105% at the end of 15 years. Find the price to yield an investor 5% eﬀective. Solution. The price is P = 30a30 j + 1050(1.05)−15 = $1135.54 where j = (1.05) 1 2 −1 Example 42.3 A 10-year accumulation bond with an initial par value(i.e. face value) of 1000 earns interest of 8% compounded semiannually. Find the price to yield an investor 10% eﬀective. Solution. The only return payment is at maturity. The price to yield 10% interest will therefore be (1.10)−10[1000(1.04)20] = $844.77 Example 42.4 Find the price which should be paid for a zero coupon bond which matures for 1000 in 10 years to yield: (a) 10% eﬀective (b) 9% eﬀective (c) Thus, a 10% reduction in the yield rate causes the price to increase by what percentage? Solution. (a) The bond is now worth 1000(1 .10)−10 = $385.54 (b) The bond is now worth 1000(1 .09)−10 = $422.41 (c) The 10% decrease in the interest rate thereby increases the price by 422.41 −385.54 385.54 = 9.6% 42 TYPES OF BONDS 383 Practice Problems Problem 42.1 A zero bond will pay $1000 at the end of 10 years, and is currently selling for $500. What is the annual return earned by the purchaser? Problem 42.2 Steve purchases a 10-year zero-coupon bond for $400 and has par value of $1,000. Find the yield rate convertible semiannually that would earned by Steve. Problem 42.3 John buys a 13-week T-bill with par value $10,000 at a simple discount rate of 7.5%. What price did he pay for the bond? Problem 42.4 An investor purchases a 26-week T-bill with face value of $10,000 and with a discount yield of dfor $9,600. (a) Find d. (b) Even though T-bills are not usually quoted on an annual eﬀective yield basis, the annual eﬀective yield can still be determined. Find the annual eﬀective rate of interest of this investment, assuming the investment period is exactly half a year. Problem 42.5 An investor purchases a 26-week T-bill with face value $10,000 for $X and earns an annual eﬀective rate of 4.17%. Determine X. Problem 42.6 An investor buys a zero-coupon bond with par value $1,000 for $493.63. The maturity date is Y years and the yield rate convertible semi-annually is 8%. Calculate Y. Problem 42.7 A 20 year bond matures for its par value of 10,000. The coupon payable semi-annually is 400. Calculate the price of the bond at a 6% yield rate convertible semi-annually. Problem 42.8 A 20 year bond matures for its par value of 10,000. The coupon payable semi-annually is 400. Calculate the price of the bond at a 6% annual eﬀective yield rate. Problem 42.9 A $1000 par value bond bearing a 5% coupon rate payable semi-annually will be redeemed at 108% at the end of 7 years. Find the price to yield an investor 5% convertible semi-annually. 384 BONDS AND RELATED TOPICS 43 The Various Pricing Formulas of a Bond In this section we consider the question of determining the purchase price which will produce a given yield rate to an investor. We have already encountered this question in the previous section. When considering the price of a bond, we make the following assumptions: (1) All obligations will be paid by the bond issuer on the speciﬁed dates of payments. Any possi- bility of default will be ignored. (2) The bond has a ﬁxed maturity date. (3) The price of the bond is decided immediately after a coupon payment date, or alternatively, at issue date if the bond is brand new. The following symbols and notation will be used in connection with bonds valuation: P = the price paid for a bond. F = The par value or face value or face amount. This amount is is usually printed on the front of the bond and is often the amount payable at the maturity date. C = the redemption value of a bond, i.e. the amount of money paid at a redemption date to the holder of the bond. r = The coupon rate is the eﬀective rate per coupon payment period used in determining the amount of the coupon. The default payment period is a half-year. For example, r= 0.035 for a 7% nominal coupon paid semi- annually. Fr = The amount of a coupon payment. g = The modiﬁed coupon rate. It is deﬁned by g= Fr C . Thus, g is the coupon rate per unit of redemption value, rather than per unit of par value. i = The yield rate of a bond, or the yield to maturity. The interest rate actually earned by the investor, assuming the bond is held until redemption or maturity. n = The number of coupon payment periods from the date of calculation until maturity date or redemption date K = the present value, computed at the yield rate, of the redemption value at the maturity date, or a redemption date, i.e. K = Cνn at the yield rate i. G = The base amount of a bond. It is deﬁned by Gi= Fr or G= Fr i . Thus, G is the amount which, if invested at the yield rate i, would produce periodic interest payments equal to the coupons on the bond. The quantities F,C,r,g, and n are given by the terms of a bond and remain ﬁxed throughout the bonds life. On the other hand, P and i will vary throughout the life of the bond. Price and yield rate have a precise inverse relationship to each other,i.e. as the price rises the yield rate falls and 43 THE VARIOUS PRICING FORMULAS OF A BOND 385 vice-versa. There are three diﬀerent quoted yields associated with a bond: (1) Nominal yield is the ratio of annualized coupon payment to par value. For example, 2 coupons per year of $3.50 on a $100 par value bond result in a 7.00% nominal yield. The reader should note that the use of the word “nominal” in this context is diﬀerent than the one used in Section 9. (2) Current yield is the ratio of the annualized coupon payment to the original price of the bond. For example, if you paid $90 per $100 of par value of the bond described above, the current yield would be $7.00 $90 = 7.78% (3) Yield to maturity is the actual annualized yield rate, or internal rate of return. Four formulas for price Like loans, the price of a bond is deﬁned to be the present value of all future payments. We shall consider four diﬀerent ways of determining the present value, or price of a bond. The formulas are derivable one from the other: there are situations where one may be more useful than another for speciﬁc applications. The ﬁrst of these is called the basic formula. It is found by writing the equation of value at time t= 0 of the time diagram given in Figure 43.1 and is given by P = Frani + Cνn i = Frani + K. Figure 43.1 This formula has a verbal interpretation: the price of a bond is equal to the present value of future coupons plus the present value of the redemption value. The second formula is the premium/discount formula, and is derived from the basic formula: P =Frani + Cνn i =Frani + C(1 −iani) =C+ (Fr −Ci)ani This says that the price of a bond is the sum of the redemption value and the present value of amortization of premium or discount paid at purchase. Amortization of premium and discount will be discussed in the next section. 386 BONDS AND RELATED TOPICS The third, the base amount formula, can also be derived from the ﬁrst formula as follows: P =Frani + Cνn i =Giani + Cνn i =G(1 −νn i ) + Cνn i =G+ (C−G)νn i This says that the price is the sum of the base amount and the present value of the diﬀerence between base amount and redemption value. The fourth, the Makeham formula, is also obtained from the basic formula: P =Cνn i + Frani =Cνn i + Cg (1 −νn i i ) =Cνn i + g i(C−Cνn i ) =K+ g i(C−K) Example 43.1 A 10-year 100 par value bond bearing a 10% coupon rate payable semi-annually, and redeemable at 105, is bought to yield 8% convertible semiannually. Find the price. Verify that all four formulas produce the same answer. Solution. We are given the following: F = 100 C = 105 r = 5% Fr = 5 g= Fr C = 1 21 i = 4% n = 20 K = Cνn = 105(1 .04)−20 = 47.92 G= Fr i = 125 Using the basic formula we ﬁnd P = Fran + K = 5a20 0.04 + 47.92 = 5(13.59031) + 47.92 = $115.87 43 THE VARIOUS PRICING FORMULAS OF A BOND 387 Using the premium/discount formula we ﬁnd P = C+ (Fr −Ci)an = 105 + (5−4.2)a20 0.04 = $115.87 Using the base amount formula we ﬁnd P = G+ (C−G)νn = 125 + (105−125)(1.04)−20 = $115.87 Using the Makeham formula we ﬁnd P = K+ g i(C−K) = 47.92 + 100 84 (105 −47.92) = $115.87 Example 43.2 For the bond of the previous example, determine the following: (a) Nominal yield based on the par value. (b) Nominal yield, based on the redemption value. (c) Current yield. (d) Yield to maturity. Solution. (a) The nominal yield is the annualized coupon value to the par value,i.e. 2Fr F = 10 100 = 10%. (b) This is the annualized modiﬁed coupon rate 2g= 2Fr C = 2 5 105 = 9.52% (c) Current yield is the ratio of annualized coupon to price 10 115.87 = 8.63% (d) Yield to maturity is the annualized actual yield which is 8% In the problems considered thus far, we have assumed that coupon payments are constant. It is possbile to have bonds with varying coupon payments. In this case, the coupons will constitute a varying annuity which can be evaluated using the approach of Sections 26 - 27. Again, the price of the bond is the present value of all future coupon payments plus the present value of the redemption value. Example 43.3 A 1000 par value 20 year bond with annual coupons and redeemable at maturity at 1050 is purchased for P to yield an annual eﬀective rate of 8.25%. The ﬁrst coupon is 75. Each subsequent coupon is 3% greater than the preceding coupon. Determine P. 388 BONDS AND RELATED TOPICS Solution. The coupon payments constitute a annuity that is varying in geometric progression. The present value of this annuity is 75 1 − (1+k 1+i )n i−k = 75 1 − ( 1.03 1.0825 )20 0.0825 −0.03 = 900.02. The present value of the redemption value is 1050(1.0825)−20 = 215.09. Thus, the price of the bond is 900.02 + 215.09 = 1115.11 It is possible that the bond yield rate is not constant over the term of the bond. We illustrate this in the following example. Example 43.4 Find the price of a 1000 par value 10-year bond with coupons of 8.4% convertible semi-annually, which will be redeemed at 1050. The bond yield rate is 10% convertible semi-annually for the ﬁrst ﬁve years and 9% convertible semi-annually for the second ﬁve years. Solution. The present value of all future coupons is 42[a10 0.05 + (1.05)−10a10 0.045] = 528.33. The present value of the redemption value is 1050(1.05)−10(1.045)−10 = 415.08. Thus, the price of the bond is 528.33 + 415.08 = 943.41 Next, we consider cases when the yield rate and coupon rate are at diﬀerent frequencies. Consider ﬁrst the case in which each coupon period contains k yield rate conversion periods. Suppose the total number of yield rate conversion periods over the term of the bond is n. There is a coupon payment of Fr paid at the end of each k yield rate conversion periods. The present value of all future coupon payments is Fran sk and the present value of the redemption value is Cvn. Thus, the price of the bond is P = Fran sk + Cvn. 43 THE VARIOUS PRICING FORMULAS OF A BOND 389 Example 43.5 Find the price of a $1000 par value 10-year bond maturing at par which has coupons at 8% con- vertible semi-annually and is bought to yield 6% convertible quarterly. Solution. The price of the bond is P = 40a40 0.015 s2 0.015 + 1000(1.015)−40 = 1145.12 The second case is when each yield rate conversion period contains m coupon conversion periods. Again, we let n be the total number of yield rate conversion periods. There is a coupon of Fr m paid at the end of each mth of a conversion period. Thus, the present value of the future coupon payments is Fra(m) n and the present value of the redemption value is Cνn. Thus, the price of the bond is P = Fra(m) n + Cνn. Example 43.6 Find the price of a $1000 par value 10-year bond maturing at par which has coupons at 8% con- vertible quarterly and is bought to yield 6% convertible semi-annually. Solution. The price of the bond is P =40a(2) 20 + 1000(1.03)−20 =40 1 −1.03−20 2[(1.03) 1 2 −1] + 1000(1.03)−20 =1153.21 390 BONDS AND RELATED TOPICS Practice Problems Problem 43.1 A 10-year $1,000 par value bond bearing a 8.4% coupon rate convertible semiannually and re- deemable at $1,050 is bought to yield 10% convertible semiannually. Find the price. Use all four formulas. Problem 43.2 For the bond of the previous problem, determine the following: (a) Nominal yield based on the par value. (b) Nominal yield, based on the redemption value. (c) Current yield. (d) Yield to maturity. Problem 43.3 A 5-year $100 par value bond bearing a 8% coupon rate payable semi-annually is selling at par value. If prevailing market rates of interest suddenly go to 10% convertible semiannually, ﬁnd the percentage change in the price of the bond. Problem 43.4 Two 1000 bonds redeemable at par at the end of the same period are bought to yield 4% convertible semiannually. One bond costs $1136.78, and has a coupon rate of 5% payable semiannually. The other bond has a coupon rate of 2.5% payable semiannually. Find the price of the second bond. Problem 43.5 A n−year $1000 par value bond bearing a 9% coupon rate payable semiannually and redeemable at $1125 is bought to yield 10% convertible semiannually. Find the purchase price of the bond if the present value of the redemption value is $225 at the given yield rate. Problem 43.6 A n−year $1000 par value bond maturing at par and with semi-annual coupon of $50 is purchased for $1110. Find the base amount G if the redemption value is $450. Problem 43.7 An investor owns a 1000 par value 10% bond with semiannual coupons. The bond will mature at par at the end of 10 years. The investor decides that an 8-year bond would be preferable. Current yield rates are 7% convertible semiannually. The investor uses the proceeds from the sale of the 10% bond to purchase a 6% bond with semiannual coupons, maturing at par at the end of 8 years. Find the par value of the 8-year bond. 43 THE VARIOUS PRICING FORMULAS OF A BOND 391 Problem 43.8 An n−year 1000 par value bond matures at par and has a coupon rate of 12% convertible semian- nually. It is bought at a price to yield 10% convertible semiannually. If the term of the bound is doubled, the price will increase by 50. Find the price of the n−year bond. Problem 43.9 A 20 year bond with a par value of $10,000 will mature in 20 years for $10,500. The coupon rate is 8% convertible semiannually. Calculate the price that Andrew would pay if he bought the bond to yield 6% convertible twice a year. Problem 43.10 After 5 years (immediately after the 10th coupon is paid), Andrew decides to sell the bond in the previous problem. Interest rates have changed such that the price of the bond at the time of the sale is priced using a yield rate of 9% convertible semiannually. Calculate the selling price. Problem 43.11 A bond matures for its par value of 1000 in one year. The bond pays a annual coupon on 80 each year. The price of the bond is 800. Which of the following are true? i. The nominal yield is 8% per annum. ii. The current yield is 10% per annum. iii. The yield to maturity is 35% per annum Problem 43.12 You paid $950 for a 7 year bond with a face value of $1,000 and semi-annual coupons of $25. What is your current yield on this bond and what is your nominal annual yield to maturity (convertible semi-annually)? Problem 43.13 A 15-year bond with a coupon of $X payable every 6 months has a face (and redemption) value of $10,000. At the nominal annual interest rate, convertible semi-annually, of 6.5%, the price of the bond is $8,576.36. What is X? Problem 43.14 ‡ A ten-year 100 par value bond pays 8% coupons semiannually. The bond is priced at 118.20 to yield an annual nominal rate of 6% convertible semiannually. Calculate the redemption value of the bond. 392 BONDS AND RELATED TOPICS Problem 43.15 ‡ Susan can buy a zero coupon bond that will pay 1000 at the end of 12 years and is currently selling for 624.60 . Instead she purchases a 6% bond with coupons payable semi-annually that will pay 1000 at the end of 10 years. If she pays X she will earn the same annual eﬀective interest rate as the zero coupon bond. Calculate X. Problem 43.16 ‡ A 30-year bond with a par value of 1000 and 12% coupons payable quarterly is selling at 850. Calculate the annual nominal yield rate convertible quarterly. Problem 43.17 ‡ An investor borrows an amount at an annual eﬀective interest rate of 5% and will repay all interest and principal in a lump sum at the end of 10 years. She uses the amount borrowed to purchase a 1000 par value 10-year bond with 8% semiannual coupons bought to yield 6% convertible semiannually. All coupon payments are reinvested at a nominal rate of 4% convertible semiannually. Calculate the net gain to the investor at the end of 10 years after the loan is repaid. Problem 43.18 ‡ Dan purchases a 1000 par value 10-year bond with 9% semiannual coupons for 925. He is able to reinvest his coupon payments at a nominal rate of 7% convertible semiannually. Calculate his nominal annual yield rate convertible semiannually over the ten-year period. Problem 43.19 ‡ You have decided to invest in BondX,an n−year bond with semi-annual coupons and the following characteristics: •Par value is 1000. •The ratio of the semi-annual coupon rate to the desired semi-annual yield rate, r i , is 1.03125. •The present value of the redemption value is 381.50. Given vn = 0.5889, what is the price of bond X? Problem 43.20 ‡ Bill buys a 10-year 1000 par value 6% bond with semi-annual coupons. The price assumes a nominal yield of 6%, compounded semi-annually. As Bill receives each coupon payment, he immediately puts the money into an account earning interest at an annual eﬀective rate of i. At the end of 10 years, immediately after Bill receives the ﬁnal coupon payment and the redemption value of the bond, Bill has earned an annual eﬀective yield of 7% on his investment in the bond. Calculate i. Problem 43.21 Two $1000 par value bonds redeemable at par at the end of the same period are bought to yield 43 THE VARIOUS PRICING FORMULAS OF A BOND 393 4% convertible quarterly. One bond costs $1098 and has a coupon rate of 5% payable quarterly. The other bond has a coupon rate of 3% payable quarterly. Find the price of the second bond to the nearest dollar. Problem 43.22 A 1000 par value n−year bond maturing at par with annual coupons of 100 is purchased for 1125. The present value of the redemption value is 500. Find n to the nearest integer. Problem 43.23 A 1000 bond with annual coupons is redeemable at par at the end of 10 years. At a purchase price of 870, the yield rate is i. The coupon rate is i−0.02. Calculate i. Problem 43.24 A bond with coupons equal to 40 sells for P.A second bond with the same maturity value and term has coupons equal to 30 and sells for Q. A third bond with the same maturity value and term has coupons equal to 80. All prices are based on the same yield rate, and all coupons are paid at the same frequency. Determine the price of the third bond in terms of P and Q. Problem 43.25 ‡ You have decided to invest in two bonds. Bond X is an n−year bond with semi-annual coupons, while bond Y is an accumulation bond redeemable in n 2 years. The desired yield rate is the same for both bonds. You also have the following information: Bond X •Par value is 1000. •The ratio of the semi-annual bond rate to the desired semi-annual yield rate, r i is 1.03125. •The present value of the redemption value is 381.50. Bond Y •Redemption value is the same as the redemption value of bond X. •Price to yield is 647.80. What is the price of bond X? Problem 43.26 You are given two n−year par value 1,000 bonds. Bond X has 14% semi-annual coupons and a price of 1,407.70 to yield i compounded semi-annually. Bond Y has 12% semi-annual coupons and a price of 1,271.80 to yield the same rate icompounded semi-annually. Calculate the price of Bond X to yield i−1%. 394 BONDS AND RELATED TOPICS Problem 43.27 Henry has a ﬁve year 1,000,000 bond with coupons at 6% convertible semi-annually. Jean buys a 10 year bond with face amount X and coupons at 6% convertible semi-annually. Both bonds are redeemable at par. Henry and Jean both buy their bonds to yield 4% compounded semi-annually and immediately sell them to an investor to yield 2% compounded semi-annually. Jean earns the same amount of proﬁt as Henry. Calculate X. Problem 43.28 On June 1, 1990, an investor buys three 14 year bonds, each with par value 1000, to yield an eﬀective annual interest rate of i on each bond. Each bond is redeemable at par. You are given (1) the ﬁrst bond is an accumulation bond priced at 195.63 (2) the second bond has 9.4% semiannual coupons and is priced at 825.72 (3) the third bond has 10% annual coupons and is priced at P. Calculate P. Problem 43.29 A 10 year bond with par value 1000 and annual coupon rate ris redeemable at 1100. You are given (1) the price to yield an eﬀective annual interest rate of 4% is P (2) the price to yield an eﬀective annual interest rate of 5% is P −81.49 (3) the price to yield an eﬀective annual interest rate of r is X. Calculate X. Problem 43.30 Patrick buys a 28 year bond with a par value of 1200 and annual coupons. The bond is redeemable at par. Patrick pays 1968 for the bond, assuming an annual eﬀective yield rate of i. The coupon rate on the bond is twice the yield rate. At the end of 7 years Patrick sells the bond for P, which produces the same annual eﬀective yield rate of i to the new buyer. Calculate P. Problem 43.31 A 10 year bond with coupons at 8% convertible quarterly will be redeemed at 1600. The bond is bought to yield 12% convertible quarterly. The purchase price is 860.40. Calculate the par value. Problem 43.32 A corporation decides to issue an inﬂation-adjusted bond with a par value of $1000 and with annual coupons at the end of each year for 10 years. The initial coupon rate is 7% and each coupon is 3% greater than the preceding coupon. The bond is redeemed for $1200 at the end of 10 years. Find the price an investor should pay to produce a yield rate of 9% eﬀective. 43 THE VARIOUS PRICING FORMULAS OF A BOND 395 Problem 43.33 A 1000 par value 10 year bond with semiannual coupons and redeemable at 1100 is purchased at 1135 to yield 12% convertible semiannually. The ﬁrst coupon is X. Each subsequent coupon is 4% greater than the preceding coupon. Determine X. Problem 43.34 A 1,000 par value 3 year bond with annual coupons of 50 for the ﬁrst year, 70 for the second year, and 90 for the third year is bought to yield a force of interest δt = 2t−1 2(t2 −t+ 1), t ≥0. Calculate the price of this bond. 396 BONDS AND RELATED TOPICS 44 Amortization of Premium or Discount Bonds can be priced at a premium, discount, or at par. If the bond’s price is higher than its par value,i.e. P > C, then the bond is said to sell at a premium and the diﬀerence P −C is called the “premium”. On the other hand, if the bond’s price is less than its par value,i.e. P < C, then the bond is said to sell at a discount and the diﬀerence C−P is called the “discount”. Thus, a discount is merely a negative premium. Using the premium/discount formula of the previous section we see that Premium = P −C = (Fr −Ci)ani = C(g−i)ani if g >i In this case, a loss in that amount incurred when the bond is redeemed. Similarly, Discount = C−P = (Ci−Fr)ani = C(i−g)ani if g <i and a gain in that amount is incurred when the bond is redeemed. Because of this proﬁt or loss at redemption, the amount of each coupon CANNOT be considered as interest income to an investor. It is necessary to divide each coupon into interest earned and principal adjustment portions similar to the separation of payments into interest and principal when discussing loans. In the same sense that a loan has an outstanding balance at any time, we can talk of the book value of a bond at any time t. We deﬁne the book value to be the present value of all future payments. If we let Bt be the book value after the tth coupon has just been paid, then the value of the remaining payments are:n−t coupons and a payment of C at the date of redemption. Hence, Bt = Fran−t + Cνn−t. Example 44.1 Find the book value immediately after the payment of the 14th coupon of a 10-year 1,000 par value bond with semiannual coupons, if r= 0.05 and the yield rate is 12% convertible semiannually. Solution. We have F = C = 1000,n = 20,r = 0.05, and i= 0.06. Thus, B14 = 1000(0.05)a6 0.06 + 1000(1.06)−6 = 950.83 Letting t= 0 in the expression of Bt we ﬁnd B0 = P. Similarly, Bn = C. Thus, on the redemption date, the book value of a bond equals the redemption value of the bond. From this it follows that when a bond is purchased at a premium, the book value of the bond will be written -down(decreased) at each coupon date, so that at the time of redemption its book value 44 AMORTIZATION OF PREMIUM OR DISCOUNT 397 equals the redemption value. This process is called amortization of premium. When it is puchased at a discount, the book value of the bond will be written-up(increased) at each coupon date, so that at the time of redemption its book value equals the redemption value. This process is called accumulation of discount. Book values are very useful in constructing bond amortization schedules. A bond amortization schedule shows the division of each bond coupon into interest earned and principal adjustment portions, together with the book value after each coupon is paid. We will denote the interest earned after the tth coupon has been made by It and the corresponding principal adjustment portion by Pt.The level coupon will continue to be denoted by Fr. Note that It =iBt−1 = i[Fran−t+1 + Cνn−t+1] =Cg(1 −νn−t+1) + iCνn−t+1 = Cg + C(i−g)νn−t+1 and Pt = Fr −It = Cg −It = C(g−i)νn−t+1. where Pt is positive in the case of a premium and negative in the case of a discount. It is easy to see that Bt = B0 − t∑ k=1 Pk. Example 44.2 A $1000 bond, redeemable at par on December 1, 1998, with 9% coupons paid semiannually. The bond is bought on June 1, 1996. Find the purchase price and construct a bond amortization schedule if the desired yield is 8% compounded semiannually. Solution. Using the premium/discount formula with C = 1000,F = 1000,r = 4.5%,and i= 4%,the purchase price on June 1, 1996 is P = 1000 + (45−40)a5 0.04 = $1022.26. Thus, the bond is purchased at a premium of $22.26. For the bond amortization schedule, we ﬁrst do the calculation of the ﬁrst line of the schedule. Each successive line of the schedule is calculated in a similar fashion. The interest earned portion of the ﬁrst coupon, i.e. on Dec 1, 1996 is I1 = iB0 = 0.04(1022.26) = 40.89. The principal adjustment portion of the ﬁrst coupon is P1 = Fr −I1 = 45 −40.89 = $4.11. 398 BONDS AND RELATED TOPICS The book value at the end of the ﬁrst period is B1 = B0 −P1 = $1018.15. Amount for Interest amortization of Book Date Coupon earned premium value June 1, 1996 0 0 0 1022.26 Dec 1, 1996 45.00 40.89 4.11 1018.15 June 1, 1997 45.00 40.73 4.27 1013.88 Dec 1, 1997 45.00 40.56 4.44 1009.44 June 1, 1998 45.00 40.38 4.62 1004.82 Dec 1, 1998 45.00 40.19 4.81 1000.01 Total 225.00 202.74 22.25 The one cent error is due to rounding The following observations can be concluded from the above amortization schedule: (1) The sum of the principal adjustment column is equal to the amount of premium/discount. (2) The sum of the interest paid column is equal to the diﬀerence between the sum of the coupons and the sum of the principal adjustment column. (3) The principal adjustment column is a geometric progression with common ration 1 + i. The following is an example of a bond bought at discount. Example 44.3 A $1000 bond, redeemable at par on December 1, 1998, with 9% coupons paid semiannually. The bond is bought on June 1, 1996. Find the purchase price and construct a bond amortization schedule if the desired yield is 10% compounded semiannually. Solution. Using the premium/discount formula with C = 1000,F = 1000,r = 4.5%,and i= 5%,the purchase price on June 1, 1996 is P = 1000 + (45−50)a5 0.05 = $978.35. Thus, the bond is purchased at a discount of $21.65. 44 AMORTIZATION OF PREMIUM OR DISCOUNT 399 Amount for Interest accumulation of Book Date Coupon earned discount value June 1, 1996 0 0 0 978.35 Dec 1, 1996 45.00 48.92 −3.92 982.27 June 1, 1997 45.00 49.11 −4.11 986.38 Dec 1, 1997 45.00 49.32 −4.32 990.70 June 1, 1998 45.00 49.54 −4.54 995.24 Dec 1, 1998 45.00 49.76 −4.76 1000.00 Total 225.00 246.65 −21.65 Much like dealing with loans, if it is desired to ﬁnd the interest earned or principal adjustment portion of any one coupon, it is not necessary to construct the entire table. Simply ﬁnd the book value at the beginning of the period in question which is equal to the price at that point computed at the original yield rate and can be determined by the methods of Section 41. Then ﬁnd that one line of the table. Example 44.4 Consider a $1000 par value two-year 8% bond with semiannual coupons bought to yield 6% con- vertible semiannually. Compute the interest portion earned and the principal adjustment portion of the 3rd coupon payment. Solution. We have F = C = 1000,n = 4,r = 0.04, and i= 0.03. Thus, the coupon payment is Fr = $40.00. The principal adjustment portion of the 3rd coupon payment is P3 = B3 −B2 = 40 a1 0.03 + 1000(1.03)−1 −(40a2 0.03 + 1000(1.03)−2) = $9.43.The interest earned portion is I3 = 40.00−9.43 = $30.57 Another method of writing up or writing down the book values of bonds is the straight line method. This does not produce results consistent with compound interest, but it is very simple to apply. In this method, the book values are linear, grading from B0 = P to Bn = C. Thus, the principal adjustment column is constant and is given by Pt = P −C n , t = 1,2,··· ,n and the interest earned portion is It = Fr −Pt, t = 1,2,··· ,n Note that Pt >0 for premium bond and Pt <0 for discount bond. 400 BONDS AND RELATED TOPICS Example 44.5 Find the book values for the bond in Example 44.2 by the straight line method. Solution. We have Pt = 22.25 5 = 4.45. Thus, B0 =1022.26 B1 =1022.26 −4.45 = 1017.81 B2 =1017.81 −4.45 = 1013.36 B3 =1013.36 −4.45 = 1008.91 B4 =1008.91 −4.45 = 1004.46 B5 =1004.46 −4.45 = 1000.01 44 AMORTIZATION OF PREMIUM OR DISCOUNT 401 Practice Problems Problem 44.1 A $1000 par value two-year 8% bond with semiannual coupons is bought to yield 6% convertible semiannually. (a) Create an amortization schedule for this transaction. (b) Find the book values of the bond using the straight line method. Problem 44.2 A $1000 par value two-year 8% bond with semiannual coupons is bought to yield 10% convertible semiannually. (a) Create an amortization schedule for this transaction. (b) Find the book values of the bond using the straight line method. Problem 44.3 For a bond of face value 1 the coupon rate is 150% of the yield rate, and the premium is p. For another bond of 1 with the same number of coupons and the same yield rate, the coupon rate is 75% of the yield rate. Find the price of the second bond. Problem 44.4 For a certain period a bond amortization schedule shows that the amount for amortization of premium is 5, and that the required interest is 75% of the coupon. Find the amount of the coupon. Problem 44.5 A 10-year bond with semi-annual coupons is bought at a discount to yield 9% convertible semian- nually. If the amount for accumulation of discount in the next-to-last coupon is 8, ﬁnd the total amount for accumulation of discount during the ﬁrst four years in the bond amortization schedule. Problem 44.6 A 1000 par value 5-year bond with a coupon rate of 10% payable semiannually and redeemable at par is bought to yield 12% convertible semiannually. Find the total of the interest paid column in the bond amortization schedule. Problem 44.7 A ﬁve year bond with a par value of 1000 will mature in 5 years for 1000. Annual coupons are payable at a rate of 6%. Create the bond amortization schedule if the bond is bought to yield 8% annually. 402 BONDS AND RELATED TOPICS Problem 44.8 A 40 year bond with a par value of 5000 is redeemable at par and pays semi-annual coupons at a rate of 7% convertible semi-annually. The bond is purchased to yield annual eﬀective rate of 6%. Calculate the amortization of the premium in the 61st coupon. Problem 44.9 A 10 year bond is redeemable at par of 100,000. The bond pays semi-annual coupons of 4000. The bond is bought to yield 6% convertible semi-annually. Calculate the premium paid for the bond. Problem 44.10 A bond is redeemable in eight years at 120% of its par value. The bond pays an annual coupon rate of 6%. Calculate the premium as a percent of the par value if the bond is purchased to yield 6% annually. Problem 44.11 Consider a three-year bond, with a $1,000 face value and a 9% coupon rate paid semi-annually, which was bought to yield 7% convertible semi-annually. Find the amount for amortization of premium during the bonds third half-year (i.e., between the second and third coupon payments). Problem 44.12 A bond of face amount 100 pays semi-annual coupons and is purchased at a premium of 36 to yield annual interest of 7% compounded semiannually. The amount for amortization of premium in the 5th coupon is 1.00. What is the term of the bond? Problem 44.13 ‡ Among a company’s assets and accounting records, an actuary ﬁnds a 15-year bond that was purchased at a premium. From the records, the actuary has determined the following: (i) The bond pays semi-annual interest. (ii) The amount for amortization of the premium in the 2nd coupon payment was 977.19. (iii) The amount for amortization of the premium in the 4th coupon payment was 1046.79. What is the value of the premium? Problem 44.14 ‡ A 10,000 par value 10-year bond with 8% annual coupons is bought at a premium to yield an annual eﬀective rate of 6%. Calculate the interest portion of the 7th coupon. Problem 44.15 ‡ A 1000 par value 5 year bond with 8% semiannual coupons was bought to yield 7.5% convertible semiannually. Determine the amount of premium amortized in the sixth coupon payment. 44 AMORTIZATION OF PREMIUM OR DISCOUNT 403 Problem 44.16 On May 1, 1985, a bond with par value 1000 and annual coupons at 5.375% was purchased to yield an eﬀective annual interest rate of 5%. On May 1, 2000, the bond is redeemable at 1100. The book value of the bond is adjusted each year so that it equals the redemption value on May 1, 2000. Calculate the amount of write-up or write-down in the book value in the year ending May 1, 1991. Problem 44.17 A bond with par value of 1000 and 6% semiannual coupons is redeemable for 1100. You are given (1) the bond is purchased at P to yield 8% convertible semiannually and (2) the amount of principal adjustment for the 16th semiannual period is 5 Calculate P. Problem 44.18 Laura buys two bonds at time 0. Bond X is a 1000 par value 14 year bond with 10% annual coupons. It is bought at a price to yield an annual eﬀective rate of 8%. Bond Y is a 14 year par value bond with 6.75% annual coupons and a face amount of F. Laura pays P for the bond to yield an annual eﬀective rate of 8%. During year 6 the writedown in premium (principal adjustment) on bond X is equal to the writeup in discount (principal adjust- ment) on bond Y. Calculate P. Problem 44.19 A 1000 par value 18 year bond with annual coupons is bought to yield an annual eﬀective rate of 5%. The amount for amortization of premium in the 10th year is 20. The book value of the bond at the end of year 10 is X. Calculate X. Problem 44.20 Thomas buys a bond at a premium of 200 to yield 6% annually. The bond pays annual coupons and is redeemable for its par value of 1000. Calculate the amount of interest in the ﬁrst coupon. Problem 44.21 Megan buys a bond that is redeemable for its par value of 20,000 after 5 years. The bond pays coupons of 800 annually. The bond is bought to yield 8% annually. Calculate the accumulation of discount in the 4th coupon. Problem 44.22 Jenna buys a bond at a premium. The principal amortized in the ﬁrst annual coupon is half that amortized in the 11th annual coupon. Calculate the yield rate used to calculate the purchase price of the bond. 404 BONDS AND RELATED TOPICS Problem 44.23 Betty buys an n−year 1000 par value bond with 6.5% annual coupons at a price of 798.48. The price assumes an annual eﬀective yield rate ofi.The total write-up in book value of the bond during the ﬁrst 3 years after purchase is 22.50. Problem 44.24 Ann purchases a 1000 face value 20-year bond, redeemable at 1200, with 5% annual coupons, at a price that yields her an annual eﬀective interest rate of 3%. Immediately upon receiving each coupon, she invests the coupon in an account that earns an annual eﬀective interest rate of 7%. Immediately after receiving the eighth coupon, Ann sells the bond at a price that yields the new purchaser an annual eﬀective interest rate of 4.75%. Determine Ann’s yield rate, as an annual eﬀective interest rate, over the 8-year period that she owns the bond. Problem 44.25 A 1000 face value 20-year 8% bond with semiannual coupons is purchased for 1014. The redemption value is 1000. The coupons are reinvested at a nominal annual rate of 6%, compounded semiannually. Determine the purchaser’s annual eﬀective yield rate over the 20- year period. Problem 44.26 An n−year 1000 par value bond with 8% annual coupons has an annual eﬀective yield of i,i >0. The book value of the bond at the end of year 3 is 1099.84 and the book value at the end of year 5 is 1082.27. Calculate the purchase price of the bond. Problem 44.27 Bryan buys a 2 n−year 1000 par value bond with 7.2% annual coupons at a price P. The price assumes an annual eﬀective yield of 12%. At the end of n years, the book value of the bond, X, is 45.24 greater than the purchase price, P. Assume νn 0.12 <0.5. Calculate X. 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 405 45 Valuation of Bonds Between Coupons Payment Dates Up to now, bond prices and book values have been calculated assuming the coupon has just been paid. In particular, we can ﬁnd a relationship between the book values of two consecutive coupon dates t and t+ 1 as follows. Let Bt and Bt+1 be the book values just after the tth and ( t+ 1)th coupons are paid and Fr the amount of the coupon. Then Bt(1 + i) −Fr =Fr (1 −νn−t i ) (1 + i) + Cνn−t(1 + i) −Fr =Fr i −Frνn−t i (1 + i) + Cνn−t−1 =Fr i −Frνn−t−1 i + Cνn−t−1 =Fr (1 −νn−t−1 i ) + Cνn−t−1 =Fran−t−1 + Cνn−t−1 =Bt+1 where we assume a constant yield rate i over the interval. Example 45.1 You buy an n−year 1,000 par value bond with 6.5% annual coupons at a price of 825.44, assuming an annual yield rate of i,i> 0. After the ﬁrst two years, the bond’s book value has changed by 23.76. Calculate i. Solution. We have B2 −B0 = B0(1 +i)2 −Fr(1 +i) −Fr −B0.This implies 23.76 = 825.44(1 +i)2 −65(1 + i) −65 −825.44 which reduces to 825.44(1 + i)2 −65(1 + i) −914.20 = 0 Solving this with the quadratic formula we ﬁnd 1 + i= 1.0925 →i= 9.25% Book values can be computed at times other than immediately after a coupon payment. In this section, we would like to analyze the book value Bt+k for 0 <k <1 where time k is measured from the last coupon payment. When buying an existing bond between its coupon dates, one must decide how to split up the 406 BONDS AND RELATED TOPICS coupon between the prior owner and the new owner. Consider the price that a buyer would pay for the bond a fractional time k through a coupon period. Assume that the buyer will obtain a yield rate equal to that of the current bond holder. The buyer will receive all of the next coupon. The current holder would expect to receive part of this coupon as interest for the period which we call accrued interest or accrued coupon and is denoted by Frk. Note that Fr0 = 0 and Fr1 = Fr, the coupon payment, where Fr1 is computed just before the coupon is paid. How should the purchase price be allocated between accrued interest (or accrued coupon) and price for the bond? The purchase price for the bond is called the ﬂat price. It is deﬁned to be the money which actually changes hands at the date of sale. It is denoted by Bf t+k. The price for the bond is the book value, which is also called the market price and is denoted by Bm t+k.The market price is the price commonly quoted in the ﬁnancial press, since the market price changes smoothly through time, while the ﬂat price ﬂuctuates due to coupon accrual. From the above deﬁnitions, it is clear that Bf t+k = Bm t+k + Frk. Example 45.2 Consider the amortization schedule of Example 44.2. Amount for Interest amortization of Book Date Coupon earned premium value June 1, 1996 0 0 0 1022.26 Dec 1, 1996 45.00 40.89 4.11 1018.15 June 1, 1997 45.00 40.73 4.27 1013.88 Dec 1, 1997 45.00 40.56 4.44 1009.44 June 1, 1998 45.00 40.38 4.62 1004.82 Dec 1, 1998 45.00 40.19 4.81 1000.01 Total 225.00 202.74 22.25 Sketch a graph describing the relationship between the ﬂat price and the market price. Solution. The graph is shown in Figure 45.1. Note that the accrued coupon at any date is equal to the vertical distance between the solid line and the dotted line 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 407 Figure 45.1 There are three methods used to ﬁnd the ﬂat price, the market price, and the accured coupon. Values on the coupon dates are known, so the diﬀerences among the three methods arise only for the interim values between coupon dates. First, we have the theoretical method. This is an exact method, based on compound interest. This method argues that the ﬂat price should be the book value Bt after the preceding coupon, accumulated by (1 + i)k, where i is the yield rate, giving the ﬂat price of Bf t+k = (1 + i)kBt. The accrued coupon is given by (See Problem 15.48) Frk = Fr [(1 + i)k −1 i ] . The market value is then the diﬀerence Bm t+k = (1 + i)kBt −Fr [(1 + i)k −1 i ] The second method is called the practical method. It is an approximate method that uses the linear approximation (1 + i)k ≈1 + ki for 0 <k <1. Thus, obtaining Bf t+k =(1 + ki)Bt Frk =kFr Bm t+k =(1 + ki)Bt −kFr = Bt + kiBt −kFr =Bt + k[Bt+1 + Fr −Bt] −kFr = (1 −k)Bt + kBt+1 408 BONDS AND RELATED TOPICS where we used the formula Bt(1 + i) −Fr = Bt+1. The third method called the semi-theoretical method is the most widely used method, and has been accepted as the standard method of calculation by the securities industry. The ﬂat price is determined as in the theoretical method, and the accrued coupon is determined as in the practical method. Bf t+k =(1 + i)kBt Frk =kFr Bm t+k =(1 + i)kBt −kFr In computing k, we use an actual count of days: k= number of days since last coupon paid number of days in the coupon period. Both the actual/actual and 30/360 methods (see Section 5) can be used. See Problem 45.12. Example 45.3 Consider a $1000 par value two-year 8% bond with semiannual coupons bought to yield 6% con- vertible semiannually. The price of the bond is computed to be $1037.17. Compute the ﬂat price, accrued interest, and market price ﬁve months after purchase of the bond. Solution. We have Fr = 1000(0.04) = $40.00 and k= 5 6 . With the theoretical method we ﬁnd Bf 5 6 =1037.17(1.03) 5 6 = $1063.04 Fr5 6 =40 [ (1.03) 5 6 −1 0.03 ] = $33.25 Bm 5 6 =1063.04 −33.25 = $1029.79 With the practical method we ﬁnd Bf 5 6 =1037.17 [ 1 + 5 6(0.03) ] = $1063.10 Fr5 6 =5 6(40) = $33.33 Bm 5 6 =1063.10 −33.33 = $1029.77 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 409 Finally, with the semi-theoretical method we have Bf 5 6 =1037.17(1.03) 5 6 = $1063.04 Fr5 6 =5 6(40) = $33.33 Bm 5 6 =1063.04 −33.33 = $1029.71 Example 45.4 A 1000 par value 5 year bond with semi-annual coupons of 60 is purchased to yield 8% convertible semi-annually. Two years and two months after purchase, the bond is sold at the ﬂat price which maintains the yield over the two years and two months. Calculate the ﬂat price using the theoretical method. Solution. The price of the bond after two years is P = 60a6 0.04 + 1000ν−6 0.04 = 1,104.84. The ﬂat price is Bf 2 6 = 1104.84(1.04) 2 6 = 1,119.38 We conclude this section by ﬁnding the premium or discount between coupon payment dates. For the premium we have the formula Premium = Bm t+k −C, i<g and for the discount we have Discount = C−Bm t+k, i>g. Thus, premiums and discounts are based on the market value or the book value and not on the ﬂat price. Example 45.5 A n−year $1000 par value bond is selling at $1010 and has a market value of $980 and acrrued coupon of $30. Find premium/discount amount. Solution. Since the premium/discount calculation is based on market value rather than ﬂat price, the bond is a discount bond. The discount amount is $20 410 BONDS AND RELATED TOPICS Practice Problems Problem 45.1 A 10 year bond with semi-annual coupons has a book value immediately after the 5th coupon of 90,000. The ﬂat price 5 months later using the theoretical method is 94,591. Calculate the semi-annual yield on the bond. Problem 45.2 A bond with semi annual coupons of 2500 has a book value immediately after the 6th coupon of 95,000. The market value using the practical method z months after the 6th coupon is 95,137.50. Calculate z if the yield rate is 7% convertible semi-annually. Problem 45.3 A 10 year bond with a par value of 100,000 and semi-annual coupons 2500 is bought at a discount to yield 6% convertible semi-annually. (a) Calculate the book value immediately after the 5th coupon. (b) Calculate the ﬂat price 4 months after the 5th coupon using the theoretical method. (c) Calculate the accrued coupon 4 months after the 5th coupon using the theoretical method. (d) Calculate the market price 4 months after the 5th coupon using the theoretical method. (e) Calculate (b)-(d) using the practical method. (f) Calculate (b)-(d) using the semi-theoretical method Problem 45.4 Which of the following are true for a bond with semi annual coupons? (I) The ﬂat price of the bond under the theoretical method will always equal the ﬂat price of the bond under the practical method. (II) The ﬂat price of the bond under the practical method will always exceed the ﬂat price of the bond under the semi-theoretical method. (III) For a bond purchased at par which matures at par with the yield rate equal to the coupon rate, the ﬂat price at any point in time will equal the par value of the bond under the semi-theoretical method. Problem 45.5 A bond pays coupons of 600 on April 30 and October 31. Calculate the accrued coupon under the practical method on July 1. Problem 45.6 A 10 year bond is redeemable at par of 100,000. The bond has semi-annual coupons of 4000. The bond is bought to yield 6% convertible semi-annually. Four months after purchase, calculate the market price based on the theoretical method less the market price based on the semi-theoretical method. 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 411 Problem 45.7 A 10-year 100 par value bond bearing a 10% coupon rate payable semi-annually, and redeemable at 105, is bought to yield 8% convertible semiannually. Using the practical method, compute the ﬂat price, accrued interest, and market price three months after purchase of the bond. Problem 45.8 Show that Bf t+k = (Bt+1 + Fr)v1−k. Problem 45.9 Arrange in increasing order of magnitude for the three interim bond price methods: (a) Flat price. (b) Market price. Problem 45.10 A 10,000 par value bond with 6% semiannual coupons is being sold 3 years and 2 months before the bond matures. The purchase will yield 8% convertible semiannually to the buyer. Determine the price of the bond. Use the practical method. Problem 45.11 ‡ A 10,000 par value bond with coupons at 8%, convertible semiannually, is being sold 3 years and 4 months before the bond matures. The purchase will yield 6% convertible semiannually to the buyer. The price at the most recent coupon date, immediately after the coupon payment, was 5640. Calculate the market price of the bond, assuming compound interest throughout. Problem 45.12 Consider a $1,000 par value bond with coupons at 8% convertible semi-annually and with maturity date 05/06/04. The bond was purchased on 11/06/01. Compute the accrued interest on 12/04/001 taking into account the Actual/Actual day-count basis. Problem 45.13 A 20-year 10% annual coupon bond has a par value of $1,000. When you originally purchased this bond, the eﬀective annual interest rate was 8%. Suppose that ﬁve years after purchase, the eﬀective annual interest rate is 10.5%. What is the absolute diﬀerence between the book and market values of the bond at that point in time? Problem 45.14 ‡ A 1000 bond with semi-annual coupons at i(2) = 6% matures at par on October 15, 2020. The bond is purchased on June 28, 2005 to yield the investor i(2) = 7%. What is the purchase price? Assume simple interest between bond coupon dates and note that: 412 BONDS AND RELATED TOPICS Date Day of the year April 15 105 June 28 179 October 15 288 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 413 46 Approximation Methods of Bonds’ Yield Rates In this section we consider the question of determining the yield rate to maturity to an investor given the purchase price of a bond. The determination of such yield rates is similar to the determination of an unknown rate of interest for an annuity, discussed in Section 20. We ﬁrst consider the yield rate for a bond purchased on a coupon payment date immediately after the coupon is paid. There are two approaches for this problem. One approach is linear interpolation where bond tables are used. We will not pursue this method any further in this book. Another approach is to develop algebraic estimation formulas for the yield rate. Three main formulas will be discussed. We start with the premium discount formula P =C+ (Fr −Ci)ani =C+ C(g−i)ani where F,r,g,C,n, and P are known and i is the unknown. Letting k= P−C C we ﬁnd (g−i)ani = k. Solving this equation for i we obtain i= g− k ani which deﬁnes i implicitly. We will develop methods for estimating i in this equation. The ﬁrst is obtained as follows: Writing the power series expansion of 1 ani (see p.209) 1 ani = 1 n [ 1 + n+ 1 2 i+ n2 −1 12 i2 + ··· ] and neglecting terms of higher than the ﬁrst degree in i we obtain the estimation i= g− k ani ≈g−k n [ 1 + n+ 1 2 i ] . Solving for i we ﬁnd i≈ g−k n 1 + n+1 2n k. (46.1) Formula (46.1) forms the basis of an even simpler method for calculating approximate yield rates called the bond salesman’s method. Noting that n+1 2n ≈1 2 for large n, this method replaces the ratio n+1 2n in formula (46.1) by 1 2 thus obtaining i≈ g−k n 1 + 1 2 k. 414 BONDS AND RELATED TOPICS Error analysis shows that this method produces less accurate results than formula (46.1). The salesman’s formula can be rewritten in the form i≈n(Cg) + C−P n 2 (P + C) = n(Fr) + C−P n 2 (P + C) = Total Interest Paid + Capital Gain/Loss Number of Periods ×Average Amount Invested Example 46.1 A $100 par value 10-year bond with 8% semiannual coupons is selling for $90. Find the yield rate convertible semi-annually. Solution. We have n= 20,P = 90,C = 100,Fr = 4 so that by the salesman’s formula we have i≈n(Fr) + C−P n 2 (P + C) = 20(4) + 100−90 0.5(20)(90 + 100) = 0.0474 so the yield rate is 9.48% convertible semi-annually The third method for estimating i is by solving the equation i= g− k ani . by using the Newton-Raphson method. In this case, we are looking for the solutions of the equation f(i) = 0 where f(i) = i−g+ k ani = i−g+ ki 1 −(1 + i)−n. The Newton-Raphson iterations are given by the formula is+1 = is −f(is) f′(is). Thus, an expression of the form i−f(i) f′(i) need to be determined. This is done as follows. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 415 i−f′(i) f(i) =i− i−g+ ki 1−(1+i)−n 1 + k ( [1−(1+i)−n]−ni(1+i)−n−1 [1−(1+i)−n]2 ) =i+ g(1 −νn)2 −i(1 −νn)2 −ki(1 −νn) (1 −νn)2 + k[(1 −νn) −niνn+1] =i+ (1 −νn)[giani −i(1 −νn) −ki] iani(1 −νn) + kiani −nkiνn+1 =i+ (1 −νn)[gani −(1 −νn) − (P C −1 ) ] ani(1 −νn) + (P C −1 ) ani −n (P C −1 ) νn+1 =i+ (1 −νn) [ gani + νn −P C ] ani(1 −νn) + (g−i)(ani)2 −n(g−i)aniνn+1 =i+ gani + νn −P C ani + (g−i)(ani)i−1 −n(g−i)i−1νn+1 =i+ i gani + νn −P C iani + (g−i)(ani) −n(g−i)νn+1 =i [ 1 + gani + νn −P C ani + (g−i)(ani) −n(g−i)νn+1 ] =i [ 1 + gani + νn −P C gani + n(i−g)νn+1 ] Thus, the Newton-Raphson iterations are given by is+1 = is [ 1 + ganis + (1 +is)−n −P C ganis + n(is −g)(1 + is)−(n+1) ] An initial starting guess can be chosen to be i0 = g−k n 1 + n+1 2n k. Example 46.2 A $100 par value 8-year bond with 6% semiannual coupons is selling for $97. Find the yield rate convertible semiannually. 416 BONDS AND RELATED TOPICS Solution. We have g = 3%, n= 16, k= P−C C = 97−100 100 = −0.03. An initial starting guess for the Newton- Raphson method is i0 = 0.03 −(−0.03/16) 1 + 16+1 32 = 0.0323912 and the iterations are given by the formulas is+1 = is ( 1 + 0.03a16 is + (1 +is)−16 −0.97 0.03a16 is + 16(is −0.03)(1 + is)−17 ) . We obtain the following values s is 2is 0 0.0323912 0.0647824 1 0.0324329 0.0648658 2 0.0324329 0.0648658 Thus, the yield rate convertible semiannually is 6 .48658%. It is worth noting that the selling price is P = 100 (0.06 2 ) a16 3.24329% + 100(1.0324329)−16 = $97 One can also detetermine the yield rate when the purchasing date is between two coupon payment dates. This situation is more complex due to the fractional period prior to the next coupon payment date. In practice, this case involves the use of the semi-theoretical method of Section 45 which leads to an implicit equation in i that can be solved by either linear interpolation or by iterations. We illustrate this in the next example. Example 46.3 A $100 par value 10-year bond with 8% semiannual coupons was issued on March 1, 2001. The market price on May 15, 2003 is $88. Find the yield rate if the bond in bought on that date. Solution. The price of the bond on March 1, 2003 immediately after a coupon payment is 100 + (4−100i)a16 i. The number of days from March 1, 2003 to May 15, 2003 is 75, while the number of days from March 1, 2003 to September 1, 2003 is 184. Using the semi-theoretical method of Section 45, we have the equation [100 + (4−100i)a16 i](1 + i) 75 184 − 75 184 ·4 = 88. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 417 Let’s solve this last equation by using linear interpolation. Let f(i) = [100 + (4−100i)a16 i](1 + i) 75 184 − 75 184 ·4 −88. By trial and error we ﬁnd f(0.05) = 23 .43 > 0 and f(0.052) = −2.45 < 0. Thus, using linear interpolation we ﬁnd i≈0.05 −23.43 × 0.052 −0.05 −2.45 −23.43 = 0.0518. So the nominal yield rate convertible semiannually is 10.36%. The price on May 15,2003 is 100 + (4−100 ×0.0518)a16 0.0518(1.0518) 75 184 − 75 184 ·4 = $87.56 a roundoﬀ error of 0.44 We ﬁnally consider ﬁnding the yield rate of a bond with coupons being reinvested at a diﬀer- ent interest rate. Consider the situation in which a bond is purchased for P,coupons of Fr are paid at the end of each period for n periods, the bond is redeemed for C at the end of n periods, and the coupons are reinvested at rate j. Denoting the yield rate (considering reinvestment) by i′, we would have the equation of value: P(1 + i′)n = Frsnj + C since both sides represent the value of the investment at the end of n periods. Example 46.4 A $100 par value 10-year bond with 8% semiannual coupons is selling for $90 and the coupons can be reinvested at only 6% convertible semi-annually. Find the yield rate taking in consideration reinvestment rates. Solution. We have 90(1 + i′)20 = 4s20 0.03 + 100 →(1 + i′)20 = 4(26.8704)+100 90 = 2.30535 →i′= 0.04265. Thus, the yield rate is 2(0 .04265) = 0 .0853 = 8 .53% convertible semi-annually. Thus, the yield rate of 9.48% found in Example 44.1 drops to 8.53% if the coupons can be reinvested at 6% 418 BONDS AND RELATED TOPICS Practice Problems Problem 46.1 A $100 par value 12-year bond with 10% semiannual coupons is selling for $110. Find the yield rate convertible semiannually. Problem 46.2 Recompute the overall yield rate to an investor purchasing the bond in the previous problem, if the coupons can be reinvested at only 7% convertible semiannually. Problem 46.3 A $100 bond with annual coupons is redeembale at par at the end of 15 years. At a purchase price of $92 the yield rate is exactly 1% more than the coupon rate. Find the yield rate on the bond. Problem 46.4 An investor buys 20-year bonds, each having semiannual coupons and each maturing at par. For each bond the purchase price produces the same yield rate. One bond has a par value of $500 and a coupon of $45. The other bond has a par value of $1000 and a coupon of $30. The dollar amount of premium on the ﬁrst bond is twice as great as the dollar amount of discount on the second bond. Find the yield rate convertible semiannually. Problem 46.5 A 20 year bond with a 20,000 par value pays semi-annual coupons of 500 and is redeemable at par. Audrey purchases the bond for 21,000. Calculate Audrey’s semi-annual yield to maturity on the bond. Use Salesman’s formula. Problem 46.6 A 10 year bond with a par value of 1000 is redeemable at par and pays annual coupons of 65. If Jamie purchased the bond for 950 and she reinvests the coupons at 4% annual eﬀective rate, calculate the actual yield that Jamie will receive taking into account the reinvestment rate. Problem 46.7 ‡ A 1000 par value 10-year bond with coupons at 5%, convertible semiannually, is selling for 1081.78. Calculate the yield rate convertible semiannually. Problem 46.8 ‡ Dan purchases a 1000 par value 10-year bond with 9% semiannual coupons for 925. He is able to reinvest his coupon payments at a nominal rate of 7% convertible semiannually. Calculate his nominal annual yield rate convertible semiannually over the ten-year period. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 419 Problem 46.9 Suppose a 10000 par bond has 6% semi-annual coupons and matures in 10 years. What is the yield rate if the price is 9000? Compute both exactly and using the Salesman’s method. Problem 46.10 A $1000 par value 10-year bond with 7% semiannual coupons was issued on October 1, 2006. The market price on April 28, 2006 is $1020. Find the yield rate if the bond in bought on that date. Problem 46.11 John buys a $1000 bond that pays coupons at a nominal interest rate of 7% compounded semian- nually. The bond is redeemable at par in 15 years. The price he pays will give him a yield of 8% compounded semiannually if held to maturity. After 5 years, he sells the bond to Peter, who desires a yield of 6% on his investment. (a) What price did John pay? (b) What price did Peter pay? (c) What yield did John realize? 420 BONDS AND RELATED TOPICS 47 Callable Bonds and Serial Bonds Callable or redeemable bonds are bonds that can be redeemed or paid oﬀ by the issuer prior to the bonds’ maturity date. The earliest such call date will generally be several years after the issue date. Call dates are usually speciﬁed by the bond issuer. When an issuer calls its bonds, it pays investors the call price (usually the face value of the bonds) together with accrued interest to date and, at that point, stops making interest payments. An issuer may choose to redeem a callable bond when current interest rates drop below the interest rate on the bond. That way the issuer can save money by paying oﬀ the bond and issuing another bond at a lower interest rate. This is similar to reﬁnancing the mortgage on your house so you can make lower monthly payments. Callable bonds are more risky for investors than non-callable bonds because an investor whose bond has been called is often faced with reinvesting the money at a lower, less attractive rate. As a result, callable bonds often have a higher annual return to compensate for the risk that the bonds might be called early. Callable bonds present a problem when calculating prices and yields, because the call date of the bond is unknown. Since the issuer has an option whether or not to call the bond, the investor should assume that the issuer will exercise that option to the disadvantage of the investor, and should calculate the yield rate and/or price accordingly. If the redemption value is the same at any call date, including the maturity date, then the following general principle will hold: (i) the call date will most likely be at the earliest date possible if the bond was sold at a premium, which occurs when the yield rate is smaller than the coupon rate (issuer would like to stop repaying the premium via the coupon payments as soon as possible). (ii) the call date will most likely be at the latest date possible if the bond was sold at a discount, which iccurs when the yield rate is larger than the coupon rate (issuer is in no rush to pay out the redemption value). Example 47.1 A 15 year 1000 par bond has 7% semiannual coupons and is callable at par after 10 years. What is the price of the bond to yield 5% for the investor? Solution. Since the yield rate is smaller than the coupan rate, the bond is selling at premium so the earliest redemption date is the most favorable for the issuer and the least favorable for the investor. The price is the smallest of 1000 + (35 −25)an0.025 for 20 ≤n≤30, which clearly occurs when n= 20. Thus, the price is 1000 + (35−25)a20 0.025 = 1,155.89 47 CALLABLE BONDS AND SERIAL BONDS 421 If the redemption values on all the redemption dates are not equal, this principle is a little harder to apply. In this case one needs to examine all call dates. The most unfavorable date will not necessarily be either the earliest or the latest possible date. The most unfavorable call date ( to the issuer) is the one that produces the smallest purchase price at the investors yield rate as illustrated in the following example. Example 47.2 A 100 par value 4% bond with semi-annual coupons is callable at the following times: 109.00, 5 to 9 years after issue 104.50, 10 to 14 years after issue 100.00, 15 years after issue (maturity date). What price should an investor pay for the callable bond if they wish to realize a yield rate of (1) 5% payable semi-annually and (2) 3% payable semi-annually? Solution. (1) Since the market rate is better than the coupon rate, the bond would have to be sold at a discount and as a result, the issuer will wait until the last possible date to redeem the bond: P = 2.00a30 2.5% + 100.00v30 0.025 = $89.53. (2) Since the coupon rate is better than the market rate, the bond would sell at a premium and as a result, the issuer will redeem at the earliest possible date for each of the three diﬀerent redemption values: P =2.00a10 1.5% + 109.00v10 0.015 = $112.37 P =2.00a20 1.5% + 104.50v20 0.015 = $111.93 P =2.00a30 1.5% + 100.00v30 0.015 = $112.01 In this case, the investor would only be willing to pay $111.93 Serial Bonds Serial bonds are bonds issued at the same time but with diﬀerent maturity dates. The price of any individual bond is found by the methods already discussed earlier in this chapter provided that the redemption date is known. The price of the entire serial bonds is just the sum of the values of the individual bonds. We next consider ﬁnding a formula for pricing an issue of serial bonds with mdiﬀerent redemption dates. Let Pt be the purchase price of the tth bond; Ct be the redemption value and Kt the present value of Ct. Then by Makeham’s formula the price of the tth bond is Pt = Kt + g i(Ct −Kt). 422 BONDS AND RELATED TOPICS Summing from t= 1 to t= m to obtain the price of the entire issue of a serial bond m∑ t=1 Pt = m∑ t=1 Kt + g i ( m∑ t=1 Ct − m∑ t=1 Kt ) or P′= K′+ g i(C′−K′) where P′= ∑m t=1 Pt, K′= ∑m t=1 Kt, and C′= ∑m t=1 Ct. Example 47.3 A $10,000 serial bond is to be redeemed in $1000 installments of principal per half-year over the next ﬁve years. Interest at the annual rate of 12% is paid semi-annually on the balance outstanding. How much should an investor pay for this bond in order to produce a yield rate of 8% convertible semi-annually? Solution. We have P′= 10∑ t=1 1000vt 0.04 + 0.06 0.04 [ 10,000 − 10∑ t=1 1000vt 0.04 ] =1000a10 0.04 + 1.5[10,000 −1000a10 0.04] =15,000 −500a10 0.04 = 10,944.58 Example 47.4 A 5000 serial bond with 10% annual coupons will be redeemed in ﬁve equal installments of 1100 beginning at the end of the 11th year and continuing through the 15th year. The bond was bought at a price P to yield 9% annual eﬀective. Determine P. Solution. We have: F = 1000,C = 1100,i = 0.09 and r= 0.10. Thus, g= 1000(0.10) 1100 = 1 11 . Hence, C′= 5500 K′= 15∑ t=11 1100vt 0.09 =1100(a15 0.09 −a10 0.09) = 1807.33 The price is P′= 1807.33 + 1 0.99(5500 −1807.33) = 5537.30 47 CALLABLE BONDS AND SERIAL BONDS 423 Practice Problems Problem 47.1 A $1000 par value bond has 8% semiannual coupons and is callable at the end of the 10th through the 15th years at par. (a) Find the price to yield 6% convertible semiannually. (b) Find the price to yield 10% convertible semiannually. Problem 47.2 A $1000 par value 8% bond with quarterly coupons is callable ﬁve years after issue. The bond matures for $1000 at the end of ten years and is sold to yield a nominal rate of 6% convertible quarterly under the assumption that the bond will not be called. Find the redemption value at the end of ﬁve years that will provide the purchaser the same yield rate. Problem 47.3 A $1000 par value 4% bond with semiannual coupons matures at the end of 10 years. The bond is callable at $1050 at the ends of years 4 through 6, at $1025 at the ends of years 7 through 9, and at $1000 at the end of year 10. Find the maximum price that an investor can pay and still be certain of a yield rate of 5% convertible semiannually. Problem 47.4 A $1000 par value 6% bond with semiannual coupons is callable at par ﬁve years after issue. It is sold to yield 7% semi-annual under the assumption that the bond will be called. The bond is not called and it matures at the end of 10 years. The bond issuer redeems the bond for 1000 + X without altering the buyer’s yield rate of 7% convertible semiannually. Find X. Problem 47.5 A 15 year bond can be called at the end of any year from 10 through 14. The bond has a par value of 1000 and an annual coupon rate of 5%. The bond is redeemable at par and can also be called at par. (a) Calculate the price an investor would pay to yield 6%. (b) Calculate the price an investor would pay to yield 4%. Problem 47.6 A 15 year bond can be called at the end of any year from 12 through 14. The bond has a par value of 1000 and an annual coupon rate of 5%. The bond is redeemable at par and can be called at 1100. (a) Calculate the price an investor would pay to yield 6%. (b) Calculate the price an investor would pay to yield 4%. 424 BONDS AND RELATED TOPICS Problem 47.7 A callable 20 year bond which matures for 1000 pays annual coupons of 80. The bond is callable in years 10 through 12 at 1100 and in years 13 through 15 at 1050. The bond is bought to yield 4% annually. Calculate the purchase price of the bond. Problem 47.8 ‡ Matt purchased a 20-year par value bond with semiannual coupons at a nominal annual rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par value X on any coupon date starting at the end of year 15 after the coupon is paid. The price guarantees that Matt will receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate X. Problem 47.9 ‡ Toby purchased a 20-year par value bond with semiannual coupons at a nominal annual rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par value 1100 on any coupon date starting at the end of year 15. What is the minimum yield that Toby could receive, expressed as a nominal annual rate of interest convertible semiannually? Problem 47.10 ‡ Sue purchased a 10-year par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value X on any coupon date starting at the end of year 5. The price guarantees that Sue will receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate X. Problem 47.11 ‡ Mary purchased a 10-year par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value 1100 on any coupon date starting at the end of year 5. What is the minimum yield that Mary could receive, expressed as a nominal annual rate of interest convertible semiannually? Problem 47.12 ‡ A 1000 par value bond with coupons at 9% payable semiannually was called for 1100 prior to maturity. The bond was bought for 918 immediately after a coupon payment and was held to call. The nominal yield rate convertible semiannually was 10%. Calculate the number of years the bond was held. 47 CALLABLE BONDS AND SERIAL BONDS 425 Problem 47.13 ‡ A 1000 par value bond pays annual coupons of 80. The bond is redeemable at par in 30 years, but is callable any time from the end of the 10th year at 1050. Based on her desired yield rate, an investor calculates the following potential purchase prices, P : Assuming the bond is called at the end of the 10th year, P = 957 Assuming the bond is held until maturity, P = 897 The investor buys the bond at the highest price that guarantees she will receive at least her desired yield rate regardless of when the bond is called. The investor holds the bond for 20 years, after which time the bond is called. Calculate the annual yield rate the investor earns. Problem 47.14 Which of the following are true: (I) If a callable bond is called or redeemed at any date other than that used to calculate the purchase price, the buyer’s yield will be higher than that used to calculate the yield rate. (II) If the redemption value of a callable bond is equal at all possible redemption dates and the bond sells at a premium, the price of the bond is calculated by assuming the redemption date is the last possible date. (III) To calculate the price of a callable bond, a buyer should calculate the price at each and every possible redemption date and take the lowest price. Problem 47.15 A 5000 par value 18-year bond has 7% semiannual coupons, and is callable at the end of the 12th through the 17th years at par. (a) Find the price to yield 6% convertible semiannually. (b) Find the price to yield 8% convertible semiannually. (c) Find the price to yield 8% convertible semiannually, if the bond pays a pre- mium of 250 if it is called. (d) Find the price to yield 6% convertible semiannually, if the bond pays a premium of 250 if it is called at the end of years 12 through 14, and a premium of 150 if it is called at the end of years 15 or 16. (It may still be called at the end of year 17 without premium, and otherwise will mature at the end of year 18.) Problem 47.16 Find the price of a $1000 issue of 5.25% bonds with annual coupons which will be redeemed in 10 annual installments at the end of the 11th through the 20th years from the issue date at 105. The bonds are bought to yield 7% eﬀective. 426 BONDS AND RELATED TOPICS Problem 47.17 ABC is issuing a $1000 bond with a maturity of 25 year and annual 10% coupon. The bond is callable with the ﬁrst year call price equal to the oﬀering price plus the coupon; thereafter the call price decreases by equal amounts to equal par at year 20; thereafter the call price is equal to par. If the bond were sold at $950 what would be the call prices for each year? Stocks and Money Market Instruments One of the reasons investors are interested in trading with stocks because of its high rate of return (about 10%). Stocks are examples of ﬁnancial instruments or securities, terms which we deﬁne next. A ﬁnancial instrument is a monetary claim that one party has on another. It is a ﬁnancial asset for the person who buys or holds it, and it is a ﬁnancial liability for the company or institution that issues it. For example, a share of Microsoft stock that you own. The stock gives you a share of Microsoft’s assets and the right to receive a share of dividends (proﬁts), if Microsoft is paying any. To Microsoft owners, a stock is an obligation to include you in their divident payments. A security is a tradeable ﬁnancial instrument, like a bond or a share of stock. A ﬁnancial market is a market in which ﬁnancial assets are traded. In addition to enabling exchange of previously issued ﬁnancial assets, ﬁnancial markets facilitate borrowing and lending by facilitating the sale by newly issued ﬁnancial assets. Examples of ﬁnancial markets include the New York Stock Exchange (resale of previously issued stock shares), the U.S. government bond market (resale of previously issued bonds), and the U.S. Treasury bills auction (sales of newly issued T−bills). A ﬁnancial institution is an institution whose primary source of proﬁts is through ﬁnancial asset transactions. Examples of such ﬁnancial institutions include discount brokers (e.g., Charles Schwab and Associates), banks, insurance companies, and complex multi−function ﬁnancial institutions such as Merrill Lynch. Stocks are ﬁnancial instruments that make the holder a co −owner of the company that issued them. They entitle the holder to a claim on the assets (and implicitly the future proﬁts) of the company. Stocks do not involve the repayment of a debt or the payment of interest. (Some stocks pay dividends, which are shares of the company’s proﬁts, but people typically hold stocks not for the dividends but for the hope of reselling them later at a higher price.) In this chapter, we introduce two types of stocks: preferred and common stocks. We also discuss trading stocks. More speciﬁcally, we will discuss buying stocks by borrowing money (buying on margin) and selling stocks that we don’t own (short sales of socks). We conclude this chapter by introducing some of the ﬁnancial instrument used in today’s ﬁnancial markets. 427 428 STOCKS AND MONEY MARKET INSTRUMENTS 48 Preferred and Common Stocks Three common types of securities which have evolved in the ﬁnancial markets: bonds, preferred stocks, and common stocks. Bonds have been the topic of the previous chapter. In this section, we discuss preferred stocks and common stocks. Preferred Stock This is a type of security which provides a ﬁxed rate of return (similar to bonds). However, unlike bonds, it is an ownership security (bonds are debt security). Generally, preferred stock has no maturity date, although on occasion preferred stock with a redemption provision is issued. The periodic payment is called a dividend, because it’s being paid to an owner. Preferred stock is typically a higher ranking stock than common stock. Preferred stock may or may not carry voting rights, and may have superior voting rights to common stock. Preferred stock may carry a dividend that is paid out prior to any dividends to common stock holders. Preferred stock may have a convertibility feature into common stock. Preferred stock holders will be paid out in assets before common stockholders and after debt holders in bankruptcy. In general, there are four diﬀerent types of preferred stock: •Cumulative preferred stock. Preferred stock on which dividends accrue in the event that the issuer does not make timely dividend payments. Most preferred stock is cumulative preferred. •Non-cumulative preferred stock. Preferred stock for which unpaid dividends do not accrue. •Participating preferred stock. This is a kind of preferred stock that pays a regular dividend and an additional dividend when the dividend on the common stock exceeds a speciﬁc amount. •Convertible preferred stock. Convertible preferred stock has a privilege similar to convertible bonds. Owners have the option to convert their preferred stock to common stock under certain conditions. Since a preferred stock has no redemption date, it pays dividend forever. Their cash ﬂows are, therefore, those of a perpetuity. Thus, the price of a preferred stock is the present value of future dividends of a perpetuity which is given by the formula P = Fr i . Example 48.1 A preferred stock pays a coupon of 50 every six months. Calculate the price of the preferred stock using a yield rate of 5% convertible semi-annually. Solution. The price of the stock is P = Fr i = 50 0.025 = 2000 48 PREFERRED AND COMMON STOCKS 429 Common stock Common stock is another type of ownership security. It does not earn a ﬁxed dividend rate as preferred stock does. Dividends on this type of stock are paid only after interest payments on all bonds and other debt and dividends on preferred stock are paid. The dividend rate is ﬂexible, and is set by the board of directors at its discretion. Value of a share of common stock is being based on the dividend discount model. This model values shares at the discounted value of future dividend payments. That is, the value of a share is the present value of all future dividends. Consider a situation where a dividend D is paid at the end of the current period. Assume that the next dividend payments follow a geometric progression with common ratio 1 + k indeﬁnitely and the stock is purchased at a yied rate of iwith −1 <k <i. Then the theoretical price of the stock is P = D lim n→∞ 1 − (1+k 1+i )n i−k = D i−k. We point out that the most common frequency of dividend payments on both preferred and commom stock in the US is quarterly. Example 48.2 A common stock is currently earning $1 per share and the dividend is assumed to increase by 5% each year forever. Find the theoretical price to earn an investor an annual eﬀective yield rate of 10%. Solution. The theoretical price is P = D 1 i−k = 1 0.10 −0.05 = 20 It is unrealistic to project constant percentage increases in dividends indeﬁnitely into the future. As corporations increase in size and become more mature, the rate of growth will generally slow down. Example 48.3 A common stock pays annual dividends at the end of each year. The earnings per share for the most recent year were 8 and are assumed to grow at a rate of 10% per year, forever. The dividend will be 0% of earnings for each of the next 10 years, and 50% of earnings thereafter. What is the theoretical price of the stock to yield 12%? 430 STOCKS AND MONEY MARKET INSTRUMENTS Solution. The theoretical price of the stock which is the present value of all future dividends is ∞∑ k=11 4(1.1)k(1.12)−k = ∞∑ k=11 4 (1.1 1.12 )k =4 (1.1 1.12 )11 [ 1 1 − (1.1 1.12 ) ] = 183.73 Example 48.4 A common stock is currently earning $4 per share and will pay $2 per share in dividends at the end of the current year. Assuming that the earnings of the corporation increase at the rate of 5% for the ﬁrst ﬁve years, 2.5% for the second ﬁve years and 0% thereafter. Assume that the corporation plans to continue to pay 50% of its earnings as dividends, ﬁnd the theoretical price to earn an investor an annual eﬀective yield rate of 10%. Solution. The theoretical price is 2 [ 1 − (1.05 1.10 )5 0.10 −0.05 ] + 2(1.05)5 (1.10)5 [ 1 − (1.025 1.10 )5 0.10 −0.025 ] + 2(1.05)5(1.025)5 (1.10)10 · 1 0.10 = $25.72 48 PREFERRED AND COMMON STOCKS 431 Practice Problems Problem 48.1 A preferred stock pays a $10 dividend at the end of the ﬁrst year, with each successive annual dividend being 5% greater than the preceding one. What level annual dividend would be equivalent if i= 12%? Problem 48.2 A preferred stock will pay a dividend of 100 today. The stock also provides that future dividends will increase at the rate of inﬂation. The stock is purchased to yield 8% and assuming an inﬂation rate of 3%. Calculate the price of the stock today (before payment of the dividend). Problem 48.3 A common stock pays annual dividends at the end of each year. The earnings per share in the year just ended were $6. Earnings are assumed to grow 8% per year in the future. The percentage of earnings paid out as a dividend will be 0% for the next ﬁve years and 50% thereafter. Find the theoretical price of the stock to yield an investor 15% eﬀective. Problem 48.4 A common stock is purchased at a price equal to 10 times current earnings. During the next 6 years the stock pays no dividends, but earnings increase 60%. At the end of 6 years the stock sold at a price equal 15 times earnings. Find the eﬀective annual yield rate earned on this investment. Problem 48.5 A company’s common stock pays a dividend of $2 each year. The next dividend will be paid in one year. If the dividend is expected to increase at 5% per year, calculate the value of the stock at a 10% yield rate. Problem 48.6 A company’s current earnings are $5 per share. They expect their earnings to increase at 10% per year. Once their earnings are $10 per share, they will payout 75% as a dividend. Calculate the value of the stock to yield 15%. Problem 48.7 Thompson Corporation currently pays a dividend of 3 on earnings of 5 per year. Thompson’s earnings are expected to increase at a rate of 6% per year with 60% of earnings paid as dividends. Calculate the theoretical price of the common stock to yield 10%. 432 STOCKS AND MONEY MARKET INSTRUMENTS Problem 48.8 A common stock is currently earning $4 per share and will pay $2 per share in dividends at the end of the current year. Assuming that the earnings of the corporation increase 5% per year indeﬁnitely and that the corporation plans to continue to pay 50% of its earnings as dividends, ﬁnd the theoretical price to earn an investor an annual eﬀective yield rate of (1) 10%, (2) 8%, and (3) 6% Problem 48.9 A common stock is purchased on January 1, 1992. It is expected to pay a dividend of 15 per share at the end of each year through December 31, 2001. Starting in 2002 dividends are expected to increase K% per year indeﬁnitely, K <8%. The theoretical price to yield an annual eﬀective rate of 8% is 200.90. Calculate K. Problem 48.10 On January 1 of each year Company ABC declares a dividend to be paid quarterly on its common shares. Currently, 2 per share is paid at the end of each calendar quarter. Future dividends are expected to increase at the rate of 5% per year. On January 1 of this year, an investor purchased some shares at X per share, to yield 12% convertible quarterly. Calculate X. Problem 48.11 ‡ The dividends of a common stock are expected to be 1 at the end of each of the next 5 years and 2 for each of the following 5 years. The dividends are expected to grow at a ﬁxed rate of 2% per year thereafter. Assume an annual eﬀective interest rate of 6%. Calculate the price of this stock using the dividend discount model. Problem 48.12 ‡ A commom stock sells for 75 per share assuming an annual eﬀective interest rate of i. Annual dividends will be paid at the end of each year forever. The ﬁrst dividend is 6, with each subsequent dividend 3% greater than the previous year’s dividend. Calculate i. 49 BUYING STOCKS 433 49 Buying Stocks In this section, we discuss the costs associated with buying stocks. One way to buy stocks is by buying on margin (i.e. buying stocks with borrowed money). When an investor is buying an asset, he is said to have a long position in the asset. If the investor is selling an asset he is said to have a short position in that asset. In the stock and bond markets, if you short an asset, it means that you borrow it, sell the asset, and then later buy it back. Short sales will be discussed in Section 50. Now, depending upon whether an investor is buying or selling a security, he will get diﬀerent price quotes. A bid price is the price at which a broker is ready to purchase the security. Thus, this is the selling price for an investor. The ask price is the price at which a broker is ready to sell the security. The ask is the buying price paid by an investor. The diﬀerence between the two prices is called the bid-ask spread (bid price is lower than the asked price, and the spread is the broker’s proﬁt). Example 49.1 MBF stock has a bid price of $10.35 and an ask price of $11.00. Assume there is a $5 brokerage commission. (a) What is the bid-ask spread? (b) What amount will you pay to buy 50 shares? (c) What amount will you receive for selling 50 shares? (d) Suppose you buy 50 shares, then immediately sell them with the bid and ask price being the same as above. What is your round-trip transaction cost? Solution. (a) The bid-ask spread is 11 .00 −10.35 = $0.65. (b) You pay 50 ×11.00 + 5 = $555.00 (c) You receive 50 ×10.35 −5 = $512.50. (d) For each stock, you buy at $11.00 and sell it an instant later for $10.35. The total loss due to the ask-bid spread: 50(11 .00 −10.35) = $32.50. In addition, you pay $5 twice. Your total transaction cost is then 32 .50 + 2(5) = $42.50 Buying on Margin When buying on margin, the investor borrows part of the purchase price of the stock from a broker or a brokerage ﬁrm and contributes the remaining portion (the initial margin requirement or simply the margin) which is kept in an account with the brokerage ﬁrm. If you buy Q shares at the price of P (per share) with initial margin percentage of IM%, then your initial margin requirement IM is given by the formula IM = Q·P ·(IM%). 434 STOCKS AND MONEY MARKET INSTRUMENTS Example 49.2 Suppose you want to buy 100 shares on margin at the price of $50 per share. Assuming that the initial margin percentage is 50%, what was your initial margin requirement? Solution. You initial margin requirement IM is IM = 100(50)(0.5) = $2500 The brokerage ﬁrm is making money on your loan in terms of interest. Moreover, your stock will be hold as the collateral against the loan. If you default, the ﬁrm will take the stock. There are risks associated to buying stock on margin. The price of your stock could always go down. By law, the brokerage ﬁrm will not be allowed to let the value of the collateral (the price of your stock) go down below a certain percentage of the loan value. This percentage is called the maintenance margin. If the stock drops below that set amount, the brokerage ﬁrm will issue a margin call on your stock. The margin call means that you will have to pay the brokerage ﬁrm the amount of money necessary to bring the brokerage ﬁrms risk down to the allowed level. If you don’t have the money, your stock will be sold to pay oﬀ the loan. If there is any money left, it will be sent to you. In most cases, there is little of your original investment remaining after the stock is sold. We deﬁne the actual margin AM by the formula AM = Market Value of Assets−Loan Market Value of Assets If AM < MM,then the investor is undermargined (he or she receives a margin call from his or her broker) and must take corrective action (pay oﬀ part of loan or deposit more collateral). If the investor fails to respond, then the broker will close the investor’s account. If AM > IM,then the investor is overmargined and can withdraw cash or buy more shares. Example 49.3 John wants to purchase 100 stocks of IBM. The current price of a stock is $100. He has $6000 on hand, so he decides to borrow $4000. The maintenance marginal is set to be 50%. (a) What is John’s liability and equity? (b) Suppose the stock falls to $70. What is the actual margin? Would a margin call be in order? (c) How far can the stock price fall before a margin call? Solution. (a) John’s liability is $4000 and his equity is $6000. So John is contributing 6000 10000 = 60% (the margin) of the purchasing price. (b) If the stock falls to $70, then John’s liability is still $4000 but his equity is reduced to $3000 so that the new margin is 3000 7000 = 43% < 50%. Since this is less than the maintenance margin, the 49 BUYING STOCKS 435 broker will issue a margin call. (c) Let P be the original price. Then the margin in this case is 100P −4000 100P . Setting this to 0.5 and solving for P we ﬁnd P = $80. Thus, anytime the stock drops below $80, John will receive a margin call Example 49.4 John wants to purchase 100 stocks of IBM. The current price of a stock is $100. He has $6000 on hand, so he decides to borrow $4000. The broker charges an annual eﬀective rate of interest of 10%. Suppose the stock goes up by 30% in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Solution. (a) The end-year-value of the shares is 100[100 + (30%)(100)] = $13 ,000. (b) The total of loan repayment and interest is 4000 + 4000(10%) = $4 ,400. (c) The rate of return is 13000 −10400 6000 = 43.3% (d) If not buying on margin, the rate of return is 13000 −10000 10000 = 30% The reason why an investor buys on margin is his wish to invest more than what the investor’s money would allow. Buying on margin could mean a huge return when stock prices go up. But there is the risk that you could lose your original investment when stock price goes down. As with any stock purchase there are risks, but when you are using borrowed money, the risk is increased. 436 STOCKS AND MONEY MARKET INSTRUMENTS Practice Problems Problem 49.1 MBF stock has a bid price of $10.35 and an ask price of $11.00. Assume there is a 0.1% brokerage commission on the bid or ask price. (a) What amount will you pay to buy 50 shares? (b) What amount will you receive for selling 50 shares? (c) Suppose you buy 50 shares, then immediately sell them with the bid and ask price being the same as above. What is your round-trip transaction cost? Problem 49.2 MBF stock has a bid price of $10.35 and an ask price of $11.00. At what price can the market-maker purchase a stock? At what price can a market-maker sell a stock? What is the market-maker proﬁt from buying and immediately selling 50 shares? Problem 49.3 Suppose the spot ask exchange rate is $2.10 = £1.00 and the spot bid exchange rate is $2.07 = £1.00. If you were to buy $5,000,000 worth of British pounds and then sell them ﬁve minutes later without the bid or ask changing, how much of your $5,000,000 would be ”eaten” by the bid-ask spread? Problem 49.4 Consider the transaction of Example 49.4 but now assume that there is no change in the stock price in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Problem 49.5 Consider the transaction of Example 49.4 but now assume that there is a decrease of 30% in the stock price in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Problem 49.6 On July 24, 2002, you bought 100 shares of Microsoft on margin at the price of $45 per share. Assuming that the initial margin percentage is 50%, what was your initial margin requirement? 49 BUYING STOCKS 437 Problem 49.7 Using Problem 49.6, suppose that On July 25, 2002, Microsoft closed at $42.83. What was your actual margin on that day? Problem 49.8 Using Problem 49.6, suppose that the maintenance margin for Microsoft is 40%. Did you receive a margin call on July 25, 2002? Problem 49.9 On December 30, 1998, you decided to bet on the January eﬀect, a well−known empirical regularity in the stock market. On that day, you bought 400 shares of Microsoft on margin at the price of $139 per share. The initial margin requirement is 55% and the maintenance margin is 30%. The annual cost of the margin loan is 6.5%. (a) Determine your initial margin requirement. (b) To what price must Microsoft fall for you to receive a margin call? (c) On January 8, 1999, Microsoft climbed to $151.25. What was the actual margin in your account on that day? (d) On January 9, 1999, you sold your Microsoft shares at the price of $150.50 per share. What was the return on your investment? (e) Recalculate your answer to part (d) assuming that you made the Microsoft stock purchase for cash instead of on margin. 438 STOCKS AND MONEY MARKET INSTRUMENTS 50 Short Sales In this section we discuss trading security over short period of time when the security is likely to decline in value. A short sale is generally a sale of a security by an investor who does not actually own the security. To deliver the security to the purchaser, the short seller will borrow the security. The short seller later ”pays oﬀ” the loan by returning the security to the lender, typically by purchasing the sold security back on the open market (hopefully at a lower price). This process of buying back the security is called covering the short because the seller is no longer at risk to movements in the security. One problem with short sales is determining the yield rate. If the transaction occurs exactly as stated, then technically the yield rate does not exist. We illustrate this in the following example. Example 50.1 An investor sells a stock short for $1000 and buys it back for $800 at the end of one year. What is the yield rate? Solution. A possible equation of value is 1000(1 + i) = 800 which yields i= −20% which is clearly unreason- able since the seller made $200 in proﬁt from the transaction. Now, if we reverse the transaction and solve the equation of value 800(1 + i) = 1000 we obtain i= 25%.But this equation states that there was $200 proﬁt on an $800 investment, but there never was an $800 investment Short sales rarely occur as illustrated above. In practice, the short seller is required to make a deposit (i.e. a collateral) of a percentage of the price at the time the short sale is made. This deposit is called the margin, and cannot be accessed by the short seller until the short sale is covered. However, the short seller will be credited with interest on the margin deposit which will increase the yield rate. Margins are usually expressed as a percentage of the value of the security. For example, govern- mental regulations in the United States require the short seller to make a deposit of 50% of the price at the time the short sale is made. In the example above, the short seller will have to deposit a margin of $500 at the time the short sale. The short seller is not allowed to earn interest on the proceeds from the short sale. Governmental regulations require that these proceeds remain in a non-interest bearing account until the short position is covered, at which time these funds will be used for the purchase necessary to cover the short. Any leftover money is proﬁt on the transaction, while a negative balance would be loss on the transaction. 50 SHORT SALES 439 If coupons or dividends are paid on the security while you are short it, you must pay them to the party you borrowed the security from. Dividend payments are often deducted from the margin account. In order to ﬁnd the yield rate on a short sale we introduce the following notation: •M = Margin deposit when sell was made ( t= 0) •S0 = Proceeds from short sale •St = Cost to repurchase stock at time t •dt = Dividend at time t •i = Periodic interest rate of margin account •j = Periodic yield rate of short sale If a short seller repurchases stock after n periods, then the equation of value at time t= n is: M(1 + j)n = M(1 + i)n + S0 −Sn − n∑ t=1 dt(1 + i)n−t. Example 50.2 ‡ Bill and Jane each sell a diﬀerent stock short for a price of 1000. For both investors, the margin requirement is 50%, and interest on the margin is credited at an annual eﬀective rate of 6%. Bill buys back his stock one year later at a price of P. At the end of the year, the stock paid a dividend of X. Jane also buys back her stock after one year, at a price of ( P −25). At the end of the year her stock paid a dividend of 2 X. Both investors earned an annual eﬀective yield of 21% on their short sales. Calculate P and X. Solution. The information for Bill gives 500(1.21) = 500(1.06) + 1000−P−X and for Jane gives 500(1.21) = 500(1.06) + 1000 −P + 25 −2X. Solving this system of two linear equations in two unknowns we ﬁnd P = 900 and X = 25 Example 50.3 Suppose Eddie sells 10 shares of a company short; the stock price is $87 per share, and margin requirement is 40%. The company pays semiannual dividends of 2 per share. Eddie’s margin account earns 3% nominal annual interest, convertible semiannually, and he buys the stock back after 18 months at a share price of 80. What is his annual eﬀective rate of return? Solution. We have M = 10(87)(0.4) = 348. Thus, 348(1 + j)3 = 348(1.015)3 + 870 −800 −20s3 0.015 Solving for j we ﬁnd j = 2.34%.Hence, the annual eﬀective rate of return is (1.0234)2 −1 = 4.73% 440 STOCKS AND MONEY MARKET INSTRUMENTS Example 50.4 On 01/01/2005, Bryan sells a stock short for a price of 800. The margin requirement is 50%, and interest on the margin is credited at an annual eﬀective interest rate of 2%. The stock pays a dividend, D, on 12/31/2005. On 01/01/2006, Bryan buys the stock back at a price of 820 to cover the short. Bryan’s yield on the short sell is −10%. Calculate D. Solution. We are given the following information: M = 400,S0 = 800,S1 = 820,i = 0.02,j = −0.10. The equation of value is 400(0.90) = 400(1.02) + 800−820 −D. Solving this equation for D we ﬁnd D= 28 Short selling by itself is a speculative endeavor and should not be entered into lightly. The short seller is counting on a signiﬁcant decline in the value of the security in order to come out ahead and such an act is istself a risky one. However, investment strategies have been developed that involve combinations of long and short positions in related securities, which have greatly reduced risk and also have a good prospect to earn a reasonable return on investment. A generic name for these transactions is hedging. Situations of this type in which a proﬁt is certain are called arbitrage. These two concepts will appear later in the book when discussing derivatives markets. Short selling has been blamed (erroneously) for many historic market crashes, and has frequently been harshly regulated or banned. 50 SHORT SALES 441 Practice Problems Problem 50.1 John sells a stock short for 200. A year later he purchases the stock for 160. The margin requirements are 50% with the margin account earning 4%. The stock that John sold paid a dividend of 6 during the year. Calculate John’s return on the stock investment. Problem 50.2 Don and Rob each sell a diﬀerent stock short. Don sells his stock short for a price of 960, and Rob sells his short for 900. Both investors buy back their securities for X at the end of one year. In addition, the required margin is 50% for both investors, and both receive 10% annual eﬀective interest on their margin accounts. No dividends are paid on either stock. Don’s yield rate on his short sale is 50% greater than Rob’s yield rate on his short sale. Calculate Don’s yield rate. Problem 50.3 Brian sells a stock short for 800 and buys it back one year later at a price of 805. The required margin on the sale is 62.5% and interest is credited on the margin deposit at an annual eﬀective rate of i. Dividends of 15 are paid during the year. Brian’s yield rate over the one year period is j.If the interest credited on the margin deposit had been 1 .25i,with everything else remaining the same, Brian’s yield rate over the one-year period would have been 1 .5j. Calculate i. Problem 50.4 An investor sells short 500 shares of stock at 10 per share and covers the short position one year later when the price of the stock has declined to 7.50. The margin requirement is 50%. Interest on the margin deposit is 8% eﬀective. Four quarterly dividends of 0.15 per share are paid. Calculate the yield rate. Problem 50.5 Andrew sells a stock short for 800. At the end of one year, Andrew purchases the stock for 760. During the year, Andrew’s stock paid a dividend of 50. Amanda sold a diﬀerent stock short 1000. At the end of the year, Amanda purchases the stock she sold short for X. Amanda’s stock paid a dividend of 25 during the year. The margin requirement for both Andrew and Amanda is 60% and they both earn 10% interest on the margin account. Andrew and Amanda both received the same return. Calculate X. Problem 50.6 Thomas sells a stock short for 1000. The margin rate is 60%. The stock does not pay dividends. After one year, Thomas purchases the stock for 880 which results in a yield of 25%. Calculate the interest rate earned on the margin account. 