Instruction: Which of the following may ﬁt the deﬁnition of interest?

Response: A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2 Marcel B. Finan Arkansas Tech University c⃝All Rights Reserved Preliminary Draft Last updated February 27, 2011 2 In memory of my parents August 1, 2008 January 7, 2009 Preface This manuscripts is designed for an introductory course in the theory of interest and annuity. This manuscript is suitablefor a junior level course in the mathematics of ﬁnance. A calculator, such as TI BA II Plus, either the solar or battery version, will be useful in solving many of the problems in this book. A recommended resource link for the use of this calculator can be found at http://www.scribd.com/doc/517593/TI-BA-II-PLUS-MANUAL. The recommended approach for using this book is to read each section, work on the embedded examples, and then try the problems. Answer keys are provided so that you check your numerical answers against the correct ones. Problems taken from previous exams will be indicated by the symbol ‡. This manuscript can be used for personal use or class use, but not for commercial purposes. If you ﬁnd any errors, I would appreciate hearing from you: mﬁnan@atu.edu This project has been supported by a research grant from Arkansas Tech University. Marcel B. Finan Russellville, Arkansas March 2009 3 4 PREFACE Contents Preface 3 The Basics of Interest Theory 9 1 The Meaning of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2 Accumulation and Amount Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3 Eﬀective Interest Rate (EIR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4 Linear Accumulation Functions: Simple Interest . . . . . . . . . . . . . . . . . . . . . . 28 5 Date Conventions Under Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . 35 6 Exponential Accumulation Functions: Compound Interest . . . . . . . . . . . . . . . . 41 7 Present Value and Discount Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Interest in Advance: Eﬀective Rate of Discount . . . . . . . . . . . . . . . . . . . . . . 56 9 Nominal Rates of Interest and Discount . . . . . . . . . . . . . . . . . . . . . . . . . . 66 10 Force of Interest: Continuous Compounding . . . . . . . . . . . . . . . . . . . . . . . 78 11 Time Varying Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 12 Equations of Value and Time Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . 100 13 Solving for the Unknown Interest Rate . . . . . . . . . . . . . . . . . . . . . . . . . . 107 14 Solving for Unknown Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 The Basics of Annuity Theory 143 15 Present and Accumulated Values of an Annuity-Immediate . . . . . . . . . . . . . . . 144 16 Annuity in Advance: Annuity Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 17 Annuity Values on Any Date: Deferred Annuity . . . . . . . . . . . . . . . . . . . . . 168 18 Annuities with Inﬁnite Payments: Perpetuities . . . . . . . . . . . . . . . . . . . . . . 176 19 Solving for the Unknown Number of Payments of an Annuity . . . . . . . . . . . . . . 184 20 Solving for the Unknown Rate of Interest of an Annuity . . . . . . . . . . . . . . . . . 192 21 Varying Interest of an Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 22 Annuities Payable at a Diﬀerent Frequency than Interest is Convertible . . . . . . . . 206 23 Analysis of Annuities Payable Less Frequently than Interest is Convertible . . . . . . . 210 5 6 CONTENTS 24 Analysis of Annuities Payable More Frequently than Interest is Convertible . . . . . . 218 25 Continuous Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 26 Varying Annuity-Immediate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 27 Varying Annuity-Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 28 Varying Annuities with Payments at a Diﬀerent Frequency than Interest is Convertible 257 29 Continuous Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Rate of Return of an Investment 275 30 Discounted Cash Flow Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 31 Uniqueness of IRR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 32 Interest Reinvested at a Diﬀerent Rate . . . . . . . . . . . . . . . . . . . . . . . . . . 292 33 Interest Measurement of a Fund: Dollar-Weighted Interest Rate . . . . . . . . . . . . 302 34 Interest Measurement of a Fund: Time-Weighted Rate of Interest . . . . . . . . . . . . 311 35 Allocating Investment Income: Portfolio and Investment Year Methods . . . . . . . . 320 36 Yield Rates in Capital Budgeting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 Loan Repayment Methods 333 37 Finding the Loan Balance Using Prospective and Retrospective Methods. . . . . . . . 334 38 Amortization Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 39 Sinking Fund Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 40 Loans Payable at a Diﬀerent Frequency than Interest is Convertible . . . . . . . . . . 365 41 Amortization with Varying Series of Payments . . . . . . . . . . . . . . . . . . . . . . 371 Bonds and Related Topics 379 42 Types of Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 43 The Various Pricing Formulas of a Bond . . . . . . . . . . . . . . . . . . . . . . . . . 384 44 Amortization of Premium or Discount . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 45 Valuation of Bonds Between Coupons Payment Dates . . . . . . . . . . . . . . . . . . 405 46 Approximation Methods of Bonds’ Yield Rates . . . . . . . . . . . . . . . . . . . . . . 413 47 Callable Bonds and Serial Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 Stocks and Money Market Instruments 427 48 Preferred and Common Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 49 Buying Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 50 Short Sales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 51 Money Market Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 CONTENTS 7 Measures of Interest Rate Sensitivity 453 52 The Eﬀect of Inﬂation on Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . 454 53 The Term Structure of Interest Rates and Yield Curves . . . . . . . . . . . . . . . . . 459 54 Macaulay and Modiﬁed Durations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 55 Redington Immunization and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . 479 56 Full Immunization and Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 An Introduction to the Mathematics of Financial Derivatives 493 57 Financial Derivatives and Related Issues . . . . . . . . . . . . . . . . . . . . . . . . . 494 58 Derivatives Markets and Risk Sharing . . . . . . . . . . . . . . . . . . . . . . . . . . . 500 59 Forward and Futures Contracts: Payoﬀ and Proﬁt Diagrams . . . . . . . . . . . . . . 504 60 Call Options: Payoﬀ and Proﬁt Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 514 61 Put Options: Payoﬀ and Proﬁt Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 523 62 Stock Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 63 Options Strategies: Floors and Caps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540 64 Covered Writings: Covered Calls and Covered Puts . . . . . . . . . . . . . . . . . . . 547 65 Synthetic Forward and Put-Call Parity . . . . . . . . . . . . . . . . . . . . . . . . . . 553 66 Spread Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 67 Collars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 68 Volatility Speculation: Straddles, Strangles, and Butterﬂy Spreads . . . . . . . . . . . 574 69 Equity Linked CDs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585 70 Prepaid Forward Contracts On Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 71 Forward Contracts on Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598 72 Futures Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 73 Understanding the Economy of Swaps: A Simple Commodity Swap . . . . . . . . . . 