442 STOCKS AND MONEY MARKET INSTRUMENTS Problem 50.7 Jenna sells short a stock for 1000. One year later she buys the stock for 1000. The stock pays a dividend of 10. Jenna earns 10% on her margin requirement of 60%. Jordan sells short a stock for 500. Jordan’s stock does not pay a dividend. Jordan earns 5% on his margin requirement of 60%. When Jordan buys the stock one year later, he earns a yield that is twice that earned by Jenna. Determine the price that Jordan paid to purchase his stock. Problem 50.8 James sells a stock short for 1000. The margin requirement is 60% and James earns 10% on the margin account. One year later he buys the stock for 790. Jack sells a stock short for 600. The margin requirement is 50% and Jack earns 10% on the margin account. One year later she sells the stock for 450. Both stocks pay the same dividend. James and Jack both earn the same yield. Determine the dividend paid on each stock. Problem 50.9 Rachel sells a stock short for 2000. The margin requirement is 50%. During the year, Rachel’s stock pays a dividend of 25. Rachel purchases the stock for 2050. Danielle sells a stock short for 1500. Her margin requirement is 60%. During the year, Danielle’s stock pays a dividend of 10. Danielle purchases the stock one year later for 1535. Danielle earns a return on her transaction that is twice the return that Rachel earns on her trans- action. Both Rachel and Danielle are paid the same interest rate on the margin account. Calculate the interest rate earned on the margin account. Problem 50.10 Adam has a yield on a short sale of the stock of 30%. The interest rate on the margin account is 10%. If the margin requirement had been doubled with no other changes to the transaction, determine the yield on Adam’s transaction. Problem 50.11 On the short sale of a stock, James realizes a yield of 22%. The margin requirement was 50% and he earned 10% on the margin. If the margin had been increased to 60%, but the interest earned on the margin is still 10%, calculate James’ yield. Problem 50.12 ‡ Theo sells a stock short with a current price of 25,000 and buys it back for X at the end of 1 year. Governmental regulations require the short seller to deposit margin of 40% at the time of the short 50 SHORT SALES 443 sale. No dividends incurred. The prevailing interest rate is an 8% annual rate, and Theo earns a 25% yield on the transaction. Calculate X. Problem 50.13 On 01/01/2005, Company XYZ ′s stock is trading at $80 per share, and Betty sells short $1000 worth of the stock. (Note: Shares of stock are bought and sold in fractional pieces; you do not have to buy or sell a whole number of shares.) Betty is required to put up a margin of $500, and it is agreed that she will receive interest on the margin at an annual eﬀective interest rate of k%. Company XYZ ′s stock pays a dividend of 1.45 per share on 12/31/2005. On 01/01/2006, Company XYZ ′sstock is trading at $76 per share, and Betty covers the short at this time for a yield of 9.75%. Calculate k. Problem 50.14 ‡ Eric and Jason each sell a diﬀerent stock short at the beginning of the year for a price of 800. The margin requirement for each investor is 50% and each will earn an annual eﬀective interest rate of 8% on his margin account. Each stock pays a dividend of 16 at the end of the year. Immediately thereafter, Eric buys back his stock at a price of (800 −2X) and Jason buys back his stock at a price of (800 +X). Eric’s annual eﬀective yield, j,on the short sale is twice Jasons annual eﬀective yield. Calculate j. Problem 50.15 ‡ Jose and Chris each sell a diﬀerent stock short for the same price. For each investor, the margin requirement is 50% and interest on the margin debt is paid at an annual eﬀective rate of 6%. Each investor buys back his stock one year later at a price of 760. Jose’s stock paid a dividend of 32 at the end of the year while Chris’s stock paid no dividends. During the 1-year period, Chris’s return on the short sale is i, which is twice the return earned by Jose. Calculate i. Problem 50.16 ‡ On January 1, 2004, Karen sold stock A short for 50 with a margin requirement of 80%. On December 31, 2004, the stock paid a dividend of 2, and an interest amount of 4 was credited to the margin account. On January 1, 2005, Karen covered the short sale at a price of X, earning a 20% return. Calculate X. Problem 50.17 Kathy and Megan each sell short the same stock. Kathy is required to maintain a margin re- quirement of 50% while Megan is required to maintain a margin requirement of 75%. Assume no dividends paid on the stock and the purchase price is the same. Which of the following are true: (I) If no interest is paid on either margin account, Kathy’s yield will be 150% of Megan’s yield. 444 STOCKS AND MONEY MARKET INSTRUMENTS (II) If the same interest rate is paid on both margin accounts, Megan’s yield will exceed 2/3 of Kathy’s yield. (III) If the same interest rate is paid on both margin accounts, Megan’s yield could never exceed Kathy’s yield. 51 MONEY MARKET INSTRUMENTS 445 51 Money Market Instruments In this section we brieﬂy introduce some of the ﬁnancial instruments used in today’s ﬁnancial mar- kets. These instruments can be analyzed with the principles developed in the earlier chapters of this book. The ﬁnancial intruments introduced fall into the two categories: Money market instru- ments and ﬁnancial derivatives. In this section we will cover money markets instruments. Basics of ﬁnancial derivatives will be introduced and discussed at a later sections. Corporations and government organizations are continually borrowing and lending money. One way for raising money is through money market instruments. Money market instruments are cash-equivalent investments that consist of short-term (usually under a year), very low-risk debt instruments. The money market is the market for buying and selling short-term loans and se- curities. The buyer of the money market instrument is the lender of money and the seller is the borrower of money. Below, we introduce few types of money market instruments that are traded in money markets. Money Market Mutual Funds A money market mutual fund is a mutual fund that attempts to keep its share price at $1. Professional money managers will take your cash and invest it in government treasury bills, savings bonds, certiﬁcates of deposit, and other safe and conservative short term commercial paper. They then turn around and pay you, the owner of the shares, your portion of the interest earned on those investments. Money market funds generally allow withdrawals on demand without penalty, making them an ideal place to keep money while considering other investment opportunities. Investment in a money market fund is neither insured nor guaranteed by neither the FDIC or by any other government agency, and there can be no guarantee that the fund will maintain a stable $1 share price. It is possible to lose money by investing in the fund. An example of money market fund investment is a commercial paper security issued by banks and large corporations. An example of a commercial paper is a promissory note. A promissory note is a written promise to repay a loan or debt under speciﬁc terms −usually at a stated time, through a speciﬁed series of payments, or upon demand. Example 51.1 On January 31, Smith borrows $500 from Brown and gives Brown a promissory note. The note states that the loan will be repaid on April 30 of the same year, with interest at 12% per annum. On March 1 Brown sells the promissory note to Jones, who pays Brown a sum of money in return for the right to collect the payment from Smith on April 30. Jones pays Brown an amount such that Jones’ yield (interest rate earned) from March 1 to the maturity date can be stated as an annual rate of interest of 15%. 446 STOCKS AND MONEY MARKET INSTRUMENTS Determine the amount that Jones paid to Brown and the yield rate Brown earned quoted on an annual basis. Assume all calculations are based on simple interest and a 365-day non-leap year. Solution. On April 30, the value of the note is AV = 500 [ 1 + (89 365 ) (0.12) ] = $514.63. The amount Jones paid to Brown on March 1 is P [ 1 + (60 365 ) (0.15) ] = $514.63 Solving for P we ﬁnd P = $502.24. The yield rate earned by Brown satisﬁes the equation 502.24 = 500 [ 1 + (29 365 ) i ] Solving for i we ﬁnd i= 5.64% Certiﬁcates of Deposit A certiﬁcate of deposit (“CD”) is like a savings account in the sense that it is an FDIC (Federal Deposit Insurance Corporation) insured investment. However, a CD has a ﬁxed term (often three months, six months, or one to ﬁve years). It is intended that the CD be held until maturity, at which time the money may be withdrawn together with the accrued interest. CDs are available at banks and savings and loan institutions. CDs are issued in a range of denom- inations with the higher denominations (e.g. $100,000) carrying the higher rates of interest than the lower denominations. CDs carry a ﬁxed yield rate and in this sense is considered more stable than a money market fund that carries a variable interest rate. However, there is the disadvantage of liquidity since penalties are imposed for early withdrawal. Example 51.2 A two-year certiﬁcate of deposit pays an annual eﬀective rate of 9%. The purchaser is oﬀered two options for prepayment penalties in the event of early withdrawal: A −a reduction in the rate of interest to 7% B −loss of three months interest. 51 MONEY MARKET INSTRUMENTS 447 In order to assist the purchaser in deciding which option to select, compute the ratio of the proceeds under Option A to those under Option B if the certiﬁcate of deposit is surrendered: (a) At the end of six months. (b) At the end of 18 months. Solution. (a) Option’s A proceeds: (1 .07)0.5 = 1.03441.Option’s B proceeds: (1 .09).25 = 1.02178.The ratio is 1.03441 1.02178 = 1.0124. So Option A is preferable. (b) Option’s A proceeds: (1 .07)1.5 = 1.10682. Option’s B proceeds: (1 .09)1.25 = 1.11374. The ratio is 1.10682 1.11374 = 0.9938. So Option B is preferable Guaranteed Investment Contracts (GIC) GICs are investment instruments issued by insurance companies, usually to large investors such as pension funds. They guarantee principal and interest over a stated period of time (like CDs), so their market value does not ﬂuctuate with movements in the interest rate. GICs are diﬀerent from CDs in that they allow additional deposits during the ﬁrst 3 months to a year. Like CDs, withdrawals are restricted. GICs pay higher rates of interest than CDs, but they are not insured by the FDIC against default. Interest earned on the initial investment may be reinvested at the guaranteed rate, or it may be reinvested at some lower ﬁxed rate. Example 51.3 An investor deposits $1,000,000 at the beginning of each year for 5 years in a 5-year GIC with an insurance company that guarantees an annual rate 6% on the fund. Determine the lump sum payable by the investor at 5 years (a) if the interest that is generated each year is reinvested at 6%; (b) if the interest that is generated each year is reinvested at 4%; Solution. (a) If the guaranteed interest is reinvested at 6% then the lump sum payable by the investor at 5 years is 448 STOCKS AND MONEY MARKET INSTRUMENTS 1,000,000¨s5 0.06 = $5,975,318.54. (b) If the guaranteed interest is reinvested at 4% then the lump sum payable by the investor at 5 years is 5,000,000 + 60,000(1.04)4 + 2 ×60,000(1.04)3 + 3 ×60,000(1.04)2 +4 ×60,000(1.04) + 5×60,000 = $5,949,463.19 Mutual Funds A mutual fund is a professionally-managed form of collective investments that pools money from many investors and invests it in stocks, bonds, short-term money market instruments, and/or other securities. In a mutual fund, the fund manager trades the fund’s underlying securities, realizing capital gains or losses, and collects the dividend or interest income. The investment proceeds are then passed along to the individual investors. The value of a share of the mutual fund, known as the net asset value per share (NAV), is calculated daily based on the total value of the fund divided by the number of shares currently issued and outstanding. Example 51.4 A mutual fund account has the balance $100 on January 1. On July 1 (exactly in the middle of the year) the investor deposits $250 into the account. The balance of the mutual fund at the end of the year is $400. (a) Calculate the exact dollar-weighted annual rate of interest for the year. (b) Suppose that the balance of the fund immediately before the deposit on July 1 is $120. Calculate the time-weighted annual rate of interest for the year. Solution. (a) 100(1 + i) + 250(1 +i) 1 2 = 400 →100(1 +i) + 250(1 +i) 1 2 −400 = 0.By the quadratic formula, (1 + i) 1 2 = 1.108495283 →(1 + i) = 1.228761792 →i= 0.228761792. (b) (1 + j) = (120/100)(400/370) = 1.297297297 so the time-weighted rate is j = 0.297297297 Mortgage Backed Securities (MBS) Mortgage-backed securities(MBS) represent an investment in mortgage loans. An MBS investor owns an interest in a pool of mortgages, which serves as the underlying assets and source of cash ﬂow for the security. Most MBSs are issued by the following three ﬁnancial companies: •Government National Mortgage Association (Ginnie Mae), a U.S. government agency. Ginnie Mae, backed by the full faith and credit of the U.S. government, guarantees that investors receive timely payments. 51 MONEY MARKET INSTRUMENTS 449 •The Federal National Mortgage Association (Fannie Mae) and the Federal Home Loan Mortgage Corporation (Freddie Mac), U.S. government-sponsored enterprises. Fannie Mae and Freddie Mac also provide certain guarantees and, while not backed by the full faith and credit of the U.S. gov- ernment, have special authority to borrow from the U.S. Treasury. Mortgage-backed securities typically carry some of the highest yields of any government or agency security. Example 51.5 A homeowner signs a 30 year mortgage that requires payments of $971.27 at the end of each month. The interest rate on the mortgage is 6% compounded monthly. If the purchase price of the house is $180,000 then what percentage down payment was required? Solution. The purchase price of the home is 971.27a360 0.005 = 162,000. The amount of down payment is 180,000 −162,000 = 18,000. So the down payment is 10% of the purchasing price 450 STOCKS AND MONEY MARKET INSTRUMENTS Practice Problems Problem 51.1 Consider again Example 51.1. Suppose instead that Jones pays Brown an amount such that Jones’ yield is 12%. Determine the amount that Jones paid. Problem 51.2 You prurchase $10,000 CD with an interest rate of 5% compounded annually and a term of one year. At year’s end, the CD will have grown to what value? Problem 51.3 ‡ A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible quarterly) 1 4% 3 5% 5 5.65% The bank does not permit early withdrawal. The certiﬁcates mature at the end of the term. During the next six years the bank will continue to oﬀer these certiﬁcates of deposit with the same terms and interest rates. An investor initially deposits $10,000 in the bank and withdraws both principal and interest at the end of six years. Calculate the maximum annual eﬀective rate of interest the investor can earn over the 6-year period. Problem 51.4 An investor deposits $100,000 in a 3-year GIC with an insurance company that guarantees an annual 4% interest rate on the fund. If the GIC speciﬁes that interest is reinvested at 4% within the GIC, determine the lump sum payable to the investor after the three-year period. Problem 51.5 An investor puts $100 into a mutual fund in the ﬁrst year and $50 in the second year. At the end of the ﬁrst year the mutual pays a dividend of $10. The investor sells the holdings in the mutual fund at the end of the second year for $180. Find the dollar weighted return. Problem 51.6 You have a mutual fund. Its balance was $10,000 on 31/12/1998. You made monthly contributions of $100 at the start of each month and the ﬁnal balance was $12,000 at 31/12/1999. What was you approximate return during the year? 51 MONEY MARKET INSTRUMENTS 451 Problem 51.7 An investor purchases 1000 worth of units in a mutual fund whose units are valued at 4.00. The investment dealer takes a 9% “front-end load” from the gross payment. One year later the units have a value of 5.00 and the fund managers claim that the “fund’s unit value has experienced a 25% growth in the past year.” When units of the fund are sold by an investor, there is a redemption fee of 1.5% of the value of the units redeemed. (a) If the investor sells all his units after one year, what is the eﬀective annual rate of interest of his investment? (b) Suppose instead that after one year the units are valued at 3.75. What is the return in this case? Problem 51.8 Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. What is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 51.9 The interest on a 30 year mortgage is 12% compounded monthly. The mortgage is being repaid by monthly payments of 700. Calculate the outstanding balance at the end of 10 years. Problem 51.10 The interest rate on a 30 year mortgage is 12% compounded monthly. Lauren is repaying the mortgage by paying monthly payments of 700. Additionally, to pay oﬀ the loan early, Lauren has made additional payments of 1000 at the end of each year. Calculate the outstanding balance at the end of 10 years. Problem 51.11 A mortgage loan is being repaid with level annual payments of 5000 at the end of the year for 20 years. The interest rate on the mortgage is 10% per year. The borrower pays 10 payments and then is unable to make payments for two years. Calculate the outstanding balance at the end of the 12th year. Problem 51.12 A 20,000 mortgage is being repaid with 20 annual installments at the end of each year. The borrower makes 5 payments, and then is temporarily unable to make payments for the next 2 years. Find an expression for the revised payment to start at the end of the 8th year if the loan is still to be repaid at the end of the original 20 years. 452 STOCKS AND MONEY MARKET INSTRUMENTS Measures of Interest Rate Sensitivity In this chapter we study the sensitivity of the interest rate. We will cover the following topics: (1) The eﬀect of inﬂation on interest rate. We will consider the eﬀect of inﬂation in the calculations of present values and accumulated values. (2) The eﬀect of the investment period on interest rates. That is, interest rate is sensitive to time. This fact is known as the term structure of interest rate. A plot of the interest rate of an investment against time to maturity will be discussed. We call such a plot a yield curve. (3) We want to measure the change in a bond’s price as interest rate change. We call this measure the duration; the two types of duration we will look at are the modiﬁed duration and the Macaulay duration. (4) Immunization is a technique that helps investors shielding their overall ﬁnancial position from exposure to interest rate ﬂuctuations. (5) Absolute matching: The idea here is to structure assets such that the resulting asset inﬂows will exactly match the liability outﬂow. 453 454 MEASURES OF INTEREST RATE SENSITIVITY 52 The Eﬀect of Inﬂation on Interest Rates What is inﬂation? Inﬂation is deﬁned as a sustained increase in the general level of prices for goods and services. It is measured as an annual percentage increase, known as rate of inﬂation . As inﬂation rises, every dollar you own buys a smaller percentage of a good or service. That is, inﬂation represents a loss of purchasing power. The opposite to inﬂation is deﬂation. Experience has shown that inﬂation has a signiﬁcant eﬀect on the rate of interest. As a matter of fact they are positively correlated in the sense that over time both tend to move in the same direction. The following example illustrates how inﬂation aﬀects the purchasing power of money. Suppose that today a gallon of milk costs $4. Then with $100 one can buy 25 gallons of milk. Suppose the $100 is invested for 2 years at an 8% annual eﬀective rate. Then at the end of the two years, you will have 100(1.08)2 = $116.64.Without the presence of inﬂation, that is assuming the cost of milk is still $4, you can now buy 29 gallons of milk. Now suppose that there has been a constant rate of inﬂation during the two years of 5% a year so that milk will cost 4(1 .05)2 = $4.41 per gallon. In this case, one can only buy 116.64 4.41 = 26.45 gallons of milk, a reduction of 3 gallons. The real rate of return, i′, is measured by solving the equation 25(1 + i′)2 = 116.64 4.41 = 25(1.08)2 (1.05)2 by taking the square root of both sides we obtain (1 + i′)(1.05) = 1.08 or 1 + ireal rate = 1 + inominal rate 1 + iinflation rate . To study the eﬀect of inﬂation on rates we introduce some notations. The rate of interest after eliminating the eﬀect of inﬂation is called the real rate of interest and will be denoted by i′, while the actual rate of interest charged in the market (i.e. the rate not adjusted for inﬂation) is called the nominal interest rate (not to be confused with the diﬀerent meaning of nominal used in Section 9) and will be denoted by i. Let r denote the rate of inﬂation. Assuming constant rate of inﬂation, the relationship between real and nominal interest rates can be described in the equation: 1 + i= (1 + i′)(1 + r), i>i ′>0,r >0. From this we can write 52 THE EFFECT OF INFLATION ON INTEREST RATES 455 1 + i′= 1 + i 1 + r (52.1) or i′= i−r 1 + r. A common approximation for the real interest rate is given by the Fisher equation: i′= i−r which states that the real interest rate is the nominal interest rate minus the expected rate of inﬂation. If inﬂation is positive, which it generally is, then the real interest rate is lower than the nominal interest rate. That’s the reason we are imposing the condition i>i ′ whenever r >0. Formula (52.1) is quite useful in performing calculations involving rates of inﬂation. For example, assume that we wish to ﬁnd the present value of a series of payments at the end of each period for nperiods in which the base payment amount at time 0 is R, but each payment is indexed to reﬂect inﬂation. If r is the periodic rate of inﬂation and i is the periodic nominal interest rate, then PV = R [ 1 + r 1 + i + (1 + r 1 + i )2 + ··· + (1 + r 1 + i )n] = R(1 + r) 1 − (1+r 1+i )n i−r . However, if we use Formula (52.1) the above result becomes PV = R [ 1 1 + i′ + 1 (1 + i′)2 + ··· + 1 (1 + i′)n ] = Rani′. The above two formulas provide a guidance in computing present values of future payments as follows: 1. If future payments are not aﬀected by inﬂation, then discount at the nominal rate of interest. 2. If future payments are adjusted to reﬂect the rate of inﬂation and the adjustment is reﬂected in the payment amount, then also discount at the nominal rate of interest. 3. If future payments are adjusted to reﬂect the rate of inﬂation but the adjustment is not reﬂected in the payment amount, the correct procedure is to discount at the real rate of interest. We next consider inﬂation in connection with accumulated values. Consider the common situation in which an investor invests A dollars for n periods at interest rate i. The value of this investment in “nominal dollars” at the end of n periods is AV = P(1 + i)n. 456 MEASURES OF INTEREST RATE SENSITIVITY However, if the rate of inﬂation is accounted for, then the purchasing power of this investment is AV = P (1 + i 1 + r )n = P(1 + i′)n. Thus, the value of this investment in “real dollars” is lower since i>i ′. Example 52.1 Money is invested for 3 years at an interest rate of 4% eﬀective. If inﬂation is 5% per year over this period, what percentage of purchasing power is lost? Solution. One dollar will accumulate to (1 .04)3 = 1.1248 over the period, but the purchasing power will only be (1.04)3 (1.05)3 = 0.9716 due to inﬂation. The loss of purchasing power per dollar of original investment is 1 −0.9716 = 2.84% Example 52.2 The real interest rate is 6% and the inﬂation rate is 4%. Allan receives a payment of 1,000 at time 1, and subsequent payments increase by 50 for 5 more years. Determine the accumulated value of these payments in nominal dollars at time 6 years. Solution. The nominal interest rate is: i= (1.06)(1.04) −1 = 10.24% The accumulated value of the cash ﬂows is: AV =1,000(1.1024)5 + 1,050(1.1024)4 + 1,100(1.1024)3 + 1,150(1.1024)2 + 1,200(1.1024) + 1,250 = 8623.08 52 THE EFFECT OF INFLATION ON INTEREST RATES 457 Practice Problems Problem 52.1 An insurance company is making annual payments under the settlement provisions of a personal injury lawsuit. A payment of $24,000 has just been made and ten more payments are due. Future payments are indexed to the Consumer Price Index which is assumed to increase at 5% per year. Find the present value of the remaining obligation if the assumed interest rate is 8%. Use the exact value of i′. Problem 52.2 The nominal rate of interest is 10% and the rate of inﬂation is 5%. A single deposit is invested for 20 years. Let Adenote the value of the investment at the end of 10 years, with the rate of inﬂation accounted for in the calculation. Let B denote the value of the investment at the end of 10 years computed at the real rate of interest. Find the ratio of A B. Problem 52.3 The nominal rate of interest is 8% and the rate of inﬂation is 5%. Calculate the real rate of interest. Problem 52.4 Mindy invests 10,000 for 10 years at a nominal rate of interest of 8%. The rate of inﬂation is 5% over the 10 year period. Calculate the value at the end of 10 years of Mindy’s investment in today’s dollars. Problem 52.5 As a result of sky-high gas prices, the cost of delivery pizza is expected to experience a constant 8% annual rate of inﬂation. In today’s dollars, Obese Oliver spends $2,000 per year on delivery pizza. His delivery pizza-eating habits will not change over the next ten years, so he will pay whatever price increases are necessary to sustain this level of consumption. If the nominal interest rate (note that “nominal” refers to both interest convertibility and its relationship to inﬂation), convertible semiannually, is 10%, ﬁnd the present value of Oliver’s delivery pizza purchases for the next ten years. Problem 52.6 Money is invested for 5 years in a savings account earnings 7% eﬀective. If the rate of inﬂation is 10%, ﬁnd the percentage of purchasing power lost during the period of investment. Problem 52.7 ‡ An insurance company has an obligation to pay the medical costs for a claimant. Average annual claims costs today are $5,000, and medical inﬂation is expected to be 7% per year. The claimant 458 MEASURES OF INTEREST RATE SENSITIVITY is expected to live an additional 20 years. Claim payments are made at yearly intervals, with the ﬁrst claim payment to be made one year from today. Find the present value of the obligation if the annual interest rate is 5%. Problem 52.8 ‡ Seth deposits X in an account today in order to fund his retirement. He would like to receive payments of 50 per year, in real terms, at the end of each year for a total of 12 years, with the ﬁrst payment occurring seven years from now. The inﬂation rate will be 0.0% for the next six years and 1.2% per annum thereafter. The annual eﬀective rate of return is 6.3%. Calculate X. 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 459 53 The Term Structure of Interest Rates and Yield Curves In the previous section we discussed a factor aﬀecting interest rates, i.e. inﬂation. In this section we discuss another factor which is the length of investment period. The term structure of interest rates refers to the way interest rates depend on time. Usually long-term market interest rates are higher than short-term market interest rates. An example of term structure of interest rates is given in the table below. Length of investment Interest rate (or Spot rate) 1 year 3% 2 year 4% 3 year 6% 4 year 7% Table 53.1 We can extend the values in this table to a continuous graph, and the resulting graphical illustration is called the yield curve corresponding to the table. The interest rates on the yield curve (or listed in the table) are often called spot rates. Economists and investors believe that the shape of the yield curve reﬂects the market’s future expectation for interest rates and the conditions for monetary policy. For example, an upward- sloping yield curve implies market’s expectation of future increases in interest rates. In some cases, short-term rates exceed long-term rates, that is the yield curve has negative slope, in this case the curve is called inverted. When all maturities have the same yield the curve is called ﬂat. Now, recall from Section 30 the formula for the net present value of a series of future payments given by NPV (i) = n∑ t=0 (1 + i)−tct. This formula is based on a single rate of interest i.With the presence of a term structure of interest rates, the net present value can be found by discounting each payment by its associated spot rate. That is, if we denote the spot rate for period t as it then the net present value is given by the equation NPV (i∗) = n∑ t=0 (1 + it)−tct. This formula can be considered as an application of computing annuity values with varying rates of interest as discussed in Section 21. Example 53.1 You are given the following term structure of interest rates: 460 MEASURES OF INTEREST RATE SENSITIVITY Length of investment Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% Find the present value of payments of $1000 at the end of each year for ﬁve years using the spot rates given above. What level yield rate would produce an equivalent value? Solution. The present value of the payments is 1000[(1.07)−1 + (1.08)−2 + (1.0875)−3 + (1.0925)−4 + (1.095)−5] = $3906.63. The equivalent level yield rate is found by solving the equation a5 i = 3 .90663 which gives i = 0.0883 = 8.83% using a ﬁnancial calculator A special type of spot rates is the forward interest rate . In order to illustrate this concept, consider a ﬁrm which needs to borrow a sizeable amount of money for two years. The ﬁrm is presented with two options. The ﬁrst option is to borrow money for 2 years at the 2-year spot rate of 10%. The second option is to borrow for one year at the one-year spot rate now of 8%, and to borrow for the second year at the one-year spot rate in eﬀect a year later. This 1-year spot rate for the next year is an example of a forward rate or more speciﬁcally a one year forward interest rate. This is what the one year spot rate will be after one year is passed. Now, if f is the forward rate then the ﬁrm will be indiﬀerent between the two options if (1.10)2 = (1.08)(1 + f) which gives f = 0.12 = 12%. So, if the ﬁrm can predict that the forward rate will be greater than 12%, it should use the ﬁrst option to borrow, but if they expect the forward rate to be less than 12%, it should use the second option to borrow. Example 53.2 Consider the term structure of interest rates given in Table 53.1. Let fn+t n represents the t−year forward rate n years from now. (a) Compute the 1-year, 2-year, and 3-year forward rate, one year from now. (b) Compute the 1-year and 2-year forward rates two years from now. (c) Compute the 1-year forward rate three years from now. 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 461 Solution. (a) The 1-year forward rate one year from now satisﬁes the equation (1.03)(1 + f2 1 ) = (1.04)2. Solving this equation we ﬁnd f2 1 = 5.01%. The 2-year forward rate, one year from now satisﬁes the equation (1.03)(1 + f3 1 )2 = (1.06)3. Solving this equation we ﬁnd f3 1 = 7.53%. The 3-year forward rate, one year from now satisﬁes the equation (1.03)(1 + f4 1 )3 = (1.07)4. Solving this equation we ﬁnd f4 1 = 8.37%. (b) The 1-year forward rate, two years from now satisﬁes the equation (1.04)2(1 + f3 2 ) = (1.06)3. Solving this equation we ﬁnd f3 2 = 10.1%. The 2-year forward rate, two years from now satisﬁes the equation (1.04)2(1 + f4 2 )2 = (1.07)4. Solving this equation we ﬁnd f4 2 = 10.00%. (c) The 1-year forward rate three years from now satisﬁes the equation (1.06)3(1 + f4 3 ) = (1.07)4. Solving this equation we ﬁnd f4 3 = 10.06% From any yield curve, one can calculate a complete set of implied forward rates. In general, the k− year forward rate n years from now satisﬁes the equation (1 + in)n(1 + fn+k n )k = (1 + in+k)n+k where it is the t−year spot rate. Note that fn+k n is the forward interest rate between the periods n and n+ k. Example 53.3 Find the present value of the remaining payments in the annuity immediate given in Example 53.1 immediately after two payments have been made assuming that f2+t 2 = 1% + it where t= 1,2,3. 462 MEASURES OF INTEREST RATE SENSITIVITY Solution. The comparison date for this calculation is at the end of two years. At that time there are three remaining annuity payments to be made. The expected forward rates are the current spot rates for 1,2, and 3 years each increased by 1%. Thus, the present value of the remaining three payments is 1000[(1.08)−1 + (1.09)−2 + (1.0975)−3] = $2524.07 Example 53.4 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% (a) Find the price of a $1000 two-year bond with annual 5% coupon using the spot rates given above. (b) Compute the yield to maturity as deﬁned in Section 43. Solution. (a) The coupon is $50 each. The price of the bond is P = 50(1.07)−1 + 1050(1.08)−2 = 946.93 (b) The yield to maturity i satisﬁes the equation 50(1 + i)−1 + 1050(1 +i)−2 = 946.93. Solving this equation we ﬁnd i= 7.975% Example 53.5 Yield rates for 6% annual coupon bonds are given to be 3.5% for 1-year bonds and 4% for 2-year bonds. Find the implied 2-year spot rate from these yield rates. Solution. Per 100 of par value we have 6a2 0.04 + 100v2 0.04 = 6 1.035 + 106 (1 + i)2 . Solving this equation for i we ﬁnd i= 4.01% 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 463 Practice Problems Problem 53.1 You are given the following term structure of interest rates: Length of investment Interest Rate 1 5.00 2 6.00 3 6.75 4 7.25 5 7.50 Table 53.2 Which of the following are true: (I) The yield curve has a positive slope. (II) The yield curve is inverted. (III) The interest rates in the table are called forward rates. Problem 53.2 Using Table 53.2, calculate the present value of a ﬁve year annuity due of 100 per year. Problem 53.3 Using Table 53.2, determine the three year deferred two year forward rate. Problem 53.4 Using Table 53.2, determine the accumulated value of a three year annuity immediate of 100 at the end of each year. Problem 53.5 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 464 MEASURES OF INTEREST RATE SENSITIVITY (a) Calculate the present value of an annuity immediate of 10 at the end of each year for 5 years using the spot rates. (b) Calculate the equivalent level rate. Problem 53.6 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 (a) Calculate the present value of an annuity due of 10 at the beginning of each year for 5 years using the spot rates. (b) Calculate the equivalent level rate. Problem 53.7 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 (a) Calculate the 2 year deferred 3 year forward rate less the three year spot rate. (b) Calculate what the 3 year spot rate is expected to be in 2 years. Problem 53.8 You are given the following term structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7% 2 8% 3 9% What is the one-year forward rate beginning two years from now? 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 465 Problem 53.9 ‡ Using the table of the previous problem, ﬁnd (a) the current price of a $1000 3-year bond with coupon rate 6% payable annually. (b) the annual eﬀective yield rate for the bond if the bond is sold at a price equal to its value. Problem 53.10 ‡ You are given the following term structure of spot interest rates: Term (in Years) Spot Interest Rate 1 5.00% 2 5.75% 3 6.25% 4 6.50% A three-year annuity-immediate will be issued a year from now with annual payments of 5000. Using the forward rates, calculate the present value of this annuity a year from now. Problem 53.11 ‡ Consider a yield curve deﬁned by the following equation: ik = 0.09 + 0.002k−0.001k2 where ik is the annual eﬀective rate of return for zero coupon bonds with maturity of k years. Let j be the one-year eﬀective rate during year 5 that is implied by this yield curve. Calculate j. Problem 53.12 ‡ Yield rates to maturity for zero coupon bonds are currently quoted at 8.5% for one-year maturity, 9.5% for two-year maturity, and 10.5% for three-year maturity. Let i be the one-year forward rate for year two implied by current yields of these bonds. Calculate i. Problem 53.13 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% 466 MEASURES OF INTEREST RATE SENSITIVITY Find the following expected forward rates: (a) 1-year deferred 2-year forward rate. (b) 2-year deferred 3-year forward rate. Problem 53.14 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% (a) Find the price of a $1000 two-year bond with annual 10% coupon using the spot rates given above. (b) Compute the yield to maturity as deﬁned in Section 41. Problem 53.15 You are given 1−year, 2−year, and 3−year spot rates of 4%, 5%, and 6%, respectively. Calculate the annual yield rate for 3 −year 5% annual coupon bonds implied by these spot rates. Problem 53.16 Consider a yield curve deﬁned by the following equation: ik = 0.09 + 0.002k−0.001k2 where ik is the annual eﬀective rate of return for zero coupon bonds with maturity of k years. (a) Calculate the 2-year spot rate implied by this yield curve. (b) Calculate the 2-year deferred 3-year forward rate implied by this yield curve. Problem 53.17 A two-year bond with annual coupons of 100 and redemption value of 1000 is priced at 1,037.41. The current two-year spot rate is 8%. Determine the current one-year spot rate that is consistent with the pricing of the bond. Problem 53.18 A 5 year bond with 6% annual coupons has a yield rate of 10% eﬀective and a 5 year bond with 8% annual coupons has a yield rate of 9% eﬀective. What is the 5 year spot rate? 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 467 Problem 53.19 ‡ Which of the following statements about zero-coupon bonds are true? (I) Zero-coupon bonds may be created by separating the coupon payments and redemption values from bonds and selling each of them separately. (II) The yield rates on stripped Treasuries at any point in time provide an immediate reading of the risk-free yield curve. (III) The interest rates on the risk-free yield curve are called forward rates. 468 MEASURES OF INTEREST RATE SENSITIVITY 54 Macaulay and Modiﬁed Durations By now, it should be clear to you that the timing of cash ﬂows is a signiﬁcant factor in the analysis of ﬁnancial instruments. In this section we develop some “indices” to measure the timing of future payments. One index known as the duration is useful as a measure of the sensitivity of a bond’s price to interest rate movements. The two types of duration we will look at are the modiﬁed duration and the Macaulay duration. The most basic index of time measuring is the term-to-maturity. For a given price and par value, among zero coupon bonds, one wishes to purchase the bond with the earliest redemption time. In the case of bonds carrying coupons, the redemption time is not enough for helping choose the bond to purchase. A better index than the term-to-maturity is the method of equated time , which was deﬁned in Section 13: Let R1,R2,··· ,Rn be a series of payments made at times 1 ,2,··· ,n. Then the weighted average of the various times of payments, where the weights are the various amounts paid is t= ∑n t=1 tRt∑n t=1 Rt . This is also known as the average term-to-maturity. Example 54.1 Consider two 10-year par value 100 bonds one with 5% coupons and the other with 10% coupons. Find the equated time for each coupon. Solution. The average term to maturity of the 5% bond is t= 1 ·5 + 2·5 + ··· + 10 ·5 + 10·100 5 + 5 +··· + 5 + 100 = 8.50. The average term to maturity of the 10% bond is t= 1 ·10 + 2·10 + ··· + 10 ·10 + 10·100 10 + 10 +··· + 10 + 100 = 7.75. This shows that the 5% bond is a longer term bond than the 10% bond An even better index that is similar to the method of equated time is obtained by replacing each payment in the above formula by its present value thus obtaining d= ∑n t=1 tνtRt∑n t=1 νtRt . 54 MACAULAY AND MODIFIED DURATIONS 469 We call d the Macaulay duration or simply duration. Example 54.2 Suppose payments of 2000, 4000, and 10000 are to be made at times 1, 2, and 4, respectively. Assume, an annual yield of 25%. (a) Find the average term to maturity using the method of equated time. (b) Find the duration of the investment. Solution. (a) The method of equated time produces the value t= 1 ·2000 + 2·4000 + 4·10000 2000 + 4000 + 10000 = 3.125. (b) The duration is given by d= 1 ·(1.25)−1 ·2000 + 2·(1.25)−2 ·4000 + 4·(1.25)−4 ·10000 (1.25)−1(2000) + (1.25)−2(4000) + (1.25)−4(10000) = 2.798 Example 54.3 Consider two bonds purchased at the redemption value of 1000, and due in 5 years. One bond has 5% annual coupon rate payable semi-annually and the other has 10% annual coupon rate payable semi-annually. Find the duration of each bond if both bonds were purchased to yield 7% compounded semi-annually. Solution. The duration of the 5% bond is d=0.5[1 ·(1.035)−1(25) + 2·(1.035)−2(25) + ··· + 10 ·(1.035)−10(25)] + 5000(1.035)−10 (1.035)−1(25) + (1.035)−2(25) + ··· + (1.035)−10(25) + 1000(1.035)−10 =12.5(Ia)10 0.035 + 5000(1.035)−10 25a10 0.035 + 1000(1.035)−10 =4.4576 The duration of the 10% bond is d=0.5[1 ·(1.035)−1(50) + 2·(1.035)−2(50) + ··· + 10 ·(1.035)−10(50)] + 5000(1.035)−10 (1.035)−1(50) + (1.035)−2(50) + ··· + (1.035)−10(50) + 1000(1.035)−10 =4.1158 470 MEASURES OF INTEREST RATE SENSITIVITY Note that dis a function of i.If i= 0 we have d= tso that the method of equated time is a special case of duration which ignores interest. Also, note that in the case of only one future payment, duration is the point in time at which that payment is made. This is clear since the summations in the numerator and the denominator have only one term each and everything cancels except the time of payment. Example 54.4 Find the duration of a 10-year zero coupon bond assuming a yield 8% eﬀective. Solution. Only one payment is involved so that d = 10. Notice that this answer is independent of the yield rate Another important fact is that d is a decreasing function of i. To see this, we have d di(d) = d di (∑n t=1 tνtRt∑n t=1 νtRt ) = −ν(∑n t=1 νtRt) (∑n t=1 t2νtRt) −(∑n t=1 tνtRt)2 (∑n t=1 νtRt)2 = −ν [∑n t=1 t2νtRt∑n t=1 νtRt − (∑n t=1 tνtRt∑n t=1 νtRt )2] Now using Cauchy-Schwartz’s inequality ( n∑ t=1 atbt)2 ≤( n∑ t=1 a2 t)( n∑ t=1 b2 t) with at = ν t 2 R 1 2 t and bt = tν t 2 R 1 2 t we ﬁnd that ∑n t=1 t2νtRt∑n t=1 νtRt − (∑n t=1 tνtRt∑n t=1 νtRt )2 ≥0 and therefore the derivative of d with respect to i is negative. We next describe a measure of how rapidly the present value of a series of future payments changes 54 MACAULAY AND MODIFIED DURATIONS 471 as the rate of interest changes. Let P(i) denote the present value of all future payments. Then by Taylor series we can write P(i+ ϵ) = P(i) + ϵ 1!P′(i) + ϵ2 2!P′′(i) + ··· . Using the ﬁrst two terms we can write P(i+ ϵ) ≈P(i) + ϵP′(i). Hence, the percentage change in P is given by P(i+ ϵ) −P(i) P(i) ≈ϵP′(i) P(i) . Thus, the percentage change of P(i) is a function of P′(i) P(i) . We deﬁne the volatility of the present value by ν = −P′(i) P(i) = −d di[ln P(i)]. The minus sign is necessary to make ν positive since P′(i) is negative ( increasing the yield rate results in a decreasing present value). Now, since P(i) = ∑n t=1(1 +i)−tRt, by taking the derivative we ﬁnd P′(i) = −∑n t=1 t(1 +i)−t−1Rt. Hence, ν = ∑n t=1 t(1 + i)−t−1Rt∑n t=1(1 + i)−tRt = 1 1 + i · ∑n t=1 t(1 + i)−tRt∑n t=1(1 + i)−tRt = d 1 + i = νd. Because of this close relationship, volatility is often called modiﬁed duration. Remark 54.1 None of the results in this section are valid if the payment amounts vary depending on the interest rate. Example 54.5 (a) Find the Macaulay duration of a ten-year, $1,000 face value, 8% annual coupon bond. Assume an eﬀective annual interest rate of 7%. (b) Find the modiﬁed duration for the bond Solution. (a) We have d=80ν+ 2(80ν2) + ··· + 10(80ν10) + 10(1000ν10) 80ν+ (80ν2) + ··· + (80ν10) + (1000ν10) =80(Ia)10 0.07 + 10000ν10 0.07 80a10 0.07 + 1000ν10 0.07 472 MEASURES OF INTEREST RATE SENSITIVITY But ν10 0.07 =(1.07)−10 = 0.508349 a10 0.07 =1 −(1.07)−10 0.07 = 7.02358 (Ia)10 0.07 =1.07a10 0.07 −10(1.07)−10 0.07 = 34.7391 Hence, d= 80(34.7392)+10000(0.508349) 80(7.02358)+1000(0.508349) = 7.3466 years (b) The modiﬁed duration is ν = νd= (1.07)−1(7.3466) = 6.8660 Example 54.6 A 10-year $100,000 mortgage will be repaid with level semi-annual payments of interest and prin- cipal. Assume that the interest rate on the mortgage is 8%, convertible semiannually, and that payments are made at the end of each half-year. Find the Macaulay duration of this mortgage. Solution. Per dollar of mortgage payment, the Macaulay duration is d=0.5ν0.04 + 2ν2 0.04 + 3ν3 0.04 + ··· + 20ν20 0.04 ν0.04 + ν2 0.04 + ν3 0.04 + ··· + ν20 0.04 =0.5(Ia)20 0.04 a20 0.04 = 4.6046 years Note that the answer is independent of the rate of interest being paid on the mortgage. Duration depends on the pattern of the level payments and not their amount Example 54.7 Find the duration of a preferred stock paying level dividends into perpetuity assuming eﬀective 8%. Solution. Per dollar dividend we have d= (Ia)∞ a∞ = 1.08/.082 1/.08 = 13.5 54 MACAULAY AND MODIFIED DURATIONS 473 Example 54.8 Find the modiﬁed duration for a share of common stock. Assume that the stock pays annual dividends, with the ﬁrst dividend of $2 payable 12 months from now, and that subsequent dividends will grow at an annual rate of 4%. Assume that the eﬀective annual interest rate is 9%. Solution. We have P(i) = D1 i−g P′(i) = −D1(i−g)−2 v= −P′(i) P(i) = 1 i−g = 1 0.09 −0.04 = 20 Example 54.9 You are the actuary for an insurance company. Your company’s liabilities include loss reserves, which are liability reserves set aside to make future claim payments on policies which the company has already sold. You believe that these liabilities, totaling $100 million on December 31, 2006, will be paid out according to the following schedule: Proportionl of Calendar year reserves paid out 2007 40% 2008 30% 2009 20% 2010 10% Find the modiﬁed duration of your company’s loss reserves. Assume that the annual interest rate is 10%, and that all losses paid during a given calendar year are paid at the mid-point of that calendar year. Solution. We have ν = ν (0.5(4ν0.5) + 1.5(3ν1.5) + 2.5(2ν2.5) + 3.5(ν3.5) 4ν0.5 + 3ν1.5 + 2ν2.5 + ν3.5 ) = 1.2796 Remark 54.2 The equation P(i+ ϵ) −P(i) P(i) = −ϵν is valid if P(i) stands for the current price of a bond. 474 MEASURES OF INTEREST RATE SENSITIVITY Example 54.10 The current price of a bond is $110 and its yield is 7%. The modiﬁed duration is 5. Estimate the price of the bond if its yield falls to 6%. Solution. The approximate percentage change in the price is −(−0.01)(5) = 5% . That is, the price of the bond increased by approximately 5%. Thus, the new price of the bond is 110 ×1.05 = $115.50 Portfolio Modiﬁed Duration Consider a portfolio consisting of n bonds. Let bond k have current price Pk(i) and modiﬁed duration νk(i). Then the current value of the portfolio is P(i) = P1(i) + P2(i) + ··· + Pn(i). Let ν denote the modiﬁed duration of the portfolio. Then, we have ν = −P′(i) P(i) = −P′ 1(i) P(i) + ( −P′ 2(i) P(i) ) + ··· + ( −P′ n(i) P(i) ) =P1(i) P(i) ( −P′ 1(i) P1(i) ) + P2(i) P(i) ( −P′ 2(i) P2(i) ) + ··· + Pn(i) P(i) ( −P′ n(i) Pn(i) ) =P1(i) P(i) ν1 + P2(i) P(i) ν2 + ··· + Pn(i) P(i) νn Thus, the modiﬁed duration of a portfolio can be calculated as the weighted average of the bonds’ modiﬁed durations, using the market values of the bonds as the weights. Example 54.11 Megan buys the following bonds: (I) Bond A with a modiﬁed duration of 7 for 1000; (II) Bond B with a modiﬁed duration of 5 for 2000; and (III) Bond C with a modiﬁed duration of 10 for 500. Calculate the modiﬁed duration of the portfolio. Solution. The modiﬁed duration of the portfolio is weighted average of the modiﬁed durations weighted by the purchase price ν = 7 ·1000 + 5·2000 + 10·500 1000 + 2000 + 500 = 6.286 54 MACAULAY AND MODIFIED DURATIONS 475 Example 54.12 An investment portfolio consists of two bonds: a two-year zero-coupon bond, and a four-year zero- coupon bond. Both bonds have redemption values of $1,000. Assume an eﬀective annual interest rate of 8%. Find the Macaulay duration of the investment portfolio. Solution. The Macaulay duration of the two-year zero-coupon is 2 with current value 1000 v2. The Macaulay duration of the four-year zero-coupon is 4 with current value 1000v4.The Macaulay duration of the portfolio is d= 2(1000ν2) + 4(1000ν4) 1000ν2 + 1000ν4 = 2.9232 years 476 MEASURES OF INTEREST RATE SENSITIVITY Practice Problems Problem 54.1 Find the modiﬁed duration of a 30-year, $1,000 face value, 6% annual coupon bond. Assume an eﬀective annual interest rate of 9%. Problem 54.2 Calculate the duration of a payment to be made in 5 years. Problem 54.3 A 10 year bond has annual coupons of 10 and matures for 100. Which of the following are true: (I) The term to maturity is 10 years. (II) The average term to maturity using the method of equated time is 7.5 years. (III) The Macaulay duration at 8% interest is 6.97. (IV) The modiﬁed duration at 8% interest is 6.45. Problem 54.4 Which of the following are true: (I) The average term to maturity under the method of equated time is always greater than the volatility. (II) The duration is a decreasing function of i. (III) A zero coupon bond will have a Macaulay duration equal to the term to maturity. Problem 54.5 (a) Calculate the duration of a 12-year annuity immediate payable using an interest rate of 5%. (b) Calculate the modiﬁed duration of the annuity. Problem 54.6 A perpetuity pays 100 immediately. Each subsequent payment is increased by inﬂation. Calculate the duration of the perpetuity using 10.25% assuming that inﬂation will be 5% annually. Problem 54.7 Calculate the duration of a perpetuity immediate of 1 less the duration of a perpetuity due of 1 at an interest rate of 10%. Problem 54.8 Calculate the modiﬁed duration of an annuity due with payments of 100 for 10 years using an interest rate of 8%. 54 MACAULAY AND MODIFIED DURATIONS 477 Problem 54.9 ‡ A bond will pay a coupon of 100 at the end of each of the next three years and will pay the face value of 1000 at the end of the three-year period. The bond’s duration (Macaulay duration) when valued using an annual eﬀective interest rate of 20% is X. Calculate X. Problem 54.10 Suppose that a 3-year ﬁnancial instrument is expected to make increasing payments to you at the end of each of the next three years. Speciﬁcally, the payments will be CF(t) = 1,000t, for t= 1,2, and 3. Assume that you purchase this ﬁnancial instrument, at time 0, at a price which provides an annual eﬀective yield of 8%. Calculate the Macaulay duration, and the modiﬁed duration of this ﬁnancial instrument. Problem 54.11 Calculate the Macaulay duration and the modiﬁed duration of a 30-year zero-coupon bond with a face value of $ 1,000. Assume that the annual yield-to-maturity is 8%. Problem 54.12 A 10-year $200,000 mortgage will be paid oﬀ with level quarterly amortization payments. Assume that the interest rate on the mortgage is 6%, convertible quarterly, and that payments are made at the end of each quarter. Find the modiﬁed duration of this mortgage. Problem 54.13 ‡ Calculate the duration of a common stock that pays dividends at the end of each year into perpetuity. Assume that the dividend increases by 2% each year and that the eﬀective rate of interest is 5%. Problem 54.14 ‡ Calculate the duration of a common stock that pays dividends at the end of each year into perpetuity. Assume that the dividend is constant, and that the eﬀective rate of interest is 10%. Problem 54.15 ‡ The current price of an annual coupon bond is 100. The derivative of the price of the bond with respect to the yield to maturity is −700. The yield to maturity is an annual eﬀective rate of 8%. Calculate the duration of the bond. Problem 54.16 ‡ Calculate the Macaulay duration of an eight-year 100 par value bond with 10% annual coupons and an eﬀective rate of interest equal to 8%. 478 MEASURES OF INTEREST RATE SENSITIVITY Problem 54.17 ‡ John purchased three bonds to form a portfolio as follows: Bond A has semiannual coupons at 4%, a duration of 21.46 years, and was purchased for 980. Bond B is a 15-year bond with a duration of 12.35 years and was purchased for 1015. Bond C has a duration of 16.67 years and was purchased for 1000. Calculate the duration of the portfolio at the time of purchase. Problem 54.18 A 10-year 1000 face value 6% annual coupon bond with redemption value C has duration equal to 6.06 using an annual eﬀective interest rate of 6%. Calculate C. Problem 54.19 You are given: (i) the annual yield rate on a zero-coupon bond with duration of 6 months is 3% (ii) the annual yield rate on a zero-coupon bond with duration of 12 months is 3.25% (iii) the annual yield rate on a zero-coupon bond with duration of 18 months is 3.5% (iv) the annual yield rate on a zero-coupon bond with duration of 24 months is 3.75% Determine the semiannual eﬀective yield rate on a 2-year 100 par value bond with 4% semiannual coupons. Problem 54.20 John purchased three bonds to form a portfolio as follows: Bond A is purchased for X and has a duration of 20 years. Bond B is purchased for Y and has a duration of 20 years. Bond C is purchased for X+ Y. The duration (in years) of the portfolio at the time of purchase is 18. Determine the duration (in years) of Bond C at the time of purchase. 55 REDINGTON IMMUNIZATION AND CONVEXITY 479 55 Redington Immunization and Convexity Most ﬁnancial institutions try to insulate their portfolios from interest rate risk movements. Some institutions (such as banks) are concerned with protecting their current net worth against interest rate movements. Other institutions (such as pension funds) may face an obligation to make pay- ments after a given number of years. These institutions are more concerned with protecting the future values of their portfolios. The common factor amongst diﬀerent investors and ﬁnancial institutions is the presence of interest rate risk because net worth ﬂuctuates with interest rate. Financial institutions try to structure their assets and liabilities in such a way that they are protected against small changes in interest rates. There are three approaches that all work toward the same goal of minimizing (or even eliminating) the risk caused by ﬂuctuating interest rates: Immunizatiom, full immunization, and dedication. In this section, we consider the ﬁrst strategy. The basic idea of immunization is that ﬁnancial institutions (i.e., banks) are vulnerable to incur- ring losses if interest rates change. As a result, they “immunize” themselves by structuring their assets and liabilities such that the modiﬁed duration of the assets equals the modiﬁed duration of the liabilities. When this occurs, a rise in the interest rate would cause the present value of assets to decline, but there would be a matching decrease in the present value of liabilities. Thus, the company is protected against small changes in interest rates. We next discuss how immunization achieves the protection required. Consider cash inﬂowsA1,A2,··· ,An generated by the assets at times 1 ,2,··· ,n. Similarly, let L1,L2,··· ,Ln be the cash outﬂows gen- erated by the liabilities at times 1 ,2,··· ,n. Let the net cash ﬂows at time t be Rt = At −Lt, t = 1,2,··· ,n. In general, banks assets and liabiltiy are roughly equal in size so that we can assume that the present value of the cash inﬂows from the assets is equal to the present value of the cash outﬂows from the liabilities. This leads to P(i) = n∑ t=1 νtRt = 0. Next, we would like P(i) to have a local minimum at i,so that small changes in the interest rate in either direction will increase the present value of the cash ﬂows. But for the function P(i) to have a minimum requires two conditions to hold. The ﬁrst is that P′(i) = 0 (i.e. the modiﬁed duration of the assets is equal to the modiﬁed duration of the liabilities ). The second condition is P′′(i) >0. The second derivative of P(i) is used to deﬁne the convexity of a series of cash ﬂows given by c= P′′(i) P(i) . 480 MEASURES OF INTEREST RATE SENSITIVITY To summarize, immunization is achieved when the following three conditions are met: (1) P(i) = 0 : The present value of cash inﬂows (assets) should be equal to the present value of cash outﬂows (liabilities). (2) P′(i) = 0 : The modiﬁed duration of the assets is equal to the modiﬁed duration of the liabilities. (3) P′′(i) >0 : The convexity of the present value of cash inﬂows (assets) should be greater than the convexity of the present value of cash outﬂows(liabilities). In other words, asset growth (decline) should be greater (less) than liability growth(decline). In practice there are some diﬃculties and limitations in implementing the immunization strategy: (a) choice of i is not always clear. (b) doesn’t work well for large changes in i. For large changes in i a technique, known as full immunization, was developed for that purpose. This concept will be covered in Section 56. (c) yield curve is assumed to change with the change ini; actually, short-term rates are more volatile than long-term rates. (d) frequent rebalancing is required in order to keep the modiﬁed duration of the assets and liabil- ities equal. (e) exact cash ﬂows may not be known and may have to be estimated. (f) assets may not have long enough maturities of duration to match liabilities. We next express the percentage change ofP(i) in terms of both the modiﬁed duration and convexity. Expanding P(i+ ϵ) as a Taylor series as far as second derivatives we ﬁnd P(i+ ϵ) ≈P(i) + ϵP′(i) + ϵ2 2 P′′(i). Thus, P(i+ ϵ) −P(i) P(i) ≈ϵP′(i) P(i) + ϵ2 2 P′′(i) P(i) = −ϵν+ ϵ2 2 c. Example 55.1 A client deposits 100,000 in a bank, with the bank agreeing to pay 8% eﬀective for two years. The client indicates that half of the account balance will be withdrawn at the end of the ﬁrst year. The bank can invest in either one year or two year zero coupon bonds. The one year bonds yield 9% and the two year bonds yield 10%. Develop an investment program based on immunization. Solution. The cash outﬂows are 50000(1 .08) at the end of the ﬁrst year and 50000(1 .08)2 at the end of the second year. 55 REDINGTON IMMUNIZATION AND CONVEXITY 481 Suppose the bank invests x in the one year bonds, and y in two year bonds. Then P(i) =x(1.09)(1 + i)−1 + y(1.10)2(1 + i)−2 −50000(1.08)(1 + i)−1 −50000(1.08)2(1 + i)−2 P′(i) = −x(1.09)(1 + i)−2 −2(1.10)2y(1 + i)−3 + 50000(1.08)(1 + i)−2 +2(50000)(1.08)2(1 + i)−3 P′′(i) =2(1.09)x(1 + i)−3 + 6(1.10)2y(1 + i)−4 −100000(1.08)(1 + i)−3 −300000(1.08)2(1 + i)−4 To immunize against interest rate risk at i= 0.08, the bank should have P(0.08) = 0,P′(0.08) = 0 and P′′(0.08) >0. The ﬁrst two conditions lead to a system in the unknowns x and y. Solving the system gives x = 49541 .28 and y = 48198 .35. Moreover, P′′(0.08) = 128 ,773.89 > 0. Thus, the bank is well protected against interest rate ﬂuctuations Example 55.2 For the assets in Example 55.1, ﬁnd the modiﬁed duration and convexity. Solution. We have PA(i) = x(1.09)(1 + i)−1 + (1.10)2y(1 + i)−2 →PA(0.08) = 50 ,000.00 + 50,000.00 = 100,000.00. Also, P′ A(i) = −x(1.09)(1 + i)−2 −2(1.10)2y(1 + i)−3 →P′ A(0.08) = −138,888.89. The modiﬁed duration is v(0.08) = −P′ A(0.08) PA(0.08) = 138,888.89 100,000.00 = 1.39. The convexity is c(0.08) = P′′ A(0.08) PA(0.08) = 342,935.54 100,000.00 = 3.43 since P′′ A(i) = 2(1.09)x(1 + i)−3 + 6(1.10)2y(1 + i)−4 →P′′ A(0.08) = 342,935.54 Example 55.3 For a 30-year home mortgage with level payments and an interest rate of 10.2% convertible monthly, ﬁnd the modiﬁed duration and the convexity of the payments. Solution. The monthly rate of interest is 0 .102/12 = 0.0085. Then for a dollar of monthly payment, we have P(0.0085) = 360∑ t=1 (1.0085)−t = (1.0085)−1 [1 −(1.0085)−360 1 −(1.0085)−1 ] = 112.0591 482 MEASURES OF INTEREST RATE SENSITIVITY and P′(0.0085) = − 360∑ t=1 t(1.0085)−t−1 = −(1.0085)−1(Ia)360 0.0085 = −11,188.69608 Thus, the modiﬁed duration is ν(0.0085) = 11,188.69608 112.0591 = 99.85. Thus, the modiﬁed duration of the payments is just under 100 months of the 360-month term of the mortgage. Now, P′′(0.0085) = 360∑ t=1 t(t+ 1)(1.0085)−t−2 = ν2 360∑ t=1 (t2 + t)νt =ν2[ 360∑ t=1 t2νt + 360∑ t=1 tνt] =ν2[ 360∑ t=1 t2νt + (Ia)360 0.0085] Hence, using a calculator we ﬁnd c(0.0085) = 1,940,079 + 11,283.80 (1.0085)2(112.0591) = 17,121 Example 55.4 A bank agrees to pay 5% compounded annually on a deposit of 100,000 made with the bank. The depositor agrees to leave the funds on deposit on these terms for 8 years. The bank can either buy 4 year zero coupon bonds or preferred stock, both yielding 5% eﬀective. How should the bank apportion its investment in order to immunize itself against interest rate risk? Solution. The present value of the cash ﬂows in terms of the amount B invested in the bonds is P(i) = B(1.05)4(1 + i)−4 + (100000 −B)(0.05) i −100000(1.05)8(1 + i)−8 since the preferred stock is assumed to pay 5% forever and the bank must repay the deposit with interest at the end of 8 years. Substitution veriﬁes that P(.05) = 0. On the other hand, P′(0.05) = −4B(1.05)4(1.05)−5 −(100000 −B)(0.05) 0.052 + 800,000(1.05)8(1.05)−9 55 REDINGTON IMMUNIZATION AND CONVEXITY 483 So setting P′(0.05) = 0 and solving for B gives B = 76470.58 as the amount to be invested in bonds with the remainder in stock. With this allocation, one ﬁnds P′′(.05) = 20(76470 .58)(1.05)−2 + 2(100000 −7647.58)(.05)−2 −720,000(1.05)−9 = 965786.7413 >0, so the portfolio is optimal Example 55.5 Find the duration and convexity of a 20 year zero coupon bond assuming that the interest rate is 7% eﬀective. Solution. The duration is 20 since the par value is repaid at the end of 20 years and this is the only payment made by the bond. Now, per dollar of par value, P(i) = (1 + i)−20, so that the convexity P′′(i) P(i) = 420 (1 + i)2 . Substituting i= 0.07 gives the convexity as 366.844 Convexity of a Portfolio Consider a portfolio consisting of nbonds. Let bond k have current price Pk(i) and convexity ck(i). Then the current value of the portfolio is P(i) = P1(i) + P2(i) + ··· + Pn(i). Let c denote the convexity of the portfolio. Then, we have c=P′′(i) P(i) = P′′ 1 (i) P(i) + P′′ 2 (i) P(i) + ··· + P′′ n(i) P(i) =P1(i) P(i) (P′′ 1 (i) P1(i) ) + P2(i) P(i) (P′′ 2 (i) P2(i) ) + ··· + Pn(i) P(i) (P′′ n(i) Pn(i) ) =P1(i) P(i) c1 + P2(i) P(i) c2 + ··· + Pn(i) P(i) cn Thus, the convexity of a portfolio can be calculated as the weighted average of the bonds’ convexities, using the market values of the bonds as the weights. 484 MEASURES OF INTEREST RATE SENSITIVITY Practice Problems Problem 55.1 ‡ Which of the following statements about immunization strategies are true? I. To achieve immunization, the convexity of the assets must equal the convexity of the liabilities. II. The full immunization technique is designed to work for any change in the interest rate. III. The theory of immunization was developed to protect against adverse eﬀects created by changes in interest rates. Problem 55.2 Nova Inc. has liabilities of 10 due in 1, 4, and 7 years. What asset incomes must the company arrange for in years 1 and 6 to immunize their cash ﬂow, assuming an annual interest rate of 10% on all transactions? Problem 55.3 A bank is required to pay 1,100 in one year. There are two investment options available with respect to how money can be invested now in order to provide for the 1,100 payback: (i) a non-interest bearing cash fund, for which x will be invested, and (ii) a two-year zero-coupon bond earning 10% per year, for which y will be invested. Based on immunization theory, develop an asset portfolio that will minimize the risk that liability cash ﬂows will exceed asset cash ﬂows. Assume the eﬀective rate of interest is equal to 10% in all calculations. Problem 55.4 A $5,000 payment is owed from Abby to Ben two years from now. Abby wants to set up an investment fund to meet this obligation, but the only investments she has available are a money market fund (currently earning 8%, but the rate changes daily), and a ﬁve-year zero-coupon bond earning 8%. Use an immunization framework to determine the amount of money Abby should invest now in each of the two investment vehicles. Assume an eﬀective annual interest rate of 8% for present value calculations. Problem 55.5 For the assets in Problem 55.4, ﬁnd the modiﬁed duration and convexity. Problem 55.6 Find the convexity of the following investments,assuming the eﬀective rate of interest is 8%: (a) A money market fund. (b) A 10-year zero coupon bond. (c) A preferred stock paying level dividends into perpetuity. 55 REDINGTON IMMUNIZATION AND CONVEXITY 485 Problem 55.7 Find the convexity of a loan repaid with equal installments over n periods if i= 0. Problem 55.8 A common stock pays dividends at the end of each year. It is assumed that each dividend is 4% greater than the prior dividend and the eﬀective rate of interest is 8%. Find the convexity of this common stock. Problem 55.9 Derive the following relationship between modiﬁed duration and convexity: d diν = ν2 −c. Problem 55.10 Suppose a company must make payments of 300 at time t = 3 and 500 at time t = 5 , and the annual interest rate is 4%. Find asset income at times t= 0 and t= 6 so that the portfolio will be immunized. 486 MEASURES OF INTEREST RATE SENSITIVITY 56 Full Immunization and Dedication In this section we discuss the two other strategies of immunization: Full immunization and dedica- tion. Full Immunization Full immunization is an extension to the Redington immunization discussed in Section 50. While Redington immunization works for small changes in i, full immunization can be applied for all changes of i. We will say that a portfolio is fully immunized if PVA(i+ ϵ) > PVL(i+ ϵ) for all ϵ> 0 or equivalently PVA(i) ≥PVL(i) for all positive i. Suppose we have one liability outﬂow Lk at time k. The concept is to hold two assets, one that will produce a cash inﬂow, A, prior to the liability outﬂow (say at time k−a) , and another that will produce a cash inﬂow subsequent to the liability outﬂow (say at time k+ b). We use the force of interest δ that is equivalent to i. The full immunization conditions for this single liability cash ﬂow are: (1) Present value of assets = Present value of liability. (2) Modiﬁed duration of assets = Modiﬁed duration of liability. (3) The asset cash ﬂows occur before and after the liability cash ﬂow. Example 56.1 An insurance company has committed to make a payment of $100,000 in 10 years. In order to fund this liability, the company has invested $27,919.74 in a 5-year zero-coupon bond and $27,919.74 in a 15-year zero-coupon bond. The annual eﬀective yield on all assets and liabilities is 6%. Determine whether the company’s position is fully immunized. Solution. Condition 1: Present value of assets = Present value of liability. We have PVL = 100,000(1.06)−10 = 55,839.48 and PVA = 27,919.74 ×2 = 55839.48, so condition (1) is met. Condition 2: Modiﬁed duration of assets = Modiﬁed duration of liability. We have Modiﬁed duration of liabilities = 10(100,000)(1.06)−10 100,000(1.06)−10 = 10 and Modiﬁed duration of assets = 5(27,919.74)+15(27,919.74) 55,839.48 = 10. 56 FULL IMMUNIZATION AND DEDICATION 487 Thus, condition (2) is true. Condition 3: The asset cash ﬂows occur before and after the liability cash ﬂow. Thus, the po- sition is fully immunized Now, conditions (1) and (2) leads to the following system P(δ) =Ae−(k−a)δ + Be−(k+b)δ −Lke−kδ = 0 P′(δ) = −A(k−a)e−(k−a)δ −B(k+ b)e−(k+b)δ + kLke−kδ = 0. The ﬁrst condition implies Lk = Aeaδ + Be−bδ. The second condition implies −A(k−a)e−(k−a)δ −B(k+ b)e−(k+b)δ + kLke−kδ =0 −Ake−(k−a)δ + Aae−(k−a)δ −Bke−(k+b)δ −Bbe−(k+b)δ + kLke−kδ =0 −k [ Ae−(k−a)δ + Be−(k+b)δ −Lke−kδ] + [Aae−(k−a)δ −Bbe−(k+b)δ] =0 Aae−(k−a)δ −Bbe−(k+b)δ =0 Aaeaδ −Bbe−bδ =0 Thus, we obtain the system Aeaδ + Be−bδ =Lk Aaeaδ =Bbe−bδ There are exceptional cases when the system has no solutions, but if the known quantities are (i) a,b (ii) B,b (iii) A,a or (iv) A,b then unique values of the other two quantities can be found. Solving the last equation for B we ﬁnd B = A (a b ) e(a+b)δ. We next show that conditions (1)-(3) achieve the required immunization. That is, we will show that for any δ′̸= δ there is a surplus, i.e. PVA(δ′) >PV L(δ′). Let S(δ′) denote the present value, at force of interest δ′ per annum, of the total-assets less the present value of the liability, i.e. the surplus function S(δ′) = PVA(δ′) −PVL(δ′) = e−kδ′ (Aeaδ′ + Be−bδ′ −Lk). 488 MEASURES OF INTEREST RATE SENSITIVITY Then we have S(δ′) =e−kδ′ [Aeaδ′ + Be−bδ′ −Lk] =e−kδ′ [Aeaδ′ + Be−bδ′ −(Aeaδ + Be−bδ)] =e−kδ′ Aeab [ ea(δ′−δ) + a be−b(δ′−δ) − ( 1 + a b )] Next, consider the function f(x) = eax + a be−bx − ( 1 + a b ) . Clearly, f(0) = 0. Moreover, since f′(x) = a(eax −e−bx), one can easily check the following f′(0) =0 f′(x) >0 for x> 0 f′(x) <0 for x< 0 Thus, f(x) >0 for all x̸= 0. This shows that S(δ′) >0 for all δ′̸= δ as required. For a portfolio with multiple liabilities, repeat the above process for each liability outﬂow. There will be two asset inﬂows for each liability outﬂow. Full immunization is like Redington immunization in the sense that portfolio must be rebalanced periodically so that to ensure that modiﬁed duration of assets is equal to modiﬁed duration of liabilities. Example 56.2 An investor has a single liability of $1000 due in 15 years’ time. The yield on zero coupon bonds of any term is currently 4% per annum, and the investor possesses cash equal to the present value of his liability, i.e. 1000(1 .04)−15 = 555.26. He wishes to invest in 10-year and 20-year zero-coupon bonds in such a way that he will make a proﬁt on any immediate change in the force of interest. How much of each security should he buy, and how large a proﬁt will he make if the rate of interest per annum immediately becomes 0.01, 0.02, 0.03, 0.05, 0.06, 0.07,or 0.08? Solution. Letting δ= ln 1.04 we ﬁnd Ae5δ + Be−5δ =1000 5Ae5δ −5Be−5δ =0 56 FULL IMMUNIZATION AND DEDICATION 489 or equivalently, we have the system of two equations A(1.04)5 + B(1.04)−5 =1000 A(1.04)5 −B(1.04)−5 =0. These equations may easily be solved, giving A = 410.96 and B = 608.32. The quantities of zero- coupon bonds which he should buy are therefore those providing $410.96 at time 10 years and $608.32 at time 20 years. The proﬁt on an immediate change in the rate of interest to iper annum is, S(i) = (1 + i)−15[410.96(1 + i)5 + 608.32(1 + i)−5 −1000]. In the following table we give the values of the liability ($1000(1 + i)−15), the assets (410 .96(1 + i)−10 + 608.32(1 + i)−20) and the proﬁt to the investor for each of the speciﬁed rates. Annual interest rate PVL PVA Present value of proﬁt 0.00 1000.00 1019.28 19.28 0.01 861.35 870.58 9.23 0.02 743.01 746.51 3.50 0.03 641.86 642.61 0.75 0.04 555.26 555.26 0.00 0.05 481.02 481.56 0.54 0.06 417.27 419.16 1.89 0.07 362.45 366.11 3.66 0.08 315.24 320.87 5.63 Dedication In this section we discuss an approach for matching liabilities and assets. This approach is called dedication or absolute matching. In this approach, an company structures an asset portfolio in such a fashion that the cash inﬂow that will be generated from assets will exactly match the cash outﬂow from the liabilities. If the above strategy is achieved, the company has full protection against any movement in interest rates. However, the following problems exist with implementing this strategy. 1. Cash ﬂows are usually not that predictable on either the asset or the liability side. 2. If the liabilities are long term in nature, it might be impossible to ﬁnd assets to exactly match the liabilities without creating reinvestment risk. 3. The yield rate on a fund structured with the major restrictions imposed by absolute matching may be less than that on a fund for which more ﬂexibility is available, and the extra return may overshadow the advantages of absolute matching in importance. 490 MEASURES OF INTEREST RATE SENSITIVITY Example 56.3 An insurance company accepts an obligation to pay 10,000 at the end of each year for 2 years. The insurance company purchases a combination of the following two bonds (both with $1,000 par and redemption values) in order to exactly match its obligation: Bond A: A 1-year 5% annual coupon bond with a yield rate of 6%. Bond B: A 2-year 8% annual coupon bond with a yield rate of 7%. (a) Find the numbers (which need not be integers) of each bond the insurer must purchase to exactly match its obligations. (b) Find the total cost to the insurer of purchasing the needed numbers of bonds. Solution. (a) We have the following chart Period Bond A Bond B Liabilities 1 1050 80 10,000 2 1080 10,000 Let a be the number of bonds A purchased and b that of bonds B. Then we have the following system of two equations in two unknowns 1050a+ 80b=10,000 1080b=10,000 Solving for a and b we ﬁnd a= 8.8183 bonds and b= 9.2593 bonds. (b) The answer is 8.8183(1050ν0.06) + 9.2593(80ν0.07 + 1080ν2 0.07) = $18161.82 56 FULL IMMUNIZATION AND DEDICATION 491 Practice Problems Problem 56.1 An insurance company has an obligation to pay $1,000,000 at the end of 10 years. It has a zero- coupon bond that matures for $413,947.55 in 5 years, and it has a zero-coupon bond that matures for $864,580.82 in 20 years. The eﬀective yield for assets and liability is 10%. (a) Determine whether the company’s position is fully immunized. (b) What is the level of surplus if the interest rate falls to 0%? (c) What is the level of surplus if the interest rate rises to 80%? Problem 56.2 ‡ A company must pay liabilities of 1000 and 2000 at the end of years 1 and 2, respectively. The only investments available to the company are the following two zero-coupon bonds: Maturity(years) Eﬀective annual yield Par 1 10% 1000 2 12% 1000 Determine the cost to the company today to match its liabilities exactly. Problem 56.3 ‡ An insurance company accepts an obligation to pay 10,000 at the end of each year for 2 years. The insurance company purchases a combination of the following two bonds at a total cost of X in order to exactly match its obligation: (i) 1-year 4% annual coupon bond with a yield rate of 5% (ii) 2-year 6% annual coupon bond with a yield rate of 5%. Calculate X. The following information applies to Problems 56.4 - 56.6 Joe must pay liabilities of 1,000 due 6 months from now and another 1,000 due one year from now. There are two available investments: •a 6-month bond with face amount of 1,000, a 8% nominal annual coupon rate convertible semi- annually, and a 6% nominal annual yield rate convertible semiannually; and •a one-year bond with face amount of 1,000, a 5% nominal annual coupon rate convertible semi- annually, and a 7% nominal annual yield rate convertible semiannually. Problem 56.4 ‡ How much of each bond should Joe purchase in order to exactly (absolutely) match the liabilities? 492 MEASURES OF INTEREST RATE SENSITIVITY Problem 56.5 ‡ What is Joe’s total cost of purchasing the bonds required to exactly (absolutely) match the liabili- ties? Problem 56.6 ‡ What is the annual eﬀective yield rate for investment in the bonds required to exactly (absolutely) match the liabilities? Problem 56.7 John wants to absolutely match a debt that he owes under which he must make payment of 1000 in one year and 2000 in two years. He can purchase the following bonds: (1) Bond A is a two year bond with annual coupons of 100 and a maturity value of 1000. (2) Bond B matures in one year for 1000 and pays a coupon of 80. Calculate the amount of Bond B that John should purchase. An Introduction to the Mathematics of Financial Derivatives What is a ﬁnancial derivative? A ﬁnancial derivative is a ﬁnancial contract (between two parties) that derives its value from the value of some underlying asset. For example, a homeowner’s insurance policy promises that in the event of a damage to your house, the insurance company will compensate you for at least part of the damage. The greater the damage, the more the insurance company will pay. Your insurance policy thus derives its value from the value of your house and therefore is a derivative. In this chapter, we introduce four ﬁnancial derivatives: forward contracts, future contracts, options (call and put) and swaps and we discuss some of the related topics. 493 494 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 57 Financial Derivatives and Related Issues In this section we introduce the concept of ﬁnancial derivatives and discuss some of the relevant topics of derivatives. What is a Financial Derivative? A ﬁnancial derivative is a contract (between two parties) that derives its value from the value of some underlying asset(s). Examples of underlying assets include ﬁnancial assets (such as stocks or bonds), commodities (such as gold or oil), market indices (such as S&P 500 or FTSE 100), interest rates (LIBOR rates) and many others. An example of a derivative is the following: Party A makes an agreement with Party B regarding the price of gold. If the price of one ounce of gold in one year is above $300, Party Aagrees to pay Party B the amount of $1. Otherwise, Party B will pay Party A$1. The agreement is a derivative since the outcome depends on the price of gold. Note that this agreement is a kind of a “bet” on the price moves of gold. In general derivatives can be viewed as bets on the price of something. The most common types of ﬁnancial derivatives that we will discuss in this book are futures, for- wards, options, and swaps. Derivatives Trading Markets Derivatives can be traded through organized exchanges such as The Chicago Board of Trade(CBOT), the New York Stock Exchange (NYSE), the New York Mercantile Exhange (NYMEX) and many oth- ers, or through contracts negotiated privately between two parties, referred to as over−the−counter (abbreviated OTC). Financial markets are usually regulated by the Securities and Exchange Com- mission (SEC) and the Commodity Futures Trading Commission (CFTC). Financial markets serve six basic functions. These functions are brieﬂy listed below: •Borrowing and Lending: Financial markets permit the transfer of funds (purchasing power) from one agent to another for either investment or consumption purposes. •Price Determination: Financial markets provide vehicles by which prices are set both for newly issued ﬁnancial assets and for the existing stock of ﬁnancial assets. •Information Aggregation and Coordination: Financial markets act as collectors and aggregators of information about ﬁnancial asset values and the ﬂow of funds from lenders to borrowers. •Risk Sharing: Financial markets allow a transfer of risk from those who undertake investments to those who provide funds for those investments. •Liquidity: Financial markets provide the holders of ﬁnancial assets with a chance to resell or liquidate these assets. •Eﬃciency: Financial markets reduce transaction costs and information costs. 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 495 Most Common Uses of Derivatives Below are some of the motives for someone to use derivatives: (1) to reduce (or hedge) exposure to risk. For example, a wheat farmer and a wheat miller could enter into a futures contract to exchange cash for wheat in the future. Both parties have reduced a future risk: for the wheat farmer, the uncertainty of the price, and for the wheat miller, the availability of wheat. (2) to speculate expected changes in future prices with the hope of making proﬁt. In this case, speculation increases the exposure to risk. The potential gain or loss can be leveraged (i.e. mag- niﬁed) relative to the initial investment. (3) to reduce transaction costs, such as commissions and other trading costs. (4) to maximize return on investments through asset management activities, tax loopholes, and regulatory restrictions. For example, a company can use derivatives to produce temporary losses to lower its taxes. We refer to this motive as regulatory arbitrage. Example 57.1 Weather derivatives are ﬁnancial instruments that can be used by organizations or individuals as part of a risk management strategy to reduce risk associated with adverse or unexpected weather conditions (rain/temperature/snow). Two types of weather derivatives are heating degree-day (HDD) and cooling degree-day (CDD) future contracts. These future contracts make payments based on whether the temperature is abnormally hot or cold. (a) Explain why a soft-drink manufacturer might be interested in such a contract. (b) The business buys these future contracts to hedge against temperature-related risk. Who is the other party accepting the risk? Solution. (a) Soft drink sales greatly depend on weather. Generally, warm weather boosts soft drink sales and cold weather reduces sales. A soft drink producer can use weather futures contracts to reduce the revenue swing caused by weather and smooth its earnings. Shareholders of a company generally want the earnings to be steady. They don’t want the management to use weather as an excuse for poor earnings or wild ﬂuctuations of earnings. (b) Anyone who can predict a weather index can enter into weather futures and make a proﬁt. However, since predicting weather is not usually easy and accurate a loss may occur Perspectives on Derivatives There are three diﬀerent user perspectives on derivatives: •The end-user perspective. End-users include corporations, investment managers, and in- vestors. End-users use derivatives in order to achieve a speciﬁc goal or goals such as managing risk, speculating, reducing costs, or avoiding regulations. 496 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES •The market-maker perspective. These are traders or intermediaries between diﬀerent end- users. They buy from end-users who want to sell (usually at a low price) and sell to end-users who want to buy (usually at a higher price.) Also, market-users might charge commissions for trading transactions. •The economic observer such as a regulator or a research economist, whose role is to watch and even sometimes regulate the markets. Traders of Derivatives There are three main traders of derivatives: hedgers, speculators and arbitrageurs. •Hedgers use derivatives to reduce risk that they face from potential future movements in market variables. Example 57.2 Suppose that it is August 16 2007 and, ﬁrm A, a company based in the US knows that it will pay £10 million on November 16, 2007 for goods it has purchased from a British ﬁrm. Firm Acan buy pounds from a bank in a three-month forward contract and thus hedge against foreign exchange risk. Suppose the agreement is to buy £1 for $1.55. (a) What would the ﬁrm A pay on November 16, 2007? (b) Suppose that on November 16, 2007 the exchange rate is $3 for £1. What would ﬁrm Apay on Novemebr 16, 2007 in the absence of hedging? Solution. (a) The ﬁrm has hedged against exchange rate volatility. The amount to be paid in US dollars on November 16, 2007 is 1 .55 ×10,000,000 = $15.5 millions. (b) The ﬁrm would have to pay 3 ×10,000,000 = $30 millions •Speculators use derivatives to bet on the future direction of a market variable. They can, for example, buy a put option on a stock if they think it will go down. If they think that the price of the stock will go up they will buy a call option. Another example of a speculator is an investor in weather derivatives who buys temperature-related future contracts. •Arbitrageurs take oﬀsetting positions in two or more instruments to lock in a riskless proﬁt if securities are inconsistently priced. Example 57.3 Suppose that ﬁrm Ais dually listed on both the NYSE and LSE. On the NYSE its share is trading at $152 and on the LSE its share is trading at £100. The exchange rate is $1.55 per pound. An arbitrageur makes a riskless proﬁt by simultaneously buying 100 shares on the NYSE and sell them on the LSE. What is the amount of this proﬁt? 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 497 Solution. He buys the shares on the NYSE for $152 a share. He sells the shares on the LSE for $155 a share. So the proﬁt per share is $3. Total proﬁt from the 100 shares is then $300 498 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 57.1 Which of the following is a derivative? (I) A bushel of corn (II) A contract to sell 100 bushels of corn (III) Short sale of a stock. Problem 57.2 The Security and Exchange Commission (SEC) holds primary responsibility for enforcing the fed- eral securities laws and regulating the securities industry. Which of the following describes the perspective of SEC on derivatives? (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.3 List four motives for using derivatives. Problem 57.4 Company A based in the US buys goods from a British company and is required to pay in British pounds for the cost of the goods. Company A buys a derivative that allows it to use the current exhange rate after six months from today. Which of the following characterizes company Auses of the derivative? (I) Hedging (II) Speculation (III) Reduced transaction costs (IV) Regulatory loop holes. Problem 57.5 You buy 100 shares of a stock through a brokerage ﬁrm. Which of the following describes your perspective of using derivatives? (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.6 You buy 100 shares of a stock through a brokerage ﬁrm. Which of the following describes brokerage ﬁrm perspective of using derivatives? 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 499 (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.7 You work for a ﬁrm who has a reimbursement program for college credits. The program reimburses 100% of costs for an “A”, 75% of costs for a “B”, 50% for a “C” and 0% for anything less. Would this program be considered an example of a derivative? Explain. Problem 57.8 Explain why the following ﬁrms might be interested in buying HDD or DCC future contracts. (a) Ski-resort owner. (b) Electric utilities. (c) Amusement park owner. Problem 57.9 Match each of the following market traders: (a) hedgers (b) speculators (c) arbitrageurs with one of the following roles: (I) proﬁt from perceived superior expectations (II) trade to eliminate or reduce future price risk (III) attempt to proﬁt when the same security or commodity is trading at diﬀerent prices in two or more markets. Problem 57.10 Explain carefully the diﬀerence between (a) hedging (b) speculation and (c) arbitrage. 500 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 58 Derivatives Markets and Risk Sharing As pointed out in the previous section, one of the basic function of derivatives markets is risk shar- ing. Risk is a major feature of economic activity. In this section we discuss the role of ﬁnancial markets as a vehicle of sharing risks. Risk-sharing is one of the most important functions of ﬁnancial markets. To illustrate this con- cept, consider auto insurance companies. These companies collect premiums for auto insurance policies. The total premiums collected are then being available to help those who get involved into car wrecks. Thus, those policyholders who had no wrecks in their records have basically lost their premiums. However, their premiums went to help those who needed it. A similar scenario occurs in the business world. Some companies proﬁt and others suﬀer. Thus, it makes sense to have a mechanism enabling companies to exchange ﬁnancial risks. Share risking mechanisms should beneﬁt everyone. Another example of risk sharing is the use of the so-called catastrophe bonds. For example, an insurance company that provides hurricane insurance for Florida residents will usually face large insurance claims in the case of a devastating hurricane happening. For that reason, the insurance company usually either issues or buys from reinsurers cat bonds enabling the company to share risks with the bondholders. Catastrophe bondholders receive interest rates at levels commensurate with the risk that they may lose all of their principal on the occurrence of a major hurricane. The sponsor issues cat bonds and typically invests the proceeds from the bond issuance in low-risk securities. The earnings on these low-risk securities, as well as insurance premiums paid to the sponsor, are used to make periodic, variable rate interest payments to investors. Sponsors of cat bonds include insurers, reinsurers, corporations, and government agencies. Diversiﬁable Risk Versus Non-Diversiﬁable Risk Risk management is one of the topics discussed in portfolio theory. The risk level of an asset can be categorized into two groups: the diversiﬁable risk and the non-diversiﬁable risk. Before we discuss the diﬀerence between diversiﬁable and non-diversiﬁable risks, we need to ﬁrst understand the term diversiﬁcation. The act of diversiﬁcation implies that an individual (or a ﬁrm) allocates his/her wealth among several types of investments rather than just one investment. In other words, diversi- ﬁcation means spreading the risk of a portfolio by investing in several diﬀerent types of investments rather than putting all the money in one investment. Diversiﬁable risk (also known as non-systematic risk) is a risk that can be reduced or eliminated by combining several diverse investments in a portfolio. Example 58.1 Suppose you are an avid Apple computer user, and you decide to invest all your money in Apple stocks. Are you making a sound decision? If not, then what is a good investment strategy? 58 DERIVATIVES MARKETS AND RISK SHARING 501 Solution. The answer is no. If you have done so, you will be very vulnerable in the last few months when Apple announced losses and a major change in management. However, you can reduce the risk of your portfolio if you invest in other computer companies such IBM, Compaq, etc. However, this is not a very well diversiﬁed portfolio because you have invested your money in the computer industry. As a result, you will still be vulnerable to changes in the computer industry. It is wise to have a portfolio consisting of shares in ﬁrms from various industries such as real estate, electronics, biomedical and so on Non-diversiﬁable risk is the part of an asset’s risk that cannot be eliminated through diver- siﬁcation. This type of risk is also known as market risk or systematic risk . Examples of non-diversiﬁable risks are natural disasters, wars, and economic factors such as inﬂation or unem- ployment levels events that can inﬂuence the entire market and usually aﬀects your entire portfolio. Example 58.2 Which of the following can be categorized as a non-diversiﬁable risk? (a) Recession. (b) The fall of a stock share. (c) Government defaults in paying bonds. Solution. Obviously only (a) and (c) might have major impact of one’s entire portfolio 502 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 58.1 Explain why a health insurance policy is a risk sharing derivative. Problem 58.2 List the two types of risk involved in a portfolio. Problem 58.3 The act of diversiﬁcation implies that an individual (or a ﬁrm) allocates his/her wealth (I) among several types of assets (II) in just one asset Problem 58.4 Determine whether each of the following events is considered a diversiﬁable or non-diversiﬁable risk for a portfolio: (a) A software ﬁrm’s best programmer quits. (b) Oil-prices on international markets suddenly increase. (c) Congress votes for a massive tax cut. (d) A low-cost foreign competitor unexpectedly enters a ﬁrm’s market. Problem 58.5 Determine whether each of the following events is considered a diversiﬁable or non-diversiﬁable risk for a portfolio: (a) A wildcat strike is declared. (b) Oil is discovered on a ﬁrm’s property. (c) The Federal Reserve institutes a restrictive monetary policy. (d) Long term interest rates rise sharply. Problem 58.6 An earthquake linked bond is called (a) callable bond (b) perpetual bond (c) cat bond (d) serial bond Problem 58.7 Decide whether the risk is a diversiﬁable risk or non-diversiﬁable risk: 58 DERIVATIVES MARKETS AND RISK SHARING 503 (a) Management risk (b) Inﬂation risk (c) Sociopolitical risk (d) Credit risk (e) Currency risk (f) Interest rate risk (g) Liquidity risk. Problem 58.8 True or false: Cat bonds are used by sponsors as a hedge against natural disasters. Problem 58.9 Which of the following is not a term that refers to non-systematic risk? (A) Non-diversiﬁable risk (B) Market risk (C) Diversiﬁable risk (D) Both (A) and (B) Problem 58.10 In 9/11, immediately after the terrorist attacks the stock market fell signiﬁcantly. This is an example of (A) Systematic risk (B) Market risk (C) Non-diversiﬁable risk (D) All of the above Problem 58.11 Recently in California the workers at Kroger supermarkets had been on strike. This strike would most likely have been seen as an example of (A) Non-systematic risk (B) Non-diversiﬁable risk (C) Diversiﬁable risk (D) None of the above Problem 58.12 Explain how cat bonds work. 504 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 59 Forward and Futures Contracts: Payoﬀ and Proﬁt Dia- grams In this section we introduce two examples of derivatives: forward and futures contracts. We also analyze the possible payoﬀs and proﬁts of these two securities. Forwards and futures are very similar as contracts but diﬀer in pricing and trading mechanisms. For example, forwards are traded over the counter (i.e. contract speciﬁcation can be customized) whereas futures are traded on exchange markets (where contracts are standardized). For the purpose of this section, we think of them as interchangeable. A more detailed discussion of futures will be covered in Section 72. A forward contract (or a futures contract ) is a commitment to purchase at a future date, known as the expiration date, the delivery date or the maturity date, a given amount of a commodity or an asset at a price agreed on today. The price ﬁxed now for future exchange is called the forward price or the delivery price. The asset or commodity on which the forward contract is based is called the underlying asset. Apart from commissions and bid-ask spreads, a forward contract requires no initial payment or premium between the two parties. At the delivery date, cash is exchanged for the asset. Examples of forward contracts include •A bond forward is an obligation to buy or sell a bond at a predetermined price and time. •An equity forward is an obligation to buy or sell an equity at a predetermined price and time. •A gold forward is an obligation to deliver a speciﬁed quantity of gold on a ﬁxed date and receive a ﬁxed delivery price. •A foreign exchange forward is an obligation to buy or sell a currency on a future date for a predetermined ﬁxed exchange rate. This type of forwards is used by ﬁrms as a protection against ﬂuctuation in foreign currency exchange rates. •An interest rate forward is used to lock-in future interest rates. For example, a ﬁrm who is planning for a loan in six months can buy forward contract that lock-in the rate at the present. This rate will be the rate that will apply when the loan is exercised in six months. The risk here is that the ﬁrm will stuck with borrowing money at the stated rate so any decline of interest rates before the six months period will cause losses for the ﬁrm. Example 59.1 Explain in words the meaning of the following: A 3-month forward contract for 1,000 tons of soybean at a forward price of $165/ton. Solution. In this contract, the buyer is committed to buy 1000 tons of soybean from the seller in three months for a price of $165 a ton 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 505 Payoﬀ on a Forward Contract Forward contracts are privately executed between two parties. The buyer of the underlying com- modity or asset is referred to as the long side whereas the seller is the short side. The obligation to buy the asset at the agreed price on the speciﬁed future date is referred to as the long position. A long position proﬁts when prices rise. The obligation to sell the asset at the agreed price on the speciﬁed future date is referred to as the short position. A short position proﬁts when prices go down. What is the payoﬀ of a forward contract on the delivery date? Let T denote the expiration date, K denote the forward price, and PT denote the spot price (or market price) at the delivery date. Then •for the long position: the payoﬀ of a forward contract on the delivery date is PT −K; •for the short position: the payoﬀ of a forward contract on the delivery date is K−PT. Figure 59.1 shows a payoﬀ diagram on a contract forward. Note that both the long and short forward payoﬀ positions break even when the spot price is equal to the forward price. Also note that a long forward’s maximum loss is the forward price whereas the maximum gain is unlimited. For a short forward, the maximum gain is the forward price and the maximum loss is unlimited. Figure 59.1 Payoﬀ diagrams show the payoﬀ of a position at expiration. These payoﬀs do not include any costs or gains earned when purchasing the assets today. Payoﬀ diagrams are widely used because they summarize the risk of the position at a glance. We have pointed out earlier that the long makes money when the price rises and the short makes money when the price falls. Example 59.2 An investor sells 20 million yen forward at a forward price of $0.0090 per yen. At expiration, the spot price is $0.0083 per yen. 506 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (a) What is the long position payoﬀ? (b) What is the short position payoﬀ? Solution. (a) At the expiration date, the long position’s payoﬀ is (0 .0083 −0.009) ×20 ·106 = −$14,000, a loss of $14,000 (b) The short position’s payoﬀ is $14 ,000 that is a proﬁt of $14,000 Example 59.3 Consider a forward contract on a stock with spot price of $25 and maturity date of 3 months from now. The forward price is $25.375. Draw the payoﬀ diagram for both the long and short forward positions after 3 months. Solution. The diagram is given in Figure 59.2. Figure 59.2 Note that both the long and short forward payoﬀ positions break even when the price of the stock at maturity is equal to the forward price, i.e., at $25.375 Example 59.4 Consider the forward contract of the previous example. Suppose the investor decides to purchase the stock outright today for $25. Is there any advantage of using the forward contract to buy the stock as opposed to buying it outright? (a) Draw the diagram for the payoﬀ to the long physical position as well as the payoﬀ diagram to the long forward. (b) If the spot price is $0 in three months then ﬁnd the long forward position’s payoﬀ. 