618 74 Interest Rate Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 75 Risk Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638 Answer Key 645 BIBLIOGRAPHY 646 8 CONTENTS The Basics of Interest Theory A component that is common to all ﬁnancial transactions is the investment of money at interest. When a bank lends money to you, it charges rent for the money. When you lend money to a bank (also known as making a deposit in a savings account), the bank pays rent to you for the money. In either case, the rent is called “interest”. In Sections 1 through 14, we present the basic theory concerning the study of interest. Our goal here is to give a mathematical background for this area, and to develop the basic formulas which will be needed in the rest of the book. 9 10 THE BASICS OF INTEREST THEORY 1 The Meaning of Interest To analyze ﬁnancial transactions, a clear understanding of the concept of interest is required. Inter- est can be deﬁned in a variety of contexts, such as the ones found in dictionaries and encyclopedias. In the most common context, interest is an amount charged to a borrower for the use of the lender’s money over a period of time. For example, if you have borrowed $100 and you promised to pay back $105 after one year then the lender in this case is making a proﬁt of $5, which is the fee for borrowing his money. Looking at this from the lender’s perspective, the money the lender is investing is changing value with time due to the interest being added. For that reason, interest is sometimes referred to as the time value of money. Interest problems generally involve four quantities: principal(s), investment period length(s), inter- est rate(s), amount value(s). The money invested in ﬁnancial transactions will be referred to as theprincipal, denoted by P.The amount it has grown to will be called the amount value and will be denoted by A.The diﬀerence I = A−P is the amount of interest earned during the period of investment. Interest expressed as a percent of the principal will be referred to as an interest rate. Interest takes into account the risk of default (risk that the borrower can’t pay back the loan). The risk of default can be reduced if the borrowers promise to release an asset of theirs in the event of their default (the asset is called collateral). The unit in which time of investment is measured is called the measurement period. The most measurement period is one year but may be longer or shorter (could be days, months, years, decades, etc.). Example 1.1 (a) The amount I owe on my credit card. (b) The amount of credit remaining on my credit card. (c) The cost of borrowing money for some period of time. (d) A fee charged on the money you’ve earned by the Federal government. Solution. The answer is (c) Example 1.2 Let A(t) denote the amount value of an investment at time t years. (a) Write an expression giving the amount of interest earned from time t to time t+ s in terms of A only. (b) Use (a) to ﬁnd the annual interest rate, i.e., the interest rate from time t years to time t+ 1 years. 1 THE MEANING OF INTEREST 11 Solution. (a) The interest earned during the time t years and t+ s years is A(t+ s) −A(t). (b) The annual interest rate is A(t+ 1) −A(t) A(t) Example 1.3 You deposit $1,000 into a savings account. One year later, the account has accumulated to $1,050. (a) What is the principal in this investment? (b) What is the interest earned? (c) What is the annual interest rate? Solution. (a) The principal is $1,000. (b) The interest earned is $1,050 - $1,000 = $50. (c) The annual interest rate is 50 1000 = 5% Interest rates are most often computed on an annual basis, but they can be determined for non- annual time periods as well. For example, a bank oﬀers you for your deposits an annual interest rate of 10% “compounded” semi-annually. What this means is that if you deposit $1000 now, then after six months, the bank will pay you 5% ×1000 = $50 so that your account balance is $1050. Six months later, your balance will be 5% ×1050 + 1050 = $1102.50.So in a period of one year you have earned $102.50. The annual interest rate is then 10.25% which is higher than the quoted 10% that pays interest semi-annually. In the next several sections, various quantitative measures of interest are analyzed. Also, the most basic principles involved in the measurement of interest are discussed. 12 THE BASICS OF INTEREST THEORY Practice Problems Problem 1.1 You invest $3,200 in a savings account on January 1, 2004. On December 31, 2004, the account has accumulated to $3,294.08. What is the annual interest rate? Problem 1.2 You borrow $12,000 from a bank. The loan is to be repaid in full in one year’s time with a payment due of $12,780. (a) What is the interest amount paid on the loan? (b) What is the annual interest rate? Problem 1.3 The current interest rate quoted by a bank on its savings accounts is 9% per year. You open an account with a deposit of $1,000. Assuming there are no transactions on the account such as depositing or withdrawing during one full year, what will be the amount value in the account at the end of the year? Problem 1.4 The simplest example of interest is a loan agreement two children might make:“I will lend you a dollar, but every day you keep it, you owe me one more penny.” Write down a formula expressing the amount value after t days. Problem 1.5 When interest is calculated on the original principal only it is calledsimple interest. Accumulated interest from prior periods is not used in calculations for the following periods. In this case, the amount value A, the principal P, the period of investment t, and the annual interest rate i are related by the formula A= P(1 + it). At what rate will $500 accumulate to $615 in 2.5 years? Problem 1.6 Using the formula of the previous problem, in how many years will 500 accumulate to 630 if the annual interest rate is 7.8%? Problem 1.7 Compounding is the process of adding accumulated interest back to the principal, so that interest is earned on interest from that moment on. In this case, we have the formula A= P(1 +i)t and we call i a yearly compound interest . You can think of compound interest as a series of back-to- back simple interest contracts. The interest earned in each period is added to the principal of the previous period to become the principal for the next period. You borrow $10,000 for three years at 5% annual interest compounded annually. What is the amount value at the end of three years? 1 THE MEANING OF INTEREST 13 Problem 1.8 Using compound interest formula, what principal does Andrew need to invest at 15% compounding annually so that he ends up with $10,000 at the end of ﬁve years? Problem 1.9 Using compound interest formula, what annual interest rate would cause an investment of $5,000 to increase to $7,000 in 5 years? Problem 1.10 Using compound interest formula, how long would it take for an investment of $15,000 to increase to $45,000 if the annual compound interest rate is 2%? Problem 1.11 You have $10,000 to invest now and are being oﬀered $22,500 after ten years as the return from the investment. The market rate is 10% compound interest. Ignoring complications such as the eﬀect of taxation, the reliability of the company oﬀering the contract, etc., do you accept the investment? Problem 1.12 Suppose that annual interest rate changes from one year to the next. Let i1 be the interest rate for the ﬁrst year, i2 the interest rate for the second year, ··· , in the interest rate for the nth year. What will be the amount value of an investment of P at the end of the nth year? Problem 1.13 Discounting is the process of ﬁnding the present value of an amount of cash at some future date. By the present value we mean the principal that must be invested now in order to achieve a desired accumulated value over a speciﬁed period of time. Find the present value of $100 in ﬁve years time if the annual compound interest is 12%. Problem 1.14 Suppose you deposit $1000 into a savings account that pays annual interest rate of 0.4% compounded quarterly (see the discussion at the end of page 11.) (a) What is the balance in the account at the end of year. (b) What is the interest earned over the year period? (c) What is the eﬀective interest rate? Problem 1.15 The process of ﬁnding the present value P of an amount A, due at the end of t years, is called discounting A.The diﬀerence A−P is called the discount on A. Notice that the discount on A is also the interest on P. For example, if $1150 is the discounted value of $1250, due at the end of 7 months, the discount on the $1250 is $100. What is the interest on $1150 for the same period of time? 14 THE BASICS OF INTEREST THEORY 2 Accumulation and Amount Functions Imagine a fund growing at interest. It would be very convenient to have a function representing the accumulated value, i.e. principal plus interest, of an invested principal at any time. Unless stated otherwise, we will assume that the change in the fund is due to interest only, that is, no deposits or withdrawals occur during the period of investment. If t is the length of time, measured in years, for which the principal has been invested, then the amount of money at that time will be denoted by A(t).This is called the amount function. Note that A(0) is just the principal P. Now, in order to compare various amount functions, it is convenient to deﬁne the function a(t) = A(t) A(0). This is called the accumulation function. It represents the accumulated value of a principal of 1 invested at time t≥0. Note that A(t) is just a constant multiple of a(t), namely A(t) = A(0)a(t). That is, A(t) is the accumulated value of an original investment of A(0). Example 2.1 Suppose that A(t) = αt2 + 10β. If X invested at time 0 accumulates to $500 at time 4, and to $1,000 at time 10, ﬁnd the amount of the original investment, X. Solution. We have A(0) = X = 10β; A(4) = 500 = 16 α+ 10β; and A(10) = 1000 = 100 α+ 10β. Using the ﬁrst equation in the second and third we obtain the following system of linear equations 16α+ X =500 100α+ X =1000. Multiply the ﬁrst equation by 100 and the second equation by 16 and subtract to obtain 1600 α+ 100X−1600α−16X = 50000 −16000 or 84X = 34000. Hence, X = 34000 84 = $404.76 What functions are possible accumulation functions? Ideally, we expect a(t) to represent the way in which money accumulates with the passage of time. Hence, accumulation functions are assumed to possess the following properties: (P1) a(0) = 1. (P2) a(t) is increasing,i.e., if t1 <t2 then a(t1) ≤a(t2).(A decreasing accumulation function implies a negative interest. For example, negative interest occurs when you start an investment with $100 and at the end of the year your investment value drops to $90. A constant accumulation function 2 ACCUMULATION AND AMOUNT FUNCTIONS 15 implies zero interest.) (P3) If interest accrues for non-integer values oft, i.e. for any fractional part of a year, then a(t) is a continuous function. If interest does not accrue between interest payment dates then a(t) possesses discontinuities. That is, the function a(t) stays constant for a period of time, but will take a jump whenever the interest is added to the account, usually at the end of the period. The graph of such an a(t) will be a step function. Example 2.2 Show that a(t) = t2 + 2t+ 1, where t ≥0 is a real number, satisﬁes the three properties of an accumulation function. Solution. (a) a(0) = 02 + 2(0) + 1 = 1. (b) a′(t) = 2t+ 2 >0 for t≥0. Thus, a(t) is increasing. (c) a(t) is continuous being a quadratic function Example 2.3 Figure 2.1 shows graphs of diﬀerent accumulation functions. Describe real-life situations where these functions can be encountered. Figure 2.1 Solution. (1) An investment that is not earning any interest. (2) The accumulation function is linear. As we shall see in Section 4, this is referred to as “simple interest”, where interest is calculated on the original principal only. Accumulated interest from prior periods is not used in calculations for the following periods. (3) The accumulation function is exponential. As we shall see in Section 6, this is referred to as “compound interest”, where the fund earns interest on the interest. (4) The graph is a step function, whose graph is horizontal line segments of unit length (the period). A situation like this can arise whenever interest is paid out at ﬁxed periods of time. If the amount of interest paid is constant per time period, the steps will all be of the same height. However, if the amount of interest increases as the accumulated value increases, then we would expect the steps to get larger and larger as time goes 16 THE BASICS OF INTEREST THEORY Remark 2.1 Properties (P2) and (P3) clearly hold for the amount function A(t). For example, since A(t) is a positive multiple of a(t) and a(t) is increasing, we conclude that A(t) is also increasing. The amount function gives the accumulated value of k invested/deposited at time 0. Then it is natural to ask what if k is not deposited at time 0, say time s> 0,then what will the accumulated value be at time t > s? For example, $100 is deposited into an account at time 2, how much does the $100 grow by time 4? Consider that a deposit of $ k is made at time 0 such that the $ k grows to $100 at time 2 (the same as a deposit of $100 made at time 2). Then A(2) = ka(2) = 100 so that k = 100 a(2) . Hence, the accumulated value of $k at time 4 (which is the same as the accumulated value at time 4 of an investment of $100 at time 2) is given by A(4) = 100 a(4) a(2) . This says that $100 invested at time 2 grows to 100 a(4) a(2) at time 4. In general, if $ k is deposited at time s, then the accumulated value of $ k at time t>s is k×a(t) a(s) , and a(t) a(s) is called the accumulation factor or growth factor. In other words, the accumulation factor a(t) a(s) gives the dollar value at time t of $1 deposited at time s. Example 2.4 It is known that the accumulation function a(t) is of the form a(t) = b(1.1)t + ct2, where b and c are constants to be determined. (a) If $100 invested at time t= 0 accumulates to $170 at time t= 3,ﬁnd the accumulated value at time t= 12 of $100 invested at time t= 1. (b) Show that a(t) is increasing. Solution. (a) By (P1), we must have a(0) = 1. Thus, b(1.1)0 + c(0)2 = 1 and this implies that b= 1. On the other hand, we have A(3) = 100a(3) which implies 170 = 100a(3) = 100[(1.1)3 + c·32] Solving for c we ﬁnd c= 0.041. Hence, a(t) = A(t) A(0) = (1.1)t + 0.041t2. It follows that a(1) = 1.141 and a(12) = 9.042428377. Now, 100 a(t) a(1) is the accumulated value of $100 investment from time t= 1 to t> 1. Hence, 100a(12) a(1) = 100 ×9.042428377 1.141 = 100(7.925002959) = 792.5002959 2 ACCUMULATION AND AMOUNT FUNCTIONS 17 so $100 at time t= 1 grows to $792.50 at time t= 12. (b) Since a(t) = (1.1)t+ 0.041t2,we have a′(t) = (1.1)tln 1.1 + 0.082t> 0 for t≥0.This shows that a(t) is increasing for t≥0 Now, let n be a positive integer. The nthperiod of time is deﬁned to be the period of time between t= n−1 and t= n. More precisely, the period normally will consist of the time interval n−1 ≤t≤n. We deﬁne the interest earned during the nth period of time by In = A(n) −A(n−1). This is illustrated in Figure 2.2. Figure 2.2 This says that interest earned during a period of time is the diﬀerence between the amount value at the end of the period and the amount value at the beginning of the period. It should be noted that In involves the eﬀect of interest over an interval of time, whereas A(n) is an amount at a speciﬁc point in time. In general, the amount of interest earned on an original investment of $ k between time s and t is I[s,t] = A(t) −A(s) = k(a(t) −a(s)). Example 2.5 Consider the amount function A(t) = t2 + 2t+ 1. Find In in terms of n. Solution. We have In = A(n) −A(n−1) = n2 + 2n+ 1 −(n−1)2 −2(n−1) −1 = 2n+ 1 Example 2.6 Show that A(n) −A(0) = I1 + I2 + ··· + In. Interpret this result verbally. Solution. We haveA(n)−A(0) = [A(1)−A(0)]+[A(2)−A(1)]+···+[A(n−1)−A(n−2)]+[A(n)−A(n−1)] = I1 + I2 + ··· + In.Hence, A(n) = A(0) + (I1 + I2 + ··· + In) so that I1 + I2 + ··· + In is the interest earned on the capital A(0). That is, the interest earned over the concatenation of n periods is the 18 THE BASICS OF INTEREST THEORY sum of the interest earned in each of the periods separately. Note that for any 0 ≤t < nwe have A(n) −A(t) = [ A(n) −A(0)] −[A(t) −A(0)] = ∑n j=1 Ij −∑t j=1 Ij = ∑n j=t+1 Ij. That is, the interest earned between time t and time n will be the total interest from time 0 to time n diminished by the total interest earned from time 0 to time t. Example 2.7 Find the amount of interest earned between time t and time n, where t < n,if Ir = r for some positive integer r. Solution. We have A(n) −A(t) = n∑ i=t+1 Ii = n∑ i=t+1 i = n∑ i=1 i− t∑ i=1 i =n(n+ 1) 2 −t(t+ 1) 2 = 1 2(n2 + n−t2 −t) where we apply the following sum from calculus 1 + 2 +··· + n= n(n+ 1) 2 2 ACCUMULATION AND AMOUNT FUNCTIONS 19 Practice Problems Problem 2.1 An investment of $1,000 grows by a constant amount of $250 each year for ﬁve years. (a) What does the graph of A(t) look like if interest is only paid at the end of each year? (b) What does the graph of A(t) look like if interest is paid continuously and the amount function grows linearly? Problem 2.2 It is known that a(t) is of the form at2 + b. If $100 invested at time 0 accumulates to $172 at time 