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 507 (c) If the spot price of the stock in three months is $25.372 then ﬁnd the long physical position’s payoﬀ and the long forward’s payoﬀ. (d) Describe the investement for the outright purchase as well as the long forward in order to own the stock. Solution. (a) The diagram is shown in Figure 59.3. Note that the value of the payoﬀ of the physical position is equal to the spot price of the stock. The payoﬀ diagram does not include costs when the stock is purchased outright. Figure 59.3 (b) The long forward’s payoﬀ is 0 −25.372 = −$25.372. (c) The payoﬀ of the long physical position is $25.372. The payoﬀ of the long forward is 25 .372 − 25.372 = $0. (d) With the outright purchase, the investor invests $25 now and owns the stock. With the forward contract, the investor invests $0 now and $25.372 after three months and owns the stock Note that the payoﬀ of the outright purchase in Figure 59.3 tells us how much money we end up with after 3 months, but does not account for the initial $25 investment for purchasing the stock outright. Thus, from the graph one can not tell whether there is an advantage to either a forward purchase or an outright purchase. In order to have a fair comparison, the initial investments for both must be the same and then account for interest rate earned over the 3 months. We illustrate this point in the next two examples. Example 59.5 Suppose the investor invests $25 in a zero-coupon bond with par value $25.372 and maturity date in three months along with the forward contract of the previous example. Note that this combined position and the outright purchase position each initially costs $25 at time 0. Suppose the bond yield rate is 5.952% payable quarterly. Show that the combined position mimics the eﬀect of buying the stock outright. 508 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. The investor pays $25 now for the bond. After three months the zero-coupon bond return is $25.372. He then uses the proceeds to pay for the forward price of $25.372 and owns the stock. This alternative is equivalent to outright purchase of the stock today for $25 Example 59.6 Suppose the investor borrows $25 to buy the physical stock which costs $25. Interest rate on the loan is 5.952% compounded quarterly. In this case, this position and the long forward position each initially costs $0 at time 0. Show that this act of borrowing mimics the eﬀect of entering into a long forward contract. Solution. The investor uses the borrowed money to buy the stock today. After three months, the investor repays $25 ×1.01488 = $25.372 for the borrowed money. Thus, the investor invested nothing ini- tially, and after three months he paid $25.372 and owned the stock. This shows that borrowing to buy the stock mimics the eﬀect of entering into a long forward contract In the above two examples, we conclude that the forward contract and the cash position are equiv- alent investments, diﬀering only in the timing of the cash ﬂows. Besides payoﬀ diagrams, one deﬁnes a proﬁt diagram. Recall that a payoﬀ diagram graphs the cash value of a position at maturity. A proﬁt diagram subtracts from the payoﬀ the future value of the initial investment in the position. Geometrically, the proﬁt diagram is a vertical shift of the payoﬀ diagram by the amount of the future value of the initial investment. Since forward contracts require no initial investment, the payoﬀ and proﬁt diagrams coincide. Example 59.7 A stock is priced at $50 and pays no dividends. The eﬀective annual interest rate is 10%. Draw the payoﬀ and proﬁt diagrams for a one year long position of the stock. Solution. The payoﬀ of a stock you buy long (and therefore you own) is just the stock market value. Let’s assume you want to ﬁnd the proﬁt after one year. Your proﬁt is the payoﬀ (market price) minus the future value of the initial investment over one year. If we denote the stock price by S then your proﬁt will be S−1.1(50) = S−55. Figure 59.4 shows both the payoﬀ and proﬁt diagrams. Notice that the proﬁt is 0 when the stock price is $55 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 509 Figure 59.4 Example 59.8 Suppose we own an index which we want to be valued at $1020 six months from today. For that we buy a forward contract with forward price of $1020 and maturity date of six months. Consider investing $1000 in a zero-coupon bond with six-month interest rate of 2%. (a) Compare the payoﬀ diagram of the combined position of the forward contract and the bond with the payoﬀ diagram of the index after six months. (b) Answer the same question for the proﬁt diagrams. Solution. (a) The payoﬀ of the forward contract and the bond in six months is forward + bond = spot price at expiration - 1020 + (1000 + 1000(0.02)) = spot price at expiration Thus, the payoﬀ diagram of the combined position is the same as the payoﬀ diagram of the index. Adding the bond shifts the payoﬀ diagram of the forward vertically upward by $1020. (b) The proﬁt diagram of the forward contract coincides with its payoﬀ’s diagram since there is no initial investment. The proﬁt of the bond is the payoﬀ minus the future value of the investment which comes out to 0. Thus, the proﬁt diagram of the combined forward and bond is the same as the proﬁt diagram of the forward Cash-Settled Forward Contracts A forward may be cash-settled, in which case the underlying asset and payment never exchange hands. Instead, the contract settles with a single payment in the amount spot price minus forward 510 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES price at delivery date. If the diﬀerence is positive, the short party pays the long party the diﬀerence. If it is negative, the long party pays the short party. Example 59.9 Two parties agree today to exchange 500,000 barrels of crude oil for $42.08 a barrel three months from today. Describe this forward when settled for cash (a) if the spot price at the delivery date is $47.36 a barrel; (b) if the spot price at the delivery date is $40.17 a barrel. Solution. If the forward were cash-settled then on the delivery date three months from today, no oil would change hands, and there would be no payment of 500 ,000 ×42.08 = $21.04 millions. (a) If the spot price at the delivery date was $47.36 a barrel then the forward would settle with a single payment of 500,000(47.36 −42.08) = $2,640,000 made by the short party to the long party. (b) If the spot price at the delivery date was $40.17 a barrel then the forward would settle with a single payment of 500,000(42.08 −40.17) = $955,000 made by the long party to the short party Credit Risk We conclude this section by mentioning the credit risk (i.e. the risk that a party will not meet its contractual obligations) associated with forward contracts. Exchange-traded contracts typically require collateral in order to minimize this risk . In over-the-counter contracts, each party bears the other’s credit risk. Example 59.10 Creditwise, which is riskier a forward contract or a futures contract? Solution. Since forward contracts are traded on over-the-counter and futures contracts are traded on ex- changes, forward contracts are more riskier than futures contracts 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 511 Practice Problems Problem 59.1 What is the diﬀerence between a long forward position and a short forward position? Problem 59.2 What is the diﬀerence between a long futures position and a short futures position? Problem 59.3 A trader enters into a short forward contract on 100 million yen. The forward exchange rate is 189.4 yen per £. How much does the trader gain or lose if the exchange rate at the end of the contract is (a) 192.0 yen per £; (b) 183.0 yen per £? Problem 59.4 A stock with no dividends has a current price of $50. The forward price for delivery in one year is $53. If there is no advantage of buying either the stock or the forward contract, what is the 1-year eﬀective interest rate? Problem 59.5 A default-free zero-coupon bond costs $91 and will pay $100 at delivery date in one year. What is the eﬀective annual interest rate? What is the payoﬀ diagram for the bond? What is the proﬁt diagram for the bond? Problem 59.6 Suppose that you enter into a long six-month forward position at a forward price of $50. What is the payoﬀ in 6 months for prices of $40, $45, $50, $55, and $60? Problem 59.7 Suppose that you enter into a short six-month forward position at a forward price of $50. What is the payoﬀ in 6 months for prices of $40, $45, $50, $55, and $60? Problem 59.8 Suppose shares of XYZ corporation are trading at $50 and currently pay no dividends. The forward price for delivery in 1 year is $55. Suppose the 1-year eﬀective annual interest rate is 10%. Draw the payoﬀ and proﬁt diagrams of this forward contract for a long position. Problem 59.9 A company agrees to buy 2,000 barrels of oil in two years; the current price is $63.50 per barrel. What is reasonable price to pay per barrel in two years? 512 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 59.10 Consider a forward contract of a stock index with maturity date six months from now and forward price of $1020. Answer the following questions if this forward contract is settled for cash. (a) Find the long and short positions payoﬀs if the spot price of the index at maturity is $1040. (b) What if the spot sprice is $960? Problem 59.11 You are a US exporter who will receive 5 millions Euro in six months time from the sale of your product in Spain. How can you hedge your foreign exchange risk ? Problem 59.12 Consider a forward contract of a stock index with maturity date three months from now and forward price of $930. Assume that the spot price of the index today is $900. What is the proﬁt or loss to a short position if the spot price of the market index rises to $920 by the expiration date? Problem 59.13 One ounce of gold today is being sold at the price of $300 per ounce. An investor borrows $300 with annual interest rate of 5% and buys one ounce of gold. The investor enters into a short forward contract to sell 1 oz. of gold in one year at $340. What is the investor’s proﬁt in this transaction? What would be his proﬁt if the original price of 1 oz. of gold is $323.81 and the loan is $323.81? Problem 59.14 An investor enters into a short forward contract to sell 100,000 British pounds for US dollars at an exchange rate of 1.50 US dollars per pound. How much does the investor gain or lose if the exchange rate at the end of the contract is (a) 1.49 and (b) 1.52? Problem 59.15 A forward contract is made for the delivery of a commodity 10 months from today at a forward price of $200 per unit of the commodity. (A) The purchaser of the forward contract pays the forward price, $200, immediately but must wait for 10 months to receive the commodity. (B) The holder of a forward contract always has the right to exchange it for an identical futures contract at any time before the delivery date. (C) The seller of the forward contract has an obligation to deliver the commodity 10 months from today in return for the spot price on the delivery date. (D) The seller of the forward contract has an obligation to deliver the commodity 10 months from today in return for $200 per unit, payable on delivery. 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 513 Problem 59.16 The main diﬀerence between forward and futures contracts is: (A) Futures contracts are traded every day (i.e., frequently) in organized exchanges prior to the delivery date speciﬁed in the contract. (B) For a futures contract there is never any obligation to deliver, or to receive, the underlying asset. (C) Forward contracts always have delivery dates further from the present than futures contracts. (D) The underlying assets in forward contracts are always physical commodities (e.g., wheat, oil or silver), while for futures contracts the underlying assets could be anything (e.g. weather indices or notional bank deposits). 514 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 60 Call Options: Payoﬀ and Proﬁt Diagrams One problem with a forward contract is the obligation of the buyer to pay the forward price at the expiration date which causes a loss if the spot price is below the forward price. So it is natural to wonder if there is a type of contract where the buyer has the option not to buy the underlying asset. The answer is aﬃrmative thanks to call options. An option is a contract to buy or sell a speciﬁc ﬁnancial product such as a security or a commodity. The contract itself is very precise. It establishes a speciﬁc price, called the strike price or exercise price at which the contract may be exercised, or acted on, and it has an expiration date. When an option expires, it no longer has value and no longer exists. There are two types of options: calls and puts. In this section we discuss call options and in the next section we discuss the put options. In a call option the call buyer expects that the price of an asset may go up. The buyer pays an upfront premium that he will never get back. He has the right to exercise the option at or before the expiration date depending on the option’s style(to be introduced below). The call seller, also known as the option’s writer, receives the premium. If the buyer decides to exercise the option, then the seller has to sell the asset at the strike price. If the buyer does not exercise the option, then the seller proﬁt is just the premium. Thus, a call option is a contract where the buyer has the right to buy but not the obligation to buy. An option style governs the time at which exercise can occur. A European call option allows the holder to exercise the option (i.e., to buy) only on the option expiration date. An American call option allows exercise at any time during the life of the option. A Bermuda call option can be exercised at certain pre-speciﬁed dates before or at the expiration date. Unless otherwise speciﬁed, we will limit our discussions to European style. For a call option, the buyer has a long position whereas the seller has a short position. Example 60.1 Trader A(call buyer) purchases a call contract to buy 100 shares of XYZ Corp from Trader B (call seller) at $50 a share. The current price is $45 a share, and Trader Apays a premium of $5 a share. (a) If the share price of XYZ stock rises to $60 a share right before expiration, then what proﬁt does Trader A make if he exercised the call by buying 100 shares from Trader B and sell them in the stock market? (b) If, however, the price of XYZ drops to $40 a share below the strike price, then Trader Awould not exercise the option. What will be his losses in this case? Solution. (a) Trader A total cost for owning the 100 shares before or on the expiration date is 50(100) + 5(100) = 5,500. The revenue from selling the shares is 60(100) = 6 ,000. Thus, Trader A makes a 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 515 proﬁt of $500 in this transaction. (b) The losses will consist basically of the future value of premium paid, i.e. the future value of 5(100) = $500 Since a buyer of a call option will exercise the option only when the spot price is higher than the strike price, the buyer’s payoﬀ is then deﬁned by the formula Buyer′s call payoﬀ = max{0,spot price at expiration −strike price}. Example 60.2 Consider a call option with a strike price of $500. (a) What would be the payoﬀ to the buyer if the spot price at the expiration date is $550? (b) What would be the payoﬀ to the buyer if the spot price at the expiration date is $450? Solution. (a) Since the spot price is larger than the strike price, the buyer will most likely exercise his option. The payoﬀ in this case is max{0,550 −500}= $50. (b) In this case, the buyer will not exercise his option and therefore the payoﬀ is max{0,450 −500}= $0 Notice that the payoﬀ does not take into consideration the cost of acquiring the position, i.e. the premium which is paid at the time the option is acquired. The payoﬀ of an option is not the money earned (or lost). The proﬁt earned by the buyer is given by the formula Buyer′s call proﬁt = Buyer′s call payoﬀ −future value of premium Example 60.3 Consider a 3-month call option with strike price of $500 and premium of $46.90. If the spot price at expiration date is $550, would you exercise the call option? How much would you make or lose? The risk-free interest rate is 1% over three months. Solution. Since spot price is higher than the strike price, you expect to exercise the call option. The future value of the premium is 46 .90(1.01) = $47.37. The buyer’s proﬁt is then max{0,550 −500}−47.37 = $2.63 516 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES A payoﬀ diagram and a proﬁt diagram of a long call with strike price K, spot price PT,and future value of premium Pc is shown in Figure 60.1. Figure 60.1 Notice that for a long call the maximum loss is the future value of the premium whereas the maximum gain is unlimited. Example 60.4 Consider a long call option (with strike price K) and a long forward contract (with a forward price K) with underlying asset a stock index. Draw the proﬁt diagram of both on the same window. (a) If the stock index is rising past K,, which of the two is more proﬁtable? (b) If the stock index falls considerably, which is more proﬁtable? Solution. Figure 60.2 plots the proﬁt for both the long forward and the long call. Figure 60.2 (a) The forward contract is more proﬁtable than the call option. (b) The call option is more proﬁtable (i.e. less loses) because the most you lose is the future 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 517 value of your premium. Notice that a call option can be thought as an insured position in the in- dex. It protects the buyer against substantial loses. The cost of this insurance is the premium paid Thus far we have considered the payoﬀ and the proﬁt from the buyer’s perspective. Next, we consider these issues from the seller’s (i.e. the writer) perspective. The writer’s payoﬀ and proﬁt are the negative of those of the buyer’s and thus they are given by Writer′s call payoﬀ = −max{0,spot price at expiration −strike price} = min{0,strike price −spot price at expiration } and Writer′s call proﬁt = Writer′s call payoﬀ + future value of premium . Example 60.5 Again, consider a 3-month call option with a strike price of $500 and a premium $46.90. Find the writer’s payoﬀ and the proﬁt if the spot price at the expiration date is $550. Assume a risk-free 3-month interest rate of 1%. Solution. Since the spot price is higher than the strike price, the writer has to sell the option. His/her payoﬀ is min{0,500 −550}= −$50 and the proﬁt is min{0,500 −550}+ 46.90(1.01) = −$2.63 The payoﬀ and proﬁt diagrams for a short call with strike price K, future value of premium Pc are shown in Figure 60.3.

Human: What is the monthly installment?

Actuarial AI: The Basics of Annuity Theory A series of payments made at equal intervals of time is called an annuity. Common examples are house rents, mortgage payments on homes, and installments payments on automobiles. An annuity where payments are guaranteed to occur for a ﬁxed period of time is called anannuity−certain. For example, mortgage payments on a home. The ﬁxed period of time for which payments are made is called term of the annuity. For example, in the case of a home mortgage a term can be either a 15−year loan or a 30 −year loan. Annuities that are not certain are called contingent annuities. For example, a pension is paid so long as the person survives. That is, regular payments are made as long as the person is alive. Pension is an example of contingent annuity also called life annuity. Unless otherwise indicated, the annuity−certain is the type of annuity we will assume in this book, and the “certain” will be dropped from the name. The interval between annuity payments is called a payment period, often just called a period. In Sections 15 through 21, we consider annuities for which the payment period and the interest conversion period are equal. Also, payments are of level amount, i.e. have the same ﬁxed monetary value for each period. In Sections 22 −29, we will discuss annuities for which payments are made more or less frequently than interest is converted and annuities with varying payments. With level annuities, we will most of the time assume payments of 1 since any other level annuity can be obtained from this by a simple multiplication. 143 144 THE BASICS OF ANNUITY THEORY 15 Present and Accumulated Values of an Annuity-Immediate An annuity under which payments of 1 are made at the end of each period for n periods is called an annuity−immediate or ordinary annuity. The cash stream represented by the annuity can be visualized on a time diagram as shown in Figure 15.1 . Figure 15.1 The ﬁrst arrow shows the beginning of the ﬁrst period, at the end of which the ﬁrst payment is due under the annuity. The second arrow indicates the last payment date −just after the payment has been made. Let idenote the interest rate per period. The present value of the annuity at time 0 will be denoted by ani or simply an. See Figure 15.2. Figure 15.2 Using the equation of value with comparison date at time t= 0, we can write an = ν+ ν2 + ··· + νn. That is, the present value of the annutiy is the sum of the present values of each of the npayments. We recognize the expression on the right −hand side as a geometric progression. Thus, multiplying both sides by ν to obtain νan = ν2 + ν3 + ··· + νn + νn+1. Subtracting this from the previous equation we ﬁnd (1 −ν)an = ν(1 −νn). Hence, an = ν·1 −νn 1 −ν = ν·1 −νn iν = 1 −(1 + i)−n i . (15.1) 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 145 Example 15.1 Calculate the present value of an annuity −immediate of amount $100 paid annually for 5 years at the rate of interest of 9%. Solution. The answer is 100a5 = 1001−(1.09)−5 0.09 ≈388.97 Formula (15.1) can be rewritten as 1 = νn + ian. This last equation is the equation of value at time t= 0 of an investment of $1 for nperiods during which an interest of iis received at the end of each period and is reinvested at the same rate i, and at the end of the nperiods the original investment of $1 is returned. Figure 15.3 describes the time diagram of this transaction. Figure 15.3 Next, we will determine the accumulated value of an annuity−immediate right after thenth payment is made. It is denoted by sn. See Figure 15.4. Figure 15.4 Writing the equation of value at the comparison date t= n we ﬁnd sn = 1 + (1 +i) + (1 +i)2 + ··· + (1 +i)n−1. That is, sn is the sum of the accumulated value of each of the n payments. Using the deﬁnition of sn and the reasoning used to establish the formula for an we can write sn =1 + (1 +i) + (1 +i)2 + ··· + (1 +i)n−1 =(1 + i)n −1 (1 + i) −1 = (1 + i)n −1 i . 146 THE BASICS OF ANNUITY THEORY This last equation is equivalent to 1 + isn = (1 + i)n. This last equation is the equation of value at time t= nof an investment of $1 for nperiods during which an interest of iis received at the end of each period and is reinvested at the same rate i, and at the end of the nperiods the original investment of $1 is returned. Figure 15.5 describes the time diagram of this transaction. Figure 15.5 Example 15.2 Calculate the future value of an annuity −immediate of amount $100 paid annually for 5 years at the rate of interest of 9%. Solution. The answer is 100s5 = 100 ×(1.09)5−1 0.09 ≈$598.47 Example 15.3 Show that am+n = am + νman = an + νnam. Interpret this result verbally. Solution. We have am + νman = 1−νm i + νm ·1−νn i = 1−νm+νm−νm+n i = 1−νm+n i = am+n. A verbal interpretation is as follows: The present value of the ﬁrst mpayments of an (m+ n)−year annuity−immediate of 1 is am.The remaining npayments have valuean at time t= m; discounted to the present, these are worth νman at time t= 0 Example 15.4 At an eﬀective annual interest rate i, you are given (1) the present value of an annuity immediate with annual payments of 1 for n years is 40 (2) the present value of an annuity immediate with annual payments of 1 for 3 n years is 70. Calculate the accumulated value of an annuity immediate with annual payments of 1 for 2 n years. Solution. Using Example 15.3 we can write a3n = a2n + ν2nan = an + νnan + ν2nan = an(1 + νn + ν2n). Hence, we obtain the quadratic equation ν2n + νn + 1 = 7 4 . Solving this equation we ﬁnd νn = 1 2 . 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 147 Again, using Example 15.3 we can write a3n = an + νna2n which implies that νna2n = 70 −40 = 30 and therefore a2n = 60. Finally, s2n = ν−2na2n = 4(60) = 240 We next establish a couple of relationships between an and sn. Theorem 15.1 With an and sn as deﬁned above we have (i) sn = (1 + i)nan. That is, the accumulated value of a principal of an after n periods is just sn. (ii) 1 an = 1 sn + i. Proof. (i) We have sn = (1+i)n−1 i = (1 + i)n ·1−(1+i)−n i = (1 + i)nan. (ii) We have 1 sn + i= i (1 + i)n −1 + i= i+ i(1 + i)n −i (1 + i)n −1 = i 1 −νn = 1 an A verbal interpretation of (ii) will be introduced when we discuss the concepts of amortization and sinking funds. Example 15.5 For a given interest rate i, an = 8.3064 and sn = 14.2068. (a) Calculate i. (b) Calculate n. Solution. (a) Using part (ii) of the previous theorem we ﬁnd i= 1 8.3064 − 1 14.2068 = 5%. (b) Using part (i) of the previous theorem we ﬁnd n= 1 ln (1 +i) ln (sn an ) = 11 The type of annuity discussed in this section involves compound interest rate. It is possible to deﬁne annuity−immediate not involving compound interest such as simple interest rate, simple discount rate, and force of interest. For example, we wish to ﬁnd the present value of an n−period annuity immediate in which each payment is invested at simple interest rate i. The present value is equal to the sum of the present value of the individual payments. Thus, we obtain an = 1 1 + i + 1 1 + 2i + ··· + 1 1 + ni. 148 THE BASICS OF ANNUITY THEORY The accumulated value of such an annuity is equal to the accumulated value of the individual payments. That is, sn = 1 + (1 +i) + (1 + 2i) + ··· + [1 + (n−1)i]. Example 15.6 Find an expression for an assuming each payment of 1 is valued at simple discount rate d. Solution. The present value is the sum of the present value of individual payments. That is, an = (1 −d) + (1−2d) + ··· + (1 −nd) = n−d(1 + 2 +··· + n) = n−n(n+ 1) 2 d 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 149 Practice Problems Problem 15.1 Consider an investment of $5,000 at 6% convertible semiannually. How much can be withdrawn each half−year to use up the fund exactly at the end of 20 years? Problem 15.2 The annual payment on a house is $18,000. If payments are made for 40 years, how much is the house worth assuming annual interest rate of 6%? Problem 15.3 If d= 0.05, calculate a12 . Problem 15.4 Calculate the present value of 300 paid at the end of each year for 20 years using an annual eﬀective interest rate of 8%. Problem 15.5 If an = x and a2n = y, express d as a function of x and y. Problem 15.6 (a) Given: a7 = 5.153, a11 = 7.036, a18 = 9.180. Find i. (b) You are given that an = 10.00 and a3n = 24.40. Determine a4n. Problem 15.7 Show that sm+n = sm + (1 +i)msn = sn + (1 +i)nsm. Interpret this result verbally. Problem 15.8 A grandmother has a granddaughter entering university next year. Her granddaughter expects to remain in school for ten years and receive a PhD. This grandmother wishes to provide $1,000 a year to her granddaughter for entertainment expenses. Assuming a 3.5% eﬀective annual interest rate, how much does the grandmother have to deposit today to provide ten annual payments starting one year from now and continuing for ten years? Problem 15.9 In the previous problem, if the PhD student saves the income from her grandmother in an account also paying 3.5% eﬀective annual interest, how much will she have when she receives the ﬁnal $1,000 payment? 150 THE BASICS OF ANNUITY THEORY Problem 15.10 Find the present value of an annuity which pays $200 at the end of each quarter −year for 12 years if the rate of interest is 6% convertible quarterly. Problem 15.11 Compare the total amount of interest that would be paid on a $3,000 loan over a 6 −year period with an eﬀective rate of interest of 7.5% per annum, under each of the following repayment plans: (a) The entire loan plus accumulated interest is paid in one lump sum at the end of 6 years. (b) Interest is paid each year as accrued, and the principal is repaid at the end of 6 years. (c) The loan is repaid with level payments at the end of each year over the 6-year period. Problem 15.12 A loan of $20,000 to purchase a car at annual rate of interest of 6% will be paid back through monthly installments over 5 years, with 1st installment to be made 1 month after the release of the loan. Problem 15.13 Over the next 20 years, you deposit money into a retirement account at the end of each year according to the following schedule: Time Amount invested each year 1 - 5 $2000 6 - 10 $3000 11 - 20 $5000 The eﬀective annual rate of interest is 9%. Find the accumulated value of your account at time 20. Problem 15.14 A family wishes to accumulate 50,000 in a college education fund by the end of 20 years. If they deposit 1,000 into the fund at the end of each of the ﬁrst 10 years, and 1 ,000 + X at the end of each of the second 10 years, ﬁnd X to the nearest unit if the fund earns 7% eﬀective. Problem 15.15 An annuity provides a payment of n at the end of each year for n years. The annual eﬀective interest rate is 1 n. What is the present value of the annuity? Problem 15.16 The cash price of a new automobile is 10,000. The purchaser is willing to ﬁnance the car at 18% convertible monthly and to make payments of 250 at the end of each month for 4 years. Find the down payment which will be necessary. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 151 Problem 15.17 You have $10,000 down payment on a $20,000 car. The dealer oﬀers you the following two options: (a) paying the balance with end −of−month payments over the next three years at i(12) = 0.12; (b) a reduction of $500 in the price of the car, the same down payment of $10,000, and bank ﬁnancing of the balance after down payment, over 3 years with end −of−month payments at i(12) = 0.18. Which option is better? Problem 15.18 Nancy has 10,000 in a bank account earning 6% compounded monthly. Calculate the amount that she can withdraw at the end of each month from the account if she wants to have zero in the account after 12 months. Problem 15.19 Megan purchased a new car for 18,000. She ﬁnances the entire purchase over 60 months at a nominal rate of 12% compounded monthly. Calculate Megan’s monthly payment. Problem 15.20 ‡ Seth, Janice, and Lori each borrow 5,000 for ﬁve years at a nominal interest rate of 12%, compounded semiannually. Seth has interest accumulated over the ﬁve years and pays all the interest and principal in a lump sum at the end of ﬁve years. Janice pays interest at the end of every six−month period as it accrues and the principal at the end of ﬁve years. Lori repays her loan with 10 level payments at the end of every six −month period. Calculate the total amount of interest paid on all three loans. Problem 15.21 If d(12) = 12%,calculate the accumulated value of 100 paid at the end of each month for 12 months. Problem 15.22 The accumulated value of an nyear annuity−immediate is four times the present value of the same annuity. Calculate the accumulated value of 100 in 2 n years. Problem 15.23 You are given the following information: (i) The present value of a 6 n−year annuity−immediate of 1 at the end of every year is 9.7578. (ii) The present value of a 6 n−year annuity−immediate of 1 at the end of every second year is 4.760. (iii) The present value of a 6 n−year annuity−immediate of 1 at the end of every third year is K. Determine K assuming an annual eﬀective interest rate of i. 152 THE BASICS OF ANNUITY THEORY Problem 15.24 ‡ Which of the following does not represent a deﬁnition of an? (a) νn [ (1+i)n−1 i ] (b) 1−νn i (c) ν+ ν2 + ··· + νn (d) ν [1−νn 1−ν ] (e) sn (1+i)n−1 Problem 15.25 ‡ To accumulate 8000 at the end of 3 n years, deposits of 98 are made at the end of each of the ﬁrst n years and 196 at the end of each of the next 2 n years. The annual eﬀective rate of interest is i. You are given (1 + i)n = 2.0. Determine i. Problem 15.26 ‡ For 10,000, Kelly purchases an annuity −immediate that pays 400 quarterly for the next 10 years. Calculate the annual nominal interest rate convertible monthly earned by Kelly’s investment. Hint: Use linear interpolation. Problem 15.27 ‡ Susan and Jeﬀ each make deposits of 100 at the end of each year for 40 years. Starting at the end of the 41st year, Susan makes annual withdrawals of X for 15 years and Jeﬀ makes annual withdrawals of Y for 15 years. Both funds have a balance of 0 after the last withdrawal. Susan’s fund earns an annual eﬀective interest rate of 8%. Jeﬀ’s fund earns an annual eﬀective interest rate of 10%. Calculate Y −X. Problem 15.28 A loan of 10,000 is being repaid by 10 semiannual payments, with the ﬁrst payment made one−half year after the loan. The ﬁrst 5 payments are K each, and the ﬁnal 5 are K+ 200 each. What is K if i(2) = 0.06? Problem 15.29 Smith makes deposits of 1,000 on the last day of each month in an account earning interest at rate i(12) = 0.12. The ﬁrst deposit is January 31, 2005 and the ﬁnal deposit is December 31, 2029. The accumulated account is used to make monthly payments of Y starting January 31, 2030 with the ﬁnal one on December 31, 2054. Find Y. Problem 15.30 A loan of $1,000 is to be repaid with annual payments at the end of each year for the next 20 years. For the ﬁrst ﬁve years the payments are k per year; the second 5 years, 2 k per year; the third 5 years, 3k per year; and the fourth 5 years, 4 k per year. Find an expression for k. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 153 Problem 15.31 ‡ Happy and ﬁnancially astute parents decide at the birth of their daughter that they will need to provide 50,000 at each of their daughter’s 18th, 19th, 20th and 21st birthdays to fund her college education. They plan to contribute X at each of their daughter’s 1st through 17th birthdays to fund the four 50,000 withdrawals. If they anticipate earning a constant 5% annual eﬀective rate on their contributions, which of the following equations of value can be used to determine X,assuming compound interest? (A) X(ν+ ν2 + ··· + ν17) = 50,000(ν+ ν2 + ν3 + ν4) (B) X[(1.05)16 + (1.05)15 + ··· + (1.05)] = 50,000(1 + ν+ ν2 + ν3) (C) X[(1.05)17 + (1.05)16 + ··· + (1.05) + 1] = 50,000(1 + ν+ ν2 + ν3) (D) X[(1.05)17 + (1.05)16 + ··· + (1.05)] = 50,000(1 + ν+ ν2 + ν3) (E) X(1 + ν+ ν2 + ··· + ν17) = 50,000(ν18 + ν19 + ν20 + ν21 + ν22) Problem 15.32 For time t> 0, the discount function is deﬁned by [a(t)]−1 = 1 1 + 0.01t. Consider a ﬁve-year annuity with payments of 1 at times t= 1,2,3,4,5. Consider the following: A calculates a5 as the sum of the present value of the individual payments. However, B accumulates the payments according to the accumulation function a(t) = 1 + 0.01t and then multiplies the result by 1 a(5) . By how much do the answers of A and B diﬀer? Remark 15.1 It can be shown that the above two processes do not produce the same answers in general for any pattern of interest other than compound interest. That’s why, it is always recommended to avoid dealing with annuities not involving compound interest, if possible. Problem 15.33 Simplify the sum ∑40 n=15 sn. Problem 15.34 A 20 year annuity pays 100 every other year beginning at the end of the second year, with additional payments of 300 each at the ends of years 3, 9, and 15. The eﬀective annual interest rate is 4%. Calculate the present value of the annuity. 154 THE BASICS OF ANNUITY THEORY Problem 15.35 An annuity pays 1 at the end of each 4 −year period for 40 years. Given a8 i = k, ﬁnd the present value of the annuity. Problem 15.36 Smith is negotiating a price for a new car. He is willing to pay $250 at the end of each month for 60 months using the 4.9% compounded monthly interest rate that he qualiﬁes for. Smith estimates that tax, title, and license for the new car will increase the negotiated price by 10%, and he estimates that he will receive $500 trade −in value for his current car. Calculate the highest negotiated price that Smith is willing to pay for the car. Problem 15.37 An account is credited interest using 6% simple interest rate from the date of each deposit into the account. Annual payments of 100 are deposited into this account. Calculate the accumulated value of the account immediately after the 20 th deposit. Problem 15.38 A homeowner signs a 30 year mortgage that requires payments of $971.27 at the end of each month. The interest rate on the mortgage is 6% compounded monthly. If the purchase price of the house is $180,000 then what percentage down payment was required? Problem 15.39 A 25 −year−old worker begins saving for retirement, making level annual deposits at the end of each year. The savings are invested at an annual eﬀective interest rate of 8%, and are of an amount that is projected to equal 1,000,000 when the worker is 65 (after the 40th deposit is made on that date). After making ﬁve annual deposits, the worker becomes unemployed for a period of time and, as a result, skips the next three annual deposits. Assuming that the account earns an 8% annual eﬀective rate in all 40 years, what amount will the worker need to deposit in each of the remaining 32 years in order to achieve the original goal of a 1,000,000 balance at age 65? Problem 15.40 Paul lends 8000 to Peter. Peter agrees to pay it back in 10 annual installments at 7% with the ﬁrst payment due in one year. After making 4 payments, Peter renegotiates to pay oﬀ the debt with 4 additional annual payments. The new payments are calculated so that Paul will get a 6.5% annual yield over the entire 8−year period. Determine how much money Peter saved by renegotiating. Problem 15.41 Mario deposits 100 into a fund at the end of each 2 year period for 20 years. The fund pays interest at an annual eﬀective rate of i. The total amount of interest earned by the fund during the 19th and 20th years is 250. Calculate the accumulated amount in Marios account at the end of year 20. 15 PRESENT AND ACCUMULATED VALUES OF AN ANNUITY-IMMEDIATE 155 Problem 15.42 Show that 1 1−ν10 = 1 s10 ( s10 + 1 i ) . Problem 15.43 If a4 = 3.2397 and s4 = 4.5731 what is the value of a8 ? Problem 15.44 Smith borrows $5,000 on January 1, 2007. He repays the loan with 20 annual payments, starting January 1, 2008. The payments in even-number year are 2 X each; the payments in odd-number years are X each. If d= 0.08, ﬁnd the total amount of all 20 payments. Problem 15.45 (a) Show that am−n = am −νmsn where 0 <n<m. (b) Show that sm−n = sm −(1 + i)man where 0 <n<m. Problem 15.46 Show that sn ≥n≥an. Thus far, we have only considered annuities where the payments are made at the end of the period, but it is possible that the circumstances may be such that the annuity is to run for a given number of periods and a portion of a period. The purpose of the remaining problems is to deﬁne terms such as an+k and sn+k where n is a positive integer and 0 <k <1. Problem 15.47 Let kbe a positive real number. Find an expression for the error involved in approximating (1+i)k−1 i by the number k. Hint: Use the Taylor series expansion of (1 + i)k around i= 0. Problem 15.48 (a) Find the interest accrued of a principal of 1 at t = n to time t = n+ k, assuming compound interest. (b) Show that i(ν+ ν2 + ν3 + ··· + νn) + [(1 +i)k −1]νn+k + νn+k = 1. (c) Let an+ 1 m denote the present value of an annuity consisting of n payments of 1 at the end of each period and a ﬁnal payment of (1+i) 1 m−1 i at the end of the 1 mth fraction of the ( n+ 1) period. Deﬁne an+ 1 m = 1−νn+ 1 m i . Show that an+ 1 m = an + νn+ 1 m [ (1 + i) 1 m −1 i ] . 156 THE BASICS OF ANNUITY THEORY Thus, an+k is the sum of the present value of an n−period annuity−immediate of 1 per period, plus a ﬁnal payment at time n+ k of (1+i)k−1 i . Problem 15.49 Show that an+k ≈an + kνn+ 1 m. Hint: Use Problems 15.47 and 15.48. Problem 15.50 Compute a5.25 if i= 5% using the following deﬁnitions: (a) The formula established in Problem 15.48. (b) A payment of 0.25 at time 5.25. (c) A payment of 0.25 at time 6. Problem 15.51 (a) Deﬁne sn+k = (1+i)n+k−1 k .Show that sn+k = (1+ i)ksn + (1+i)k−1 i .Thus, sn+k can be interprested as the accumulated value of an n−period annuity−immediate at time t= n+ k with an additional payment of (1+i)k−1 i at time t= n+ k. (b) Show that sn+k ≈(1 + i)ksn + k. 16 ANNUITY IN ADVANCE: ANNUITY DUE 157 16 Annuity in Advance: Annuity Due An annuity−due is an annuity for which the payments are made at the beginning of the payment periods. The cash stream represented by the annuity can be visualized on a time diagram as shown in Figure 16.1 . Figure 16.1 The ﬁrst arrow shows the beginning of the ﬁrst period at which the ﬁrst payment is made under the annuity. The second arrow appears n periods after arrow 1, one period after the last payment is made. Let idenote the interest rate per period. The present value of the annuity at time 0 will be denoted by ¨an. To determine ¨an, we can proceed as in the case of determining an. In this case, we use the time diagram shown in Figure 16.2 Figure 16.2 Considering the equation of value at time t= 0 we can write ¨an = 1 + ν+ ν2 + ··· + νn−1. That ¨an is is equal to the sum of the present values of each of the n payments. We recognize the expression on the right −hand side as a geometric progression. Thus, multiplying both sides by ν to obtain ν¨an = ν+ ν2 + ν3 + ··· + νn−1 + νn. Subtracting this from the previous equation we ﬁnd (1 −ν)¨an = (1 −νn). 158 THE BASICS OF ANNUITY THEORY Hence, ¨an = 1 −νn 1 −ν . Since 1 −ν = d, we have ¨an = 1 −νn d = 1 −(1 + i)−n d = 1 −(1 −d)n d . Example 16.1 Find ¨a8 if the eﬀective rate of discount is 10%. Solution. Since d= 0.10, we have ν = 1 −d= 0.9. Hence, ¨a8 = 1−(0.9)8 0.1 = 5.6953279 Example 16.2 What amount must you invest today at 6% interest rate compounded annually so that you can withdraw $5,000 at the beginning of each year for the next 5 years? Solution. The answer is 5000¨a5 = 5000 ·1−(1.06)−5 0.06(1.06)−1 = $22,325.53 Now, let ¨sn denote the accumulated value of an annuity −due at the end of n periods. To de- termine ¨sn we proceed in a way analogous to determining sn.We consider the time diagram shown in Figure 16.3. Figure 16.3 Writing the equation of value at time t= n we ﬁnd ¨sn =(1 + i) + (1 +i)2 + ··· + (1 +i)n−1 + (1 +i)n =(1 + i)(1 + i)n −1 (1 + i) −1 =(1 + i)n −1 iν = (1 + i)n −1 d 16 ANNUITY IN ADVANCE: ANNUITY DUE 159 Example 16.3 What amount will accumulate if we deposit $5,000 at the beginning of each year for the next 5 years? Assume an interest of 6% compounded annually. Solution. The answer is 5000¨s5 = 5000 · (1.06)5−1 0.06(1.06)−1 = $29,876.59 The following theorem provides a couple of relationships between ¨an and ¨sn Theorem 16.1 (a) ¨sn = (1 + i)n¨an (b) 1 ¨an = 1 ¨sn + d Proof. (a) We have ¨sn = (1 + i)n −1 d = (1 + i)n ·1 −(1 + i)−n d = (1 + i)n¨an (b) 1 ¨sn + d= d (1 + i)n −1 + d(1 + i)n −1 (1 + i)n −1 =d+ d[(1 + i)n −1] (1 + i)n −1 = d(1 + i)n (1 + i)n −1 = d 1 −(1 + i)−n = 1 ¨an The result in (a) says that if the present value at time 0, ¨ an, is accumulated forward to time n, then you will have its future value, ¨sn. Example 16.4 For a given interest rate i, ¨an = 8.3064 and ¨sn = 14.2068. (a) Calculate d. (b) Calculate n. Solution. (a) Using part (b) of the previous theorem we ﬁnd d= 1 8.3064 − 1 14.2068 = 5%. (b) Using part (a) of the previous theorem we ﬁnd n= 1 ln (1 +i) ln (¨sn ¨an ) = 10.463 It is possible to relate the annuity-immediate and the annuity-due as shown in the following theorem. 160 THE BASICS OF ANNUITY THEORY Theorem 16.2 (a) ¨an = (1 + i)an. (b) ¨sn = (1 + i)sn. Proof. The two results can be obtained directly by using the time diagram shown in Figure 16.4 Figure 16.4 An algebraic justiﬁcation is given next. (a) Since d= i i+1 , we have ¨an = 1−(1+i)−n d = (1 + i) ·1−(1+i)−n i = (1 + i)an. (b) Since d= i i+1 , we have: ¨sn = (1+i)n−1 d = (1 + i) ·(1+i)n−1 i = (1 + i)sn An annuity−due starts one period earlier than an annuity −immediate and as a result, earns one more period of interest, hence more proﬁt is made. Example 16.5 Over the next 30 years, you deposit money into a retirement account at the beginning of each year. The ﬁrst 10 payments are 200 each. The remaining 20 payments are 300 each. The eﬀective annual rate of interest is 9%. Find the present value of these payments. Solution. The time diagram of this situation is shown in Figure 16.5 Figure 16.5 The answer is 300¨a30 −100¨a10 = 1.09(300a30 −100a10 ) = $2659.96 More relationships between annuity-immediate and annuity-due are given below. 16 ANNUITY IN ADVANCE: ANNUITY DUE 161 Theorem 16.3 (a) ¨an = 1 + an−1 . (b) sn = ¨sn−1 + 1. Proof. (a) We have ¨an = 1−(1+i)−n d = i+1 i [1 −(1 + i)−n] = 1−(1+i)−n+1+i i = 1 + an−1 . The result has the following verbal interpretation: An additional payment of 1 at time 0 results in an−1 becoming n payments that now commence at the beginning of each year whose present value is ¨an. (b) We have ¨sn−1 = (1+i)n−1−1 d = i+1 i [(1 + i)n−1 −1] = (1+i)n−1−i i = sn −1 An interpretation of (b) is as follows: A withdrawal of 1 at time n results in sn becoming n−1 payments that commence at the beginning of each year (starting at t= 1) whose accumulated value at time t= n is ¨sn−1 Remark 16.1 Most compound interest tables do not include values of annuities −due. Thus, the formulas of Theorems 16.2 - 16.3 must be used in ﬁnding numerical values for annuities −due. Example 16.6 An investor wishes to accumulate $3000 at the end of 15 years in a fund which earns 8% eﬀective. To accomplish this, the investor plans to make deposits at the beginning of each year, with the ﬁnal payment to be made one year prior to the end of the investment period. How large should each deposit be? Solution. Let R be the payment at the beginning of each year. The accumulated value of the investment at the end of the investment period is to be $3000, so the equation of value at time t = 15 is 3000 = R¨s14 . Solving for R we ﬁnd R= 3000 ¨s14 = 3000 s15 −1 = $114.71 Example 16.7 Show that ¨an = an + 1 −νn. Interpret the result verbally. Solution. We have ¨an = (1 + i)an = an + ian = an + i·1−νn i = an + 1 −νn. An additional payment of 1 at time 0 results in an becoming n+ 1 payments that now commence at the beginning of the year and whose present value is ¨an + νn 162 THE BASICS OF ANNUITY THEORY Example 16.8 Show that ¨sn = sn −1 + (1 +i)n. Interpret the result verbally. Solution. We have ¨sn = (1 + i)sn = sn + i·(1+i)n−1 i = sn −1 + (1 +i)n. An additional payment of 1 at time n results in ¨sn becoming n+ 1 payments that now commence at the beginning of the year and whose accumulated value at time t= n is sn + (1 +i)n Remark 16.2 The names annuity−immediate and annuity-due are used traditionally, although they do not seem to be logical. The ﬁrst payment of an annuity−immediate is not made immediately at the beginning of the ﬁrst payment period, it is due at the end of it. 16 ANNUITY IN ADVANCE: ANNUITY DUE 163 Practice Problems Problem 16.1 An 8-year annuity due has a present value of $1,000. If the eﬀective annual interest rate is 5%, then what is the value of the periodic payment? Problem 16.2 An 8-year annuity due has a future value of $1,000. Find the periodic payment of this annuity if the eﬀective annual interest rate is 5%. Problem 16.3 A 5-year annuity due has periodic cash ﬂows of $100 each year. Find the accumulated value of this annuity if the eﬀective annual interest rate is 8%. Problem 16.4 A 5-year annuity due has periodic cash ﬂows of $100 each year. Find the present value of this annuity if the eﬀective annual interest rate is 8%. Problem 16.5 Calculate the accumulated value immediately after the last payment of a 20 year annuity due of annual payments of 500 per year. The annual eﬀective interest rate is 7%. Problem 16.6 Megan wants to buy a car in 4 years for 18,000. She deposits X at the beginning of each month for four years into an account earning 6% compounded monthly. Calculate X. Problem 16.7 Kathy wants to accumulate a sum of money at the end of 10 years to buy a house. In order to accomplish this goal, she can deposit 80 per month at the beginning of the month for the next ten years or 81 per month at the end of the month for the next ten years. Calculate the annual eﬀective rate of interest earned by Kathy. Problem 16.8 An annuity pays $500 every six months for 5 years. The annual rate of interest is 8% convertible semiannually. Find each of the following: (a) The PV of the annuity six months (one period) before the ﬁrst payment, (b) the PV of the annuity on the day of the ﬁrst payment, (c) the FV of the annuity on the day of the last payment, (d) and the FV of the annuity six months after the last payment. 164 THE BASICS OF ANNUITY THEORY Problem 16.9 You will retire in 30 years. At the beginning of each month until you retire, you will invest X earning interest at 9% convertible monthly. Starting at year 30, you will withdraw $4,000 at the beginning of each month for the next 15 years. Also, starting at year 30, your fund will only earn interest at 6% convertible monthly. Find X such that your account will be empty after the last withdrawal. Problem 16.10 ‡ Chuck needs to purchase an item in 10 years. The item costs 200 today, but its price increases by 4% per year. To ﬁnance the purchase, Chuck deposits 20 into an account at the beginning of each year for 6 years. He deposits an additional X at the beginning of years 4, 5, and 6 to meet his goal. The annual eﬀective interest rate is 10%. Calculate X. Problem 16.11 Which of the following are true: (i) ¨s9 + 1 = s10 (ii) a10 −a3 = v2a7 (iii) a5 (1 + i) −1 = a4 ? Problem 16.12 Find the present value of payments of $200 every six months starting immediately and continuing through four years from the present, and $100 every six months thereafter through ten years from the present, if i(2) = 0.06. Problem 16.13 A worker aged 40 wishes to accumulate a fund for retirement by depositing $1,000 at the beginning of each year for 25 years. Starting at age 65 the worker plans to make 15 annual withdrawals at the beginning of each year. Assuming all payments are certain to be made, ﬁnd the amount of each withdrawal starting at age 65 to the nearest dollar, if the eﬀective rate of interest is 8% during the ﬁrst 25 years but only 7% thereafter. Problem 16.14 Which one is greater sn −an or ¨sn −¨an? Problem 16.15 A (2n−1)−payment annuity-immediate has payments 1,2,··· ,n −1,n,n −1,··· ,2,1.Show that the present value of the annuity one payment period before the ﬁrst payment is an ·¨an. 16 ANNUITY IN ADVANCE: ANNUITY DUE 165 Problem 16.16 Jeﬀ makes payments at the end of each year into an account for 10 years. The present value of Jeﬀ’s payments is 5,000. Ryan makes a payment equal to Jeﬀ’s payment at the beginning of each year for 11 years into the same account. The present value of Ryan’s payments is 5,900. Calculate the amount of Jeﬀ’s payment. Problem 16.17 A person deposits 100 at the beginning of each year for 20 years. Simple interest at an annual rate of i is credited to each deposit from the date of deposit to the end of the twenty year period. The total amount thus accumulated is 2,840. If instead, compound interest had been credited at an eﬀective annual rate of i,what would the accumulated value of these deposits have been at the end of twenty years? Problem 16.18 You plan to accumulate 100,000 at the end of 42 years by making the following deposits: X at the beginning of years 1-14 No deposits at the beginning of years 15-32; and Y at the beginning of years 33-42. The annual eﬀective interest rate is 7%. Suppose X−Y = 100. Calculate Y. Problem 16.19 ‡ Jim began saving money for his retirement by making monthly deposits of 200 into a fund earning 6% interest compounded monthly. The ﬁrst deposit occurred on January 1, 1985. Jim became unemployed and missed making deposits 60 through 72. He then continued making monthly deposits of 200. How much did Jim accumulate in his fund on December 31, 1999 ? Problem 16.20 ‡ An investor accumulates a fund by making payments at the beginning of each month for 6 years. Her monthly payment is 50 for the ﬁrst 2 years, 100 for the next 2 years, and 150 for the last 2 years. At the end of the 7th year the fund is worth 10,000. The annual eﬀective interest rate is i, and the monthly eﬀective interest rate is j. Which of the following formulas represents the equation of value for this fund accumulation? (a) ¨s24 i(1 + i)[(1 + i)4 + 2(1 +i)2 + 3] = 200 (b) ¨s24 i(1 + j)[(1 + j)4 + 2(1 +j)2 + 3] = 200 (c) ¨s24 j(1 + i)[(1 + i)4 + 2(1 +i)2 + 3] = 200 (d) s24 j(1 + i)[(1 + i)4 + 2(1 +i)2 + 3] = 200 (e) s24 i(1 + j)[(1 + j)4 + 2(1 +j)2 + 3] = 200 166 THE BASICS OF ANNUITY THEORY Problem 16.21 ‡ Carol and John shared equally in an inheritance. Using his inheritance, John immediately bought a 10-year annuity-due with an annual payment of 2,500 each. Carol put her inheritance in an investment fund earning an annual eﬀective interest rate of 9%. Two years later, Carol bought a 15-year annuity-immediate with annual payment of Z. The present value of both annuities was determined using an annual eﬀective interest rate of 8%. Calculate Z. Problem 16.22 ‡ Jerry will make deposits of 450 at the end of each quarter for 10 years. At the end of 15 years, Jerry will use the fund to make annual payments of Y at the beginning of each year for 4 years, after which the fund is exhausted. The annual eﬀective rate of interest is 7% . Determine Y. Problem 16.23 If a(t) = 1 log2 (t+2)−log2 (t+1) , ﬁnd an expression for ¨ an by directly taking the present value of the payments. Problem 16.24 The accumulated value of an nyear annuity-due is four times the present value of the same annuity. Calculate the accumulated value of 100 in 2 n years. Problem 16.25 Show that sn ·¨an >n2 for i> 0 and n> 1. Problem 16.26 ‡ A man turns 40 today and wishes to provide supplemental retirement income of 3000 at the begin- ning of each month starting on his 65th birthday. Starting today, he makes monthly contributions of X to a fund for 25 years. The fund earns a nominal rate of 8% compounded monthly. Each 1000 will provide for 9.65 of income at the beginning of each month starting on his 65th birthday until the end of his life. Calculate X. Problem 16.27 An annuity pays 3 at the beginning of each 3 year period for 30 years. Find the accumulated value of the annuity just after the ﬁnal payment, using i(2) = 0.06. Problem 16.28 A worker aged 30 wishes to accumulate a fund for retirement by depositing $3,000 at the beginning 16 ANNUITY IN ADVANCE: ANNUITY DUE 167 of each year for 35 years. Starting at age 65 the worker plans to make 20 equal annual withdrawals at the beginning of each year. Assuming that all payments are certain to be made, ﬁnd the amount of each withdrawal starting at age 65, if the annual eﬀective rate of interest is 9% during the ﬁrst 35 years but only 6% thereafter. Problem 16.29 Irene deposits 100 at the beginning of each year for 20 years into an account in which each deposit earns simple interest at a rate of 10% from the time of the deposit. Other than these deposits Irene makes no other deposits or withdrawals from the account until exactly 25 years after the ﬁrst deposit was made, at which time she withdraws the full amount in the account. Determine the amount of Irene’s withdrawal. Problem 16.30 Show that ¨s2n ¨sn + ¨sn ¨s2n −¨s3n ¨s2n = 1. Problem 16.31 If ¨ap = x and sq = y, show that ap+q = νx+y 1+iy. 168 THE BASICS OF ANNUITY THEORY 17 Annuity Values on Any Date: Deferred Annuity Evaluating annuities thus far has always been done at the beginning of the term( either on the date of, or one period before the ﬁrst payment) or at the end of the term (either on the date of, or one period after the last payment). In this section, we shall now consider evaluating the (1) present value of an annuity more than one period before the ﬁrst payment date, (2) accumulated value of an annuity more than one period after the last payment date, (3) current value of an annuity between the ﬁrst and last payment dates. We will assume that the evaluation date is always an integral number of periods from each payment date. (1) Present values more than one period before the ﬁrst payment date Consider the question of ﬁnding the present value of an annuity −immediate with periodic interest rate iand m+ 1 periods before the ﬁrst payment date. Figure 17.1 shows the time diagram for this case where “?” indicates the present value to be found. Figure 17.1 The present value of an n−period annuity−immediate m+ 1 periods before the ﬁrst payment date (called a deferred annuity since payments do not begin until some later period) is the present value at time m discounted for m time periods, that is, vman. It is possible to express this answer strictly in terms of annuity values. Indeed, am+n −am = 1 −νm+n i −1 −νm i = νm −νm+n i = νm1 −νn i = νman. Such an expression is convenient for calculation, if interest tables are being used. Example 17.1 Exactly 3 years from now is the ﬁrst of four $200 yearly payments for an annuity-immediate, with an eﬀective 8% rate of interest. Find the present value of the annuity. Solution. The answer is 200ν2a4 = 200(a6 −a2 ) = 200(4.6229 −1.7833) = $567.92 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 169 The deferred −annuity introduced above uses annuity −immediate. It is possible to work with a deferred annuity−due. In this case, one can easily see that the present value is given by νm¨an = ¨am+n −¨am. Example 17.2 Calculate the present value of an annuity −due paying annual payments of 1200 for 12 years with the ﬁrst payment two years from now. The annual eﬀective interest rate is 6%. Solution. The answer is 1200(1.06)−2¨a12 = 1200(¨a14 −¨a2 ) = 1200(9.8527 −1.9434) ≈9,491.16 (2) Accumulated values more than 1 period after the last payment date Consider the question of ﬁnding the accumulated value of an annuity −immediate with periodic interest rate i and m periods after the last payment date. Figure 17.2 shows the time diagram for this case where “?” indicates the sought accumulated value. Figure 17.2 The accumulated value of an n−period annuity−immediate m periods after the last payment date is the accumulated value at time naccumulated for mtime periods, that is, (1 + i)msn.Notice that sm+n −sm =(1 + i)m+n −1 i −(1 + i)m −1 i =(1 + i)m+n −(1 + i)m i = (1 + i)m(1 + i)n −1 i = (1 + i)msn Example 17.3 For four years, an annuity pays $200 at the end of each year with an eﬀective 8% rate of interest. Find the accumulated value of the annuity 3 years after the last payment. Solution. The answer is 200(1 + 0.08)3s4 = 200(s7 −s3 ) = 200(8.9228 −3.2464) = $1135.28 It is also possible to work with annuities −due instead of annuities −immediate. The reader should verify that (1 + i)m¨sn = ¨sm+n −¨sm. 170 THE BASICS OF ANNUITY THEORY Example 17.4 A monthly annuity−due pays 100 per month for 12 months. Calculate the accumulated value 24 months after the ﬁrst payment using a nominal rate of 4% compounded monthly. Solution. The answer is 100 ( 1 + 0.04 12 )12 ¨s12 0.04 12 = 1,276.28 (3) Current value between the ﬁrst and last payment date Next, we consider the question of ﬁnding the present value of an n−period annuity −immediate after the payment at the end of mth period where 1 ≤m≤n.Figure 17.3 shows the time diagram for this case. Figure 17.3 The current value of an n−period annuity−immediate immediately upon the mth payment date is the present value at time 0 accumulated for m time periods which is equal to the accumulated value at time n discounted for n−m time periods, that is, (1 + i)man = νn−msn. One has the following formula, (1 + i)man = νn−msn = sm + an−m. To see this, (1 + i)man =(1 + i)m ·1 −(1 + i)−n i =(1 + i)m −(1 + i)m−n i =(1 + i)m −1 i + 1 −(1 + i)m−n i =sm + an−m. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 171 Example 17.5 For four years, an annuity pays $200 at the end of each half −year with an 8% rate of interest convertible semiannually. Find the current value of the annuity immediately upon the 5th payment (i.e., middle of year 3). Solution. The answer is 200(1.04)5a8 0.04 = 200(s5 0.04 + a3 0.04) = 200(5.4163 + 2.7751) = $1,638.28 For annuity−due we have a similar formula for the current value (1 + i)m¨an = νn−m¨sn = ¨sm + ¨an−m. Example 17.6 Calculate the current value at the end of 2 years of a 10 year annuity due of $100 per year using a discount rate of 6%. Solution. We have (1 + i)−1 = 1 −d= 1 −0.06 = 0.94 and i= 0.06 0.94 . Thus, 100(.94)−2¨a10 = $870.27 Up to this point, we have assumed that the date is an integral number of periods. In the case the date is not an integral number of periods from each payment date, the value of an annuity is found by ﬁnding the value on a date which is an integral number of periods from each payment date and then the value on this date is either accumulated or discounted for the fractional period to the actual evaluation date. We illustrate this situation in the next example. Example 17.7 An annuity−immediate pays $1000 every six months for three years. Calculate the present value of this annuity two months before the ﬁrst payment using a nominal interest rate of 12% compounded semiannually. Solution. The present value at time t= 0 is 1000a6 0.06 = 10001 −(1.06)−6 0.06 = $4917.32. Let j be the interest rate per 2-month. Then 1 + j = (1 + 0.06) 1 3 . The present value two months before the ﬁrst payment is made is 4917.32(1.06) 2 3 = $5112.10 172 THE BASICS OF ANNUITY THEORY Practice Problems Problem 17.1 For four years, an annuity pays $200 at the end of each half-year with an 8% rate of interest convertible semiannually. Find the current value of the annuity three months after the 5th payment (i.e., nine months into year 3). Problem 17.2 Which of the following is equal to the current value shown on the time diagram? (i) s3 + a4 (ii) ¨s3 + a4 (iii) ¨a6 (1 + i)3 (iv) v4s6 (v) ¨s2 + (1 +i)a4 Problem 17.3 Calculate the present value of an annuity immediate with 20 annual payments of 500 if the ﬁrst payment of the annuity immediate starts at the end of the ﬁfth year. The annual eﬀective interest rate is 8%. Problem 17.4 Calculate the current value at the end of 5 years of an annuity due paying annual payments of 1200 for 12 years. The annual eﬀective interest rate is 6%. Problem 17.5 A monthly annuity due pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. Problem 17.6 A monthly annuity immediate pays 100 per month for 12 months. Calculate the accumulated value 12 months after the last payment using a nominal rate of 4% compounded monthly. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 173 Problem 17.7 Annuities X and Y provide the following payments: End of Year Annuity X Annuity Y 1 - 10 1 K 11 - 20 2 0 21 - 30 1 K Annuities X and Y have equal present values and an annual eﬀective interest rate of i such that v10 = 0.5. Determine K. Problem 17.8 Payments of $100 per quarter are made from June 7, Z through December 7, Z+ 11, inclusive. If the nominal interest convertible quarterly is 6%: (a) Find the present value on September 7, Z−1. (b) Find the current value on March 7, Z+ 8. (c) Find the accumulated value on June 7, Z+ 12. Problem 17.9 Find the current value to the nearest dollar on January 1 of an annuity which pays $2,000 every six months for ﬁve years. The ﬁrst payment is due on the next April 1 and the rate of interest is 9% convertible semiannually. Problem 17.10 John buys a series of payments. The ﬁrst payment of 50 is in six years. Annual payments of 50 are made thereafter until 14 total payments have been made. Calculate the price John should pay now to realize an annual eﬀective return of 7%. Problem 17.11 Which of the following are true? (i) ¨a10 −¨a3 = a9 −a2 (ii) ν3¨a3 = ν2a3 (iii) ν8s12 = ¨a3 + ¨s9 Problem 17.12 A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the ﬁfth year and to continue thereafter for as long as necessary. The eﬀective rate of discount is 5%. Find the amount of the ﬁnal payment if it is to be larger than the regular payments. 174 THE BASICS OF ANNUITY THEORY Problem 17.13 Using an annual eﬀective interest rate i> 0, you are given: (i) The present value of 2 at the end of each year for 2 n years, plus an additional 1 at the end of each of the ﬁrst n years, is 36. (ii) The present value of an n−year deferred annuity-immediate paying 2 per year for n years is 6. Calculate i. Problem 17.14 Show that ∑15 t=10(¨st −st) = s16 −s10 −6. Problem 17.15 It is known that a7 a11 = a3 +sx ay+sz . Find x,y, and z. Problem 17.16 Simplify a15 (1 + ν15 + ν30) to one symbol. Problem 17.17 The present value of an annuity-immediate which pays $200 every 6 months during the next 10 years and $100 every 6 months during the following 10 years is $4,000. The present value of a 10-year deferred annuity-immediate which pays $250 every 6 months for 10 years is $2,500. Find the present value of an annuity-immediate which pays $200 every 6 months during the next 10 years and $300 every 6 months during the following 10 years. Problem 17.18 ‡ At an annual eﬀective interest rate of i,i> 0, both of the following annuities have a present value of X : (i) a 20-year annuity-immediate with annual payments of 55; (ii) a 30-year annuity-immediate with annual payments that pays 30 per year for the ﬁrst 10 years, 60 per year for the second 10 years, and 90 per year for the ﬁnal 10 years. Calculate X. Problem 17.19 Tom borrows 100 at an annual eﬀective interest rate of 4% and agrees to repay it with 30 annual installments. The amount of each payment in the last 20 years is set at twice that in the ﬁrst 10 years. At the end of 10 years, Tom has the option to repay the entire loan with a ﬁnal payment X, in addition to the regular payment. This will yield the lender an annual eﬀective rate of 4.5% over the 10 year period. Calculate X. 17 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 175 Problem 17.20 You are given an annuity-immediate with 11 annual payments of 100 and a ﬁnal larger payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all payments is 1000. Using an annual eﬀective interest rate of 1%, calculate the present value at the beginning of the ninth year of all remaining payments. Round your answer to the nearest integer. Problem 17.21 The proceeds of a 10,000 death beneﬁt are left on deposit with an insurance company for seven years at an eﬀective annual interest rate of 5%. The balance at the end of seven years is paid to the beneﬁciary in 120 equal monthly payments of X, with the ﬁrst payment made immediately. During the payout period, interest is credited at an annual eﬀective rate of 3%. Problem 17.22 The present value today of a 20 year annuity-immediate paying 500 every 6 months but with the ﬁrst payment deferred tyears has a present value of 5,805.74. If i(12) = .09 is the interest rate used to calculate the present value, ﬁnd t. Problem 17.23 John buys an enterprise that sells dental equipment in the amount of 60 million dollars. He decides to ﬁnance the purchase by making 20 semiannual payments with the ﬁrst payment in two years. Find the value of each payment if the nominal interest rate is 12% compounded semiannually. Problem 17.24 Peter wants to accumulate $20,000 in ﬁve years by making monthly payment at the end of each month for the ﬁrst three years into a savings account that pays annual interest rate of 9% com- pounded monthly and then leave the accumulated amount in the account for the remaining two years. What is the value of his regular monthly payment? 176 THE BASICS OF ANNUITY THEORY 18 Annuities with Inﬁnite Payments: Perpetuities A perpetuity is an annuity whose term is inﬁnite, i.e., an annuity whose payments continue forever with the ﬁrst payment occurs either immediately (perpetuity −due) or one period from now (perpetuity−immediate). Thus, accumulated values of perpetuities do not exist. Let us determine the present value of a perpetuity −immediate at the time one period before the ﬁrst payment, where a payment of 1 is made at the end of each period. The present value will be denoted by a∞. A time diagram of this case is given in Figure 18.1 Figure 18.1 Using the equation of value at time t= 0 we ﬁnd a∞ =ν+ ν2 + ··· =inﬁnite geometric progression with ν <1 = ν 1 −ν = ν iν = 1 i. The verbal interpretation of this formula is as follows: If the periodic eﬀective rate of interest is i then one can invest a principal of 1 i for one period and obtain a balance of 1 + 1 i at the end of the ﬁrst period $1. A payment of $1 is made and the remaining balance of 1 i is reinvested for the next period. This process continues forever. Now, since an = 1−νn i and limn→∞νn = 0 for 0 <ν <1 we have a∞ = lim n→∞ an = 1 i. Example 18.1 Suppose a company issues a stock that pays a dividend at the end of each year of $10 indeﬁnitely, and the the companies cost of capital is 6%. What is the the value of the stock at the beginning of the year? Solution. The answer is 10 ·a∞ = 10 · 1 0.06 = $166.67 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 177 Analogously to perpetuity−immediate, we may deﬁne a perpetuity −due to be an inﬁnite sequence of equal payments where each payment is made at the beginning of the period. Let ¨ a∞ denote the present value of a perpetuity−due at the time of ﬁrst payment is made. A time diagram describing this case is given in Figure 18.2. Figure 18.2 The equation of value at time t= 0 is ¨a∞ = 1 + ν+ ν2 + ··· = 1 1 −ν = 1 d = lim n→∞ ¨an. This formula can be obtained from ﬁnding the present value of a payment of $1 at the beginning of the ﬁrst period and an annutiy −immediate. Indeed, 1 + a∞ = 1 + 1 i = 1 + i i = 1 d = ¨a∞. Example 18.2 What would you be willing to pay for an inﬁnite stream of $37 annual payments (cash inﬂows) beginning now if the interest rate is 8% per annum? Solution. The answer is 37¨a∞ = 37 0.08(1.08)−1 = $499.50 Remark 18.1 We pointed out at the beginning of the section that accumulated values for perpetuities do not exist since payments continue forever. We can argue mathematically as follows: If a perpetuity−immediate has an accumulated value, denoted by s∞, then we expect to have s∞ = limn→∞sn. But lim n→∞ sn = lim n→∞ (1 + i)n −1 i does not exist since 1 + i> 1 and limn→∞(1 + i)n = ∞. 178 THE BASICS OF ANNUITY THEORY Perpetuities are useful in providing verbal explanations of identities. For example, the formula an = 1 −νn i = 1 i −νn i = a∞ −νna∞. can be interpreted as the diﬀerence between payments for two perpetuities each paying 1 at the end of each period; the ﬁrst payment of the ﬁrst perpetuity is one period from now and the ﬁrst payment of the second perpetuity is n+ 1 periods from now. The present value of the ﬁrst perpetuity is 1 i and that of the second perpetuity is νn i . An identical relationship holds for perpetuity due. That is ¨an = 1 −νn d = 1 d −νn d = ¨a∞ −νn¨a∞. Example 18.3 You can receive one of the following two sets of cash ﬂows. Under Option A,you will receive $5,000 at the end of each of the next 10 years. Under Option B, you will receive X at the beginning of each year, forever. The annual eﬀective rate of interest is 10%. Find the value of X such that you are indiﬀerent between these two options. Solution. The equation of value at time t= 0 is 5000a10 =X d = X (1 i + 1 ) 30722.84 =11X X =$2,792.99 Similar to deferred annuities, one can discuss deferred perpetuities. The present value P0 of a deferred perpetuity −immediate with periodic payment of 1 that starts in n periods time, with a ﬁrst cash ﬂow at the beginning of period n+ 1, is given by the equation of value at time t= n (1 + i)nP0 = a∞. Example 18.4 Fifty dollars is paid at the end of each year forever starting six years from now. Assume the annual eﬀective rate of interest is 10%, ﬁnd the present value of the investment. Solution. The deferred period is t= 5. The answer is 50(1 + i)−5a∞ = 50(1.1)−5 · 1 0.1 = $310.46 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 179 Example 18.5 A deferred perpetuity-immediate paying $2000 monthly is bought for a sum of $229,433.67. Find the deferred period if the interest rate compounded monthly is 6%. Solution. The monthly interest rate is 0.06 12 = 0 .005. Let n be the deferred period, i.e. the time it takes $229,433.67 to accumulate to 2000 a∞. Thus, n satisﬁes the equation 2000a∞ = 229,433.67(1.005)n ⇒2000 · 1 0.005 = 229,433.67(1.005)n. Solving this equation for n we ﬁnd n= 111.45 months 180 THE BASICS OF ANNUITY THEORY Practice Problems Problem 18.1 What would you be willing to pay for an inﬁnite stream of $37 annual payments (cash inﬂows) beginning one year from today if the interest rate is 8%? Problem 18.2 Which of the following are equal to a3 . (i) ν+ ν2 + ν3 (ii) ¨a4 −1 (iii) a∞(1 −ν3) Problem 18.3 An annuity−due pays 100 at the beginning of each year forever. The eﬀective annual rate of interest is 25%. What is the present value of the annuity? Problem 18.4 A special perpetuity pays $200 at the end of each year for the ﬁrst 10 years and $100 at the end of each year thereafter. If the eﬀective annual rate of interest is 10%, ﬁnd the present value of the perpetuity. Problem 18.5 ‡ A perpetuity −immediate pays X per year. Brian receives the ﬁrst n payments, Colleen receives the next n payments, and Jeﬀ receives the remaining payments. Brian’s share of the present value of the original perpetuity is 40%, and Jeﬀ’s share is K. Calculate K. Problem 18.6 A sum P is used to buy a deferred perpetuity −due of 1 payable annually. The annual eﬀective interest rate is i> 0. Find an expression for the deferred period. Problem 18.7 Deposits of $1,000 are placed in a fund at the beginning of each year for the next 20 years. After 30 years annual payments commence and continue forever, with the ﬁrst payment at the end of the 30th year. Find an expression for the amount of each payment. Problem 18.8 A benefactor leaves an inheritance to four charities, A,B,C, and D.The total inheritance is a series of level payments at the end of each year forever. During the ﬁrst n years A,B, and C share each payment equally. All payments after n years revert to D. If the present values of the shares of A,B,C, and D are all equal, ﬁnd (1 + i)n. 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 181 Problem 18.9 A level perpetuity−immediate is to be shared by A,B,C, and D. Areceives the ﬁrst n payments. B the second n payments, C the third n payments, and D all payments thereafter. It is known that the ratio of the present value of C′s share to A′s share is 0.49. Find the ratio of the present value of B′s share to D′s share. Problem 18.10 Adam buys a perpetuity due of 1,000 per month for 100,000. Calculate the annual eﬀective rate of interest used to calculate the price of this perpetuity. Problem 18.11 The present value of a perpetuity immediate where the payment is P is 1,000 less than the present value of a perpetuity due where the payment is P. Calculate P. Problem 18.12 A trust has been established such that RJ will receive a perpetuity of 1000 a year with the ﬁrst payment at the end of 5 years. Calculate the present value of the perpetuity at a discount rate of d= 8%. Problem 18.13 Julie, Chris, and Allen will share an annual perpetuity immediate of 1200. Julie will receive the ﬁrst 9 payments. Chris will receive the next 16 payments. Allen will receive all remaining payments. At an annual eﬀective interest rate of 5%, order the value of each person’s share of the perpetuity. Problem 18.14 John is receiving annual payments from a perpetuity immediate of 12,000. Krista is receiving annual payments from a perpetuity due of 10,000. If the present value of each perpetuity is equal, calculate i. Problem 18.15 The present value of an annual perpetuity immediate of 150 is equal to the present value of an annual perpetuity immediate that pays 100 at the end of the ﬁrst 20 years and 200 at the end of year 21 and each year thereafter. Calculate i. Problem 18.16 ‡ An estate provides a perpetuity with payments of X at the end of each year. Seth, Susan, and Lori share the perpetuity such that Seth receives the payments of X for the ﬁrst n years and Susan receives the payments of X for the next m years, after which Lori receives all the remaining payments of X.Which of the following represents the diﬀerence between the present value of Seth’s 182 THE BASICS OF ANNUITY THEORY and Susan’s payments using a constant rate of interest? (a) X[an −νnam] (b) X[¨an −νn¨am] (c) X[an −νn+1am] (d) X[an −νn−1am] (e) X[van −νn+1am] Problem 18.17 ‡ Which of the following are characteristics of all perpetuities?(including non −level payments) I. The present value is equal to the ﬁrst payment divided by the annual eﬀective interest rate. II. Payments continue forever. III. Each payment is equal to the interest earned on the principal. Problem 18.18 ‡ A perpetuity paying 1 at the beginning of each 6-month period has a present value of 20 . A second perpetuity pays X at the beginning of every 2 years. Assuming the same annual eﬀective interest rate, the two present values are equal. Determine X. Problem 18.19 What is the present value of receiving $30 at the end of each year forever, starting 9 years from now? Assume an annual rate of interest of 7% Problem 18.20 Jim buys a perpetuity of 100 per year, with the ﬁrst payment 1 year from now. The price for the perpetuity is 975.61, based on a nominal yield of i compounded semiannually. Immediately after the second payment is received, the perpetuity is sold for 1642.04, to earn for the buyer a nominal yield of j compounded semiannually. Calculate i−j. Problem 18.21 Victor wants to purchase a perpetuity paying 100 per year with the ﬁrst payment due at the end of year 11. He can purchase it by either (1) paying 90 per year at the end of each year for 10 years, or (2) paying K per year at the end of each year for the ﬁrst 5 years and nothing for the next 5 years. Calculate K. Problem 18.22 A perpetuity pays 1 at the end of every year plus an additional 1 at the end of every second year. The present value of the perpetuity is K for i≥0. Determine K. 18 ANNUITIES WITH INFINITE PAYMENTS: PERPETUITIES 183 Problem 18.23 The accumulated value just after the last payment under a 12-year annuity-immediate of 1000 per year, paying interest at the rate of 5% per annum eﬀective, is to be used to purchase a perpetuity at an interest rate of 6%, with ﬁrst payment to be made 1 year after the last payment under the annuity. Determine the size of the payments under the perpetuity. 184 THE BASICS OF ANNUITY THEORY 19 Solving for the Unknown Number of Payments of an An- nuity In this section we consider the question of ﬁnding the number of payments n given the regular payment R, the interest per period i and either the present value or the accumulated value of an annuity. We will assume an annuity−immediate. A similar calculation applies for annuity −due. Let P be the present value of an annuity −immediate. A time diagram is given in Figure 19.1 Figure 19.1 The equation of value at time t= 0 is P = Rani, where P,R, and i are known quantities and n is the unknown. Solving this equation for n we ﬁnd P = Rani = R (1 −νn i ) ⇒νn = 1 −i (P R ) ⇒n= ln [ 1 −i (P R )] ln ν . This last expression is not necessary a positive integer. Thus, the equation P = Ran is replaced by P = Ran+k where n is a positive integer and 0 <k <1. In this case, using the results of Problems 15.47-15.48 we can write P = Ran+ki = R (1 −νn+k i ) ⇒νn+k = 1 −i (P R ) which implies n+ k= ln [ 1 −i (P R )] ln ν . It follows that, for the annuity to have the present value P, nregular payments of Rmust be made and an additional one payment in the amount of R [(1 + i)k −1 i ] to be made at time t= n+ k, that is, at the fractional period k of the (n+ 1)th period. In practice, the last smaller payment is made either at the same time as the last regular payment making the last payment larger than the regular payment (such a payment is called a balloon 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 185 payment) or at the end of the period following the last regular payment. In this case the smaller payment is called drop payment. The following examples illustrate how to deal with the question of ﬁnding the unknown number of payments. Example 19.1 An investment of $80,000 is to be used to make payments of $5000at the end of every six months for as long as possible. If the fund earns a nominal interest rate of 12% compounded semiannually, ﬁnd how many regular payments can be made and ﬁnd the amount of the smaller payment: (a) to be paid along the last regular payment, (b) to be paid six months after the last regular payment, (c) to be paid during the six months following the last regular payment. Solution. We ﬁrst solve the equation 8000 = 5000 an+k0.06 as follows 80000 = 5000an+k0.06 = 5000 [1 −(1.06)−(n+k) i ] ⇒1 −(1.06)−(n+k) = 80000 5000 (0.06) = 0.96 from which we ﬁnd 1 −0.96 = 0.04 = (1.06)−(n+k) ⇒n+ k= −ln 0.04 ln 1.06 = 55.242. Thus, n= 55 and k= 0.242. (a) Let X be the amount of the smaller payment to be be made at the end of the 55th period. A time diagram of this situation is given in Figure 19.2. Figure 19.2 Note that every period on the time diagram consists of six months. The equation of value at time t= 0 is 80000 = 5000a55 0.06 + X(1.06)−55 ⇒80000 = 79952.715 + (0.0405674)X ⇒X = $1165.59. Thus, in this case, we have 54 payments of $5000 each, and a last payment of 5000 + 1165 .59 = 6165.59 186 THE BASICS OF ANNUITY THEORY (b) This situation is illustrated in Figure 19.3. Figure 19.3 The equation of value at time t= 0 is 80000 = 5000a55 0.06 + Y(1.06)−56 ⇒80000 = 79952.715 + (0.0382712)Y ⇒Y = $1235.53. In this case, we have 55 payments of $5000 each and a last payment of $1235.53 six months after the last regular payment. (c) In this case we have 55 payments of $5000 each and one last payment of 5000 [(1 + 0.06)0.242 −1 0.06 ] = $1182.13 to be made 44 days (0 .242 ×365 2 = 44.165) after the last regular payment A similar type of calculation can be done with known accumulated value instead of present value as illustrated in the next example. Example 19.2 In order to accumulate $5,000, you will deposit $100 at the end of each month for as long as necessary. Interest is 15% compounded monthly. (a) Find the number of regular payments and the fractional period that are required to accumulate the $5,000? (b) If a ﬁnal fractional payment will be added to the last regular payment, what must this fractional payment be? (c) If a ﬁnal fractional payment will be made one month after the last regular payment, what must this fractional payment be? (d) If the fractional payment to be made during the month following the last regular payment. Solution. (a) The monthly interest rate is 0.15 12 = 0.0125. The equation of value at time t= n+ k is 5000 = 100sn+k0.0125 = 100 ·(1 + i)n+k −1 i = 100 ·(1.0125)n+k −1 0.0125 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 187 Solving for n+ k we ﬁnd n+ k= ln ( 1 + 5000(0.0125) 100 ) ln 1.0125 = 39.08 Thus, there will be 39 regular payments and a payment at the fractional time 0 .08 = 0.08 ×365 12 = 2.433, i.e., two days after the last regular payment. (b) Let X be the fractional payment added to the last regular payment. The time diagram of this case is shown in Figure 19.4. Figure 19.4 The equation of value at time t= 39 is 5000 = 100s39 0.0125 + X. Solving for x we ﬁnd X = $13.38. (c) Let Y be such a payment. The time diagram of this case is shown in Figure 19.5. Figure 19.5 The equation of value at time t= 40 is 5000 = 100 s39 (1.0125) + Y. Solving for Y we ﬁnd Y = −48.96. What does this negative answer mean? If 39 regular payments are made at the end of each month, then at time 39 (measured in months) the account will have accumulated 100 s39 0.0125 = 4986.62. This is extremely close to the 5000 that we wanted in the ﬁrst place. If no additional payment is made from time 39 to time 40, then interest alone will make the account have 4986 .62(1.0125) = 5048.96. This means that there will be no additional payment required at time 40. In fact, $48.96 would have to be withdrawn from the account to make the account have only $5000 at time 40. (d) In tis case, 39 regular payments of $100 each to be made and one additional payment in the amount of 100 [(1 + 0.0125)0.08 −1 0.0125 ] = $7.95 to be made 2 days after the last regular payment 188 THE BASICS OF ANNUITY THEORY Example 19.3 An investment of $10,000 is to be used to make payments of $700 at the end of every year for as long as possible. If the eﬀective annual rate if interest is 6%, ﬁnd: (a) the number of regular payments the fund will make, (b) the size of the smaller ﬁnal payment if the payment will be made in addition to the last regular payment, (c) the size of the smaller ﬁnal payment if the payment will be made one year after the last regular payment. (d) the size of the smaller ﬁnal payment if the payment is to be made during the year following the last regular payment. Solution. (a) We have 10000 = 700an or 10000 = 700 ·1−(1+0.06)−n 0.06 . Solving for n we ﬁnd n= ln ( 1 −10000(0.06) 700 ) −ln 1.06 = 33.40 Thus, there will be 33 regular payments. (b) Let x be the size of the payment. The equation of value at t= 0 is 10000 = 700a33 + v33x or 10000(1.06)33 = 700s33 + x. Solving for x we ﬁnd x= 265.68. (c) Let x be the size of the payment. The equation of value at time t= 0 is 10000 = 700a33 + v34x. Solving for x we ﬁnd x= 281.62. (d) The equation of value at time t= 0 is 700a33+k = 10000 or 0.07a33+k = 1 where 0 < k <1. (See Problem 15.29). Thus, we have 0 .071−v33+k i = 1 ⇒v33+k = 1 −6 7 = 1 7 ⇒ 33 + k= ln 7 ln 1.06 ≈33.395 ⇒k= 0.395. In this case, the equation of value at time t= 0 is 700a33 + (1.06)−33.395x= 10000. Solving for x we ﬁnd x = $271.87. That is, the ﬁnal irregular payment of $271.87 is paid at time 33.395 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 189 Practice Problems Problem 19.1 Allison pays $27,506.28 for an annuity due today. This annuity will pay her $1,750 at the beginning of each month for n months with the ﬁrst payment coming today. If i(12) = 12%, ﬁnd n. Round your answer to the nearest integer. Problem 19.2 In order to accumulate $2,000, you will deposit $100 at the end of each month for as long as necessary. Interest is i(12) = 4%. (a) How many regular payments are required to accumulate the $2,000? (b) If a ﬁnal fractional payment will be added to the last regular payment, what must this fractional payment be? Problem 19.3 ‡ The present values of the following three annuities are equal: 1. perpetuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of 7.25%; 2. 50-year annuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of j%; 3. n−year annuity-immediate paying 1 each year, calculated at an annual eﬀective interest rate of (j−1)%. Calculate n. Problem 19.4 Jenna is the beneﬁciary of a fund of 20,000 that pays her 1,000 at the end of each month. The fund earns 6% compounded monthly. The ﬁnal payment to exhaust the fund will be a balloon payment. Calculate the amount of the balloon payment. Problem 19.5 Jordan inherits 50,000. This inheritance is invested in a fund earning an annual rate of interest of 6%. He withdraws 5,000 per year beginning immediately. Once Jordan can no longer withdraw a full 5,000, he will withdraw a ﬁnal payment one year after the prior regular payment. Calculate the ﬁnal payment. Problem 19.6 A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the 5th year, and to continue thereafter for as long as necessary. Find the time and amount of the ﬁnal payment if the ﬁnal payment is to be larger than the regular payments. Assume i= 4.5%. 190 THE BASICS OF ANNUITY THEORY Problem 19.7 A fund of 2,000 is to be accumulated by nannual payments of 50 by the end of each year, followed by n annual payments of 100 by the end of each year, plus a smaller ﬁnal payment made 1 year after the last regular payment. If the eﬀective rate of interest is 4.5%, ﬁnd nand the amount of the ﬁnal irregular payment. Problem 19.8 One annuity pays 4 at the end of each year for 36 years. Another annuity pays 5 at the end of each year for 18 years. The present values of both annuities are equal at eﬀective rate of interest i.If an amount of money invested at the same rate i will double in n years, ﬁnd n. Problem 19.9 A fund earning 8% eﬀective is being accumulated with payments of 500 at the beginning of each year for 20 years. Find the maximum number of withdrawals of 1,000 which can be made at the end of each year under the condition that once withdrawals start they must continue through the end of the 20-year period. Problem 19.10 A borrower has the following two options for repaying a loan: (i) Sixty monthly payments of 100 at the end of each month. (ii) A single payment of 6,000 at the end of K months. Interest is at the nominal annual rate of 12% convertible monthly. The two options have the same present value. Find K to the nearest integer. Problem 19.11 ‡ An annuity pays 1 at the end of each year fornyears. Using an annual eﬀective interest rate of i,the accumulated value of the annuity at time ( n+ 1) is 13.776 . It is also known that (1 + i)n = 2.476. Calculate n. Problem 19.12 Ten annual deposits of 1,000 each are made to account A, starting on January 1, 1986. Annual deposits of 500 each are made to account B indeﬁnitely, also starting on January 1, 1986. Interest on both accounts is i= 0.05, with interest credited on December 31. On what date will the balance in account B be larger than the balance in account A? Assume the only transactions are deposits and interest credited to the accounts on December 31. Problem 19.13 ‡ You are given an annuity-immediate with 11 annual payments of 100 and a ﬁnal payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all payments is 1,000. Calculate the ﬁnal payment. 19 SOLVING FOR THE UNKNOWN NUMBER OF PAYMENTS OF AN ANNUITY 191 Problem 19.14 A loan of 10,000 is being repaid with payment of 300 at the end of each month for as long as necessary plus an additional payment at the time of the last regular payment. What is the amount of the additional payment using an interest rate of 9% compounded monthly. Problem 19.15 A loan of 10,000 is being repaid with payments of 500 starting one month after the loan is made and lasting as long as necessary. A ﬁnal smaller payment is made one month after the last regular payment of 500. What is the amount of the additional smaller payment using an interest rate of 12% compounded monthly? Problem 19.16 You are given an annuity immediate with 11 annual payments of 100 and a ﬁnal balloon payment at the end of 12 years. At an annual eﬀective interest rate of 3.5%, the present value at time 0 of all the payments is 1,000. Using an annual eﬀective interest rate of 1%, calculate the present value at the beginning of the ninth year of all remaining payments. Problem 19.17 On the ﬁrst day of every January, April, July and October Smith deposits 100 in an account earning i(4) = 0.16.He continues the deposits until he accumulates a suﬃcient balance to begin withdrawals of 200 every 3 months, starting three months after the ﬁnal deposit such that he can make twice as many withdrawals as he made deposits. How many deposits are needed? Problem 19.18 On the ﬁrst day of the month, starting January 1, 1995, Smith deposits 100 in an account earning i(12) = 0.09,with interest credited the last day of each month. In addition, Smith deposits 1,000 in the account every December 31. Around what date does the account ﬁrst exceed 100,000? 192 THE BASICS OF ANNUITY THEORY 20 Solving for the Unknown Rate of Interest of an Annuity In this section we consider the problem of ﬁnding the interest rate given the number of payments and either the present value or the accumulated value of the payments. That is, we want to solve an equation like an = k or sn = k for i. We consider three methods in determining an unknown rate of interest. The ﬁrst is to solve for i by algebraic techniques. For example, the expression k= an = ν+ ν2 + ··· + νn is a polynomial in ν of degree n. If the roots of this polynomial can be determined algebraically, then iis immediately determined. That is, i= ν−1 −1.This method is generally practical only for small values of n. Example 20.1 If a2 = 1.75, ﬁnd an exact expression for i> 0. Solution. We have a2 = 1.75 ⇔ν+ ν2 = 1.75 ⇔ν2 + ν−1.75 = 0 ⇔ν = −1 + √ 8 2 . Since (1 + i)−1 = −1+ √ 8 2 , solving this for i we ﬁnd i= 2√ 8−1 −1 ≈0.09384 Example 20.2 Solve s3 i = 3.31 for i. Solution. We have (1 + i)3 −1 i = 3.31. This reduces to i3 + 3i2 −0.31i= 0 and can be factored to i(i+ 3.1)(i−0.1) = 0. i= 0 is clearly an extraneous solution and is not the correct answer. Hence, since i >0 the correct answer is i= 0.1 = 10% 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 193 Example 20.3 The Honors Society decides to set up a scholarship for university students. They deposit $1,000 in an account for 11 years with the ﬁrst payment a year from now, and then transfer the accumulated value to a perpetuity-immediate paying $500 each year to a deserving student. Assuming the account and the perpetuity are at the same eﬀective annual interest rate i, ﬁnd i. Solution. The accumulated value in the account after the last deposit is 100 s11 . The present value of the perpetuity at that time is 500 a∞; the accumulated value and the value of the perpetuity must be equal. 1000s11 =500a∞ 2 ·(1 + i)11 −1 i =1 i (1 + i)11 =1.5 i=1.5 1 11 −1 ≈3.75% The second method for determining the unknown rate is to use linear interpolation (see Section 13). Example 20.4 An annuity−immediate pays you $100 in each of the next eight years. Its present value is $680. Using linear interpolation of the interest table, ﬁnd an estimate of the annual eﬀective rate i. i s8 a8 2.5% 8.7361 7.1701 3% 8.8923 7.0197 3.5% 9.0517 6.8740 4% 9.2142 6.7327 4.5% 9.3800 6.5959 5% 9.5491 6.4632 6% 9.8975 6.2098 7% 10.2598 5.9713 Solution. From the given table we see that a8 0.035 = 6.8740 and a8 0.04 = 6.7327. Using linear interpolation (See Section 13), we ﬁnd i≈0.035 + (0.04 −0.035) · 6.80 −6.8740 6.7327 −6.8740 = 3.762% 194 THE BASICS OF ANNUITY THEORY Example 20.5 At what interest rate, convertible quarterly, is $16,000 the present value of $1,000 paid at the end of every quarter for ﬁve years? Solution. Let j = i(4) 4 so that the equation of value at time t= 0 becomes 1000a20 j = 16,000 or a20 j = 16. From the tables of interest we see that a20 0.02 = 16.3514 and a20 0.0250 = 15.5892.Thus, using linear interpolation we ﬁnd j = 0.02 + (0.0250 −0.0200) · 16 −16.3514 15.5892 −16.3514 = 0.0223 which gives i(4) = 4(0.0223) = 0.0892 = 8.92% i a20 1.0% 18.0456 1.5% 17.1686 2.0% 16.3514 2.5% 15.5892 The third method is to use the Newton-Raphson iteration method. This method is used to approxi- mate the zeros of the equation f(x) = 0 where f is a diﬀerentiable function. The idea of the method is to start with an initial guess x0 and then ﬁnd the equation of the tangent line at x0.We next ﬁnd the x−intercept of this line say x1, which is closer to the real solution than x0. See Figure 20.1. One can easily see that x1 = x0 −f(x0) f′(x0). Next, we ﬁnd the equation of the tangent line at x= x1 and ﬁnd the x−intercept of this line. Say, x2 = x1 −f(x1) f′(x1). The iteration continues so that we generate a sequence x0,x1,··· ,xn,··· with xn+1 = xn −f(xn) f′(xn) 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 195 xn converging to the solution of f(x) = 0. Figure 20.1 In our context, we want to solve an = k for i using the above mentioned method. Equivalently, we want to solve the equation f(i) = 1 −(1 + i)−n i −k= 0. The iteration formula in this case is given by is+1 =is − [ [1 −(1 + is)−n]i−1 s −k n(1 + is)−n−1i−1 s + [1 −(1 + is)−n]i−2 s ] =is [ 1 + 1 −(1 + is)−n −kis 1 −(1 + is)−n−1{1 + is(n+ 1)} ] The next question regarding this third method is the selection of the initial guessi0.From the power series of (1 + i)−n we ﬁnd an =1 i(1 −(1 + i)−n) =1 i ( 1 − [ 1 −ni+ (−n)(−n−1) 2! i2 + ··· ]) =n [ 1 −n+ 1 2! i+ (n+ 1)(n+ 2) 3! i2 −··· ] 196 THE BASICS OF ANNUITY THEORY Thus, 1 an = 1 n [ 1 − (n+ 1 2! i−(n+ 1)(n+ 2) 3! i2 + ··· )]−1 = 1 n [ 1 + (n+ 1 2! i−(n+ 1)(n+ 2) 3! i2 + ··· ) + (n+ 1 2! i−(n+ 1)(n+ 2) 3! i2 + ··· )2 + ··· ] = 1 n [ 1 + n+ 1 2! i+ n2 −1 12 i2 −n(n2 −1) 24 i3 + ··· ] . The rate of convergence of this series is much faster than the one for the series expansion of an. Now, for the initial guess we use the approximation 1 k ≈1 n [ 1 + n+ 1 2 i0 ] . Solving for i0 we ﬁnd i0 = 2(n−k) k(n+ 1). In practice, the iterations will be carried out until is+1 = is to the required degree of accuracy. Example 20.6 Rework Example 20.5 using Newton-Raphson iterations Solution. Our starting value is i0 = 2(n−k) k(n+1) = 2(20−16) 16(20+1) = 0.0238. Finding several iterations we obtain i1 =0.0238 [ 1 + 1 −(1.0238)−20 −16(0.0238) 1 −(1.0238)−21{1 + 0.0238 ×21} ] = 0.022246 i2 =0.022246 [ 1 + 1 −(1.022246)−20 −16(0.022246) 1 −(1.022246)−21{1 + 0.022246 ×21} ] = 0.0222623 i3 =0.0222623 [ 1 + 1 −(1.0222623)−20 −16(0.0222623) 1 −(1.0222623)−21{1 + 0.0222623 ×21} ] = 0.0222623 Thus, a more accurate interest rate to Example 20.5 is i(4) = 4(0.0222623) = 8.9049% Next, consider the problem of ﬁnding i solution to the equation sni = k. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 197 For this, we let f(i) = (1 + i)n −1 i −k. The Newton-Raphson iterations are given by is+1 = is [ 1 + (1 + is)n −1 −kis (1 + is)n−1{1 −is(n−1)}−1 ] and the initial guess is i0 ≈2(k−n) k(n−1). See Problem 20.20 198 THE BASICS OF ANNUITY THEORY Practice Problems Problem 20.1 (a) The accumulated value of a ﬁve-year annuity-immediate with semiannual payments of $2,000 is $25,000. Use linear interpolation of the interest table to ﬁnd an estimate for i. (b) The present value of a 11-year annuity-due with annual payments of $2,000 is $18,000. Use linear interpolation of the interest table to ﬁnd an estimate for i. Problem 20.2 ‡ To accumulate 8,000 at the end of 3 n years, deposits of 98 are made at the end of each of the ﬁrst n years and 196 at the end of each of the next 2 n years. The annual eﬀective rate of interest is i. You are given (1 + i)n = 2.0. Determine i. Problem 20.3 ‡ For 10,000, Kelly purchases an annuity-immediate that pays 400 quarterly for the next 10 years. Calculate the annual nominal interest rate convertible monthly earned by Kelly’s investment. Problem 20.4 ‡ A 10-year loan of $20,000 is to be repaid with payments at the end of each year. It can be repaid under the following two options: (X) Equal annual payments at the annual eﬀective rate of interest of 8%; (Y) Installments of $2,000 each year plus interest on the unpaid balance at an annual eﬀective rate of i. The sum of the payments under option (X) equals the sum of the payments under option (Y). Determine i. Problem 20.5 Given a∞ = X and ¨a∞ = 1.25X. Calculate the interest rate used. Problem 20.6 A perpetuity pays 1 at the beginning of every year. The present value is 10. Calculate the annual eﬀective rate of interest earned by the perpetuity. Problem 20.7 If ¨a2 ·s2 = 4.05, calculate i. Problem 20.8 A deferred perpetuity pays 500 annually beginning at the end of year 5. The present value of the deferred perpetuity is 4,992. Calculate the annual eﬀective interest rate used to calculate the present value. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 199 Problem 20.9 An annuity due which pays 100 per month for 12 years has a present value of 7,908. Calculate the annual eﬀective interest rate used to determine the present value. Problem 20.10 An annuity which pays 200 at the end of each quarter for 5 years has a present value of 3,600. Calculate the nominal rate of interest compounded quarterly. Problem 20.11 An annuity immediate pays 750 per year for 15 years. The accumulated value of the annuity after 15 years is 15,000. Calculate the annual eﬀective rate of interest used to calculate the accumulated value. Problem 20.12 Adam buys a perpetuity due of 1000 per month for 100,000. Calculate the annual eﬀective rate of interest used to calculate the price of this perpetuity. Problem 20.13 A beneﬁciary receives a 10,000 life insurance beneﬁt. If the beneﬁciary uses the proceeds to buy a 10-year annuity-immediate, the annual payout will be 1,538. If a 20-year annuity-immediate is purchased, the annual payout will be 1,072. Both calculations are based on an annual eﬀective interest rate of i. Find i. Problem 20.14 If ¨sn = 11 and sn+1 = 2an+1 , ﬁnd i. Problem 20.15 If sn = 10 and s2n = 24, ﬁnd νn. Problem 20.16 ‡ Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual eﬀective interest rate of i. The accumulated amount in the account at the end of 40 years is X,which is 5 times the accumulated amount in the account at the end of 20 years. Calculate X. Problem 20.17 The following two investment plans result in the same accumulated amount at the end of 10 years. 1. You lend $10,000, to be repaid in 10 equal installments at the end of each year, at the annual rate of interest of 6%. The installments are deposited to a savings account paying 3% eﬀective annual rate of interest. 2. You deposit $10,000 to an account paying the annual eﬀective rate of interest i. Find i. 200 THE BASICS OF ANNUITY THEORY Problem 20.18 ‡ At an annual eﬀective interest rate of i,i> 0%, the present value of a perpetuity paying 10 at the end of each 3-year period, with the ﬁrst payment at the end of year 3, is 32. At the same annual eﬀective rate of i,the present value of a perpetuity paying 1 at the end of each 4-month period, with ﬁrst payment at the end of 4 months, is X. Calculate X. Problem 20.19 A fund of $17,000 is to be accumulated at the end of ﬁve years with payments at the end of each half-year. The ﬁrst ﬁve payments are $1000 each, while the second ﬁve are $2000 each. Find the nominal rate of interest convertible semiannually earned on the fund. Problem 20.20 Consider solving the equation sn = k. (a) Show that sn = n ( 1 + n−1 2! i+ (n−1)(n−2) 3! i2 + ··· ) . (b) Show that [sn]−1 = 1 n ( 1 −n−1 2 i+ ··· ) . (c) Show that the Newton-Raphson iterations are given by is+1 = is [ 1 + (1 + is)n −1 −kis (1 + is)n−1{1 −(n−1)is}−1 ] . (d) Show that an initial guess for Newton-Raphson method is given by i0 ≈2(n−k) k(n−1). Problem 20.21 Use the Newton-Raphson iteration method with three iterations to estimate the rate of interest if s20 i = 25. Problem 20.22 Eric receives 12000 from a life insurance policy. He uses the fund to purchase two diﬀerent annuities, each costing 6000. The ﬁrst annuity is a 24 year annuity immediate paying K per year to himself. The second annuity is an 8 year annuity immediate paying 2 K per year to his son. Both annuities are based on an annual eﬀective interest rate of i> 0. Determine i. Problem 20.23 Jeﬀ deposits 100 at the end of each year for 13 years into Fund X. Antoinette deposits 100 at the end of each year for 13 years into Fund Y. Fund X earns an annual eﬀective rate of 15% for the ﬁrst 5 years and an annual eﬀective rate of 6% thereafter. Fund Y earns an annual eﬀective rate of i. At the end of 13 years, the accumulated value of Fund X equals the accumulated value of Fund Y. Calculate i. 20 SOLVING FOR THE UNKNOWN RATE OF INTEREST OF AN ANNUITY 201 Problem 20.24 Dottie receives payments of X at the end of each year for nyears. The present value of her annuity is 493. Sam receives payments of 3 X at the end of each year for 2 n years. The present value of his annuity is 2748. Both present values are calculated at the same annual eﬀective interest rate. Determine νn. Problem 20.25 ‡ A discount electronics store advertises the following ﬁnancing arrangement: “We don’t oﬀer you confusing interest rates. We’ll just divide your total cost by 10 and you can pay us that amount each month for a year.” The ﬁrst payment is due on the date of sale and the remaining eleven payments at monthly intervals thereafter. Calculate the eﬀective annual interest rate the store’s customers are paying on their loans. 202 THE BASICS OF ANNUITY THEORY 21 Varying Interest of an Annuity In this section we consider situations in which interest can vary each period, but compound interest is still in eﬀect. Let ik denote the rate of interest applicable from time k−1 to time k. We consider ﬁrst the present value of an n−period annuity-immediate. We consider the following two variations. The ﬁrst is when ik is applicable only for period k regardless of when the payment is made. That is, the rate ik is used only in period kfor discounting all payments. In this case the present value is given by an = (1 + i1)−1 + (1 +i1)−1(1 + i2)−1 + ··· + (1 +i1)−1(1 + i2)−1 ···(1 + in)−1. The second variation is when the rate ik is used as the eﬀective rate for each period i≤k when a payment is made at time k. In this case, the present value is an = (1 + i1)−1 + (1 +i2)−2 + ··· + (1 +in)−n. Present value of annuity-due can be obtained from present value of annuity-immediate by using the formula ¨an = 1 + an−1 . We now turn to accumulated values. We will consider an annuity-due. Again we consider two diﬀerent situations. If ik is applicable only for period k regardless of when the payment is made, then the accumulated value is given by ¨sn = (1 + i1)(1 + i2) ···(1 + in) + ··· + (1 +in−1)(1 + in) + (1 +in). For ik applicable for all periods i≥k, the accumulated value is ¨sn = (1 + i1)n + (1 +i2)n−1 + ··· + (1 +in). Accumulated values of annuity-immediate can be obtained from the accumulated values of annuity- due using the formula sn+1 = ¨sn + 1. Example 21.1 Find the accumulated value of a 12-year annuity-immediate of $500 per year, if the eﬀective rate of interest (for all money) is 8% for the ﬁrst 3 years, 6% for the following 5 years, and 4% for the last 4 years. 21 VARYING INTEREST OF AN ANNUITY 203 Solution. The accumulated value of the ﬁrst 3 payments to the end of year 3 is 500s3 0.08 = 500(3.2464) = $1623.20. The accumulated value of the ﬁrst 3 payments to the end of year 8 at 6% and then to the end of year 12 at 4% is 1623.20(1.06)5(1.04)4 = $2541.18. The accumulated value of payments 4, 5, 6, 7, and 8 at 6% to the end of year 8 is 500s5 0.06 = 500(5.6371) = $2818.55. The accumulated value of payments 4, 5, 6, 7, and 8 to the end of year 12 at 4% is 2818.55(1.04)4 = $3297.30. The accumulated value of payments 9, 10, 11, and 12 to the end of year 12 at 4% is 500s4 0.04 = 500(4.2465) = $2123.23. The accumulated value of the 12-year annuity immediate is 2541.18 + 3297.30 + 2123.23 = $7961.71 Example 21.2 How much must a person deposit now into a special account in order to withdraw $1,000 at the end of each year for the next ﬁfteen years, if the eﬀective rate of interest is equal to 7% for the ﬁrst ﬁve years, and equal to 9% for the last ten years? Solution. The answer is PV =1000(a5 0.07 + a10 0.09(1.07)−5) =1000(4.1002 + 4.5757) = $8675.90 Example 21.3 Find s5 , if δt = 0.02t for 0 ≤t≤5. Solution. s5 is equal to the sum of the accumulated value of the individual payments. That is, s5 = e ∫5 1 δtdt + e ∫5 2 δtdt + ··· + 1 Since e ∫5 r δtdt = e ∫5 r 0.02tdt = e.25−0.01r2 we ﬁnd s5 = e0.24 + e0.21 + e0.16 + e0.09 + 1 ≈5.7726 204 THE BASICS OF ANNUITY THEORY Practice Problem Problem 21.1 James deposits 1,000 into an account at the end of each year for the next 6 years. The account earns 5% interest. James also deposits 1,000 at the end of each of years 7 through 10 into another account earning 4%. Calculate the total amount James will have in both accounts at the end of ten years. Problem 21.2 A fund earns 5% during the next six years and 4% during years 7 through 10. James deposits 1,000 into the account now. Calculate his accumulated value after 10 years. Problem 21.3 A fund earns 5% during the next six years and 4% during years 7 through 10. James deposits 1,000 into the account at the end of each year for the next ten years. Calculate his accumulated value after 10 years. Problem 21.4 A perpetuity pays $1,200 at the beginning of each year with the ﬁrst payment being made imme- diately. The trust funding the perpetuity will earn an annual eﬀective interest rate of 10% for the ﬁrst 10 years, 8% for the second 10 years and 5% thereafter. Calculate the amount needed to fund the perpetuity immediately before the ﬁrst payment. Problem 21.5 You are given: (i) X is the current value at time 2 of a 20-year annuity-due of 1 per annum. (ii) The annual eﬀective interest rate for year t is 1 8+t. Find X. Problem 21.6 Fund Apays interest of 4% on all money deposited in the ﬁrst 5 years and 5% on all money deposited thereafter. Fund B pays interest of 4% during the ﬁrst 5 years and 5% thereafter without regard to the date the deposit was made. (a) If Kevin deposits 500 into each fund now, how much will he have after 10 years in each fund? (b) Heather deposits 100 at the start of each year for 10 years into Fund B. Lisa deposits 100 at the end of each year into Fund A.Who will have more after 10 years and how much more will that person have? 21 VARYING INTEREST OF AN ANNUITY 205 Problem 21.7 A loan P is to be repaid by 10 annual payments beginning 6 months from the date of the loan. The ﬁrst payment is to be half as large as the others. For the ﬁrst 4 1 2 years interest is ieﬀective; for the remainder of the term interest is j eﬀective. Find an expression for the ﬁrst payment. Problem 21.8 Find the present value of an annuity-immediate which pays 1 at the end of each half-year for ﬁve years, if the rate of interest is 8% convertible semiannually for the ﬁrst three years and 7% convertible semiannually for the last two years. Problem 21.9 Find the present value of an annuity-immediate which pays 1 at the end of each half-year for ﬁve years, if the payments for the ﬁrst three years are discounted at 8% convertible semiannually and the payments for the last two years are discounted at 7% convertible semiannually. Problem 21.10 Given that δt = 1 20−t, t≥0, ﬁnd s10 . Problem 21.11 What is the cost of an annuity of $300 per year for 15 years where the future interest rate for the ﬁrst 5 years is 4%, the rate for the next 5 years is 5%, and the rate for the last 5 years is 6%. Problem 21.12 A perpetuity pays $200 each year, with the ﬁrst payment scheduled one year from now. Assume that the eﬀective annual interest rate for the next ﬁve years is 8%, and thereafter it is 5%. Find the present value of the perpetuity. 206 THE BASICS OF ANNUITY THEORY 22 Annuities Payable at a Diﬀerent Frequency than Interest is Convertible In this section we address annuities for which payment period and the interest conversion period diﬀer and for which the payments are level amount. For example, an annuity-due with monthly payment of $2,000 and with interest rate say 12% convertible quarterly. The approach we use in this section for solving this type of problem consists of the following two steps: (i) Find the rate of interest convertible at the same frequency as payments are made, which is equivalent to the given rate of interest. In the example above, we need to ﬁnd the interest rate convertible monthly which is equivalent to the rate of 12% convertible quarterly. In this case, we ﬁnd i(12) = 12 [( 1 + 0.12 4 )4 12 −1 ] = 11.9%. (ii) Using this new interest rate, ﬁnd the value of the annuity using the techniques discussed in the previous sections. We illustrate this approach in the following examples. Example 22.1 You want to accumulate $250,000. You intend to do this by making deposits of $2,000 into an investment account at the beginning of each month. The account earns 12% interest, convertible quarterly. How many months will it take to reach your goal? (Round to the nearest month.) Solution. We are given an interest rate of 12% per quarter. Let j be the equivalent rate of interest per month, which is the payment period. We have (1 + j)12 = ( 1 + 0.12 4 )4 . Solving for j we ﬁnd j = (1.03) 1 3 −1 = 0.9901634%.Thus, 250,000 = 2000¨snj →(1+j)n−1 j j+1 = 125 →(1 +j)n = 2.22557 → n= ln 2.22557 ln 1.009901634 = 81.195 ≈81 months Example 22.2 A loan of $3,000 is to be repaid with quarterly installments at the end of each quarter for 5 years. If the rate of interest charged on the loan is 10% convertible semi-annually, ﬁnd the amount of each quarterly payment. Solution. We are given an interest rate of 5% per half-year. Letjbe the equivalent rate of interest per quarter, which is the payment period. We have (1 + j)4 = (1 + 0.05)2 →j = (1.05) 1 2 −1 = 0.024695. If R denote the quarterly payment then the equation of value at time t= 0 is Ra20 j = 3000.Solving for R we ﬁnd R= 3000 15.6342 = $191.89 22 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE207 Example 22.3 The nominal rate of interest, convertible semiannually, is 6%. An annuity-immediate pays $50 each month for ﬁve years. Find the accumulated value of this annuity at the time of the last payment. Solution. We are given an interest rate of 3% per half-year. Let j be the equivalent rate of interest per month, which is the payment period. We have (1 + j)12 = (1 + 0.03)2 →j = (1.03) 1 6 −1 = 0.49386%. The accumulated value is 50s60 j = 50 ·(1+j)60−1 j = $3,481.91 Example 22.4 At what annual eﬀective rate of interest will payments of $100 at the end of every quarter accumulate to $2500 at the end of ﬁve years? Solution. Let j be the eﬀective rate per quarter. We are given that 100 s20 j = 2500 or (1 +j)20 −1 = 25j.Let f(j) = (1 +j)20 −25j−1.By trial and error we ﬁnd f(0.02) = −0.014053 and f(0.024) = 0.006938. Using linear interpolation we ﬁnd j ≈0.02 + 0.014053 × 0.024 −0.02 0.006938 + 0.014053 = 0.02268. Let i be the annual eﬀective rate of interest. Then i= (1 + 0.02268)4 −1 = 9.39%. Alternatively, we can use the Newton-Raphson iterations to estimate j. Letting j0 = 2(k−n) k(n−1) = 2(25−20) 25(20−1) ≈0.021053 be the starting value of the interations, the ﬁrst three iterations of Newton- Raphson method are j1 =0.021053 [ 1 + 1.02105320 −1 −25(0.021053) (1.021053)19(1 −0.021053 ×19) −1 ] = 0.02288 j2 =0.02288 [ 1 + 1.0228820 −1 −25(0.02288) (1.02288)19(1 −0.02288 ×19) −1 ] = 0.02277 j3 =0.02277 [ 1 + 1.0227720 −1 −25(0.02277) (1.02277)19(1 −0.02277 ×19) −1 ] = 0.02285 j4 =0.02285 [ 1 + 1.0228520 −1 −25(0.02285) (1.02285)19(1 −0.02285 ×19) −1 ] = 0.02285 Thus, j = 2.285% and consequently i= (1.02285)4 −1 = 0.0946 = 9.46% 208 THE BASICS OF ANNUITY THEORY Practice Problems Problem 22.1 Calculate the present value of an annuity due that pays 500 per month for 10 years. The annual eﬀective interest rate is 6%. Problem 22.2 Calculate the present value of an annuity immediate of 100 per quarter for 6 years using a nominal interest rate of 9% compounded monthly. Problem 22.3 Calculate the accumulated value of an annuity which pays 1,000 at the beginning of each year for 10 years. Use an interest rate of i(12) = 0.08. Problem 22.4 A perpetuity pays 1,000 at the end of each quarter. Calculate the present value using an annual eﬀective interest rate of 10%. Problem 22.5 A perpetuity pays 100 at the beginning of each quarter. Calculate the present value using a force of interest of 0.06. Problem 22.6 A perpetuity due pays 6,000 at the start of each year. Calculate the present value using i(4) = 0.06. Problem 22.7 Find the accumulated value at the end of four years of an investment fund in which $100 is deposited at the beginning of each quarter for the ﬁrst two years and $200 is deposited at the beginning of each quarter for the second two years, if the fund earns 12% convertible monthly. Problem 22.8 A 20-year annuity-due pays $1,000 each year. If i(4) = 8%, ﬁnd the present value of this annuity three years before the ﬁrst payment. Problem 22.9 Find the accumulated value 18 years after the ﬁrst payment is made of an annuity-immediate on which there are 8 payments of $2,000 each made at two-year intervals. The nominal interest rate is 7% convertible semiannually. Round your answer to the nearest dollar. 22 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE209 Problem 22.10 Find the present value of a ten-year annuity which pays $400 at the beginning of each quarter for the ﬁrst 5 years, increasing to $600 per quarter thereafter. The annual eﬀective rate of interest is 12%. Round to the nearest dollar. Problem 22.11 A sum of $100 is placed into a fund at the beginning of every other year for eight years. If the fund balance at the end of eight years is $520, ﬁnd the rate of simple interest earned by the fund. Problem 22.12 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using an annual eﬀective interest rate of 12%. Problem 22.13 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using a nominal interest rate of 12% compounded monthly. Problem 22.14 An annuity pays 100 at the end of each quarter for 10 years. Calculate the present value of the annuity using a nominal interest rate of 12% compounded six times per year. Problem 22.15 A 30−year annuity pays $1,000 every quarter. Ifi(12) = 6%,what is the present value of this annuity six months prior to the ﬁrst payment? Problem 22.16 You deposit $2,500 into an account at the beginning of every month for 15 years. The interest rate on the account is i(2) = 8%. Find the accumulated value in the account nine months after the last deposit. Problem 22.17 A perpetuity will make annual payments of $3,000, with the ﬁrst payment occurring 5 months from now. The interest rate is 12% convertible monthly. Find the present value of this annuity. Problem 22.18 You want to accumulate $2,000,000 over the next 30 years. You intend to do this by making deposits of X into an investment account at the end of each month, for 30 years. The account earns 12% convertible semi-annually. Find X. 210 THE BASICS OF ANNUITY THEORY 23 Analysis of Annuities Payable Less Frequently than In- terest is Convertible In this section we analyze annuities where the compounding frequency is larger than the payment frequency. We start with the case of annuity −immediate. Let kbe the number of interest conversion periods in one payment period. Consider an annuity−immediate that pays 1 at the end of each payment period. Let i be the rate per conversion period and n the total number of conversion periods for the term of the annuity. We will assume that each payment period contains an integral number of interest conversion periods so that n and k are positive in- tegers and also we assume that n is divisible by k. The total number of annuity payments made is then n k. Let Ldenote the present value of an annuity−immediate which pays 1 at the end of each k interest conversion periods for a total of n interest conversion periods. A time diagram of this sitatuation is shown in Figure 32.1. Figure 23.1 Using the equation of value at time t= 0 we ﬁnd L=νk + ν2k + ·+ ν n k·k =νk(1 + νk + ν2k + ·+ ν(n k−1)k) =νk ×1 −(νk) n k 1 −νk = νk1 −νn 1 −νk = 1 −νn (1 + i)k −1 = 1−νn i (1+i)k−1 i = an sk . The accumulated value of this annuity immediately after the last payment is made is (1 + i)nan sk = sn sk . 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE211 Remark 23.1 The annuity of making a payment of $1 at the end of each k interest conversion periods for a total of n k payments is equivalent to the annuity that consists of a periodic payment of 1 sk for n periods. A time diagram illustrating this annuity is given in Figure 23.2. Figure 23.2 Example 23.1 Find the present value and the accumulated value of an annuity −immediate with payments of $1,500 every 4 years from years 4 to 40. The interest earned is 8% converted annually. Solution. With i= 0.08,the present value is 1500 a40 s4 = 1500 ·11.9246 4.5061 = $3,969.48 and the accumulated value is (1.08)40(3969.48) = $86,235.05 Next, we consider the case of an annuity−due. Let ¨Lbe the present value of an annuity which pays 1 at the beginning of each k interest conversion periods for a total of nconversion interest periods. Then we have the following time diagram. Figure 23.3 Using the equation of value at time t= 0 we ﬁnd 212 THE BASICS OF ANNUITY THEORY ¨L=1 + νk + ν2k + ·+ νn−k =1 + (νk) + (νk)2 + ·+ (νk) n−k k =1 −(νk) n−k k +1 1 −νk = 1 −νn 1 −νk = 1−νn i 1−νk i = an ak . The accumulated value of this annuity k interest conversion periods after the last payment is (1 + i)nan ak = sn ak . Remark 23.2 It is easy to see that the annuity described above is equivalent to the annuity that consists of n payments of 1 ak at the end of each interest conversion period. Example 23.2 An annuity pays 1,000 at time 0, time 3, time 6, etc until 20 payments have been made. Calculate the present value and the accumulated value using an annual eﬀective interest rate of 6%. Solution. With i= 0.06,the present value is 1000 a60 a3 = 1000 ·16.16142771 2.67301194 = 6046.15 and the accumulated value is (1.06)60(6046.15) = 199,448.48 In the case of a perpetuity −immediate, the present value is νk + ν2k + ··· = νk 1 −νk = 1 (1 + i)k −1 = 1 i·sk = lim n→∞ an sk . Example 23.3 The present value of a perpetuity paying 1 at the end of every 3 years is 125 91 . Find i. 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE213 Solution. We have 1 i·s3 = 125 91 .Thus, 1 (1+i)3−1 = 125 91 →(1 +i)3 = (6 5 )3 →1 +i= 6 5 →i= 6 5 −1 = 0.2 = 20% In the case of a perpetuity-due, the present value is 1 + νk + ν2k + ··· = 1 1 −νk =1 i · i 1 −(1 + i)−k = 1 i·ak = lim n→∞ an ak . Example 23.4 A perpetuity paying 1 at the beginning of each year has a present value of 20. If this perpetuity is exchanged for another perpetuity paying Rat the beginning of every 2 years with the same eﬀective annual rate as the ﬁrst perpetuity, ﬁnd R so that the present values of the two perpetuities are equal. Solution. An equation of value now for the ﬁrst perpetuity is 1 d = 20 implying i = 1 19 . Also, given that 20 = R ia2 = R 1−ν2 = R (1−ν)(1+ν) = R (0.05)(1.95) →R= (20)(0.0975) = 1.95 Another problem that comes under the category of annuities payable less frequently than inter- est is convertible is the problem of ﬁnding the value of a series of payments at a given force of interest δ.We illustrate this in the next example. Example 23.5 Find an expression for the present value of an annuity on which payments are 100 per quarter for 5 years, just before the ﬁrst payment is made, if δ= 0.08. Solution. Let j be the rate per quarter equivalent to δ. Then (1 + j)4 = e0.08 or 1 + j = e0.02. Thus, PV = 100¨a20 j = 100 [ 1−(1+j)−20 1−(1+j)−1 ] = 100 [ 1−e−0.40 1−e−0.02 ] In some cases, the number of conversion periods in a payment period is not an integral, i.e. k >1 but k is not an integer. In this case, we handle this problem using the basic principles, i.e. to ﬁnd the present value or the accumulated value we write a sum of present values or accumulated values of the individual payments. An illustration of this type is shown next. 214 THE BASICS OF ANNUITY THEORY Example 23.6 Find an expression for the present value of an annuity on which payments are 1 at the beginning of each 4−month period for 12 years, assuming a rate of interest per 3 −month period i. Solution. Let i be the eﬀective rate of interest per 3-month period and j the eﬀective interest rate per 4−month. Then (1 + j)3 = (1 + i)4 ⇒1 + j = (1 + i) 4 3 . The present value is given by ¨a36 j = 1−(1+j)−36 1−(1+j)−1 = 1−ν48 1−ν 4 3 We conclude this section by pointing out that the approach discussed in this section can be general- ized to ﬁnding annuities payable less frequently than interest is convertible on any date, as discussed in Section 17. Example 23.7 Find the present value of an annuity−immediate in which there are a total of 5 payments of 1, the ﬁrst to be made at the end of seven years, and the remaining four payments at three-year intervals, at an annual eﬀective rate of 6%. Solution. The time diagram for this example is given in Figure 23.4. Figure 23.4 The present value is given by ν4 a15 s3 = (1.06)−4 1 −(1.06)−15 (1.06)3 −1 = $2.416 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE215 Practice Problems Problem 23.1 An annuity pays 1,000 at time 0, time 3, time 6, etc until 20 payments have been made. Calculate the accumulated value immediately after the last payment using an annual eﬀective interest rate of 6%. Problem 23.2 Give an expression of the the present value, 3 years before the ﬁrst payment is made, of an annuity on which there are payments of 200 every 4 months for 12 years: (a) expressed as an annuity-immediate; (b) expressed as an annuity-due. Assume monthly rate of interest. Problem 23.3 Find an expression for the present value of an annuity-due of 600 per annum payable semiannually for 10 years, if d(12) = 0.09. Problem 23.4 An annuity pays 500 at the start of each year for 15 years. Calculate the accumulated value of the annuity after 15 years assuming a constant force of interest of 5%. Problem 23.5 A perpetuity pays 100 at the end of every third year. The ﬁrst payment is in three years. Calculate the present value of the perpetuity using a constant force of interest of 10%. Problem 23.6 Find an expression for the present value of an annuity in which there are a total of rpayments of 1, the ﬁrst to be made at the end of seven years, and the remaining payments at three-year intervals, at an annual eﬀective rate i, expressed as (a) annuity-immediate (b) annuity-due. Problem 23.7 A 30-year annuity-immediate with 2,000 payable at the end of every 6 months and a 30-year annuity- immediate with 10,000 payable at the end of every 6 years are to be replaced by a perpetuity paying R every 3 months. You are given that i(4) = .08. Find R. 216 THE BASICS OF ANNUITY THEORY Problem 23.8 The payments you received from a 20 year annuity −immediate paying 500 every 6 months have been left to accumulate in a fund and are now worth 40,000. If i(12) = 0.06 is the rate earned by your fund, calculate how long it is since the last annuity payment was made. Problem 23.9 A perpetuity of $1,000 payable at the end of every 6 months and a perpetuity of $10,000 payable at the end of every 6 years are to be replaced by a 30 year annuity paying R every 3 months. You are given that i(4) = 0.08. Find R. Problem 23.10 An investment of $1,000 is used to make payments of $100 at the end of each year for as along as possible with a smaller payment to be made at the time of the last regular payment. If interest is 7% convertible semiannually, ﬁnd the number of payments and the amount of the total ﬁnal payment. Problem 23.11 Show that the present value at time 0 of 1 payable at times 7, 11, 15, 19, 23, and 27, where the eﬀective rate per annum is i, is given by a28 −a4 s3 + a1 . Problem 23.12 ‡ Fence posts set in soil last 9 years and cost $Y each while fence posts set in concrete last 15 years and cost $(Y + X). The posts will be needed for 35 years. Find an expression of X such that a fence builder would be indiﬀerent between the two types of posts? Problem 23.13 A perpetuity of $750 payable at the end of every year and a perpetuity of $750 payable at the end of every 20 years are to be replaced by an annuity of R payable at the end of every year for 30 years. If i(2) = 0.04, ﬁnd an expression for R. Problem 23.14 Given that δt = 2 10+t, t≥0, ﬁnd a4 . Problem 23.15 At a nominal rate of interest i, convertible semiannually, the present value of a series of payments of 1 at the end of every 2 years forever, is 5.89. Calculate i. 23 ANALYSIS OF ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CONVERTIBLE217 Problem 23.16 Gus deposits 25 into a bank account at the beginning of each 3-year period for 18 years (i.e. there is no deposit in year 18). The account credits interest at an annual eﬀective rate of interest of i. The accumulated amount in the account after 18 years is X, which is four times as large as the accumulated amount in the account after 9 years (excluding the deposit made at time t = 9). Calculate X. Problem 23.17 Find the present value of an annuity−due in which there are a total of 5 payments of 1, the ﬁrst to be made at the end of seven years, and the remaining four payments at three-year intervals, at an annual eﬀective rate of 6%. Problem 23.18 The present value today of a 20 year annuity −immediate paying 500 every 6 months but with the ﬁrst payment deferred tyears has a present value of 5,805.74. If i(12) = 0.09 is the interest rate used to calculate the present value, ﬁnd t. Problem 23.19 A 20−year annuity−due pays $1,000 each year. If i(4) = 8%, ﬁnd the present value of this annuity three years before the ﬁrst payment. 218 THE BASICS OF ANNUITY THEORY 24 Analysis of Annuities Payable More Frequently than In- terest is Convertible In a discussion parallel to the one in section 23, we discuss annuities payable more frequently than interest is convertible. We ﬁrst consider the case of an annuity-immediate. Let m denote the number of payments per one interest conversion period. Let n be the total number of conversion periods in the term of the annuity. Then the total number of payments for the term of the annuity is mn.Let ibe the interest rate per conversion period. We will assume that the number of payments per conversion period is an integral number. Payments of 1 are being made per interest conversion period with 1 m being made at the end of each mth of an interest conversion period. The present value of such an annuity will be denoted by a(m) n . We have the following time diagram. Figure 24.1 The formula for the present value is derived as follows a(m) n = 1 m [ ν 1 m + ν 2 m + ··· + νn−1 m + νn ] = ν 1 m m [ 1 −(ν 1 m)mn 1 −ν 1 m ] = 1 −νn m [ (1 + i) 1 m −1 ]= 1 −νn i(m) . The accumulated value of this annuity immediately after the last payment is made is given by s(m) n = (1 + i)na(m) n = (1 + i)n −1 i(m) . Formulas of a(m) n and s(m) n in terms of an and sn are a(m) n = i i(m) an = s(m) 1 an and s(m) n = i i(m) sn = s(m) 1 sn. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE219 One consideration which is important in practice involves the proper coeﬃcients for annuity payable mthly. Each payment made is of amount 1 m,while the coeﬃcient of the symbol a(m) n is 1. In general, the proper coeﬃcient is the total amount paid during one interest conversion period, and not the amount of each actual payment. The total amount paid during one interest conversion period is called periodic rent of the annuity. Example 24.1 A loan of $3,000 is to be repaid with quarterly installments at the end of each quarter for ﬁve years. If the rate of interest charged is 10% converted semiannually, ﬁnd the amount of each quarterly payment. Solution. Let Rbe the amount of quarterly payment. In one interest conversion period there are two payments. Then 2Ra10 (2) 0.05 = 3000 →R= 1500 1−(1.05)−10 2[(1.05)0.5−1] = 191.89 Example 24.2 What is the accumulated value following the last payment of $100 deposited at the end of each month for 30 years into an account that earns an eﬀective annual rate of 10%? Solution. Each conversion period contains 12 payment periods. The term of the loan is 30 interest conversion periods. Thus, 12 ·100 · (1.1)30−1 12[(1.1) 1 12 −1] = 206,284.33 Example 24.3 The accumulated amount of an annuity-immediate of $R per year payable quarterly for seven years is $3317.25. Find R if i(1) = 0.05. Solution. Each R is paid quarterly at R 4 . The equation of value at time t= 7 years is Rs(4) 7 = 3317.25. But s(4) 7 = i i(4) s7 . On the other hand, ( 1 + i(4) 4 )4 = 1.05 ⇒i(4) = 4[(1.05) 1 4 −1] = 0.0491. 220 THE BASICS OF ANNUITY THEORY Hence, we obtain 3317.25 = R 0.05 0.0491 (8.1420). Solving for R; we ﬁnd R= $400 In the case of annuity-due, the amount 1 m is payable at the beginning of the mth period of an interest conversion period for a total of n interest conversion period. The present value is denoted by ¨a(m) n . A time diagram of this case is shown in Figure 24.2. Figure 24.2 The present value is determined as follows: ¨a(m) n = 1 m[1 + ν 1 m + ν 2 m + ··· + νn−1 m] = 1 m [ 1 −(ν 1 m)mn 1 −ν 1 m ] = 1 −νn m[1 −(1 + i)−1 m] = 1 −νn d(m) . The accumulated value one mth of an interest conversion period after the last payment is made is given by ¨s(m) n = (1 + i)n¨a(m) n = (1 + i)n −1 d(m) . It follows from the formulas of an,sn,¨a(m) n , and ¨s(m) n that ¨a(m) n = i d(m) an = ¨s(m) 1 an and ¨s(m) n = i d(m) sn = ¨s(m) 1 sn. Example 24.4 What is the present value of the payment of $150 per quarter for 10 years if the eﬀective annual interest rate is 6% and the ﬁrst payment is due today. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE221 Solution. The answer is 600·¨a10 (4) 0.06 = 600·a10 0.06 · i d(4) = 600 ·7.3601 ·1.037227 = $4,580.46 Example 24.5 Express ¨a(12) n in terms of a(2) n with an adjustment factor. Solution. We have ¨a(12) n = i d(12) an and a(2) n = i i(2) an. Thus, ¨a(12) n = i(2) d(12) a(2) n The following identities involving an,sn,¨a(m) n , and ¨s(m) n are analogous to the identity involving an,sn,¨an, and ¨sn discussed in earlier sections. Theorem 24.1 (a) 1 a(m) n = 1 s(m) n + i(m) (b) 1 ¨a(m) n = 1 ¨s(m) n + d(m) (c) ¨a(m) n = (1 + i) 1 ma(m) n = [ i i(m) + i m ] an (d) ¨s(m) n = (1 + i) 1 ms(m) n = [ i i(m) + i m ] sn (e) ¨a(m) n = 1 m + a(m) n−1 m (f) ¨s(m) n = s(m) n+ 1 m −1 m Proof. See Problem 24.10 Next, consider an inﬁnite payment of 1 m at the end of mth of an interest conversion period. Let a(m) ∞ denote the present value. A time diagram describing this case is shown in Figure 24.3. 222 THE BASICS OF ANNUITY THEORY The present value is determined as follows. a(m) ∞ = 1 m[ν 1 m + ν 2 m + ν 3 m + ···] = 1 m ∞∑ p=1 ν p m = 1 m ν 1 m 1 −ν 1 m = 1 m(1 + i) 1 m(1 −ν 1 m) = 1 m((1 + i) 1 m −1) = 1 i(m) Figure 24.3 Example 24.6 An annuity pays 100 at the end of each month forever. Calculate the present value of the annuity using an annual eﬀective interest rate of 8%. Solution. The present value is 12 ·100a(12) ∞ = 1200 12[(1.08) 1 12 −1] = 15,542.36 Now, consider an inﬁnite payment of 1 m at the beginning of mth of an interest conversion pe- riod. Let the present value be denoted by ¨a(m) ∞ . Figure 24.4 describes this situation. Figure 24.4 The present value is determined as follows. ¨a(m) ∞ = 1 m[1 + ν 1 m + ν 2 m + ν 3 m + ···] = 1 m ∞∑ p=0 ν p m = 1 m[1 −ν 1 m] = 1 d(m) , ν 1 m <1 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE223 Example 24.7 At what annual eﬀective rate of interest is the present value of a series of payments of $1 for every six months forever, with the ﬁrst payment made now, equal to $10? Solution. We have 10 = 2¨a(2) ∞ = 2 d(2) →d(2) = 0.2 →i= (1 −0.1)−2 −1 = 23.46% In some cases, each conversion period does not contain an integral number of payments, i.e. m> 1 but mis not an integer. In this case, we handle this problem using the basic principles, i.e. to ﬁnd the present value or the accumulated value we write a sum of present values or accumulated values of the individual payments. An illustration of this type is shown next. Example 24.8 Find an expression for the present value of an annuity which pays 1 at the beginning of each 3-month period for 12 years, assuming a rate of interest per 4-month period. Solution. Working in years the annual payments are 4 ×1 = $4. Thus, the present value is 4¨a(4) 12 = 4 ·1 −(1 + i)−12 d(4) . But 1 + i= ( 1 −d(4) 4 )−4 ⇒d(4) = 4[1 −(1 + i)−1 4 ]. Thus 4¨a(4) 10 =4 1 −(1 + i)−12 4[1 −(1 + i)−1 4 ] =1 −ν12 1 −ν 1 4 =1 −ν48 j 1 −νj =1 −ν36 k 1 −ν 3 4 k where j is the eﬀective interest rate per 3-month and k is the eﬀective interest rate per 4-month Finally, it is possible to generalize the approach to ﬁnding annuity values on any date, as discussed in Section 17, to annuities payable more frequently than interest is convertible. 224 THE BASICS OF ANNUITY THEORY Example 24.9 Payments of $400 at the end of each month are made over a ten-year period. Find expressions for (a) the present value of these payments two years prior to the ﬁrst payment; (b) the accumulated value three years after the ﬁnal payment. Use symbols based on an eﬀective rate of interest i. Solution. (a) The present value is given by 12 ·400νa(12) 10 . (b) The accumulated value is given by 12 ·400s(12) 10 (1 + i)3 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE225 Practice Problems Problem 24.1 The present value of an annuity is denoted by 120a(12) n .What is the amount of the monthly payment of this annuity? Problem 24.2 The present value of an annuity that pays 100 at the end of each year for n years using an annual eﬀective interest rate of 10.25% is 1,000. Calculate the present value of an annuity that pays 100 at the end of every six months for n years using the same interest rate. (Note: You should work this problem without ﬁnding n.) Problem 24.3 Suppose you deposit $200 per month, at the end of each month, in a savings account paying the eﬀective annual rate of interest 6%. How much will be in the account after ten years? (i.e. just after your 120th deposit?) Problem 24.4 A loan of $150,000 is repaid by equal installments at the end of each month, for 25 years. Given that the nominal annual rate of interest, convertible semiannually, is 8%, compute the amount of monthly installment. Problem 24.5 Suppose you deposit $300 at the end of every three months for 10 years into a bank account paying the annual eﬀective rate of interest 4%. What is the accumulated amount after 10 years? Problem 24.6 An annuity pays 100 at the end of each month for 20 years. Using a nominal rate of interest of 4% compounded quarterly, calculate the current value of the annuity at the end of the 3rd year. Problem 24.7 Which of the following are true? (i) an = i(m) i a(m) n (ii) s(m) n = (1 + i)na(m) n (iii) a(∞) ∞ = ¨a(∞) ∞ Problem 24.8 If 3a(2) n = 2a(2) 2n = 45s(2) 1 , ﬁnd i. 226 THE BASICS OF ANNUITY THEORY Problem 24.9 A sum of 10,000 is used to buy a deferred perpetuity-due paying $500 every six months forever. Find an expression for the deferred period expressed as a function of d, the annual discount rate. Problem 24.10 Prove the following identities: (a) 1 a(m) n = 1 s(m) n + i(m) (b) 1 ¨a(m) n = 1 ¨s(m) n + d(m) (c) ¨a(m) n = (1 + i) 1 ma(m) n = [ i i(m) + i m ] an (d) ¨s(m) n = (1 + i) 1 ms(m) n = [ i i(m) + i m ] sn (e) ¨a(m) n = 1 m + a(m) n−1 m (f) ¨s(m) n = s(m) n+ 1 m −1 m Problem 24.11 Find the present value of a ten-year annuity which pays $400 at the beginning of each quarter for the ﬁrst 5 years, increasing to $600 per quarter thereafter. The annual eﬀective rate of interest is 12%. Round to the nearest dollar. Use the approach developed in this section. Problem 24.12 A family wishes to provide an annuity of $100 at the end of each month to their daughter now entering college. The annuity will be paid for only nine months each year for four years. Prove that the present value one month before the ﬁrst payment is 1200¨a4 a(12) 9/12 . Problem 24.13 The nominal rate of interest, convertible semiannually, is 6%. An annuity−immediate pays $50 each month for ﬁve years. Find the accumulated value of this annuity at the time of the last payment. Problem 24.14 You deposit $2,500 into an account at the beginning of every month for 15 years. The interest rate on the account is i(2) = 8%. Find the accumulated value in the account nine months after the last deposit. 24 ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLY THAN INTEREST IS CONVERTIBLE227 Problem 24.15 You want to accumulate $1,000,000. You intend to do this by making deposits of $5,000 into an investment account at the end of each month, until your account balance equals $1,000,000. The account earns 8% convertible quarterly. Determine the number of monthly deposits you will need to make to achieve your goal. 228 THE BASICS OF ANNUITY THEORY 25 Continuous Annuities In this section we consider annuities with a ﬁnite term and an inﬁnte frequency of payments . Formulas corresponding to such annuities are useful as approximations corresponding to annuities payable with great frequency such as daily. Consider an annuity in which a very small payment dtis made at time tand these small payments are payable continuously for n interest conversion periods. Let i denote the periodic interest rate. Then the total amount paid during each period is ∫ k k−1 dt= [t]k k−1 = $1. Let an denote the present value of an annuity payable continuously forninterest conversion periods so that 1 is the total amount paid during each interest conversion period. Then the present value can be found as follows: an = ∫ n 0 νtdt= νt ln ν ⏐⏐⏐⏐ n 0 =νn −1 ln ν = νn −1 −ln (1 +i) =1 −νn δ . It is easy to see the following an = lim m→∞ a(m) n = lim m→∞ 1 −νn i(m) = 1 −νn δ since lim m→∞ i(m) = δ. Similarly, an = lim m→∞ ¨a(m) n = lim m→∞ 1 −νn d(m) = 1 −νn δ since lim m→∞ d(m) = δ. Moreover, an = i δan = d δ¨an = 1 −e−nδ δ . 25 CONTINUOUS ANNUITIES 229 Example 25.1 Starting four years from today, you will receive payment at the rate of $1,000 per annum, payable continuously, with the payment terminating twelve years from today. Find the present value of this continuous annuity if δ= 5%. Solution. The present value is PV = 1000v4 ·a8 = 1000e−0.20 ·1−e−0.40 0.05 = $5,398.38 The next example exhibit the situation when the constant force above is replaced by a variable force of interest. Example 25.2 Find an expression for an if δt = 1 1+t. Solution. The present value of the paymentdtat the exact time twith variable force of interest ise− ∫t 0 δrdrdt= e−ln (1+t)dt= (1 + t)−1dt. Hence, an = ∫ n 0 dt 1 + t = ln (1 + t)|n 0 = ln (1 +n) Next, let sn denote the accumulated value at the end of the term of an annuity payable continuously for ninterest conversion periods so that 1 is the total amound paid during each interest conversion period. Then sn =(1 + i)nan = ∫ n 0 (1 + i)n−tdt = −(1 + i)n−t ln (1 +i) ⏐⏐⏐⏐ n 0 =(1 + i)n −1 δ . It is easy to see that sn = lim m→∞ s(m) n = lim m→∞ ¨s(m) n = enδ −1 δ = i δsn = d δ¨sn. Example 25.3 Find the force of interest at which the accumulated value of a continuous payment of 1 every year for 8 years will be equal to four times the accumulated value of a continuous payment of 1 every year for four years. 230 THE BASICS OF ANNUITY THEORY Solution. We have s8 =4s4 e8δ −1 δ =4 ·e4δ −1 δ e8δ −4e4δ + 3 =0 (e4δ −3)(e4δ −1) =0 If e4δ = 3 then δ= ln 3 4 ≈0.275 = 27.5%. If e4δ = 1 then δ= 0, an extraneous solution Example 25.4 The annual eﬀective rate is i = 5% . Find the accumulated value after six years of an annuity that oﬀers payments of $1,000 per annum, convertible continuously for ﬁve years. After ﬁve years, payments terminate, but the balance still earns interest during the sixth year. Solution. The accumulated value is AV = 1000 s5 (1.05). But δ = ln (1 + i) = ln 1 .05 = 0 .0487902. Thus, AV = 1000 ·(1.05)5−1 0.0487902 ·(1.05) = $5,945.78 The present value of a perpetuity payable continuously with total of 1 per period is given by a∞ = lim n→∞ an = 1 δ. Example 25.5 A perpetuity paid continuously at a rate of 100 per year has a present value of 800. Calculate the annual eﬀective interest rate used to calculate the present value. Solution. The equation of value at time t= 0 is 800 = 100 δ = 100 ln (1 +i). Thus, i= e 1 8 −1 = 13.3% 25 CONTINUOUS ANNUITIES 231 Practice Problems Problem 25.1 Calculate the present value of a continuous annuity of 1,000 per annum for 8 years at: (a) An annual eﬀective interest rate of 4%; (b) A constant force of interest of 4%. Problem 25.2 Find the force of interest at which s20 = 3s10 . Problem 25.3 If an = 4 and sn = 12, ﬁnd δ. Problem 25.4 There is $40,000 in a fund which is accumulating at 4% per annum convertible continuously. If money is withdrawn continuously at the rate of $2,400 per annum, how long will the fund last? Problem 25.5 Annuity A oﬀers to pay you $100 per annum, convertible continuously, for the next ﬁve years. Annuity B oﬀers you $ X at the end of each year for ten years. The annual eﬀective interest rate i is 8%. Find X such that you are indiﬀerent between the two annuities. Problem 25.6 Given δ= 0.1. Evaluate a(12) 10 a10 . Problem 25.7 You are given d dtst = (1.02)2t. Calculate δ. Problem 25.8 Given an = n−4 and δ= 10%, ﬁnd ∫n 0 atdt. Problem 25.9 Payments will be made to you at a continuous rate of $100 per year for the next ﬁve years. The eﬀective rate of interest over this period is i = 0.06. Calculate the present value of this payment stream. Problem 25.10 Show that d dtat = νt s1 . 232 THE BASICS OF ANNUITY THEORY Problem 25.11 Show that an <a(m) n <an <¨a(m) n <¨an. Hint: See Example 10.15. Problem 25.12 Find an expression for t, 0 <t< 1, such that 1 paid at time t is equivalent to 1 paid continuously between 0 and 1. Problem 25.13 A bank makes payments continuously at a rate of $400 a year. The payments are made between 5 and 7 years. Find the current value of these payments at time 2 years using an annual rate of discount of 4%. Problem 25.14 A company makes payments continuously at a rate of $200 per year. The payments are made between 2 and 7 years. Find the accumulated value of these payments at time 10 years using an annual rate of interest of 6.5%. Problem 25.15 Lauren is being paid a continuous perpetuity payable at a rate of 1000 per year. Calculate the present value of the perpetuity assuming d(12) = 0.12. Problem 25.16 If i = 0.04, calculate the accumulated value of a continuous annuity payable at a rate of 100 per year for 10 years. Problem 25.17 If δ= 0.06, calculate the present value of a continuous annuity of 1 payable for 20 years. Problem 25.18 You are given ∫n 0 atdt= 100. Calculate an. Problem 25.19 An n−year continuous annuity that pays at a rate of $748 per year has a present value of $10,000 when using an interest rate of 6%, compounded continuously. Determine n. Problem 25.20 Which of the following are true? (I) ( an −d δ ) (1 + i) = an−1 . 25 CONTINUOUS ANNUITIES 233 (II) The present value of a 10 year annuity immediate paying 10 per month for the ﬁrst eight months of each year is 120 a10 a(12) 8/12 . (III) The present value of a perpetuity paying one at the end of each year, except paying nothing every fourth year, is s3 is4 . Problem 25.21 Payments of $3650 per year are made continuously over a ﬁve-year period. Find the present value of this continuous annuity two years prior to the ﬁrst payment. Given the nominal rate i(365) = 8%. 234 THE BASICS OF ANNUITY THEORY 26 Varying Annuity-Immediate Thus far in this book, all annuities that we have considered had level series of payments, that is, payments are all equal in values. In this and the next three sections we consider annuities with a varying series of payments. In this section, we assume that the payment period and interest conversion period coincide. Annuities with varying payments will be called varying annuities. Any type of annuities can be evaluated by taking the present value or the accumulated value of each payment seperately and adding the results. There are, however, several types of varying annuities for which relatively simple compact expressions are possible. The only general types we will study vary in either arithmetic progression or geometric progression. Payments Varying in an Arithmetic Progression First, let us assume that payments vary in arithmetic progression. That is, the ﬁrst payment is P and then the payments increase by Q thereafter, continuing for n years as shown in the time diagram of Figure 26.1. Figure 26.1 If PV is the present value for this annuity-immediate, then PV = Pν + (P + Q)ν2 + (P + 2Q)ν3 + ··· + [P + (n−1)Q]νn. (26.1) Multiplying Equation (26.1) by (1 + i), we obtain (1 + i)PV = P + (P + Q)ν+ (P + 2Q)ν2 + ··· + [P + (n−1)Q]νn−1. (26.2) Subtracting Equation(26.1) from Equation (26.2) we obtain iPV = P(1 −νn) + (ν+ ν2 + ··· + νn)Q−nνnQ or PV = Pan + Q[an −nνn] i . (26.3) The accumulated value of these payments at time n is AV = (1 + i)nPV = Psn + Q[sn −n] i . (26.4) 26 VARYING ANNUITY-IMMEDIATE 235 Two special cases of the above varying annuity often occur in practice. The ﬁrst of these is the increasing annuity where P = Q= 1 as shown in Figure 26.2. Figure 26.2 The present value of such an annuity is (Ia)n = an + an −nνn i = (1 + i)an −nνn i = ¨an −nνn i . (26.5) The accumulated value at time n is given by (Is)n = (1 + i)n(Ia)n = ¨sn −n i = sn+1 −(n+ 1) i . Example 26.1 The following payments are to be received: $500 at the end of the ﬁrst year, $520 at the end of the second year, $540 at the end of the third year and so on, until the ﬁnal payment is $800. Using an annual eﬀective interest rate of 2% (a) determine the present value of these payments at time 0; (b) determine the accumulated value of these payments at the time of the last payment. Solution. In n years the payment is 500 + 20( n−1). So the total number of payments is 16. The given payments can be regarded as the sum of a level annuity immediate of $480 and an increasing annuity-immediate $20,$40,··· ,$320. (a) The present value at time t= 0 is 480a16 + 20(Ia)16 = 480(13.5777) + 20(109.7065) = $8,711.43. (b) The accumulated value at time t= 16 is 480s16 + 20(Is)16 = 480(18.6393) + 20(150.6035) = $11,958.93 Example 26.2 Show that (Ia)n = ∑n−1 t=0 νtan−t. 236 THE BASICS OF ANNUITY THEORY Solution. We have (Ia)n = n−1∑ t=0 νtan−t = n−1∑ t=0 νt1 −νn−t i =1 i n−1∑ t=0 νt −νn i n−1∑ t=0 1 =¨an i −nνn i = ¨an −nνn i . The second special case is the decreasing annuity-immediate where P = n and Q = −1 as shown in Figure 26.3. Figure 26.3 In this case, the present value one year before the ﬁrst payment (i.e., at time t= 0) is given by (Da)n = nan −an −nνn i = n−nνn −an + nνn i = n−an i . (26.6) The accumulated value at time n is given by (Ds)n = (1 + i)n(Da)n = n(1 + i)n −sn i = (n+ 1)an −(Ia)n. Example 26.3 John receives $400 at the end of the ﬁrst year, $350 at the end of the second year, $300 at the end of the third year and so on, until the ﬁnal payment of $50. Using an annual eﬀective rate of 3.5%, calculate the present value of these payments at time 0. Solution. In year nthe payment is 400−50(n−1).Since the ﬁnal payment is 50 we must have 400−50(n−1) = 50. Solving for n we ﬁnd n= 8. Thus, the present value is 50(Da)8 = 50 ·8 −a8 0.035 = $1,608.63 26 VARYING ANNUITY-IMMEDIATE 237 Example 26.4 Calculate the accumulated value in Example 26.3. Solution. The answer is 50(Ds)8 = 50 · 8(1.035)8−s8 0.035 = $2,118.27 Example 26.5 Show that (Ia)n + (Da)n = (n+ 1)an. Solution. We have (Ia)n + (Da)n =an + an −nνn i + n−an i =an + nan = (n+ 1)an Besides varying annuities immediate, it is also possible to have varying perpetuity −immediate. Consider a perpetuity−immediate with payments that form an arithmetic progression (and of course P >0 and Q >0). The present value for such a perpetuity with the ﬁrst payment at the end of the ﬁrst period is PV = lim n→∞ [ Pan + Q[an −nνn] i ] == P lim n→∞ an + Q lim n→∞ [an −nνn] i =P lim n→∞ an + Q[limn→∞an −limn→∞nνn] i =Pa∞ + Qa∞ i = P i + Q i2 since a∞ = 1 i and limn→∞nνn = 0 (by L’Hopital’s rule). For the special case P = Q= 1 we ﬁnd (Ia)∞ = 1 i + 1 i2 . Example 26.6 Find the present value of a perpetuity-immediate whose successive payments are 1, 2, 3, 4, ··· at an eﬀective rate of 6%. Solution. We have (Ia)∞ = 1 i + 1 i2 = 1 0.06 + 1 0.062 = $294.44 238 THE BASICS OF ANNUITY THEORY Payments Varying in a Geometric Progression Next, we consider payments varying in a geometric progression. Consider an annuity-immediate with a term of n periods where the interest rate is i per period, and where the ﬁrst payment is 1 and successive payments increase in geometric progression with common ratio 1 + k. The present value of this annuity is ν+ ν2(1 + k) + ν3(1 + k)2 + ··· + νn(1 + k)n−1 =ν·1 −[(1 + k)ν]n 1 −(1 + k)ν = 1 1 + i · 1 − (1+k 1+i )n 1 − (1+k 1+i ) = 1 − (1+k 1+i )n i−k provided that k ̸= i. If k = i then the original sum consists of a sum of n terms of ν which equals to nν. Example 26.7 The ﬁrst of 30 payments of an annuity occurs in exactly one year and is equal to $500. The payments increase so that each payment is 5% greater than the preceding payment. Find the present value of this annuity with an annual eﬀective rate of interest of 8%. Solution. The present value is given by PV = 500 ·1 − (1.05 1.08 )30 0.08 −0.05 = $9,508.28 For an annuity-immediate with a term ofnperiods where the interest rate is iper period, and where the ﬁrst payment is 1 and successive payments decrease in geometric progression with common ratio 1 −k. The present value of this annuity is ν+ ν2(1 −k) + ν3(1 −k)2 + ··· + νn(1 −k)n−1 =ν·1 −[(1 −k)ν]n 1 −(1 −k)ν = 1 1 + i · 1 − (1−k 1+i )n 1 − (1−k 1+i ) = 1 − (1−k 1+i )n i+ k 26 VARYING ANNUITY-IMMEDIATE 239 provided that k̸= i. If k= i then the original sum becomes ν+ ν2(1 −i) + ν3(1 −i)2 + ··· + νn(1 −i)n−1 = 1 2i [ 1 − (1 −i 1 + i )n] Finally, we consider a perpetuity with payments that form a geometric progression where 0 < 1 +k <1 +i.The present value for such a perpetuity with the ﬁrst payment at the end of the ﬁrst period is ν+ ν2(1 + k) + ν3(1 + k)2 + ··· = ν 1 −(1 + k)ν = 1 i−k. Observe that the value for these perpetuities cannot exist if 1 + k≥1 + i. Example 26.8 What is the present value of a stream of annual dividends, which starts at 1 at the end of the ﬁrst year, and grows at the annual rate of 2%, given that the rate of interest is 6% ? Solution. The present value is 1 i−k = 1 0.06−0.02 = 25 240 THE BASICS OF ANNUITY THEORY Practice Problems Problem 26.1 Smith receives $400 in 1 year, $800 in 2 years, $1,200 in 3 years and so on until the ﬁnal payment of $4,000. Using an eﬀective interest rate of 6%, determine the present value of these payments at time 0. Problem 26.2 Find an expression for the present value at time 0 of payments of $55 at time 1 year, $60 at time 2 years, $65 at time 3 years and so on, up to the last payment at time 20 years. Problem 26.3 Find an expression for the accumulated value at time 20 years of payments of $5 at time 1 year, $10 at time 2 years, $15 at time 3 years, and so on, up to $100 at time 20 years. Problem 26.4 Find the present value of a perpetuity-immediate whose successive payments are $5 ,$10,$15,··· assuming an annual eﬀective rate of interest of 5%. Problem 26.5 An annuity pays $100 at the end of one month. It pays $110 at the end of the second month. It pays $120 at the end of the third month. The payments continue to increase by $10 each month until the last payment is made at the end of the 36th month. Find the present value of the annuity at 9% compounded monthly. Problem 26.6 An annual annuity-immediate pays $100 at the end of the ﬁrst year. Each subsequent payment is 5% greater than the preceding payment. The last payment is at the end of the 20th year. Calculate the accumulated value at: (a) an annual eﬀective interest rate of 4%; (b) an annual eﬀective interest rate of 5%. Problem 26.7 A perpetuity pays $100 at the end of the ﬁrst year. Each subsequent annual payment increases by $50. Calculate the present value at an annual eﬀective interest rate of 10%. Problem 26.8 A small business pays you an annual proﬁt at the end of each year for 20 years. The payments grow at an annual rate of 2.5%, the ﬁrst payment is $10,000. What is the present value of this stream of payments at the annual eﬀective rate of interest 6%? 26 VARYING ANNUITY-IMMEDIATE 241 Problem 26.9 What is the present value of a 30 year immediate annuity at 10% interest where the ﬁrst payment is $100 and each payment thereafter is increased by 5%? Problem 26.10 A perpetuity-immediate has annual payments of 1, 3, 5, 7, ··· .If the present values of the 6th and 7th payments are equal, ﬁnd the present value of the perpetuity. Problem 26.11 If X is the present value of a perpetuity of 1 per year with the ﬁrst payment at the end of the 2nd year and 20X is the present value of a series of annual payments 1, 2, 3, ··· with the ﬁrst payment at the end of the 3rd year, ﬁnd d. Problem 26.12 There are two perpetuities-immediate. The ﬁrst has level payments of p at the end of each year. The second is increasing such that the payments are q,2q,3q,··· .Find the rate of interest that will make the diﬀerence in present value between these perpetuities a) zero; b) a maximum. Problem 26.13 An annuity pays 10 at the end of the ﬁrst year, 20 at the end of the second year, 30 at the end of the third year, etc. The last payment is made at the end of the 12th year. Calculate the accumulated value of this annuity immediately after the last payment using an annual eﬀective rate of 4%. Problem 26.14 A 20 year annuity-immediate pays 500 + 50t at the end of year t. Calculate the present value of this annuity using an annual eﬀective rate of 6%. Problem 26.15 An annuity pays 10 at the end of year 2, and 9 at the end of year 4. The payments continue decreasing by 1 each two year period until 1 is paid at the end of year 20. Calculate the present value of the annuity at an annual eﬀective interest rate of 5%. Problem 26.16 Find an expression for the present value of a perpetuity under which a payment of 1 is made at the end of the 1st year, 2 at the end of the 2nd year, increasing until a payment of n is made at the end of the nth year, and thereafter payments are level at n per year forever. 242 THE BASICS OF ANNUITY THEORY Problem 26.17 Find an expression for the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of n, and then decrease by 1 each period up to a ﬁnal payment of 1. Problem 26.18 An 11-year annuity has a series of payments 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, with the ﬁrst payment made at the end of the second year. The present value of this annuity is 25 at interest rate i. A 12-year annuity has a series of payments 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, with the ﬁrst payment made at the end of the ﬁrst year. Calculate the present value of the 12-year annuity at interest rate i. Problem 26.19 Olga buys a 5-year increasing annuity for X. Olga will receive 2 at the end of the ﬁrst month, 4 at the end of the second month, and for each month thereafter the payment increases by 2 . The nominal interest rate is 9% convertible quarterly. Calculate X. Problem 26.20 An annuity-immediate has semiannual payments of 800, 750, 700, ··· , 350, at i(2) = 0 .16. If a10 0.08 = A, ﬁnd the present value of the annuity in terms of A. Problem 26.21 ‡ A perpetuity costs 77.1 and makes annual payments at the end of the year. The perpetuity pays 1 at the end of year 2, 2 at the end of year 3, ··· ,n at the end of year ( n+ 1).After year (n+ 1),the payments remain constant at n. The annual eﬀective interest rate is 10.5%. Calculate n. Problem 26.22 ‡ A perpetuity-immediate pays 100 per year. Immediately after the ﬁfth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the ﬁrst year. Each subsequent annual payment will be 8% greater than the preceding payment. The annual eﬀective rate of interest is 8%. Calculate X. Problem 26.23 ‡ A perpetuity-immediate pays 100 per year. Immediately after the ﬁfth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the ﬁrst year. Each subsequent annual payment will be 8% greater than the preceding payment. Immediately after the 10th payment of the 25-year annuity, the annuity will be exchanged for a perpetuity-immediate paying Y per year. The annual eﬀective rate of interest is 8%. Calculate Y. 26 VARYING ANNUITY-IMMEDIATE 243 Problem 26.24 ‡ Mike buys a perpetuity-immediate with varying annual payments. During the ﬁrst 5 years, the payment is constant and equal to 10. Beginning in year 6, the payments start to increase. For year 6 and all future years, the current year’s payment is K% larger than the previous year’s payment. At an annual eﬀective interest rate of 9.2%, the perpetuity has a present value of 167.50. Calculate K, given K <9.2. Problem 26.25 ‡ A company deposits 1000 at the beginning of the ﬁrst year and 150 at the beginning of each subsequent year into perpetuity. In return the company receives payments at the end of each year forever. The ﬁrst payment is 100. Each subsequent payment increases by 5%. Calculate the company’s yield rate for this transaction. Problem 26.26 ‡ Megan purchases a perpetuity-immediate for 3250 with annual payments of 130. At the same price and interest rate, Chris purchases an annuity-immediate with 20 annual payments that begin at amount P and increase by 15 each year thereafter. Calculate P. Problem 26.27 ‡ The present value of a 25-year annuity-immediate with a ﬁrst payment of 2500 and decreasing by 100 each year thereafter is X. Assuming an annual eﬀective interest rate of 10%, calculate X. Problem 26.28 ‡ The present value of a series of 50 payments starting at 100 at the end of the ﬁrst year and increasing by 1 each year thereafter is equal to X. The annual eﬀective rate of interest is 9%. Calculate X. Problem 26.29 ‡ An annuity-immediate pays 20 per year for 10 years, then decreases by 1 per year for 19 years. At an annual eﬀective interest rate of 6%, the present value is equal to X. Calculate X. Problem 26.30 ‡ At an annual eﬀective interest rate of i, the present value of a perpetuity-immediate starting with a payment of 200 in the ﬁrst year and increasing by 50 each year thereafter is 46,530. Calculate i. Problem 26.31 ‡ An insurance company has an obligation to pay the medical costs for a claimant. Average annual claims costs today are $5,000, and medical inﬂation is expected to be 7% per year. The claimant is expected to live an additional 20 years. Claim payments are made at yearly intervals, with the ﬁrst claim payment to be made one year from today. Find the present value of the obligation if the annual interest rate is 5%. 244 THE BASICS OF ANNUITY THEORY Problem 26.32 ‡ Sandy purchases a perpetuity-immediate that makes annual payments. The ﬁrst payment is 100, and each payment thereafter increases by 10. Danny purchases a perpetuity-due which makes annual payments of 180. Using the same annual eﬀective interest rate, i> 0,the present value of both perpetuities are equal. Calculate i. Problem 26.33 ‡ Joe can purchase one of two annuities: Annuity 1: A 10-year decreasing annuity-immediate, with annual payments of 10 ,9,8,··· ,1. Annuity 2: A perpetuity-immediate with annual payments. The perpetuity pays 1 in year 1, 2 in year 2, 3 in year 3, ··· , and 11 in year 11 . After year 11, the payments remain constant at 11. At an annual eﬀective interest rate of i,the present value of Annuity 2 is twice the present value of Annuity 1 . Calculate the value of Annuity 1. Problem 26.34 ‡ Mary purchases an increasing annuity-immediate for 50,000 that makes twenty annual payments as follows: i) P,2P,··· ,10P in years 1 through 10; and ii) 10(1.05)P,10(1.05)2P,··· ,10(1.05)10P in years 11 through 20. The annual eﬀective interest rate is 7% for the ﬁrst 10 years and 5% thereafter. Calculate P. Problem 26.35 Find the present value of an annuity-immediate such that payments start at 1, each payment thereafter increases by 1 until reaching 10, and then remain at that level until 25 payments in total are made. Problem 26.36 Find an expression of the present value of an annuity that pays 10 at the end of the ﬁfth year and decreases by 1 thereafter until reaching 0. Problem 26.37 Annual deposits are made into a fund at the beginning of each year for 10 years. The ﬁrst ﬁve deposits are $1000 each and deposits increase by 5% per year thereafter. If the fund earns 8% eﬀective, ﬁnd the accumulated value at the end of 10 years. Problem 26.38 Find the present value of a 20-year annuity-immediate with ﬁrst payment of $600 and each subse- quent payment is 5% greater than the preceding payment. The annual eﬀective rate of interest is 10.25%. 26 VARYING ANNUITY-IMMEDIATE 245 Problem 26.39 A perpetuity pays 200 at the end of each of the ﬁrst two years, 300 at the end of years 3 and 4, 400 at the end of years 5 and 6, etc. Calculate the present value of this perpetuity if the annual eﬀective rate of interest is 10%. Problem 26.40 Janice is receiving a perpetuity of 100 at the end of each year. Megan is receiving a perpetuity that pays 10 at the end of the ﬁrst year, 20 at the end of the second year, 30 at the end of the third year, etc. The present values of the two perpetuities are equal. Calculate the annual eﬀective interest rate used to determine the present values. Problem 26.41 Gloria borrows 100,000 to be repaid over 30 years. You are given (1) Her ﬁrst payment is X at the end of year 1 (2) Her payments increase at the rate of 100 per year for the next 19 years and remain level for the following 10 years (3) The eﬀective rate of interest is 5% per annum Calculate X. Problem 26.42 You are given a perpetual annuity immediate with annual payments increasing in geometric pro- gression, with a common ratio of 1.07. The annual eﬀective interest rate is 12%. The ﬁrst payment is 1. Calculate the present value of this annuity. Problem 26.43 An annuity immediate pays 10 at the ends of years 1 and 2, 9 at the ends of years 3 and 4, etc., with payments decreasing by 1 every second year until nothing is paid. The eﬀective annual rate of interest is 5%. Calculate the present value of this annuity immediate. Problem 26.44 Barbara purchases an increasing perpetuity with payments occurring at the end of every 2 years. The ﬁrst payment is 1, the second one is 2, the third one is 3, etc. The price of the perpetuity is 110. Calculate the annual eﬀective interest rate. Problem 26.45 You are given (Ia)n−1 = K d , where d is the annual eﬀective discount rate. Calculate K. Problem 26.46 Francois purchases a 10 year annuity immediate with annual payments of 10 X. Jacques purchases 246 THE BASICS OF ANNUITY THEORY a 10 year decreasing annuity immediate which also makes annual payments. The payment at the end of year 1 is equal to 50. At the end of year 2, and at the end of each year through year 10, each subsequent payment is reduced over what was paid in the previous year by an amount equal to X. At an annual eﬀective interest rate of 7.072%, both annuities have the same present value. Calculate X, where X <5. Problem 26.47 ‡ 1000 is deposited into Fund X,which earns an annual eﬀective rate of 6%. At the end of each year, the interest earned plus an additional 100 is withdrawn from the fund. At the end of the tenth year the fund is depleted. The annual withdrawals of interest and principal are deposited into Fund Y, which earns an annual eﬀective rate of 9%. Determine the accumulated value of Fund Y at the end of year 10. Problem 26.48 You are given two series of payments. Series A is a perpetuity with payments of 1 at the end of each of the ﬁrst 2 years, 2 at the end of each of the next 2 years, 3 at the end of each of the next 2 years, and so on. Series B is a perpetuity with payments of K at the end of each of the ﬁrst 3 years, 2K at the end of the next 3 years, 3 K at the end of each of the next 3 years, and so on. The present values of the two series of payments are equal. Calculate K. Problem 26.49 John deposits 100 at the end of each year for 20 years into a fund earning an annual eﬀective interest rate of 7%. Mary makes 20 deposits into a fund at the end of each year for 20 years. The ﬁrst 10 deposits are 100 each, while the last 10 deposits are 100 + X each. The fund earns an annual eﬀective interest rate of 8% during the ﬁrst 10 years and 6% annual eﬀective interest thereafter. At the end of 20 years, the amount in John’s fund equals the amount in Mary’s fund. Calculate X. Problem 26.50 An annuity-immediate pays an initial beneﬁt of 1 per year, increasing by 10.25% every four years. The annuity is payable for 40 years. Using an annual eﬀective interest rate of 5%, determine an expression for the present value of this annuity. Problem 26.51 An increasing perpetuity with annual payments has a present value of 860 at an annual eﬀective discount rate, d.The initial payment at the end of the ﬁrst year is 3 and each subsequent payment is 2 more than its preceding payment. Calculate d. 26 VARYING ANNUITY-IMMEDIATE 247 Problem 26.52 A 10-year increasing annuity-immediate paying 5 in the ﬁrst year and increasing by 5 each year thereafter has a present value of G. A 10-year decreasing annuity-immediate paying y in the ﬁrst year and decreasing by y 10 each year thereafter has a present value of G. Both present values are calculated using an annual eﬀective interest rate of 4%. Calculate y to the nearest integer. Problem 26.53 A loan of 10,000 is being repaid by 10 semiannual payments, with the ﬁrst payment made one-half year after the loan. The ﬁrst 5 payments are K each, and the ﬁnal 5 are K + 200 each. Given i(2) = 0.6, determine K. Problem 26.54 An annuity immediate has the following payment pattern: 1 ,2,··· ,n −1,n,n −1,··· ,2,1. The present value of this annuity using an annual eﬀective interest rate of 9% is 129.51. Determine n. Problem 26.55 An increasing 25-year annuity immediate has an initial payment in year one of 4 and each subsequent annual payment is 3 more than the preceding one. Find the present value of this annuity using an annual eﬀective interest rate of 7%. Problem 26.56 A 10-year annuity-immediate has a ﬁrst payment of 2,000. Each subsequent annual payment is 100 less than the preceding payment. (The payments are 2,000; 1,900; 1,800; etc.) At an annual eﬀective interest rate of 5%, what is the accumulated value of this stream of payments on the date of the ﬁnal payment? Problem 26.57 A payment of 100 purchases an increasing perpetuity with a ﬁrst payment of 10 at the end of the ﬁrst year. Each annual payment thereafter is 5% greater than the prior year’s payment. What is the annual interest rate for this stream of payments? Problem 26.58 You are given a perpetuity, with annual payments as follows: (1) Payments of 1 at the end of the ﬁrst year and every three years thereafter. (2) Payments of 2 at the end of the second year and every three years thereafter. (3) Payments of 3 at the end of the third year and every three years thereafter. (4) The interest rate is 5% convertible semi-annually. Calculate the present value of this perpetuity. 248 THE BASICS OF ANNUITY THEORY Problem 26.59 (a) You just paid $50 for a share of stock which pays annual dividends. You expect to receive the ﬁrst dividend payment of $5 one year from now. After that, you expect dividends to increase by $X each year, forever. The price you paid implies that you would be happy with an annual return of 12%. What is X? (b) What value $Y could you sell your share for if the new buyer has the same desired annual return but expects the annual dividend increase to be X+ 1? Problem 26.60 A 30-year annuity-immediate pays 20,000 in the ﬁrst year. Thereafter, the payments decrease 500 each year. If d(4) is 6%, what is the present value of this annuity? Problem 26.61 You are about to buy shares in a company. The ﬁrst annual dividend, of $3 per share, is expected to be paid 3 years from now. Thereafter, you expect that the dividend will grow at the rate of 10% per year for ever. Assuming your desired return is 12%, what are you willing to pay per share for this stock? Problem 26.62 A perpetuity with annual payments is payable beginning 10 years from now. The ﬁrst payment is 50. Each annual payment thereafter is increased by 10 until a payment of 150 is reached. Subsequent payments remain level at 150. This perpetuity is purchased by means of 10 annual premiums, with the ﬁrst premium of P due immediately. Each premium after the ﬁrst is 105% of the preceding one. The annual eﬀective interest rates are 5% during the ﬁrst 9 years and 3% thereafter. Calculate P. Problem 26.63 Show that (Da)n = ∑n t=1 at. Problem 26.64 Simplify ∑10 t=1(t+ 1)νt. 27 VARYING ANNUITY-DUE 249 27 Varying Annuity-Due In this section, we examine the case of an increasing annuity-due. Consider an annuity with the ﬁrst payment is P at the beginning of year 1 and then the payments increase by Qthereafter, continuing for n years. A time diagram of this situation is given in Figure 27.1 Figure 27.1 The present value for this annuity-due is PV = P + (P + Q)ν+ (P + 2Q)ν2 + ··· + [P + (n−1)Q]νn−1. (27.1) Multiplying Equation (28.1) by ν, we obtain νPV = Pν + (P + Q)ν2 + (P + 2Q)ν3 + ··· + [P + (n−1)Q]νn. (27.2) Subtracting Equation(28.2) from Equation (28.1) we obtain (1 −ν)PV = P(1 −νn) + (ν+ ν2 + ··· + νn)Q−nνnQ or PV = P¨an + Q[an −nνn] d . The accumulated value of these payments at time n is AV = (1 + i)nPV = P¨sn + Q[sn −n] d . In the special case when P = Q= 1 we ﬁnd (I¨a)n = ¨an −nνn d and (I¨s)n = ¨sn −n d = sn+1 −(n+ 1) d . 250 THE BASICS OF ANNUITY THEORY Example 27.1 Determine the present value and future value of payments of $75 at time 0, $80 at time 1 year, $85 at time 2 years, and so on up to $175 at time 20 years. The annual eﬀective rate is 4%. Solution. The present value is 70¨ a21 + 5(I¨a)21 = $1 ,720.05 and the future value is (1 .04)21(1,720.05) = $3919.60 In the case of a decreasing annuity-due where P = n and Q = −1, the present value at time 0 is (D¨a)n = n−an d and the accumulated value at time n is (D¨s)n = (1 + i)n(D¨a)n = n(1 + i)n −sn d . Example 27.2 Calculate the present value and the accumulated value of a series of payments of $100 now, $90 in 1 year, $80 in 2 years, and so on, down to $10 at time 9 years using an annual eﬀective interest rate of 3%. Solution. The present value is 10(D¨a)10 = 10·10−8.530203 0.03/1.03 = $504.63 and the accumulated value is (1.03)10(504.63) = $678.18 Next, we consider a perpetuity-due with payments that form an arithmetic progression (and of course P >0 and Q >0). The present value for such a perpetuity with the ﬁrst payment at time 0 is PV = lim n→∞ [ P¨an + Q[an −nνn] d ] =P lim n→∞ ¨an + Q lim n→∞ [an −nνn] d =P lim n→∞ ¨an + Q[limn→∞an −limn→∞nνn] d =P¨a∞ + Qa∞ d = P d + Q(1 + i) i2 27 VARYING ANNUITY-DUE 251 since a∞ = 1 i, ¨a∞ = 1 d and limn→∞nνn = 0 (by L’Hopital’s rule). In the special case P = Q= 1 we ﬁnd (I¨a)∞ = 1 d2 . Example 27.3 Determine the present value at time 0 of payments of $10 paid at time 0, $20 paid at time 1 year, $30 paid at time 2 years, and so on, assuming an annual eﬀective rate of 5%. Solution. The answer is 10(I¨a)∞ = 10 d2 = 10 (1.05 0.05 )2 = $4,410.00 Next, we consider payments varying in a geometric progression. Consider an annuity-due with a term of nperiods where the interest rate is iper period, and where the ﬁrst payment is 1 at time 0 and successive payments increase in geometric progression with common ratio 1 + k. The present value of this annuity is 1 + ν(1 + k) + ν2(1 + k)2 + ··· + νn−1(1 + k)n−1 =1 −[(1 + k)ν]n 1 −(1 + k)ν = 1 − (1+k 1+i )n 1 − (1+k 1+i ) =(1 + i) 1 − (1+k 1+i )n i−k provided that k ̸= i. If k = i then the original sum consists of a sum of n terms of 1 which equals to n. Example 27.4 An annual annuity due pays $1 at the beginning of the ﬁrst year. Each subsequent payment is 5% greater than the preceding payment. The last payment is at the beginning of the 10th year. Calculate the present value at: (a) an annual eﬀective interest rate of 4%; (b) an annual eﬀective interest rate of 5%. Solution. (a) PV = (1.04) [ 1−(1.05 1.04 ) 10] (0.04−0.05) = $10.44. (b) Since i= k, PV = n= $10.00 252 THE BASICS OF ANNUITY THEORY For an annuity-due withnpayments where the ﬁrst payment is 1 at time 0 and successive payments decrease in geometric progression with common ratio 1 −k. The present value of this annuity is 1 + ν(1 −k) + ν2(1 −k)2 + ··· + νn−1(1 −k)n−1 =1 −[(1 −k)ν]n 1 −(1 −k)ν = 1 − (1−k 1+i )n 1 − (1−k 1+i ) =(1 + i) 1 − (1−k 1+i )n i+ k provided that k̸= i. If k= i then the original sum is 1 + ν(1 −i) + ν2(1 −i)2 + ··· + νn−1(1 −i)n−1 = 1 2d [ 1 − (1 −i 1 + i )n] . Example 27.5 ‡ Matthew makes a series of payments at the beginning of each year for 20 years. The ﬁrst payment is 100. Each subsequent payment through the tenth year increases by 5% from the previous payment. After the tenth payment, each payment decreases by 5% from the previous payment. Calculate the present value of these payments at the time the ﬁrst payment is made using an annual eﬀective rate of 7%. Solution. The present value at time 0 of the ﬁrst 10 payments is 100 [ 1 − (1.05 1.07 )10 0.07 −0.05 ] ·(1.07) = 919.95. The value of the 11th payment is 100(1 .05)9(0.95) = 147 .38. The present value of the last ten payments is 147.38 [ 1 − (0.95 1.07 )10 0.07 + 0.05 ] ·(1.07)(1.07)−10 = 464.71. The total present value of the 20 payments is 919 .95 + 464.71 = 1384.66 Finally, the present value of a perpetuity with ﬁrst payment of 1 at time 0 and successive pay- ments increase in geometric progression with common ration 1 + k is 1 + ν(1 + k) + ν2(1 + k)2 + ··· = 1 1 −(1 + k)ν = 1 + i i−k. Observe that the value for these perpetuities cannot exist if 1 + k≥1 + i. 27 VARYING ANNUITY-DUE 253 Example 27.6 Perpetuity A has the following sequence of annual payments beginning on January 1, 2005: 1,3,5,7,··· Perpetuity B is a level perpetuity of 1 per year, also beginning on January 1, 2005. Perpetuity C has the following sequence of annual payments beginning on January 1, 2005: 1,1 + r,(1 + r)2,··· On January 1, 2005, the present value of Perpetuity A is 25 times as large as the present value of Perpetuity B, and the present value of Perpetuity A is equal to the present value of Perpetuity C. Based on this information, ﬁnd r. Solution. The present value of Perpetuity A is 1 d + 2(1+i) i2 . The present value of Perpetuity B is 1 d. The present value of Perpetuity C is 1+i i−r. We are told that 1 + i i + 2(1 + i) i2 = 25(1 + i) i . This is equivalent to 12i2 + 11i−1 = 0. Solving for i we ﬁnd i= 1 12 . Also, we are told that 1 + i i + 2(1 + i) i2 = 1 + i i−r or 25(12)(1 + 1 12) = 1 + 1 12 1 12 −r. Solving for r we ﬁnd r= 0.08 = 8% 254 THE BASICS OF ANNUITY THEORY Practice Problems Problem 27.1 A 20 year increasing annuity due pays 100 at the start of year 1, 105 at the start of year 2, 110 at the start of year 3, etc. In other words, each payment is 5% greater than the prior payment. Calculate the present value of this annuity at an annual eﬀective rate of 5%. Problem 27.2 A perpetuity pays 1000 at the beginning of the ﬁrst year. Each subsequent payment is increased by inﬂation. If inﬂation is assumed to be 4% per year, calculate the present value of this perpetuity using an annual eﬀective interest rate of 10%. Problem 27.3 A perpetuity pays 1000 at the beginning of the ﬁrst year. Each subsequent payment is increased by inﬂation. If inﬂation is assumed to be 4% per year, calculate the present value of this perpetuity using an annual eﬀective interest rate of 4%. Problem 27.4 A 10-year decreasing annuity-due makes a payment of 100 at the beginning of the ﬁrst year, and each subsequent payment is 5 less then the previous. What is the accumulated value of this annuity at time 10 (one year after the ﬁnal payment)? The eﬀective annual rate of interest is 5%. Problem 27.5 A 10 year annuity due pays 1,000 as the ﬁrst payment. Each subsequent payment is 50 less than the prior payment. Calculate the current value of the annuity at the end of 5 years using an annual eﬀective rate of 8%. Problem 27.6 You buy an increasing perpetuity-due with annual payments starting at 5 and increasing by 5 each year until the payment reaches 100. The payments remain at 100 thereafter. The annual eﬀective interest rate is 7.5%. Determine the present value of this perpetuity. Problem 27.7 Debbie receives her ﬁrst annual payment of 5 today. Each subsequent payment decreases by 1 per year until time 4 years. After year 4, each payment increases by 1 until time 8 years. The annual interest rate is 6%. Determine the present value. 27 VARYING ANNUITY-DUE 255 Problem 27.8 Determine the accumulated value at time 20 years of payments of $10 at time 0, $20 at time 1 year, $30 at time 2 years, and so on, up to $200 at time 19 years. The annual eﬀective rate of interest is 4%. Problem 27.9 Perpetuity X has payments of 1,2,3,··· at the beginning of each year. Perpetuity Y has payments of q,q, 2q,2q,3q,3q,··· at the beginning of each year. The present value of X is equal to the present value of Y at an eﬀective annual interest rate of 10%. Calculate q. Problem 27.10 Kendra receives $900 now, $970 in 1 year, $1040 in 2 years, $1,110 in 3 years, and so on, until the ﬁnal payment of $1600. Using an annual eﬀective rate of interest of 9%, ﬁnd (a) the present value of these payments at time 0; (b) the accumulated value at time 11 years. Problem 27.11 Mary is saving money for her retirement. She needs $750,000 in 10 years to purchase a retirement apartment in Florida. She invests X, now, X−5,000 in 1 year, X−10,000 in 2 years, and so on, down to X−45,000 in 9 years. Using an annual eﬀective rate of interest of 5%, ﬁnd X. Problem 27.12 Chris makes annual deposits into a bank account at the beginning of each year for 20 years. Chris’ initial deposit is equal to 100, with each subsequent deposit k% greater than the previous year’s deposit. The bank credits interest at an annual eﬀective rate of 5%. At the end of 20 years, the accumulated amount in Chris’ account is equal to 7276.35. Given k >5, calculate k. Problem 27.13 You are given: (i) The present value of an annuity-due that pays 300 every 6 months during the ﬁrst 15 years and 200 every 6 months during the second 15 years is 6000. (ii) The present value of a 15-year deferred annuity-due that pays 350 every 6 months for 15 years is 1580. (iii) The present value of an annuity-due that pays 100 every 6 months during the ﬁrst 15 years and 200 every 6 months during the next 15 years is X. The same interest rate is used in all calculations. Determine X. Problem 27.14 A 20 year annuity-due with annual payments has a ﬁrst payment of 100 and each subsequent annual payment is 5% more than its preceding payment. Calculate the present value of this annuity at a nominal interest rate of 10% compounded semiannually. 256 THE BASICS OF ANNUITY THEORY Problem 27.15 The payments of a perpetuity-due are three years apart. The initial payment is 7 and each sub- sequent payment is 5 more than its previous payment. Find the price (present value at time 0) of this perpetuity-due, using an annual eﬀective discount rate of 6%. Problem 27.16 You are considering the purchase of a share ofXYZ stock. It pays a quarterly dividend. You expect to receive the ﬁrst dividend, of $2.50, 3 months from now. You expect the dividend to increase by 6% per year, forever, starting with the 5th dividend. If the desired return on your investment is 15%, what is the amount your are willing to pay for a share of XYZ stock? Problem 27.17 You plan to make annual deposits at the start of each year for 25 years into your retirement fund. The ﬁrst 5 deposits are to be 2,000 each. Thereafter, you plan to increase the deposits by X per year. Assuming the fund can earn an eﬀective annual interest rate of 6%, what must X be for you to have accumulated 250,000 at the end of the 25 years? 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE257 28 Varying Annuities with Payments at a Diﬀerent Fre- quency than Interest is Convertible In Sections 26 and 27, we discussed varying annuities where the payment period coincide with the interest conversion period. In this section we consider varying annuities with payments made more or less frequently than interest is convertible. We will limit our discussion to increasing annuities. Decreasing annuities can be handled in a similar fashion. Varying Annuities Payable Less Frequently Than Interest is Convertible Consider annuities payable less frequently than interest is convertible. We let k be the number of interest conversion periods in one payment period, n the term of the annuity measured in interest conversion periods, and i the rate of interest per conversion period. It follows that the number of payments over the term of the annuity is given by n k which we assume is a positive integer. Let PV be the present value of a generalized increasing annuity-immediate with payments 1, 2, 3, etc. occurring at the end of each interval of k interest conversion periods. Then, PV = νk + 2ν2k + 3ν3k + ··· + (n k −1 ) νn−k + n kνn. (28.1) Multiply Equation (28.1) by (1 + i)k to obtain (1 + i)kPV = 1 + 2νk + 3ν2k + ··· + (n k −1 ) νn−2k + n kνn−k. (28.2) Now subtracting Equation (28.1) from Equation (28.2) we ﬁnd PV [(1 + i)k −1] = 1 + νk + ν2k + ··· + νn−k −n knun = an ak −n kνn. Hence, PV = an ak −n kνn isk . The accumulated value at time t= n is AV = (1 + i)nPV = (1 + i)n an ak −n kνn isk . Example 28.1 A 10-year annuity-immediate pays 1 in two years, 2 in four years, 3 in six years, 4 in eight years, and 5 in ten years. Using an eﬀective annual interest rate 5%, ﬁnd the present value and the accumulated value of this annuity. 258 THE BASICS OF ANNUITY THEORY Solution. The present value of this annuity at time 0 year is PV = a10 a2 −5(1.05)−10 0.05s2 = 7.721735 1.85941 −5(1.05)−10 (1.05)2 −1 = 10.568. The accumulated value is 10.568(1.05)10 = $17.21 Now, consider an increasing annuity-due with payments 1, 2, 3, etc. occurring at the start of each interval of k interest conversion periods. Then PV = 1 + 2νk + 3ν2k + ··· + (n k −1 ) νn−2k + n kνn−k. (28.3) Multiply Equation (28.3) by νk to obtain νkPV = νk + 2ν2k + 3ν3k + ··· + (n k −1 ) νn−k + n kνn. (28.4) Now subtracting Equation (28.4) from Equation (28.3) we ﬁnd PV [1 −νk] = 1 + νk + ν2k + ··· + νn−k −n kνn = an ak −n kνn. Hence, PV = an ak −n kνn iak . The accumulated value at time t= n is AV = (1 + i)n an ak −n kνn iak . Example 28.2 Find the present value and the accumulated value of a 5-year annuity in which payments are made at the beginning of each half-year, with the ﬁrst payment of $50, the second payment of $100, the third payment of $150, and so on. Interest is 10% convertible quarterly. Solution. The present value is PV = 50 a20 a2 −10ν20 0.025a2 = 15.58916 1.92742 −10(1.025)−20 1 −(1.025)−2 = $2060.15. 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE259 The accumulated value at time t= 20 is (2060.15)(1.025)20 = $3375.80 Varying Annuities Payable More Frequently Than Interest is Convertible We next consider increasing annuities where payments are made more frequently than interest is convertible. We examine two diﬀerent types. The ﬁrst is the increasing mthly annuity where the increase occurs once per conversion period. The second is the mthly increasing mthly annuity where the increase takes place with each mthly payment. Annuities Payable mthly Consider ﬁrst an increasing annuity-immediate with constant payments during each interest con- version period with increases occurring only once per interest conversion period. Let the annuity be payable mthly with each payment in the ﬁrst period equal to 1 m, each payment in the second period equal to 2 m,··· , each payment in the nth period equal to n m as shown in Figure 28.1. Figure 28.1 Let i be the rate per one conversion period and i(m) be the nominal interest rate payable m times per interest conversion period. Making the mthly payments of k m in period k is the same as making one payment equal to the accumulated value of the mthly payments at the end of the period, that is, one payment of k msmj = k m ·(1 + j)m −1 j = k i i(m) where j = i(m) m . The time diagram of this is given in Figure 28.2. Figure 28.2 260 THE BASICS OF ANNUITY THEORY Consequently, the increasing annuity payable mthly is the same as an increasing annuity with P = Q= i i(m) which implies the present value (Ia)(m) n = i i(m) (Ia)n = ¨an −nνn i(m) . The accumulated value is (Is)(m) n = (1 + i)n(Ia)(m) n = ¨sn −n i(m) . Example 28.3 Determine the present value and the accumulated value of 1 at the end of each quarter in the ﬁrst year, 2 at the end of each quarter in the second year, 3 at the end of each quarter in the third year, 4 at the end of each quarter in the fourth year, and 5 at the end of each quarter in the ﬁfth year. The annual eﬀective interest rate is 4% Solution. The factor ( Ia)(4) 5 values a payment of 1/4 at the end of each quarter during the ﬁrst year, 2/4 at the end of each quarter during the second year, and so on. So, the present value is 4(Ia)(4) 5 = 4 · ¨a5 −5ν5 i(4) . But i(4) =4[(1.04) 1 4 −1] = 3.9414% ¨a5 =1 −(1.04)−5 0.04/1.04 = 4.629895 (Ia)(4) 5 =4.629895 −5(1.04)−5 0.039414 = 13.199996 4(Ia)(4) 5 =52.80. The accumulated value is (Is)(4) 5 = (1.04)5(Ia)(4) 5 = 1.045 ×52.80 = 64.24 In the case of an annuity-due, one can easily show that (I¨a)(m) n = ¨an −nνn d(m) and (I¨s)(m) n = (1 + i)n(I¨a)(m) n = ¨sn −n d(m) . 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE261 Example 28.4 Determine the present value and accumulated value of 10 at the start of each quarter in the ﬁrst year, 20 at the start of each quarter in the second year, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The present value is 40(I¨a)(4) 5 =40 ×¨a5 −5ν5 d(4) =40 ×4.629895 −5(1.04)−5 0.039029 =533.20 where d(4) = 4[1 −(1.04)−1 4 ] = 3.9029%. The accumulated value is (1.04)5 ×533.20 = 648.72 mthly Increasing mthly Payable Annuities Consider next a situation in which payments vary within each interest conversion period as shown in Figure 28.3. Such an annuity is referred to as mthly increasing mthly annuity. Figure 28.3 Denoting the present value of such an annuity by ( I(m)a)(m) n we ﬁnd (I(m)a)(m) n = 1 m2 [ν 1 m + 2ν 2 m + ··· + mnν mn m ]. (28.5) Multiply both sided by (1 + i) 1 m to obtain (1 + i) 1 m(I(m)a)(m) n = 1 m2 [1 + 2ν 1 m + ··· + mnνn−1 m]. (28.6) 262 THE BASICS OF ANNUITY THEORY Subtract Equation (28.5) from Equation (28.6) to obtain (I(m)a)(m) n [(1 + i) 1 m −1] = 1 m2 [1 + ν 1 m + ··· + νn−1 m −nmνn] = 1 m[¨a(m) n −nνn] (I(m)a)(m) n = ¨a(m) n −nνn m[(1 + i) 1 m −1] =¨a(m) n −nνn i(m) where ¨a(m) n = 1 −νn d(m) = i d(m) an. The accumulated value is (I(m)s)(m) n = (1 + i)n(I(m)a)(m) n = ¨s(m) n −n i(m) where ¨s(m) n = (1 + i)m −1 d(m) . Example 28.5 Determine the present value and the accumulated value of 1 at the end of the ﬁrst quarter, 2 at the end of the second quarter, 3 at the end of the third quarter, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The factor ( I(4)a)(4) 5 values a payment of 1 16 at the end of the ﬁrst quarter, 2 16 at the end of the second quarter, and so on. So the present value is 16(I(4)a)(4) 5 = 16 ×¨a(4) 5 −5ν5 i(4) = 16 ×4.56257 −5(1.04)−5 0.039414 = 183.87 and the accumulated value is (1.04)5 ×183.87 = 223.71 In the case of an annuity-due, it can be easily shown that (I(m)¨a)(m) n = ¨a(m) n −nνn d(m) 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE263 and (I(m)¨s)(m) n = (1 + i)n(I(m)¨a)(m) n = ¨s(m) n −n d(m) . Example 28.6 Determine the present value and the accumulated value of 1 at the start of the ﬁrst quarter, 2 at the start of the second quarter, 3 at the start of the third quarter, and so on for 5 years. The annual eﬀective interest rate is 4%. Solution. The present value is 16(I(5)¨a)(4) 5 = 16 ×¨a(4) 5 −5ν5 d(4) = 4.56257 −5(1.04)−5 0.03903 = 185.68 and the accumulated value is (1.04)5 ×185.68 = 225.91 Other Considerations Types of annuities like the one discussed above but that vary in geometric progression can be handled by expressing the annuity value as a summation of the present value or accumulated value of each payment. This summation is a geometric progression which can be directly evaluated. We illutrate this in the next example. Example 28.7 Find the accumulated value at the end of eight years of an annuity in which payments are made at the beginning of each quarter for four years. The ﬁrst payment is $3,000, and each of the other payments is 95% of the previous payment. Interest is credited at 6% convertible semiannually. Solution. We count periods in quarters of a year. If j is the interest rate per quarter then 1 + j = (1.03) 1 2 . The accumulated value at t= 8 ×4 = 32 is 3000[(1.03)16 + (0.95)(1.03)15.5 + (0.95)2(1.03)15 + ··· + (0.95)15(1.03)8.5] =3000(1.03)16 [ 1 + ( 0.95√ 1.03 ) + ( 0.95√ 1.03 )2 + ··· + ( 0.95√ 1.03 )15] =3000(1.03)16 · 1 − ( 0.95√ 1.03 )16 1 − ( 0.95√ 1.03 ) = $49,134.18 264 THE BASICS OF ANNUITY THEORY Example 28.8 Find an expression for the present value of a perpetuity which pays 1 at the end of the third year, 2 at the end of the sixth year, 3 at the end of the ninth year and so on. Solution. The present value satisﬁes the equation PV = ν3 + 2ν6 + 3ν9 + ··· Thus, PV (1 −ν3) = ν3 + ν6 + ν9 + ··· = ν3 1 −ν3 or PV = ν3 (1 −ν3)2 Some of the problems can be solved using the basic principles. Example 28.9 Scott deposits 1 at the beginning of each quarter in year 1, 2 at the beginning of each quarter in year 2, ··· , 8 at the beginning of each quarter in year 8. One quarter after the last deposit Scott withdraws the accumulated value of the fund and uses it to buy a perpetuity immediate with level payments of X at the end of each year. All calculations assume a nominal interest rate of 10% per annum compounded quarterly. Calculate X. Solution. Regard the annuity as an annuity-due with 8 annual payments: k¨a4 0.025 where 1 ≤k ≤8. The annual eﬀective rate of interest is 1 + i= (1.025)4. The accumulated value at time t= 32 is ¨a4 0.025(I¨s)8 i = ¨a4 0.025 [s9 i−9](1+i) i = 196.77. On the other hand, we have X (1.025)4 −1 = 196.77. Solving this equation for X we ﬁnd X = 20.43 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE265 Practice Problems Problem 28.1 (a) Find the total sum of all the payments in ( Ia)(12) 2 . (b) Find the total sum of all the payments in ( I(12)a)(12) 2 . Problem 28.2 Find an expression for ( I(m)a)(m) ∞ . Problem 28.3 An annuity pays $100 at the end of each month in the ﬁrst year, $200 at the end of each month in the second year, and continues to increase until it pays 1000 at the end of each month during the 10th year. Calculate the present value of the annuity at an annual eﬀective interest rate of 6%. Problem 28.4 A monthly annuity due pays $1 at the beginning of the ﬁrst month. Each subsequent payment increases by $1. The last payment is made at the beginning of the 240th month. Calculate the accumulated value of the annuity at the end of the 240th month using: (a) an interest rate of 6% compounded monthly; (b) an annual eﬀective interest rate of 6%. Problem 28.5 A 20 year annuity pays 10 at the beginning of each quarter during the ﬁrst year, 20 at the beginning of each quarter during the second year, etc with 200 being paid at the beginning of each quarter during the last year. Calculate the present value of the annuity assuming an annual eﬀective interest rate of 12%. Problem 28.6 A 20 year annuity pays 10 at the beginning of each quarter during the ﬁrst year, 20 at the beginning of each quarter during the second year, etc with 200 being paid at the beginning of each quarter during the last year. Calculate the accumulated value of the annuity assuming a nominal interest rate of 6% compounded monthly. Problem 28.7 Show that the present value of a perpetuity on which payments are 1 at the end of the 5th and 6th years, 2 at the end of the 7th and 8th years, 3 at the end of the 9th and 10th years, etc, is ν4 i−νd. 266 THE BASICS OF ANNUITY THEORY Problem 28.8 A perpetuity has payments at the end of each four-year period. The ﬁrst payment at the end of four years is 1. Each subsequent paymet is 5 more than the previous payment. It is known that v4 = 0.75. Calculate the present value of this perpetuity. Problem 28.9 A perpetuity provides payments every six months starting today. The ﬁrst payment is 1 and each payment is 3% greater than the immediately preceding payment. Find the present value of the perpetuity if the eﬀective rate of interest is 8% per annum. Problem 28.10 Find the accumulated value at the end of ten years of an annuity in which payments are made at the beginning of each half-year for ﬁve years. The ﬁrst payment is $2,000, and each of the other payments is 98% of the previous payment. Interest is credited at 10% convertible quarterly. Problem 28.11 Find the present value of payments of 5 now, 10 in 6 months, 15 in one year, 20 in 18 months, and so on for 6 years. The nominal discount rate is 12% convertible semiannually. Problem 28.12 Marlen invests $ X now in order to receive $5 in 2 months, $10 in 4 months, $15 in 6 months, and so on. The payments continue for 10 years. The annual eﬀective rate of interest is 8%. Determine X. Problem 28.13 A 10-year annuity has the following schedule of payments: $100 on each January 1, $200 on each April 1, $300 on each July 1, and $400 on each October 1. Show that the present value of this annuity on January 1 just before the ﬁrst payment is made is 1600¨a10 (I(4)¨a)(4) 1 . Problem 28.14 Calculate the accumulated value at time 15 years of payments of 35 at the start of every quarter during the ﬁrst year, 70 at the start of every quarter during the second year, 105 at the start of every quarter during the third year, and so on for 15 years. The nominal interest is 12% convertible quarterly. Problem 28.15 A perpetuity pays 1000 immediately. The second payment is 97% of the ﬁrst payment and is made at the end of the fourth year. Each subsequent payment is 97% of the previous payment and is paid four years after the previous payment. Calculate the present value of this annuity at an annual eﬀective rate of 8%. 28 VARYING ANNUITIES WITH PAYMENTS AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE267 Problem 28.16 A 20 year annuity pays 10 at the end of the ﬁrst quarter, 20 at the end of the second quarter, etc with each payment increasing by 10 until the last quarterly payment is made at the end of the 20th year. Calculate the present value of the annuity assuming a nominal interest rate of 6% compounded monthly. Problem 28.17 A 20 year annuity pays 10 at the end of the ﬁrst quarter, 20 at the end of the second quarter, etc with each payment increasing by 10 until the last quarterly payment is made at the end of the 20th year. Calculate the present value of the annuity assuming an annual eﬀective interest rate of 12%. Problem 28.18 Joan has won a lottery that pays 1000 per month in the ﬁrst year, 1100 per month in the second year, 1200 per month in the third year, and so on. Payments are made at the end of each month for 10 years. Using an eﬀective interest rate of 3% per annum, calculate the present value of this prize. Round your answer to the nearest integer. Problem 28.19 On his 65th birthday Smith would like to purchase a ten-year annuity-immediate that pays 6000 per month for the ﬁrst year, 5500 per month for the second year, 5000 per month for the third year, and so on. Using an annual eﬀective interest rate of 5%, how much will Smith pay for such an annuity? Problem 28.20 On Susan’s 65 th birthday, she elects to receive her retirement beneﬁt over 20 years at the rate of 2,000 at the end of each month. The monthly beneﬁt increases by 4% each year. Assuming an annual eﬀective interest rate of 8.16%, and letting j denote the equivalent monthly eﬀective interest rate, ﬁnd the present value on Susan’s 65 th birthday of her retirement beneﬁt. 