3, ﬁnd the accumulated value at time 10 of $100 invested at time 5. Problem 2.3 Consider the amount function A(t) = t2 + 2t+ 3. (a) Find the the corresponding accumulation function. (b) Find In in terms of n. Problem 2.4 Find the amount of interest earned between time t and time n, where t < n,if Ir = 2r for some positive integer r.Hint: Recall the following sum from Calculus: ∑n i=0 ari = a1−rn+1 1−r , r̸= 1. Problem 2.5 $100 is deposited at time t= 0 into an account whose accumulation function is a(t) = 1 + 0.03 √ t. (a) Find the amount of interest generated at time 4, i.e., between t= 0 and t= 4. (b) Find the amount of interest generated between time 1 and time 4. Problem 2.6 Suppose that the accumulation function for an account is a(t) = (1 + 0.5it).You invest $500 in this account today. Find i if the account’s value 12 years from now is $1,250. Problem 2.7 Suppose that a(t) = 0.10t2 + 1.The only investment made is $300 at time 1. Find the accumulated value of the investment at time 10. Problem 2.8 Suppose a(t) = at2 + 10b.If $X invested at time 0 accumulates to $1,000 at time 10, and to $2,000 at time 20, ﬁnd the original amount of the investment X. Problem 2.9 Show that the function f(t) = 225 −(t−10)2 cannot be used as an amount function for t> 10. 20 THE BASICS OF INTEREST THEORY Problem 2.10 For the interval 0 ≤t≤10, determine the accumulation function a(t) that corresponds to A(t) = 225 −(t−10)2. Problem 2.11 Suppose that you invest $4,000 at time 0 into an investment account with an accumulation function of a(t) = αt2 + 4β. At time 4, your investment has accumulated to $5,000. Find the accumulated value of your investment at time 10. Problem 2.12 Suppose that an accumulation function a(t) is diﬀerentiable and satisﬁes the property a(s+ t) = a(s) + a(t) −a(0) for all non-negative real numbers s and t. (a) Using the deﬁnition of derivative as a limit of a diﬀerence quotient, show that a′(t) = a′(0). (b) Show that a(t) = 1 + it where i= a(1) −a(0) = a(1) −1. Problem 2.13 Suppose that an accumulation function a(t) is diﬀerentiable and satisﬁes the property a(s+ t) = a(s) ·a(t) for all non-negative real numbers s and t. (a) Using the deﬁnition of derivative as a limit of a diﬀerence quotient, show that a′(t) = a′(0)a(t). (b) Show that a(t) = (1 + i)t where i= a(1) −a(0) = a(1) −1. Problem 2.14 Consider the accumulation functions as(t) = 1 + it and ac(t) = (1 + i)t where i> 0. Show that for 0 <t< 1 we have ac(t) ≈a1(t). That is (1 + i)t ≈1 + it. Hint: Expand (1 + i)t as a power series. Problem 2.15 Consider the amount function A(t) = A(0)(1 + i)t. Suppose that a deposit 1 at time t = 0 will increase to 2 in ayears, 2 at time 0 will increase to 3 in byears, and 3 at time 0 will increase to 15 in c years. If 6 will increase to 10 in n years, ﬁnd an expression for n in terms of a,b, and c. 2 ACCUMULATION AND AMOUNT FUNCTIONS 21 Problem 2.16 For non-negative integer n, deﬁne in = A(n) −A(n−1) A(n−1) . Show that (1 + in)−1 = A(n−1) A(n) . Problem 2.17 (a) For the accumulation function a(t) = (1 + i)t, show that a′(t) a(t) = ln (1 +i). (b) For the accumulation function a(t) = 1 + it, show that a′(t) a(t) = i 1+it. Problem 2.18 Deﬁne δt = a′(t) a(t) . Show that a(t) = e ∫t 0 δrdr. Hint: Notice that d dr(ln a(r)) = δr. Problem 2.19 Show that, for any amount function A(t), we have A(n) −A(0) = ∫ n 0 A(t)δtdt. Problem 2.20 You are given that A(t) = at2 + bt+ c, for 0 ≤t ≤2, and that A(0) = 100 ,A(1) = 110 , and A(2) = 136. Determine δ1 2 . Problem 2.21 Show that if δt = δfor all tthen in = a(n)−a(n−1) a(n−1) = eδ−1.Letting i= eδ−1,show that a(t) = (1+i)t. Problem 2.22 Suppose that a(t) = 0 .1t2 + 1. At time 0, $1,000 is invested. An additional investment of $X is made at time 6. If the total accumulated value of these two investments at time 8 is $18,000, ﬁnd X. 22 THE BASICS OF INTEREST THEORY 3 Eﬀective Interest Rate (EIR) Thus far, interest has been deﬁned by Interest = Accumulated value −Principal. This deﬁnition is not very helpful in practical situations, since we are generally interested in compar- ing diﬀerent ﬁnancial situations to ﬁgure out the most proﬁtable one. In this section we introduce the ﬁrst measure of interest which is developed using the accumulation function. Such a measure is referred to as the eﬀective rate of interest: The eﬀective rate of interest is the amount of money that one unit invested at the beginning of a period will earn during the period, with interest being paid at the end of the period. If i is the eﬀective rate of interest for the ﬁrst time period then we can write i= a(1) −a(0) = a(1) −1. where a(t) is the accumulation function. Remark 3.1 We assume that the principal remains constant during the period; that is, there is no contribution to the principal or no part of the principal is withdrawn during the period. Also, the eﬀective rate of interest is a measure in which interest is paid at the end of the period compared to discount interest rate (to be discussed in Section 8) where interest is paid at the beginning of the period. We can write i in terms of the amount function: i= a(1) −a(0) = a(1) −a(0) a(0) = A(1) −A(0) A(0) = I1 A(0). Thus, we have the following alternate deﬁnition: The eﬀective rate of interest for a period is the amount of interest earned in one period divided by the principal at the beginning of the period. One can deﬁne the eﬀective rate of interest for any period: The eﬀective rate of interest in the nth period is deﬁned by in = A(n) −A(n−1) A(n−1) = In A(n−1) where In = A(n) −A(n−1).Note that In represents the amount of growth of the function A(t) in the nth period whereas in is the rate of growth (based on the amount in the fund at the beginning of the period). Thus, the eﬀective rate of interest in is the ratio of the amount of interest earned during the period to the amount of principal invested at the beginning of the period. Note that i1 = i= a(1) −1 and for any accumulation function, it must be true that a(1) = 1 + i. 3 EFFECTIVE INTEREST RATE (EIR) 23 Example 3.1 Assume that A(t) = 100(1.1)t. Find i5. Solution. We have i5 = A(5) −A(4) A(4) = 100(1.1)5 −100(1.1)4 100(1.1)4 = 0.1 Now, using the deﬁnition of in and solving for A(n) we ﬁnd A(n) = A(n−1) + inA(n−1) = (1 + in)A(n−1). Thus, the fund at the end of the nth period is equal to the fund at the beginning of the period plus the interest earned during the period. Note that the last equation leads to A(n) = (1 + i1)(1 + i2) ···(1 + in)A(0). Example 3.2 If A(4) = 1000 and in = 0.01n, ﬁnd A(7). Solution. We have A(7) =(1 + i7)A(6) =(1 + i7)(1 + i6)A(5) =(1 + i7)(1 + i6)(1 + i5)A(4) = (1.07)(1.06)(1.05)(1000) = 1,190.91 Note that in can be expressed in terms of a(t) : in = A(n) −A(n−1) A(n−1) = A(0)a(n) −A(0)a(n−1) A(0)a(n−1) = a(n) −a(n−1) a(n−1) . Example 3.3 Suppose that a(n) = 1 + in, n≥1. Show that in is decreasing as a function of n. Solution. We have in = a(n) −a(n−1) a(n−1) = [1 + in−(1 + i(n−1))] 1 + i(n−1) = i 1 + i(n−1). Since in+1 −in = i 1 + in − i 1 + i(n−1) = − i2 (1 + in)(1 + i(n−1)) <0 we conclude that as n increases in decreases 24 THE BASICS OF INTEREST THEORY Example 3.4 Show that if in = i for all n≥1 then a(n) = (1 + i)n. Solution. We have a(n) = A(n) A(0) = (1 + i1)(1 + i2) ···(1 + in) = (1 + i)n Remark 3.2 In all of the above discussion the interest rate is associated with one complete period; this will be contrasted later with rates −called “nominal”−that are stated for one period, but need to be applied to fractional parts of the period. Most loans and ﬁnancial products are stated with nominal rates such as a nominal rate that is compounded daily, or monthly, or semiannually, etc. To compare these loans, one compare their equivalent eﬀective interest rates. Nominal rates will be discussed in more details in Section 9. We pointed out in the previous section that a decreasing accumulated function leads to negative interest rate. We illustrate this in the next example. Example 3.5 You buy a house for $100,000. A year later you sell it for $80,000. What is the eﬀective rate of return on your investment? Solution. The eﬀective rate of return is i= 80,000 −100,000 100,000 = −20% which indicates a 20% loss of the original value of the house 3 EFFECTIVE INTEREST RATE (EIR) 25 Practice Problems Problem 3.1 Consider the accumulation function a(t) = t2 + t+ 1. (a) Find the eﬀective interest rate i. (b) Find in. (c) Show that in is decreasing. Problem 3.2 If $100 is deposited into an account, whose accumulation function is a(t) = 1 + 0.03 √ t, at time 0, ﬁnd the eﬀective rate for the ﬁrst period(between time 0 and time 1) and second period (between time 1 and time 2). Problem 3.3 Assume that A(t) = 100 + 5t. (a) Find i5. (b) Find i10. Problem 3.4 Assume that A(t) = 225 −(t−10)2, 0 ≤t≤10. Find i6. Problem 3.5 An initial deposit of 500 accumulates to 520 at the end of one year and 550 at the end of the second year. Find i1 and i2. Problem 3.6 A fund is earning 5% simple interest (See Problem 1.5). Calculate the eﬀective interest rate in the 6th year. Problem 3.7 Given A(5) = 2500 and i= 0.05. (a) What is A(7) assuming simple interest (See Problem 1.5)? (b) What is a(10)? Answer: (a) 2700 (b) 1.5 Problem 3.8 If A(4) = 1200,A(n) = 1800, and i= 0.06. (a) What is A(0) assuming simple interest? (b) What is n? 26 THE BASICS OF INTEREST THEORY Problem 3.9 John wants to have $800. He may obtain it by promising to pay $900 at the end of one year; or he may borrow $1,000 and repay $1,120 at the end of the year. If he can invest any balance over $800 at 10% for the year, which should he choose? Problem 3.10 Given A(0) = $1,500 and A(15) = $2,700. What is i assuming simple interest? Problem 3.11 You invest $1,000 now, at an annual simple interest rate of 6%. What is the eﬀective rate of interest in the ﬁfth year of your investment? Problem 3.12 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 3.13 Suppose a(t) = 1.12t −0.05 √ t. (a) How much interest will be earned during the 5th year on an initial investment of $12? (b) What is the eﬀective annual interest rate during the 5th year? Problem 3.14 Assume that A(t) = 100 + 5t, where t is in years. (a) Find the principal. (b) How much is the investment worth after 5 years? (c) How much is earned on this investment during the 5th year? Problem 3.15 If $64 grows to $128 in four years at a constant eﬀective annual interest rate, how much will $10,000 grow to in three years at the same rate of interest? Problem 3.16 Suppose that in = 5% for all n≥1. How long will it take an investment to triple in value? 3 EFFECTIVE INTEREST RATE (EIR) 27 Problem 3.17 You have $1000 that you want to deposit in a savings account. Bank Acomputes the amount value of your investment using the amount function A1(t) = 1 + 0.049twhereas Bank B uses the amount function A2(t) = (1.4)12t. Where should you put your money? Problem 3.18 Consider the amount function A(t) = 12(1.01)4t. (a) Find the principal. (b) Find the eﬀective annual interest rate. Problem 3.19 Given i5 = 0.1 and A(4) = 146.41. Find A(5). Problem 3.20 Given i5 = 0.1 and I5 = 14.641. Find A(4). Problem 3.21 Suppose that in = 0.01n for n≥1. Show that In = 0.01n(1.01)(1.02) ···[1 + 0.01(n−1)]A(0). Problem 3.22 If A(3) = 100 and in = 0.02n, ﬁnd A(6). 28 THE BASICS OF INTEREST THEORY 4 Linear Accumulation Functions: Simple Interest Accumulation functions of two common types of interest are discussed next. The accumulation func- tion of “simple” interest is covered in this section and the accumulation function of “compound” interest is discussed in Section 6. Consider an investment of 1 such that the interest earned in each period is constant and equals to i. Then, at the end of the ﬁrst period, the accumulated value is a(1) = 1 +i,at the end of the second period it is a(2) = 1 + 2i and at the end of the nth period it is a(n) = 1 + in, n ≥0. Thus, the accumulation function is a linear function. The accruing of interest according to this function is called simple interest. Note that the eﬀective rate of interest i= a(1)−1 is also called the simple interest rate. We next show that for a simple interest rate i, the eﬀective interest rate in is decreasing . Indeed, in = a(n) −a(n−1) a(n−1) = [1 + in−(1 + i(n−1))] 1 + i(n−1) = i 1 + i(n−1), n≥1 and in+1 −in = i 1 + in − i 1 + i(n−1) = − i2 (1 + in)(1 + i(n−1)) <0. Thus, even though the rate of simple interest is constant over each period of time, the eﬀective rate of interest per period is not constant −it is decreasing from each period to the next and converges to 0 in the long run. Because of this fact, simple interest is less favorable to the investor as the number of periods increases. Example 4.1 A fund is earning 5% simple interest. Calculate the eﬀective interest rate in the 6th year. Solution. The eﬀective interest rate in the 6th year is i6 which is given by i6 = i 1 + i(n−1) = 0.05 1 + 0.05(5) = 4% Remark 4.1 For simple interest, the absolute amount of interest earned in each time interval, i.e., In = a(n) − a(n−1) is constant whereas in is decreasing in value as n increases; in Section 6 we will see that under compound interest, it is the relative amount of interest that is constant, i.e. in = a(n)−a(n−1) a(n−1) . 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 29 The accumulation function for simple interest has been deﬁned for integral values of n ≥0. In order for this function to have the graph shown in Figure 2.1(2), we need to extend a(n) for nonintegral values of n. This is equivalent to crediting interest proportionally over any fraction of a period. If interest accrued only for completed periods with no credit for fractional periods, then the accumulation function becomes a step function as illustrated in Figure 2.1(4). Unless stated otherwise, it will be assumed that interest is allowed to accrue over fractional periods under simple interest. In order to deﬁne a(t) for real numbers t≥0 we will redeﬁne the rate of simple interest in such a way that the previous deﬁnition is a consequence of this general assumption. The general assumption states the following: Under simple interest, the interest earned by an initial investment of $1 in all time periods of length t+ s is equal to the sum of the interest earned for periods of lengths t and s. Symbolically, a(t+ s) −a(0) = [a(t) −a(0)] + [a(s) −a(0)] or a(t+ s) = a(t) + a(s) −a(0) (4.1) for all non-negative real numbers t and s. Note that the deﬁnition assumes the rule is to hold for periods of any non-negative length, not just of integer length. Are simple interest accumulation functions the only ones which preserve property (4.1)? Suppose that a(t) is a diﬀerentiable function satisfying property (4.1). Then a′(t) = lim s→0 a(t+ s) −a(t) s = lim s→0 a(t) + a(s) −a(0) −a(t) s = lim s→0 a(s) −a(0) s =a′(0), a constant Thus the time derivative of a(t) is shown to be constant. We know from elementary calculus that a(t) must have the form a(t) = a′(0)t+ C where C is a constant; and we can determine that constant by assigning to t the particular value 0, so that C = a(0) = 1. 30 THE BASICS OF INTEREST THEORY Thus, a(t) = 1 + a′(0)t. Letting t= 1 and deﬁning i1 = i= a(1) −a(0) we can write a(t) = 1 + it, t ≥0. Consequently, simple interest accumulation functions are the only ones which preserve property (4.1). It is important to notice that the above derivation does not depend ontbeing a nonnegative integer, and is valid for all nonnegative real numbers t. Example 4.2 You invest $100 at time 0, at an annual simple interest rate of 10%. Find the accumulated value after 6 months. Solution. The accumulated function for simple interest is a continuous function. Thus, A(0.5) = 100[1 + 0.1(0.5)] = $105 Remark 4.2 Simple interest is in general inconvenient for use by banks. For if such interest is paid by a bank, then at the end of each period, depositors will withdraw the interest earned and the original deposit and immediately redeposit the sum into a new account with a larger deposit. This leads to a higher interest earning for the next investment year. We illustrate this in the next example. Example 4.3 Consider the following investments by John and Peter. John deposits $100 into a savings account paying 6% simple interest for 2 years. Peter deposits $100 now with the same bank and at the same simple interest rate. At the end of the year, he withdraws his balance and closes his account. He then reinvests the total money in a new savings account oﬀering the same rate. Who has the greater accumulated value at the end of two years? Solution. John’s accumulated value at the end of two years is 100(1 + 0.06 ×2) = $112. Peter’s accumulated value at the end of two years is 100(1 + 0.06)2 = $112.36. 