268 THE BASICS OF ANNUITY THEORY 29 Continuous Varying Annuities In this section we look at annuities in which payments are being made continuously at a varying rate. Consider an annuity for n interest conversion periods in which payments are being made contin- uously at the rate f(t) at exact moment t and the interest rate is variable with variable force of interest δt. Then f(t)e− ∫t 0 δrdrdt is the present value of the payment f(t)dt made at exact moment t. Hence, the present value of this n−period continuous varying annuity is PV = ∫ n 0 f(t)e− ∫t 0 δrdrdt. (29.1) Example 29.1 Find an expression for the present value of a continuously increasing annuity with a term of nyears if the force of interest is δ and if the rate of payment at time t is t2 per annum. Solution. Using integration by parts process, we ﬁnd ∫ n 0 t2e−δtdt= −t2 δe−δt ]n 0 + 2 δ ∫ n 0 te−δtdt = −n2 δ e−δn − [2t δ2 e−δt ]n 0 + 2 δ2 ∫ n 0 e−δtdt = −n2 δ e−δn −2n δ2 e−δn − [2 δ3 e−δt ]n 0 = −n2 δ e−δn −2n δ2 e−δn − 2 δ3 e−δn + 2 δ3 = 2 δ3 −e−δn [n2 δ + 2n δ2 + 2 δ3 ] Under compound interest, i.e., δt = ln (1 +i), formula (29.1) becomes PV = ∫ n 0 f(t)νtdt. 29 CONTINUOUS VARYING ANNUITIES 269 Under compound interest and with f(t) = t (an increasing annuity), the present value is (Ia)n = ∫ n 0 tνtdt = tνt ln ν ⏐⏐⏐⏐ n 0 − ∫ n 0 νt ln νdt = tνt ln ν ⏐⏐⏐⏐ n 0 − νt (ln ν)2 ⏐⏐⏐⏐ n 0 = −nνn δ −νn δ2 + 1 δ2 =1 −νn δ2 −nνn δ =an −nνn δ and the accumulated value at time n years is (Is)n = (1 + i)n(Ia)n = sn −n δ . The above two formulas can be derived from the formulas of ( I(m)a)(m) n and (I(m)s)(m) n . Indeed, (Ia)n = lim m→∞ (I(m)a)(m) n = lim m→∞ ¨a(m) n −nνn i(m) = an −nνn δ and (Is)n = lim m→∞ (I(m)s)(m) n = lim m→∞ ¨s(m) n −n i(m) = sn −n δ . Example 29.2 Sam receives continuous payments at an annual rate of 8 t + 5 from time 0 to 10 years. The continuously compounded interest rate is 9%. (a) Determine the present value at time 0. (b) Determine the accumulated value at time 10 years. Solution. (a) The payment stream can be split into two parts so that the present value is 8(Ia)10 + 5a10 . 270 THE BASICS OF ANNUITY THEORY Since i=e0.09 −1 = 9.4174% a10 =1 −(1.094174)−10 0.09 = 6.59370 (Ia)10 =6.59370 −10(1.094174)−10 0.09 = 28.088592 we obtain 8(Ia)10 + 5a10 = 8 ×28.088592 + 5×6.59370 = 257.68. (b) The accumulated value at time 10 years is 257.68 ×(1.094174)10 = 633.78 Example 29.3 If δ= 0.06, calculate the present value of 10 year continuous annuity payable at a rate of t at time t. Solution. The present value is (Ia)10 = 1 −ν10 δ2 −10ν10 δ = 1 −e−10×0.06 0.062 −10e−10×0.06 0.06 = 33.86 For a continuous payable continuously increasing perpetuity (where f(t) = t), the present value at time 0 is (Ia)∞ = lim n→∞ an −nνn δ = lim n→∞ 1−(1+i)−n δ −n(1 + i)−n δ = 1 δ2 . Example 29.4 Determine the present value of a payment stream that pays a rate of 5 t at time t. The payments start at time 0 and they continue indeﬁnitely. The annual eﬀective interest rate is 7%. Solution. The present value is 5(Ia)∞ = 5 [ln (1.07)]2 = 1,092.25 We conclude this section by considering the case of a continuously decreasing continuously payable stream in which a continuous payment is received from time 0 to time nyears. The rate of payment 29 CONTINUOUS VARYING ANNUITIES 271 at time t is f(t) = n−t, and the force of interest is constant at δ. The present value is (Da)n =nan −(Ia)n =n1 −νn δ −an −nνn δ =n−an δ Example 29.5 Otto receives a payment at an annual rate of 10 −t from time 0 to time 10 years. The force of interest is 6%. Determine the present value of these payments at time 0. Solution. Since i= e0.06 −1 = 6.184% a10 = 1 −(1.06184)−10 0.06 = 7.5201 the present value is then (Da)10 = 10 −7.5201 0.06 = 41.33 Example 29.6 Using the information from the previous example, determine the accumulated value at time 10 years of the payments received by Otto. Solution. The accumulated value at time 10 years is (1.06184)10 ×(Da)10 = (1.06184)10 ×41.33 = 75.31 272 THE BASICS OF ANNUITY THEORY Practice Problems Problem 29.1 Calculate the present value at an annual eﬀective interest rate of 6% of a 10 year continuous annuity which pays at the rate of t2 per period at exact moment t. Problem 29.2 Calculate the accumulated value at a constant force of interest of 5% of a 20 year continuous annuity which pays at the rate of t+1 2 per period at exact moment t. Problem 29.3 Evaluate (Ia)∞ if δ= 0.08. Problem 29.4 Payments under a continuous perpetuity are made at the periodic rate of (1 + k)t at time t. The annual eﬀective rate of interest is i, where 0 < k < i.Find an expression for the present value of the perpetuity. Problem 29.5 A perpetuity is payable continuously at the annual rate of 1 + t2 at time t. If δ = 0.05, ﬁnd the present value of the perpetuity. Problem 29.6 A one-year deferred continuous varying annuity is payable for 13 years. The rate of payment at time tis t2 −1 per annum, and the force of interest at time tis (1 +t)−1.Find the present value of the annuity. Problem 29.7 If an = a and a2n = b, express (Ia)n in terms of a and b. Problem 29.8 A 10 year continuous annuity pays at a rate of t2 + 2t+ 1 at time t. Calculate the present value of this annuity if δt = (1 + t)−1. Problem 29.9 A 30 year continuous annuity pays at a rate of t at time t. If δ = 0.10, calculate the current value of the annuity after ten years. Problem 29.10 A 10 year continuous annuity pays at a rate of t at time t. If νt = .94t, calculate the accumulated value of the annuity after ten years. 29 CONTINUOUS VARYING ANNUITIES 273 Problem 29.11 Nancy receives a payment stream from time 2 to time 7 years that pays an annual rate of 2 t−3 at time t. The force of interest is constant at 6% over the period. Calculate the accumulated value at time 7 years. Problem 29.12 Find the ratio of the total payments made under ( Ia)10 during the second half of the term of the annuity to those made during the ﬁrst half. Problem 29.13 ‡ Payments are made to an account at a continuous rate of (8 k+ tk), where 0 ≤t ≤10. Interest is credited at a force of interest δt = 1 8+t. After 10 years, the account is worth 20,000. Calculate k. Problem 29.14 Show that (Ia)n <(I(m)a)(m) n <(Ia)(m) n whenever nδ <1. Hint: Use Problem 25.11. Problem 29.15 An investment fund is started with an initial deposit of 1 at time 0. New deposits are made continuously at the annual rate 1 + t at time t over the next n years. The force of interest at time t is given by δt = (1 + t)−1. Find the accumulated value in the fund at the end of n years. Problem 29.16 Money was deposited continuously at the rate of 5,000 per annum into a fund for 7 years. If the force of interest is 0.05 from t= 0 to t= 10, what will the fund balance be after 10 years? Problem 29.17 A 5−year annuity makes payments continuously at a ratef(t) = e ( t2 24 +t ) .The force of interest varies continuously at a rate of δt = t 12 . Find the present value of this annuity. 274 THE BASICS OF ANNUITY THEORY Rate of Return of an Investment In this chapter, we present a number of important results for using the theory of interest in more complex “real-world” contexts than considered in the earlier chapters. In this chapter, the concept of yield rates is introduced: A yield rate (also known as the internal rate of return) is that interest rate at which the present value of all returns from the investment is equal to the present value of all contributions into the investment. Thus, yield rates are solutions to the equation NPV (i) = 0 where NPV stands for the net present value and is the diﬀerence between the present value of all returns and that of all contributions. Techniques for measuring the internal rate of return will be discussed. Yield rates are used as an index to see how favorable or unfavorable a particular transaction is. When lower, yield rates favor the borrower and when higher they favor the lender. We point out here, as always the case thus far, that the eﬀect of taxes will be ignored throughout the discussions. 275 276 RATE OF RETURN OF AN INVESTMENT 30 Discounted Cash Flow Technique In this section we consider the question of proﬁtability of investment projects that involve cash ﬂows. More speciﬁcally, we look at the present values of projects consisting of cash ﬂows. Such projects can be considered as a generalization of annuities with any pattern of payments and/or withdrawals compared to annuities previously discussed where the annuities consisted of regular series of payments. The process of ﬁnding the present value of the type of projects consisting of a stream of cash ﬂows will be referred to as the discounted cash ﬂow technique, abbreviated by DCF. We examine two DCF measures for assessing a project: net present value and internal rate of return. Consider an investment project with contributions (outﬂow) C0,C1,··· ,Cn at time t0 <t1 <t2 < ··· <tn and returns (inﬂow) R0,R1,··· ,Rn at the same point of time. Let ct = Rt−Ct be the net change ( or net cash ﬂow) in the account at time t which can be positive or negative. From the vantage point of the lender, if ct >0, then there is a net cash deposit into the investment at time t. If ct < 0 then there is a net cash withdrawal from the investment at time t. Thus, if at time 1 a deposit of $1,000 is made and at the same time a withdrawal of $2,000 is made then c1 = 1000 −2000 = −1,000 so there is a net cash withdrawal of $1,000 from the investment at that time. We have adopted the vantage point of the lender. From the vantage point of the borrower the signs in the above discussion are switched. The following example illustrates these deﬁnitions. Example 30.1 In order to develop a new product and place it in the market for sale, a company has to invest $80,000 at the beginning of the year and then $10,000 for each of the next three years. The product is made available for sale in the fourth year. For that, a contribution of $20,000 must be made in the fourth year. The company incurs maintenance expenses of $2000 in each of the next ﬁve years. The project is expected to provide an investment return of $12,000 at the end of the fourth year, $30,000 at the end of the ﬁfth year, $40,000 at the end of the sixth year, $35,000 at the end of the seventh year, $25,000 at the end of the eigth year, $15,000 at the end of the ninth year and $8,000 at the end of the tenth year. After the tenth year, the product is withdrawn from the market. Create a chart to describe the cash ﬂows of this project. Solution. Table 30.1 describes the cash ﬂows of this project 30 DISCOUNTED CASH FLOW TECHNIQUE 277 Year Contributions Returns ct 0 80,000 0 −80,000 1 10,000 0 −10,000 2 10,000 0 −10,000 3 10,000 0 −10,000 4 20,000 12,000 −8,000 5 2,000 30,000 28,000 6 2,000 40,000 38,000 7 2,000 35,000 33,000 8 2,000 25,000 23,000 9 2,000 15,000 13,000 10 0 8,000 8,000 Table 30.1 Now, suppose that the rate of interest per period is i.This interest rate is sometimes referred to as the required return of the investment or the cost of capital. Using the discounted cash ﬂow technique, the net present value at rate i of the investment is deﬁned by NPV (i) = n∑ k=0 νtkctk, ν = (1 + i)−1. Thus, the net present value of a series of cash ﬂows is the present value of cash inﬂows minus the present value of cash outﬂows. Expressed in another way, NPV is the sum of the present values of the net cash ﬂows over n periods. The value of NPV (i) can be positive, negative, or zero depending on how large or small the value of i is. Example 30.2 Find the net present value of the investment discussed in the previous example. Solution. The net present value is NPV (i) = −80,000 −10,000ν−10,000ν2 −10,000ν3 −8,000ν4 + 28,000ν5 + 38,000ν6 + 33,000ν7 + 23,000ν8 + 13,000ν9 + 8,000ν10 Notice that NPV (0.03) = 1488.04 >0, NPV(0.032180) = 0, and NPV (0.04) = −5122.13 <0 The value of i for which NPV (i) = 0 is called the yield rate (or internal rate of return ); 278 RATE OF RETURN OF AN INVESTMENT that is, the yield rate is the rate of interest at which the present value of contributions from the investment is equal to the present value of returns into the investment . Stated diﬀerently, the yield rate is the rate that causes investment to break even. According to the eqution NPV (i) = 0 , yield rates are the same from either the borrower’s or lender’s perspective. They are totally determined by the cash ﬂows deﬁned in the transaction and their timing. Yield rates are often used to measure how favorable or unfavorable a transaction might be. From a lender’s perspective, a higher yield rate makes a transaction more favorable. From a borrower’s perspective, a lower yield rate makes a transaction more favorable. Based on ﬁnancial factors alone, projects with positive NPV are considered acceptable. Projects with NPV (i) < 0 should be rejected. When NPV (i) = 0, the investment would neither gain nor lose value; We should be indiﬀerent in the decision whether to accept or reject the project. Example 30.3 Determine the net present value of the investment project with the following cash ﬂow, at a cost of capital of 4.8%. Time 0 1 2 3 4 5 Flow −1000 70 70 70 70 1070 Solution. The answer is NPV (i) = −1000 + 70a5 + 1000ν5 = 95.78 Example 30.4 Assume that cash ﬂows from the construction and sale of an oﬃce building is as follows. Year 0 1 2 Amount −150000 −100000 300000 Given a 7% cost of capital, create a present value worksheet and show the net present value, NPV. Solution. t vt ct PV 0 1.0 −150000 −150000 1 1 1.07 = 0.935 −100000 −93500 2 1 1.072 = 0.873 300000 261900 NPV= 18400 30 DISCOUNTED CASH FLOW TECHNIQUE 279 Finding the yield rate may require the use of various approximation methods, since the equations that have to be solved may be polynomials of high degree. Example 30.5 An investment project has the following cash ﬂows: Year Contributions Returns 0 100 0 1 200 0 2 10 60 3 10 80 4 10 100 5 5 120 6 0 60 (a) Calculate the net present value at 15%. (b) Calculate the internal rate of return (i.e. the yield rate) on this investment. Solution. (a) NPV (0.15) = −100 −200(1.15)−1 + 50(1.15)−2 + 70(1.15)−3 + 90(1.15)−4 + 115(1.15)−5 + 60(1.15)−6 = −$55.51. (b) We must solve the equation −100 −200ν+ 50ν2 + 70ν3 + 90ν4 + 115ν5 + 60ν6 = 0.Using linear interpolation with the points (7% ,4.53) and (8%,−4.52) one ﬁnds i≈0.07 −4.53 × 0.08 −0.07 −4.52 −4.53 = 7.50% Yield rate needs not be unique as illustrated in the following example. Example 30.6 In exchange for receiving $230 at the end of one year, an investor pays $100 immediately and pays $132 at the end of two years. Find the yield rate. Solution. The net present value is NPV (i) = −100 −132ν2 + 230ν. To ﬁnd the IRR we must solve 132 ν2 − 230ν+ 100 = 0. Solving this equation by the quadratic formula, we ﬁnd i= 10% or i= 20% Consider the situation where a bank makes a loan to an individual. In this case the bank acts as a lender or investor, and the individual acts as the borrower. Payments that are received by the investor (cash inﬂows) are taken to be positive cash ﬂows. Payments that are made by the investor (cash outﬂows) are considered to be negative. For the borrower, the cash ﬂows will be of the opposite sign. 280 RATE OF RETURN OF AN INVESTMENT Example 30.7 A bank lends Diane $8,000 now. She repays $600 at the end of each quarter for 5 years. Find the annual eﬀective rate of return. Solution. Let j be the eﬀective IRR per quarter. Then j satisﬁes the equation −8000 + 600a20 j = 0 or 3(1 −(1 + j)−20) −40j = 0. Using linear interpolation with the points (4% ,0.030839) and (4.5%,−0.04392) we ﬁnd j ≈0.04 −0.030839 × 0.045 −0.04 −0.04392 −0.030839 = 0.042. Thus, the annual rate of return is i= 1.0424 −1 = 17.89% Yield rates need not be positive. If a yield rate is 0, then the investor(lender) received no return on investment. If a yield rate is negative, then the investor(lender) lost money on the investment. We will assume that such a negative yield rate satisﬁes −1 <i< 0.A yield rate i< −1 implies full loss of the investment. Example 30.8 Find the yield rates of Example 30.1 Solution. Using a scientiﬁc calculator one ﬁnds the two real solutions i= −1.714964 and i= 0.032180.Since, i> −1, the only yield rate is i= 3.218% 30 DISCOUNTED CASH FLOW TECHNIQUE 281 Practice Problems Problem 30.1 Assuming a nominal rate of interest, convertible quarterly, of 12%, ﬁnd the net present value of a project that requires an investment of $100,000 now, and returns $16,000 at the end of each of years 4 through 10. Problem 30.2 The internal rate of return for an investment with contributions of $3,000 at time 0 and $1,000 at time 1 and returns of $2,000 at time 1 and $4,000 at time 2 can be expressed as 1 n. Find n. Problem 30.3 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. Find NPV (0.10). Problem 30.4 An investment project has the following cash ﬂows: Year Contributions Returns 0 100,000 0 1 5,000 0 2 4,000 10,000 3 2,000 10,000 4 0 20,000 5 0 40,000 6 0 60,000 7 0 80,000 (a) Calculate the net present value at 15%. (b) Calculate the internal rate of return on this investment. Problem 30.5 You are the President of XYZ Manufacturing Company. You are considering building a new factory. The factory will require an investment of 100,000 immediately. It will also require an additional investment of 15,000 at the beginning of year 2 to initiate production. Finally, maintenance costs for the factory will be 5,000 per year at the beginning of years 3 through 6. The factory is expected to generate proﬁts of 10,000 at the end of year one, 15,000 at the end of year two, 20,000 at the end of year 3, and 30,000 at the end of years 4 through 6. Calculate the internal rate of return on the potential factory. 282 RATE OF RETURN OF AN INVESTMENT Problem 30.6 An investment project has the following cash ﬂows: Year Contributions Returns 0 100,000 0 1 0 10,000 2 0 20,000 3 0 30,000 4 0 20,000 5 0 10,000 Calculate the internal rate of return on this investment. Problem 30.7 What is the internal rate of return of a project that requires an investment of $1,000,000 now, and returns $150,000 at the end of each of years 1 through 15? Problem 30.8 Assuming an annual eﬀective discount rate of 6%, what is the net present value of a project that requires an investment of $75,000 now, and returns 2 ,000(13 −t) at times t = 1 ,2,··· ,12 (i.e, $24,000 at time 1, $22,000 at time 2, $20,000 at time 3, etc.)? Problem 30.9 What is the internal rate of return on a project that requires a $10,000 investment now, and returns $4,000 four years from now and $12,000 eight years from now? Problem 30.10 A project has the following out-ﬂows and in-ﬂows: out-ﬂows: 12,000 at t= 0, 6,000 at t= 3, 9,000 at t= 6 and 12,000 at t= 9 in-ﬂows: 3,000 at t= 2 through t= 14 plus 12,000 at t= 15. If your desired internal rate of return was 8%, would you accept this project? Problem 30.11 A company pays $120,000 to purchase a property. The company pays $3,000 at the end of each of the next 6 months to renovate the property. At the end of the eigth month, the company sells the property for $150,000. Find the net present value of this project for the company at an annual eﬀective rate of 8%. Problem 30.12 What is the project’s monthly internal rate of return of the previous problem? 30 DISCOUNTED CASH FLOW TECHNIQUE 283 Problem 30.13 A project requires an initial investment of $50,000. The project will generate net cash ﬂows of $15,000 at the end of the ﬁrst year, $40,000 at the end of the second year, and $10,000 at the end of the third year. The project’s cost of capital is 13%. According to the DCF technique, should you invest in the project? Problem 30.14 A ten-year investment project requires an initial investment of $100,000 at inception and mainte- nance expenses at the beginning of each year. The maintenance expense for the ﬁrst year is $3,000, and is anticipated to increase 6% each year thereafter. Projected annual returns from the project are $30,000 at the end of the ﬁrst year decreasing 4% per year thereafter. Find c6 = R6 −C6. Problem 30.15 Determine the net present value for a project that costs $104,000 and would yield after −tax cash ﬂows of $16,000 the ﬁrst year, $18,000 the second year, $21,000 the third year, $23,000 the fourth year, $27,000 the ﬁfth year, and $33,000 the sixth year. Your ﬁrm’s cost of capital is 12%. Problem 30.16 Determine the net present value for a project that costs $253,494.00 and is expected to yield after−tax cash ﬂows of $29,000 per year for the ﬁrst ten years, $37,000 per year for the next ten years, and $50,000 per year for the following ten years. Your ﬁrm’s cost of capital is 12%. Problem 30.17 An investment project has the following cash ﬂows: Year Contributions Returns 0 10,000 0 1 5,000 0 2 1,000 0 3 1,000 0 4 1,000 0 5 1,000 0 6 1,000 8,000 7 1,000 9,000 8 1,000 10,000 9 1,000 11,000 10 0 12,000 Find the yield rate of this investment project. 284 RATE OF RETURN OF AN INVESTMENT Problem 30.18 Suppose a sum of money could be invested in project A which pays a 10% eﬀective rate for six years, or be invested in project B which pays an 8% eﬀective rate for 12 years. (a) Which option we choose as an investor? (b) Find the eﬀective rate of interest necessary for the six years after projectAends so that investing in project B would be equivalent. Problem 30.19 ‡ Project P requires an investment of 4000 at time 0. The investment pays 2000 at time 1 and 4000 at time 2. Project Qrequires an investment of X at time 2. The investment pays 2000 at time 0 and 4000 at time 1. The net present values of the two projects are equal at an interest rate of 10%. Calculate X. Problem 30.20 X corporation must decide whether to introduce a new product line. The new product will have startup costs, operational costs, and incoming cash ﬂows over six years. This project will have an immediate (t= 0) cash outﬂow of $100,000 (which might include machinery, and employee training costs). Other cash outﬂows for years 1-6 are expected to be $5,000 per year. Cash inﬂows are expected to be $30,000 per year for years 1-6. All cash ﬂows are after-tax, and there are no cash ﬂows expected after year 6. The required rate of return is 10%. Determine the net present value at this rate. Problem 30.21 An investment account is established on which it is estimated that 8% can be earned over the next 20 years. If the interest each year is subject to income tax at a 25% tax rate, ﬁnd the percentage reduction in the accumulated interest at the end of 20 years. 31 UNIQUENESS OF IRR 285 31 Uniqueness of IRR As pointed out in Example 30.4, an internal rate of return may not be unique. In this section, we look at conditions under which the equation ∑n i=0 ctiνti = 0 has exactly one solution. For the ease of the analysis we choose time to be equally spaced so that NPV (i) = ∑n t=0 ctνt = 0, that is, the net present value function is a polynomial of degree n in ν. By the Fundamental Theorem of Algebra, NPV (i) = 0 has n roots, counting repeated roots and complex roots. Example 31.1 Consider an investment with C0 = 1,C1 = 0,C2 = 1.36,R0 = 0,R1 = 2.3,R2 = 0.Find the internal rate of return. Solution. We have c0 = −1,c1 = 2.3,c2 = −1.36. The internal rate of return is a solution to the equation −1 + 2.3ν−1.36ν2 = 0. Solving this equation using the quadratic formula we ﬁnd ν = 2.3±i √ 0.0111 2.72 so no real root for i Example 31.2 Consider an investment with C0 = 1,C1 = 0,C2 = 1.32,R0 = 0,R1 = 2.3,R2 = 0.Find the internal rate of return. Solution. We have c0 = −1,c1 = 2.3,c2 = −1.32. Thus, the equation of value at time t = 0 is −1 + 2.3ν− 1.32ν2 = 0.Solving this equation using the quadratic formula we ﬁndν = −2.3±0.1 −2.64 so either ν = 0.833 or ν = 0.91. Hence, either i= 20% or i= 10% so two internal rates of return Example 31.3 Consider an investment with C0 = 1 ,C1 = 0 ,C2 = 1 .2825,R0 = 0 ,R1 = 2 .3,R2 = 0 . Find the internal rate of return. Solution. We have c0 = −1,c1 = 2.3,c2 = −1.2825. Thus, the equation of value at time t= 0 is −1 + 2.3ν− 1.2825ν2 = 0. Solving this equation we ﬁnd ν = −2.3±0.4 −2.565 so either ν = 0.74 or ν = 1.053. Hence, either i= 35% or i= −5% What we would like to do next is to formulate conditions on the ct that guarantee a unique yield rate i> −1. A ﬁrst set of conditions is given in the following theorem. 286 RATE OF RETURN OF AN INVESTMENT Theorem 31.1 Let k be an integer such that 0 <k <n. Suppose that either (i) ct ≤0 for 0 ≤t≤k and ct ≥0 for k+ 1 ≤t≤n; or (ii) ct ≥0 for 0 ≤t≤k and ct ≤0 for k+ 1 ≤t≤n. Then there is a unique interest rate i> −1 such that NPV (i) = 0. This theorem states that a unique internal rate of return exists for a transaction in which the net payments are all of one sign for the ﬁrst portion of the transaction and then have the opposite sign for the remainder of the transaction. To prove the theorem, we require the concept of Descartes’ Rule of signs which is a technique for determining the number of positive or negative roots of a polynomial: If the terms of a single-variable polynomial with real coeﬃcients and a nonzero constant term are ordered by descending variable exponent, then the number of positive roots counting multiplicity of a polynomial f(x) is either equal to the number of sign changes between adjacent nonzero coeﬃcients, or less than it by a multiple of 2. The number of negative zeros of a polynomial f(x) counting multiplicity is equal to the number of positive zeros of f(−x), so it is the number of changes of signs of f(−x) or that number decreased by an even integer. Complex roots always come in pairs. That’s why the number of positive or number of negative roots must decrease by two. Example 31.4 Use Descartes’ rule of sign to ﬁnd the possible number of positive roots and negative roots of the polynomial f(x) = x2 + x+ 1. Solution. The polynomial f(x) has no sign change between adjacent coeﬃcients. Thus, f(x) has no positive roots. The polynomial f(−x) = x2 −x+ 1 has two sign changes between adjacent coeﬃcients so that f(x) has either two negative roots or zero negative roots. Thus, the two possibilities are either two negative roots or two complex roots Example 31.5 Use Descartes’ rule of sign to ﬁnd the possible number of positive roots and negative roots of the polynomial f(x) = x3 + x2 −x−1. Solution. The polynomial f(x) has one sign change between the second and third terms. Therefore it has at most one positive root. On the other hand, f(−x) = −x3 + x2 + x−1 has two sign changes, so f(x) has 2 or 0 negative roots. Since complex roots occur in conjugate, the two possibilities are either one positive root and two negative roots or one positive root and two complex roots. Thus, f(x) 31 UNIQUENESS OF IRR 287 has exactly one positive root Proof of Theorem 31.1. According to either (i) and (ii), the coeﬃcients of NPV (i) changes signs once. So by Descartes’ rule of signs, there will be at most one positive real root. If such a root exists, then ν >0 implies i> −1. For such values of i we deﬁne the function f(i) = (1 + i)kNPV (i) = k∑ t=0 ct(1 + i)k−t + n∑ t=k+1 ct(1 + i)k−t. Taking the ﬁrst derivative we ﬁnd f′(i) = k−1∑ t=0 ct(k−t)(1 + i)k−t−1 + n∑ t=k+1 ct(k−t)(1 + i)k−t−1. If (i) is satisﬁed then ct(k−t) ≤0 for 0 ≤t≤n.Hence, f′(i) ≤0 and the function f(i) is decreasing. If (ii) is satisﬁed then f′(i) ≥0 and f(i) is increasing. In case (i), the function f(i) is decreasing. Moreover, as i→−1+ the ﬁrst sum in f(i) approaches ck whereas the second sum approaches ∞. Hence, lim i→−1+ f(i) = ∞. Similarly, as i approaches +∞ the ﬁrst sum in f(i) approaches −∞whereas the second sum approaches 0. Hence, lim i→∞f(i) = −∞. Since f is decreasing, there is a unique i >−1 such that NPV (i) = 0 . A similar argument holds for case (ii) Example 31.6 A project requires an initial investment of $10,000 and it produces net cash ﬂows of $10,000 one year from now and $2,000 two years from now. Show that there is a unique internal rate of return. Solution. The net present value is NPV (i) = −10,000+10 ,000ν+2,000ν2.The result follows from Theorem 31.1(i) with k= 1 The following theorem establishes uniqueness of yield rates under a broader set of conditions than the one given in the previous theorem. Theorem 31.2 288 RATE OF RETURN OF AN INVESTMENT Let i> −1 be a solution to NPV (i) = 0. Suppose that B0 =c0 >0 B1 =c0(1 + i) + c1 >0 B2 =c0(1 + i)2 + c1(1 + i) + c2 >0 ... Bn−1 =c0(1 + i)n−1 + c1(1 + i)n−2 + ··· + cn−1 >0 Then (i) Bn = c0(1 + i)n + c1(1 + i)n−1 + ··· + cn = 0 (ii) i is unique. Proof. (i) We are given that ∑n t=0 ctνt = 0.This is an nth degree polynomial in ν and could be written as an nth degree polynomial in i by simply multiplying both sides by (1 + i)n to obtain c0(1 + i)n + c1(1 + i)n−1 + ··· + cn = 0. Thus, the assumption of (i) is established. (ii) Suppose that j satisﬁes NPV (j) = 0 and j >i.Then we have B′ 0 =c0 = B0 >0 B′ 1 =B′ 0(1 + j) + c1 = c0(1 + j) + c1 >B1 >0 B′ 2 =B′ 1(1 + j) + c2 = c0(1 + j)2 + c1(1 + j) + c2 >B2 >0 ... B′ n−1 =c0(1 + j)n−1 + c1(1 + j)n−2 + ··· + cn−1 >Bn−1 >0 B′ n =B′ n−1(1 + j) + cn = c0(1 + j)n + c1(1 + j)n−1 + ··· + cn >Bn = 0 The last inequality shows that j is not a yield rate, a contradiction. A similar argument holds for −1 <j <i. Hence, we conclude that j = i Remark 31.1 Note that Bt is the outstanding balance at timet.According to the theorem above, if the outstanding balance is positive throughout the period of investment, then the yield rate will be unique. Also, note that c0 > 0 and cn = −Bn−1(1 + i) < 0, but that ct for t = 1 ,2,··· ,n −1 may be either positive, negative, or zero. 31 UNIQUENESS OF IRR 289 Example 31.7 Show that we cannot guarantee uniqueness of the yield rate if c0 and cn have the same sign. Solution. The conclusion of uniqueness depends on the outstanding balance being positive at all time during the investment. If c0 >0 and cn >0 then the outstanding balance must be negative prior to t= n. The negative outstanding balance implies no guarantee of uniqueness. Similarly, if c0 < 0 and cn <0, the outstanding balance at the start of the investment is negative so that no guarantee of uniqueness of yield rate The discussion of this section focused on the existence of yield rates. However, it is possible that no yield rate exists or all yield rates are imaginary (See Example 31.1). Example 31.8 An investor is able to borrow $1,000 at 8% eﬀective for one year and immediately invest the $1,000 at 12% eﬀective for the same year. Find the investor’s yield rate on this transaction. Solution. The proﬁt at the end of the year is $40, but there is no yield rate, since the net investment is zero 290 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 31.1 Use Descartes’ rule of signs to determine the maximum number of real zeroes of the polynomial f(x) = x5 −x4 + 3x3 + 9x2 −x+ 5. Problem 31.2 Consider the following transaction: Payments of $100 now and $108.15 two years from now are equivalent to a payment of $208 one year from now. Find the absolute value of the diﬀerence of the two rates that result from the equation of value at time t= 0. Problem 31.3 An investor pays $100 immediately and $X at the end of two years in exchange for $200 at the end of one year. Find the range of X such that two yield rates exist which are equal in absolute value but opposite in sign. Problem 31.4 What is the yield rate on a transaction in which an investor makes payments of $100 immediately and $101 at the end of two years, in exchange for a payment of $200 at the end of one year? Problem 31.5 ‡ A company deposits 1,000 at the beginning of the ﬁrst year and 150 at the beginning of each subsequent year into perpetuity. In return the company receives payments at the end of each year forever. The ﬁrst payment is 100. Each subsequent payment increases by 5%. Calculate the company’s yield rate for this transaction. Problem 31.6 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. What is the maximum number of possible yield rates using Descartes’ rule of signs? Problem 31.7 An investment costs $85 and it will pay $100 in 5 years. Determine the internal rate of return for this investment. Problem 31.8 A project requires an initial investment of $50,000. The project will generate net cash ﬂows of $15,000 at the end of the ﬁrst year, $40,000 at the end of the second year, and $10,000 at the end of the third year. Show that an internal rate of return exists and is unique. 31 UNIQUENESS OF IRR 291 Problem 31.9 Suppose payments of 100 now and 109.20 two years from now are to be made in return for receiving 209 one year from now. Determine the internal rate of return. Problem 31.10 What is the internal rate of return (in terms of an eﬀective annual interest rate) on a project that requires a $1,000 investment now, and provides returns to the investor of $500 one year from now and $800 two years from now? Problem 31.11 Suppose a project requires you to invest $10,000 now and $20,000 one year from now. The project returns $29,000 six months from now. Find all the yield rates (internal rates of return) of this project. Express these yield rates as annual eﬀective rates. Problem 31.12 For the previous problem, ﬁnd the range of annual interest rates which will produce a net present value greater than zero. Problem 31.13 Two growing perpetuities, each with annual payments, have the same yield rate (internal rate of return). The ﬁrst perpetuity has an initial payment of $50 one year from now, and each subsequent annual payment increases by $5. The present value of this ﬁrst perpetuity is $2,000. The second perpetuity (a perpetuity-due) has an initial payment of $100 now, and each subsequent annual payment increases by 3%. Find the present value, now, of the second perpetuity. 292 RATE OF RETURN OF AN INVESTMENT 32 Interest Reinvested at a Diﬀerent Rate In this section we consider transactions where interest may be reinvested at a rate which may or may not be equal to the original investment rate. We analyze two situations in which reinvestment rates are directly taken into consideration. First, consider an investment of 1 for n periods at rate i where the interest is reinvested at rate j. The time diagram of this situation is shown in Figure 32.1. Figure 32.1 The accumulated value at the end of the n periods is equal to the principal plus the accumulated value of the annuity-immediate with periodic payments of i at the end of each period and periodic rate j. That is, AV = 1 + isnj. Note that in the case i= j,the formula reduces to the familiar formula AV = (1 +i)n.That is, for compound interest, the reinvestment rate is equal to the original rate. Example 32.1 100 is invested in a Fund A now. Fund A will pay interest at 10% each year. When the interest is paid in Fund A, it will be immediately removed and invested in Fund B paying 8 %. Calculate the sum of the amounts in Fund A and Fund B after 10 years. Solution. At the end of 10 years, Fund A will just have the original principal: $100. Fund B will have the accumulated value of the interest payments at the rate 8%: 10s10 0.08 = 10(1.0810 −1) 0.08 = 144.87. So the total amount of money in both funds is about $244.87 Example 32.2 Brown deposits 5000 in a 5-year investment that pays interest quarterly at 8%, compounded quar- terly. Upon receipt of each interest payment, Brown reinvests the interest in an account that earns 6%, compounded quarterly. Determine Brown’s yield rate over the 5 year period, as a nominal interest rate compounded quar- terly. 32 INTEREST REINVESTED AT A DIFFERENT RATE 293 Solution. The equation of value after 20 quarters is 5000 ( 1 + i(4) 4 )20 = 5000 + 5000(0.08)s20 0.015. Solving this equation we ﬁnd i(4) = 4 [1 + 0.08s20 0.015] 1 20 −1 = 7.68% Second, consider an investment of 1 at the end of each period for n periods at rate i where the interest is reinvested at rate j. The time diagram of this situation is shown in Figure 32.2. Figure 32.2 The accumulated value at the end of the n periods is the sum of the annuity payments and the accumulated value of the interest,i.e. AV = n+ i(Is)n−1 j = n+ i [snj −n j ] . Note that the above formula simpliﬁes to the familiar result AV = sn when j = i. Example 32.3 Payments of $1,000 are invested at the end of each year for 10 years. The payments earn interest at 7% eﬀective, and the interest is reinvested at 5% eﬀective. Find the (a) amount in the fund at the end of 10 years, (b) purchase price an investor(a buyer) must pay for a yield rate (to the seller) of 8% eﬀective. Solution. (a) The amount in the fund at the end of 10 years is 1000 ( 10 + 0.07 [s10 0.05 −10 0.05 ]) =1000 ( 10 + 0.07 [12.5779 −10 0.05 ]) =$13,609.06. (b) The purchase price is 13 ,609.06(1.08)−10 = $6,303.63 294 RATE OF RETURN OF AN INVESTMENT Example 32.4 Esther invests 100 at the end of each year for 12 years at an annual eﬀective interest rate of i.The interest payments are reinvested at an annual eﬀective rate of 5%. The accumulated value at the end of 12 years is 1748.40. Calculate i. Solution. The equation of value at time t= 12 is 1748.40 = 1200 + 100i [s12 0.05 −12 0.05 ] . Solving this equation for i we ﬁnd i= [1748.40 −1200](0.05) 100[s12 0.05 −12] = 7% If the payments of 1 are made at the beginning of each period (instead of at the end), the accumu- lated value at the end of the n periods is AV = n+ i(Is)nj = n+ i [sn+1 j −(n+ 1) j ] . Example 32.5 David pays 1,000 at the beginning of each year into a fund which earns 6%. Any interest earned is reinvested at 8%. Calculate the total that David will have at the end of 7 years. Solution. Note that payment are made at the beginning of the year (annuity-due). The answer is AV = 7,000 + 1000(0.06) [ 1.088−1 0.08 −8 0.08 ] = 8,977.47 An important consideration to a lender (investor) is the rate of repayment by the borrower. A faster rate of repayment by the borrower, results in a higher yield rate for the investor. This is illustrated in the following example. Example 32.6 Three loan repayment plans are described for a $3,000 loan over a 6-year period with an eﬀective rate of interest of 7.5%. If the repayments to the lender can be reinvested at an eﬀective rate of 32 INTEREST REINVESTED AT A DIFFERENT RATE 295 6%, ﬁnd the yield rates (for the lender) (a) if the entire loan plus accumulated interest is paid in one lump sum at the end of 6 years; (b) if interest is paid each year as accrued and the principal is repaid at the end of 6 years; (c) if the loan is repaid by level payments over the 6-year period. Solution. (a) The lump sum is 3000(1.075)6 = $4629.90.Since there is no repayment to reinvest during the 6- year period, the yield rate is obviously 7.5%. This answer can be also found by solving the equation 3000(1 + i)6 = 4629.90. (b) At the end of each year during the 5-year period, the payment is 3000(0.075) = $225. The accumulated value of all payments at the end of the 6-year period is 3000 + 225 s6 0.06 = 3000+225(6.9753) = $4569.44.To ﬁnd the yield rate, we solve the equation 3000(1+i)6 = 4569.44 → i= 0.07265 = 7.265% (c) Each year during the 6-year period, the payment isRwhere 3000 = Ra6 0.075 →R= 639.13.The accumulated value of all payments at the end of the 6-year period is 639.13s6 0.06 = 639.13(6.9753) = $4458.12. To ﬁnd the yield rate, we solve the equation 3000(1 + i)6 = 4458.12 →i = 0.06825 = 6.825% 296 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 32.1 Vanessa invests $5,000 at the end of each of the next 15 years and her investment earns interest at an annual eﬀective rate of 10%. The interest that she receives at the end of each year is reinvested and earns interest at an annual eﬀective rate of 5%. Calculate the accumulated value at time 15 years. Problem 32.2 Lauren has 100,000 invested in a fund earning 5%. Each year the interest is paid to Lauren who is able to invest the interest at 8%. Calculate the amount that Lauren will have at the end of 10 years. Problem 32.3 Julie has a sum of money, S, invested in a fund which earns 10%. Each year the fund pays the interest earned to Julie. Julie can only reinvest the interest at an annual eﬀective interest rate of 6%. After 20 years, Julie has 100,000 total including the amount in the fund plus the reinvested interest. Calculate S. Problem 32.4 Megan invests 500 at the end of each year. The investment earns 8% per year which is paid to Megan who reinvests the interest at 6%. Calculate how much Megan will have after 5 years. Problem 32.5 Thomas invests X into Fund 1 at the beginning of each year for 10 years. Fund 1 pays interest annually into Fund 2. Fund 1 earns 7% annually while Fund 2 earns 6% annually. After 10 years, Thomas has a total of 50,000. Calculate X. Problem 32.6 Chris is investing 1,000 at the beginning of each year into Fund A. Fund A earns interest at a nominal interest rate of 12% compounded monthly. Fund A pays Chris interest monthly. Chris reinvests that interest in Fund B earning an annual eﬀective rate of 8%. Calculate the total amount in Fund A and Fund B after 10 years. Problem 32.7 John invests 100 at the end of year one, 200 at the end of year two, etc until he invests 1,000 at the end of year ten. The investment goes into a bank account earning 4%. At the end of each year, the interest is paid into a second bank account earning 3%. Calculate the total amount John will have after 10 years. 32 INTEREST REINVESTED AT A DIFFERENT RATE 297 Problem 32.8 Kathy pays 1,000 at the end of each year into Fund A which earns interest at an annual eﬀective interest rate of i.At the end of each year, the interest earned is transferred to Fund B earning 10% interest. After 10 years. Kathy has 15,947.52. Calculate i. Problem 32.9 It is desired to accumulate a fund of 1,000 at the end of 10 years by equal deposits at the beginning of each year. If the deposits earn interest at 8% eﬀective, but the interest can only be reinvested at only 4% eﬀective, show that the deposit necessary is 1000 2s11 0.04 −12. Problem 32.10 A loan of 10,000 is being repaid with payments of 1,000 at the end of each year for 20 years. If each payment is immediately reinvested at 5% eﬀective, ﬁnd the yield rate of this investment. Problem 32.11 An investor purchases a 5-year ﬁnancial instrument having the following features: (i) The investor receives payments of 1,000 at the end of each year for 5 years. (ii) These payments earn interest at an eﬀective rate of 4% per annum. At the end of the year, this interest is reinvested at the eﬀective rate of 3% per annum. Find the purchase price to the investor to produce a yield rate of 4%. Problem 32.12 An investor deposits 1,000 at the beginning of each year for ﬁve years in a fund earning 5% eﬀective. The interest from this fund can be reinvested at only 4% eﬀective. Show that the total accumulated value at the end of ten years is 1250(s11 0.04 −s6 0.04 −1) Problem 32.13 A invests 2,000 at an eﬀective interest rate of 17% for 10 years. Interest is payable annually and is reinvested at an eﬀective rate of 11%. At the end of 10 years the accumulated interest is 5,685.48. B invests 150 at the end of each year for 20 years at an eﬀective interest rate of 14%. Interest is payable annually and is reinvested at an eﬀective rate of 11%. Find B’s accumulated interest at the end of 20 years. Problem 32.14 ‡ Victor invests 300 into a bank account at the beginning of each year for 20 years. The account pays out interest at the end of every year at an annual eﬀective interest rate of i%. The interest is reinvested at an annual eﬀective rate of (i 2 ) %. The yield rate on the entire investment over the 20 year period is 8% annual eﬀective. Determine i. 298 RATE OF RETURN OF AN INVESTMENT Problem 32.15 ‡ Sally lends 10,000 to Tim. Tim agrees to pay back the loan over 5 years with monthly payments payable at the end of each month. Sally can reinvest the monthly payments from Tim in a savings account paying interest at 6%, compounded monthly. The yield rate earned on Sally’s investment over the ﬁve-year period turned out to be 7.45%, compounded semi-annually. What nominal rate of interest, compounded monthly, did Sally charge Tim on the loan? Problem 32.16 Paul invests 1,000 at the beginning of each year for 10 years into a Fund A earning Y% each year. At the end of the year, any interest earned is moved to Fund B which earns 5%. Combining both Funds, Paul has a total of 11,924.08 at the end of 10 years. Determine Y. Problem 32.17 Lisa invests 1200 at the beginning of each year for 8 years into an account earning 8%. The interest earned each year is transferred to an account earning 6%. At the end of 8 years, the total amount is paid out. Calculate the amount an investor would pay now for the ﬁnal payout if the investor wanted a return of 10%. Problem 32.18 Amy invests 1000 at an eﬀective annual rate of 14% for 10 years. Interest is payable annually and is reinvested at an annual eﬀective rate of i. At the end of 10 years the accumulated interest is 2341.08. Bob invests 150 at the end of each year for 20 years at an annual eﬀective rate of 15%. Interest is payable annually and is reinvested at an annual eﬀective rate of i. Find Bob’s accumulated interest at the end of 20 years. Problem 32.19 ‡ An investor pays $100,000 today for a 4-year investment that returns cash ﬂows of $60,000 at the end of each of years 3 and 4. The cash ﬂows can be reinvested at 4.0% per annum eﬀective. If the rate of interest at which the investment is to be valued is 5.0%, what is the net present value of this investment today? Problem 32.20 ‡ An investor wishes to accumulate 10,000 at the end of 10 years by making level deposits at the beginning of each year. The deposits earn a 12% annual eﬀective rate of interest paid at the end of each year. The interest is immediately reinvested at an annual eﬀective interest rate of 8%. Calculate the level deposit. 32 INTEREST REINVESTED AT A DIFFERENT RATE 299 Problem 32.21 ‡ Payments of X are made at the beginning of each year for 20 years. These payments earn interest at the end of each year at an annual eﬀective rate of 8%. The interest is immediately reinvested at an annual eﬀective rate of 6%. At the end of 20 years, the accumulated value of the 20 payments and the reinvested interest is 5600. Calculate X. Problem 32.22 A deposit of 1 is made at the end of each year for 30 years into a bank account that pays interest at the end of each year at j per annum. Each interest payment is reinvested to earn an annual eﬀective interest rate of j 2 .The accumulated value of these interest payments at the end of 30 years is 72.88. Determine j. Problem 32.23 ‡ Susan invests Z at the end of each year for seven years at an annual eﬀective interest rate of 5%. The interest credited at the end of each year is reinvested at an annual eﬀective rate of 6%. The accumulated value at the end of seven years is X. Lori invests Z at the end of each year for 14 years at an annual eﬀective interest rate of 2.5%. The interest credited at the end of each year is reinvested at an annual eﬀective rate of 3%. The accumulated value at the end of 14 years is Y. Calculate Y/X. Problem 32.24 ‡ Jason deposits 3960 into a bank account at t= 0.The bank credits interest at the end of each year at a force of interest δt = 1 8 + t. Interest can be reinvested at an annual eﬀective rate of 7%. The total accumulated amount at time t= 3 is equal to X. Calculate X. Problem 32.25 ‡ Eric deposits 12 into a fund at time 0 and an additional 12 into the same fund at time 10. The fund credits interest at an annual eﬀective rate of i. Interest is payable annually and reinvested at an annual eﬀective rate of 0 .75i. At time 20 the accumulated amount of the reinvested interest payments is equal to 64. Calculate i,i> 0. Problem 32.26 ‡ At time t = 0, Sebastian invests 2000 in a fund earning 8% convertible quarterly, but payable an- nually. He reinvests each interest payment in individual separate funds each earning 9% convertible quarterly, but payable annually. The interest payments from the separate funds are accumulated in 300 RATE OF RETURN OF AN INVESTMENT a side fund that guarantees an annual eﬀective rate of 7%. Determine the total value of all funds at t= 10. Problem 32.27 A loan of 1000 is being repaid in 10 years by semiannual installments of 50, plus interest on the unpaid balance at 4% per annum compounded semiannually. The installments and interest payments are reinvested at 5% per annum compounded semiannually. Calculate the annual eﬀective yield rate of the loan. Problem 32.28 John invests a total of 10,000. He purchases an annuity with payments of 1000 at the beginning of each year for 10 years at an eﬀective annual interest rate of 8%. As annuity payments are received, they are reinvested at an eﬀective annual interest rate of 7%. The balance of the 10,000 is invested in a 10-year certiﬁcate of deposit with a nominal annual interest rate of 9%, compounded quarterly. Calculate the annual eﬀective yield rate on the entire 10,000 investment over the 10 year period. Problem 32.29 Brown deposits $1000 at the beginning of each year for 5 years, but makes no additional deposits in years 6 through 10. The fund pays interest annually at an annual eﬀective interest rate of 5%. Interest is reinvested at only a 4% annual eﬀective interest rate. What is Brown’s total accumulated value at the end of 10 years? Problem 32.30 You have $20,000 to invest. Alternative A allows you to invest your money at 8% compounded annually for 12 years. Alternative B pays you simple interest at the end of each year of 6%. At what rate must you reinvest the interest in alternative B to have the same accumulation after 12 years? Problem 32.31 Henry invests 2000 at the beginning of the year in a fund which credits interest at an annual eﬀective rate of 9%. Henry reinvests each interest payment in a separate fund, accumulating at an annual eﬀective rate of 8%. The interest payments from this fund accumulate in a bank account that guarantees an annual eﬀective rate of 7%. Determine the sum of the principal and interest at the end of 10 years. Problem 32.32 Jim borrowed 10,000 from Bank X at an annual eﬀective rate of 8%. He agreed to repay the bank with ﬁve level annual installments at the end of each year. At the same time, he also borrowed 32 INTEREST REINVESTED AT A DIFFERENT RATE 301 15,000 from Bank Y at an annual eﬀective rate of 7.5%. He agreed to repay this loan with ﬁve level annual installments at the end of each year. He lent the 25,000 to Wayne immediately in exchange for four annual level repayments at the end of each year, at an annual eﬀective rate of 8.5%. Jim can only reinvest the proceeds at an annual eﬀective rate of 6%. Immediately after repaying the loans to the banks in full, determine how much Jim has left. 302 RATE OF RETURN OF AN INVESTMENT 33 Interest Measurement of a Fund: Dollar-Weighted Inter- est Rate An investment fund usually is a ﬁrm that invests the pooled funds of investors for a fee. In this section we examine a method of determining the interest rate earned by an investment fund. In practice, it is common for a fund to be incremented with new principal deposits, decremented with principal withdrawals, and incremented with interest earnings many times throughout a pe- riod. Since these occurrences are often at irregular intervals, we devise notation for the purpose of obtaining the eﬀective rate of interest i over one measurement period: A = the amount in the fund at the beginning of the period,i.e. t= 0. B = the amount in the fund at the end of the period, i.e. t= 1. I = the amount of interest earned during the period. ct = the net amount of principal contributed at time t (inﬂow less outﬂow at time t) where 0 ≤t≤1. C = ∑ tct = total net amount of principal contributed during the period (if negative, this is an indication of net withdrawal). The amount at the end of a period is equal to the amount at the beginning of the period plus the net principal contributions plus the interest earned: B = A+ C+ I. (33.1) To be consistent with the deﬁnition of the eﬀective rate we will assume that all the interest earned I is received at the end of the period. Assuming compound interest i throughout the period, the exact equation of value for the interest earned over the period 0 ≤t≤1 is I = iA+ ∑ 0≤t≤1 ct[(1 + i)1−t −1]. (33.2) That is, the amount of interest earned is the sum of amounts of interest earned by each individual contribution, plus the amount of interest earned on the balance at the beginning. Note that (1 + i)1−t −1 is the eﬀective rate for period from t to 1,i.e., A(1) −A(t) A(t) = A(1) A(t) −1 = (1 + i)1−t −1. Substituting equation(33.2) into equation (33.1) we ﬁnd the equation of value B = A(1 + i) + ∑ 0≤t≤1 ct(1 + i)1−t. (33.3) 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE303 The interest isatisfying equation (33.3) is called the dollar-weighted rate of interest. Note that the required return of the investment, cost of capital and dollar weighted rate of return stand for the same quantity. Example 33.1 At the beginning of a year, an investment fund was established with an initial deposit of $3,000. At the end of six months, a new deposit of $1,500 was made. Withdrawals of $500 and $800 were made at the end of four months and eight months respectively. The amount in the fund at the end of the year is $3,876. Set up the equation of value to calculate the dollar-weighted rate of interest. Solution. The equation of value is 3,000(1 + i) + 1,500(1 + i)0.5 −500(1 + i)(1−4 12 ) −800(1 + i)(1−8 12 ) = 3,876 Finding ifrom the equation (33.2) requires approximation methods and this is not an easy problem. However, in practice one uses a simple interest approximation (1 + i)1−t −1 = 1 + (1−t)i−1 = (1 −t)i obtaining I ≈iA+ ∑ 0≤t≤1 ct(1 −t)i which leads to the approximation i≈ I A+ ∑ 0≤t≤1 ct(1 −t). (33.4) It is shown that this approximation is very good as long as the ct’s are small in relation to Awhich is often the case in practice. However, if the c′ ts are not small in relation to A, then the error can become signiﬁcant. The denominator in the above approximation is commonly called the exposure associated with i. Alternatively, by using equation (33.3) one ﬁnds i by solving the equation B = A(1 + i) + ∑ 0≤t≤1 ct[1 + (1−t)i]. Example 33.2 Using Example 33.1, ﬁnd the approximate eﬀective rate of interest earned by the fund during the year using the dollar-weighted rate of return formula. 304 RATE OF RETURN OF AN INVESTMENT Solution. The interest earned I is computed to be I = 3876 + 500 + 800−(3000 + 1500) = 676. The exposure associated with i is 3000 + (−500) ( 1 −1 3 ) + (1500) ( 1 −1 2 ) + (−800) ( 1 −2 3 ) = 3150. Thus, the approximate eﬀective rate is 676 3150 = 0.2146 = 21.46%. Alternatively, i can be found by solving the equation 3,000(1 + i) + 1,500(1 + 0.5i) −500(1 + 2 3i) −800(1 + i 3) = 3,876 giving i= 21.46% Formula (33.4) is in a form which can be directly calculated. However, it is tedious to use it because of the summation term in the denominator. Therefore, estimating the summation is fa- vored. One way for doing that is to assume that the net principal contributions C occur at time k that can be approximated by the arithmetic weighted average k= 1 C ∑ 0≤t≤1 t·ct obtaining a simpler approximation i≈ I A+ (1 −k)C = I A+ (1 −k)(B−A−I) = I kA+ (1 −k)B−(1 −k)I. Example 33.3 A fund has 100,000 on January 1 and 125,000 on December 31. Interest earned by the fund during the year totaled 10,000. Calculate the net yield earned by the fund assuming that the net contributions occurred on April 1. 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE305 Solution. Using the formula, i≈ I [kA+(1−k)B−(1−k)I] , we have i≈ 10,000 [(1/4)100,000 + (3/4)125,000 −(3/4)10,000] = 10,000 111,250 = 8.98876404% One useful special case is when the principal deposits or withdrawals occur uniformly throughout the period. Thus, on average, we might assume that net principal contributions occur at timek= 1 2 , in which case we have i≈ I 0.5A+ 0.5B−0.5I = 2I A+ B−I. (33.5) This formula has been used for many years by insurance company regulators in both Canada and the US to allow them to get a feel for the investment returns a particular company is obtaining. Example 33.4 You have a mutual fund. Its balance was $10,000 on 31/12/1998. You made monthly contributions of $100 at the start of each month and the ﬁnal balance was $12,000 at 31/12/1999. What was you approximate return during the year? Solution. We are given that A = $10,000, B= $12,000, and C = $1,200. Thus, I = B−(A+ C) = $800 and i= 2(800) 10,000 + 12,000 −800 = 7.54717% Example 33.5 Find the eﬀective rate of interest earned during a calendar year by an insurance company with the following data: Assets, beginning of year 10,000,000 Premium income 1,000,000 Gross investment income 530,000 Policy beneﬁts 420,000 Investment expenses 20,000 Other expenses 180,000 306 RATE OF RETURN OF AN INVESTMENT Solution. We are given A=10,000,000 B =10,000,000 + 1,000,000 + 530,000 −(420,000 + 20,000 + 180,000) =10,910,000 I =530,000 −20,000 = 510,000 Thus, i= 2I A+ B−I = 2(510,000) 10,000,000 + 10,910,000 −510,000 = 5% 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE307 Practice Problems Problem 33.1 A fund has 10,000 at the start of the year. During the year $5,000 is added to the fund and $2,000 is removed. The interest earned during the year is $1,000. Which of the following are true? (I) The amount in the fund at the end of the year is $14,000. (II) If we assume that any deposits and withdrawals occur uniformly throughout the year, i is approximately 8.33%. (III) If the deposit was made on April 1 and the withdrawal was made on August 1, then i is approximately 7.74%. Problem 33.2 The funds of an insurance company at the beginning of the year were $500,000 and at the end of the year $680,000. Gross interest earned was $60,000, against which there were investment expenses of $5,000. Find the net eﬀective rate of interest yielded by the fund. Problem 33.3 A fund earning 4% eﬀective has a balance of $1,000 at the beginning of the year. If $200 is added to the fund at the end of three months and if $300 is withdrawn from the fund at the end of nine months, ﬁnd the ending balance under the assumption that simple interest approximation is used. Problem 33.4 At the beginning of the year an investment fund was established with an initial deposit of $1,000. A new deposit of $500 was made at the end of four months. Withdrawals of $200 and $100 were made at the end of six months and eight months, respectively. The amount in the fund at the end of the year is $1,272. Find the approximate eﬀective rate of interest earned by the fund during the year using the dollar-weighted rate of return formula. Problem 33.5 ‡ An association had a fund balance of 75 on January 1 and 60 on December 31. At the end of every month during the year, the association deposited 10 from membership fees. There were withdrawals of 5 on February 28, 25 on June 30, 80 on October 15, and 35 on October 31. Calculate the dollar-weighted rate of return for the year. Problem 33.6 ‡ An insurance company earned a simple rate of interest of 8% over the last calendar year based on the following information: 308 RATE OF RETURN OF AN INVESTMENT Assets, beginning of year 25,000,000 Sales revenue X Net investment income 2,000,000 Salaries paid 2,200,000 Other expenses paid 750,000 All cash ﬂows occur at the middle of the year. Calculate the eﬀective yield rate. Problem 33.7 ‡ At the beginning of the year, an investment fund was established with an initial deposit of 1,000. A new deposit of 1,000 was made at the end of 4 months. Withdrawals of 200 and 500 were made at the end of 6 months and 8 months, respectively. The amount in the fund at the end of the year is 1,560. Calculate the dollar-weighted yield rate earned by the fund during the year. Problem 33.8 On January 1, an investment account is worth $500,000. On April 1, the account value has increased to $530,000, and $120,000 of new principal is deposited. On August 1, the account value has decreased to $575,000, and $250,000 is withdrawn. On January 1 of the following year, the account value is $400,000. Compute the yield rate using the dollar-weighted method. Problem 33.9 A fund has 10,000 at the beginning of the year and 12,000 at the end of the year. Net contributions of 1,000 were made into the fund during year. Calculate the net yield earned by the fund assuming that the net contributions were contributed uniformly throughout the year. Problem 33.10 A fund has 100,000 on January 1 and 125,000 on December 31. Interest earned by the fund during the year totaled 10,000. The net yield earned by the fund during the year was 9.6385%. Two contributions were made to the fund during the year and there were no withdrawals. The contributions were for equal amounts made two months apart. Determine the date of the ﬁrst contribution. Problem 33.11 An investor fund has a balance on January 1 of $273,000 and a balance on December 31 of $372,000. The amount of interest earned during the year was $18,000 and the computed yield rate on the fund was 6%. What was the average date for contributions to and withdrawals from the fund? 33 INTEREST MEASUREMENT OF A FUND: DOLLAR-WEIGHTED INTEREST RATE309 Problem 33.12 You deposited 5,000 into a fund on January 1, 2002. On March 1,2002, you withdrew 1,000 from the fund. On August 1, 2002, you deposited 700 into the fund. On October 1, 2002, you deposited another 750 into the fund. If your annual dollar-weighted rate of return on your investment was 12.00%, what was your fund worth on December 31, 2002? Problem 33.13 At the beginning of the year a fund has deposits of 10,000. A deposit of 1000 is made after 3 months and a withdrawal of 2000 is made after 9 months. The amount in the fund at the end of the year is 9500. Find the dollar weighted rate of interest (a) using formula (33.3) (b) using formula (33.4) (c) using formula (33.5). Problem 33.14 Deposits of 10000 are made into an investment fund at time zero and 1. The fund balance at time 2 is 21000. Compute the eﬀective yield using the dollar weighted method. Problem 33.15 An investor puts $100 into a mutual fund in the ﬁrst year and $50 in the second year. At the end of the ﬁrst year the mutual pays a dividend of $10. The investor sells the holdings in the mutual fund at the end of the second year for $180. Find the dollar weighted return. Problem 33.16 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the exact dollar weighted rate of return( i.e. using (33.3)). Problem 33.17 A fund has $2,000,000 at the beginning of the year. On June 1 and September 1, two withdrawals of $100,000 each were made. On November 1, a contribution of $20,000 was made. The value in the fund at the end of the year is $1,900,000. Find (a) the exact dollar weighted rate of return; (b) the dollar weigthed rate of return using formula (33.4). Problem 33.18 The following table shows the history of account balances, deposits, and withdrawals for an account 310 RATE OF RETURN OF AN INVESTMENT during 2008: Date Value before dep/withd Dep/Withd January 1,2008 10,000 0 March 1,2008 9,500 3,000 June 1,2008 14,000 −2,000 September 1,2008 12,000 −2,000 January 1,2009 12,000 What is this account’s dollar-weighted rate of return?(Assume 30-day months.) Problem 33.19 On January 1, 2000, the balance in account is $25200. On April 1, 2000, $500 are deposited in this account and on July 1, 2001, a withdraw of $1000 is made. The balance in the account on October 1, 2001 is $25900. What is the annual rate of interest in this account according with the dollarweighted method? Problem 33.20 You invest $10,000 in a fund on January 1, 2008. On May 1, 2008, you withdraw $3,000 from the fund. On November 1, 2008, you deposit $4,000 into the fund. On December 31, 2008, your fund is worth $13,500. What was the annual dollar-weighted rate of return on your investment? Problem 33.21 You invest $5,000 in a fund on 1/1/07. On 3/1/07, you withdraw X from the fund. On 7/1/07, you deposit $2,500 into the fund. On 9/1/07, you withdraw $1,000 from the fund. On 12/31/07, your fund is worth $5,700. The annual dollar −weighted rate of return on your investment was 12.0%. Find X. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST311 34 Interest Measurement of a Fund: Time-Weighted Rate of Interest The dollar-weighted rate of interest depends on the precise timing and amount of the cash ﬂows. In practice, professional fund managers who direct investment funds have no control over the timing or amounts of the external cash ﬂows. Therefore, if we are comparing the performance of diﬀerent fund managers, the dollar weighted rate of interest doesn’t always provide a fair comparison. In this section, we consider an alternative measure of performance that does not depend on the size or the timing of the cash ﬂows, namely the time-weighted rate of interest also known as the time-weighted rate of return. Consider the following one-year investment. An amount X is invested in a fund at the beginning of the year. In six months, the fund is worth X 2 at which time the investor can decide to add to the fund, to withdraw from the fund, or to do nothing. At the end of the year, the fund balance is double the balance at six months. We next examine three situations related to the above investment: Consider three investors A,B, and C. Investor A initially invests X = $1,000. At the end of six months his investment is worth $500. He decideds not to withdraw from the fund or deposits into the fund. His fund at the end of the year worths $1,000. Since the interest earned is I = 0, the dollar-weighted rate of interest is i= 0. Next, investor B invests initially X = $1 ,000. Again, after six months the balance is $500. At the end of six months the investor withdraws half the fund balance (i.e. $250). His balance in the fund at the end of the year is $500. The equation of value at t = 1 for this transaction is 1000(1 +i) −250(1 +i) 1 2 = 500.Solving this quadratic equation we ﬁnd (1 + i) 1 2 = 0.84307 and this implies a yield rate of i= −0.2892 or −28.92%. Finally, investor C invests initially X = $1 ,000 and at six months deposits an amount equal to the fund balance(i.e. $500). Hence, his balance at the end of the year is $2,000. The equation of value at time t = 1 is 1000(1 + i) + 500(1 + i) 1 2 = 2000. Solving this quadratic equation we ﬁnd (1 + i) 1 2 = 1.18614. Hence, the yield rate in this case is i= 0.4069 or 40.69%. The yield rate for investor C is so much better than the yield rate for either investor A or investor B, because of the decisions made by each investor. The dollar-weighted rate of interest measures both the behavior of the fund and the skills of the investor. If the investor did not perform any transactions, then the dollar-weighted rate of interest would be close to 0%. Now how can we evaluate the decisions made by the fund manager? In the illustration above, we ﬁnd the yield rate for the ﬁrst six months to be j1 = −50% and for the second six months to be j2 = 100%. If i is the yield rate for the entire year then 1 + i= (1 + j1)(1 + j2) = 1 312 RATE OF RETURN OF AN INVESTMENT which implies that i = 0 and this is regardless of when cash is deposited or withdrawn. This indicates that the manager did a poor job of maintaining the fund. Yield rates computed by considering only changes in interest rate over time (which is what was previously done to evaluate the fund manager’s performance) are called time-weighted rates of interest. We can generalize the above approach as follows. Suppose m−1 deposits or withdrawals are made during a year at times t1,t2,··· ,tm−1 (so that no contributions at t0 = 0 and tm = 1). Thus, the year is divided into m subintervals. For k = 1,2,··· ,m we let jk be the yield rate over the kth subinterval. For k = 1,··· ,m −1, let Ctk be the net contribution at exact time tk and Btk the value of the fund before the contribution at time tk. Note that C0 = Cm = 0 and B0 is the initial investment and B1 is the value of the fund at the end of the year. The yield rate of the fund from time tk−1 to time tk satisﬁes the equation of value Btk = (1 + jk)(Btk−1 + Ctk−1) or 1 + jk = Btk Btk−1 + Ctk−1 , k = 1,2,··· ,m. The overall yield rate i for the entire year is given by 1 + i= (1 + j1)(1 + j2) ···(1 + jm) or i= (1 + j1)(1 + j2) ···(1 + jm) −1. We call i the time-weighted rate of return. Example 34.1 Suppose that an investor makes a series of payments and withdrawals, as follows: Date Flow Balance before Balance after 01/01/2003 0 100,000 100,000 30/06/2003 +1,000,000 74,681 1,074,681 31/12/2003 0 1,474,081 1,474,081 (a) Compute the dollar-weighted rate of interest that the investor has realized. (b) Compute the time-weigted rate of interest. Solution. (a) We assume that the contribution of $1,000,000 occurred exactly in the middle of the year. We 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST313 have A = $100 ,000, B = $1 ,474,081, C = $1 ,000,000. Hence, I = 1 ,474,081 −1,000,000 − 100,000 = 374,081 and i= 2(374,081) 100,000 + 1,474,081 −374,081 = 62.35%. (b) We have i= (74,681 100,000 )(1,474,081 1,074,681 ) −1 = 2.44% It follows from the previous example that the time-weighted method does not provide a valid measure of yield rate as does the dollar weighted method. However, the time-weighted method does provide a better indicator of underlying investment performance than the dollar-weighted method. Example 34.2 Your balance at time 0 is $2,000. At time t= 1 3 ,the balance has grown to $2,500 and a contribution of $1,000 is made. At time t= 2 3 , the balance has dropped to $3,000 and a withdrawal of $1,500 is made. At the end of the year, the fund balance is $2,000. What is the time weighted rate of return? Solution. We create the following chart k tk Btk Ctk jk 0 0.0000 $2,000 $0 0% 1 1/3 $2,500 $1,000 25.00% 2 2/3 $3,000 -$1,500 -14.29% 3 1 $2,000 $0 33.33% The time weighted rate of return is i= (1.25)(.8571)(1.3333) −1 = .428464 Example 34.3 You are given the following information about an investment account: Date June 30,01 Sep. 30,01 Dec. 31,01 March 31,02 June 30,02 Bk 12,000 10,000 14,000 13,000 X Deposit 2,000 2,000 Withdrawal 2,000 2,000 If the eﬀective annual dollar-weighted rate of return from June 30, 2001 to June 30, 2002 was exactly 10%, what was the eﬀective annual time-weighted rate of return during that same period? 314 RATE OF RETURN OF AN INVESTMENT Solution. First, we ﬁnd X : X = 10,000(1.1) + 2,000(1.1)0.75 −2,000(1.1)0.5 + 2,000(1.1)0.25 = 13,098.81 The time-weighted rate of return is then i= 10,000 12,000 −2,000 · 14,000 10,000 + 2,000 · 13,000 14,000 −2,000 · 13,098.81 13,000 −2,000 −1 = 10.37% 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST315 Practice Problems Problem 34.1 Which of the following are true? (I) Time-weighted method provides better indicator for the fund underlying performance than the dollar-weighted method. (II) Time weighted rates of interest will always be higher than dollar weighted rates of interest. (III) Dollar weighted rate of interest provide better indicators of underlying investment performance than do time weighted rates of interest. (IV) Dollar weighted rates of interest provide a valid measure of the actual investment results. Problem 34.2 A fund has 1,000 at beginning of the year. Half way through the year, the fund value has increased to 1,200 and an additional 1,000 is invested. At the end of the year, the fund has a value of 2,100. (a) Calculate the exact dollar weighted rate of return using compound interest. (b) Calculate the estimated dollar weighted rate of return using the simple interest assumptions. (c) Calculate the time weighted rate of return. Problem 34.3 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the time weighted rate of return. Problem 34.4 A fund has 100,000 at the start of the year. Six months later, the fund has a value of 75,000 at which time, Stuart adds an additional 75,000 to the fund. At the end of the year, the fund has a value of 200,000. Calculate the exact dollar weighted rate of return. Problem 34.5 A fund has 10,000 at the start of the year. Six months later, the fund has a value of 15,000 at which time, Stuart removes 5,000 from the fund. At the end of the year, the fund has a value of 10,000. Calculate the exact dollar weighted rate of return less the time weighted rate of return. Problem 34.6 ‡ You are given the following information about an investment account: 316 RATE OF RETURN OF AN INVESTMENT Date Value Immediately Deposit Before Deposit January 1 10 July 1 12 X December 31 X Over the year, the time-weighted return is 0%, and the dollar-weighted (money-weighted) return is Y. Determine Y. Problem 34.7 ‡ An investor deposits 50 in an investment account on January 1 . The following summarizes the activity in the account during the year: Date Value Immediately Deposit Before Deposit March 15 40 20 June 1 80 80 October 1 175 75 On June 30, the value of the account is 157.50 . On December 31, the value of the account is X. Using the time-weighted method, the equivalent annual eﬀective yield during the ﬁrst 6 months is equal to the (time-weighted) annual eﬀective yield during the entire 1-year period. Calculate X. Problem 34.8 On January 1, an investment account is worth $100,000. On May 1, the account value has increased to $112,000, and $30,000 of new principal is deposited. On November 1, the account value has decreased to $125,000, and $42,000 is withdrawn. On January 1 of the following year, the account value is $100,000. Compute the yield rate using (a) the dollar-weighted method and (b) the time-weighted method. Problem 34.9 In Problem 34.8, change May 1 to June 1 and November 1 to October 1. (a) Would the yield rate change when computed by the dollar-weighted method? (b) Would the yield rate change when computed by the time-weighted method? Problem 34.10 In Problem 34.8, assume that everything is unchanged except that an additional $5,000 is withdrawn on July 1. (a) Would the yield rate change when computed by the dollar-weighted method? (b) Explain why the yield rate cannot be computed by the time-weighted method. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST317 Problem 34.11 A mutual fund account has the balance $100 on January 1. On July 1 (exactly in the middle of the year) the investor deposits $250 into the account. The balance of the mutual fund at the end of the year is $400. (a) Calculate the exact dollar-weighted annual rate of interest for the year. (b) Suppose that the balance of the fund immediately before the deposit on July 1 is $120. Calculate the time-weighted annual rate of interest for the year. Problem 34.12 You are given the following information about two funds: Fund X: January 1, 2005: Balance 50,000 May 1, 2005: Deposit 24,000; Pre-deposit Balance 50,000 November 1, 2005: Withdrawal 36,000; Pre-withdrawal Balance 77,310 December 31, 2005: Balance 43,100 Fund Y: January 1, 2005: Balance 100,000 July 1, 2005: Withdrawal 15,000; Pre-withdrawal Balance 105,000 December 31, 2005: Balance F. Fund Y’s time-weighted rate of return in 2005 is equal to Fund X’s dollar-weighted rate of return in 2005. Calculate F. Problem 34.13 You invested 10,000 in a fund on January 1, 2000. On April 1, 2000, your fund was worth 11,000, and you added 2,000 to it. On June 1, 2000, your fund was worth 12,000 and you withdrew 2,000 from the fund. On December 31, 2000, your fund was worth 9,500. What was the time-weighted rate of return on your investment from January 1, 2000 to December 31, 2000? Problem 34.14 Deposits of $1000 are made into an investment fund at time 0 and 1. The fund balance is $1200 at time 1 and $2200 at time 2. (a) Compute the annual eﬀective yield rate computed by a dollar-weighted method. (b) Compute the annual eﬀective yield rate which is equivalent to that produced by a time-weighted method. Problem 34.15 Let A be the fund balance on January 1, B the fund balance on June 30, and C the balance on December 31. 318 RATE OF RETURN OF AN INVESTMENT (a) If there are no deposits or withdrawals, show that yield rates computed by the dollar-weighted method and the time-weighted method are both equal. (b) If there was a single deposit of D immediately after the June 30 balance was calculated, ﬁnd expressions for the dollar-weighted and time-weighted yield rates. (c) If there was a single deposit of D immediately before the June 30 balance was calculated, ﬁnd expressions for the dollar-weighted and time-weighted yield rates. (d) Compare the dollar-weighted yield rates in (b) and (c). (e) Compare the time-weighted yield rates in (b) and (c). Problem 34.16 100 is deposited into an investment account on January 1, 1998. You are given the following information on investment activity that takes place during the year: April 19,1998 October 30,1998 Value immediately prior to deposit 95 105 deposit 2X X The amount in the account on January 1, 1999 is 115. During 1998, the dollar-weighted return is 0% and the time-weighted return is Y. Calculate Y. Problem 34.17 An investment manager’s portfolio begins the year with a value of 100,000. Eleven months through the year a withdrawal of 50,000 is made and the value of the portfolio after the withdrawal is 57,000. At the end of the year the value of the portfolio is 60,000. Find the time-weighted yield rate less the dollar-weighted yield rate. Problem 34.18 On January 1, 1997 an investment account is worth 100,000. On April 1, 1997 the value has increased to 103,000 and 8,000 is withdrawn. On January 1, 1999 the account is worth 103,992. Assuming a dollar weighted method for 1997 and a time weighted method for 1998 the annual eﬀective interest rate was equal to x for both 1997 and 1998. Calculate x. Problem 34.19 On January 1, 1999, Luciano deposits 90 into an investment account. On April 1, 1999, when the amount in Luciano’s account is equal to X, a withdrawal of W is made. No further deposits or withdrawals are made to Luciano’s account for the remainder of the year. On December 31, 1999, the amount in Luciano’s account is 85. The dollar weighted return over the 1 year period is 20%. The time weighted return over the 1 year period is 16%. Calculate X. 34 INTEREST MEASUREMENT OF A FUND: TIME-WEIGHTED RATE OF INTEREST319 Problem 34.20 You are given the following information about the activity in two diﬀerent investment accounts. Account K Activity Date Fund value before activity Deposit Withdrawal January 1,1999 100 July 1,1999 125 X October 1,1999 110 2X December 31,1999 125 Account L Actyivity Date Fund value before activity Deposit Withdrawal January 1,1999 100 July 1,1999 125 X December 31,1999 105.8 During 1999 the dollar weighted return for investment account K equals the time weighted return for investment account L, which equals i. Calculate i. Problem 34.21 The following table shows the history of account balances, deposits, and withdrawals for an account during a particular year: Date Value before dep/withd Dep/Withd January 1,X 1,000 0 May 1,X 1,050 1,100 September 1,X 2,250 −900 January 1,X+1 16,00 What is the relationship among this account’s dollar-weighted rate of return (DW), its time-weighted rate of return (TW), and its internal rate of return (IRR)?Assume 30-day months. 320 RATE OF RETURN OF AN INVESTMENT 35 Allocating Investment Income: Portfolio and Investment Year Methods Consider an investment fund that involves many investors with seperate accounts. An example would be a pension fund for retirement in which each plan participant has an account but with commingled pool of assets. Each investor owns shares in the fund. In this section we consider the issue of allocating investment income to the various accounts. Two common methods are in use: the portfolio method and the investment year method also known as the new money method. The Portfolio Method Under the portfolio method an average rate based on the earnings of the entire fund is computed and credited to each account. All new deposits will earn this same portfolio rate. There are disadvantages of this method during time of ﬂuctuating interest rates. For example, sup- pose that the market rates have been rising signiﬁcantly during the recent years while the average portfolio rate is being lowered. In this case, there is a higher tendency for less new deposits to the fund and more withdrawals from the fund. The investment year method is an attempt to address this problem and to allocate investment income in a more equitable manner. Investment Year Method (IYM) We will describe this method on an annual basis, although typically smaller time intervals would be used. Under the investment year method, an investors’s contribution will be credited during the year with the interest rate that was in eﬀect at the time of the contribution. This interest rate is often referred to as the new money rate. For example, a portion of the IYM chart for a fund might look something like the one shown in Table 35.1. Suppose we invested $1,000 on January 1, 1994 and $500 on January 1, 1995. Then our total accumulation on Jan. 1, 2003 would be (1.064)(1.068)(1.071)(1.069)(1.073)(1.051)(1.049)(1.048)(1.047)1000 = 1688.75 from our 1994 contribution, plus (1.069)(1.070)(1.070)(1.074)(1.050)(1.049)(1.048)(1.047)500 = 794.31 from our 1995 contribution, for a total of 1688.75 + 794.31 = 2483.06. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS321 Purchase Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 1994 6.4% 6.8% 7.1% 6.9% 7.3% 5.1% 4.9% 4.8% 4.7% 1995 6.9% 7.0% 7.0% 7.4% 5.0% 4.9% 4.8% 4.7% 1996 7.1% 7.3% 7.3% 5.5% 5.4% 4.8% 4.7% 1997 7.0% 7.4% 5.4% 5.2% 4.6% 4.7% 1998 7.2% 5.7% 5.5% 4.5% 4.4% 1999 5.8% 5.1% 4.3% 4.7% 2000 5.0% 4.1% 4.6% 2001 4.0% 4.5% 2002 4.1% Table 35.1 Notice that from the chart above, investments made between 1994-1996 all earned the same rate of return both in 2001 (4.8%) and 2002 (4.7%). This illustrates the principle that under the investment year method, funds on investment longer than a certain ﬁxed period (5 years in our example) are assumed to grow at the overall yield rate for the fund (the portfolio rate), regardless of when the funds were invested. To better indicate the switch from the investment year method to the portfolio method, the data from the chart above would typically be displayed by a two-dimensional table such as the one in Table 35.2 Ifyis the calendar year of deposit, and mis the number of years for which the investment year method is applicable, then the rate of interest credited for the tth year of investment is denoted as iy t for t= 1,2,··· ,m. For t>m, the portfolio method is applicable, and the portfolio rate of interest credited for calendar year y is denoted as iy. The investment year method is more complicated than the portfolio method, but it was deemed necessary to attract new deposits and to discourage withdrawals during periods of rising interest rates. We illustrate the use of the investment year method in the following example. Example 35.1 You are given the following table of interest rates (in percentages). A person deposits 1,000 on January 1, 1997. Find the accumulated amount on January 1, 2000 (a) under the investment year method; (b) under the portfolio method; (c) when the balance is withdrawn at the end of every year and is reinvested at the new money rate. 322 RATE OF RETURN OF AN INVESTMENT y iy 1 iy 2 iy 3 iy 4 iy 5 iy+5 Portfolio Year 1992 8.25 8.25 8.4 8.5 8.5 8.35 1997 1993 8.5 8.7 8.75 8.9 9.0 8.6 1998 1994 9.0 9.0 9.1 9.1 9.2 8.85 1999 1995 9.0 9.1 9.2 9.3 9.4 9.1 2000 1996 9.25 9.35 9.5 9.55 9.6 9.35 2001 1997 9.5 9.5 9.6 9.7 9.7 1998 10.0 10.0 9.9 9.8 1999 10.0 9.8 9.7 2000 9.5 9.5 2001 9.0 Table 35.2 Solution. (a) The sequence of interest rates beginning from a given year of investment runs horizontally through the row for that year and then down the last column of rates. An investment made at the beginning of 1992 would earn the investment year rate of 8.25% in 1992, the money rate of 8.25% in 1993, 8.4% in 1994, 8.5% in 1995 and 1996. Starting in 1997, an investment made at the beginning of 1992 would earn the portfolio average rates. The 1992 investment would earn the portfolio rate of 8.35% in 1997, the portfolio rate of 8.6% in 1998, and so on. An investment made at the beginning of 2001 would earn the investment year rate of 9% in 2001- 2005. Starting the year 2006, the investment made in 2001 would earn the portfolio rate of 8.35% in 2006, the portfolio rate of 8.6% in 2007, and so on. (a) The accumulated value is 1000 ·(1.095)(1.095)(1.096) = 1,314.13. (b) The accumulated value is Q= 1000 ·(1.0835)(1.086)(1.0885) = 1,280.82. (c) The accumulated value is 1000 ·(1.095)(1.1)(1.1) = 1,324.95 Example 35.2 An investment of 1000 is made on January 1996 in an investment fund crediting interest according to Table 35.2. How much interest is credited from January 1, 1999 to January 1, 2001? Solution. The accumulated value on January 1, 1999 is 1000(1.0925)(1.0935)(1.095) = 1308.14. The accumulated value on January 1, 2001 is 1000(1.0925)(1.0935)(1.095)(1.0955)(1.096) = 1570.64. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS323 Thus, the total amount of interest credited is 1570.64 −1308.14 = 262.50 Example 35.3 Using Table 35.2, ﬁnd (a) the interest rates credited in calendar year 2000 for deposits made in 1994 - 2000; (b) the new money rates credited in the ﬁrst year of investment for deposits made in 1992 - 2001. Solution. (a) 9.5% credited for new deposits made in 2000, 9.8% for deposits made in 1999, 9.9% for deposits made in 1998, 9.7% for deposits made in 1997, 9.6% for deposits made in 1996, 9.1% for deposits made in 1994 - 1995. Thus, the interest credited in the calendar year 2000 appear on an upwardly diagonal within the table. (b) The new money rates credited in the ﬁrst year of investment appear in the ﬁrst column of the table 324 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 35.1 Using Table 35.2, what were the interest rates credited in calendar year 2001 for deposits made in 2001 and before 2001? Problem 35.2 If 100 is invested in the fund at the beginning of 2002, ﬁnd the accumulated amount at the end of 2003 using: (a) The investment year method (b) The portfolio method (c) The investment year method if the amount is withdrawn and then reinvested at the end of 2002. y iy 1 iy 2 iy 3 iy 4 iy+4 Portfolio Year 1998 7.0 6.5 6.0 5.8 5.9 2002 1999 6.4 6.1 5.8 5.9 6.0 2003 2000 6.2 6.0 5.9 6.0 6.2 2004 2001 6.1 5.9 6.1 6.4 6.6 2005 2002 6.0 6.1 6.3 6.6 2003 6.4 6.6 6.7 2004 6.8 7.0 2005 7.5 Problem 35.3 The following table lists the interest rate credited under an investment year method of crediting interest. y iy 1 iy 2 iy 3 iy 4 iy 5 iy+5 1999 7.0 6.75 6.5 6.25 6.0 5.5 2000 6.0 5.5 5.25 5.1 5.0 2001 5.0 4.8 4.6 4.3 2002 4.0 3.75 3.5 2003 3.0 3.2 2004 4.0 Becky invests $1,000 on January 1, 1999 and an additional $500 on January 1, 2003. How much money does Becky have on December 31, 2004? Use the following chart of interest rates to answer Problems 35.4 through 35.6. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS325 y iy 1 iy 2 iy 3 iy 4 iy+4 Calendar year of portfolio rate y+ 4 1997 0.0650 0.0625 0.0600 0.0575 0.0550 2001 1998 0.0600 0.0575 0.0550 0.0525 0.0500 2002 1999 0.0500 0.0475 0.0460 0.0450 0.0450 2003 2000 0.0450 0.0440 0.0430 0.0420 0.0410 2004 2001 0.0400 0.0390 0.0380 0.0370 2002 0.0300 0.0300 0.0325 2003 0.0300 0.0325 2004 0.0300 Problem 35.4 Jordan invests 1,000 in a fund on January 1, 1998. The fund uses the investment year method of determining interest rates. Calculate the amount that Jordan will have at the end of 2004. Problem 35.5 Jenna invests 1,000 in a fund on January 1, 2002. The fund uses the portfolio method of determining interest rates. Calculate the amount that Jenna will have at the end of 2004. Problem 35.6 James invests 1,000 into a fund at the beginning of each year from 2002 to 2004. The fund pays interest using the investment year interest rate. Calculate the amount of money that James will have at the end of 2004. Use the following chart of interest rates to answer Questions 35.7 through 35.10. y iy 1 iy 2 iy 3 iy+3 1995 3.7% 3.6% 3.5% 6.0% 1996 3.2% 3.1% 3.0% 5.5% 1997 2.7% 2.6% 2.5% 5.0% 1998 2.2% 2.1% 2.0% 4.5% 1999 1.7% 1.6% 1.5% 4.0% Problem 35.7 A deposit of $100 is made at the beginning of 1997. How much interest was credited during 1998? What is the accumulated value at the end of 2002? 326 RATE OF RETURN OF AN INVESTMENT Problem 35.8 How much interest is credited in the calendar years 1997 through 1999 inclusive to a deposit of $100 made at the beginning of 1995? Problem 35.9 What were the interest rates credited in calendar year 1999 for deposits made in 1999, 1998, 1997, 1996, and 1995? Problem 35.10 What were the new money rates credited in the ﬁrst year of investment for deposits made in 1995, 1996, 1997, and so on? Problem 35.11 The following table shows the annual eﬀective interest rates credited to an investment fund by calendar year of investment. The investment year method applies for the ﬁrst two years, after which a portfolio method is used. y iy 1 iy 2 iy+2 1995 t 5.5% 4.5% 1996 6.0% 6.1% 5.0% 1997 7.0% t+ 2.5% 5.5% An investment of $100 is made at the beginning of 1995 and 1997. The total amount of interest credited by the fund during the year 1998 is $13.81. Find t. Problem 35.12 It is known that 1 + iy t = (1.08 + 0.005t)1+0.01y for integral t, 0 ≤t≤5,and integral y, 0 ≤y≤10. If $1,000 is invested for three years beginning in year y = 5, ﬁnd the equivalent eﬀective rate of interest. Problem 35.13 For a certain portfolio, iy t = (1.06 + 0.004t)1+0.01y for 1 ≤t≤5 and 0 ≤y≤10. (a) What is the accumulated value at the end of year 4 of a single investment of 1,000 at the beginning of year 2? (b) What is the accumulated value at the end of year 4 of a deposit of 1,000 at the beginning of each of years 2 through 4? Problem 35.14 ‡ The following table shows the annual eﬀective interest rates being credited by an investment account, by calendar year of investment. The investment year method is applicable for the ﬁrst 3 years, after which a portfolio rate is used. 35 ALLOCATING INVESTMENT INCOME: PORTFOLIO AND INVESTMENT YEAR METHODS327 y iy 1 iy 2 iy 3 iy+3 1990 10% 10% t% 8% 1991 12% 5% 10% (t−1)% 1992 8% ( t−2)% 12% 6% 1993 9% 11% 6% 9% 1994 7% 7% 10% 10% An investment of 100 is made at the beginning of years 1990, 1991, and 1992. The total amount of interest credited by the fund during the year 1993 is equal to 28.40. Calculate t. Problem 35.15 Find ¨s5 from the data in Table 35.2 assuming the ﬁrst payment is made in calendar year 1995. Problem 35.16 You are given the following table of interest rates: y iy 1 iy 2 iy+5 Portfolio Year 2004 9.00% 10.00% 11.00% 2006 2005 7.00% 8.00% 2006 5.00% 1000 is invested at the beginning of calendar years 2004, 2005, and 2006. What is the total amount of interest credited for calendar year 2006? Problem 35.17 You are given: y iy 1 iy 2 iy 3 iy+5 2001 5.0 5.5 6.0 3.0 2002 4.5 5.0 5.5 2.5 2003 3.0 3.5 4.0 2.0 2004 4.0 X 5.0 2005 5.0 5.5 6.0 2006 4.5 5.0 5.5 The accumulated value on 1/1/2007 of a deposit on 1/1/2001 is 9.24% greater than the accumulated value on 1/1/2007 of the same amount deposited on 1/1/2004. Determine X. 328 RATE OF RETURN OF AN INVESTMENT 36 Yield Rates in Capital Budgeting One of the use of yield rates is in capital budgeting. An investor always is faced with the need to allocate an amount of capital among various alternative investments in order to achieve the highest possible level of return on the investments. This process of making such ﬁnancial decisions is referred to as capital budgeting. In this section we discuss brieﬂy this concept. Moreover, we will assume that risk is non-existent in the alternative investments being compared. The two major approaches to capital budgeting that are encountered in practice are the yield rate method and the net present value method. In the yield rate method the investor computes the yield rate(s) for each alternative investment by solving the equation NPV (i) = 0.Then these rates are compared to an interest preference rate set by the investor. This is usually the minimum rate of return acceptable by the investor. All yield rates that are lower than the interest preference rate are rejected. The yield rates that are higher than the interest preference rateare the only one considered. They are ranked from highest to lowest and are selected in descending order until the amount of capital available for investment is exhausted. In the net present value method or the NPV method, the investor computes NPV (i) for each alternative investment, where i is the interest preference rate. Negative values of NPV (i) are rejected and only the positive values are considered since the present value of returns is larger than that of the contributions. Capital is then allocated among those investments with positive NPV (i) in such a manner that the total present value of returns from the investment (computed at the interest preference rate) minus the contributions to the investment is maximized. It has been shown in Finance Theory that the NPV method usually provides better decisions than other methods when making capital investments. Consequently, it is the more popular evaluation method of capital budgeting projects. Example 36.1 Consider the investment project given in the table below. (a) Find the yield rate of this project. (b) Assuming an interest preference rate of 3%, would you accept this project when using the yield rate method? the net present value method? Solution. (a) Solving the equation NPV (i) = −80,000 −10,000ν−10,000ν2 −10,000ν3 −8,000ν4 + 28,000ν5 + 38,000ν6 + 33,000ν7 + 23,000ν8 + 13,000ν9 + 8,000ν10 = 0 we ﬁnd the two solutions i= −1.714964 <−1 and i= 0.032180. Thus, the yield rate is 3.218%. (b) Using the yield rate method, since 3 .218% > 3%, the investor would accept this project 36 YIELD RATES IN CAPITAL BUDGETING 329 for consideration. Using the net present value method the investor would also accept it, since NPV (0.03) = 1488.04 >0 Year Contributions Returns ct 0 80,000 0 −80,000 1 10,000 0 −10,000 2 10,000 0 −10,000 3 10,000 0 −10,000 4 20,000 12,000 −8,000 5 2,000 30,000 28,000 6 2,000 40,000 38,000 7 2,000 35,000 33,000 8 2,000 25,000 23,000 9 2,000 15,000 13,000 10 0 8,000 8,000 Example 36.2 Repeat the same problem as above but with an interest preference rate of 4%. Solution. Using the yield rate method, since 4% >3.218%, the investor would reject this project. Using the net present value method the investor would also reject it, since NPV (0.04) = −5122.13 <0 330 RATE OF RETURN OF AN INVESTMENT Practice Problems Problem 36.1 An investment project has the following cash ﬂows: Year Contributions Returns 0 100 0 1 200 0 2 10 60 3 10 80 4 10 100 5 5 120 6 0 60 (a) Using the net present value method with an interest preference rate of 15%, should the invest- ment be accepted or rejected? (b) Answer the same question when using the yield rate method. Problem 36.2 An investor enters into an agreement to contribute $7,000 immediately and $1,000 at the end of two years in exchange for the receipt of $4,000 at the end of one year and $5,500 at the end of three years. (a) Calculate NPV (0.09) and NPV (0.10). (b) If an investor’s interest preference rate is 12%, should the investment be accepted or rejected? Problem 36.3 A used car can be purchased for $5000 cash or for $2400 down and $1500 at the end of each of the next two years. Should a purchaser with an interest preference rate of 10% pay cash or ﬁnance the car? Problem 36.4 Consider an investment in which a person makes payments of $100 immediately and $132 at the end of one year. What method would be better to use, the yield rate method or the net present value method if the preference interest rate is 15%? Problem 36.5 Consider an investment in which a person makes payments of $100 immediately and $101 at the end of two years in exchange for a payment of $200 at the end of one year. Explain why the yield rate method is not applicable in this case. 36 YIELD RATES IN CAPITAL BUDGETING 331 Problem 36.6 A borrower needs $800. The funds can be obtained in two ways: (i) By promising to pay $900 at the end of the period. (ii) By borrowing $1000 and repaying $1120 at the end of the period. If the interest preference rate for the period is 10%, which option should be chosen? 332 RATE OF RETURN OF AN INVESTMENT Loan Repayment Methods Loan is an arrangement in which a lender gives money or property to a borrower, and the borrower agrees to return the property or repay the money, usually along with interest, at some future point(s) in time. Various methods of repaying a loan are possible. We will consider two of them: The amortization method and the sinking fund method. The amortization method: In this method the borrower makes installment payments to the lender. Usually these payments are at a regularly spaced periodic intervals; the progressive reduction of the amount owed is described as the amortization of the loan. Examples include car loan, home mortgage repayment. The sinking fund method: In this method the loan will be repaid by a single lump sum payment at the end of the term of the loan. The borrower pays interest on the loan in installments over this period. However, the borrower may prepare himself for the repayment by making deposits to a fund called a sinking fund to accumulate the repayment amount. 333 334 LOAN REPAYMENT METHODS 37 Finding the Loan Balance Using Prospective and Retro- spective Methods. When using the amortization method, the payments form an annuity whose present value is equal to the original amount of the loan. In this section, we want to determine the unpaid balance, also referred to as the outstanding loan balance or unpaid principal at any time after the inception of the loan. There are two approaches used in ﬁnding the amount of the outstanding balance: the prospective and the retrospective method. With the prospective method, the outstanding loan balance at any point in time is equal to the present value at that date of the remaining payments. With the retrospective method, the outstanding loan balance at any point in time is equal to the original amount of the loan accumulated to that date less the accumulated value at that date of all payments previously made. In general, the two approaches are equivalent. At the time of inception of the loan we have the following equality Present Value of All Payments = Amount of Loan Accumulate each side of the equation to the date at which the outstanding loan balance is desired, obtaining Current Value of Payments = Accumulated Value of Loan Amount But payments can be divided into past and future payments giving Accumulated Value of Past Payments + Present Value of Future Payments = Accumulated Value of Loan Amount or Present Value of Future Payments = Accumulated Value of Loan Amount −Accumulated Value of Past Payments. But the left side of this equation represents the prospective approach and the right side represents the retrospective approach. We can prove that the two methods are equivalent algebraically as follows. LetBp t and Br t denote the outstanding loan balances at time t using the prospective and retrospective methods respectively. We denote the initial loan by L. 