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 31 Thus, Peter has a greater accumulated value at the end of two years Simple interest is very useful for approximating compound interest, a concept to be discussed in Section 6, for a short time period such as a fraction of a year. To be more speciﬁc, we will see that the accumulation function for compound interest i is given by the formula a(t) = (1 + i)t. Using the binomial theorem we can write the series expansion of a(t) obtaining (1 + i)t = 1 + it+ t(t−1) 2! i2 + t(t−1)(t−2) 3! i3 + ··· . Thus, for 0 <t< 1 we can write the approximation (1 + i)t ≈1 + it. Example 4.4 $10,000 is invested for four months at 12.6% compounded annually, that is A(t) = 10000(1 + 0.126)t, where interest is computed using a quadratic to approximate an exact calculation. Find the accumulated value. Solution. We want to estimate A(1/3) = 10000(1 + 0.126) 1 3 using the ﬁrst three terms of the series expansion of (1 + i)t discussed above. That is, A(1/3) =10000(1.126) 1 3 ≈10000 ( 1 + 1 3 ×0.126 + (1 3 )(−2 3 ) 2! (0.126)2 ) = $10,402.36 Remark 4.3 Under simple interest, what is the accumulated value at time t > sof 1 deposited at time s? The SOA/CAS approach is diﬀerent than the approach discussed in Section 2 right after Remark 2.1. According to the SOA/CAS, “simple interest” is generally understood to mean that the linear function starts all over again from the date of each deposit or withdrawal. We illustrate this approach in the next example Example 4.5 Suppose you make a deposit of $100 at time t = 0. A year later, you make a withdrawal of $50. Assume annual simple interest rate of 10%, what is the accumulated value at time t= 2 years? Solution. With the SOA/CAS recommended approach for simple interest, the answer is 100(1 + 0.1 ×2) −50(1 + 0.1 ×1) = $65 32 THE BASICS OF INTEREST THEORY Practice Problems Problem 4.1 You invest $100 at time 0, at an annual simple interest rate of 9%. (a) Find the accumulated value at the end of the ﬁfth year. (b) How much interest do you earn in the ﬁfth year? Problem 4.2 At what annual rate of simple interest will $500 accumulate to $615 in 2 1 2 years? Problem 4.3 In how many years will $500 accumulate to $630 at 7.8% annual simple interest? Problem 4.4 What principal will earn interest of 100 in 7 years at a simple interest rate of 6%? Problem 4.5 What simple interest rate is necessary for $10,000 to earn $100 interest in 15 months? Problem 4.6 At a certain rate of simple interest $1,000 will accumulate to $1110 after a certain period of time. Find the accumulated value of $500 at a rate of simple interest three fourths as great over twice as long a period of time. Problem 4.7 At time 0, you invest some money into an account earning 5.75% simple interest. How many years will it take to double your money? Problem 4.8 You invest $1,000 now, at an annual simple interest rate of 6%. What is the eﬀective rate of interest in the ﬁfth year of your investment? Problem 4.9 Suppose that the accumulation function for an account is a(t) = 1 + 3it.At time 0, you invest $100 in this account. If the value in the account at time 10 is $420, what is i? Problem 4.10 You have $260 in a bank savings account that earns simple interest. You make no subsequent deposits in the account for the next four years, after which you plan to withdraw the entire account balance and buy the latest version of the iPod at a cost of $299. Find the minimum rate of simple interest that the bank must oﬀer so that you will be sure to have enough money to make the purchase in four years. 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 33 Problem 4.11 The total amount of a loan to which interest has been added is $20,000. The term of the loan was four and one-half years. If money accumulated at simple interest at a rate of 6%, what was the amount of the loan? Problem 4.12 If ik is the rate of simple interest for period k, where k = 1 ,2,··· ,n, show that a(n) −a(0) = i1 + i2 + ··· + in. Be aware that in is not the eﬀective interest rate of the nth period as deﬁned in the section! Problem 4.13 A fund is earning 5% simple interest. The amount in the fund at the end of the 5th year is $10,000. Calculate the amount in the fund at the end of 7 years. Problem 4.14 Simple interest of i = 4% is being credited to a fund. The accumulated value at t = n−1 is a(n−1).The accumulated value at t= nis a(n) = 1 + 0.04n.Find nso that the accumulated value of investing a(n−1) for one period with an eﬀective interest rate of 2.5% is the same as a(n). Problem 4.15 A deposit is made on January 1, 2004. The investment earns 6% simple interest. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 4.16 Consider an investment with nonzero interest rate i. If i5 is equal to i10, show that interest is not computed using simple interest. Problem 4.17 Smith has just ﬁled his income tax return and is expecting to receive, in 60 days, a refund check of 1000. (a) The tax service that helped him ﬁll out his return oﬀers to buy Smith’s refund check for 850. What annual simple interest rate is implied? (b) Smith negotiates and sells his refund check for 900. What annual simple interest rate does this correspond to? (c) Smith deposits the 900 in an account which earns simple interest at annual rate 9%. How many days would it take from the time of his initial deposit of 900 for the account to reach 1000? Problem 4.18 Let i be a simple interest rate and suppose that i6 = 0.04. Calculate i. 34 THE BASICS OF INTEREST THEORY Problem 4.19 A fund is earning 5% simple interest. If in = 0.04, calculate n. Problem 4.20 Suppose that an account earns simple interest with annual interest rate of i. If an investment of k is made at time s years, what is the accumulated value at time t>s years? Note that the answer is diﬀerent from ka(t) a(s) . Problem 4.21 Suppose A(5) = $2,500 and i= 0.05. (a) What is A(7) assuming simple interest? (b) What is a(10) assuming simple interest? Problem 4.22 If A(4) = $1,200 and A(n) = $1,800, (a) what is A(0), assuming a simple interest of 6%? (b) what is n if i= 0.06 assuming simple interest? Problem 4.23 What is A(15), if A(0) = $1,100, simple interest is assumed and in = 0.01n? 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 35 5 Date Conventions Under Simple Interest In the simple interest problems encountered thus far, the length of the investment has been an integral number of years. What happens if the time is given in days. In this section we discuss three techniques for counting the number of days in a period of investment or between two dates. In all three methods, time= # of days between two dates # of days in a year . In what follows, it is assumed, unless stated otherwise, that in counting days interest is not credited for both the starting date and the ending date, but for only one of these dates. Exact Simple Interest: The “actual/actual” method is to use the exact number of days for the period of investment and to use 365 days in a nonleap year and 366 for a leap year (a year divisible by 4). Simple interest computed with this method is called exact simple interest. For this method, it is important to know the number of days in each month. In counting days between two dates, the last, but not the ﬁrst, date is included. Example 5.1 Suppose that $2,500 is deposited on March 8 and withdrawn on October 3 of the same year, and that the interest rate is 5%. Find the amount of interest earned, if it is computed using exact simple interest. Assume non leap year. Solution. From March 8 (not included) to October 3 (included) there are 23+30+31+30+31+31+30+3 = 209 days. Thus, the amount of interest earned using exact simple interest is 2500(0 .05) ·209 365 = $71.58 Ordinary Simple Interest: This method is also known as “30/360”. The 30/360 day counting scheme was invented in the days before computers to make the computations easier. The premise is that for the purposes of computation, all months have 30 days, and all years have 12 ×30 = 360 days. Simple interest computed with this method is called ordinary simple interest. The Public Securities Association (PSA) publishes the following rules for calculating the number of days between any two dates from M1/D1/Y1 to M2/D2/Y2: •If D1 (resp. D2) is 31, change D1 (resp. D2)to 30. •If M1 (resp. M2) is 2, and D1 (resp. D2) is 28 (in a non-leap year) or 29, then change D1 (resp. D2) to 30. Then the number of days, N is: N = 360(Y2 −Y1) + 30(M2 −M1) + (D2 −D1). 