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.335 Suppose we want to repay the loan by level payments of P for nperiods at a periodic interest of i.. Then P satiﬁes the equation P = L an . For 0 <t<n, the outstanding loan balance at time t computed after making the tth payment is Bp t = Pan−t by the prospective method and Br t = L(1 + i)t −Pst by the retrospective method. We will show that Br t = Bp t. Indeed, Br t =L(1 + i)t −Pst =Pan(1 + i)t −Pst =P [(1 + i)t −(1 + i)−(n−t) −(1 + i)t + 1 i ] =P1 −νn−t i = Pan−t = Bp t. Example 37.1 ‡ A loan is being repaid with 25 annual payments of 300 each. With the 10th payment, the borrower pays an extra 1,000, and then repays the balance over 10 years with a revised annual payment. The eﬀective rate of interest is 8%. Calculate the amount of the revised annual payment. Solution. The balance after 10 years, prospectively, is Bp 10 = 300a15 = $2,567.84. If the borrower pays an additional $1,000, the balance becomes $1,567.84. An equation of value for the revised payment, denoted by R, is Ra10 = 1,567.84 or R= 1,567.84 a10 = $233.66 336 LOAN REPAYMENT METHODS Example 37.2 A loan is being repaid with 16 quarterly payments, where the ﬁrst 8 payments are each $200 and the last 8 payments are each $400. If the nominal rate of interest convertible quarterly is 10%, use both the prospective method and the retrospective method to ﬁnd the outstanding loan balance immediately after the ﬁrst six payments are made. Solution. With the prospective method, we have Bp 6 =200(ν+ ν2) + 400ν2(ν+ ν2 + ··· + ν8) =400(ν+ ν2 + ··· + ν10) −200(ν+ ν2) =400a10 −200a2 =400(8.7521) −200(1.9274) = $3,115.36. With the retrospective method, we have that the original loan amount is L=200(ν+ ν2 + ··· + ν8) + 400ν8(ν+ ν2 + ··· + ν8) =400(ν+ ν2 + ··· + ν16) −200(ν+ ν2 + ··· + ν8) =400a16 −200a8 =400(13.0550) −200(7.1701) = $3,787.98 The outstanding loan balance is Br 6 = 3787.98(1.025)6 −200s6 = 4392.90 −200(6.3877) = $3,115.36 Example 37.3 Megan is buying a car for 30,000 using a 60-month car loan with an interest rate of 9% compounded monthly. For the ﬁrst two years, Megan makes the required payment. Beginning with the ﬁrst payment in the third year, Megan begins paying twice the required payment. Megan will completely pay oﬀ her loan by making a smaller ﬁnal payment. Determine the total number of payments that Megan will make. Solution. The monthly eﬀective interest rate is 0.09 12 = 0.0075. The original required payment is P such that 30000 = Pa60 0.0075 = P(1−1.0075−60) 0.0075 →P = 30000×0.0075 (1−1.0075−60) = 622.7506568. At the end of the ﬁrst 2 years, Megan’s outstanding loan balance, prospectively, is 622.7506568a36 0.0075 = 622.7506568(1 −1.0075−36) 0.0075 = 19583.51862 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.337 From this point on, Megan pays 2P = 1245.501314 every month. Then 19583.51862 = 1245.501314an0.0075 = 1245.501314(1−1.0075−n) 0.0075 ,where nis the number of months until the loan is repaid. Thus, 0.1179255197 = (1 −1.0075−n) →1.0075−n = 0.8820744803 →n = −ln 0.8820744803 ln 1.0075 = 16.79315649. So Megan will make 16 full payments and 1 smaller payment for a total of 17 additional payments and 17 + 24 = 41 total payments over the course of repaying the loan Example 37.4 A loan of 1,000 is being repaid with quarterly payments at the end of each quarter for 5 years, at 6% convertible quarterly. Find the outstanding loan balance at the end of the 2nd year. Solution. The quarterly eﬀective interest rate is 0.06 4 = 0.015. Let P be the quarterly payment. Then 1000 = Pa20 0.015 = P(1−1.015−20) 0.015 →P = 1000×0.015 (1−1.015−20) = 58.24573587. The outstanding loan balance at the end of the 2nd year, prospectively, is 58 .24573587a12 0.015 = 58.24573587(1−1.015−12) 0.015 = $635.32 338 LOAN REPAYMENT METHODS Practice Problems Problem 37.1 A loan is being repaid with level annual payments of $1,000. Calculate the outstanding balance of the loan if there are 12 payments left. The next payment will be paid one year from now and the eﬀective annual interest rate is 5%. Problem 37.2 A loan of 10,000 is being repaid will 20 non-level annual payments. The interest rate on the loan is an annual eﬀective rate of 6%. The loan was originated 4 years ago. Payments of 500 at the end of the ﬁrst year, 750 at the end of the second year, 1,000 at the end of the third year and 1,250 at the end of the fourth year have been paid. Calculate the outstanding balance immediately after the fourth payment. Problem 37.3 Calculate the outstanding balance to the loan in the previous problem one year after the fourth payment immediately before the ﬁfth payment. Problem 37.4 Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. What is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 37.5 If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 37.6 A loan 20,000 is being repaid with annual payments of 2,000 at the end of each year. The interest rate charged on the loan is an annual eﬀective rate of 8%. Calculate the outstanding balance of the loan immediately after the 5th payment. Problem 37.7 The interest rate on a 30 year mortgage is 12% compounded monthly. The mortgage is being repaid by monthly payments of 700. Calculate the outstanding balance at the end of 10 years. Problem 37.8 The interest rate on a 30 year mortgage is 12% compounded monthly. Lauren is repaying the mortgage by paying monthly payments of 700. Additionally, to pay oﬀ the loan early, Lauren has made additional payments of 1,000 at the end of each year. Calculate the outstanding balance at the end of 10 years. 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.339 Problem 37.9 A loan is being repaid with 20 payments of 1,000. The total interest paid during the life of the loan is 5,000. Calculate the amount of the loan. Problem 37.10 A loan of 10,000 is being repaid by installments of 2,000 at the end of each year, and a smaller ﬁnal payment made one year after the last regular payment. Interest is at the eﬀective rate of 12%. Find the amount of outstanding loan balance remaining when the borrower has made payments equal to the amount of the loan. Problem 37.11 A loan is being repaid by quarterly installments of 1,500 at the end of each quarter, at 10% con- vertible quarterly. If the loan balance at the end of the ﬁrst year is 12,000, ﬁnd the original loan balance. Problem 37.12 A loan is being repaid by 15 annual payments at the end of each year. The ﬁrst 5 installments are 4,000 each, the next 5 are 3,000 each, and the ﬁnal 5 are 2,000 each. Find expressions for the outstanding loan balance immediately after the second 3,000 installment. (a) prospectively; (b) retrospectively. Problem 37.13 A loan is to be repaid with level installments payable at the end of each half-year for 3 1 2 years, at a nominal rate of interest of 8% convertible semiannually. After the fourth payment the outstanding loan balance is 5,000. Find the initial amount of the loan. Problem 37.14 A 20,000 loan is to be repaid with annual payments at the end of each year for 12 years. If (1 + i)4 = 2, ﬁnd the outstanding balance immediately after the fourth payment. Problem 37.15 ‡ A 20-year loan of 1,000 is repaid with payments at the end of each year. Each of the ﬁrst ten payments equals 150% of the amount of interest due. Each of the last ten payments is X. The lender charges interest at an annual eﬀective rate of 10%. Calculate X. Problem 37.16 ‡ A loan is amortized over ﬁve years with monthly payments at a nominal interest rate of 9% com- pounded monthly. The ﬁrst payment is 1,000 and is to be paid one month from the date of the loan. Each succeeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made. 340 LOAN REPAYMENT METHODS Problem 37.17 A mortgage loan is being repaid with level annual payments of 5,000 at the end of the year for 20 years. The interest rate on the mortgage is 10% per year. The borrower pays 10 payments and then is unable to make payments for two years. Calculate the outstanding balance at the end of the 12th year. Problem 37.18 ‡ An investor took out a loan of 150,000 at 8% compounded quarterly, to be repaid over 10 years with quarterly payments of 5483.36 at the end of each quarter. After 12 payments, the interest rate dropped to 6% compounded quarterly. The new quarterly payment dropped to 5134.62. After 20 payments in total, the interest rate on the loan increased to 7% compounded quarterly. The investor decided to make an additional payment of X at the time of his 20th payment. After the additional payment was made, the new quarterly payment was calculated to be 4265.73, payable for ﬁve more years. Determine X. Problem 37.19 ‡ A small business takes out a loan of 12,000 at a nominal rate of 12%, compounded quarterly, to help ﬁnance its start-up costs. Payments of 750 are made at the end of every 6 months for as long as is necessary to pay back the loan. Three months before the 9th payment is due, the company reﬁnances the loan at a nominal rate of 9%, compounded monthly. Under the reﬁnanced loan, payments of R are to be made monthly, with the ﬁrst monthly payment to be made at the same time that the 9th payment under the old loan was to be made. A total of 30 monthly payments will completely pay oﬀ the loan. Determine R. Problem 37.20 A loan of $1,000 is being repaid with quarterly payments at the end of each quarter for 5 years, at 6% convertible quarterly. Find the outstanding loan balance at the end of the second year. Problem 37.21 A 20,000 mortgage is being repaid with 20 annual installments at the end of each year. The borrower makes 5 payments, and then is temporarily unable to make payments for the next 2 years. Find an expression for the revised payment to start at the end of the 8th year if the loan is still to be repaid at the end of the original 20 years. Problem 37.22 A loan of 1 was originally scheduled to be repaid by 25 equal annual payments at the end of each 37 FINDING THE LOAN BALANCE USING PROSPECTIVE AND RETROSPECTIVE METHODS.341 year. An extra payment K with each of the 6th through the 10th scheduled payments will be suﬃcient to repay the loan 5 years earlier than under the original schedule. Show that K = a20 −a15 a25 a5 . Problem 37.23 Bob takes out a loan of 1000 at an annual eﬀective interest rate of i. You are given: (a) The ﬁrst payment is made at the end of year 6. (b) Ten equal annual payments are made to repay the loan in full at the end of 15 years. (c) The outstanding principal after the payment made at the end of year 10 is 908.91. Calculate the outstanding principal at the end of year 5. Problem 37.24 The original amount of an inheritance was just suﬃcient at 3.5% eﬀective to pay $10,000 at the end of each year for 10 years. The payments were made as scheduled for the ﬁrst ﬁve years even though the fund actually earned 5% eﬀective. How much excess interest was in the fund at the end of the ﬁfth year? 342 LOAN REPAYMENT METHODS 38 Amortization Schedules When a loan is being repaid with the amortization method, each payment is partially a repayment of principal and partially a payment of interest. Determining the amount of each for a payment can be important (for income tax purposes, for example). An amortization schedule is a table which shows the division of each payment into principal and interest, together with the outstanding loan balance after each payment is made. Consider a loan of an at interest rate i per period being repaid with payments of 1 at the end of each period for n periods. At the end of period 1 (i.e. after the ﬁrst payment), the interest paid is ian = 1 −νn so that the principal repaid is νn,and the outstanding loan balance is an −νn = an−1 . Next, at the end of the second period, the interest paid is ian−1 = 1 −νn−1 so that the principal repaid is νn−1,and the outstanding loan balance is an−1 −νn−1 = an−2 .Continuing this process, we see that at the end of period k, the interest paid is ian−k+1 = 1 −νn−k+1 and the principal repaid is νn−k+1. The outstanding loan balance is an−k+1 −νn−k+1 = an−k = Bp k. The amortization table is shown below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 an 1 1 ian = 1 −νn νn an −νn = an−1 2 1 ian−1 = 1 −νn−1 νn−1 an−1 −νn−1 = an−2 ... ... ... ... ... k 1 ian−k+1 = 1 −νn−k+1 νn−k+1 an−k+1 −νn−k+1 = an−k ... ... ... ... ... n−1 1 ia2 = 1 −ν2 ν2 a2 −ν2 = a1 n 1 ia1 = 1 −ν ν a1 −ν = 0 Total n n−an an Observe each of the following from this table: First, it should be noted that the outstanding loan balance agrees with that obtained by the prospective method. Second, the sum of the principal repayments equals to the original amount of the loan. Third, the sum of interest payments is equal to the diﬀerence between the sum of the total payments and the sum of the principal repayments. Fourth, the sum of principal repayments is a geometric progression with common ration (1 + i). Example 38.1 Create an amortization schedule for a loan of $1,000 repaid over four years if the annual eﬀective rate of interest is 8%. 38 AMORTIZATION SCHEDULES 343 Solution. If R is the periodic payment then R= 1000 a4 = $301.92. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 1000.00 1 301.92 80.00 221.92 778.08 2 301.92 62.25 239.67 538.41 3 301.92 43.07 258.85 279.56 4 301.92 22.36 279.56 0 Table 38.1 In the previous example, the last line exactly balances. That is, after the last payment is made the loan is paid oﬀ. In practice, there will be rounding errors as the table is generated line by line, and the last line may not lead to a zero balance. Standard practice is to adjust the last payment so that it is exactly equal to the amount of interest for the ﬁnal period plus the outstanding loan balance at the beginning of the ﬁnal period, in order to bring the outstanding loan balance to 0. Example 38.2 Create an amortization schedule for a loan of $10,000 repaid over three years if the annual eﬀective rate of interest is 7%. Solution. If R is the periodic payment then R= 10,000 a3 = $3,810.52. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 10,000.00 1 3,810.52 700.00 3,110.52 6,889.48 2 3,810.52 482.26 3,328.26 3,561.22 3 3,810.50 249.28 3,561.22 0 It should be noted that if it is desired to ﬁnd the amount of principal and interest in one particular payment, it is not necessary to construct the entire amortization schedule. The outstanding loan balance at the beginning of the period in question can be determined by either the retrospective or the prospective methods, and then that one line of the schedule can be created. 344 LOAN REPAYMENT METHODS Example 38.3 Jones borrows $20,000 from Smith and agrees to repay the loan with equal quarterly installments of principal and interest at 10% convertible quarterly over eight years. At the end of three years, Smith sells the right to receive future payments to Collins at a price which produces a yield rate of 12% convertible quarterly for Smith. Find the total amount of interest received (a) by Collins, and (b) by Smith. Solution. (a) Each quarterly payment by Jones is 20000 a32 0.025 = $915.37. Total payments by Jones to Collins over the last ﬁve years are (20)915.37 = $18307.40. After three years, the price Collins pays to Smith is 915.37a20 0.03 = 915.37(14.8775) = $13618.42. Total amount of interest received by Collins is 18307.40 −13618.42 = $4688.98. (b) Total payments by Jones to Smith over the ﬁrst three years are (12)915.37 = $10984.44. After three years, the outstanding loan balance is 915.37a20 0.025 = 915.37(15.5892) = $14269.89. Recall from (a) that the price Collins pays to Smith after three years is $13618.42. Total amount of interest received by Smith is 13618.42 + 10984.44 −20000 = $4602.86 Let It denote the amount of interest paid in the tth installment; Pt be the amount of principal and Bt be the loan balance. 38 AMORTIZATION SCHEDULES 345 Example 38.4 Show that for n= 100 and with a periodic payment of R, we have P11 + P12 + ··· + P50 = B10 −B50 and I11 + I12 + ··· + I50 = 40R−(B10 −B50). Solution. Let P = P11 + P12 + ··· + P50. Then P =R 50∑ k=11 ν100−k+1 = Rν101 50∑ k=11 (1 + i)k =Rv101( 50∑ k=0 (1 + i)k − 10∑ k=0 (1 + i)k) =R(1 + i)−50 −(1 + i)−90 i =Ra90 −Ra50 = B10 −B50 Now, let I = I11 + I12 + ··· + I50 then I = ∑50 j=11 R−∑50 j=11 Pj = 40R−(B10 −B50) Amortization schedules of perpetuities do not exist since the entire payment represents interest and therefore the loan balance remains unchanged. Example 38.5 A $5,000 loan is being repaid by payments of $X at the end of each half year for as long as necessary until a smaller ﬁnal payment is made. The nominal rate of interest convertible semiannually is 14%. (a) If X = $400 ﬁnd the amount of principal and the interest in the sixth payment. (b) If X = $350, ﬁnd the principal in the sixth payment, and interpret this. Solution. (a) The outstanding balance at the beginning of the sixth half-year is Br 5 = 5000(1.07)5 −400s5 = 7012.76 −400(5.7507) = $4712.48. The interest in the sixth payment is (0 .07)(4712.48) = $329.87 and the principal is 400 −329.87 = $70.13. (b) The outstanding balance at the beginning of the sixth half-year is Br 5 = 5000(1.07)5 −350s5 = 7012.76 −350(5.7507) = $5000.01. 346 LOAN REPAYMENT METHODS If X ≤350, then the loan will never be paid oﬀ, because every payment will count only toward interest Thus far, in creating an amortization schedule we have assumed a constant rate of interest, the conversion period and payment period are the same, and the payments are leveled. Example 38.2 is an example where the payments are not all leveled. It is possible to create an amortization schedule with varying interest rate as illustrated in the next example. Example 38.6 An amount is invested at an annual eﬀective rate of interest i which is exactly suﬃcient to pay 1 at the end of each year for n years. In the ﬁrst year, the fund earns rate i and 1 is paid at the end of the year. However, in the second year, the fund earns rate j > i.If X is the revised payment which could be made at the end of years 2 to n, then ﬁnd X assuming that (a) the rate reverts back to i again after this one year, (b) the rate earned remains at j for the rest of the n−year period. Solution. (a) From the amortization table, the balance at the end of the ﬁrst year is an−1 i,and therefore the balance after two years must be (1 + j)an−1 i −X. However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j)an−1 i −X =Xan−2 i X(1 + an−2 i) =(1 + j)an−1 i X(1 + i)an−1 i =(1 + j)an−1 i X =1 + j 1 + i. (b) As in (a), the balance at the end of two years is (1 + j)an−1 i−X. However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j)an−1 i −X =Xan−2 j X(1 + an−2 j) =(1 + j)an−1 i X(1 + j)an−1 j =(1 + j)an−1 i X = an−1 i an−1 j Situations where payment period and interest conversion period do not coincide are best handled 38 AMORTIZATION SCHEDULES 347 from ﬁrst principles rather than from developing formulas. With an amortization schedule in which payments are made at a diﬀerent frequency than interest is convertible, the following two-step procedure can be followed: (1) Find the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. (2) Construct the amortization schedule with the rate of interest in step 1. Example 38.7 A 25-year mortgage for $100,000 bears an interest rate of 5.5% compounded semi-annually. The payments are made at the end of each month. Find the size of the payment. Solution. The monthly interest rate is j = (1.0275) 1 6 −1 = .00453168. The size of each payment is R= 100,000 a300 j = 610.39 Example 38.8 A loan of $3,000 is being amortized by 20 quarterly payments. Payments 11 and 12 are not made. At the designated time of the 12th payment, the loan is renegotiated so that the 13th payment is $N and payments 14, 16, 18, and 20 are each $40 more that the preceding payment. If the rate of interest is 8% convertible semiannually, ﬁnd the value of N which would provide that the loan be completely repaid after 20 quarters. Solution. The quarterly eﬀective rate of interest isj = (1.02) 1 2 −1.The original quareterly payment isR= 3000 a20 j . The outstanding balance at time t = 10, by the prospective method, is Ra10 j. The outstanding balance at time t= 12 is Ra10 j(1 + j)2 = 3000a10 j a20 j (1 + j)2 = 1712.46. Now, the new payments stream starting at time t= 13 are: N,N + 40,N + 40,N + 80,N + 80,N + 120,N + 120,N + 160. By the prospective method 1712.46 = Na8 j + 40(ν2 j + ν3 j + 2ν4 j + 2ν5 j + 3ν6 j + 3ν7 j + 4ν8 j). Solving this equation for N we ﬁnd N = 155.73 348 LOAN REPAYMENT METHODS Practice Problems Problem 38.1 A loan of 10,000 is being repaid with annual payments of 1500 for 11 years. Calculate the amount of principal paid over the life of the loan. Problem 38.2 A loan of 10000 is being repaid with annual payments of 1500 for 11 years. Calculate the amount of interest paid over the life of the loan. Problem 38.3 A loan is being repaid with level annual payments based on an annual eﬀective interest rate of 8%. If the amount of principal in the 10th payment is 100, calculate the amount of principal in the 5th payment. Problem 38.4 A loan is being repaid with level annual payments based on an annual eﬀective interest rate of 8%. If the outstanding balance immediately after the 10th payment is 1000, calculate the amount of interest in the 11th payment. Problem 38.5 A loan of 10,000 is being repaid with annual payments of 1500 for n years. The total principal paid in the ﬁrst payment is 685.58. Calculate the interest rate on the loan. Problem 38.6 For a loan with level annual payments, the principal repaid by the 10th payment is 10,000 while the principle repaid by the 11th payment is 11,000. Calculate the principal repaid by the 15th payment. Problem 38.7 A 60-month loan is to be repaid with level payments of 1000 at the end of each month. The interest in the last payment is 7.44. Calculate the total interest paid over the life of the loan. Problem 38.8 A 60-month loan is to be repaid with level payments of 1000 at the end of each month. The principal in the ﬁrst payment is 671.21. Calculate the eﬀective annual interest rate. Problem 38.9 Jenna is repaying a 120-month loan with interest compounded monthly at 12%. Calculate the payment in which the absolute value of the diﬀerence between the interest paid and the principal repaid is minimized. 38 AMORTIZATION SCHEDULES 349 Problem 38.10 A loan is being repaid with level payments at the end of each year for 20 years. The principal repaid in the 10th payment is 1000 and the principal repaid in the 15th payment is 1200. Calculate the amount of the loan. Problem 38.11 A loan of 10,000 is being repaid with annual payments over 10 years at 8% compound monthly. Create the amortization table for this loan. Problem 38.12 A loan is being repaid with quarterly installments of 1,000 at the end of each quarter for 5 years at 12% convertible quarterly. Find the amount of principal in the 6th installment. Problem 38.13 Consider a loan which is being repaid with installments of 1 at the end of each period for nperiods. Find an expression at issue for the present value of the interest which will be paid over the life of the loan. Problem 38.14 A loan of 10,000 is being repaid with 20 installments at the end of each year, at 10% eﬀective. Show that the amount of interest in the 11th installment is 1000 1 + ν10 . Problem 38.15 A loan is being repaid with 20 installments at the end of each year at 9% eﬀective. In what installment are the principal and interest portions most nearly equal to each other? Problem 38.16 A loan is being repaid with a series of payments at the end of each quarter, for 5 years. If the amount of principal in the 3rd payment is 100, ﬁnd the amount of principal in the last 5 payments. Interest is at the rate of 10% convertible quarterly. Problem 38.17 A loan is being repaid with installments of 1 at the end of each year for 20 years. Interest is at eﬀective rate i for the ﬁrst 10 years, and eﬀective rate j for the second 10 years. Find expressions for (a) the amount of interest paid in the 5th installment; (b) the amount of principal repaid in the 15th installment. 350 LOAN REPAYMENT METHODS Problem 38.18 A loan of 25,000 is being repaid with annual payments of 2,243 at the end of the year. The interest rate on the loan 7.5%. Calculate the interest in the 5th payment. Problem 38.19 ‡ Seth borrows X for four years at an annual eﬀective interest rate of 8%, to be repaid with equal payments at the end of each year. The outstanding loan balance at the end of the second year is 1076.82 and at the end of the third year is 559.12 . Calculate the principal repaid in the ﬁrst payment. Problem 38.20 ‡ A bank customer takes out a loan of 500 with a 16% nominal interest rate convertible quarterly. The customer makes payments of 20 at the end of each quarter. Calculate the amount of principal in the fourth payment. Problem 38.21 ‡ A loan is repaid with level annual payments based on an annual eﬀective interest rate of 7%. The 8th payment consists of 789 of interest and 211 of principal. Calculate the amount of interest paid in the 18th payment. Problem 38.22 ‡ Kevin takes out a 10-year loan of L, which he repays by the amortization method at an annual eﬀective interest rate of i.Kevin makes payments of 1000 at the end of each year . The total amount of interest repaid during the life of the loan is also equal to L. Calculate the amount of interest repaid during the ﬁrst year of the loan. Problem 38.23 ‡ Ron is repaying a loan with payments of 1 at the end of each year for n years. The amount of interest paid in period tplus the amount of principal repaid in period t+ 1 equals X. Calculate X. Problem 38.24 A 35-year loan is to be repaid with equal installments at the end of each year. The amount of interest paid in the 8th installment is 135. The amount of interest paid in the 22nd installment is 108. Calculate the amount of interest paid in the 29th installment. Problem 38.25 A 1000 loan is repaid with equal payments at the end of each year for 20 years. The principal portion of the 13th payment is 1.5 times the principal portion of the 5th payment. Calculate the total amount of interest paid on the loan. 38 AMORTIZATION SCHEDULES 351 Problem 38.26 A loan is being amortized with payments at the end of each quarter for 25 years. If the amount of principal repaid in the third payment is $100, ﬁnd the total amount of principal repaid in the forty payments consisting of payments eleven through ﬁfty. Interest is at the rate of 8% convertible quarterly. Problem 38.27 A loan is being amortized using an interest rate of 8% convertible quarterly with level payments at the end of each quarter. If the amount of principal repaid in the third payment is $100, ﬁnd the total amount of principal repaid in the ﬁve payments consisting of payments eleven through ﬁfteen. Problem 38.28 A 100,000 loan is to be repaid by 30 equal payments at the end of each year. The outstanding balance is amortized at 4%. In addition to the annual payments, the borrower must pay an origination fee at the time the loan is made. The fee is 2% of the loan but does not reduce the loan balance. When the second payment is due, the borrower pays the remaining loan balance. Determine the yield to the lender considering the origination fee and the early pay-oﬀ of the loan. Problem 38.29 A loan of 10,000 is amortized by equal annual payments for 30 years at an annual eﬀective interest rate of 5%. Determine the year in which the interest portion of the payment is most nearly equal to one-third of the payment. Problem 38.30 A loan of 1000 at a nominal rate of 12% convertible monthly is to be repaid by six monthly payments with the ﬁrst payment due at the end of 1 month. The ﬁrst three payments are X each, and the ﬁnal three payments are 3X each. Determine the sum of the principal repaid in the third payment and the interest paid in the ﬁfth payment. Problem 38.31 Carl borrows $10,000 at 12% compounded monthly, and Carl will repay the loan with 60 monthly amortization payments beginning at the end of the ﬁrst month. Find: (a) B20 (b) B30 (c) ∑30 j=21 Pj (d) ∑30 j=21 Ij Problem 38.32 ‡ A bank customer borrows X at an annual eﬀective rate of 12.5% and makes level payments at the end of each year for nyears. The interest portion of the ﬁnal payment is 153.86. The total principal repaid as of time ( n−1) is 6009.12. The principal repaid in the ﬁrst payment is Y. Calculate Y. 352 LOAN REPAYMENT METHODS Problem 38.33 A loan is to be amortized by n level annual payments of X where n> 5. You are given (1) The amount of interest in the ﬁrst payment is 604.00 (2) The amount of interest in the third payment is 593.75 (3) The amount of interest in the ﬁfth payment is 582.45 Calculate X. Problem 38.34 A loan is to be repaid by annual installments of X at the end of each year for 10 years. You are given (1) the total principal repaid in the ﬁrst 3 years is 290.35 (2) the total principal repaid in the last 3 years is 408.55 Calculate the total amount of interest paid during the life of the loan. Problem 38.35 Iggy borrows X for 10 years at an annual eﬀective rate of 6%. If he pays the principal and accumulated interest in one lump sum at the end of 10 years, he would pay 356.54 more in interest than if he repaid the loan with 10 level payments at the end of each year. Calculate X. Problem 38.36 A loan is being amortized by means of level monthly payments at an annual eﬀective interest rate of 8%. The amount of principal repaid in the 12th payment is 1000 and the amount of principal repaid in the tth payment is 3700. Calculate t. Problem 38.37 John is repaying a loan with payments of $3,000 at the end of every year over an unknown period of time. if the amount on interest in the third installment is $2,000, ﬁnd the amount of principal in the sixth installment. Assume the interest is 10% convertible quarterly. 39 SINKING FUND METHOD 353 39 Sinking Fund Method An alternative for repaying a loan in installments by the amortization method, a borrower can accumulate a fund which will exactly repay the loan in one lump sum at the end of a speciﬁed period of time. This fund is called a sinking fund. It is generally required that the borrower periodically pay interest on the loan, sometimes referred to as a service. In some cases payments into a sinking fund may vary irregularly at the discretion of the borrower. However, we will be dealing with sinking funds with regular contributions. Because the balance in the sinking fund could be applied against the loan at any point, the net amount of the loan is equal to the original amount of the loan minus the sinking fund balance. We next show that if the rate of interest paid on the loan equals the rate of interest earned on the sinking fund, then the amortization method and the sinking fund method are equivalent. To see this, Suppose that the eﬀective annual rate on the loan is i, and the sinking fund earns the same rate. Suppose that the amount of the loan is 1 , and the loan term is n periods. With the amortization method, the payment at the end of each period is 1 an .With the sinking fund method, to accumulate the amount of 1 in the sinking fund, the borrower deposits of 1 sn at the end of each year for n years. At the same moment the borrower also pays i per period to the lender. That is, payments of size i (interest) plus payments of size 1 sn are required. But we know from Section 15 that 1 sn + i= 1 an . Thus, the two methods are equivalent. As was seen in Section 38, a way to visualize the activity of an amortization method is the creation of an amortization schedule. The same idea can be used with sinking funds. The following example illustrates the creation of a sinking fund schedule. Example 39.1 Create a sinking fund schedule for a loan of $1000 repaid over four years if the annual eﬀective rate of interest is 8% Solution. If R is the sinking fund deposit then R= 1000 s4 = $221.92. 354 LOAN REPAYMENT METHODS Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1000.00 1 80.00 221.92 0 221.92 778.08 2 80.00 221.92 17.75 461.59 538.41 3 80.00 221.92 36.93 720.44 279.56 4 80.00 221.92 57.64 1000.00 0 Table 39.1 Notice from this table that the amount in the sinking fund after the tth payment is found by multiplying the sinking fund deposit by stj where j is the sinking fund rate of interest. Comparing Table 38.1 and Table 39.1 we notice the following: (1) For each period, Interest paid + Sinking Fund deposit (in Table 39.1) = Payment amount (in Table 38.1). (2) For each period, Interest paid −Interest earned on sinking fund (in Table 39.1) = Interest paid (in Table 38.1). (3) For each period, Sinking fund deposit + Interest earned on sinking fund (in Table 39.1) = principal repaid (in Table 38.1). (4) For each period, the Net amount of loan (in Table 39.1) = Outstanding loan balance in (Table 38.1) Notice that the net amount of loan concept plays the same role for the sinking fund method that the outsanding loan balance does for the amortization method. Next, we consider the situation in which the interest rate on the loan and the interest rate earned on the sinking fund diﬀers. The rate on the loan is denoted by i, and the rate on the sinking fund is denoted by j. Usually, j is less than i because the sinking fund wouldn’t normally be riskier than the loan. In this case, an amount of i will be deducted from the sinking fund deposit and the remaining amount will be invested in the sinking fund at a rate of j. Let ani&j be the present value of an annuity which pays 1 at the end of each period for n periods with i and j as previously deﬁned. If a loan of 1 is made, by the amortization method the amount of the loan will be repaid by interest payments of iani&j at the end of each year for nyears together with yearly deposits of 1 −iani&j into a sinking fund which earns interest at the eﬀective annual rate of interest j. The sinking ﬁnd should accumulate ani&j at the end of n years. That is, (1 −iani&j)snj = ani&j or ani&j = snj 1 + isnj 39 SINKING FUND METHOD 355 which is equivalent to 1 ani&j = 1 snj + i. In other words, loan of 1 can be repaid by interest payments of i to the lender and depositing 1 snj into the sinking fund at the end of each year for n years. Now since 1 anj = 1 snj + j it follows that 1 ani&j = 1 anj + (i−j) or ani&j = anj 1 + (i−j)anj . It should be noted that if i= j, then ani&j = ani. Example 39.2 ‡ John borrows 10,000 for 10 years at an annual eﬀective interest rate of 10%. He can repay this loan using the amortization method with payments of 1,627.45 at the end of each year. Instead, John repays the 10,000 using a sinking fund that pays an annual eﬀective interest rate of 14%. The deposits to the sinking fund are equal to 1,627.45 minus the interest on the loan and are made at the end of each year for 10 years. Determine the balance in the sinking fund immediately after repayment of the loan. Solution. Under the amortization method, the periodic installment is 1627.45. The periodic interest payment on the loan is 0.1(10000) = 1000.Thus, deposits into the sinking fund are 1627 .45 −1000 = 627.45. Thus, the amount in sinking fund immediately after repayment of the loan is 627.45s10 0.14−10,000 = 2130 In general, the sinking fund schedule at two rates of interest is identical to the sinking fund schedule at one rate of interest equal to the rate of interest earned on the sinking fund, except that a constant addition of (i−j) times the amount of the original loan is added to the interest paid column. We illustrate this in the next example. Example 39.3 Create a sinking fund schedule for a loan of $1,000 repaid over four years if the annual eﬀective rate of interest is 10% and the sinking fund interest of 8% 356 LOAN REPAYMENT METHODS Solution. If R is the sinking fund deposit then R= 1,000 s4 0.08 = $221.92. Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1000.00 1 100.00 221.92 0 221.92 778.08 2 100.00 221.92 17.75 461.59 538.41 3 100.00 221.92 36.93 720.44 279.56 4 100.00 221.92 57.64 1000 0 Note that this table is identical to Table 39.1 except each entry in the interest paid column in the table above is equal to each entry in the interest paid column in Table 39.1 increased by a constant equals to 20 which is (0 .10 −0.08) ×1000 The following is an important example whose results are to keep in mind. Example 39.4 A loan of 1 yields the lender rate iper period for nperiods, while the borrower replaces the capital in a sinking fund earning rate j per period. Find expressions for the following if 1 ≤t≤n: (a) Periodic interest paid to the lender. (b) Periodic sinking fund deposit. (c) Interest earned on sinking fund during the tth period. (d) Amount in sinking fund at end of the tth period. (e) Net amount of loan at the end of the tth period. (f) Net interest paid in period t. (g) Principal repaid in period t. Solution. (a) The yield rate is i, so the lender must be receiving an amount of i each period. (b) The capital is 1, so the borrower is depositing ( snj)−1 regularly into the sinking fund. (c) At the beginning of the tth period the balance in the sinking fund is ( st−1 j/snj); during the period it earns interest in the amount of j· st−1 j snj payable at the end of the period, i.e., at time t. (d) The amount in sinking fund at end of the tth period is stj snj . 39 SINKING FUND METHOD 357 (e) The net amount of loan at the end of the tth period is the excess of 1 over the balance in the sinking fund, i.e., 1 − stj snj . (f) The net interest paid in the tth period is the excess of interest paid over interest earned, i.e. i−j st−1 j snj . (g) By (e) above, the change in the amount of the loan between the ( t−1)th and the tth payment is ( 1 −stj snj ) − ( 1 − st−1 j snj ) = (1 + i)t−1 snj In creating the sinking fund schedule the payment period and the conversion interest period can be diﬀerent. These cases can be handled from basic principles as illustrated in the next two examples. Remember to use the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. Example 39.5 John borrows $5,000 for 10 years at 10% convertible quarterly. John does not pay interest currently and will pay all accrued interest at the end of 10 years together with the principal. Find the annual sinking fund deposit necessary to liquidate the loan at the end of 10 years if the sinking fund earns 7% convertible semi-annually. Solution. Let j be the annual interest rate. Then j = (1.025)4 −1. The loan balance at the end of 10 years is 5000(1 + j)10 = 5000(1.025)40 = 13425.32. Hence, the annual sinking fund deposit is R= 13425.32 s10 (1.035)2−1 = 966.08 Example 39.6 Create the sinking fund table for the following: a 3 year loan of $10,000, with interest payable semiannually at the nominal interest rate of 8.00% is to be retired by a sinking fund funded by quarterly deposits earning an eﬀective nominal interest rate of 6.00% compounded semiannually. Solution. Every six months the interest paid on the loan is 0 .04(10,000) = 400 . The quarterly rate on the sinking fund is j = 1.030.5 −1. The quarterly sinking fund deposit is D= 10,000 s12 j = 767.28. 358 LOAN REPAYMENT METHODS Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 10,000.00 0.25 0 767.28 0 767.28 9,232.72 0.50 400.00 767.28 11.42 1,545.98 8,454.02 0.75 0 767.28 23.02 2,336.27 7,663.73 1.00 400.00 767.28 34.79 3,138.33 6,861.67 1.25 0 767.28 46.73 3,952.33 6,047.67 1.50 400.00 767.28 58.85 4,778.45 5,221.55 1.75 0 767.28 71.15 5,616.88 4,383.12 2.00 400.00 767.28 83.63 6,467.78 3,532.22 2.25 0 767.28 96.30 7,331.36 2,668.64 2.50 400.00 767.28 109.16 8,207.79 1,792.21 2.75 0 767.28 122.21 9,097.27 902.73 3.00 400.00 767.28 135.45 10,000.00 0.00 39 SINKING FUND METHOD 359 Practice Problems Problem 39.1 A loan of 10,000 is being repaid with annual payments for 10 years using the sinking fund method. The loan charges 10% interest and the sinking fund earns 8%. (a) Calculate the interest payment that is paid annually to service the loan. (b) Calculate the sinking fund payment made annually. (c) Calculate the amount in the sinking fund immediately after the deposit made at the end of 5 years. (d) Create the sinking fund schedule for the loan. Problem 39.2 If the loan in Problem 39.1 was repaid using the amortization method, but the annual payment was equal to the sum of the interest payment and the sinking fund deposit, calculate the interest rate under the amortization method. Problem 39.3 A loan of 20,000 is being repaid with annual payments for 5 years using the sinking fund method. The loan charges 10% interest compounded twice a year. The sinking fund earns 8% compounded monthly. Calculate the interest payment that is paid annually to service the loan and the sinking fund deposit. Problem 39.4 A loan of 20,000 is being repaid with monthly payments for 5 years using the sinking fund method. The loan charges 10% interest compounded twice a year. The sinking fund earns an annual eﬀective interest rate of 8%. (a) Calculate the interest payment that is paid monthly to service the loan and the sinking fund deposit paid monthly. (b) Calculate the amount in the sinking fund immediately after the 30th payment. Problem 39.5 Julie agrees to repay a loan of 10,000 using the sinking fund method over 10 years. The loan charges an annual eﬀective interest rate of 7% while the sinking fund earns 6%. Calculate the amount paid into the sinking fund each year less the amount of interest paid on the loan each year. Problem 39.6 Kathy can take out a loan of 50,000 with Bank A or Bank B. With Bank A, she must repay the loan with 60 monthly payments using the amortization method with interest at 7% compounded 360 LOAN REPAYMENT METHODS monthly. With Bank B, she can repay the loan with 60 monthly payments using the sinking fund method. The sinking fund will earn 6.5% compounded monthly. What interest rate can Bank B charge on the loan so that Kathy’s payment will be the same under either option? Problem 39.7 Lauren is repaying a loan of 100,000 using the sinking fund method. At the end of each year she pays 7,000 into a sinking fund earning 8%. At the end of 5 years, Lauren pays oﬀ the loan using the sinking fund plus an additional payment of X. Calculate X. Problem 39.8 Lauren is repaying a loan of 100,000 using the sinking fund method. At the end of each year she pays 7,000 into a sinking fund earning 8%. At the end of year Y, Lauren will have suﬃcient money in the sinking fund to repay the loan. Calculate Y. Problem 39.9 Ryan takes out a loan of 100,000 and agrees to repay it over 10 years using the sinking fund method. Ryan agrees to pay interest to the lender at the end of each year. The interest rate is .01(11 −t) in year t. The sinking fund will earn 5% per year. Determine the amount in the sinking fund after 10 years if the total payment made by Ryan at the end of each year is 12,000. Problem 39.10 On a loan of 10,000, interest at 9% eﬀective must be paid at the end of each year. The borrower also deposits X at the beginning of each year into a sinking fund earning 7% eﬀective. At the end of 10 years the sinking fund is exactly suﬃcient to pay oﬀ the loan. Calculate X. Problem 39.11 A borrower is repaying a loan with 10 annual payments of 1,000. Half of the loan is repaid by the amortization method at 5% eﬀective. The other half of the loan is repaid by the sinking fund method, in which the lender receives 5% eﬀective on the investment and the sinking fund accumulates at 4% eﬀective. Find the amount of the loan. Problem 39.12 A borrows 12,000 for 10 years, and agrees to make semiannual payments of 1,000. The lender receives 12% convertible semiannually on the investment each year for the ﬁrst 5 years and 10% convertible semiannually for the second 5 years. The balance of each payment is invested in a sinking fund earning 8% convertible semiannually. Find the amount by which the sinking fund is short of repaying the loan at the end of the 10 years. 39 SINKING FUND METHOD 361 Problem 39.13 A borrower takes out a loan of 3000 for 10 years at 8% convertible semiannually. The borrower replaces one-third of the principal in a sinking fund earning 5% convertible semiannually, and the other two-thirds in a sinking fund earning 7% convertible semiannually. Find the total semiannual payment. Problem 39.14 A payment of 36,000 is made at the end of each year for 31 years to repay a loan of 400,000. If the borrower replaces the capital by means of a sinking fund earning 3% eﬀective, ﬁnd the eﬀective rate paid to the lender on the loan. Problem 39.15 A loan of 30,000 is to be repaid using the sinking fund method over 6 years. The interest on the loan is paid at the end of each year and the interest rate is 10%. The sinking fund payment is made at the beginning of each year with the sinking fund earning 6%. Calculate the amount paid into the sinking fund each year. Problem 39.16 You have two equivalent ways to repay a loan of $100,000 over 20 years. (a) Using the sinking fund method, where the rate of interest on the loan is 8% and the rate of interest earned by the sinking fund is 6%. (b) Using the amortization method where the rate of interest is k. Find k. Problem 39.17 ‡ A 12-year loan of 8000 is to be repaid with payments to the lender of 800 at the end of each year and deposits of X at the end of each year into a sinking fund. Interest on the loan is charged at an 8% annual eﬀective rate. The sinking fund annual eﬀective interest rate is 4%. Calculate X. Problem 39.18 ‡ A 20-year loan of 20,000 may be repaid under the following two methods: i) amortization method with equal annual payments at an annual eﬀective rate of 6.5% ii) sinking fund method in which the lender receives an annual eﬀective rate of 8% and the sinking fund earns an annual eﬀective rate of j Both methods require a payment of X to be made at the end of each year for 20 years. Calculate j. Problem 39.19 ‡ Lori borrows 10,000 for 10 years at an annual eﬀective interest rate of 9%. At the end of each year, she pays the interest on the loan and deposits the level amount necessary to repay the principal to a sinking fund earning an annual eﬀective interest rate of 8%. The total payments made by Lori over the 10-year period is X. Calculate X. 362 LOAN REPAYMENT METHODS Problem 39.20 John borrows 10,000 for 10 years and uses a sinking fund to repay the principal. The sinking fund deposits earn an annual eﬀective interest rate of 5%. The total required payment for both the interest and the sinking fund deposit made at the end of each year is 1445.04. Calculate the annual eﬀective interest rate charged on the loan. Problem 39.21 Joe repays a loan of 10,000 by establishing a sinking fund and making 20 equal payments at the end of each year. The sinking fund earns 7% eﬀective annually. Immediately after the ﬁfth payment, the yield on the sinking fund increases to 8% eﬀective annually. At that time, Joe adjusts his sinking fund payment to X so that the sinking fund will accumulate to 10,000 20 years after the original loan date. Determine X. Problem 39.22 A corporation borrows 10,000 for 25 years, at an eﬀective annual interest rate of 5%. A sinking fund is used to accumulate the principal by means of 25 annual deposits earning an eﬀective annual interest rate of 4%. Calculate the sum of the net amount of interest paid in the 13 th installment and the increment in the sinking fund for the ninth year. Problem 39.23 A loan of 1,000 is taken out at an annual eﬀective interest rate of 5%. Level annual interest payments are made at the end of each year for 10 years, and the principal amount is repaid at the end of 10 years. At the end of each year, the borrower makes level annual payments to a sinking fund that earns interest at an annual eﬀective rate of 4%. At the end of 10 years the sinking fund accumulates to the loan principal. Calculate the diﬀerence between the interest payment on the loan and the interest earned by the sinking fund in the ﬁfth year. Problem 39.24 John borrows 10,000 for 10 years at an annual eﬀective interest rate ofi.He accumulates the amount necessary to repay the loan by using a sinking fund. He makes 10 payments of X at the end of each year, which includes interest on the loan and the payment into the sinking fund, which earns an annual eﬀective rate of 8%. If the annual eﬀective rate of the loan had been 2 i,his total annual payment would have been 1.5X. Calculate i. Problem 39.25 Jason and Margaret each take out a 17 year loan of L.Jason repays his loan using the amortization method, at an annual eﬀective interest rate of i. He makes an annual payment of 500 at the end of each year. Margaret repays her loan using the sinking fund method. She pays interest annually, also at an annual eﬀective interest rate of i. In addition, Margaret makes level annual deposits at 39 SINKING FUND METHOD 363 the end of each year for 17 years into a sinking fund. The annual eﬀective rate on the sinking fund is 4.62% and she pays oﬀ the loan after 17 years. Margaret’s total payment each year is equal to 10% of the original loan amount. Calculate L. Problem 39.26 A 10 year loan of 10,000 is to be repaid with payments at the end of each year consisting of interest on the loan and a sinking fund deposit. Interest on the loan is charged at a 12% annual eﬀective rate. The sinking fund’s annual eﬀective interest rate is 8%. However, beginning in the sixth year the annual eﬀective interest rate on the sinking fund drops to 6%. As a result, the annual payment to the sinking fund is then increased by X. Calculate X. Problem 39.27 NTL Corp. has just taken out a 15 year, 200,000 loan on which it has to make semi-annual interest payments of 7,000. As part of the loan agreement, NTL Corp. needs to make a deposit at the end of every 6 months into a sinking fund in order to retire the loan at the end of the 15th year. NTL Corp. plans to make semi-annual deposits of 4,929.98 into the sinking fund. (a) What eﬀective annual interest rate is NTL Corp. assuming it can earn on the sinking fund? (b) What is the net interest cost in the 6 months following the 10th sinking fund deposit? Problem 39.28 E Corp has a $100,000 loan outstanding on which it has been paying semi-annual interest payments of $4,000. E Corp has also been accumulating deposits made at the end of every 6 months in a sinking fund earning 5% eﬀective annual interest so that it can retire the loan at the end of the 15th year. The lender has oﬀered to accept 103% of the sinking fund balance immediately after the 29th deposit in exchange for the outstanding loan balance and the remaining loan interest payment. At what eﬀective annual rate was the lender calculating the present value of the remaining amounts in order to make this oﬀer equivalent in value? Problem 39.29 A borrower takes out a loan of $2,000 for two years. Create a sinking fund schedule if the lender re- ceives 10% eﬀective on the loan and if the borrower replaces the amount of the loan with semiannual deposits in a sinking fund earning 8% convertible quarterly. Problem 39.30 John borrows $10,000 for ﬁve years at 12% convertible semi-annually. John replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Find the total dollar amount that John must pay over the ﬁve-year period to completely repay the loan. 364 LOAN REPAYMENT METHODS Problem 39.31 A borrower is repaying a loan of 300,000 by the sinking fund method. The sinking fund earns an annual eﬀective interest rate of 6.75%. Payments of $22,520 are made at the end of each year for 20 years to repay the loan. These payments consist of both the interest payment to the lender and also the sinking fund deposit. What is the annual eﬀective interest rate paid to the lender of the loan? 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE365 40 Loans Payable at a Diﬀerent Frequency than Interest is Convertible In Sections 37 - 39, we considered loans where the payment period coincides with the interest conversion period. In this section we examine loans where the payments are made at a diﬀerent frequency than interest is convertible. The same observations made about the amortization method and the sinking fund method apply to the loans considered in this section. We will restrict our discussion to payments made at the end of an interest conversion period. A similar argument holds for payments made at the beginning of an interest conversion period. Consider ﬁrst the amortization schedule of a loan with payments made less frequently than interest is convertible. Consider a loan of an sk at interest rate i per period being repaid with payments of 1 at the end of each k interest conversion periods for a total of n interest conversion periods. Thus, the total number of payments is n k which we assume is an integral number. At the end of k interest conversion periods a payment of 1 is made, the interest paid is [(1 + i)k − 1]an sk = 1−νn so that the principal repaid is νn,and the outstanding loan balance is an sk −νn = an−k sk . Next, at the end of 2 k interest conversion periods, the interest paid is [(1 + i)k −1] an−k sk = 1 −νn−k so that the principal repaid is νn−k, and the outstanding loan balance is an−k sk −νn−k = an−2k sk . Continuing this process, we see that at the end of period mk, the interest paid is [(1 + i)k − 1] an−(m−1)k sk = 1 −νn−(m−1)k and the principal repaid is νn−(m−1)k.The outstanding loan balance is an−(m−1)k sk −νn−(m−1)k = an−mk sk . The amortization table is shown below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 an sk k 1 [(1 + i)k −1]an sk = 1 −νn νn an sk −νn = an−k sk 2k 1 [(1 + i)k −1] an−k sk = 1 −νn−k νn−k an−k sk −νn−k = an−2k sk ... ... ... ... ... mk 1 [(1 + i)k −1] an−(m−1)k sk = 1 −νn−(m−1)k νn−(m−1)k an−(m−1)k sk −νn−(m−1)k = an−mk sk ... ... ... ... ... n−k 1 [(1 + i)k −1] a2k sk = 1 −ν2k ν2k a2k sk −ν2k = ak sk n 1 [(1 + i)k −1] ak sk = 1 −νk νk ak sk −νk = 0 Total n k n k −an sk an sk 366 LOAN REPAYMENT METHODS Note that the principal repaid column is a geometric progression with common ratio (1 + i)k. Example 40.1 A loan of $15,000 is to be repaid by means of six annual payments of $3719.18 each. The nominal interest rate is 12% compounded monthly. Create an amortization schedule for this transaction. Solution. The schedule is given below. Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 15,000 12 3719.18 1902.37 1816.81 13,183.19 24 3719.18 1671.96 2047.22 11,135.97 36 3719.18 1412.32 2306.86 8829.11 48 3719.18 1119.75 2599.43 6229.68 60 3719.18 790.08 2929.10 3300.58 72 3719.18 418.60 3300.58 0 Next, consider a loan of a(m) n at interest rate i per period being repaid with payments of 1 m at the end of each mth of an interest conversion period for a total of n interest conversion periods. Thus, the total number of payments is mn which we assume is an integral number. At the end of the ﬁrst mth of an interest conversion period a payment of 1 m is made, the interest paid is i i(m) a(m) n = 1 m(1 −νn) so that the principal repaid is 1 mνn, and the outstanding loan balance is a(m) n −1 mνn = a(m) n−1 m .Next, at the end of second mth of an interest conversion period, the interest paid is i i(m) a(m) n−1 m = 1 m(1 −νn−1 m) so that the principal repaid is 1 mνn−1 m, and the outstanding loan balance is a(m) n−1 m −1 mνn−1 m = a(m) n−2 m .Continuing this process, we see that at the end of period t n, the interest paid is i i(m) a(m) n−(t−1) m = 1 m(1 −νn−t−1 m ) and the principal repaid is 1 mνn−(t−1) m .The outstanding loan balance is a(m) n−t−1 m −1 mνn−t−1 m = a(m) n−t m . The amortization table is shown below. 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE367 Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 a(m) n 1 m 1 m i i(m) a(m) n = 1 m(1 −νn) 1 mνn a(m) n −1 mνn = a(m) n−1 m 2 m 1 m i i(m) a(m) n−1 m = 1 m(1 −νn−1 m) 1 mνn−1 m a(m) n−1 m −1 mνn−1 m = a(m) n−2 m ... ... ... ... ... t m 1 m i i(m) a(m) n−(t−1) m = 1 m(1 −νn−t−1 m ) 1 mνn−(t−1) m a(m) n−t−1 m −1 mνn−t−1 m = a(m) n−t m ... ... ... ... ... n−1 m 1 m i i(m) a(m) 2 m = 1 m(1 −ν 2 m) 1 mν 2 m a(m) n−2 m −1 mνn−2 m = a(m) n−1 m n 1 m i i(m) a(m) 1 m = 1 m(1 −ν 1 m) 1 mν 1 m a(m) n−1 m −1 mνn−1 m = a(m) n−1 m −1 mν 1 m = 0 Total n n−a(m) n a(m) n Note that the principal repaid column is a geometric progression with common ratio (1 + i) 1 m. Example 40.2 A debt is being amortized by means of monthly payments at an annual eﬀective rate of interest of 11%. If the amount of principal in the third payment is $1000, ﬁnd the amount of principal in the 33rd payment. Solution. Recall that the principal repaid column is a geometric progression with common ratio (1+ i) 1 m.The interval of time from the 3rd payment to the 33rd payment is 33−3 12 = 2.5 years. Thus, the principal in the 33rd payment is 1000(1.11)2.5 = $1298.10 Example 40.3 A loan of $10,000 is to be repaid by means of twelve monthly payments. The nominal interest rate is 12% compounded quarterly. Create an amortization schedule for this transaction. Solution. Let Rbe the amount of monthly payment. Then Rsatisﬁes the equation 10000 = 3Ra4 (3) 0.03.Solving this equation for R we ﬁnd R= $887.94. Moreover, we have i(3) 3 a4 (3) 0.03 = 0.02970490211 3 = 99.02. The schedule is given below. 368 LOAN REPAYMENT METHODS Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 10,000 1 3 887.94 99.02 788.92 9211.08 2 3 887.94 91.20 796.74 8414.34 1 887.94 83.32 804.62 7609.72 4 3 887.94 75.35 812.59 6797.13 5 3 887.94 67.30 820.64 5976.49 2 887.94 59.18 828.76 5147.73 7 3 887.94 50.97 836.97 4310.76 8 3 887.94 42.68 845.26 3465.50 3 887.94 34.31 853.63 2611.87 10 3 887.94 25.86 862.08 1749.79 11 3 887.94 17.33 870.61 879.18 4 887.94 8.71 879.23 0 It is not recommended that the reader depends upon the memorization of formulas in amortization and sinking fund schedules. It is rather more preferable to create these schedules using the basic principles. We illustrate the use of basic principles in creating a sinking fund schedule in the next example. Example 40.4 John borrows $2000 for two years at an annual eﬀective interest rate of 10%. He replaces the principal by means of semiannual deposits for two years in a sinking fund that earns 8% convertible quarterly. Create a sinking fund schedule for this transaction. Solution. The interest payment on the loan is 2000 ×10% = $200 at the end of each year. If R is the semiannual deposit in the sinking fund then Rs8 0.02 s2 0.02 = 2000. Solving this equation for R we ﬁnd R= 2000s2 0.02 s8 0.02 = $470.70. The sinking fund schedule is given below. 40 LOANS PAYABLE AT A DIFFERENT FREQUENCY THAN INTEREST IS CONVERTIBLE369 Interest earned Interest Sinking fund on sinking Amount in Net amount Period paid deposit fund sinking fund of loan 0 1/4 0 0 0 0 2000.00 1/2 0 470.70 0 470.70 1529.30 3/4 0 0 9.41 480.11 1519.89 1 200.00 470.70 9.60 960.41 1039.59 1 1/4 0 0 19.21 979.62 1020.38 1 1/2 0 470.70 19.59 1469.91 530.09 1 3/4 0 0 29.40 1499.31 500.69 2 200.00 470.70 29.99 2000.00 0 370 LOAN REPAYMENT METHODS Practice Problems Problem 40.1 A debt is being amortized by means of annual payments at a nominal interest rate of 12% com- pounded semiannually. If the amount of principal in the third payment is $1000, ﬁnd the amount of principal in the 33rd payment. Problem 40.2 A borrower is repaying a loan by making quarterly payments of $500 over 10 years. Portion of each payment is principal and the rest is interest. Assume an annual eﬀective rate of 8%, how much interest is paid by the borrower over the 10-year period? Problem 40.3 A lender receives payments of $3000 at the end of every year over an unknown number of years. If the interest portion of the third payment is $2000, ﬁnd the amount of principal in the sixth payment. Assume a nominal interest rate of 10% payable quarterly. Problem 40.4 A borrows $10,000 for ﬁve years at 12% convertible semiannually. A replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Find the total dollar amount which A must pay over the ﬁve-year period to completely repay the loan. Problem 40.5 A borrows $5000 for 10 years at 10% convertible quarterly. A does not pay interest currently and will pay all accrued interest at the end of 10 years together with the principal. Find the annual sinking fund deposit necessary to liquidate the loan at the end of 10 years if the sinking fund earns 7% convertible quarterly. Problem 40.6 A borrows $10,000 for ﬁve years at 12% convertible semiannually. A replaces the principal by means of deposits at the end of every year for ﬁve years into a sinking fund which earns 8% eﬀective. Create a sinking fund schedule for this loan. 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 371 41 Amortization with Varying Series of Payments In this section, we consider amortization methods with more general patterns of payments, not necessarily leveled. We will keep assuming that the interest conversion period and the payment period are equal and coincide. Consider a loan L to be repaid with n periodic payments (that include principal and interest) R1,R2,··· ,Rn. The equation of value at time t= 0 is L= n∑ t=1 νtRt. In most cases the series of payments Rt follows some regular pattern encountered with annuties so that the results of Section 26 can be used. Example 41.1 A borrower is repaying a loan with payments at the end of each year for 10 years, such that the payment the ﬁrst year is $200, the second year is $190, and so forth, until the 10th year it is $110. Find an expression of the amount of the loan. Solution. The amount of the loan is L= 100a10 + 10(Da)10 For the type of installments considered in this section, amortization schedules can be constructed from ﬁrst principles as discussed in Section 38. Furthermore, the outstanding loan balance column can be found retrospectively or prospectively as in Section 37, from which the remaining columns of interest paid and principal repaid can be found. Example 41.2 A loan is repaid with payments which start at $200 the ﬁrst year and increase by $50 per year until a payment of $1000 is made, at which time payments cease. If the interest is 4% eﬀective, ﬁnd the amount of interest and principal paid in the fourth payment. Solution. The fourth payment is R4 = $350.00.The outstanding loan balance after making the third payment is Bp 3 = 300a14 + 50(Ia)14 . 372 LOAN REPAYMENT METHODS The interest paid after the fourth payment is I4 =iBp 3 = 300(1 −ν14) + 50(¨a14 −14ν14) =300(1 −0.57747) + 50(10.98565 −8.08465) =$271.81 The principal portion of the fourth payment is P4 = R4 −I4 = 350.00 −271.81 = $78.19 One common pattern of periodic installments is when the borrower makes level payments of princi- pal. Clearly, this leads to successive total payments (consisting of interest and principal) to decrease due to the decrease in successive outstanding loan balance (which in turn results in a decrease in successive interest paid). We illustrate this in the following example. Example 41.3 A borrower is repaying a $1000 loan with 10 equal payments of principal. Interest at 6% compounded semiannually is paid on the outstanding balance each year. Find the price to yield an investor 10% convertible semiannually. Solution. The period principal payment is $100 each. The periodic interest payments are :30 ,27,24,··· ,9,3. Thus, the price at time t= 0 is just the present value of the principal and interest paid. That is, Price = 100a10 0.05 + 3(Da)10 0.05 = 908.87 It is possible that when using varying payments, the loan payment for a period is less than the interest charged over that period. This leads to an increase in the outstanding balance which in turn means a negative principal. In ﬁnance, such a case is referred to as negative amortization or deferred interest . Credit card interest rates and payment plans are examples of negative amortization. Most credit cards carry high interest rates yet require a low minimum payment. Example 41.4 A loan of 1000 is being repaid with annual payments over 10 years. The payments in the last ﬁve years are 5 times the payments in the ﬁrst 5 years. If i= 0.08,calculate the principal amortized in the ﬁfth payment. Solution. Let K be the size of the payment in the ﬁrst 5 years. Then 1000 = K(1 −1.08−5) 0.08 + 5K(1.08−5)(1 −1.08−5) 0.08 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 373 or 1000 = 17.57956685K →K = 56.88422296. The outstanding loan balance after making the 4th payment is 56.88422296 1.08 + 5(56.88422296)(1.08−1)(1 −1.08−5) 0.08 = 1104.16228. The outstanding loan balance after making the 5th payment is 5(56.88422296)(1 −1.08−5) 0.08 = 1135.61104. The principal amortized in the ﬁfth payment is 1104 .16228 −1135.61104 = −$31.45 Next, consider a varying series of payments with the sinking fund method. We will assume that the interest paid to the lender is constant each period so that only the sinking fund deposits vary. Let the varying payments be denoted by R1,R2,··· ,Rn and assume that i ̸= j. Let L denote the amount of the loan. Then the sinking fund deposit for the tth period is Rt −iL. Since the accumulated value of the sinking fund at the end of n periods must be L, we have L=(R1 −iL)(1 + j)n−1 + (R2 −iL)(1 + j)n−2 + ··· + (Rn −iL) = n∑ t=1 Rt(1 + j)n−t −iLsnj where i is rate of interest paid on the loan and j rate of interest earned on the sinking fund. Solving for L we obtain L= ∑n t=1 Rt(1 + j)n−t 1 + isnj = ∑n t=1 νt jRt 1 + (i−j)anj . Note that if i= jthen L= ∑n t=1 Rtνt.Thus, the amortization method and the sinking fund method are equivalent. Example 41.5 A borrower is repaying a loan at 5% eﬀective with payments at the end of each year for 10 years, such that the payment the ﬁrst year is $200, the second year is $190, and so forth, until the 10th year it is $110. Find the amount of the loan if the borrower pays 6% eﬀective on the loan and accumulates a sinking fund to replace the amount fo the loan at 5% eﬀective. Solution. We have L= 100a10 0.05 + 10(Da)10 0.05 1 + (0.06 −0.05)a10 0.05 = 1227.83 1 + (0.01)(7.7217) = $1139.82 374 LOAN REPAYMENT METHODS Practice Problems Problem 41.1 Aborrows $10,000 from B and agrees to repay it with a series of 10 installments at the end of each year such that each installment is 20% greater than the preceding installment. The rate of interest on the loan is 10% eﬀective. Find the amount of principal repaid in the ﬁrst three installments. Problem 41.2 John borrows 10,000 from Tony and agrees to repay the loan with 10 equal annual installments of principal plus interest on the unpaid balance at 5% eﬀective. Immediately after the loan is made, Tony sells the loan to Chris for a price which is equal to the present value of John’s future payments calculated at 4%. Calculate the price that Chris will pay for the loan. Problem 41.3 A loan is being repaid with 10 payments of 1000 t at the end of year t. (In other words, the ﬁrst payment is 1000, the second payment is 2000, etc.) The interest rate on the loan is 6%. Calculate the outstanding principal immediately after the third payment. Problem 41.4 A loan of 100,000 is being repaid with annual payments at the end of each year for 10 years. The interest rate on the loan is 10.25%. Each annual payment increases by 5% over the previous annual payment. Calculate the principal in the ﬁfth payment. Problem 41.5 ‡ A 30-year loan of 1000 is repaid with payments at the end of each year. Each of the ﬁrst ten payments equals the amount of interest due. Each of the next ten payments equals 150% of the amount of interest due. Each of the last ten payments is X.The lender charges interest at an annual eﬀective rate of 10%. Calculate X. Problem 41.6 ‡ A loan is repaid with 20 increasing annual installments of 1,2,3, ··· ,20. The payments begin one year after the loan is made. Find the principal contained in the 10th payment, if the annual interest rate is 4%. Problem 41.7 In order to repay a school loan, a payment schedule of 200 at the end of the year for the ﬁrst 5 years, 1200 at the end of the year for the next 5 years, and 2200 at the end of the year for the ﬁnal 5 years is agreed upon. If interest is at the annual eﬀective rate of i= 6%, what is the loan value? 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 375 Problem 41.8 Create an amortization schedule of the previous problem for the ﬁrst seven years. Problem 41.9 A borrows $20,000 from B and agrees to repay it with 20 equal annual installments of principal in addition to the interest on the unpaid balance at 3% eﬀective. After 10 years B sells the right to future payments to C, at a price which yields C 5% eﬀective over the next 5 years and 4% eﬀective over the ﬁnal 5 years. Find the price which C should pay to the nearest dollar. Problem 41.10 A loan of 3000 at an eﬀective quarterly interest rate ofj = .02 is amortized by means of 12 quarterly payments , beginning one quarter after the loan is made. Each payment consists of a principal repayment of 250 plus interest due on the previous quarter’s outstanding balance. Construct the amortization schedule. Problem 41.11 A loan is being repaid with 10 annual payments. The ﬁrst payment is equal to the interest due only, the second payment is twice the ﬁrst, the third payment is three time the ﬁrst, and so forth. Prove that at the rate of interest on the loan is: ( Ia)10 = a∞. Problem 41.12 A loan is being repaid with 10 payments. The ﬁrst payment is 10, the second 9, the third is 8, and so forth with the tenth payment being 1. Find an expression for the amount of interest in the sixth payment. Problem 41.13 A has money invested at eﬀective rate i. At the end of the ﬁrst year A withdraws 162.5% of the interest earned; at the end of the second year Awithdraws 325% of the interest earned, and so forth with the withdrawal factor increasing in arithmetic progression. At the end of 16 years the fund is exhausted. Find i. Problem 41.14 A loan is amortized over ﬁve years with monthly payments at a nominal interest rate of 9% com- pounded monthly. The ﬁrst payment is 1000 and is to be paid one month from the date of the loan. Each succeeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made. Problem 41.15 June borrows 20,000 from April and agrees to repay it with a series of ten installments at the end 376 LOAN REPAYMENT METHODS of each year, such that each installment is 15% greater than the preceding installment. The rate of interest on the loan is 10% annual eﬀective. Let Pj denote the amount of principal repaid in year j. Calculate P1 + P2. Problem 41.16 Smith borrows 30,000 at an annual eﬀective interest rate of 10%. He agrees to make annual payments at the end of each year for 9 years, and an additional balloon payment, X, at the end of the tenth year. Each of the ﬁrst 9 payments equals 25% more than what is owed in interest at the time of the payment. The balloon payment, X, equals the amount needed to pay oﬀ the loan. Determine X. Problem 41.17 A loan is being repaid by the amortization method using 10 semiannual payments with the ﬁrst payment due six months after the loan is made. The ﬁrst payment is 50 and each subsequent payment is 50 more than its previous payment. The interest rate on the loan is 10%, compounded semiannually. Determine the amount of principle repaid in the seventh payment. Problem 41.18 A 10 year loan with an eﬀective annual interest rate of 5% is to be repaid with the following payments (1) 100 at the end of the second year (2) 200 at the end of the fourth year (3) 300 at the end of the sixth year (4) 400 at the end of the eighth year (5) 500 at the end of the tenth year Calculate the amount of interest included in the second payment. Problem 41.19 John takes out a 10 year loan. The loan is repaid by making 10 annual repayments at the end of each year. The ﬁrst loan repayment is equal to X, with each subsequent repayment 10.16% greater than the previous year’s repayment. The annual eﬀective interest rate being charged on the loan is 8%. The amount of interest repaid during the ﬁrst year is equal to 892.20. Calculate X. Problem 41.20 Don takes out a 10 year loan of L which he repays with annual payments at the end of each year using the amortization method. Interest on the loan is charged at an annual eﬀective rate of i.Don repays the loan with a decreasing series of payments. He repays 1000 in year one, 900 in year two, 800 in year three, ··· , and 100 in year ten. The amount of principal repaid in year three is equal to 600. Calculate L. 41 AMORTIZATION WITH VARYING SERIES OF PAYMENTS 377 Problem 41.21 ‡ Betty borrows 19,800 from Bank X.Betty repays the loan by making 36 equal payments of principal at the end of each month. She also pays interest on the unpaid balance each month at a nominal rate of 12%, compounded monthly. Immediately after the 16th payment is made, Bank X sells the rights to future payments to Bank Y. Bank Y wishes to yield a nominal rate of 14%, compounded semi-annually, on its investment. What price does Bank X receive? Problem 41.22 Two loans for equal amounts are repaid at an eﬀective interest rate of 4%. Loan A is repaid with 30 equal annual payments. Loan B is to be repaid by 30 annual payments, each containing equal principal amounts and an interest amount based on the unpaid balance. Payments are made at the end of each year. The annual payment for Loan A ﬁrst exceeds the annual payment for Loan B with the kth payment. Find k. Problem 41.23 A borrows $2000 at an eﬀective rate of interest of 10% per annum and agrees to repay the loan with payments at the end of each year. The ﬁrst payment is to be $400 and each payment thereafter is to be 4% greater than the preceding payment, with a smaller ﬁnal payment made one year after the last regular payment. (a) Find the outstanding loan balance at the end of three years. (b) Find the principal repaid in the third payment. 378 LOAN REPAYMENT METHODS Bonds and Related Topics This chapter is about bonds and the valuation of bonds. We will learn how to determine the price of a bond that should be paid by an investor for a desired given yield rate, determine the yield rate given the price, how a bond is amortized, and ﬁnally determine the value of a bond on a given date after it has been purchased. 379 380 BONDS AND RELATED TOPICS 42 Types of Bonds Perhaps the simplest way for a company or a government agency to raise cash is for them to sell bonds to the public. A bond is an interest bearing security which promises to pay a stated amount (or amounts) of money at some future date (or dates). The company or government branch which is issuing the bond outlines how much money it would like to borrow and speciﬁes a length of time, along with an interest rate it is willing to pay. Investors who then lend the requested money to the issuer become the issuer’s creditors through the bonds that they hold. The term of the bond is the length of time from the date of issue until the date of ﬁnal payment. The date of the ﬁnal payment is called the maturity date. Bonds with an inﬁnite term are called perpetuals. The detailed conditions of the bond may permit the bond to be called at some date prior to the maturity date, at which time the issuer (i.e., the lender) will repay the commitment, possibly with some additional amounts. Such a bond is callable. Any date prior to, or including, the maturity date on which a bond may be redeemed is termed a redemption date. The par valueor face valueof a bond is the amount that the issuer agrees to repay the bondholder by the maturity date. There are several classiﬁcation of bonds: •Accumulation bond is one in which the redemption price includes the original loan plus accu- mulated interest. Examples of such bonds are the Series E Savings bonds. •Bonds with coupons are periodic payments of interest made by the issuer of the bond prior to redemption. It is called a “coupon” because some bonds literally have coupons attached to them. Holders receive interest by stripping oﬀ the coupons and redeeming them. This is less common today as more records are kept electronically. In what follows, we we use the term “bond” we mean bonds with coupons. Zero coupon bonds are bonds that pay no periodic interest payments. It just pays a lump sum at redemption date. •Registered and Unregistered bonds. A registered bond is a bond issued with the name of the owner printed on the face of the certiﬁcate. If the owner decides to sell the bond, the change must be reported to the borrower. The coupon payments are paid by the borrower to the owners of record on each coupon payment date. An unregistered bond or bearer bond is one in which the lender is not listed in the records of the borrower. In this case, the bond belongs to whomever has legal possession of it. Again, these bonds are occasionally called coupon bonds, due to the physically attached coupons. •Fixed-rate and ﬂoated-rate bonds. A ﬁxed-rate bond is a bond that has a ﬁxed rate over the term of the bond. On the other hand, a bond that has a ﬂuctuated interest rate over the term of the bond is called a ﬂoating-rate bond. •Mortgage and debenture bonds. A mortgage bond is a more secured bond backed by a 42 TYPES OF BONDS 381 collateral such as a mortgage on a property. Mortgage bonds are backed by real estate or physical equipment that can be liquidated. A debenture bond is an unsecured bond issued by a civil or governmental corporation or agency and backed only by the credit standing of the issuer. •Income or adjustment bonds. They are a type of high risk bonds in which coupons are paid only if the borrower has earned suﬃcient income to pay them. •Junk bonds. A junk bond is a high-risk bond of default in payments. The risk that a bond issuer does not pay the coupon or principal payments is called default risk. Because of this risk, these bonds typically pay higher yields than better quality bonds in order to make them attractive to investors. •Convertible bond. A convertible bond is a bond that can be converted into the common stock of the company at the option of the bond owner. The owner of the bond is compensated with the ability to convert the bond to common stock, usually at a substantial discount to the stock’s market value. •Serial bonds. A set of bonds issued at the same time but having diﬀerent maturity dates. These are used when a borrower is in need of a large amount of money. •Treasury bonds. Issued by the US Treasury. Terms of seven or more years. •Treasury bills. Short term debt with maturities of 13, 26, or 52 weeks. T-bills yields are com- puted as rates of discount. These yields are computed on a simple discount basis. The basis for counting time periods for T-bills is actual/360. •Municipal bonds. These are bonds issued by state and local governments to ﬁnance large, long−term capital projects (e.g., hospitals, highways, schools). Example 42.1 A 10 year bond matures for its par value of 5000. The price of the bond is 4320.48 at an 8% yield convertible semi-annually. Calculate the coupon rate convertible semi-annually. Coupon payment is deﬁned to be the product of the face value and the coupon periodic rate. Also, the price of a bond is the present value of all future payments. Solution. Let k be the semiannual eﬀective coupon rate. Then 4320.48 = 5000 1.0420 + 5000k(1 −1.04−20) 0.04 or 5000k(1 −1.04−20) 0.04 = 2038.545269 Thus, k= 2038.545269 0.04 [5000(1 −1.04−20)] = 0.029999 382 BONDS AND RELATED TOPICS the coupon rate convertible semi-annually is 2 k= 0.059999 = 6.00% Example 42.2 A $1000 par value bond bearing a 6% coupon rate payable semi-annually will be redeemed at 105% at the end of 15 years. Find the price to yield an investor 5% eﬀective. Solution. The price is P = 30a30 j + 1050(1.05)−15 = $1135.54 where j = (1.05) 1 2 −1 Example 42.3 A 10-year accumulation bond with an initial par value(i.e. face value) of 1000 earns interest of 8% compounded semiannually. Find the price to yield an investor 10% eﬀective. Solution. The only return payment is at maturity. The price to yield 10% interest will therefore be (1.10)−10[1000(1.04)20] = $844.77 Example 42.4 Find the price which should be paid for a zero coupon bond which matures for 1000 in 10 years to yield: (a) 10% eﬀective (b) 9% eﬀective (c) Thus, a 10% reduction in the yield rate causes the price to increase by what percentage? Solution. (a) The bond is now worth 1000(1 .10)−10 = $385.54 (b) The bond is now worth 1000(1 .09)−10 = $422.41 (c) The 10% decrease in the interest rate thereby increases the price by 422.41 −385.54 385.54 = 9.6% 42 TYPES OF BONDS 383 Practice Problems Problem 42.1 A zero bond will pay $1000 at the end of 10 years, and is currently selling for $500. What is the annual return earned by the purchaser? Problem 42.2 Steve purchases a 10-year zero-coupon bond for $400 and has par value of $1,000. Find the yield rate convertible semiannually that would earned by Steve. Problem 42.3 John buys a 13-week T-bill with par value $10,000 at a simple discount rate of 7.5%. What price did he pay for the bond? Problem 42.4 An investor purchases a 26-week T-bill with face value of $10,000 and with a discount yield of dfor $9,600. (a) Find d. (b) Even though T-bills are not usually quoted on an annual eﬀective yield basis, the annual eﬀective yield can still be determined. Find the annual eﬀective rate of interest of this investment, assuming the investment period is exactly half a year. Problem 42.5 An investor purchases a 26-week T-bill with face value $10,000 for $X and earns an annual eﬀective rate of 4.17%. Determine X. Problem 42.6 An investor buys a zero-coupon bond with par value $1,000 for $493.63. The maturity date is Y years and the yield rate convertible semi-annually is 8%. Calculate Y. Problem 42.7 A 20 year bond matures for its par value of 10,000. The coupon payable semi-annually is 400. Calculate the price of the bond at a 6% yield rate convertible semi-annually. Problem 42.8 A 20 year bond matures for its par value of 10,000. The coupon payable semi-annually is 400. Calculate the price of the bond at a 6% annual eﬀective yield rate. Problem 42.9 A $1000 par value bond bearing a 5% coupon rate payable semi-annually will be redeemed at 108% at the end of 7 years. Find the price to yield an investor 5% convertible semi-annually. 384 BONDS AND RELATED TOPICS 43 The Various Pricing Formulas of a Bond In this section we consider the question of determining the purchase price which will produce a given yield rate to an investor. We have already encountered this question in the previous section. When considering the price of a bond, we make the following assumptions: (1) All obligations will be paid by the bond issuer on the speciﬁed dates of payments. Any possi- bility of default will be ignored. (2) The bond has a ﬁxed maturity date. (3) The price of the bond is decided immediately after a coupon payment date, or alternatively, at issue date if the bond is brand new. The following symbols and notation will be used in connection with bonds valuation: P = the price paid for a bond. F = The par value or face value or face amount. This amount is is usually printed on the front of the bond and is often the amount payable at the maturity date. C = the redemption value of a bond, i.e. the amount of money paid at a redemption date to the holder of the bond. r = The coupon rate is the eﬀective rate per coupon payment period used in determining the amount of the coupon. The default payment period is a half-year. For example, r= 0.035 for a 7% nominal coupon paid semi- annually. Fr = The amount of a coupon payment. g = The modiﬁed coupon rate. It is deﬁned by g= Fr C . Thus, g is the coupon rate per unit of redemption value, rather than per unit of par value. i = The yield rate of a bond, or the yield to maturity. The interest rate actually earned by the investor, assuming the bond is held until redemption or maturity. n = The number of coupon payment periods from the date of calculation until maturity date or redemption date K = the present value, computed at the yield rate, of the redemption value at the maturity date, or a redemption date, i.e. K = Cνn at the yield rate i. G = The base amount of a bond. It is deﬁned by Gi= Fr or G= Fr i . Thus, G is the amount which, if invested at the yield rate i, would produce periodic interest payments equal to the coupons on the bond. The quantities F,C,r,g, and n are given by the terms of a bond and remain ﬁxed throughout the bonds life. On the other hand, P and i will vary throughout the life of the bond. Price and yield rate have a precise inverse relationship to each other,i.e. as the price rises the yield rate falls and 43 THE VARIOUS PRICING FORMULAS OF A BOND 385 vice-versa. There are three diﬀerent quoted yields associated with a bond: (1) Nominal yield is the ratio of annualized coupon payment to par value. For example, 2 coupons per year of $3.50 on a $100 par value bond result in a 7.00% nominal yield. The reader should note that the use of the word “nominal” in this context is diﬀerent than the one used in Section 9. (2) Current yield is the ratio of the annualized coupon payment to the original price of the bond. For example, if you paid $90 per $100 of par value of the bond described above, the current yield would be $7.00 $90 = 7.78% (3) Yield to maturity is the actual annualized yield rate, or internal rate of return. Four formulas for price Like loans, the price of a bond is deﬁned to be the present value of all future payments. We shall consider four diﬀerent ways of determining the present value, or price of a bond. The formulas are derivable one from the other: there are situations where one may be more useful than another for speciﬁc applications. The ﬁrst of these is called the basic formula. It is found by writing the equation of value at time t= 0 of the time diagram given in Figure 43.1 and is given by P = Frani + Cνn i = Frani + K. Figure 43.1 This formula has a verbal interpretation: the price of a bond is equal to the present value of future coupons plus the present value of the redemption value. The second formula is the premium/discount formula, and is derived from the basic formula: P =Frani + Cνn i =Frani + C(1 −iani) =C+ (Fr −Ci)ani This says that the price of a bond is the sum of the redemption value and the present value of amortization of premium or discount paid at purchase. Amortization of premium and discount will be discussed in the next section. 386 BONDS AND RELATED TOPICS The third, the base amount formula, can also be derived from the ﬁrst formula as follows: P =Frani + Cνn i =Giani + Cνn i =G(1 −νn i ) + Cνn i =G+ (C−G)νn i This says that the price is the sum of the base amount and the present value of the diﬀerence between base amount and redemption value. The fourth, the Makeham formula, is also obtained from the basic formula: P =Cνn i + Frani =Cνn i + Cg (1 −νn i i ) =Cνn i + g i(C−Cνn i ) =K+ g i(C−K) Example 43.1 A 10-year 100 par value bond bearing a 10% coupon rate payable semi-annually, and redeemable at 105, is bought to yield 8% convertible semiannually. Find the price. Verify that all four formulas produce the same answer. Solution. We are given the following: F = 100 C = 105 r = 5% Fr = 5 g= Fr C = 1 21 i = 4% n = 20 K = Cνn = 105(1 .04)−20 = 47.92 G= Fr i = 125 Using the basic formula we ﬁnd P = Fran + K = 5a20 0.04 + 47.92 = 5(13.59031) + 47.92 = $115.87 43 THE VARIOUS PRICING FORMULAS OF A BOND 387 Using the premium/discount formula we ﬁnd P = C+ (Fr −Ci)an = 105 + (5−4.2)a20 0.04 = $115.87 Using the base amount formula we ﬁnd P = G+ (C−G)νn = 125 + (105−125)(1.04)−20 = $115.87 Using the Makeham formula we ﬁnd P = K+ g i(C−K) = 47.92 + 100 84 (105 −47.92) = $115.87 Example 43.2 For the bond of the previous example, determine the following: (a) Nominal yield based on the par value. (b) Nominal yield, based on the redemption value. (c) Current yield. (d) Yield to maturity. Solution. (a) The nominal yield is the annualized coupon value to the par value,i.e. 2Fr F = 10 100 = 10%. (b) This is the annualized modiﬁed coupon rate 2g= 2Fr C = 2 5 105 = 9.52% (c) Current yield is the ratio of annualized coupon to price 10 115.87 = 8.63% (d) Yield to maturity is the annualized actual yield which is 8% In the problems considered thus far, we have assumed that coupon payments are constant. It is possbile to have bonds with varying coupon payments. In this case, the coupons will constitute a varying annuity which can be evaluated using the approach of Sections 26 - 27. Again, the price of the bond is the present value of all future coupon payments plus the present value of the redemption value. Example 43.3 A 1000 par value 20 year bond with annual coupons and redeemable at maturity at 1050 is purchased for P to yield an annual eﬀective rate of 8.25%. The ﬁrst coupon is 75. Each subsequent coupon is 3% greater than the preceding coupon. Determine P. 388 BONDS AND RELATED TOPICS Solution. The coupon payments constitute a annuity that is varying in geometric progression. The present value of this annuity is 75 1 − (1+k 1+i )n i−k = 75 1 − ( 1.03 1.0825 )20 0.0825 −0.03 = 900.02. The present value of the redemption value is 1050(1.0825)−20 = 215.09. Thus, the price of the bond is 900.02 + 215.09 = 1115.11 It is possible that the bond yield rate is not constant over the term of the bond. We illustrate this in the following example. Example 43.4 Find the price of a 1000 par value 10-year bond with coupons of 8.4% convertible semi-annually, which will be redeemed at 1050. The bond yield rate is 10% convertible semi-annually for the ﬁrst ﬁve years and 9% convertible semi-annually for the second ﬁve years. Solution. The present value of all future coupons is 42[a10 0.05 + (1.05)−10a10 0.045] = 528.33. The present value of the redemption value is 1050(1.05)−10(1.045)−10 = 415.08. Thus, the price of the bond is 528.33 + 415.08 = 943.41 Next, we consider cases when the yield rate and coupon rate are at diﬀerent frequencies. Consider ﬁrst the case in which each coupon period contains k yield rate conversion periods. Suppose the total number of yield rate conversion periods over the term of the bond is n. There is a coupon payment of Fr paid at the end of each k yield rate conversion periods. The present value of all future coupon payments is Fran sk and the present value of the redemption value is Cvn. Thus, the price of the bond is P = Fran sk + Cvn. 43 THE VARIOUS PRICING FORMULAS OF A BOND 389 Example 43.5 Find the price of a $1000 par value 10-year bond maturing at par which has coupons at 8% con- vertible semi-annually and is bought to yield 6% convertible quarterly. Solution. The price of the bond is P = 40a40 0.015 s2 0.015 + 1000(1.015)−40 = 1145.12 The second case is when each yield rate conversion period contains m coupon conversion periods. Again, we let n be the total number of yield rate conversion periods. There is a coupon of Fr m paid at the end of each mth of a conversion period. Thus, the present value of the future coupon payments is Fra(m) n and the present value of the redemption value is Cνn. Thus, the price of the bond is P = Fra(m) n + Cνn. Example 43.6 Find the price of a $1000 par value 10-year bond maturing at par which has coupons at 8% con- vertible quarterly and is bought to yield 6% convertible semi-annually. Solution. The price of the bond is P =40a(2) 20 + 1000(1.03)−20 =40 1 −1.03−20 2[(1.03) 1 2 −1] + 1000(1.03)−20 =1153.21 390 BONDS AND RELATED TOPICS Practice Problems Problem 43.1 A 10-year $1,000 par value bond bearing a 8.4% coupon rate convertible semiannually and re- deemable at $1,050 is bought to yield 10% convertible semiannually. Find the price. Use all four formulas. Problem 43.2 For the bond of the previous problem, determine the following: (a) Nominal yield based on the par value. (b) Nominal yield, based on the redemption value. (c) Current yield. (d) Yield to maturity. Problem 43.3 A 5-year $100 par value bond bearing a 8% coupon rate payable semi-annually is selling at par value. If prevailing market rates of interest suddenly go to 10% convertible semiannually, ﬁnd the percentage change in the price of the bond. Problem 43.4 Two 1000 bonds redeemable at par at the end of the same period are bought to yield 4% convertible semiannually. One bond costs $1136.78, and has a coupon rate of 5% payable semiannually. The other bond has a coupon rate of 2.5% payable semiannually. Find the price of the second bond. Problem 43.5 A n−year $1000 par value bond bearing a 9% coupon rate payable semiannually and redeemable at $1125 is bought to yield 10% convertible semiannually. Find the purchase price of the bond if the present value of the redemption value is $225 at the given yield rate. Problem 43.6 A n−year $1000 par value bond maturing at par and with semi-annual coupon of $50 is purchased for $1110. Find the base amount G if the redemption value is $450. Problem 43.7 An investor owns a 1000 par value 10% bond with semiannual coupons. The bond will mature at par at the end of 10 years. The investor decides that an 8-year bond would be preferable. Current yield rates are 7% convertible semiannually. The investor uses the proceeds from the sale of the 10% bond to purchase a 6% bond with semiannual coupons, maturing at par at the end of 8 years. Find the par value of the 8-year bond. 43 THE VARIOUS PRICING FORMULAS OF A BOND 391 Problem 43.8 An n−year 1000 par value bond matures at par and has a coupon rate of 12% convertible semian- nually. It is bought at a price to yield 10% convertible semiannually. If the term of the bound is doubled, the price will increase by 50. Find the price of the n−year bond. Problem 43.9 A 20 year bond with a par value of $10,000 will mature in 20 years for $10,500. The coupon rate is 8% convertible semiannually. Calculate the price that Andrew would pay if he bought the bond to yield 6% convertible twice a year. Problem 43.10 After 5 years (immediately after the 10th coupon is paid), Andrew decides to sell the bond in the previous problem. Interest rates have changed such that the price of the bond at the time of the sale is priced using a yield rate of 9% convertible semiannually. Calculate the selling price. Problem 43.11 A bond matures for its par value of 1000 in one year. The bond pays a annual coupon on 80 each year. The price of the bond is 800. Which of the following are true? i. The nominal yield is 8% per annum. ii. The current yield is 10% per annum. iii. The yield to maturity is 35% per annum Problem 43.12 You paid $950 for a 7 year bond with a face value of $1,000 and semi-annual coupons of $25. What is your current yield on this bond and what is your nominal annual yield to maturity (convertible semi-annually)? Problem 43.13 A 15-year bond with a coupon of $X payable every 6 months has a face (and redemption) value of $10,000. At the nominal annual interest rate, convertible semi-annually, of 6.5%, the price of the bond is $8,576.36. What is X? Problem 43.14 ‡ A ten-year 100 par value bond pays 8% coupons semiannually. The bond is priced at 118.20 to yield an annual nominal rate of 6% convertible semiannually. Calculate the redemption value of the bond. 392 BONDS AND RELATED TOPICS Problem 43.15 ‡ Susan can buy a zero coupon bond that will pay 1000 at the end of 12 years and is currently selling for 624.60 . Instead she purchases a 6% bond with coupons payable semi-annually that will pay 1000 at the end of 10 years. If she pays X she will earn the same annual eﬀective interest rate as the zero coupon bond. Calculate X. Problem 43.16 ‡ A 30-year bond with a par value of 1000 and 12% coupons payable quarterly is selling at 850. Calculate the annual nominal yield rate convertible quarterly. Problem 43.17 ‡ An investor borrows an amount at an annual eﬀective interest rate of 5% and will repay all interest and principal in a lump sum at the end of 10 years. She uses the amount borrowed to purchase a 1000 par value 10-year bond with 8% semiannual coupons bought to yield 6% convertible semiannually. All coupon payments are reinvested at a nominal rate of 4% convertible semiannually. Calculate the net gain to the investor at the end of 10 years after the loan is repaid. Problem 43.18 ‡ Dan purchases a 1000 par value 10-year bond with 9% semiannual coupons for 925. He is able to reinvest his coupon payments at a nominal rate of 7% convertible semiannually. Calculate his nominal annual yield rate convertible semiannually over the ten-year period. Problem 43.19 ‡ You have decided to invest in BondX,an n−year bond with semi-annual coupons and the following characteristics: •Par value is 1000. •The ratio of the semi-annual coupon rate to the desired semi-annual yield rate, r i , is 1.03125. •The present value of the redemption value is 381.50. Given vn = 0.5889, what is the price of bond X? Problem 43.20 ‡ Bill buys a 10-year 1000 par value 6% bond with semi-annual coupons. The price assumes a nominal yield of 6%, compounded semi-annually. As Bill receives each coupon payment, he immediately puts the money into an account earning interest at an annual eﬀective rate of i. At the end of 10 years, immediately after Bill receives the ﬁnal coupon payment and the redemption value of the bond, Bill has earned an annual eﬀective yield of 7% on his investment in the bond. Calculate i. Problem 43.21 Two $1000 par value bonds redeemable at par at the end of the same period are bought to yield 43 THE VARIOUS PRICING FORMULAS OF A BOND 393 4% convertible quarterly. One bond costs $1098 and has a coupon rate of 5% payable quarterly. The other bond has a coupon rate of 3% payable quarterly. Find the price of the second bond to the nearest dollar. Problem 43.22 A 1000 par value n−year bond maturing at par with annual coupons of 100 is purchased for 1125. The present value of the redemption value is 500. Find n to the nearest integer. Problem 43.23 A 1000 bond with annual coupons is redeemable at par at the end of 10 years. At a purchase price of 870, the yield rate is i. The coupon rate is i−0.02. Calculate i. Problem 43.24 A bond with coupons equal to 40 sells for P.A second bond with the same maturity value and term has coupons equal to 30 and sells for Q. A third bond with the same maturity value and term has coupons equal to 80. All prices are based on the same yield rate, and all coupons are paid at the same frequency. Determine the price of the third bond in terms of P and Q. Problem 43.25 ‡ You have decided to invest in two bonds. Bond X is an n−year bond with semi-annual coupons, while bond Y is an accumulation bond redeemable in n 2 years. The desired yield rate is the same for both bonds. You also have the following information: Bond X •Par value is 1000. •The ratio of the semi-annual bond rate to the desired semi-annual yield rate, r i is 1.03125. •The present value of the redemption value is 381.50. Bond Y •Redemption value is the same as the redemption value of bond X. •Price to yield is 647.80. What is the price of bond X? Problem 43.26 You are given two n−year par value 1,000 bonds. Bond X has 14% semi-annual coupons and a price of 1,407.70 to yield i compounded semi-annually. Bond Y has 12% semi-annual coupons and a price of 1,271.80 to yield the same rate icompounded semi-annually. Calculate the price of Bond X to yield i−1%. 394 BONDS AND RELATED TOPICS Problem 43.27 Henry has a ﬁve year 1,000,000 bond with coupons at 6% convertible semi-annually. Jean buys a 10 year bond with face amount X and coupons at 6% convertible semi-annually. Both bonds are redeemable at par. Henry and Jean both buy their bonds to yield 4% compounded semi-annually and immediately sell them to an investor to yield 2% compounded semi-annually. Jean earns the same amount of proﬁt as Henry. Calculate X. Problem 43.28 On June 1, 1990, an investor buys three 14 year bonds, each with par value 1000, to yield an eﬀective annual interest rate of i on each bond. Each bond is redeemable at par. You are given (1) the ﬁrst bond is an accumulation bond priced at 195.63 (2) the second bond has 9.4% semiannual coupons and is priced at 825.72 (3) the third bond has 10% annual coupons and is priced at P. Calculate P. Problem 43.29 A 10 year bond with par value 1000 and annual coupon rate ris redeemable at 1100. You are given (1) the price to yield an eﬀective annual interest rate of 4% is P (2) the price to yield an eﬀective annual interest rate of 5% is P −81.49 (3) the price to yield an eﬀective annual interest rate of r is X. Calculate X. Problem 43.30 Patrick buys a 28 year bond with a par value of 1200 and annual coupons. The bond is redeemable at par. Patrick pays 1968 for the bond, assuming an annual eﬀective yield rate of i. The coupon rate on the bond is twice the yield rate. At the end of 7 years Patrick sells the bond for P, which produces the same annual eﬀective yield rate of i to the new buyer. Calculate P. Problem 43.31 A 10 year bond with coupons at 8% convertible quarterly will be redeemed at 1600. The bond is bought to yield 12% convertible quarterly. The purchase price is 860.40. Calculate the par value. Problem 43.32 A corporation decides to issue an inﬂation-adjusted bond with a par value of $1000 and with annual coupons at the end of each year for 10 years. The initial coupon rate is 7% and each coupon is 3% greater than the preceding coupon. The bond is redeemed for $1200 at the end of 10 years. Find the price an investor should pay to produce a yield rate of 9% eﬀective. 43 THE VARIOUS PRICING FORMULAS OF A BOND 395 Problem 43.33 A 1000 par value 10 year bond with semiannual coupons and redeemable at 1100 is purchased at 1135 to yield 12% convertible semiannually. The ﬁrst coupon is X. Each subsequent coupon is 4% greater than the preceding coupon. Determine X. Problem 43.34 A 1,000 par value 3 year bond with annual coupons of 50 for the ﬁrst year, 70 for the second year, and 90 for the third year is bought to yield a force of interest δt = 2t−1 2(t2 −t+ 1), t ≥0. Calculate the price of this bond. 396 BONDS AND RELATED TOPICS 44 Amortization of Premium or Discount Bonds can be priced at a premium, discount, or at par. If the bond’s price is higher than its par value,i.e. P > C, then the bond is said to sell at a premium and the diﬀerence P −C is called the “premium”. On the other hand, if the bond’s price is less than its par value,i.e. P < C, then the bond is said to sell at a discount and the diﬀerence C−P is called the “discount”. Thus, a discount is merely a negative premium. Using the premium/discount formula of the previous section we see that Premium = P −C = (Fr −Ci)ani = C(g−i)ani if g >i In this case, a loss in that amount incurred when the bond is redeemed. Similarly, Discount = C−P = (Ci−Fr)ani = C(i−g)ani if g <i and a gain in that amount is incurred when the bond is redeemed. Because of this proﬁt or loss at redemption, the amount of each coupon CANNOT be considered as interest income to an investor. It is necessary to divide each coupon into interest earned and principal adjustment portions similar to the separation of payments into interest and principal when discussing loans. In the same sense that a loan has an outstanding balance at any time, we can talk of the book value of a bond at any time t. We deﬁne the book value to be the present value of all future payments. If we let Bt be the book value after the tth coupon has just been paid, then the value of the remaining payments are:n−t coupons and a payment of C at the date of redemption. Hence, Bt = Fran−t + Cνn−t. Example 44.1 Find the book value immediately after the payment of the 14th coupon of a 10-year 1,000 par value bond with semiannual coupons, if r= 0.05 and the yield rate is 12% convertible semiannually. Solution. We have F = C = 1000,n = 20,r = 0.05, and i= 0.06. Thus, B14 = 1000(0.05)a6 0.06 + 1000(1.06)−6 = 950.83 Letting t= 0 in the expression of Bt we ﬁnd B0 = P. Similarly, Bn = C. Thus, on the redemption date, the book value of a bond equals the redemption value of the bond. From this it follows that when a bond is purchased at a premium, the book value of the bond will be written -down(decreased) at each coupon date, so that at the time of redemption its book value 44 AMORTIZATION OF PREMIUM OR DISCOUNT 397 equals the redemption value. This process is called amortization of premium. When it is puchased at a discount, the book value of the bond will be written-up(increased) at each coupon date, so that at the time of redemption its book value equals the redemption value. This process is called accumulation of discount. Book values are very useful in constructing bond amortization schedules. A bond amortization schedule shows the division of each bond coupon into interest earned and principal adjustment portions, together with the book value after each coupon is paid. We will denote the interest earned after the tth coupon has been made by It and the corresponding principal adjustment portion by Pt.The level coupon will continue to be denoted by Fr. Note that It =iBt−1 = i[Fran−t+1 + Cνn−t+1] =Cg(1 −νn−t+1) + iCνn−t+1 = Cg + C(i−g)νn−t+1 and Pt = Fr −It = Cg −It = C(g−i)νn−t+1. where Pt is positive in the case of a premium and negative in the case of a discount. It is easy to see that Bt = B0 − t∑ k=1 Pk. Example 44.2 A $1000 bond, redeemable at par on December 1, 1998, with 9% coupons paid semiannually. The bond is bought on June 1, 1996. Find the purchase price and construct a bond amortization schedule if the desired yield is 8% compounded semiannually. Solution. Using the premium/discount formula with C = 1000,F = 1000,r = 4.5%,and i= 4%,the purchase price on June 1, 1996 is P = 1000 + (45−40)a5 0.04 = $1022.26. Thus, the bond is purchased at a premium of $22.26. For the bond amortization schedule, we ﬁrst do the calculation of the ﬁrst line of the schedule. Each successive line of the schedule is calculated in a similar fashion. The interest earned portion of the ﬁrst coupon, i.e. on Dec 1, 1996 is I1 = iB0 = 0.04(1022.26) = 40.89. The principal adjustment portion of the ﬁrst coupon is P1 = Fr −I1 = 45 −40.89 = $4.11. 398 BONDS AND RELATED TOPICS The book value at the end of the ﬁrst period is B1 = B0 −P1 = $1018.15. Amount for Interest amortization of Book Date Coupon earned premium value June 1, 1996 0 0 0 1022.26 Dec 1, 1996 45.00 40.89 4.11 1018.15 June 1, 1997 45.00 40.73 4.27 1013.88 Dec 1, 1997 45.00 40.56 4.44 1009.44 June 1, 1998 45.00 40.38 4.62 1004.82 Dec 1, 1998 45.00 40.19 4.81 1000.01 Total 225.00 202.74 22.25 The one cent error is due to rounding The following observations can be concluded from the above amortization schedule: (1) The sum of the principal adjustment column is equal to the amount of premium/discount. (2) The sum of the interest paid column is equal to the diﬀerence between the sum of the coupons and the sum of the principal adjustment column. (3) The principal adjustment column is a geometric progression with common ration 1 + i. The following is an example of a bond bought at discount. Example 44.3 A $1000 bond, redeemable at par on December 1, 1998, with 9% coupons paid semiannually. The bond is bought on June 1, 1996. Find the purchase price and construct a bond amortization schedule if the desired yield is 10% compounded semiannually. Solution. Using the premium/discount formula with C = 1000,F = 1000,r = 4.5%,and i= 5%,the purchase price on June 1, 1996 is P = 1000 + (45−50)a5 0.05 = $978.35. Thus, the bond is purchased at a discount of $21.65. 44 AMORTIZATION OF PREMIUM OR DISCOUNT 399 Amount for Interest accumulation of Book Date Coupon earned discount value June 1, 1996 0 0 0 978.35 Dec 1, 1996 45.00 48.92 −3.92 982.27 June 1, 1997 45.00 49.11 −4.11 986.38 Dec 1, 1997 45.00 49.32 −4.32 990.70 June 1, 1998 45.00 49.54 −4.54 995.24 Dec 1, 1998 45.00 49.76 −4.76 1000.00 Total 225.00 246.65 −21.65 Much like dealing with loans, if it is desired to ﬁnd the interest earned or principal adjustment portion of any one coupon, it is not necessary to construct the entire table. Simply ﬁnd the book value at the beginning of the period in question which is equal to the price at that point computed at the original yield rate and can be determined by the methods of Section 41. Then ﬁnd that one line of the table. Example 44.4 Consider a $1000 par value two-year 8% bond with semiannual coupons bought to yield 6% con- vertible semiannually. Compute the interest portion earned and the principal adjustment portion of the 3rd coupon payment. Solution. We have F = C = 1000,n = 4,r = 0.04, and i= 0.03. Thus, the coupon payment is Fr = $40.00. The principal adjustment portion of the 3rd coupon payment is P3 = B3 −B2 = 40 a1 0.03 + 1000(1.03)−1 −(40a2 0.03 + 1000(1.03)−2) = $9.43.The interest earned portion is I3 = 40.00−9.43 = $30.57 Another method of writing up or writing down the book values of bonds is the straight line method. This does not produce results consistent with compound interest, but it is very simple to apply. In this method, the book values are linear, grading from B0 = P to Bn = C. Thus, the principal adjustment column is constant and is given by Pt = P −C n , t = 1,2,··· ,n and the interest earned portion is It = Fr −Pt, t = 1,2,··· ,n Note that Pt >0 for premium bond and Pt <0 for discount bond. 400 BONDS AND RELATED TOPICS Example 44.5 Find the book values for the bond in Example 44.2 by the straight line method. Solution. We have Pt = 22.25 5 = 4.45. Thus, B0 =1022.26 B1 =1022.26 −4.45 = 1017.81 B2 =1017.81 −4.45 = 1013.36 B3 =1013.36 −4.45 = 1008.91 B4 =1008.91 −4.45 = 1004.46 B5 =1004.46 −4.45 = 1000.01 44 AMORTIZATION OF PREMIUM OR DISCOUNT 401 Practice Problems Problem 44.1 A $1000 par value two-year 8% bond with semiannual coupons is bought to yield 6% convertible semiannually. (a) Create an amortization schedule for this transaction. (b) Find the book values of the bond using the straight line method. Problem 44.2 A $1000 par value two-year 8% bond with semiannual coupons is bought to yield 10% convertible semiannually. (a) Create an amortization schedule for this transaction. (b) Find the book values of the bond using the straight line method. Problem 44.3 For a bond of face value 1 the coupon rate is 150% of the yield rate, and the premium is p. For another bond of 1 with the same number of coupons and the same yield rate, the coupon rate is 75% of the yield rate. Find the price of the second bond. Problem 44.4 For a certain period a bond amortization schedule shows that the amount for amortization of premium is 5, and that the required interest is 75% of the coupon. Find the amount of the coupon. Problem 44.5 A 10-year bond with semi-annual coupons is bought at a discount to yield 9% convertible semian- nually. If the amount for accumulation of discount in the next-to-last coupon is 8, ﬁnd the total amount for accumulation of discount during the ﬁrst four years in the bond amortization schedule. Problem 44.6 A 1000 par value 5-year bond with a coupon rate of 10% payable semiannually and redeemable at par is bought to yield 12% convertible semiannually. Find the total of the interest paid column in the bond amortization schedule. Problem 44.7 A ﬁve year bond with a par value of 1000 will mature in 5 years for 1000. Annual coupons are payable at a rate of 6%. Create the bond amortization schedule if the bond is bought to yield 8% annually. 402 BONDS AND RELATED TOPICS Problem 44.8 A 40 year bond with a par value of 5000 is redeemable at par and pays semi-annual coupons at a rate of 7% convertible semi-annually. The bond is purchased to yield annual eﬀective rate of 6%. Calculate the amortization of the premium in the 61st coupon. Problem 44.9 A 10 year bond is redeemable at par of 100,000. The bond pays semi-annual coupons of 4000. The bond is bought to yield 6% convertible semi-annually. Calculate the premium paid for the bond. Problem 44.10 A bond is redeemable in eight years at 120% of its par value. The bond pays an annual coupon rate of 6%. Calculate the premium as a percent of the par value if the bond is purchased to yield 6% annually. Problem 44.11 Consider a three-year bond, with a $1,000 face value and a 9% coupon rate paid semi-annually, which was bought to yield 7% convertible semi-annually. Find the amount for amortization of premium during the bonds third half-year (i.e., between the second and third coupon payments). Problem 44.12 A bond of face amount 100 pays semi-annual coupons and is purchased at a premium of 36 to yield annual interest of 7% compounded semiannually. The amount for amortization of premium in the 5th coupon is 1.00. What is the term of the bond? Problem 44.13 ‡ Among a company’s assets and accounting records, an actuary ﬁnds a 15-year bond that was purchased at a premium. From the records, the actuary has determined the following: (i) The bond pays semi-annual interest. (ii) The amount for amortization of the premium in the 2nd coupon payment was 977.19. (iii) The amount for amortization of the premium in the 4th coupon payment was 1046.79. What is the value of the premium? Problem 44.14 ‡ A 10,000 par value 10-year bond with 8% annual coupons is bought at a premium to yield an annual eﬀective rate of 6%. Calculate the interest portion of the 7th coupon. Problem 44.15 ‡ A 1000 par value 5 year bond with 8% semiannual coupons was bought to yield 7.5% convertible semiannually. Determine the amount of premium amortized in the sixth coupon payment. 44 AMORTIZATION OF PREMIUM OR DISCOUNT 403 Problem 44.16 On May 1, 1985, a bond with par value 1000 and annual coupons at 5.375% was purchased to yield an eﬀective annual interest rate of 5%. On May 1, 2000, the bond is redeemable at 1100. The book value of the bond is adjusted each year so that it equals the redemption value on May 1, 2000. Calculate the amount of write-up or write-down in the book value in the year ending May 1, 1991. Problem 44.17 A bond with par value of 1000 and 6% semiannual coupons is redeemable for 1100. You are given (1) the bond is purchased at P to yield 8% convertible semiannually and (2) the amount of principal adjustment for the 16th semiannual period is 5 Calculate P. Problem 44.18 Laura buys two bonds at time 0. Bond X is a 1000 par value 14 year bond with 10% annual coupons. It is bought at a price to yield an annual eﬀective rate of 8%. Bond Y is a 14 year par value bond with 6.75% annual coupons and a face amount of F. Laura pays P for the bond to yield an annual eﬀective rate of 8%. During year 6 the writedown in premium (principal adjustment) on bond X is equal to the writeup in discount (principal adjust- ment) on bond Y. Calculate P. Problem 44.19 A 1000 par value 18 year bond with annual coupons is bought to yield an annual eﬀective rate of 5%. The amount for amortization of premium in the 10th year is 20. The book value of the bond at the end of year 10 is X. Calculate X. Problem 44.20 Thomas buys a bond at a premium of 200 to yield 6% annually. The bond pays annual coupons and is redeemable for its par value of 1000. Calculate the amount of interest in the ﬁrst coupon. Problem 44.21 Megan buys a bond that is redeemable for its par value of 20,000 after 5 years. The bond pays coupons of 800 annually. The bond is bought to yield 8% annually. Calculate the accumulation of discount in the 4th coupon. Problem 44.22 Jenna buys a bond at a premium. The principal amortized in the ﬁrst annual coupon is half that amortized in the 11th annual coupon. Calculate the yield rate used to calculate the purchase price of the bond. 404 BONDS AND RELATED TOPICS Problem 44.23 Betty buys an n−year 1000 par value bond with 6.5% annual coupons at a price of 798.48. The price assumes an annual eﬀective yield rate ofi.The total write-up in book value of the bond during the ﬁrst 3 years after purchase is 22.50. Problem 44.24 Ann purchases a 1000 face value 20-year bond, redeemable at 1200, with 5% annual coupons, at a price that yields her an annual eﬀective interest rate of 3%. Immediately upon receiving each coupon, she invests the coupon in an account that earns an annual eﬀective interest rate of 7%. Immediately after receiving the eighth coupon, Ann sells the bond at a price that yields the new purchaser an annual eﬀective interest rate of 4.75%. Determine Ann’s yield rate, as an annual eﬀective interest rate, over the 8-year period that she owns the bond. Problem 44.25 A 1000 face value 20-year 8% bond with semiannual coupons is purchased for 1014. The redemption value is 1000. The coupons are reinvested at a nominal annual rate of 6%, compounded semiannually. Determine the purchaser’s annual eﬀective yield rate over the 20- year period. Problem 44.26 An n−year 1000 par value bond with 8% annual coupons has an annual eﬀective yield of i,i >0. The book value of the bond at the end of year 3 is 1099.84 and the book value at the end of year 5 is 1082.27. Calculate the purchase price of the bond. Problem 44.27 Bryan buys a 2 n−year 1000 par value bond with 7.2% annual coupons at a price P. The price assumes an annual eﬀective yield of 12%. At the end of n years, the book value of the bond, X, is 45.24 greater than the purchase price, P. Assume νn 0.12 <0.5. Calculate X. 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 405 45 Valuation of Bonds Between Coupons Payment Dates Up to now, bond prices and book values have been calculated assuming the coupon has just been paid. In particular, we can ﬁnd a relationship between the book values of two consecutive coupon dates t and t+ 1 as follows. Let Bt and Bt+1 be the book values just after the tth and ( t+ 1)th coupons are paid and Fr the amount of the coupon. Then Bt(1 + i) −Fr =Fr (1 −νn−t i ) (1 + i) + Cνn−t(1 + i) −Fr =Fr i −Frνn−t i (1 + i) + Cνn−t−1 =Fr i −Frνn−t−1 i + Cνn−t−1 =Fr (1 −νn−t−1 i ) + Cνn−t−1 =Fran−t−1 + Cνn−t−1 =Bt+1 where we assume a constant yield rate i over the interval. Example 45.1 You buy an n−year 1,000 par value bond with 6.5% annual coupons at a price of 825.44, assuming an annual yield rate of i,i> 0. After the ﬁrst two years, the bond’s book value has changed by 23.76. Calculate i. Solution. We have B2 −B0 = B0(1 +i)2 −Fr(1 +i) −Fr −B0.This implies 23.76 = 825.44(1 +i)2 −65(1 + i) −65 −825.44 which reduces to 825.44(1 + i)2 −65(1 + i) −914.20 = 0 Solving this with the quadratic formula we ﬁnd 1 + i= 1.0925 →i= 9.25% Book values can be computed at times other than immediately after a coupon payment. In this section, we would like to analyze the book value Bt+k for 0 <k <1 where time k is measured from the last coupon payment. When buying an existing bond between its coupon dates, one must decide how to split up the 406 BONDS AND RELATED TOPICS coupon between the prior owner and the new owner. Consider the price that a buyer would pay for the bond a fractional time k through a coupon period. Assume that the buyer will obtain a yield rate equal to that of the current bond holder. The buyer will receive all of the next coupon. The current holder would expect to receive part of this coupon as interest for the period which we call accrued interest or accrued coupon and is denoted by Frk. Note that Fr0 = 0 and Fr1 = Fr, the coupon payment, where Fr1 is computed just before the coupon is paid. How should the purchase price be allocated between accrued interest (or accrued coupon) and price for the bond? The purchase price for the bond is called the ﬂat price. It is deﬁned to be the money which actually changes hands at the date of sale. It is denoted by Bf t+k. The price for the bond is the book value, which is also called the market price and is denoted by Bm t+k.The market price is the price commonly quoted in the ﬁnancial press, since the market price changes smoothly through time, while the ﬂat price ﬂuctuates due to coupon accrual. From the above deﬁnitions, it is clear that Bf t+k = Bm t+k + Frk. Example 45.2 Consider the amortization schedule of Example 44.2. Amount for Interest amortization of Book Date Coupon earned premium value June 1, 1996 0 0 0 1022.26 Dec 1, 1996 45.00 40.89 4.11 1018.15 June 1, 1997 45.00 40.73 4.27 1013.88 Dec 1, 1997 45.00 40.56 4.44 1009.44 June 1, 1998 45.00 40.38 4.62 1004.82 Dec 1, 1998 45.00 40.19 4.81 1000.01 Total 225.00 202.74 22.25 Sketch a graph describing the relationship between the ﬂat price and the market price. Solution. The graph is shown in Figure 45.1. Note that the accrued coupon at any date is equal to the vertical distance between the solid line and the dotted line 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 407 Figure 45.1 There are three methods used to ﬁnd the ﬂat price, the market price, and the accured coupon. Values on the coupon dates are known, so the diﬀerences among the three methods arise only for the interim values between coupon dates. First, we have the theoretical method. This is an exact method, based on compound interest. This method argues that the ﬂat price should be the book value Bt after the preceding coupon, accumulated by (1 + i)k, where i is the yield rate, giving the ﬂat price of Bf t+k = (1 + i)kBt. The accrued coupon is given by (See Problem 15.48) Frk = Fr [(1 + i)k −1 i ] . The market value is then the diﬀerence Bm t+k = (1 + i)kBt −Fr [(1 + i)k −1 i ] The second method is called the practical method. It is an approximate method that uses the linear approximation (1 + i)k ≈1 + ki for 0 <k <1. Thus, obtaining Bf t+k =(1 + ki)Bt Frk =kFr Bm t+k =(1 + ki)Bt −kFr = Bt + kiBt −kFr =Bt + k[Bt+1 + Fr −Bt] −kFr = (1 −k)Bt + kBt+1 408 BONDS AND RELATED TOPICS where we used the formula Bt(1 + i) −Fr = Bt+1. The third method called the semi-theoretical method is the most widely used method, and has been accepted as the standard method of calculation by the securities industry. The ﬂat price is determined as in the theoretical method, and the accrued coupon is determined as in the practical method. Bf t+k =(1 + i)kBt Frk =kFr Bm t+k =(1 + i)kBt −kFr In computing k, we use an actual count of days: k= number of days since last coupon paid number of days in the coupon period. Both the actual/actual and 30/360 methods (see Section 5) can be used. See Problem 45.12. Example 45.3 Consider a $1000 par value two-year 8% bond with semiannual coupons bought to yield 6% con- vertible semiannually. The price of the bond is computed to be $1037.17. Compute the ﬂat price, accrued interest, and market price ﬁve months after purchase of the bond. Solution. We have Fr = 1000(0.04) = $40.00 and k= 5 6 . With the theoretical method we ﬁnd Bf 5 6 =1037.17(1.03) 5 6 = $1063.04 Fr5 6 =40 [ (1.03) 5 6 −1 0.03 ] = $33.25 Bm 5 6 =1063.04 −33.25 = $1029.79 With the practical method we ﬁnd Bf 5 6 =1037.17 [ 1 + 5 6(0.03) ] = $1063.10 Fr5 6 =5 6(40) = $33.33 Bm 5 6 =1063.10 −33.33 = $1029.77 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 409 Finally, with the semi-theoretical method we have Bf 5 6 =1037.17(1.03) 5 6 = $1063.04 Fr5 6 =5 6(40) = $33.33 Bm 5 6 =1063.04 −33.33 = $1029.71 Example 45.4 A 1000 par value 5 year bond with semi-annual coupons of 60 is purchased to yield 8% convertible semi-annually. Two years and two months after purchase, the bond is sold at the ﬂat price which maintains the yield over the two years and two months. Calculate the ﬂat price using the theoretical method. Solution. The price of the bond after two years is P = 60a6 0.04 + 1000ν−6 0.04 = 1,104.84. The ﬂat price is Bf 2 6 = 1104.84(1.04) 2 6 = 1,119.38 We conclude this section by ﬁnding the premium or discount between coupon payment dates. For the premium we have the formula Premium = Bm t+k −C, i<g and for the discount we have Discount = C−Bm t+k, i>g. Thus, premiums and discounts are based on the market value or the book value and not on the ﬂat price. Example 45.5 A n−year $1000 par value bond is selling at $1010 and has a market value of $980 and acrrued coupon of $30. Find premium/discount amount. Solution. Since the premium/discount calculation is based on market value rather than ﬂat price, the bond is a discount bond. The discount amount is $20 410 BONDS AND RELATED TOPICS Practice Problems Problem 45.1 A 10 year bond with semi-annual coupons has a book value immediately after the 5th coupon of 90,000. The ﬂat price 5 months later using the theoretical method is 94,591. Calculate the semi-annual yield on the bond. Problem 45.2 A bond with semi annual coupons of 2500 has a book value immediately after the 6th coupon of 95,000. The market value using the practical method z months after the 6th coupon is 95,137.50. Calculate z if the yield rate is 7% convertible semi-annually. Problem 45.3 A 10 year bond with a par value of 100,000 and semi-annual coupons 2500 is bought at a discount to yield 6% convertible semi-annually. (a) Calculate the book value immediately after the 5th coupon. (b) Calculate the ﬂat price 4 months after the 5th coupon using the theoretical method. (c) Calculate the accrued coupon 4 months after the 5th coupon using the theoretical method. (d) Calculate the market price 4 months after the 5th coupon using the theoretical method. (e) Calculate (b)-(d) using the practical method. (f) Calculate (b)-(d) using the semi-theoretical method Problem 45.4 Which of the following are true for a bond with semi annual coupons? (I) The ﬂat price of the bond under the theoretical method will always equal the ﬂat price of the bond under the practical method. (II) The ﬂat price of the bond under the practical method will always exceed the ﬂat price of the bond under the semi-theoretical method. (III) For a bond purchased at par which matures at par with the yield rate equal to the coupon rate, the ﬂat price at any point in time will equal the par value of the bond under the semi-theoretical method. Problem 45.5 A bond pays coupons of 600 on April 30 and October 31. Calculate the accrued coupon under the practical method on July 1. Problem 45.6 A 10 year bond is redeemable at par of 100,000. The bond has semi-annual coupons of 4000. The bond is bought to yield 6% convertible semi-annually. Four months after purchase, calculate the market price based on the theoretical method less the market price based on the semi-theoretical method. 45 VALUATION OF BONDS BETWEEN COUPONS PAYMENT DATES 411 Problem 45.7 A 10-year 100 par value bond bearing a 10% coupon rate payable semi-annually, and redeemable at 105, is bought to yield 8% convertible semiannually. Using the practical method, compute the ﬂat price, accrued interest, and market price three months after purchase of the bond. Problem 45.8 Show that Bf t+k = (Bt+1 + Fr)v1−k. Problem 45.9 Arrange in increasing order of magnitude for the three interim bond price methods: (a) Flat price. (b) Market price. Problem 45.10 A 10,000 par value bond with 6% semiannual coupons is being sold 3 years and 2 months before the bond matures. The purchase will yield 8% convertible semiannually to the buyer. Determine the price of the bond. Use the practical method. Problem 45.11 ‡ A 10,000 par value bond with coupons at 8%, convertible semiannually, is being sold 3 years and 4 months before the bond matures. The purchase will yield 6% convertible semiannually to the buyer. The price at the most recent coupon date, immediately after the coupon payment, was 5640. Calculate the market price of the bond, assuming compound interest throughout. Problem 45.12 Consider a $1,000 par value bond with coupons at 8% convertible semi-annually and with maturity date 05/06/04. The bond was purchased on 11/06/01. Compute the accrued interest on 12/04/001 taking into account the Actual/Actual day-count basis. Problem 45.13 A 20-year 10% annual coupon bond has a par value of $1,000. When you originally purchased this bond, the eﬀective annual interest rate was 8%. Suppose that ﬁve years after purchase, the eﬀective annual interest rate is 10.5%. What is the absolute diﬀerence between the book and market values of the bond at that point in time? Problem 45.14 ‡ A 1000 bond with semi-annual coupons at i(2) = 6% matures at par on October 15, 2020. The bond is purchased on June 28, 2005 to yield the investor i(2) = 7%. What is the purchase price? Assume simple interest between bond coupon dates and note that: 412 BONDS AND RELATED TOPICS Date Day of the year April 15 105 June 28 179 October 15 288 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 413 46 Approximation Methods of Bonds’ Yield Rates In this section we consider the question of determining the yield rate to maturity to an investor given the purchase price of a bond. The determination of such yield rates is similar to the determination of an unknown rate of interest for an annuity, discussed in Section 20. We ﬁrst consider the yield rate for a bond purchased on a coupon payment date immediately after the coupon is paid. There are two approaches for this problem. One approach is linear interpolation where bond tables are used. We will not pursue this method any further in this book. Another approach is to develop algebraic estimation formulas for the yield rate. Three main formulas will be discussed. We start with the premium discount formula P =C+ (Fr −Ci)ani =C+ C(g−i)ani where F,r,g,C,n, and P are known and i is the unknown. Letting k= P−C C we ﬁnd (g−i)ani = k. Solving this equation for i we obtain i= g− k ani which deﬁnes i implicitly. We will develop methods for estimating i in this equation. The ﬁrst is obtained as follows: Writing the power series expansion of 1 ani (see p.209) 1 ani = 1 n [ 1 + n+ 1 2 i+ n2 −1 12 i2 + ··· ] and neglecting terms of higher than the ﬁrst degree in i we obtain the estimation i= g− k ani ≈g−k n [ 1 + n+ 1 2 i ] . Solving for i we ﬁnd i≈ g−k n 1 + n+1 2n k. (46.1) Formula (46.1) forms the basis of an even simpler method for calculating approximate yield rates called the bond salesman’s method. Noting that n+1 2n ≈1 2 for large n, this method replaces the ratio n+1 2n in formula (46.1) by 1 2 thus obtaining i≈ g−k n 1 + 1 2 k. 414 BONDS AND RELATED TOPICS Error analysis shows that this method produces less accurate results than formula (46.1). The salesman’s formula can be rewritten in the form i≈n(Cg) + C−P n 2 (P + C) = n(Fr) + C−P n 2 (P + C) = Total Interest Paid + Capital Gain/Loss Number of Periods ×Average Amount Invested Example 46.1 A $100 par value 10-year bond with 8% semiannual coupons is selling for $90. Find the yield rate convertible semi-annually. Solution. We have n= 20,P = 90,C = 100,Fr = 4 so that by the salesman’s formula we have i≈n(Fr) + C−P n 2 (P + C) = 20(4) + 100−90 0.5(20)(90 + 100) = 0.0474 so the yield rate is 9.48% convertible semi-annually The third method for estimating i is by solving the equation i= g− k ani . by using the Newton-Raphson method. In this case, we are looking for the solutions of the equation f(i) = 0 where f(i) = i−g+ k ani = i−g+ ki 1 −(1 + i)−n. The Newton-Raphson iterations are given by the formula is+1 = is −f(is) f′(is). Thus, an expression of the form i−f(i) f′(i) need to be determined. This is done as follows. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 415 i−f′(i) f(i) =i− i−g+ ki 1−(1+i)−n 1 + k ( [1−(1+i)−n]−ni(1+i)−n−1 [1−(1+i)−n]2 ) =i+ g(1 −νn)2 −i(1 −νn)2 −ki(1 −νn) (1 −νn)2 + k[(1 −νn) −niνn+1] =i+ (1 −νn)[giani −i(1 −νn) −ki] iani(1 −νn) + kiani −nkiνn+1 =i+ (1 −νn)[gani −(1 −νn) − (P C −1 ) ] ani(1 −νn) + (P C −1 ) ani −n (P C −1 ) νn+1 =i+ (1 −νn) [ gani + νn −P C ] ani(1 −νn) + (g−i)(ani)2 −n(g−i)aniνn+1 =i+ gani + νn −P C ani + (g−i)(ani)i−1 −n(g−i)i−1νn+1 =i+ i gani + νn −P C iani + (g−i)(ani) −n(g−i)νn+1 =i [ 1 + gani + νn −P C ani + (g−i)(ani) −n(g−i)νn+1 ] =i [ 1 + gani + νn −P C gani + n(i−g)νn+1 ] Thus, the Newton-Raphson iterations are given by is+1 = is [ 1 + ganis + (1 +is)−n −P C ganis + n(is −g)(1 + is)−(n+1) ] An initial starting guess can be chosen to be i0 = g−k n 1 + n+1 2n k. Example 46.2 A $100 par value 8-year bond with 6% semiannual coupons is selling for $97. Find the yield rate convertible semiannually. 416 BONDS AND RELATED TOPICS Solution. We have g = 3%, n= 16, k= P−C C = 97−100 100 = −0.03. An initial starting guess for the Newton- Raphson method is i0 = 0.03 −(−0.03/16) 1 + 16+1 32 = 0.0323912 and the iterations are given by the formulas is+1 = is ( 1 + 0.03a16 is + (1 +is)−16 −0.97 0.03a16 is + 16(is −0.03)(1 + is)−17 ) . We obtain the following values s is 2is 0 0.0323912 0.0647824 1 0.0324329 0.0648658 2 0.0324329 0.0648658 Thus, the yield rate convertible semiannually is 6 .48658%. It is worth noting that the selling price is P = 100 (0.06 2 ) a16 3.24329% + 100(1.0324329)−16 = $97 One can also detetermine the yield rate when the purchasing date is between two coupon payment dates. This situation is more complex due to the fractional period prior to the next coupon payment date. In practice, this case involves the use of the semi-theoretical method of Section 45 which leads to an implicit equation in i that can be solved by either linear interpolation or by iterations. We illustrate this in the next example. Example 46.3 A $100 par value 10-year bond with 8% semiannual coupons was issued on March 1, 2001. The market price on May 15, 2003 is $88. Find the yield rate if the bond in bought on that date. Solution. The price of the bond on March 1, 2003 immediately after a coupon payment is 100 + (4−100i)a16 i. The number of days from March 1, 2003 to May 15, 2003 is 75, while the number of days from March 1, 2003 to September 1, 2003 is 184. Using the semi-theoretical method of Section 45, we have the equation [100 + (4−100i)a16 i](1 + i) 75 184 − 75 184 ·4 = 88. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 417 Let’s solve this last equation by using linear interpolation. Let f(i) = [100 + (4−100i)a16 i](1 + i) 75 184 − 75 184 ·4 −88. By trial and error we ﬁnd f(0.05) = 23 .43 > 0 and f(0.052) = −2.45 < 0. Thus, using linear interpolation we ﬁnd i≈0.05 −23.43 × 0.052 −0.05 −2.45 −23.43 = 0.0518. So the nominal yield rate convertible semiannually is 10.36%. The price on May 15,2003 is 100 + (4−100 ×0.0518)a16 0.0518(1.0518) 75 184 − 75 184 ·4 = $87.56 a roundoﬀ error of 0.44 We ﬁnally consider ﬁnding the yield rate of a bond with coupons being reinvested at a diﬀer- ent interest rate. Consider the situation in which a bond is purchased for P,coupons of Fr are paid at the end of each period for n periods, the bond is redeemed for C at the end of n periods, and the coupons are reinvested at rate j. Denoting the yield rate (considering reinvestment) by i′, we would have the equation of value: P(1 + i′)n = Frsnj + C since both sides represent the value of the investment at the end of n periods. Example 46.4 A $100 par value 10-year bond with 8% semiannual coupons is selling for $90 and the coupons can be reinvested at only 6% convertible semi-annually. Find the yield rate taking in consideration reinvestment rates. Solution. We have 90(1 + i′)20 = 4s20 0.03 + 100 →(1 + i′)20 = 4(26.8704)+100 90 = 2.30535 →i′= 0.04265. Thus, the yield rate is 2(0 .04265) = 0 .0853 = 8 .53% convertible semi-annually. Thus, the yield rate of 9.48% found in Example 44.1 drops to 8.53% if the coupons can be reinvested at 6% 418 BONDS AND RELATED TOPICS Practice Problems Problem 46.1 A $100 par value 12-year bond with 10% semiannual coupons is selling for $110. Find the yield rate convertible semiannually. Problem 46.2 Recompute the overall yield rate to an investor purchasing the bond in the previous problem, if the coupons can be reinvested at only 7% convertible semiannually. Problem 46.3 A $100 bond with annual coupons is redeembale at par at the end of 15 years. At a purchase price of $92 the yield rate is exactly 1% more than the coupon rate. Find the yield rate on the bond. Problem 46.4 An investor buys 20-year bonds, each having semiannual coupons and each maturing at par. For each bond the purchase price produces the same yield rate. One bond has a par value of $500 and a coupon of $45. The other bond has a par value of $1000 and a coupon of $30. The dollar amount of premium on the ﬁrst bond is twice as great as the dollar amount of discount on the second bond. Find the yield rate convertible semiannually. Problem 46.5 A 20 year bond with a 20,000 par value pays semi-annual coupons of 500 and is redeemable at par. Audrey purchases the bond for 21,000. Calculate Audrey’s semi-annual yield to maturity on the bond. Use Salesman’s formula. Problem 46.6 A 10 year bond with a par value of 1000 is redeemable at par and pays annual coupons of 65. If Jamie purchased the bond for 950 and she reinvests the coupons at 4% annual eﬀective rate, calculate the actual yield that Jamie will receive taking into account the reinvestment rate. Problem 46.7 ‡ A 1000 par value 10-year bond with coupons at 5%, convertible semiannually, is selling for 1081.78. Calculate the yield rate convertible semiannually. Problem 46.8 ‡ Dan purchases a 1000 par value 10-year bond with 9% semiannual coupons for 925. He is able to reinvest his coupon payments at a nominal rate of 7% convertible semiannually. Calculate his nominal annual yield rate convertible semiannually over the ten-year period. 46 APPROXIMATION METHODS OF BONDS’ YIELD RATES 419 Problem 46.9 Suppose a 10000 par bond has 6% semi-annual coupons and matures in 10 years. What is the yield rate if the price is 9000? Compute both exactly and using the Salesman’s method. Problem 46.10 A $1000 par value 10-year bond with 7% semiannual coupons was issued on October 1, 2006. The market price on April 28, 2006 is $1020. Find the yield rate if the bond in bought on that date. Problem 46.11 John buys a $1000 bond that pays coupons at a nominal interest rate of 7% compounded semian- nually. The bond is redeemable at par in 15 years. The price he pays will give him a yield of 8% compounded semiannually if held to maturity. After 5 years, he sells the bond to Peter, who desires a yield of 6% on his investment. (a) What price did John pay? (b) What price did Peter pay? (c) What yield did John realize? 420 BONDS AND RELATED TOPICS 47 Callable Bonds and Serial Bonds Callable or redeemable bonds are bonds that can be redeemed or paid oﬀ by the issuer prior to the bonds’ maturity date. The earliest such call date will generally be several years after the issue date. Call dates are usually speciﬁed by the bond issuer. When an issuer calls its bonds, it pays investors the call price (usually the face value of the bonds) together with accrued interest to date and, at that point, stops making interest payments. An issuer may choose to redeem a callable bond when current interest rates drop below the interest rate on the bond. That way the issuer can save money by paying oﬀ the bond and issuing another bond at a lower interest rate. This is similar to reﬁnancing the mortgage on your house so you can make lower monthly payments. Callable bonds are more risky for investors than non-callable bonds because an investor whose bond has been called is often faced with reinvesting the money at a lower, less attractive rate. As a result, callable bonds often have a higher annual return to compensate for the risk that the bonds might be called early. Callable bonds present a problem when calculating prices and yields, because the call date of the bond is unknown. Since the issuer has an option whether or not to call the bond, the investor should assume that the issuer will exercise that option to the disadvantage of the investor, and should calculate the yield rate and/or price accordingly. If the redemption value is the same at any call date, including the maturity date, then the following general principle will hold: (i) the call date will most likely be at the earliest date possible if the bond was sold at a premium, which occurs when the yield rate is smaller than the coupon rate (issuer would like to stop repaying the premium via the coupon payments as soon as possible). (ii) the call date will most likely be at the latest date possible if the bond was sold at a discount, which iccurs when the yield rate is larger than the coupon rate (issuer is in no rush to pay out the redemption value). Example 47.1 A 15 year 1000 par bond has 7% semiannual coupons and is callable at par after 10 years. What is the price of the bond to yield 5% for the investor? Solution. Since the yield rate is smaller than the coupan rate, the bond is selling at premium so the earliest redemption date is the most favorable for the issuer and the least favorable for the investor. The price is the smallest of 1000 + (35 −25)an0.025 for 20 ≤n≤30, which clearly occurs when n= 20. Thus, the price is 1000 + (35−25)a20 0.025 = 1,155.89 47 CALLABLE BONDS AND SERIAL BONDS 421 If the redemption values on all the redemption dates are not equal, this principle is a little harder to apply. In this case one needs to examine all call dates. The most unfavorable date will not necessarily be either the earliest or the latest possible date. The most unfavorable call date ( to the issuer) is the one that produces the smallest purchase price at the investors yield rate as illustrated in the following example. Example 47.2 A 100 par value 4% bond with semi-annual coupons is callable at the following times: 109.00, 5 to 9 years after issue 104.50, 10 to 14 years after issue 100.00, 15 years after issue (maturity date). What price should an investor pay for the callable bond if they wish to realize a yield rate of (1) 5% payable semi-annually and (2) 3% payable semi-annually? Solution. (1) Since the market rate is better than the coupon rate, the bond would have to be sold at a discount and as a result, the issuer will wait until the last possible date to redeem the bond: P = 2.00a30 2.5% + 100.00v30 0.025 = $89.53. (2) Since the coupon rate is better than the market rate, the bond would sell at a premium and as a result, the issuer will redeem at the earliest possible date for each of the three diﬀerent redemption values: P =2.00a10 1.5% + 109.00v10 0.015 = $112.37 P =2.00a20 1.5% + 104.50v20 0.015 = $111.93 P =2.00a30 1.5% + 100.00v30 0.015 = $112.01 In this case, the investor would only be willing to pay $111.93 Serial Bonds Serial bonds are bonds issued at the same time but with diﬀerent maturity dates. The price of any individual bond is found by the methods already discussed earlier in this chapter provided that the redemption date is known. The price of the entire serial bonds is just the sum of the values of the individual bonds. We next consider ﬁnding a formula for pricing an issue of serial bonds with mdiﬀerent redemption dates. Let Pt be the purchase price of the tth bond; Ct be the redemption value and Kt the present value of Ct. Then by Makeham’s formula the price of the tth bond is Pt = Kt + g i(Ct −Kt). 422 BONDS AND RELATED TOPICS Summing from t= 1 to t= m to obtain the price of the entire issue of a serial bond m∑ t=1 Pt = m∑ t=1 Kt + g i ( m∑ t=1 Ct − m∑ t=1 Kt ) or P′= K′+ g i(C′−K′) where P′= ∑m t=1 Pt, K′= ∑m t=1 Kt, and C′= ∑m t=1 Ct. Example 47.3 A $10,000 serial bond is to be redeemed in $1000 installments of principal per half-year over the next ﬁve years. Interest at the annual rate of 12% is paid semi-annually on the balance outstanding. How much should an investor pay for this bond in order to produce a yield rate of 8% convertible semi-annually? Solution. We have P′= 10∑ t=1 1000vt 0.04 + 0.06 0.04 [ 10,000 − 10∑ t=1 1000vt 0.04 ] =1000a10 0.04 + 1.5[10,000 −1000a10 0.04] =15,000 −500a10 0.04 = 10,944.58 Example 47.4 A 5000 serial bond with 10% annual coupons will be redeemed in ﬁve equal installments of 1100 beginning at the end of the 11th year and continuing through the 15th year. The bond was bought at a price P to yield 9% annual eﬀective. Determine P. Solution. We have: F = 1000,C = 1100,i = 0.09 and r= 0.10. Thus, g= 1000(0.10) 1100 = 1 11 . Hence, C′= 5500 K′= 15∑ t=11 1100vt 0.09 =1100(a15 0.09 −a10 0.09) = 1807.33 The price is P′= 1807.33 + 1 0.99(5500 −1807.33) = 5537.30 47 CALLABLE BONDS AND SERIAL BONDS 423 Practice Problems Problem 47.1 A $1000 par value bond has 8% semiannual coupons and is callable at the end of the 10th through the 15th years at par. (a) Find the price to yield 6% convertible semiannually. (b) Find the price to yield 10% convertible semiannually. Problem 47.2 A $1000 par value 8% bond with quarterly coupons is callable ﬁve years after issue. The bond matures for $1000 at the end of ten years and is sold to yield a nominal rate of 6% convertible quarterly under the assumption that the bond will not be called. Find the redemption value at the end of ﬁve years that will provide the purchaser the same yield rate. Problem 47.3 A $1000 par value 4% bond with semiannual coupons matures at the end of 10 years. The bond is callable at $1050 at the ends of years 4 through 6, at $1025 at the ends of years 7 through 9, and at $1000 at the end of year 10. Find the maximum price that an investor can pay and still be certain of a yield rate of 5% convertible semiannually. Problem 47.4 A $1000 par value 6% bond with semiannual coupons is callable at par ﬁve years after issue. It is sold to yield 7% semi-annual under the assumption that the bond will be called. The bond is not called and it matures at the end of 10 years. The bond issuer redeems the bond for 1000 + X without altering the buyer’s yield rate of 7% convertible semiannually. Find X. Problem 47.5 A 15 year bond can be called at the end of any year from 10 through 14. The bond has a par value of 1000 and an annual coupon rate of 5%. The bond is redeemable at par and can also be called at par. (a) Calculate the price an investor would pay to yield 6%. (b) Calculate the price an investor would pay to yield 4%. Problem 47.6 A 15 year bond can be called at the end of any year from 12 through 14. The bond has a par value of 1000 and an annual coupon rate of 5%. The bond is redeemable at par and can be called at 1100. (a) Calculate the price an investor would pay to yield 6%. (b) Calculate the price an investor would pay to yield 4%. 424 BONDS AND RELATED TOPICS Problem 47.7 A callable 20 year bond which matures for 1000 pays annual coupons of 80. The bond is callable in years 10 through 12 at 1100 and in years 13 through 15 at 1050. The bond is bought to yield 4% annually. Calculate the purchase price of the bond. Problem 47.8 ‡ Matt purchased a 20-year par value bond with semiannual coupons at a nominal annual rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par value X on any coupon date starting at the end of year 15 after the coupon is paid. The price guarantees that Matt will receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate X. Problem 47.9 ‡ Toby purchased a 20-year par value bond with semiannual coupons at a nominal annual rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par value 1100 on any coupon date starting at the end of year 15. What is the minimum yield that Toby could receive, expressed as a nominal annual rate of interest convertible semiannually? Problem 47.10 ‡ Sue purchased a 10-year par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value X on any coupon date starting at the end of year 5. The price guarantees that Sue will receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate X. Problem 47.11 ‡ Mary purchased a 10-year par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value 1100 on any coupon date starting at the end of year 5. What is the minimum yield that Mary could receive, expressed as a nominal annual rate of interest convertible semiannually? Problem 47.12 ‡ A 1000 par value bond with coupons at 9% payable semiannually was called for 1100 prior to maturity. The bond was bought for 918 immediately after a coupon payment and was held to call. The nominal yield rate convertible semiannually was 10%. Calculate the number of years the bond was held. 47 CALLABLE BONDS AND SERIAL BONDS 425 Problem 47.13 ‡ A 1000 par value bond pays annual coupons of 80. The bond is redeemable at par in 30 years, but is callable any time from the end of the 10th year at 1050. Based on her desired yield rate, an investor calculates the following potential purchase prices, P : Assuming the bond is called at the end of the 10th year, P = 957 Assuming the bond is held until maturity, P = 897 The investor buys the bond at the highest price that guarantees she will receive at least her desired yield rate regardless of when the bond is called. The investor holds the bond for 20 years, after which time the bond is called. Calculate the annual yield rate the investor earns. Problem 47.14 Which of the following are true: (I) If a callable bond is called or redeemed at any date other than that used to calculate the purchase price, the buyer’s yield will be higher than that used to calculate the yield rate. (II) If the redemption value of a callable bond is equal at all possible redemption dates and the bond sells at a premium, the price of the bond is calculated by assuming the redemption date is the last possible date. (III) To calculate the price of a callable bond, a buyer should calculate the price at each and every possible redemption date and take the lowest price. Problem 47.15 A 5000 par value 18-year bond has 7% semiannual coupons, and is callable at the end of the 12th through the 17th years at par. (a) Find the price to yield 6% convertible semiannually. (b) Find the price to yield 8% convertible semiannually. (c) Find the price to yield 8% convertible semiannually, if the bond pays a pre- mium of 250 if it is called. (d) Find the price to yield 6% convertible semiannually, if the bond pays a premium of 250 if it is called at the end of years 12 through 14, and a premium of 150 if it is called at the end of years 15 or 16. (It may still be called at the end of year 17 without premium, and otherwise will mature at the end of year 18.) Problem 47.16 Find the price of a $1000 issue of 5.25% bonds with annual coupons which will be redeemed in 10 annual installments at the end of the 11th through the 20th years from the issue date at 105. The bonds are bought to yield 7% eﬀective. 426 BONDS AND RELATED TOPICS Problem 47.17 ABC is issuing a $1000 bond with a maturity of 25 year and annual 10% coupon. The bond is callable with the ﬁrst year call price equal to the oﬀering price plus the coupon; thereafter the call price decreases by equal amounts to equal par at year 20; thereafter the call price is equal to par. If the bond were sold at $950 what would be the call prices for each year? Stocks and Money Market Instruments One of the reasons investors are interested in trading with stocks because of its high rate of return (about 10%). Stocks are examples of ﬁnancial instruments or securities, terms which we deﬁne next. A ﬁnancial instrument is a monetary claim that one party has on another. It is a ﬁnancial asset for the person who buys or holds it, and it is a ﬁnancial liability for the company or institution that issues it. For example, a share of Microsoft stock that you own. The stock gives you a share of Microsoft’s assets and the right to receive a share of dividends (proﬁts), if Microsoft is paying any. To Microsoft owners, a stock is an obligation to include you in their divident payments. A security is a tradeable ﬁnancial instrument, like a bond or a share of stock. A ﬁnancial market is a market in which ﬁnancial assets are traded. In addition to enabling exchange of previously issued ﬁnancial assets, ﬁnancial markets facilitate borrowing and lending by facilitating the sale by newly issued ﬁnancial assets. Examples of ﬁnancial markets include the New York Stock Exchange (resale of previously issued stock shares), the U.S. government bond market (resale of previously issued bonds), and the U.S. Treasury bills auction (sales of newly issued T−bills). A ﬁnancial institution is an institution whose primary source of proﬁts is through ﬁnancial asset transactions. Examples of such ﬁnancial institutions include discount brokers (e.g., Charles Schwab and Associates), banks, insurance companies, and complex multi−function ﬁnancial institutions such as Merrill Lynch. Stocks are ﬁnancial instruments that make the holder a co −owner of the company that issued them. They entitle the holder to a claim on the assets (and implicitly the future proﬁts) of the company. Stocks do not involve the repayment of a debt or the payment of interest. (Some stocks pay dividends, which are shares of the company’s proﬁts, but people typically hold stocks not for the dividends but for the hope of reselling them later at a higher price.) In this chapter, we introduce two types of stocks: preferred and common stocks. We also discuss trading stocks. More speciﬁcally, we will discuss buying stocks by borrowing money (buying on margin) and selling stocks that we don’t own (short sales of socks). We conclude this chapter by introducing some of the ﬁnancial instrument used in today’s ﬁnancial markets. 427 428 STOCKS AND MONEY MARKET INSTRUMENTS 48 Preferred and Common Stocks Three common types of securities which have evolved in the ﬁnancial markets: bonds, preferred stocks, and common stocks. Bonds have been the topic of the previous chapter. In this section, we discuss preferred stocks and common stocks. Preferred Stock This is a type of security which provides a ﬁxed rate of return (similar to bonds). However, unlike bonds, it is an ownership security (bonds are debt security). Generally, preferred stock has no maturity date, although on occasion preferred stock with a redemption provision is issued. The periodic payment is called a dividend, because it’s being paid to an owner. Preferred stock is typically a higher ranking stock than common stock. Preferred stock may or may not carry voting rights, and may have superior voting rights to common stock. Preferred stock may carry a dividend that is paid out prior to any dividends to common stock holders. Preferred stock may have a convertibility feature into common stock. Preferred stock holders will be paid out in assets before common stockholders and after debt holders in bankruptcy. In general, there are four diﬀerent types of preferred stock: •Cumulative preferred stock. Preferred stock on which dividends accrue in the event that the issuer does not make timely dividend payments. Most preferred stock is cumulative preferred. •Non-cumulative preferred stock. Preferred stock for which unpaid dividends do not accrue. •Participating preferred stock. This is a kind of preferred stock that pays a regular dividend and an additional dividend when the dividend on the common stock exceeds a speciﬁc amount. •Convertible preferred stock. Convertible preferred stock has a privilege similar to convertible bonds. Owners have the option to convert their preferred stock to common stock under certain conditions. Since a preferred stock has no redemption date, it pays dividend forever. Their cash ﬂows are, therefore, those of a perpetuity. Thus, the price of a preferred stock is the present value of future dividends of a perpetuity which is given by the formula P = Fr i . Example 48.1 A preferred stock pays a coupon of 50 every six months. Calculate the price of the preferred stock using a yield rate of 5% convertible semi-annually. Solution. The price of the stock is P = Fr i = 50 0.025 = 2000 48 PREFERRED AND COMMON STOCKS 429 Common stock Common stock is another type of ownership security. It does not earn a ﬁxed dividend rate as preferred stock does. Dividends on this type of stock are paid only after interest payments on all bonds and other debt and dividends on preferred stock are paid. The dividend rate is ﬂexible, and is set by the board of directors at its discretion. Value of a share of common stock is being based on the dividend discount model. This model values shares at the discounted value of future dividend payments. That is, the value of a share is the present value of all future dividends. Consider a situation where a dividend D is paid at the end of the current period. Assume that the next dividend payments follow a geometric progression with common ratio 1 + k indeﬁnitely and the stock is purchased at a yied rate of iwith −1 <k <i. Then the theoretical price of the stock is P = D lim n→∞ 1 − (1+k 1+i )n i−k = D i−k. We point out that the most common frequency of dividend payments on both preferred and commom stock in the US is quarterly. Example 48.2 A common stock is currently earning $1 per share and the dividend is assumed to increase by 5% each year forever. Find the theoretical price to earn an investor an annual eﬀective yield rate of 10%. Solution. The theoretical price is P = D 1 i−k = 1 0.10 −0.05 = 20 It is unrealistic to project constant percentage increases in dividends indeﬁnitely into the future. As corporations increase in size and become more mature, the rate of growth will generally slow down. Example 48.3 A common stock pays annual dividends at the end of each year. The earnings per share for the most recent year were 8 and are assumed to grow at a rate of 10% per year, forever. The dividend will be 0% of earnings for each of the next 10 years, and 50% of earnings thereafter. What is the theoretical price of the stock to yield 12%? 430 STOCKS AND MONEY MARKET INSTRUMENTS Solution. The theoretical price of the stock which is the present value of all future dividends is ∞∑ k=11 4(1.1)k(1.12)−k = ∞∑ k=11 4 (1.1 1.12 )k =4 (1.1 1.12 )11 [ 1 1 − (1.1 1.12 ) ] = 183.73 Example 48.4 A common stock is currently earning $4 per share and will pay $2 per share in dividends at the end of the current year. Assuming that the earnings of the corporation increase at the rate of 5% for the ﬁrst ﬁve years, 2.5% for the second ﬁve years and 0% thereafter. Assume that the corporation plans to continue to pay 50% of its earnings as dividends, ﬁnd the theoretical price to earn an investor an annual eﬀective yield rate of 10%. Solution. The theoretical price is 2 [ 1 − (1.05 1.10 )5 0.10 −0.05 ] + 2(1.05)5 (1.10)5 [ 1 − (1.025 1.10 )5 0.10 −0.025 ] + 2(1.05)5(1.025)5 (1.10)10 · 1 0.10 = $25.72 48 PREFERRED AND COMMON STOCKS 431 Practice Problems Problem 48.1 A preferred stock pays a $10 dividend at the end of the ﬁrst year, with each successive annual dividend being 5% greater than the preceding one. What level annual dividend would be equivalent if i= 12%? Problem 48.2 A preferred stock will pay a dividend of 100 today. The stock also provides that future dividends will increase at the rate of inﬂation. The stock is purchased to yield 8% and assuming an inﬂation rate of 3%. Calculate the price of the stock today (before payment of the dividend). Problem 48.3 A common stock pays annual dividends at the end of each year. The earnings per share in the year just ended were $6. Earnings are assumed to grow 8% per year in the future. The percentage of earnings paid out as a dividend will be 0% for the next ﬁve years and 50% thereafter. Find the theoretical price of the stock to yield an investor 15% eﬀective. Problem 48.4 A common stock is purchased at a price equal to 10 times current earnings. During the next 6 years the stock pays no dividends, but earnings increase 60%. At the end of 6 years the stock sold at a price equal 15 times earnings. Find the eﬀective annual yield rate earned on this investment. Problem 48.5 A company’s common stock pays a dividend of $2 each year. The next dividend will be paid in one year. If the dividend is expected to increase at 5% per year, calculate the value of the stock at a 10% yield rate. Problem 48.6 A company’s current earnings are $5 per share. They expect their earnings to increase at 10% per year. Once their earnings are $10 per share, they will payout 75% as a dividend. Calculate the value of the stock to yield 15%. Problem 48.7 Thompson Corporation currently pays a dividend of 3 on earnings of 5 per year. Thompson’s earnings are expected to increase at a rate of 6% per year with 60% of earnings paid as dividends. Calculate the theoretical price of the common stock to yield 10%. 432 STOCKS AND MONEY MARKET INSTRUMENTS Problem 48.8 A common stock is currently earning $4 per share and will pay $2 per share in dividends at the end of the current year. Assuming that the earnings of the corporation increase 5% per year indeﬁnitely and that the corporation plans to continue to pay 50% of its earnings as dividends, ﬁnd the theoretical price to earn an investor an annual eﬀective yield rate of (1) 10%, (2) 8%, and (3) 6% Problem 48.9 A common stock is purchased on January 1, 1992. It is expected to pay a dividend of 15 per share at the end of each year through December 31, 2001. Starting in 2002 dividends are expected to increase K% per year indeﬁnitely, K <8%. The theoretical price to yield an annual eﬀective rate of 8% is 200.90. Calculate K. Problem 48.10 On January 1 of each year Company ABC declares a dividend to be paid quarterly on its common shares. Currently, 2 per share is paid at the end of each calendar quarter. Future dividends are expected to increase at the rate of 5% per year. On January 1 of this year, an investor purchased some shares at X per share, to yield 12% convertible quarterly. Calculate X. Problem 48.11 ‡ The dividends of a common stock are expected to be 1 at the end of each of the next 5 years and 2 for each of the following 5 years. The dividends are expected to grow at a ﬁxed rate of 2% per year thereafter. Assume an annual eﬀective interest rate of 6%. Calculate the price of this stock using the dividend discount model. Problem 48.12 ‡ A commom stock sells for 75 per share assuming an annual eﬀective interest rate of i. Annual dividends will be paid at the end of each year forever. The ﬁrst dividend is 6, with each subsequent dividend 3% greater than the previous year’s dividend. Calculate i. 49 BUYING STOCKS 433 49 Buying Stocks In this section, we discuss the costs associated with buying stocks. One way to buy stocks is by buying on margin (i.e. buying stocks with borrowed money). When an investor is buying an asset, he is said to have a long position in the asset. If the investor is selling an asset he is said to have a short position in that asset. In the stock and bond markets, if you short an asset, it means that you borrow it, sell the asset, and then later buy it back. Short sales will be discussed in Section 50. Now, depending upon whether an investor is buying or selling a security, he will get diﬀerent price quotes. A bid price is the price at which a broker is ready to purchase the security. Thus, this is the selling price for an investor. The ask price is the price at which a broker is ready to sell the security. The ask is the buying price paid by an investor. The diﬀerence between the two prices is called the bid-ask spread (bid price is lower than the asked price, and the spread is the broker’s proﬁt). Example 49.1 MBF stock has a bid price of $10.35 and an ask price of $11.00. Assume there is a $5 brokerage commission. (a) What is the bid-ask spread? (b) What amount will you pay to buy 50 shares? (c) What amount will you receive for selling 50 shares? (d) Suppose you buy 50 shares, then immediately sell them with the bid and ask price being the same as above. What is your round-trip transaction cost? Solution. (a) The bid-ask spread is 11 .00 −10.35 = $0.65. (b) You pay 50 ×11.00 + 5 = $555.00 (c) You receive 50 ×10.35 −5 = $512.50. (d) For each stock, you buy at $11.00 and sell it an instant later for $10.35. The total loss due to the ask-bid spread: 50(11 .00 −10.35) = $32.50. In addition, you pay $5 twice. Your total transaction cost is then 32 .50 + 2(5) = $42.50 Buying on Margin When buying on margin, the investor borrows part of the purchase price of the stock from a broker or a brokerage ﬁrm and contributes the remaining portion (the initial margin requirement or simply the margin) which is kept in an account with the brokerage ﬁrm. If you buy Q shares at the price of P (per share) with initial margin percentage of IM%, then your initial margin requirement IM is given by the formula IM = Q·P ·(IM%). 434 STOCKS AND MONEY MARKET INSTRUMENTS Example 49.2 Suppose you want to buy 100 shares on margin at the price of $50 per share. Assuming that the initial margin percentage is 50%, what was your initial margin requirement? Solution. You initial margin requirement IM is IM = 100(50)(0.5) = $2500 The brokerage ﬁrm is making money on your loan in terms of interest. Moreover, your stock will be hold as the collateral against the loan. If you default, the ﬁrm will take the stock. There are risks associated to buying stock on margin. The price of your stock could always go down. By law, the brokerage ﬁrm will not be allowed to let the value of the collateral (the price of your stock) go down below a certain percentage of the loan value. This percentage is called the maintenance margin. If the stock drops below that set amount, the brokerage ﬁrm will issue a margin call on your stock. The margin call means that you will have to pay the brokerage ﬁrm the amount of money necessary to bring the brokerage ﬁrms risk down to the allowed level. If you don’t have the money, your stock will be sold to pay oﬀ the loan. If there is any money left, it will be sent to you. In most cases, there is little of your original investment remaining after the stock is sold. We deﬁne the actual margin AM by the formula AM = Market Value of Assets−Loan Market Value of Assets If AM < MM,then the investor is undermargined (he or she receives a margin call from his or her broker) and must take corrective action (pay oﬀ part of loan or deposit more collateral). If the investor fails to respond, then the broker will close the investor’s account. If AM > IM,then the investor is overmargined and can withdraw cash or buy more shares. Example 49.3 John wants to purchase 100 stocks of IBM. The current price of a stock is $100. He has $6000 on hand, so he decides to borrow $4000. The maintenance marginal is set to be 50%. (a) What is John’s liability and equity? (b) Suppose the stock falls to $70. What is the actual margin? Would a margin call be in order? (c) How far can the stock price fall before a margin call? Solution. (a) John’s liability is $4000 and his equity is $6000. So John is contributing 6000 10000 = 60% (the margin) of the purchasing price. (b) If the stock falls to $70, then John’s liability is still $4000 but his equity is reduced to $3000 so that the new margin is 3000 7000 = 43% < 50%. Since this is less than the maintenance margin, the 49 BUYING STOCKS 435 broker will issue a margin call. (c) Let P be the original price. Then the margin in this case is 100P −4000 100P . Setting this to 0.5 and solving for P we ﬁnd P = $80. Thus, anytime the stock drops below $80, John will receive a margin call Example 49.4 John wants to purchase 100 stocks of IBM. The current price of a stock is $100. He has $6000 on hand, so he decides to borrow $4000. The broker charges an annual eﬀective rate of interest of 10%. Suppose the stock goes up by 30% in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Solution. (a) The end-year-value of the shares is 100[100 + (30%)(100)] = $13 ,000. (b) The total of loan repayment and interest is 4000 + 4000(10%) = $4 ,400. (c) The rate of return is 13000 −10400 6000 = 43.3% (d) If not buying on margin, the rate of return is 13000 −10000 10000 = 30% The reason why an investor buys on margin is his wish to invest more than what the investor’s money would allow. Buying on margin could mean a huge return when stock prices go up. But there is the risk that you could lose your original investment when stock price goes down. As with any stock purchase there are risks, but when you are using borrowed money, the risk is increased. 436 STOCKS AND MONEY MARKET INSTRUMENTS Practice Problems Problem 49.1 MBF stock has a bid price of $10.35 and an ask price of $11.00. Assume there is a 0.1% brokerage commission on the bid or ask price. (a) What amount will you pay to buy 50 shares? (b) What amount will you receive for selling 50 shares? (c) Suppose you buy 50 shares, then immediately sell them with the bid and ask price being the same as above. What is your round-trip transaction cost? Problem 49.2 MBF stock has a bid price of $10.35 and an ask price of $11.00. At what price can the market-maker purchase a stock? At what price can a market-maker sell a stock? What is the market-maker proﬁt from buying and immediately selling 50 shares? Problem 49.3 Suppose the spot ask exchange rate is $2.10 = £1.00 and the spot bid exchange rate is $2.07 = £1.00. If you were to buy $5,000,000 worth of British pounds and then sell them ﬁve minutes later without the bid or ask changing, how much of your $5,000,000 would be ”eaten” by the bid-ask spread? Problem 49.4 Consider the transaction of Example 49.4 but now assume that there is no change in the stock price in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Problem 49.5 Consider the transaction of Example 49.4 but now assume that there is a decrease of 30% in the stock price in one year period. (a) What is the end-year-value of the shares? (b) Find the total of loan repayment and interest. (c) What is the rate of return? (d) What is the rate of return if not buying on margin? Problem 49.6 On July 24, 2002, you bought 100 shares of Microsoft on margin at the price of $45 per share. Assuming that the initial margin percentage is 50%, what was your initial margin requirement? 49 BUYING STOCKS 437 Problem 49.7 Using Problem 49.6, suppose that On July 25, 2002, Microsoft closed at $42.83. What was your actual margin on that day? Problem 49.8 Using Problem 49.6, suppose that the maintenance margin for Microsoft is 40%. Did you receive a margin call on July 25, 2002? Problem 49.9 On December 30, 1998, you decided to bet on the January eﬀect, a well−known empirical regularity in the stock market. On that day, you bought 400 shares of Microsoft on margin at the price of $139 per share. The initial margin requirement is 55% and the maintenance margin is 30%. The annual cost of the margin loan is 6.5%. (a) Determine your initial margin requirement. (b) To what price must Microsoft fall for you to receive a margin call? (c) On January 8, 1999, Microsoft climbed to $151.25. What was the actual margin in your account on that day? (d) On January 9, 1999, you sold your Microsoft shares at the price of $150.50 per share. What was the return on your investment? (e) Recalculate your answer to part (d) assuming that you made the Microsoft stock purchase for cash instead of on margin. 438 STOCKS AND MONEY MARKET INSTRUMENTS 50 Short Sales In this section we discuss trading security over short period of time when the security is likely to decline in value. A short sale is generally a sale of a security by an investor who does not actually own the security. To deliver the security to the purchaser, the short seller will borrow the security. The short seller later ”pays oﬀ” the loan by returning the security to the lender, typically by purchasing the sold security back on the open market (hopefully at a lower price). This process of buying back the security is called covering the short because the seller is no longer at risk to movements in the security. One problem with short sales is determining the yield rate. If the transaction occurs exactly as stated, then technically the yield rate does not exist. We illustrate this in the following example. Example 50.1 An investor sells a stock short for $1000 and buys it back for $800 at the end of one year. What is the yield rate? Solution. A possible equation of value is 1000(1 + i) = 800 which yields i= −20% which is clearly unreason- able since the seller made $200 in proﬁt from the transaction. Now, if we reverse the transaction and solve the equation of value 800(1 + i) = 1000 we obtain i= 25%.But this equation states that there was $200 proﬁt on an $800 investment, but there never was an $800 investment Short sales rarely occur as illustrated above. In practice, the short seller is required to make a deposit (i.e. a collateral) of a percentage of the price at the time the short sale is made. This deposit is called the margin, and cannot be accessed by the short seller until the short sale is covered. However, the short seller will be credited with interest on the margin deposit which will increase the yield rate. Margins are usually expressed as a percentage of the value of the security. For example, govern- mental regulations in the United States require the short seller to make a deposit of 50% of the price at the time the short sale is made. In the example above, the short seller will have to deposit a margin of $500 at the time the short sale. The short seller is not allowed to earn interest on the proceeds from the short sale. Governmental regulations require that these proceeds remain in a non-interest bearing account until the short position is covered, at which time these funds will be used for the purchase necessary to cover the short. Any leftover money is proﬁt on the transaction, while a negative balance would be loss on the transaction. 50 SHORT SALES 439 If coupons or dividends are paid on the security while you are short it, you must pay them to the party you borrowed the security from. Dividend payments are often deducted from the margin account. In order to ﬁnd the yield rate on a short sale we introduce the following notation: •M = Margin deposit when sell was made ( t= 0) •S0 = Proceeds from short sale •St = Cost to repurchase stock at time t •dt = Dividend at time t •i = Periodic interest rate of margin account •j = Periodic yield rate of short sale If a short seller repurchases stock after n periods, then the equation of value at time t= n is: M(1 + j)n = M(1 + i)n + S0 −Sn − n∑ t=1 dt(1 + i)n−t. Example 50.2 ‡ Bill and Jane each sell a diﬀerent stock short for a price of 1000. For both investors, the margin requirement is 50%, and interest on the margin is credited at an annual eﬀective rate of 6%. Bill buys back his stock one year later at a price of P. At the end of the year, the stock paid a dividend of X. Jane also buys back her stock after one year, at a price of ( P −25). At the end of the year her stock paid a dividend of 2 X. Both investors earned an annual eﬀective yield of 21% on their short sales. Calculate P and X. Solution. The information for Bill gives 500(1.21) = 500(1.06) + 1000−P−X and for Jane gives 500(1.21) = 500(1.06) + 1000 −P + 25 −2X. Solving this system of two linear equations in two unknowns we ﬁnd P = 900 and X = 25 Example 50.3 Suppose Eddie sells 10 shares of a company short; the stock price is $87 per share, and margin requirement is 40%. The company pays semiannual dividends of 2 per share. Eddie’s margin account earns 3% nominal annual interest, convertible semiannually, and he buys the stock back after 18 months at a share price of 80. What is his annual eﬀective rate of return? Solution. We have M = 10(87)(0.4) = 348. Thus, 348(1 + j)3 = 348(1.015)3 + 870 −800 −20s3 0.015 Solving for j we ﬁnd j = 2.34%.Hence, the annual eﬀective rate of return is (1.0234)2 −1 = 4.73% 440 STOCKS AND MONEY MARKET INSTRUMENTS Example 50.4 On 01/01/2005, Bryan sells a stock short for a price of 800. The margin requirement is 50%, and interest on the margin is credited at an annual eﬀective interest rate of 2%. The stock pays a dividend, D, on 12/31/2005. On 01/01/2006, Bryan buys the stock back at a price of 820 to cover the short. Bryan’s yield on the short sell is −10%. Calculate D. Solution. We are given the following information: M = 400,S0 = 800,S1 = 820,i = 0.02,j = −0.10. The equation of value is 400(0.90) = 400(1.02) + 800−820 −D. Solving this equation for D we ﬁnd D= 28 Short selling by itself is a speculative endeavor and should not be entered into lightly. The short seller is counting on a signiﬁcant decline in the value of the security in order to come out ahead and such an act is istself a risky one. However, investment strategies have been developed that involve combinations of long and short positions in related securities, which have greatly reduced risk and also have a good prospect to earn a reasonable return on investment. A generic name for these transactions is hedging. Situations of this type in which a proﬁt is certain are called arbitrage. These two concepts will appear later in the book when discussing derivatives markets. Short selling has been blamed (erroneously) for many historic market crashes, and has frequently been harshly regulated or banned. 50 SHORT SALES 441 Practice Problems Problem 50.1 John sells a stock short for 200. A year later he purchases the stock for 160. The margin requirements are 50% with the margin account earning 4%. The stock that John sold paid a dividend of 6 during the year. Calculate John’s return on the stock investment. Problem 50.2 Don and Rob each sell a diﬀerent stock short. Don sells his stock short for a price of 960, and Rob sells his short for 900. Both investors buy back their securities for X at the end of one year. In addition, the required margin is 50% for both investors, and both receive 10% annual eﬀective interest on their margin accounts. No dividends are paid on either stock. Don’s yield rate on his short sale is 50% greater than Rob’s yield rate on his short sale. Calculate Don’s yield rate. Problem 50.3 Brian sells a stock short for 800 and buys it back one year later at a price of 805. The required margin on the sale is 62.5% and interest is credited on the margin deposit at an annual eﬀective rate of i. Dividends of 15 are paid during the year. Brian’s yield rate over the one year period is j.If the interest credited on the margin deposit had been 1 .25i,with everything else remaining the same, Brian’s yield rate over the one-year period would have been 1 .5j. Calculate i. Problem 50.4 An investor sells short 500 shares of stock at 10 per share and covers the short position one year later when the price of the stock has declined to 7.50. The margin requirement is 50%. Interest on the margin deposit is 8% eﬀective. Four quarterly dividends of 0.15 per share are paid. Calculate the yield rate. Problem 50.5 Andrew sells a stock short for 800. At the end of one year, Andrew purchases the stock for 760. During the year, Andrew’s stock paid a dividend of 50. Amanda sold a diﬀerent stock short 1000. At the end of the year, Amanda purchases the stock she sold short for X. Amanda’s stock paid a dividend of 25 during the year. The margin requirement for both Andrew and Amanda is 60% and they both earn 10% interest on the margin account. Andrew and Amanda both received the same return. Calculate X. Problem 50.6 Thomas sells a stock short for 1000. The margin rate is 60%. The stock does not pay dividends. After one year, Thomas purchases the stock for 880 which results in a yield of 25%. Calculate the interest rate earned on the margin account. 442 STOCKS AND MONEY MARKET INSTRUMENTS Problem 50.7 Jenna sells short a stock for 1000. One year later she buys the stock for 1000. The stock pays a dividend of 10. Jenna earns 10% on her margin requirement of 60%. Jordan sells short a stock for 500. Jordan’s stock does not pay a dividend. Jordan earns 5% on his margin requirement of 60%. When Jordan buys the stock one year later, he earns a yield that is twice that earned by Jenna. Determine the price that Jordan paid to purchase his stock. Problem 50.8 James sells a stock short for 1000. The margin requirement is 60% and James earns 10% on the margin account. One year later he buys the stock for 790. Jack sells a stock short for 600. The margin requirement is 50% and Jack earns 10% on the margin account. One year later she sells the stock for 450. Both stocks pay the same dividend. James and Jack both earn the same yield. Determine the dividend paid on each stock. Problem 50.9 Rachel sells a stock short for 2000. The margin requirement is 50%. During the year, Rachel’s stock pays a dividend of 25. Rachel purchases the stock for 2050. Danielle sells a stock short for 1500. Her margin requirement is 60%. During the year, Danielle’s stock pays a dividend of 10. Danielle purchases the stock one year later for 1535. Danielle earns a return on her transaction that is twice the return that Rachel earns on her trans- action. Both Rachel and Danielle are paid the same interest rate on the margin account. Calculate the interest rate earned on the margin account. Problem 50.10 Adam has a yield on a short sale of the stock of 30%. The interest rate on the margin account is 10%. If the margin requirement had been doubled with no other changes to the transaction, determine the yield on Adam’s transaction. Problem 50.11 On the short sale of a stock, James realizes a yield of 22%. The margin requirement was 50% and he earned 10% on the margin. If the margin had been increased to 60%, but the interest earned on the margin is still 10%, calculate James’ yield. Problem 50.12 ‡ Theo sells a stock short with a current price of 25,000 and buys it back for X at the end of 1 year. Governmental regulations require the short seller to deposit margin of 40% at the time of the short 50 SHORT SALES 443 sale. No dividends incurred. The prevailing interest rate is an 8% annual rate, and Theo earns a 25% yield on the transaction. Calculate X. Problem 50.13 On 01/01/2005, Company XYZ ′s stock is trading at $80 per share, and Betty sells short $1000 worth of the stock. (Note: Shares of stock are bought and sold in fractional pieces; you do not have to buy or sell a whole number of shares.) Betty is required to put up a margin of $500, and it is agreed that she will receive interest on the margin at an annual eﬀective interest rate of k%. Company XYZ ′s stock pays a dividend of 1.45 per share on 12/31/2005. On 01/01/2006, Company XYZ ′sstock is trading at $76 per share, and Betty covers the short at this time for a yield of 9.75%. Calculate k. Problem 50.14 ‡ Eric and Jason each sell a diﬀerent stock short at the beginning of the year for a price of 800. The margin requirement for each investor is 50% and each will earn an annual eﬀective interest rate of 8% on his margin account. Each stock pays a dividend of 16 at the end of the year. Immediately thereafter, Eric buys back his stock at a price of (800 −2X) and Jason buys back his stock at a price of (800 +X). Eric’s annual eﬀective yield, j,on the short sale is twice Jasons annual eﬀective yield. Calculate j. Problem 50.15 ‡ Jose and Chris each sell a diﬀerent stock short for the same price. For each investor, the margin requirement is 50% and interest on the margin debt is paid at an annual eﬀective rate of 6%. Each investor buys back his stock one year later at a price of 760. Jose’s stock paid a dividend of 32 at the end of the year while Chris’s stock paid no dividends. During the 1-year period, Chris’s return on the short sale is i, which is twice the return earned by Jose. Calculate i. Problem 50.16 ‡ On January 1, 2004, Karen sold stock A short for 50 with a margin requirement of 80%. On December 31, 2004, the stock paid a dividend of 2, and an interest amount of 4 was credited to the margin account. On January 1, 2005, Karen covered the short sale at a price of X, earning a 20% return. Calculate X. Problem 50.17 Kathy and Megan each sell short the same stock. Kathy is required to maintain a margin re- quirement of 50% while Megan is required to maintain a margin requirement of 75%. Assume no dividends paid on the stock and the purchase price is the same. Which of the following are true: (I) If no interest is paid on either margin account, Kathy’s yield will be 150% of Megan’s yield. 444 STOCKS AND MONEY MARKET INSTRUMENTS (II) If the same interest rate is paid on both margin accounts, Megan’s yield will exceed 2/3 of Kathy’s yield. (III) If the same interest rate is paid on both margin accounts, Megan’s yield could never exceed Kathy’s yield. 51 MONEY MARKET INSTRUMENTS 445 51 Money Market Instruments In this section we brieﬂy introduce some of the ﬁnancial instruments used in today’s ﬁnancial mar- kets. These instruments can be analyzed with the principles developed in the earlier chapters of this book. The ﬁnancial intruments introduced fall into the two categories: Money market instru- ments and ﬁnancial derivatives. In this section we will cover money markets instruments. Basics of ﬁnancial derivatives will be introduced and discussed at a later sections. Corporations and government organizations are continually borrowing and lending money. One way for raising money is through money market instruments. Money market instruments are cash-equivalent investments that consist of short-term (usually under a year), very low-risk debt instruments. The money market is the market for buying and selling short-term loans and se- curities. The buyer of the money market instrument is the lender of money and the seller is the borrower of money. Below, we introduce few types of money market instruments that are traded in money markets. Money Market Mutual Funds A money market mutual fund is a mutual fund that attempts to keep its share price at $1. Professional money managers will take your cash and invest it in government treasury bills, savings bonds, certiﬁcates of deposit, and other safe and conservative short term commercial paper. They then turn around and pay you, the owner of the shares, your portion of the interest earned on those investments. Money market funds generally allow withdrawals on demand without penalty, making them an ideal place to keep money while considering other investment opportunities. Investment in a money market fund is neither insured nor guaranteed by neither the FDIC or by any other government agency, and there can be no guarantee that the fund will maintain a stable $1 share price. It is possible to lose money by investing in the fund. An example of money market fund investment is a commercial paper security issued by banks and large corporations. An example of a commercial paper is a promissory note. A promissory note is a written promise to repay a loan or debt under speciﬁc terms −usually at a stated time, through a speciﬁed series of payments, or upon demand. Example 51.1 On January 31, Smith borrows $500 from Brown and gives Brown a promissory note. The note states that the loan will be repaid on April 30 of the same year, with interest at 12% per annum. On March 1 Brown sells the promissory note to Jones, who pays Brown a sum of money in return for the right to collect the payment from Smith on April 30. Jones pays Brown an amount such that Jones’ yield (interest rate earned) from March 1 to the maturity date can be stated as an annual rate of interest of 15%. 446 STOCKS AND MONEY MARKET INSTRUMENTS Determine the amount that Jones paid to Brown and the yield rate Brown earned quoted on an annual basis. Assume all calculations are based on simple interest and a 365-day non-leap year. Solution. On April 30, the value of the note is AV = 500 [ 1 + (89 365 ) (0.12) ] = $514.63. The amount Jones paid to Brown on March 1 is P [ 1 + (60 365 ) (0.15) ] = $514.63 Solving for P we ﬁnd P = $502.24. The yield rate earned by Brown satisﬁes the equation 502.24 = 500 [ 1 + (29 365 ) i ] Solving for i we ﬁnd i= 5.64% Certiﬁcates of Deposit A certiﬁcate of deposit (“CD”) is like a savings account in the sense that it is an FDIC (Federal Deposit Insurance Corporation) insured investment. However, a CD has a ﬁxed term (often three months, six months, or one to ﬁve years). It is intended that the CD be held until maturity, at which time the money may be withdrawn together with the accrued interest. CDs are available at banks and savings and loan institutions. CDs are issued in a range of denom- inations with the higher denominations (e.g. $100,000) carrying the higher rates of interest than the lower denominations. CDs carry a ﬁxed yield rate and in this sense is considered more stable than a money market fund that carries a variable interest rate. However, there is the disadvantage of liquidity since penalties are imposed for early withdrawal. Example 51.2 A two-year certiﬁcate of deposit pays an annual eﬀective rate of 9%. The purchaser is oﬀered two options for prepayment penalties in the event of early withdrawal: A −a reduction in the rate of interest to 7% B −loss of three months interest. 51 MONEY MARKET INSTRUMENTS 447 In order to assist the purchaser in deciding which option to select, compute the ratio of the proceeds under Option A to those under Option B if the certiﬁcate of deposit is surrendered: (a) At the end of six months. (b) At the end of 18 months. Solution. (a) Option’s A proceeds: (1 .07)0.5 = 1.03441.Option’s B proceeds: (1 .09).25 = 1.02178.The ratio is 1.03441 1.02178 = 1.0124. So Option A is preferable. (b) Option’s A proceeds: (1 .07)1.5 = 1.10682. Option’s B proceeds: (1 .09)1.25 = 1.11374. The ratio is 1.10682 1.11374 = 0.9938. So Option B is preferable Guaranteed Investment Contracts (GIC) GICs are investment instruments issued by insurance companies, usually to large investors such as pension funds. They guarantee principal and interest over a stated period of time (like CDs), so their market value does not ﬂuctuate with movements in the interest rate. GICs are diﬀerent from CDs in that they allow additional deposits during the ﬁrst 3 months to a year. Like CDs, withdrawals are restricted. GICs pay higher rates of interest than CDs, but they are not insured by the FDIC against default. Interest earned on the initial investment may be reinvested at the guaranteed rate, or it may be reinvested at some lower ﬁxed rate. Example 51.3 An investor deposits $1,000,000 at the beginning of each year for 5 years in a 5-year GIC with an insurance company that guarantees an annual rate 6% on the fund. Determine the lump sum payable by the investor at 5 years (a) if the interest that is generated each year is reinvested at 6%; (b) if the interest that is generated each year is reinvested at 4%; Solution. (a) If the guaranteed interest is reinvested at 6% then the lump sum payable by the investor at 5 years is 448 STOCKS AND MONEY MARKET INSTRUMENTS 1,000,000¨s5 0.06 = $5,975,318.54. (b) If the guaranteed interest is reinvested at 4% then the lump sum payable by the investor at 5 years is 5,000,000 + 60,000(1.04)4 + 2 ×60,000(1.04)3 + 3 ×60,000(1.04)2 +4 ×60,000(1.04) + 5×60,000 = $5,949,463.19 Mutual Funds A mutual fund is a professionally-managed form of collective investments that pools money from many investors and invests it in stocks, bonds, short-term money market instruments, and/or other securities. In a mutual fund, the fund manager trades the fund’s underlying securities, realizing capital gains or losses, and collects the dividend or interest income. The investment proceeds are then passed along to the individual investors. The value of a share of the mutual fund, known as the net asset value per share (NAV), is calculated daily based on the total value of the fund divided by the number of shares currently issued and outstanding. Example 51.4 A mutual fund account has the balance $100 on January 1. On July 1 (exactly in the middle of the year) the investor deposits $250 into the account. The balance of the mutual fund at the end of the year is $400. (a) Calculate the exact dollar-weighted annual rate of interest for the year. (b) Suppose that the balance of the fund immediately before the deposit on July 1 is $120. Calculate the time-weighted annual rate of interest for the year. Solution. (a) 100(1 + i) + 250(1 +i) 1 2 = 400 →100(1 +i) + 250(1 +i) 1 2 −400 = 0.By the quadratic formula, (1 + i) 1 2 = 1.108495283 →(1 + i) = 1.228761792 →i= 0.228761792. (b) (1 + j) = (120/100)(400/370) = 1.297297297 so the time-weighted rate is j = 0.297297297 Mortgage Backed Securities (MBS) Mortgage-backed securities(MBS) represent an investment in mortgage loans. An MBS investor owns an interest in a pool of mortgages, which serves as the underlying assets and source of cash ﬂow for the security. Most MBSs are issued by the following three ﬁnancial companies: •Government National Mortgage Association (Ginnie Mae), a U.S. government agency. Ginnie Mae, backed by the full faith and credit of the U.S. government, guarantees that investors receive timely payments. 51 MONEY MARKET INSTRUMENTS 449 •The Federal National Mortgage Association (Fannie Mae) and the Federal Home Loan Mortgage Corporation (Freddie Mac), U.S. government-sponsored enterprises. Fannie Mae and Freddie Mac also provide certain guarantees and, while not backed by the full faith and credit of the U.S. gov- ernment, have special authority to borrow from the U.S. Treasury. Mortgage-backed securities typically carry some of the highest yields of any government or agency security. Example 51.5 A homeowner signs a 30 year mortgage that requires payments of $971.27 at the end of each month. The interest rate on the mortgage is 6% compounded monthly. If the purchase price of the house is $180,000 then what percentage down payment was required? Solution. The purchase price of the home is 971.27a360 0.005 = 162,000. The amount of down payment is 180,000 −162,000 = 18,000. So the down payment is 10% of the purchasing price 450 STOCKS AND MONEY MARKET INSTRUMENTS Practice Problems Problem 51.1 Consider again Example 51.1. Suppose instead that Jones pays Brown an amount such that Jones’ yield is 12%. Determine the amount that Jones paid. Problem 51.2 You prurchase $10,000 CD with an interest rate of 5% compounded annually and a term of one year. At year’s end, the CD will have grown to what value? Problem 51.3 ‡ A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible quarterly) 1 4% 3 5% 5 5.65% The bank does not permit early withdrawal. The certiﬁcates mature at the end of the term. During the next six years the bank will continue to oﬀer these certiﬁcates of deposit with the same terms and interest rates. An investor initially deposits $10,000 in the bank and withdraws both principal and interest at the end of six years. Calculate the maximum annual eﬀective rate of interest the investor can earn over the 6-year period. Problem 51.4 An investor deposits $100,000 in a 3-year GIC with an insurance company that guarantees an annual 4% interest rate on the fund. If the GIC speciﬁes that interest is reinvested at 4% within the GIC, determine the lump sum payable to the investor after the three-year period. Problem 51.5 An investor puts $100 into a mutual fund in the ﬁrst year and $50 in the second year. At the end of the ﬁrst year the mutual pays a dividend of $10. The investor sells the holdings in the mutual fund at the end of the second year for $180. Find the dollar weighted return. Problem 51.6 You have a mutual fund. Its balance was $10,000 on 31/12/1998. You made monthly contributions of $100 at the start of each month and the ﬁnal balance was $12,000 at 31/12/1999. What was you approximate return during the year? 51 MONEY MARKET INSTRUMENTS 451 Problem 51.7 An investor purchases 1000 worth of units in a mutual fund whose units are valued at 4.00. The investment dealer takes a 9% “front-end load” from the gross payment. One year later the units have a value of 5.00 and the fund managers claim that the “fund’s unit value has experienced a 25% growth in the past year.” When units of the fund are sold by an investor, there is a redemption fee of 1.5% of the value of the units redeemed. (a) If the investor sells all his units after one year, what is the eﬀective annual rate of interest of his investment? (b) Suppose instead that after one year the units are valued at 3.75. What is the return in this case? Problem 51.8 Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. What is the outstanding balance at the end of 10 years immediately after the 120th payment? Problem 51.9 The interest on a 30 year mortgage is 12% compounded monthly. The mortgage is being repaid by monthly payments of 700. Calculate the outstanding balance at the end of 10 years. Problem 51.10 The interest rate on a 30 year mortgage is 12% compounded monthly. Lauren is repaying the mortgage by paying monthly payments of 700. Additionally, to pay oﬀ the loan early, Lauren has made additional payments of 1000 at the end of each year. Calculate the outstanding balance at the end of 10 years. Problem 51.11 A mortgage loan is being repaid with level annual payments of 5000 at the end of the year for 20 years. The interest rate on the mortgage is 10% per year. The borrower pays 10 payments and then is unable to make payments for two years. Calculate the outstanding balance at the end of the 12th year. Problem 51.12 A 20,000 mortgage is being repaid with 20 annual installments at the end of each year. The borrower makes 5 payments, and then is temporarily unable to make payments for the next 2 years. Find an expression for the revised payment to start at the end of the 8th year if the loan is still to be repaid at the end of the original 20 years. 452 STOCKS AND MONEY MARKET INSTRUMENTS Measures of Interest Rate Sensitivity In this chapter we study the sensitivity of the interest rate. We will cover the following topics: (1) The eﬀect of inﬂation on interest rate. We will consider the eﬀect of inﬂation in the calculations of present values and accumulated values. (2) The eﬀect of the investment period on interest rates. That is, interest rate is sensitive to time. This fact is known as the term structure of interest rate. A plot of the interest rate of an investment against time to maturity will be discussed. We call such a plot a yield curve. (3) We want to measure the change in a bond’s price as interest rate change. We call this measure the duration; the two types of duration we will look at are the modiﬁed duration and the Macaulay duration. (4) Immunization is a technique that helps investors shielding their overall ﬁnancial position from exposure to interest rate ﬂuctuations. (5) Absolute matching: The idea here is to structure assets such that the resulting asset inﬂows will exactly match the liability outﬂow. 453 454 MEASURES OF INTEREST RATE SENSITIVITY 52 The Eﬀect of Inﬂation on Interest Rates What is inﬂation? Inﬂation is deﬁned as a sustained increase in the general level of prices for goods and services. It is measured as an annual percentage increase, known as rate of inﬂation . As inﬂation rises, every dollar you own buys a smaller percentage of a good or service. That is, inﬂation represents a loss of purchasing power. The opposite to inﬂation is deﬂation. Experience has shown that inﬂation has a signiﬁcant eﬀect on the rate of interest. As a matter of fact they are positively correlated in the sense that over time both tend to move in the same direction. The following example illustrates how inﬂation aﬀects the purchasing power of money. Suppose that today a gallon of milk costs $4. Then with $100 one can buy 25 gallons of milk. Suppose the $100 is invested for 2 years at an 8% annual eﬀective rate. Then at the end of the two years, you will have 100(1.08)2 = $116.64.Without the presence of inﬂation, that is assuming the cost of milk is still $4, you can now buy 29 gallons of milk. Now suppose that there has been a constant rate of inﬂation during the two years of 5% a year so that milk will cost 4(1 .05)2 = $4.41 per gallon. In this case, one can only buy 116.64 4.41 = 26.45 gallons of milk, a reduction of 3 gallons. The real rate of return, i′, is measured by solving the equation 25(1 + i′)2 = 116.64 4.41 = 25(1.08)2 (1.05)2 by taking the square root of both sides we obtain (1 + i′)(1.05) = 1.08 or 1 + ireal rate = 1 + inominal rate 1 + iinflation rate . To study the eﬀect of inﬂation on rates we introduce some notations. The rate of interest after eliminating the eﬀect of inﬂation is called the real rate of interest and will be denoted by i′, while the actual rate of interest charged in the market (i.e. the rate not adjusted for inﬂation) is called the nominal interest rate (not to be confused with the diﬀerent meaning of nominal used in Section 9) and will be denoted by i. Let r denote the rate of inﬂation. Assuming constant rate of inﬂation, the relationship between real and nominal interest rates can be described in the equation: 1 + i= (1 + i′)(1 + r), i>i ′>0,r >0. From this we can write 52 THE EFFECT OF INFLATION ON INTEREST RATES 455 1 + i′= 1 + i 1 + r (52.1) or i′= i−r 1 + r. A common approximation for the real interest rate is given by the Fisher equation: i′= i−r which states that the real interest rate is the nominal interest rate minus the expected rate of inﬂation. If inﬂation is positive, which it generally is, then the real interest rate is lower than the nominal interest rate. That’s the reason we are imposing the condition i>i ′ whenever r >0. Formula (52.1) is quite useful in performing calculations involving rates of inﬂation. For example, assume that we wish to ﬁnd the present value of a series of payments at the end of each period for nperiods in which the base payment amount at time 0 is R, but each payment is indexed to reﬂect inﬂation. If r is the periodic rate of inﬂation and i is the periodic nominal interest rate, then PV = R [ 1 + r 1 + i + (1 + r 1 + i )2 + ··· + (1 + r 1 + i )n] = R(1 + r) 1 − (1+r 1+i )n i−r . However, if we use Formula (52.1) the above result becomes PV = R [ 1 1 + i′ + 1 (1 + i′)2 + ··· + 1 (1 + i′)n ] = Rani′. The above two formulas provide a guidance in computing present values of future payments as follows: 1. If future payments are not aﬀected by inﬂation, then discount at the nominal rate of interest. 2. If future payments are adjusted to reﬂect the rate of inﬂation and the adjustment is reﬂected in the payment amount, then also discount at the nominal rate of interest. 3. If future payments are adjusted to reﬂect the rate of inﬂation but the adjustment is not reﬂected in the payment amount, the correct procedure is to discount at the real rate of interest. We next consider inﬂation in connection with accumulated values. Consider the common situation in which an investor invests A dollars for n periods at interest rate i. The value of this investment in “nominal dollars” at the end of n periods is AV = P(1 + i)n. 456 MEASURES OF INTEREST RATE SENSITIVITY However, if the rate of inﬂation is accounted for, then the purchasing power of this investment is AV = P (1 + i 1 + r )n = P(1 + i′)n. Thus, the value of this investment in “real dollars” is lower since i>i ′. Example 52.1 Money is invested for 3 years at an interest rate of 4% eﬀective. If inﬂation is 5% per year over this period, what percentage of purchasing power is lost? Solution. One dollar will accumulate to (1 .04)3 = 1.1248 over the period, but the purchasing power will only be (1.04)3 (1.05)3 = 0.9716 due to inﬂation. The loss of purchasing power per dollar of original investment is 1 −0.9716 = 2.84% Example 52.2 The real interest rate is 6% and the inﬂation rate is 4%. Allan receives a payment of 1,000 at time 1, and subsequent payments increase by 50 for 5 more years. Determine the accumulated value of these payments in nominal dollars at time 6 years. Solution. The nominal interest rate is: i= (1.06)(1.04) −1 = 10.24% The accumulated value of the cash ﬂows is: AV =1,000(1.1024)5 + 1,050(1.1024)4 + 1,100(1.1024)3 + 1,150(1.1024)2 + 1,200(1.1024) + 1,250 = 8623.08 52 THE EFFECT OF INFLATION ON INTEREST RATES 457 Practice Problems Problem 52.1 An insurance company is making annual payments under the settlement provisions of a personal injury lawsuit. A payment of $24,000 has just been made and ten more payments are due. Future payments are indexed to the Consumer Price Index which is assumed to increase at 5% per year. Find the present value of the remaining obligation if the assumed interest rate is 8%. Use the exact value of i′. Problem 52.2 The nominal rate of interest is 10% and the rate of inﬂation is 5%. A single deposit is invested for 20 years. Let Adenote the value of the investment at the end of 10 years, with the rate of inﬂation accounted for in the calculation. Let B denote the value of the investment at the end of 10 years computed at the real rate of interest. Find the ratio of A B. Problem 52.3 The nominal rate of interest is 8% and the rate of inﬂation is 5%. Calculate the real rate of interest. Problem 52.4 Mindy invests 10,000 for 10 years at a nominal rate of interest of 8%. The rate of inﬂation is 5% over the 10 year period. Calculate the value at the end of 10 years of Mindy’s investment in today’s dollars. Problem 52.5 As a result of sky-high gas prices, the cost of delivery pizza is expected to experience a constant 8% annual rate of inﬂation. In today’s dollars, Obese Oliver spends $2,000 per year on delivery pizza. His delivery pizza-eating habits will not change over the next ten years, so he will pay whatever price increases are necessary to sustain this level of consumption. If the nominal interest rate (note that “nominal” refers to both interest convertibility and its relationship to inﬂation), convertible semiannually, is 10%, ﬁnd the present value of Oliver’s delivery pizza purchases for the next ten years. Problem 52.6 Money is invested for 5 years in a savings account earnings 7% eﬀective. If the rate of inﬂation is 10%, ﬁnd the percentage of purchasing power lost during the period of investment. Problem 52.7 ‡ An insurance company has an obligation to pay the medical costs for a claimant. Average annual claims costs today are $5,000, and medical inﬂation is expected to be 7% per year. The claimant 458 MEASURES OF INTEREST RATE SENSITIVITY is expected to live an additional 20 years. Claim payments are made at yearly intervals, with the ﬁrst claim payment to be made one year from today. Find the present value of the obligation if the annual interest rate is 5%. Problem 52.8 ‡ Seth deposits X in an account today in order to fund his retirement. He would like to receive payments of 50 per year, in real terms, at the end of each year for a total of 12 years, with the ﬁrst payment occurring seven years from now. The inﬂation rate will be 0.0% for the next six years and 1.2% per annum thereafter. The annual eﬀective rate of return is 6.3%. Calculate X. 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 459 53 The Term Structure of Interest Rates and Yield Curves In the previous section we discussed a factor aﬀecting interest rates, i.e. inﬂation. In this section we discuss another factor which is the length of investment period. The term structure of interest rates refers to the way interest rates depend on time. Usually long-term market interest rates are higher than short-term market interest rates. An example of term structure of interest rates is given in the table below. Length of investment Interest rate (or Spot rate) 1 year 3% 2 year 4% 3 year 6% 4 year 7% Table 53.1 We can extend the values in this table to a continuous graph, and the resulting graphical illustration is called the yield curve corresponding to the table. The interest rates on the yield curve (or listed in the table) are often called spot rates. Economists and investors believe that the shape of the yield curve reﬂects the market’s future expectation for interest rates and the conditions for monetary policy. For example, an upward- sloping yield curve implies market’s expectation of future increases in interest rates. In some cases, short-term rates exceed long-term rates, that is the yield curve has negative slope, in this case the curve is called inverted. When all maturities have the same yield the curve is called ﬂat. Now, recall from Section 30 the formula for the net present value of a series of future payments given by NPV (i) = n∑ t=0 (1 + i)−tct. This formula is based on a single rate of interest i.With the presence of a term structure of interest rates, the net present value can be found by discounting each payment by its associated spot rate. That is, if we denote the spot rate for period t as it then the net present value is given by the equation NPV (i∗) = n∑ t=0 (1 + it)−tct. This formula can be considered as an application of computing annuity values with varying rates of interest as discussed in Section 21. Example 53.1 You are given the following term structure of interest rates: 460 MEASURES OF INTEREST RATE SENSITIVITY Length of investment Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% Find the present value of payments of $1000 at the end of each year for ﬁve years using the spot rates given above. What level yield rate would produce an equivalent value? Solution. The present value of the payments is 1000[(1.07)−1 + (1.08)−2 + (1.0875)−3 + (1.0925)−4 + (1.095)−5] = $3906.63. The equivalent level yield rate is found by solving the equation a5 i = 3 .90663 which gives i = 0.0883 = 8.83% using a ﬁnancial calculator A special type of spot rates is the forward interest rate . In order to illustrate this concept, consider a ﬁrm which needs to borrow a sizeable amount of money for two years. The ﬁrm is presented with two options. The ﬁrst option is to borrow money for 2 years at the 2-year spot rate of 10%. The second option is to borrow for one year at the one-year spot rate now of 8%, and to borrow for the second year at the one-year spot rate in eﬀect a year later. This 1-year spot rate for the next year is an example of a forward rate or more speciﬁcally a one year forward interest rate. This is what the one year spot rate will be after one year is passed. Now, if f is the forward rate then the ﬁrm will be indiﬀerent between the two options if (1.10)2 = (1.08)(1 + f) which gives f = 0.12 = 12%. So, if the ﬁrm can predict that the forward rate will be greater than 12%, it should use the ﬁrst option to borrow, but if they expect the forward rate to be less than 12%, it should use the second option to borrow. Example 53.2 Consider the term structure of interest rates given in Table 53.1. Let fn+t n represents the t−year forward rate n years from now. (a) Compute the 1-year, 2-year, and 3-year forward rate, one year from now. (b) Compute the 1-year and 2-year forward rates two years from now. (c) Compute the 1-year forward rate three years from now. 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 461 Solution. (a) The 1-year forward rate one year from now satisﬁes the equation (1.03)(1 + f2 1 ) = (1.04)2. Solving this equation we ﬁnd f2 1 = 5.01%. The 2-year forward rate, one year from now satisﬁes the equation (1.03)(1 + f3 1 )2 = (1.06)3. Solving this equation we ﬁnd f3 1 = 7.53%. The 3-year forward rate, one year from now satisﬁes the equation (1.03)(1 + f4 1 )3 = (1.07)4. Solving this equation we ﬁnd f4 1 = 8.37%. (b) The 1-year forward rate, two years from now satisﬁes the equation (1.04)2(1 + f3 2 ) = (1.06)3. Solving this equation we ﬁnd f3 2 = 10.1%. The 2-year forward rate, two years from now satisﬁes the equation (1.04)2(1 + f4 2 )2 = (1.07)4. Solving this equation we ﬁnd f4 2 = 10.00%. (c) The 1-year forward rate three years from now satisﬁes the equation (1.06)3(1 + f4 3 ) = (1.07)4. Solving this equation we ﬁnd f4 3 = 10.06% From any yield curve, one can calculate a complete set of implied forward rates. In general, the k− year forward rate n years from now satisﬁes the equation (1 + in)n(1 + fn+k n )k = (1 + in+k)n+k where it is the t−year spot rate. Note that fn+k n is the forward interest rate between the periods n and n+ k. Example 53.3 Find the present value of the remaining payments in the annuity immediate given in Example 53.1 immediately after two payments have been made assuming that f2+t 2 = 1% + it where t= 1,2,3. 462 MEASURES OF INTEREST RATE SENSITIVITY Solution. The comparison date for this calculation is at the end of two years. At that time there are three remaining annuity payments to be made. The expected forward rates are the current spot rates for 1,2, and 3 years each increased by 1%. Thus, the present value of the remaining three payments is 1000[(1.08)−1 + (1.09)−2 + (1.0975)−3] = $2524.07 Example 53.4 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% (a) Find the price of a $1000 two-year bond with annual 5% coupon using the spot rates given above. (b) Compute the yield to maturity as deﬁned in Section 43. Solution. (a) The coupon is $50 each. The price of the bond is P = 50(1.07)−1 + 1050(1.08)−2 = 946.93 (b) The yield to maturity i satisﬁes the equation 50(1 + i)−1 + 1050(1 +i)−2 = 946.93. Solving this equation we ﬁnd i= 7.975% Example 53.5 Yield rates for 6% annual coupon bonds are given to be 3.5% for 1-year bonds and 4% for 2-year bonds. Find the implied 2-year spot rate from these yield rates. Solution. Per 100 of par value we have 6a2 0.04 + 100v2 0.04 = 6 1.035 + 106 (1 + i)2 . Solving this equation for i we ﬁnd i= 4.01% 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 463 Practice Problems Problem 53.1 You are given the following term structure of interest rates: Length of investment Interest Rate 1 5.00 2 6.00 3 6.75 4 7.25 5 7.50 Table 53.2 Which of the following are true: (I) The yield curve has a positive slope. (II) The yield curve is inverted. (III) The interest rates in the table are called forward rates. Problem 53.2 Using Table 53.2, calculate the present value of a ﬁve year annuity due of 100 per year. Problem 53.3 Using Table 53.2, determine the three year deferred two year forward rate. Problem 53.4 Using Table 53.2, determine the accumulated value of a three year annuity immediate of 100 at the end of each year. Problem 53.5 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 464 MEASURES OF INTEREST RATE SENSITIVITY (a) Calculate the present value of an annuity immediate of 10 at the end of each year for 5 years using the spot rates. (b) Calculate the equivalent level rate. Problem 53.6 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 (a) Calculate the present value of an annuity due of 10 at the beginning of each year for 5 years using the spot rates. (b) Calculate the equivalent level rate. Problem 53.7 You are given the following yield curve Length of investment Interest Rate 1 0.040 2 0.045 3 0.048 4 0.050 5 0.051 (a) Calculate the 2 year deferred 3 year forward rate less the three year spot rate. (b) Calculate what the 3 year spot rate is expected to be in 2 years. Problem 53.8 You are given the following term structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7% 2 8% 3 9% What is the one-year forward rate beginning two years from now? 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 465 Problem 53.9 ‡ Using the table of the previous problem, ﬁnd (a) the current price of a $1000 3-year bond with coupon rate 6% payable annually. (b) the annual eﬀective yield rate for the bond if the bond is sold at a price equal to its value. Problem 53.10 ‡ You are given the following term structure of spot interest rates: Term (in Years) Spot Interest Rate 1 5.00% 2 5.75% 3 6.25% 4 6.50% A three-year annuity-immediate will be issued a year from now with annual payments of 5000. Using the forward rates, calculate the present value of this annuity a year from now. Problem 53.11 ‡ Consider a yield curve deﬁned by the following equation: ik = 0.09 + 0.002k−0.001k2 where ik is the annual eﬀective rate of return for zero coupon bonds with maturity of k years. Let j be the one-year eﬀective rate during year 5 that is implied by this yield curve. Calculate j. Problem 53.12 ‡ Yield rates to maturity for zero coupon bonds are currently quoted at 8.5% for one-year maturity, 9.5% for two-year maturity, and 10.5% for three-year maturity. Let i be the one-year forward rate for year two implied by current yields of these bonds. Calculate i. Problem 53.13 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% 466 MEASURES OF INTEREST RATE SENSITIVITY Find the following expected forward rates: (a) 1-year deferred 2-year forward rate. (b) 2-year deferred 3-year forward rate. Problem 53.14 You are given the following term of structure of spot interest rates: Term (in Years) Spot Interest Rate 1 7.00% 2 8.00% 3 8.75% 4 9.25% 5 9.50% (a) Find the price of a $1000 two-year bond with annual 10% coupon using the spot rates given above. (b) Compute the yield to maturity as deﬁned in Section 41. Problem 53.15 You are given 1−year, 2−year, and 3−year spot rates of 4%, 5%, and 6%, respectively. Calculate the annual yield rate for 3 −year 5% annual coupon bonds implied by these spot rates. Problem 53.16 Consider a yield curve deﬁned by the following equation: ik = 0.09 + 0.002k−0.001k2 where ik is the annual eﬀective rate of return for zero coupon bonds with maturity of k years. (a) Calculate the 2-year spot rate implied by this yield curve. (b) Calculate the 2-year deferred 3-year forward rate implied by this yield curve. Problem 53.17 A two-year bond with annual coupons of 100 and redemption value of 1000 is priced at 1,037.41. The current two-year spot rate is 8%. Determine the current one-year spot rate that is consistent with the pricing of the bond. Problem 53.18 A 5 year bond with 6% annual coupons has a yield rate of 10% eﬀective and a 5 year bond with 8% annual coupons has a yield rate of 9% eﬀective. What is the 5 year spot rate? 53 THE TERM STRUCTURE OF INTEREST RATES AND YIELD CURVES 467 Problem 53.19 ‡ Which of the following statements about zero-coupon bonds are true? (I) Zero-coupon bonds may be created by separating the coupon payments and redemption values from bonds and selling each of them separately. (II) The yield rates on stripped Treasuries at any point in time provide an immediate reading of the risk-free yield curve. (III) The interest rates on the risk-free yield curve are called forward rates. 468 MEASURES OF INTEREST RATE SENSITIVITY 54 Macaulay and Modiﬁed Durations By now, it should be clear to you that the timing of cash ﬂows is a signiﬁcant factor in the analysis of ﬁnancial instruments. In this section we develop some “indices” to measure the timing of future payments. One index known as the duration is useful as a measure of the sensitivity of a bond’s price to interest rate movements. The two types of duration we will look at are the modiﬁed duration and the Macaulay duration. The most basic index of time measuring is the term-to-maturity. For a given price and par value, among zero coupon bonds, one wishes to purchase the bond with the earliest redemption time. In the case of bonds carrying coupons, the redemption time is not enough for helping choose the bond to purchase. A better index than the term-to-maturity is the method of equated time , which was deﬁned in Section 13: Let R1,R2,··· ,Rn be a series of payments made at times 1 ,2,··· ,n. Then the weighted average of the various times of payments, where the weights are the various amounts paid is t= ∑n t=1 tRt∑n t=1 Rt . This is also known as the average term-to-maturity. Example 54.1 Consider two 10-year par value 100 bonds one with 5% coupons and the other with 10% coupons. Find the equated time for each coupon. Solution. The average term to maturity of the 5% bond is t= 1 ·5 + 2·5 + ··· + 10 ·5 + 10·100 5 + 5 +··· + 5 + 100 = 8.50. The average term to maturity of the 10% bond is t= 1 ·10 + 2·10 + ··· + 10 ·10 + 10·100 10 + 10 +··· + 10 + 100 = 7.75. This shows that the 5% bond is a longer term bond than the 10% bond An even better index that is similar to the method of equated time is obtained by replacing each payment in the above formula by its present value thus obtaining d= ∑n t=1 tνtRt∑n t=1 νtRt . 54 MACAULAY AND MODIFIED DURATIONS 469 We call d the Macaulay duration or simply duration. Example 54.2 Suppose payments of 2000, 4000, and 10000 are to be made at times 1, 2, and 4, respectively. Assume, an annual yield of 25%. (a) Find the average term to maturity using the method of equated time. (b) Find the duration of the investment. Solution. (a) The method of equated time produces the value t= 1 ·2000 + 2·4000 + 4·10000 2000 + 4000 + 10000 = 3.125. (b) The duration is given by d= 1 ·(1.25)−1 ·2000 + 2·(1.25)−2 ·4000 + 4·(1.25)−4 ·10000 (1.25)−1(2000) + (1.25)−2(4000) + (1.25)−4(10000) = 2.798 Example 54.3 Consider two bonds purchased at the redemption value of 1000, and due in 5 years. One bond has 5% annual coupon rate payable semi-annually and the other has 10% annual coupon rate payable semi-annually. Find the duration of each bond if both bonds were purchased to yield 7% compounded semi-annually. Solution. The duration of the 5% bond is d=0.5[1 ·(1.035)−1(25) + 2·(1.035)−2(25) + ··· + 10 ·(1.035)−10(25)] + 5000(1.035)−10 (1.035)−1(25) + (1.035)−2(25) + ··· + (1.035)−10(25) + 1000(1.035)−10 =12.5(Ia)10 0.035 + 5000(1.035)−10 25a10 0.035 + 1000(1.035)−10 =4.4576 The duration of the 10% bond is d=0.5[1 ·(1.035)−1(50) + 2·(1.035)−2(50) + ··· + 10 ·(1.035)−10(50)] + 5000(1.035)−10 (1.035)−1(50) + (1.035)−2(50) + ··· + (1.035)−10(50) + 1000(1.035)−10 =4.1158 470 MEASURES OF INTEREST RATE SENSITIVITY Note that dis a function of i.If i= 0 we have d= tso that the method of equated time is a special case of duration which ignores interest. Also, note that in the case of only one future payment, duration is the point in time at which that payment is made. This is clear since the summations in the numerator and the denominator have only one term each and everything cancels except the time of payment. Example 54.4 Find the duration of a 10-year zero coupon bond assuming a yield 8% eﬀective. Solution. Only one payment is involved so that d = 10. Notice that this answer is independent of the yield rate Another important fact is that d is a decreasing function of i. To see this, we have d di(d) = d di (∑n t=1 tνtRt∑n t=1 νtRt ) = −ν(∑n t=1 νtRt) (∑n t=1 t2νtRt) −(∑n t=1 tνtRt)2 (∑n t=1 νtRt)2 = −ν [∑n t=1 t2νtRt∑n t=1 νtRt − (∑n t=1 tνtRt∑n t=1 νtRt )2] Now using Cauchy-Schwartz’s inequality ( n∑ t=1 atbt)2 ≤( n∑ t=1 a2 t)( n∑ t=1 b2 t) with at = ν t 2 R 1 2 t and bt = tν t 2 R 1 2 t we ﬁnd that ∑n t=1 t2νtRt∑n t=1 νtRt − (∑n t=1 tνtRt∑n t=1 νtRt )2 ≥0 and therefore the derivative of d with respect to i is negative. We next describe a measure of how rapidly the present value of a series of future payments changes 54 MACAULAY AND MODIFIED DURATIONS 471 as the rate of interest changes. Let P(i) denote the present value of all future payments. Then by Taylor series we can write P(i+ ϵ) = P(i) + ϵ 1!P′(i) + ϵ2 2!P′′(i) + ··· . Using the ﬁrst two terms we can write P(i+ ϵ) ≈P(i) + ϵP′(i). Hence, the percentage change in P is given by P(i+ ϵ) −P(i) P(i) ≈ϵP′(i) P(i) . Thus, the percentage change of P(i) is a function of P′(i) P(i) . We deﬁne the volatility of the present value by ν = −P′(i) P(i) = −d di[ln P(i)]. The minus sign is necessary to make ν positive since P′(i) is negative ( increasing the yield rate results in a decreasing present value). Now, since P(i) = ∑n t=1(1 +i)−tRt, by taking the derivative we ﬁnd P′(i) = −∑n t=1 t(1 +i)−t−1Rt. Hence, ν = ∑n t=1 t(1 + i)−t−1Rt∑n t=1(1 + i)−tRt = 1 1 + i · ∑n t=1 t(1 + i)−tRt∑n t=1(1 + i)−tRt = d 1 + i = νd. Because of this close relationship, volatility is often called modiﬁed duration. Remark 54.1 None of the results in this section are valid if the payment amounts vary depending on the interest rate. Example 54.5 (a) Find the Macaulay duration of a ten-year, $1,000 face value, 8% annual coupon bond. Assume an eﬀective annual interest rate of 7%. (b) Find the modiﬁed duration for the bond Solution. (a) We have d=80ν+ 2(80ν2) + ··· + 10(80ν10) + 10(1000ν10) 80ν+ (80ν2) + ··· + (80ν10) + (1000ν10) =80(Ia)10 0.07 + 10000ν10 0.07 80a10 0.07 + 1000ν10 0.07 472 MEASURES OF INTEREST RATE SENSITIVITY But ν10 0.07 =(1.07)−10 = 0.508349 a10 0.07 =1 −(1.07)−10 0.07 = 7.02358 (Ia)10 0.07 =1.07a10 0.07 −10(1.07)−10 0.07 = 34.7391 Hence, d= 80(34.7392)+10000(0.508349) 80(7.02358)+1000(0.508349) = 7.3466 years (b) The modiﬁed duration is ν = νd= (1.07)−1(7.3466) = 6.8660 Example 54.6 A 10-year $100,000 mortgage will be repaid with level semi-annual payments of interest and prin- cipal. Assume that the interest rate on the mortgage is 8%, convertible semiannually, and that payments are made at the end of each half-year. Find the Macaulay duration of this mortgage. Solution. Per dollar of mortgage payment, the Macaulay duration is d=0.5ν0.04 + 2ν2 0.04 + 3ν3 0.04 + ··· + 20ν20 0.04 ν0.04 + ν2 0.04 + ν3 0.04 + ··· + ν20 0.04 =0.5(Ia)20 0.04 a20 0.04 = 4.6046 years Note that the answer is independent of the rate of interest being paid on the mortgage. Duration depends on the pattern of the level payments and not their amount Example 54.7 Find the duration of a preferred stock paying level dividends into perpetuity assuming eﬀective 8%. Solution. Per dollar dividend we have d= (Ia)∞ a∞ = 1.08/.082 1/.08 = 13.5 54 MACAULAY AND MODIFIED DURATIONS 473 Example 54.8 Find the modiﬁed duration for a share of common stock. Assume that the stock pays annual dividends, with the ﬁrst dividend of $2 payable 12 months from now, and that subsequent dividends will grow at an annual rate of 4%. Assume that the eﬀective annual interest rate is 9%. Solution. We have P(i) = D1 i−g P′(i) = −D1(i−g)−2 v= −P′(i) P(i) = 1 i−g = 1 0.09 −0.04 = 20 Example 54.9 You are the actuary for an insurance company. Your company’s liabilities include loss reserves, which are liability reserves set aside to make future claim payments on policies which the company has already sold. You believe that these liabilities, totaling $100 million on December 31, 2006, will be paid out according to the following schedule: Proportionl of Calendar year reserves paid out 2007 40% 2008 30% 2009 20% 2010 10% Find the modiﬁed duration of your company’s loss reserves. Assume that the annual interest rate is 10%, and that all losses paid during a given calendar year are paid at the mid-point of that calendar year. Solution. We have ν = ν (0.5(4ν0.5) + 1.5(3ν1.5) + 2.5(2ν2.5) + 3.5(ν3.5) 4ν0.5 + 3ν1.5 + 2ν2.5 + ν3.5 ) = 1.2796 Remark 54.2 The equation P(i+ ϵ) −P(i) P(i) = −ϵν is valid if P(i) stands for the current price of a bond. 474 MEASURES OF INTEREST RATE SENSITIVITY Example 54.10 The current price of a bond is $110 and its yield is 7%. The modiﬁed duration is 5. Estimate the price of the bond if its yield falls to 6%. Solution. The approximate percentage change in the price is −(−0.01)(5) = 5% . That is, the price of the bond increased by approximately 5%. Thus, the new price of the bond is 110 ×1.05 = $115.50 Portfolio Modiﬁed Duration Consider a portfolio consisting of n bonds. Let bond k have current price Pk(i) and modiﬁed duration νk(i). Then the current value of the portfolio is P(i) = P1(i) + P2(i) + ··· + Pn(i). Let ν denote the modiﬁed duration of the portfolio. Then, we have ν = −P′(i) P(i) = −P′ 1(i) P(i) + ( −P′ 2(i) P(i) ) + ··· + ( −P′ n(i) P(i) ) =P1(i) P(i) ( −P′ 1(i) P1(i) ) + P2(i) P(i) ( −P′ 2(i) P2(i) ) + ··· + Pn(i) P(i) ( −P′ n(i) Pn(i) ) =P1(i) P(i) ν1 + P2(i) P(i) ν2 + ··· + Pn(i) P(i) νn Thus, the modiﬁed duration of a portfolio can be calculated as the weighted average of the bonds’ modiﬁed durations, using the market values of the bonds as the weights. Example 54.11 Megan buys the following bonds: (I) Bond A with a modiﬁed duration of 7 for 1000; (II) Bond B with a modiﬁed duration of 5 for 2000; and (III) Bond C with a modiﬁed duration of 10 for 500. Calculate the modiﬁed duration of the portfolio. Solution. The modiﬁed duration of the portfolio is weighted average of the modiﬁed durations weighted by the purchase price ν = 7 ·1000 + 5·2000 + 10·500 1000 + 2000 + 500 = 6.286 54 MACAULAY AND MODIFIED DURATIONS 475 Example 54.12 An investment portfolio consists of two bonds: a two-year zero-coupon bond, and a four-year zero- coupon bond. Both bonds have redemption values of $1,000. Assume an eﬀective annual interest rate of 8%. Find the Macaulay duration of the investment portfolio. Solution. The Macaulay duration of the two-year zero-coupon is 2 with current value 1000 v2. The Macaulay duration of the four-year zero-coupon is 4 with current value 1000v4.The Macaulay duration of the portfolio is d= 2(1000ν2) + 4(1000ν4) 1000ν2 + 1000ν4 = 2.9232 years 476 MEASURES OF INTEREST RATE SENSITIVITY Practice Problems Problem 54.1 Find the modiﬁed duration of a 30-year, $1,000 face value, 6% annual coupon bond. Assume an eﬀective annual interest rate of 9%. Problem 54.2 Calculate the duration of a payment to be made in 5 years. Problem 54.3 A 10 year bond has annual coupons of 10 and matures for 100. Which of the following are true: (I) The term to maturity is 10 years. (II) The average term to maturity using the method of equated time is 7.5 years. (III) The Macaulay duration at 8% interest is 6.97. (IV) The modiﬁed duration at 8% interest is 6.45. Problem 54.4 Which of the following are true: (I) The average term to maturity under the method of equated time is always greater than the volatility. (II) The duration is a decreasing function of i. (III) A zero coupon bond will have a Macaulay duration equal to the term to maturity. Problem 54.5 (a) Calculate the duration of a 12-year annuity immediate payable using an interest rate of 5%. (b) Calculate the modiﬁed duration of the annuity. Problem 54.6 A perpetuity pays 100 immediately. Each subsequent payment is increased by inﬂation. Calculate the duration of the perpetuity using 10.25% assuming that inﬂation will be 5% annually. Problem 54.7 Calculate the duration of a perpetuity immediate of 1 less the duration of a perpetuity due of 1 at an interest rate of 10%. Problem 54.8 Calculate the modiﬁed duration of an annuity due with payments of 100 for 10 years using an interest rate of 8%. 54 MACAULAY AND MODIFIED DURATIONS 477 Problem 54.9 ‡ A bond will pay a coupon of 100 at the end of each of the next three years and will pay the face value of 1000 at the end of the three-year period. The bond’s duration (Macaulay duration) when valued using an annual eﬀective interest rate of 20% is X. Calculate X. Problem 54.10 Suppose that a 3-year ﬁnancial instrument is expected to make increasing payments to you at the end of each of the next three years. Speciﬁcally, the payments will be CF(t) = 1,000t, for t= 1,2, and 3. Assume that you purchase this ﬁnancial instrument, at time 0, at a price which provides an annual eﬀective yield of 8%. Calculate the Macaulay duration, and the modiﬁed duration of this ﬁnancial instrument. Problem 54.11 Calculate the Macaulay duration and the modiﬁed duration of a 30-year zero-coupon bond with a face value of $ 1,000. Assume that the annual yield-to-maturity is 8%. Problem 54.12 A 10-year $200,000 mortgage will be paid oﬀ with level quarterly amortization payments. Assume that the interest rate on the mortgage is 6%, convertible quarterly, and that payments are made at the end of each quarter. Find the modiﬁed duration of this mortgage. Problem 54.13 ‡ Calculate the duration of a common stock that pays dividends at the end of each year into perpetuity. Assume that the dividend increases by 2% each year and that the eﬀective rate of interest is 5%. Problem 54.14 ‡ Calculate the duration of a common stock that pays dividends at the end of each year into perpetuity. Assume that the dividend is constant, and that the eﬀective rate of interest is 10%. Problem 54.15 ‡ The current price of an annual coupon bond is 100. The derivative of the price of the bond with respect to the yield to maturity is −700. The yield to maturity is an annual eﬀective rate of 8%. Calculate the duration of the bond. Problem 54.16 ‡ Calculate the Macaulay duration of an eight-year 100 par value bond with 10% annual coupons and an eﬀective rate of interest equal to 8%. 478 MEASURES OF INTEREST RATE SENSITIVITY Problem 54.17 ‡ John purchased three bonds to form a portfolio as follows: Bond A has semiannual coupons at 4%, a duration of 21.46 years, and was purchased for 980. Bond B is a 15-year bond with a duration of 12.35 years and was purchased for 1015. Bond C has a duration of 16.67 years and was purchased for 1000. Calculate the duration of the portfolio at the time of purchase. Problem 54.18 A 10-year 1000 face value 6% annual coupon bond with redemption value C has duration equal to 6.06 using an annual eﬀective interest rate of 6%. Calculate C. Problem 54.19 You are given: (i) the annual yield rate on a zero-coupon bond with duration of 6 months is 3% (ii) the annual yield rate on a zero-coupon bond with duration of 12 months is 3.25% (iii) the annual yield rate on a zero-coupon bond with duration of 18 months is 3.5% (iv) the annual yield rate on a zero-coupon bond with duration of 24 months is 3.75% Determine the semiannual eﬀective yield rate on a 2-year 100 par value bond with 4% semiannual coupons. Problem 54.20 John purchased three bonds to form a portfolio as follows: Bond A is purchased for X and has a duration of 20 years. Bond B is purchased for Y and has a duration of 20 years. Bond C is purchased for X+ Y. The duration (in years) of the portfolio at the time of purchase is 18. Determine the duration (in years) of Bond C at the time of purchase. 55 REDINGTON IMMUNIZATION AND CONVEXITY 479 55 Redington Immunization and Convexity Most ﬁnancial institutions try to insulate their portfolios from interest rate risk movements. Some institutions (such as banks) are concerned with protecting their current net worth against interest rate movements. Other institutions (such as pension funds) may face an obligation to make pay- ments after a given number of years. These institutions are more concerned with protecting the future values of their portfolios. The common factor amongst diﬀerent investors and ﬁnancial institutions is the presence of interest rate risk because net worth ﬂuctuates with interest rate. Financial institutions try to structure their assets and liabilities in such a way that they are protected against small changes in interest rates. There are three approaches that all work toward the same goal of minimizing (or even eliminating) the risk caused by ﬂuctuating interest rates: Immunizatiom, full immunization, and dedication. In this section, we consider the ﬁrst strategy. The basic idea of immunization is that ﬁnancial institutions (i.e., banks) are vulnerable to incur- ring losses if interest rates change. As a result, they “immunize” themselves by structuring their assets and liabilities such that the modiﬁed duration of the assets equals the modiﬁed duration of the liabilities. When this occurs, a rise in the interest rate would cause the present value of assets to decline, but there would be a matching decrease in the present value of liabilities. Thus, the company is protected against small changes in interest rates. We next discuss how immunization achieves the protection required. Consider cash inﬂowsA1,A2,··· ,An generated by the assets at times 1 ,2,··· ,n. Similarly, let L1,L2,··· ,Ln be the cash outﬂows gen- erated by the liabilities at times 1 ,2,··· ,n. Let the net cash ﬂows at time t be Rt = At −Lt, t = 1,2,··· ,n. In general, banks assets and liabiltiy are roughly equal in size so that we can assume that the present value of the cash inﬂows from the assets is equal to the present value of the cash outﬂows from the liabilities. This leads to P(i) = n∑ t=1 νtRt = 0. Next, we would like P(i) to have a local minimum at i,so that small changes in the interest rate in either direction will increase the present value of the cash ﬂows. But for the function P(i) to have a minimum requires two conditions to hold. The ﬁrst is that P′(i) = 0 (i.e. the modiﬁed duration of the assets is equal to the modiﬁed duration of the liabilities ). The second condition is P′′(i) >0. The second derivative of P(i) is used to deﬁne the convexity of a series of cash ﬂows given by c= P′′(i) P(i) . 480 MEASURES OF INTEREST RATE SENSITIVITY To summarize, immunization is achieved when the following three conditions are met: (1) P(i) = 0 : The present value of cash inﬂows (assets) should be equal to the present value of cash outﬂows (liabilities). (2) P′(i) = 0 : The modiﬁed duration of the assets is equal to the modiﬁed duration of the liabilities. (3) P′′(i) >0 : The convexity of the present value of cash inﬂows (assets) should be greater than the convexity of the present value of cash outﬂows(liabilities). In other words, asset growth (decline) should be greater (less) than liability growth(decline). In practice there are some diﬃculties and limitations in implementing the immunization strategy: (a) choice of i is not always clear. (b) doesn’t work well for large changes in i. For large changes in i a technique, known as full immunization, was developed for that purpose. This concept will be covered in Section 56. (c) yield curve is assumed to change with the change ini; actually, short-term rates are more volatile than long-term rates. (d) frequent rebalancing is required in order to keep the modiﬁed duration of the assets and liabil- ities equal. (e) exact cash ﬂows may not be known and may have to be estimated. (f) assets may not have long enough maturities of duration to match liabilities. We next express the percentage change ofP(i) in terms of both the modiﬁed duration and convexity. Expanding P(i+ ϵ) as a Taylor series as far as second derivatives we ﬁnd P(i+ ϵ) ≈P(i) + ϵP′(i) + ϵ2 2 P′′(i). Thus, P(i+ ϵ) −P(i) P(i) ≈ϵP′(i) P(i) + ϵ2 2 P′′(i) P(i) = −ϵν+ ϵ2 2 c. Example 55.1 A client deposits 100,000 in a bank, with the bank agreeing to pay 8% eﬀective for two years. The client indicates that half of the account balance will be withdrawn at the end of the ﬁrst year. The bank can invest in either one year or two year zero coupon bonds. The one year bonds yield 9% and the two year bonds yield 10%. Develop an investment program based on immunization. Solution. The cash outﬂows are 50000(1 .08) at the end of the ﬁrst year and 50000(1 .08)2 at the end of the second year. 55 REDINGTON IMMUNIZATION AND CONVEXITY 481 Suppose the bank invests x in the one year bonds, and y in two year bonds. Then P(i) =x(1.09)(1 + i)−1 + y(1.10)2(1 + i)−2 −50000(1.08)(1 + i)−1 −50000(1.08)2(1 + i)−2 P′(i) = −x(1.09)(1 + i)−2 −2(1.10)2y(1 + i)−3 + 50000(1.08)(1 + i)−2 +2(50000)(1.08)2(1 + i)−3 P′′(i) =2(1.09)x(1 + i)−3 + 6(1.10)2y(1 + i)−4 −100000(1.08)(1 + i)−3 −300000(1.08)2(1 + i)−4 To immunize against interest rate risk at i= 0.08, the bank should have P(0.08) = 0,P′(0.08) = 0 and P′′(0.08) >0. The ﬁrst two conditions lead to a system in the unknowns x and y. Solving the system gives x = 49541 .28 and y = 48198 .35. Moreover, P′′(0.08) = 128 ,773.89 > 0. Thus, the bank is well protected against interest rate ﬂuctuations Example 55.2 For the assets in Example 55.1, ﬁnd the modiﬁed duration and convexity. Solution. We have PA(i) = x(1.09)(1 + i)−1 + (1.10)2y(1 + i)−2 →PA(0.08) = 50 ,000.00 + 50,000.00 = 100,000.00. Also, P′ A(i) = −x(1.09)(1 + i)−2 −2(1.10)2y(1 + i)−3 →P′ A(0.08) = −138,888.89. The modiﬁed duration is v(0.08) = −P′ A(0.08) PA(0.08) = 138,888.89 100,000.00 = 1.39. The convexity is c(0.08) = P′′ A(0.08) PA(0.08) = 342,935.54 100,000.00 = 3.43 since P′′ A(i) = 2(1.09)x(1 + i)−3 + 6(1.10)2y(1 + i)−4 →P′′ A(0.08) = 342,935.54 Example 55.3 For a 30-year home mortgage with level payments and an interest rate of 10.2% convertible monthly, ﬁnd the modiﬁed duration and the convexity of the payments. Solution. The monthly rate of interest is 0 .102/12 = 0.0085. Then for a dollar of monthly payment, we have P(0.0085) = 360∑ t=1 (1.0085)−t = (1.0085)−1 [1 −(1.0085)−360 1 −(1.0085)−1 ] = 112.0591 482 MEASURES OF INTEREST RATE SENSITIVITY and P′(0.0085) = − 360∑ t=1 t(1.0085)−t−1 = −(1.0085)−1(Ia)360 0.0085 = −11,188.69608 Thus, the modiﬁed duration is ν(0.0085) = 11,188.69608 112.0591 = 99.85. Thus, the modiﬁed duration of the payments is just under 100 months of the 360-month term of the mortgage. Now, P′′(0.0085) = 360∑ t=1 t(t+ 1)(1.0085)−t−2 = ν2 360∑ t=1 (t2 + t)νt =ν2[ 360∑ t=1 t2νt + 360∑ t=1 tνt] =ν2[ 360∑ t=1 t2νt + (Ia)360 0.0085] Hence, using a calculator we ﬁnd c(0.0085) = 1,940,079 + 11,283.80 (1.0085)2(112.0591) = 17,121 Example 55.4 A bank agrees to pay 5% compounded annually on a deposit of 100,000 made with the bank. The depositor agrees to leave the funds on deposit on these terms for 8 years. The bank can either buy 4 year zero coupon bonds or preferred stock, both yielding 5% eﬀective. How should the bank apportion its investment in order to immunize itself against interest rate risk? Solution. The present value of the cash ﬂows in terms of the amount B invested in the bonds is P(i) = B(1.05)4(1 + i)−4 + (100000 −B)(0.05) i −100000(1.05)8(1 + i)−8 since the preferred stock is assumed to pay 5% forever and the bank must repay the deposit with interest at the end of 8 years. Substitution veriﬁes that P(.05) = 0. On the other hand, P′(0.05) = −4B(1.05)4(1.05)−5 −(100000 −B)(0.05) 0.052 + 800,000(1.05)8(1.05)−9 55 REDINGTON IMMUNIZATION AND CONVEXITY 483 So setting P′(0.05) = 0 and solving for B gives B = 76470.58 as the amount to be invested in bonds with the remainder in stock. With this allocation, one ﬁnds P′′(.05) = 20(76470 .58)(1.05)−2 + 2(100000 −7647.58)(.05)−2 −720,000(1.05)−9 = 965786.7413 >0, so the portfolio is optimal Example 55.5 Find the duration and convexity of a 20 year zero coupon bond assuming that the interest rate is 7% eﬀective. Solution. The duration is 20 since the par value is repaid at the end of 20 years and this is the only payment made by the bond. Now, per dollar of par value, P(i) = (1 + i)−20, so that the convexity P′′(i) P(i) = 420 (1 + i)2 . Substituting i= 0.07 gives the convexity as 366.844 Convexity of a Portfolio Consider a portfolio consisting of nbonds. Let bond k have current price Pk(i) and convexity ck(i). Then the current value of the portfolio is P(i) = P1(i) + P2(i) + ··· + Pn(i). Let c denote the convexity of the portfolio. Then, we have c=P′′(i) P(i) = P′′ 1 (i) P(i) + P′′ 2 (i) P(i) + ··· + P′′ n(i) P(i) =P1(i) P(i) (P′′ 1 (i) P1(i) ) + P2(i) P(i) (P′′ 2 (i) P2(i) ) + ··· + Pn(i) P(i) (P′′ n(i) Pn(i) ) =P1(i) P(i) c1 + P2(i) P(i) c2 + ··· + Pn(i) P(i) cn Thus, the convexity of a portfolio can be calculated as the weighted average of the bonds’ convexities, using the market values of the bonds as the weights. 484 MEASURES OF INTEREST RATE SENSITIVITY Practice Problems Problem 55.1 ‡ Which of the following statements about immunization strategies are true? I. To achieve immunization, the convexity of the assets must equal the convexity of the liabilities. II. The full immunization technique is designed to work for any change in the interest rate. III. The theory of immunization was developed to protect against adverse eﬀects created by changes in interest rates. Problem 55.2 Nova Inc. has liabilities of 10 due in 1, 4, and 7 years. What asset incomes must the company arrange for in years 1 and 6 to immunize their cash ﬂow, assuming an annual interest rate of 10% on all transactions? Problem 55.3 A bank is required to pay 1,100 in one year. There are two investment options available with respect to how money can be invested now in order to provide for the 1,100 payback: (i) a non-interest bearing cash fund, for which x will be invested, and (ii) a two-year zero-coupon bond earning 10% per year, for which y will be invested. Based on immunization theory, develop an asset portfolio that will minimize the risk that liability cash ﬂows will exceed asset cash ﬂows. Assume the eﬀective rate of interest is equal to 10% in all calculations. Problem 55.4 A $5,000 payment is owed from Abby to Ben two years from now. Abby wants to set up an investment fund to meet this obligation, but the only investments she has available are a money market fund (currently earning 8%, but the rate changes daily), and a ﬁve-year zero-coupon bond earning 8%. Use an immunization framework to determine the amount of money Abby should invest now in each of the two investment vehicles. Assume an eﬀective annual interest rate of 8% for present value calculations. Problem 55.5 For the assets in Problem 55.4, ﬁnd the modiﬁed duration and convexity. Problem 55.6 Find the convexity of the following investments,assuming the eﬀective rate of interest is 8%: (a) A money market fund. (b) A 10-year zero coupon bond. (c) A preferred stock paying level dividends into perpetuity. 55 REDINGTON IMMUNIZATION AND CONVEXITY 485 Problem 55.7 Find the convexity of a loan repaid with equal installments over n periods if i= 0. Problem 55.8 A common stock pays dividends at the end of each year. It is assumed that each dividend is 4% greater than the prior dividend and the eﬀective rate of interest is 8%. Find the convexity of this common stock. Problem 55.9 Derive the following relationship between modiﬁed duration and convexity: d diν = ν2 −c. Problem 55.10 Suppose a company must make payments of 300 at time t = 3 and 500 at time t = 5 , and the annual interest rate is 4%. Find asset income at times t= 0 and t= 6 so that the portfolio will be immunized. 486 MEASURES OF INTEREST RATE SENSITIVITY 56 Full Immunization and Dedication In this section we discuss the two other strategies of immunization: Full immunization and dedica- tion. Full Immunization Full immunization is an extension to the Redington immunization discussed in Section 50. While Redington immunization works for small changes in i, full immunization can be applied for all changes of i. We will say that a portfolio is fully immunized if PVA(i+ ϵ) > PVL(i+ ϵ) for all ϵ> 0 or equivalently PVA(i) ≥PVL(i) for all positive i. Suppose we have one liability outﬂow Lk at time k. The concept is to hold two assets, one that will produce a cash inﬂow, A, prior to the liability outﬂow (say at time k−a) , and another that will produce a cash inﬂow subsequent to the liability outﬂow (say at time k+ b). We use the force of interest δ that is equivalent to i. The full immunization conditions for this single liability cash ﬂow are: (1) Present value of assets = Present value of liability. (2) Modiﬁed duration of assets = Modiﬁed duration of liability. (3) The asset cash ﬂows occur before and after the liability cash ﬂow. Example 56.1 An insurance company has committed to make a payment of $100,000 in 10 years. In order to fund this liability, the company has invested $27,919.74 in a 5-year zero-coupon bond and $27,919.74 in a 15-year zero-coupon bond. The annual eﬀective yield on all assets and liabilities is 6%. Determine whether the company’s position is fully immunized. Solution. Condition 1: Present value of assets = Present value of liability. We have PVL = 100,000(1.06)−10 = 55,839.48 and PVA = 27,919.74 ×2 = 55839.48, so condition (1) is met. Condition 2: Modiﬁed duration of assets = Modiﬁed duration of liability. We have Modiﬁed duration of liabilities = 10(100,000)(1.06)−10 100,000(1.06)−10 = 10 and Modiﬁed duration of assets = 5(27,919.74)+15(27,919.74) 55,839.48 = 10. 56 FULL IMMUNIZATION AND DEDICATION 487 Thus, condition (2) is true. Condition 3: The asset cash ﬂows occur before and after the liability cash ﬂow. Thus, the po- sition is fully immunized Now, conditions (1) and (2) leads to the following system P(δ) =Ae−(k−a)δ + Be−(k+b)δ −Lke−kδ = 0 P′(δ) = −A(k−a)e−(k−a)δ −B(k+ b)e−(k+b)δ + kLke−kδ = 0. The ﬁrst condition implies Lk = Aeaδ + Be−bδ. The second condition implies −A(k−a)e−(k−a)δ −B(k+ b)e−(k+b)δ + kLke−kδ =0 −Ake−(k−a)δ + Aae−(k−a)δ −Bke−(k+b)δ −Bbe−(k+b)δ + kLke−kδ =0 −k [ Ae−(k−a)δ + Be−(k+b)δ −Lke−kδ] + [Aae−(k−a)δ −Bbe−(k+b)δ] =0 Aae−(k−a)δ −Bbe−(k+b)δ =0 Aaeaδ −Bbe−bδ =0 Thus, we obtain the system Aeaδ + Be−bδ =Lk Aaeaδ =Bbe−bδ There are exceptional cases when the system has no solutions, but if the known quantities are (i) a,b (ii) B,b (iii) A,a or (iv) A,b then unique values of the other two quantities can be found. Solving the last equation for B we ﬁnd B = A (a b ) e(a+b)δ. We next show that conditions (1)-(3) achieve the required immunization. That is, we will show that for any δ′̸= δ there is a surplus, i.e. PVA(δ′) >PV L(δ′). Let S(δ′) denote the present value, at force of interest δ′ per annum, of the total-assets less the present value of the liability, i.e. the surplus function S(δ′) = PVA(δ′) −PVL(δ′) = e−kδ′ (Aeaδ′ + Be−bδ′ −Lk). 488 MEASURES OF INTEREST RATE SENSITIVITY Then we have S(δ′) =e−kδ′ [Aeaδ′ + Be−bδ′ −Lk] =e−kδ′ [Aeaδ′ + Be−bδ′ −(Aeaδ + Be−bδ)] =e−kδ′ Aeab [ ea(δ′−δ) + a be−b(δ′−δ) − ( 1 + a b )] Next, consider the function f(x) = eax + a be−bx − ( 1 + a b ) . Clearly, f(0) = 0. Moreover, since f′(x) = a(eax −e−bx), one can easily check the following f′(0) =0 f′(x) >0 for x> 0 f′(x) <0 for x< 0 Thus, f(x) >0 for all x̸= 0. This shows that S(δ′) >0 for all δ′̸= δ as required. For a portfolio with multiple liabilities, repeat the above process for each liability outﬂow. There will be two asset inﬂows for each liability outﬂow. Full immunization is like Redington immunization in the sense that portfolio must be rebalanced periodically so that to ensure that modiﬁed duration of assets is equal to modiﬁed duration of liabilities. Example 56.2 An investor has a single liability of $1000 due in 15 years’ time. The yield on zero coupon bonds of any term is currently 4% per annum, and the investor possesses cash equal to the present value of his liability, i.e. 1000(1 .04)−15 = 555.26. He wishes to invest in 10-year and 20-year zero-coupon bonds in such a way that he will make a proﬁt on any immediate change in the force of interest. How much of each security should he buy, and how large a proﬁt will he make if the rate of interest per annum immediately becomes 0.01, 0.02, 0.03, 0.05, 0.06, 0.07,or 0.08? Solution. Letting δ= ln 1.04 we ﬁnd Ae5δ + Be−5δ =1000 5Ae5δ −5Be−5δ =0 56 FULL IMMUNIZATION AND DEDICATION 489 or equivalently, we have the system of two equations A(1.04)5 + B(1.04)−5 =1000 A(1.04)5 −B(1.04)−5 =0. These equations may easily be solved, giving A = 410.96 and B = 608.32. The quantities of zero- coupon bonds which he should buy are therefore those providing $410.96 at time 10 years and $608.32 at time 20 years. The proﬁt on an immediate change in the rate of interest to iper annum is, S(i) = (1 + i)−15[410.96(1 + i)5 + 608.32(1 + i)−5 −1000]. In the following table we give the values of the liability ($1000(1 + i)−15), the assets (410 .96(1 + i)−10 + 608.32(1 + i)−20) and the proﬁt to the investor for each of the speciﬁed rates. Annual interest rate PVL PVA Present value of proﬁt 0.00 1000.00 1019.28 19.28 0.01 861.35 870.58 9.23 0.02 743.01 746.51 3.50 0.03 641.86 642.61 0.75 0.04 555.26 555.26 0.00 0.05 481.02 481.56 0.54 0.06 417.27 419.16 1.89 0.07 362.45 366.11 3.66 0.08 315.24 320.87 5.63 Dedication In this section we discuss an approach for matching liabilities and assets. This approach is called dedication or absolute matching. In this approach, an company structures an asset portfolio in such a fashion that the cash inﬂow that will be generated from assets will exactly match the cash outﬂow from the liabilities. If the above strategy is achieved, the company has full protection against any movement in interest rates. However, the following problems exist with implementing this strategy. 1. Cash ﬂows are usually not that predictable on either the asset or the liability side. 2. If the liabilities are long term in nature, it might be impossible to ﬁnd assets to exactly match the liabilities without creating reinvestment risk. 3. The yield rate on a fund structured with the major restrictions imposed by absolute matching may be less than that on a fund for which more ﬂexibility is available, and the extra return may overshadow the advantages of absolute matching in importance. 490 MEASURES OF INTEREST RATE SENSITIVITY Example 56.3 An insurance company accepts an obligation to pay 10,000 at the end of each year for 2 years. The insurance company purchases a combination of the following two bonds (both with $1,000 par and redemption values) in order to exactly match its obligation: Bond A: A 1-year 5% annual coupon bond with a yield rate of 6%. Bond B: A 2-year 8% annual coupon bond with a yield rate of 7%. (a) Find the numbers (which need not be integers) of each bond the insurer must purchase to exactly match its obligations. (b) Find the total cost to the insurer of purchasing the needed numbers of bonds. Solution. (a) We have the following chart Period Bond A Bond B Liabilities 1 1050 80 10,000 2 1080 10,000 Let a be the number of bonds A purchased and b that of bonds B. Then we have the following system of two equations in two unknowns 1050a+ 80b=10,000 1080b=10,000 Solving for a and b we ﬁnd a= 8.8183 bonds and b= 9.2593 bonds. (b) The answer is 8.8183(1050ν0.06) + 9.2593(80ν0.07 + 1080ν2 0.07) = $18161.82 56 FULL IMMUNIZATION AND DEDICATION 491 Practice Problems Problem 56.1 An insurance company has an obligation to pay $1,000,000 at the end of 10 years. It has a zero- coupon bond that matures for $413,947.55 in 5 years, and it has a zero-coupon bond that matures for $864,580.82 in 20 years. The eﬀective yield for assets and liability is 10%. (a) Determine whether the company’s position is fully immunized. (b) What is the level of surplus if the interest rate falls to 0%? (c) What is the level of surplus if the interest rate rises to 80%? Problem 56.2 ‡ A company must pay liabilities of 1000 and 2000 at the end of years 1 and 2, respectively. The only investments available to the company are the following two zero-coupon bonds: Maturity(years) Eﬀective annual yield Par 1 10% 1000 2 12% 1000 Determine the cost to the company today to match its liabilities exactly. Problem 56.3 ‡ An insurance company accepts an obligation to pay 10,000 at the end of each year for 2 years. The insurance company purchases a combination of the following two bonds at a total cost of X in order to exactly match its obligation: (i) 1-year 4% annual coupon bond with a yield rate of 5% (ii) 2-year 6% annual coupon bond with a yield rate of 5%. Calculate X. The following information applies to Problems 56.4 - 56.6 Joe must pay liabilities of 1,000 due 6 months from now and another 1,000 due one year from now. There are two available investments: •a 6-month bond with face amount of 1,000, a 8% nominal annual coupon rate convertible semi- annually, and a 6% nominal annual yield rate convertible semiannually; and •a one-year bond with face amount of 1,000, a 5% nominal annual coupon rate convertible semi- annually, and a 7% nominal annual yield rate convertible semiannually. Problem 56.4 ‡ How much of each bond should Joe purchase in order to exactly (absolutely) match the liabilities? 492 MEASURES OF INTEREST RATE SENSITIVITY Problem 56.5 ‡ What is Joe’s total cost of purchasing the bonds required to exactly (absolutely) match the liabili- ties? Problem 56.6 ‡ What is the annual eﬀective yield rate for investment in the bonds required to exactly (absolutely) match the liabilities? Problem 56.7 John wants to absolutely match a debt that he owes under which he must make payment of 1000 in one year and 2000 in two years. He can purchase the following bonds: (1) Bond A is a two year bond with annual coupons of 100 and a maturity value of 1000. (2) Bond B matures in one year for 1000 and pays a coupon of 80. Calculate the amount of Bond B that John should purchase. An Introduction to the Mathematics of Financial Derivatives What is a ﬁnancial derivative? A ﬁnancial derivative is a ﬁnancial contract (between two parties) that derives its value from the value of some underlying asset. For example, a homeowner’s insurance policy promises that in the event of a damage to your house, the insurance company will compensate you for at least part of the damage. The greater the damage, the more the insurance company will pay. Your insurance policy thus derives its value from the value of your house and therefore is a derivative. In this chapter, we introduce four ﬁnancial derivatives: forward contracts, future contracts, options (call and put) and swaps and we discuss some of the related topics. 493 494 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 57 Financial Derivatives and Related Issues In this section we introduce the concept of ﬁnancial derivatives and discuss some of the relevant topics of derivatives. What is a Financial Derivative? A ﬁnancial derivative is a contract (between two parties) that derives its value from the value of some underlying asset(s). Examples of underlying assets include ﬁnancial assets (such as stocks or bonds), commodities (such as gold or oil), market indices (such as S&P 500 or FTSE 100), interest rates (LIBOR rates) and many others. An example of a derivative is the following: Party A makes an agreement with Party B regarding the price of gold. If the price of one ounce of gold in one year is above $300, Party Aagrees to pay Party B the amount of $1. Otherwise, Party B will pay Party A$1. The agreement is a derivative since the outcome depends on the price of gold. Note that this agreement is a kind of a “bet” on the price moves of gold. In general derivatives can be viewed as bets on the price of something. The most common types of ﬁnancial derivatives that we will discuss in this book are futures, for- wards, options, and swaps. Derivatives Trading Markets Derivatives can be traded through organized exchanges such as The Chicago Board of Trade(CBOT), the New York Stock Exchange (NYSE), the New York Mercantile Exhange (NYMEX) and many oth- ers, or through contracts negotiated privately between two parties, referred to as over−the−counter (abbreviated OTC). Financial markets are usually regulated by the Securities and Exchange Com- mission (SEC) and the Commodity Futures Trading Commission (CFTC). Financial markets serve six basic functions. These functions are brieﬂy listed below: •Borrowing and Lending: Financial markets permit the transfer of funds (purchasing power) from one agent to another for either investment or consumption purposes. •Price Determination: Financial markets provide vehicles by which prices are set both for newly issued ﬁnancial assets and for the existing stock of ﬁnancial assets. •Information Aggregation and Coordination: Financial markets act as collectors and aggregators of information about ﬁnancial asset values and the ﬂow of funds from lenders to borrowers. •Risk Sharing: Financial markets allow a transfer of risk from those who undertake investments to those who provide funds for those investments. •Liquidity: Financial markets provide the holders of ﬁnancial assets with a chance to resell or liquidate these assets. •Eﬃciency: Financial markets reduce transaction costs and information costs. 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 495 Most Common Uses of Derivatives Below are some of the motives for someone to use derivatives: (1) to reduce (or hedge) exposure to risk. For example, a wheat farmer and a wheat miller could enter into a futures contract to exchange cash for wheat in the future. Both parties have reduced a future risk: for the wheat farmer, the uncertainty of the price, and for the wheat miller, the availability of wheat. (2) to speculate expected changes in future prices with the hope of making proﬁt. In this case, speculation increases the exposure to risk. The potential gain or loss can be leveraged (i.e. mag- niﬁed) relative to the initial investment. (3) to reduce transaction costs, such as commissions and other trading costs. (4) to maximize return on investments through asset management activities, tax loopholes, and regulatory restrictions. For example, a company can use derivatives to produce temporary losses to lower its taxes. We refer to this motive as regulatory arbitrage. Example 57.1 Weather derivatives are ﬁnancial instruments that can be used by organizations or individuals as part of a risk management strategy to reduce risk associated with adverse or unexpected weather conditions (rain/temperature/snow). Two types of weather derivatives are heating degree-day (HDD) and cooling degree-day (CDD) future contracts. These future contracts make payments based on whether the temperature is abnormally hot or cold. (a) Explain why a soft-drink manufacturer might be interested in such a contract. (b) The business buys these future contracts to hedge against temperature-related risk. Who is the other party accepting the risk? Solution. (a) Soft drink sales greatly depend on weather. Generally, warm weather boosts soft drink sales and cold weather reduces sales. A soft drink producer can use weather futures contracts to reduce the revenue swing caused by weather and smooth its earnings. Shareholders of a company generally want the earnings to be steady. They don’t want the management to use weather as an excuse for poor earnings or wild ﬂuctuations of earnings. (b) Anyone who can predict a weather index can enter into weather futures and make a proﬁt. However, since predicting weather is not usually easy and accurate a loss may occur Perspectives on Derivatives There are three diﬀerent user perspectives on derivatives: •The end-user perspective. End-users include corporations, investment managers, and in- vestors. End-users use derivatives in order to achieve a speciﬁc goal or goals such as managing risk, speculating, reducing costs, or avoiding regulations. 496 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES •The market-maker perspective. These are traders or intermediaries between diﬀerent end- users. They buy from end-users who want to sell (usually at a low price) and sell to end-users who want to buy (usually at a higher price.) Also, market-users might charge commissions for trading transactions. •The economic observer such as a regulator or a research economist, whose role is to watch and even sometimes regulate the markets. Traders of Derivatives There are three main traders of derivatives: hedgers, speculators and arbitrageurs. •Hedgers use derivatives to reduce risk that they face from potential future movements in market variables. Example 57.2 Suppose that it is August 16 2007 and, ﬁrm A, a company based in the US knows that it will pay £10 million on November 16, 2007 for goods it has purchased from a British ﬁrm. Firm Acan buy pounds from a bank in a three-month forward contract and thus hedge against foreign exchange risk. Suppose the agreement is to buy £1 for $1.55. (a) What would the ﬁrm A pay on November 16, 2007? (b) Suppose that on November 16, 2007 the exchange rate is $3 for £1. What would ﬁrm Apay on Novemebr 16, 2007 in the absence of hedging? Solution. (a) The ﬁrm has hedged against exchange rate volatility. The amount to be paid in US dollars on November 16, 2007 is 1 .55 ×10,000,000 = $15.5 millions. (b) The ﬁrm would have to pay 3 ×10,000,000 = $30 millions •Speculators use derivatives to bet on the future direction of a market variable. They can, for example, buy a put option on a stock if they think it will go down. If they think that the price of the stock will go up they will buy a call option. Another example of a speculator is an investor in weather derivatives who buys temperature-related future contracts. •Arbitrageurs take oﬀsetting positions in two or more instruments to lock in a riskless proﬁt if securities are inconsistently priced. Example 57.3 Suppose that ﬁrm Ais dually listed on both the NYSE and LSE. On the NYSE its share is trading at $152 and on the LSE its share is trading at £100. The exchange rate is $1.55 per pound. An arbitrageur makes a riskless proﬁt by simultaneously buying 100 shares on the NYSE and sell them on the LSE. What is the amount of this proﬁt? 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 497 Solution. He buys the shares on the NYSE for $152 a share. He sells the shares on the LSE for $155 a share. So the proﬁt per share is $3. Total proﬁt from the 100 shares is then $300 498 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 57.1 Which of the following is a derivative? (I) A bushel of corn (II) A contract to sell 100 bushels of corn (III) Short sale of a stock. Problem 57.2 The Security and Exchange Commission (SEC) holds primary responsibility for enforcing the fed- eral securities laws and regulating the securities industry. Which of the following describes the perspective of SEC on derivatives? (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.3 List four motives for using derivatives. Problem 57.4 Company A based in the US buys goods from a British company and is required to pay in British pounds for the cost of the goods. Company A buys a derivative that allows it to use the current exhange rate after six months from today. Which of the following characterizes company Auses of the derivative? (I) Hedging (II) Speculation (III) Reduced transaction costs (IV) Regulatory loop holes. Problem 57.5 You buy 100 shares of a stock through a brokerage ﬁrm. Which of the following describes your perspective of using derivatives? (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.6 You buy 100 shares of a stock through a brokerage ﬁrm. Which of the following describes brokerage ﬁrm perspective of using derivatives? 57 FINANCIAL DERIVATIVES AND RELATED ISSUES 499 (I) The end-user perspective (II) The market-maker perspective (III) The economic observer. Problem 57.7 You work for a ﬁrm who has a reimbursement program for college credits. The program reimburses 100% of costs for an “A”, 75% of costs for a “B”, 50% for a “C” and 0% for anything less. Would this program be considered an example of a derivative? Explain. Problem 57.8 Explain why the following ﬁrms might be interested in buying HDD or DCC future contracts. (a) Ski-resort owner. (b) Electric utilities. (c) Amusement park owner. Problem 57.9 Match each of the following market traders: (a) hedgers (b) speculators (c) arbitrageurs with one of the following roles: (I) proﬁt from perceived superior expectations (II) trade to eliminate or reduce future price risk (III) attempt to proﬁt when the same security or commodity is trading at diﬀerent prices in two or more markets. Problem 57.10 Explain carefully the diﬀerence between (a) hedging (b) speculation and (c) arbitrage. 500 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 58 Derivatives Markets and Risk Sharing As pointed out in the previous section, one of the basic function of derivatives markets is risk shar- ing. Risk is a major feature of economic activity. In this section we discuss the role of ﬁnancial markets as a vehicle of sharing risks. Risk-sharing is one of the most important functions of ﬁnancial markets. To illustrate this con- cept, consider auto insurance companies. These companies collect premiums for auto insurance policies. The total premiums collected are then being available to help those who get involved into car wrecks. Thus, those policyholders who had no wrecks in their records have basically lost their premiums. However, their premiums went to help those who needed it. A similar scenario occurs in the business world. Some companies proﬁt and others suﬀer. Thus, it makes sense to have a mechanism enabling companies to exchange ﬁnancial risks. Share risking mechanisms should beneﬁt everyone. Another example of risk sharing is the use of the so-called catastrophe bonds. For example, an insurance company that provides hurricane insurance for Florida residents will usually face large insurance claims in the case of a devastating hurricane happening. For that reason, the insurance company usually either issues or buys from reinsurers cat bonds enabling the company to share risks with the bondholders. Catastrophe bondholders receive interest rates at levels commensurate with the risk that they may lose all of their principal on the occurrence of a major hurricane. The sponsor issues cat bonds and typically invests the proceeds from the bond issuance in low-risk securities. The earnings on these low-risk securities, as well as insurance premiums paid to the sponsor, are used to make periodic, variable rate interest payments to investors. Sponsors of cat bonds include insurers, reinsurers, corporations, and government agencies. Diversiﬁable Risk Versus Non-Diversiﬁable Risk Risk management is one of the topics discussed in portfolio theory. The risk level of an asset can be categorized into two groups: the diversiﬁable risk and the non-diversiﬁable risk. Before we discuss the diﬀerence between diversiﬁable and non-diversiﬁable risks, we need to ﬁrst understand the term diversiﬁcation. The act of diversiﬁcation implies that an individual (or a ﬁrm) allocates his/her wealth among several types of investments rather than just one investment. In other words, diversi- ﬁcation means spreading the risk of a portfolio by investing in several diﬀerent types of investments rather than putting all the money in one investment. Diversiﬁable risk (also known as non-systematic risk) is a risk that can be reduced or eliminated by combining several diverse investments in a portfolio. Example 58.1 Suppose you are an avid Apple computer user, and you decide to invest all your money in Apple stocks. Are you making a sound decision? If not, then what is a good investment strategy? 58 DERIVATIVES MARKETS AND RISK SHARING 501 Solution. The answer is no. If you have done so, you will be very vulnerable in the last few months when Apple announced losses and a major change in management. However, you can reduce the risk of your portfolio if you invest in other computer companies such IBM, Compaq, etc. However, this is not a very well diversiﬁed portfolio because you have invested your money in the computer industry. As a result, you will still be vulnerable to changes in the computer industry. It is wise to have a portfolio consisting of shares in ﬁrms from various industries such as real estate, electronics, biomedical and so on Non-diversiﬁable risk is the part of an asset’s risk that cannot be eliminated through diver- siﬁcation. This type of risk is also known as market risk or systematic risk . Examples of non-diversiﬁable risks are natural disasters, wars, and economic factors such as inﬂation or unem- ployment levels events that can inﬂuence the entire market and usually aﬀects your entire portfolio. Example 58.2 Which of the following can be categorized as a non-diversiﬁable risk? (a) Recession. (b) The fall of a stock share. (c) Government defaults in paying bonds. Solution. Obviously only (a) and (c) might have major impact of one’s entire portfolio 502 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Practice Problems Problem 58.1 Explain why a health insurance policy is a risk sharing derivative. Problem 58.2 List the two types of risk involved in a portfolio. Problem 58.3 The act of diversiﬁcation implies that an individual (or a ﬁrm) allocates his/her wealth (I) among several types of assets (II) in just one asset Problem 58.4 Determine whether each of the following events is considered a diversiﬁable or non-diversiﬁable risk for a portfolio: (a) A software ﬁrm’s best programmer quits. (b) Oil-prices on international markets suddenly increase. (c) Congress votes for a massive tax cut. (d) A low-cost foreign competitor unexpectedly enters a ﬁrm’s market. Problem 58.5 Determine whether each of the following events is considered a diversiﬁable or non-diversiﬁable risk for a portfolio: (a) A wildcat strike is declared. (b) Oil is discovered on a ﬁrm’s property. (c) The Federal Reserve institutes a restrictive monetary policy. (d) Long term interest rates rise sharply. Problem 58.6 An earthquake linked bond is called (a) callable bond (b) perpetual bond (c) cat bond (d) serial bond Problem 58.7 Decide whether the risk is a diversiﬁable risk or non-diversiﬁable risk: 58 DERIVATIVES MARKETS AND RISK SHARING 503 (a) Management risk (b) Inﬂation risk (c) Sociopolitical risk (d) Credit risk (e) Currency risk (f) Interest rate risk (g) Liquidity risk. Problem 58.8 True or false: Cat bonds are used by sponsors as a hedge against natural disasters. Problem 58.9 Which of the following is not a term that refers to non-systematic risk? (A) Non-diversiﬁable risk (B) Market risk (C) Diversiﬁable risk (D) Both (A) and (B) Problem 58.10 In 9/11, immediately after the terrorist attacks the stock market fell signiﬁcantly. This is an example of (A) Systematic risk (B) Market risk (C) Non-diversiﬁable risk (D) All of the above Problem 58.11 Recently in California the workers at Kroger supermarkets had been on strike. This strike would most likely have been seen as an example of (A) Non-systematic risk (B) Non-diversiﬁable risk (C) Diversiﬁable risk (D) None of the above Problem 58.12 Explain how cat bonds work. 504 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 59 Forward and Futures Contracts: Payoﬀ and Proﬁt Dia- grams In this section we introduce two examples of derivatives: forward and futures contracts. We also analyze the possible payoﬀs and proﬁts of these two securities. Forwards and futures are very similar as contracts but diﬀer in pricing and trading mechanisms. For example, forwards are traded over the counter (i.e. contract speciﬁcation can be customized) whereas futures are traded on exchange markets (where contracts are standardized). For the purpose of this section, we think of them as interchangeable. A more detailed discussion of futures will be covered in Section 72. A forward contract (or a futures contract ) is a commitment to purchase at a future date, known as the expiration date, the delivery date or the maturity date, a given amount of a commodity or an asset at a price agreed on today. The price ﬁxed now for future exchange is called the forward price or the delivery price. The asset or commodity on which the forward contract is based is called the underlying asset. Apart from commissions and bid-ask spreads, a forward contract requires no initial payment or premium between the two parties. At the delivery date, cash is exchanged for the asset. Examples of forward contracts include •A bond forward is an obligation to buy or sell a bond at a predetermined price and time. •An equity forward is an obligation to buy or sell an equity at a predetermined price and time. •A gold forward is an obligation to deliver a speciﬁed quantity of gold on a ﬁxed date and receive a ﬁxed delivery price. •A foreign exchange forward is an obligation to buy or sell a currency on a future date for a predetermined ﬁxed exchange rate. This type of forwards is used by ﬁrms as a protection against ﬂuctuation in foreign currency exchange rates. •An interest rate forward is used to lock-in future interest rates. For example, a ﬁrm who is planning for a loan in six months can buy forward contract that lock-in the rate at the present. This rate will be the rate that will apply when the loan is exercised in six months. The risk here is that the ﬁrm will stuck with borrowing money at the stated rate so any decline of interest rates before the six months period will cause losses for the ﬁrm. Example 59.1 Explain in words the meaning of the following: A 3-month forward contract for 1,000 tons of soybean at a forward price of $165/ton. Solution. In this contract, the buyer is committed to buy 1000 tons of soybean from the seller in three months for a price of $165 a ton 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 505 Payoﬀ on a Forward Contract Forward contracts are privately executed between two parties. The buyer of the underlying com- modity or asset is referred to as the long side whereas the seller is the short side. The obligation to buy the asset at the agreed price on the speciﬁed future date is referred to as the long position. A long position proﬁts when prices rise. The obligation to sell the asset at the agreed price on the speciﬁed future date is referred to as the short position. A short position proﬁts when prices go down. What is the payoﬀ of a forward contract on the delivery date? Let T denote the expiration date, K denote the forward price, and PT denote the spot price (or market price) at the delivery date. Then •for the long position: the payoﬀ of a forward contract on the delivery date is PT −K; •for the short position: the payoﬀ of a forward contract on the delivery date is K−PT. Figure 59.1 shows a payoﬀ diagram on a contract forward. Note that both the long and short forward payoﬀ positions break even when the spot price is equal to the forward price. Also note that a long forward’s maximum loss is the forward price whereas the maximum gain is unlimited. For a short forward, the maximum gain is the forward price and the maximum loss is unlimited. Figure 59.1 Payoﬀ diagrams show the payoﬀ of a position at expiration. These payoﬀs do not include any costs or gains earned when purchasing the assets today. Payoﬀ diagrams are widely used because they summarize the risk of the position at a glance. We have pointed out earlier that the long makes money when the price rises and the short makes money when the price falls. Example 59.2 An investor sells 20 million yen forward at a forward price of $0.0090 per yen. At expiration, the spot price is $0.0083 per yen. 506 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES (a) What is the long position payoﬀ? (b) What is the short position payoﬀ? Solution. (a) At the expiration date, the long position’s payoﬀ is (0 .0083 −0.009) ×20 ·106 = −$14,000, a loss of $14,000 (b) The short position’s payoﬀ is $14 ,000 that is a proﬁt of $14,000 Example 59.3 Consider a forward contract on a stock with spot price of $25 and maturity date of 3 months from now. The forward price is $25.375. Draw the payoﬀ diagram for both the long and short forward positions after 3 months. Solution. The diagram is given in Figure 59.2. Figure 59.2 Note that both the long and short forward payoﬀ positions break even when the price of the stock at maturity is equal to the forward price, i.e., at $25.375 Example 59.4 Consider the forward contract of the previous example. Suppose the investor decides to purchase the stock outright today for $25. Is there any advantage of using the forward contract to buy the stock as opposed to buying it outright? (a) Draw the diagram for the payoﬀ to the long physical position as well as the payoﬀ diagram to the long forward. (b) If the spot price is $0 in three months then ﬁnd the long forward position’s payoﬀ. 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 507 (c) If the spot price of the stock in three months is $25.372 then ﬁnd the long physical position’s payoﬀ and the long forward’s payoﬀ. (d) Describe the investement for the outright purchase as well as the long forward in order to own the stock. Solution. (a) The diagram is shown in Figure 59.3. Note that the value of the payoﬀ of the physical position is equal to the spot price of the stock. The payoﬀ diagram does not include costs when the stock is purchased outright. Figure 59.3 (b) The long forward’s payoﬀ is 0 −25.372 = −$25.372. (c) The payoﬀ of the long physical position is $25.372. The payoﬀ of the long forward is 25 .372 − 25.372 = $0. (d) With the outright purchase, the investor invests $25 now and owns the stock. With the forward contract, the investor invests $0 now and $25.372 after three months and owns the stock Note that the payoﬀ of the outright purchase in Figure 59.3 tells us how much money we end up with after 3 months, but does not account for the initial $25 investment for purchasing the stock outright. Thus, from the graph one can not tell whether there is an advantage to either a forward purchase or an outright purchase. In order to have a fair comparison, the initial investments for both must be the same and then account for interest rate earned over the 3 months. We illustrate this point in the next two examples. Example 59.5 Suppose the investor invests $25 in a zero-coupon bond with par value $25.372 and maturity date in three months along with the forward contract of the previous example. Note that this combined position and the outright purchase position each initially costs $25 at time 0. Suppose the bond yield rate is 5.952% payable quarterly. Show that the combined position mimics the eﬀect of buying the stock outright. 508 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Solution. The investor pays $25 now for the bond. After three months the zero-coupon bond return is $25.372. He then uses the proceeds to pay for the forward price of $25.372 and owns the stock. This alternative is equivalent to outright purchase of the stock today for $25 Example 59.6 Suppose the investor borrows $25 to buy the physical stock which costs $25. Interest rate on the loan is 5.952% compounded quarterly. In this case, this position and the long forward position each initially costs $0 at time 0. Show that this act of borrowing mimics the eﬀect of entering into a long forward contract. Solution. The investor uses the borrowed money to buy the stock today. After three months, the investor repays $25 ×1.01488 = $25.372 for the borrowed money. Thus, the investor invested nothing ini- tially, and after three months he paid $25.372 and owned the stock. This shows that borrowing to buy the stock mimics the eﬀect of entering into a long forward contract In the above two examples, we conclude that the forward contract and the cash position are equiv- alent investments, diﬀering only in the timing of the cash ﬂows. Besides payoﬀ diagrams, one deﬁnes a proﬁt diagram. Recall that a payoﬀ diagram graphs the cash value of a position at maturity. A proﬁt diagram subtracts from the payoﬀ the future value of the initial investment in the position. Geometrically, the proﬁt diagram is a vertical shift of the payoﬀ diagram by the amount of the future value of the initial investment. Since forward contracts require no initial investment, the payoﬀ and proﬁt diagrams coincide. Example 59.7 A stock is priced at $50 and pays no dividends. The eﬀective annual interest rate is 10%. Draw the payoﬀ and proﬁt diagrams for a one year long position of the stock. Solution. The payoﬀ of a stock you buy long (and therefore you own) is just the stock market value. Let’s assume you want to ﬁnd the proﬁt after one year. Your proﬁt is the payoﬀ (market price) minus the future value of the initial investment over one year. If we denote the stock price by S then your proﬁt will be S−1.1(50) = S−55. Figure 59.4 shows both the payoﬀ and proﬁt diagrams. Notice that the proﬁt is 0 when the stock price is $55 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 509 Figure 59.4 Example 59.8 Suppose we own an index which we want to be valued at $1020 six months from today. For that we buy a forward contract with forward price of $1020 and maturity date of six months. Consider investing $1000 in a zero-coupon bond with six-month interest rate of 2%. (a) Compare the payoﬀ diagram of the combined position of the forward contract and the bond with the payoﬀ diagram of the index after six months. (b) Answer the same question for the proﬁt diagrams. Solution. (a) The payoﬀ of the forward contract and the bond in six months is forward + bond = spot price at expiration - 1020 + (1000 + 1000(0.02)) = spot price at expiration Thus, the payoﬀ diagram of the combined position is the same as the payoﬀ diagram of the index. Adding the bond shifts the payoﬀ diagram of the forward vertically upward by $1020. (b) The proﬁt diagram of the forward contract coincides with its payoﬀ’s diagram since there is no initial investment. The proﬁt of the bond is the payoﬀ minus the future value of the investment which comes out to 0. Thus, the proﬁt diagram of the combined forward and bond is the same as the proﬁt diagram of the forward Cash-Settled Forward Contracts A forward may be cash-settled, in which case the underlying asset and payment never exchange hands. Instead, the contract settles with a single payment in the amount spot price minus forward 510 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES price at delivery date. If the diﬀerence is positive, the short party pays the long party the diﬀerence. If it is negative, the long party pays the short party. Example 59.9 Two parties agree today to exchange 500,000 barrels of crude oil for $42.08 a barrel three months from today. Describe this forward when settled for cash (a) if the spot price at the delivery date is $47.36 a barrel; (b) if the spot price at the delivery date is $40.17 a barrel. Solution. If the forward were cash-settled then on the delivery date three months from today, no oil would change hands, and there would be no payment of 500 ,000 ×42.08 = $21.04 millions. (a) If the spot price at the delivery date was $47.36 a barrel then the forward would settle with a single payment of 500,000(47.36 −42.08) = $2,640,000 made by the short party to the long party. (b) If the spot price at the delivery date was $40.17 a barrel then the forward would settle with a single payment of 500,000(42.08 −40.17) = $955,000 made by the long party to the short party Credit Risk We conclude this section by mentioning the credit risk (i.e. the risk that a party will not meet its contractual obligations) associated with forward contracts. Exchange-traded contracts typically require collateral in order to minimize this risk . In over-the-counter contracts, each party bears the other’s credit risk. Example 59.10 Creditwise, which is riskier a forward contract or a futures contract? Solution. Since forward contracts are traded on over-the-counter and futures contracts are traded on ex- changes, forward contracts are more riskier than futures contracts 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 511 Practice Problems Problem 59.1 What is the diﬀerence between a long forward position and a short forward position? Problem 59.2 What is the diﬀerence between a long futures position and a short futures position? Problem 59.3 A trader enters into a short forward contract on 100 million yen. The forward exchange rate is 189.4 yen per £. How much does the trader gain or lose if the exchange rate at the end of the contract is (a) 192.0 yen per £; (b) 183.0 yen per £? Problem 59.4 A stock with no dividends has a current price of $50. The forward price for delivery in one year is $53. If there is no advantage of buying either the stock or the forward contract, what is the 1-year eﬀective interest rate? Problem 59.5 A default-free zero-coupon bond costs $91 and will pay $100 at delivery date in one year. What is the eﬀective annual interest rate? What is the payoﬀ diagram for the bond? What is the proﬁt diagram for the bond? Problem 59.6 Suppose that you enter into a long six-month forward position at a forward price of $50. What is the payoﬀ in 6 months for prices of $40, $45, $50, $55, and $60? Problem 59.7 Suppose that you enter into a short six-month forward position at a forward price of $50. What is the payoﬀ in 6 months for prices of $40, $45, $50, $55, and $60? Problem 59.8 Suppose shares of XYZ corporation are trading at $50 and currently pay no dividends. The forward price for delivery in 1 year is $55. Suppose the 1-year eﬀective annual interest rate is 10%. Draw the payoﬀ and proﬁt diagrams of this forward contract for a long position. Problem 59.9 A company agrees to buy 2,000 barrels of oil in two years; the current price is $63.50 per barrel. What is reasonable price to pay per barrel in two years? 512 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES Problem 59.10 Consider a forward contract of a stock index with maturity date six months from now and forward price of $1020. Answer the following questions if this forward contract is settled for cash. (a) Find the long and short positions payoﬀs if the spot price of the index at maturity is $1040. (b) What if the spot sprice is $960? Problem 59.11 You are a US exporter who will receive 5 millions Euro in six months time from the sale of your product in Spain. How can you hedge your foreign exchange risk ? Problem 59.12 Consider a forward contract of a stock index with maturity date three months from now and forward price of $930. Assume that the spot price of the index today is $900. What is the proﬁt or loss to a short position if the spot price of the market index rises to $920 by the expiration date? Problem 59.13 One ounce of gold today is being sold at the price of $300 per ounce. An investor borrows $300 with annual interest rate of 5% and buys one ounce of gold. The investor enters into a short forward contract to sell 1 oz. of gold in one year at $340. What is the investor’s proﬁt in this transaction? What would be his proﬁt if the original price of 1 oz. of gold is $323.81 and the loan is $323.81? Problem 59.14 An investor enters into a short forward contract to sell 100,000 British pounds for US dollars at an exchange rate of 1.50 US dollars per pound. How much does the investor gain or lose if the exchange rate at the end of the contract is (a) 1.49 and (b) 1.52? Problem 59.15 A forward contract is made for the delivery of a commodity 10 months from today at a forward price of $200 per unit of the commodity. (A) The purchaser of the forward contract pays the forward price, $200, immediately but must wait for 10 months to receive the commodity. (B) The holder of a forward contract always has the right to exchange it for an identical futures contract at any time before the delivery date. (C) The seller of the forward contract has an obligation to deliver the commodity 10 months from today in return for the spot price on the delivery date. (D) The seller of the forward contract has an obligation to deliver the commodity 10 months from today in return for $200 per unit, payable on delivery. 59 FORWARD AND FUTURES CONTRACTS: PAYOFF AND PROFIT DIAGRAMS 513 Problem 59.16 The main diﬀerence between forward and futures contracts is: (A) Futures contracts are traded every day (i.e., frequently) in organized exchanges prior to the delivery date speciﬁed in the contract. (B) For a futures contract there is never any obligation to deliver, or to receive, the underlying asset. (C) Forward contracts always have delivery dates further from the present than futures contracts. (D) The underlying assets in forward contracts are always physical commodities (e.g., wheat, oil or silver), while for futures contracts the underlying assets could be anything (e.g. weather indices or notional bank deposits). 514 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES 60 Call Options: Payoﬀ and Proﬁt Diagrams One problem with a forward contract is the obligation of the buyer to pay the forward price at the expiration date which causes a loss if the spot price is below the forward price. So it is natural to wonder if there is a type of contract where the buyer has the option not to buy the underlying asset. The answer is aﬃrmative thanks to call options. An option is a contract to buy or sell a speciﬁc ﬁnancial product such as a security or a commodity. The contract itself is very precise. It establishes a speciﬁc price, called the strike price or exercise price at which the contract may be exercised, or acted on, and it has an expiration date. When an option expires, it no longer has value and no longer exists. There are two types of options: calls and puts. In this section we discuss call options and in the next section we discuss the put options. In a call option the call buyer expects that the price of an asset may go up. The buyer pays an upfront premium that he will never get back. He has the right to exercise the option at or before the expiration date depending on the option’s style(to be introduced below). The call seller, also known as the option’s writer, receives the premium. If the buyer decides to exercise the option, then the seller has to sell the asset at the strike price. If the buyer does not exercise the option, then the seller proﬁt is just the premium. Thus, a call option is a contract where the buyer has the right to buy but not the obligation to buy. An option style governs the time at which exercise can occur. A European call option allows the holder to exercise the option (i.e., to buy) only on the option expiration date. An American call option allows exercise at any time during the life of the option. A Bermuda call option can be exercised at certain pre-speciﬁed dates before or at the expiration date. Unless otherwise speciﬁed, we will limit our discussions to European style. For a call option, the buyer has a long position whereas the seller has a short position. Example 60.1 Trader A(call buyer) purchases a call contract to buy 100 shares of XYZ Corp from Trader B (call seller) at $50 a share. The current price is $45 a share, and Trader Apays a premium of $5 a share. (a) If the share price of XYZ stock rises to $60 a share right before expiration, then what proﬁt does Trader A make if he exercised the call by buying 100 shares from Trader B and sell them in the stock market? (b) If, however, the price of XYZ drops to $40 a share below the strike price, then Trader Awould not exercise the option. What will be his losses in this case? Solution. (a) Trader A total cost for owning the 100 shares before or on the expiration date is 50(100) + 5(100) = 5,500. The revenue from selling the shares is 60(100) = 6 ,000. Thus, Trader A makes a 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 515 proﬁt of $500 in this transaction. (b) The losses will consist basically of the future value of premium paid, i.e. the future value of 5(100) = $500 Since a buyer of a call option will exercise the option only when the spot price is higher than the strike price, the buyer’s payoﬀ is then deﬁned by the formula Buyer′s call payoﬀ = max{0,spot price at expiration −strike price}. Example 60.2 Consider a call option with a strike price of $500. (a) What would be the payoﬀ to the buyer if the spot price at the expiration date is $550? (b) What would be the payoﬀ to the buyer if the spot price at the expiration date is $450? Solution. (a) Since the spot price is larger than the strike price, the buyer will most likely exercise his option. The payoﬀ in this case is max{0,550 −500}= $50. (b) In this case, the buyer will not exercise his option and therefore the payoﬀ is max{0,450 −500}= $0 Notice that the payoﬀ does not take into consideration the cost of acquiring the position, i.e. the premium which is paid at the time the option is acquired. The payoﬀ of an option is not the money earned (or lost). The proﬁt earned by the buyer is given by the formula Buyer′s call proﬁt = Buyer′s call payoﬀ −future value of premium Example 60.3 Consider a 3-month call option with strike price of $500 and premium of $46.90. If the spot price at expiration date is $550, would you exercise the call option? How much would you make or lose? The risk-free interest rate is 1% over three months. Solution. Since spot price is higher than the strike price, you expect to exercise the call option. The future value of the premium is 46 .90(1.01) = $47.37. The buyer’s proﬁt is then max{0,550 −500}−47.37 = $2.63 516 AN INTRODUCTION TO THE MATHEMATICS OF FINANCIAL DERIVATIVES A payoﬀ diagram and a proﬁt diagram of a long call with strike price K, spot price PT,and future value of premium Pc is shown in Figure 60.1. Figure 60.1 Notice that for a long call the maximum loss is the future value of the premium whereas the maximum gain is unlimited. Example 60.4 Consider a long call option (with strike price K) and a long forward contract (with a forward price K) with underlying asset a stock index. Draw the proﬁt diagram of both on the same window. (a) If the stock index is rising past K,, which of the two is more proﬁtable? (b) If the stock index falls considerably, which is more proﬁtable? Solution. Figure 60.2 plots the proﬁt for both the long forward and the long call. Figure 60.2 (a) The forward contract is more proﬁtable than the call option. (b) The call option is more proﬁtable (i.e. less loses) because the most you lose is the future 60 CALL OPTIONS: PAYOFF AND PROFIT DIAGRAMS 517 value of your premium. Notice that a call option can be thought as an insured position in the in- dex. It protects the buyer against substantial loses. The cost of this insurance is the premium paid Thus far we have considered the payoﬀ and the proﬁt from the buyer’s perspective. Next, we consider these issues from the seller’s (i.e. the writer) perspective. The writer’s payoﬀ and proﬁt are the negative of those of the buyer’s and thus they are given by Writer′s call payoﬀ = −max{0,spot price at expiration −strike price} = min{0,strike price −spot price at expiration } and Writer′s call proﬁt = Writer′s call payoﬀ + future value of premium . Example 60.5 Again, consider a 3-month call option with a strike price of $500 and a premium $46.90. Find the writer’s payoﬀ and the proﬁt if the spot price at the expiration date is $550. Assume a risk-free 3-month interest rate of 1%. Solution. Since the spot price is higher than the strike price, the writer has to sell the option. His/her payoﬀ is min{0,500 −550}= −$50 and the proﬁt is min{0,500 −550}+ 46.90(1.01) = −$2.63 The payoﬀ and proﬁt diagrams for a short call with strike price K, future value of premium Pc are shown in Figure 60.3.