36 THE BASICS OF INTEREST THEORY For exampe, the number of days from February 25 to March 5 of the same year is 10 days. Like the exact simple interest, the ending date is counted and not the starting date. Example 5.2 Jack borrows 1,000 from the bank on January 28, 1996 at a rate of 15% simple interest per year. How much does he owe on March 5, 1996? Use ordinary simple interest. Solution. The amount owed at time t is A(t) = 1000(1 + 0 .15t). Using ordinary simple interest with Y1 = Y2 = 1996,M1 = 1,M2 = 3, and D1 = 28 and D2 = 5 we ﬁnd t = 37 360 and the amount owed on March 5, 1996 is 1,000 ( 1 + 0.15 × 37 360 ) = $1,015.42 Banker’s Rule: This method is also known as “actual/360”. This method uses the exact number of days for the period of investment and that the calendar year has 360 days. Simple interest computed with this method is called Banker’s rule. The number of days between two dates is found in the same way as for exact simple interest. In this method, we also count the last day but not the ﬁrst day. Example 5.3 Jack borrows 1,000 from the bank on January 1, 1996 at a rate of 15% simple interest per year. How much does he owe on January 17, 1996?Use Banker’s rule. Solution. Jack owes 1000 ( 1 + 0.15 × 16 360 ) = $1,006.67 Example 5.4 If an investment was made on the date the United States entered World War II, i.e., December 7, 1941, and was terminated at the end of the war on August 8, 1945, for how many days was the money invested under: 1. the “actual/actual” basis? 2. the “30/360” basis? Solution. 1. From December 7, 1941(not included) to December 31, 1941 (included) there were 24 days. From January 1, 1942 to December 31, 1944 ( including a leap year) there were 3(365) + 1 = 1096 days. 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 37 From January 1, 1945 to August 8, 1945(included) the numbers of days is 31 + 28 + 31 + 30 + 31 + 30 + 31 + 8 = 220 days. The total number of days is 24 + 1096 + 220 = 1340. 2. We have 360(1945 −1941) + 30(8−12) + (8−7) = 1321 . Remark 5.1 If the time is given in months, reduce it to a fraction of a year on the basis of 12 months to the year, without changing to days. Example 5.5 A merchant is oﬀered $50 discount for cash payment of a $1200 bill due in two months. If he pays cash, at what rate may he consider his money to be earning interest in the next two months? Solution. The merchant would pay now $1150 in place of $1200 in two months. To ﬁnd the interest rate under which $1150 is the present value of $1200, due in two months, use the formula I = Pit which by substitution becomes 50 = 1150i 6 . Solving for i we ﬁnd i= 26.087% We end this section by pointing out that the methods discussed above do not only apply for simple interest rate problems but also to compound interest rate problems. Compound interest rates are introduced in the next section. Unless otherwise stated, in later sections we will assume always the actual/actual method is in use. 38 THE BASICS OF INTEREST THEORY Practice Problems Problem 5.1 Find the amount of interest that $2,000 deposited on June 17 will earn, if the money is withdrawn on September 10 in the same year and if the simple rate of interest is 8% using (a) exact simple interest, (b) ordinary simple interest, and (c) Banker’s rule. Assume non-leap year. Problem 5.2 A sum of 10,000 is invested for the months of July and August at 6% simple interest. Find the amount of interest earned: 1. Assuming exact simple interest (Assume non-leap year). 2. Assuming ordinary simple interest. 3. Assuming the Banker’s Rule. Problem 5.3 Show that the Banker’s Rule is always more favorable to the lender than is exact simple interest. Problem 5.4 (a) Show that the Banker’s Rule is usually more favorable to the lender than is ordinary simple interest. (b) Give an example in (a) for which the opposite relationship holds. Problem 5.5 Suppose that $2,500 is deposited on March 8 and withdrawn on October 3 of the same year, and that the interest rate is 5%. Find the amount of interest earned, if it is computed using (a) exact simple interest (Assume non-leap year), (b) ordinary simple interest, (c) the Banker’s Rule. Problem 5.6 The sum of $ 5,000 is invested for the months of April, May, and June at 7% simple interest. Find the amount of interest earned (a) assuming exact simple interest in a non-leap year; (b) assuming exact simple interest in a leap year (with 366 days); (c) assuming ordinary simple interest; (d) assuming the Banker’s Rule. Problem 5.7 Fund A calculates interest using exact simple interest (actual/actual). Fund B calculates interest 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 39 using ordinary simple interest (30/360). Fund C calculates interest using the Banker’s Rule (ac- tual/360). All Funds earn 5% simple interest and have the same amount of money deposited on January 1, 2005. Order the Funds based on the amount in the funds on March 1, 2005 from smallest to largest. Problem 5.8 Suppose you lend $60 to your sister on Sept 14, at an annual rate of simple interest of 10%, to be repaid on Dec 25. How much does she have to pay you back? Use actual/360 time measurement. Problem 5.9 John borrows $60 from Eddie. If he repays Eddie $63 after 5 weeks, what simple interest has John paid? Use actual/actual for days counting. Assume non-leap year. Problem 5.10 Henry invests 1000 on January 15 in an account earning simple interest at an annual eﬀective rate of 10%. On November 25 of the same year, Henry withdraws all his money. How much money will Henry withdraw if the bank counts days: (a) Using exact simple interest (ignoring February 29th) (b) Using ordinary simple interest (c) Using Banker’s Rule Problem 5.11 Let Ie denote the interest earned using the exact simple interest method and Ib the interest earned using the Banker’s rule. Find the ratio Ie Ib . Problem 5.12 Suppose that the interest earned using the Banker’s rule is $40.58. What is the interest earned using the exact simple interest method? Problem 5.13 Show that Ib = Ie + Ie 72 and Ie = Ib −Ib 73 . Thus, the Banker’s rule is more favorable to the investor than the actual/actual rule. Problem 5.14 On January 1, 2000, you invested $1,000. Your investment grows to $1,400 by December 31, 2007. What was the exact simple interest rate at which you invested? Problem 5.15 A 3% discount is oﬀered for cash payment of a $2500 bill, due at the end of 90 days. At what rate of simple interest earned over the 90 days is cash payment is made? Use the Banker’s rule. 40 THE BASICS OF INTEREST THEORY Problem 5.16 The terms of payment of a certain debt are: net cash in 90 days or 2% discount for cash in 30 days. At what rate of simple interest is earned if the discount is taken advantage of? 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 41 6 Exponential Accumulation Functions: Compound Interest Simple interest has the property that the interest earned is not invested to earn additional interest. In contrast, compound interest has the property that the interest earned at the end of one period is automatically invested in the next period to earn additional interest. We next ﬁnd the accumulation function for compound interest. Starting with an investment of 1 and with compound interest rate iper period. At the end of the ﬁrst period, the accumulated value is 1 + i. At the end of the second period, the accumulated value is (1 + i) + i(1 + i) = (1 + i)2. Continuing this way, we ﬁnd that the accumulated value after tperiods is given by the exponential function a(t) = (1 + i)t, for integral t ≥0. Interest accruing according to this function is called compound interest. We call i the rate of compound interest. From the above accumulation function, we can write in = a(n) −a(n−1) a(n−1) = (1 + i)n −(1 + i)n−1 (1 + i)n−1 = i. Thus, the eﬀective rate of interest for compound interest is constant. The accumulation function for compound interest has been deﬁned for nonnegative integers. In order to extend the domain to fractional periods we note that the compound interest accumulation function a(t) = (1 + i)t satisﬁes the property (1 + i)t+s = (1 + i)t ·(1 + i)s. Thus, under compound interest we will require the accumulation function to satisfy the property a(t+ s) = a(t) ·a(s), t,s ≥0. (6.1) This formula says that under compound interest the amount of interest earned by an initial invest- ment of 1 over t+speriods is equal to the amount of interest earned if the investment is terminated at the end of t periods and the accumulated value at that point is immediately reinvested for an additional s periods. Are compound interest accumulation functions the only ones which preserve property (6.1)? Assum- ing a(t) is diﬀerentiable and satisfying the above property, then from the deﬁnition of the derivative 42 THE BASICS OF INTEREST THEORY we have a′(t) = lim s→0 a(t+ s) −a(t) s = lim s→0 a(t)a(s) −a(t) s =a(t) lim s→0 a(s) −a(0) s =a(t)a′(0) Hence, a′(t) a(t) = (ln a(t))′= a′(0) a constant for all t≥0. Hence, ln a(t) = a′(0)t+ C Since a(0) = 1, we obtain C = 0 so ln a(t) = a′(0)t. Setting t= 1 and recalling that a(1) −a(0) = i1 = i yields a′(0) = ln (1 +i) so a(t) = (1 + i)t and this is valid for all t≥0. We thus see the graphical distinction between simple and compound interest: the graph of an accumulation function under simple interest is a straight line −a linear function; the graph of an accumulation function under compound interest is an exponential function. We next present a comparison result between simple interest and compound interest. Theorem 6.1 Let 0 <i< 1. We have (a) (1 + i)t <1 + it for 0 <t< 1, (b) (1 + i)t = 1 + it for t= 0 or t= 1, (c) (1 + i)t >1 + it for t> 1. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 43 Proof. (a) Suppose 0 < t <1. Let f(i) = (1 + i)t −1 −it. Then f(0) = 0 and f′(i) = t(1 + i)t−1 −t. Since i> 0, it follows that 1 + i> 1, and since t< 1 we have t−1 <0. Hence, (1 + i)t−1 <1 and therefore f′(i) = t(1 + i)t−1 −t< 0 for 0 <t< 1. It follows that f(i) <0 for 0 <t< 1. (b) Follows by substitution. (c) Suppose t >1. Let g(i) = 1 + it−(1 + i)t. Then g(0) = 0 and g′(i) = t−t(1 + i)t−1. Since i >0, we have 1 + i >1. Since t >1 we have t−1 > 0. Hence (1 + i)t−1 > 1 and therefore g′(i) = t−t(1 + i)t−1 <t −t= 0. We conclude that g(i) <0. This establishes part (c) Thus, simple and compound interest produce the same result over one measurement period. Com- pound interest produces a larger return than simple interest for periods greater than 1 and smaller return for periods smaller than 1. See Figure 6.1. Figure 6.1 It is worth observing the following (1) With simple interest, the absolute amount of growth is constant, that is, for a ﬁxed s the diﬀerence a(t+ s) −a(t) = a(s) −1 does not depend on t. (2) With compound interest, the relative rate of growth is constant, that is, for a ﬁxed s the ratio [a(t+s)−a(t)] a(t) = a(s) −1 does not depend on t. Example 6.1 It is known that $600 invested for two years will earn $264 in interest. Find the accumulated value of $2,000 invested at the same rate of compound interest for three years. Solution. We are told that 600(1 + i)2 = 600 + 264 = 864 . Thus, (1 + i)2 = 1.44 and solving for i we ﬁnd i = 0 .2. Thus, the accumulated value of investing $2,000 for three years at the rate i = 20% is 2,000(1 + 0.2)3 = $3,456 44 THE BASICS OF INTEREST THEORY Example 6.2 At an annual compound interest rate of 5%, how long will it take you to triple your money? (Provide an answer in years, to three decimal places.) Solution. We must solve the equation (1 + 0.05)t = 3. Thus, t= ln 3 ln 1.05 ≈22.517 Example 6.3 At a certain rate of compound interest, 1 will increase to 2 in ayears, 2 will increase to 3 in byears, and 3 will increase to 15 in c years. If 6 will increase to 10 in n years, ﬁnd an expression for n in terms of a,b, and c. Solution. If the common rate is i, the hypotheses are that 1(1 + i)a =2 →ln 2 = aln (1 +i) 2(1 + i)b =3 →ln 3 2 = bln (1 +i) 3(1 + i)c =15 →ln 5 = cln (1 +i) 6(1 + i)n =10 →ln 5 3 = nln (1 +i) But ln 5 3 = ln 5−ln 3 = ln 5−(ln 2 + ln 1.5). Hence, nln (1 +i) = cln (1 +i) −aln (1 +i) −bln (1 +i) = (c−a−b) ln (1 +i) and this implies n= c−a−b We conclude this section by noting that the three counting days techniques discussed in Section 5 for simple interest applies as well for compound interest. Example 6.4 Christina invests 1000 on April 1 in an account earning compound interest at an annual eﬀective rate of 6%. On June 15 of the same year, Christina withdraws all her money. Assume non-leap year, how much money will Christina withdraw if the bank counts days: (a) Using actual/actual method. (b) Using 30/360 method. (c) Using actual/360 method. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 45 Solution. (a) The number of days is 29 + 31 + 15 = 75. Thus, the amount of money withdrawn is 1000(1 + 0.06) 75 365 = $1,012.05. (b) Using the 30/360, we ﬁnd 1000(1 + 0.06) 74 360 = $1,012.05. (c) Using the actual/360 method, we ﬁnd 1000(1 + 0.06) 75 360 = $1,012.21 46 THE BASICS OF INTEREST THEORY Practice Problems Problem 6.1 If $4,000 is invested at an annual rate of 6.0% compounded annually, what will be the ﬁnal value of the investment after 10 years? Problem 6.2 Jack has deposited $1,000 into a savings account. He wants to withdraw it when it has grown to $2,000. If the interest rate is 4% annual interest compounded annually, how long will he have to wait? Problem 6.3 At a certain rate of compound interest, $250 deposited on July 1, 2005 has to accumulate to $275 on January 1, 2006. Assuming the interest rate does not change and there are no subsequent deposits, ﬁnd the account balance on January 1, 2008. Problem 6.4 You want to triple your money in 25 years. What is the annual compound interest rate necessary to achieve this? Problem 6.5 You invest some money in an account earning 6% annual compound interest. How long will it take to quadruple your account balance? (Express your answer in years to two decimal places.) Problem 6.6 An amount of money is invested for one year at a rate of interest of 3% per quarter. Let D(k) be the diﬀerence between the amount of interest earned on a compound interest basis, and on a simple interest basis for quarter k, where k= 1,2,3,4. Find the ratio of D(4) to D(3). Problem 6.7 Show that the ratio of the accumulated value of 1 invested at rateifor nperiods, to the accumulated value of 1 invested at ratejfor nperiods, where i>j, is equal to the accumulated value of 1 invested for n periods at rate r.Find an expression for r as a function of i and j. Problem 6.8 At a certain rate of compound interest an investment of $1,000 will grow to $1,500 at the end of 12 years. Determine its value at the end of 5 years. Problem 6.9 At a certain rate of compound interest an investment of $1,000 will grow to $1,500 at the end of 12 years. Determine precisely when its value is exactly $1,200. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 47 Problem 6.10 ‡ Bruce and Robbie each open up new bank accounts at time 0. Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns the same annual eﬀective interest rate. The amount of interest earned in Bruce’s account during the 11th year is equal to X. The amount of interest earned in Robbie’s account during the 17th year is also equal to X. Calculate X. Problem 6.11 Given A(5) = $7,500 and A(11) = $9,000. What is A(0) assuming compound interest? Problem 6.12 If $200 grows to $500 overnyears, what will $700 grow to over 3nyears? Assuming same compound annual interest rate. Problem 6.13 Suppose that your sister repays you $62 two months after she borrows $60 from you. What eﬀective annual rate of interest have you earned on the loan? Assume compound interest. Problem 6.14 An investor puts 100 into Fund X and 100 into Fund Y. Fund Y earns compound interest at the annual rate of j >0,and Fund X earns simple interest at the annual rate of 1 .05j.At the end of 2 years, the amount in Fund Y is equal to the amount in Fund X. Calculate the amount in Fund Y at the end of 5 years. Problem 6.15 Fund A is invested at an eﬀective annual interest rate of 3%. Fund B is invested at an eﬀective annual interest rate of 2.5%. At the end of 20 years, the total in the two funds is 10,000. At the end of 31 years, the amount in Fund A is twice the amount in Fund B. Calculate the total in the two funds at the end of 10 years. Problem 6.16 Carl puts 10,000 into a bank account that pays an annual eﬀective interest rate of 4% for ten years. If a withdrawal is made during the ﬁrst ﬁve and one-half years, a penalty of 5% of the withdrawal amount is made. Carl withdraws K at the end of each of years 4, 5, 6, 7. The balance in the account at the end of year 10 is 10,000. Calculate K. Problem 6.17 ‡ Joe deposits 10 today and another 30 in ﬁve years into a fund paying simple interest of 11% per 48 THE BASICS OF INTEREST THEORY year. Tina will make the same two deposits, but the 10 will be deposited nyears from today and the 30 will be deposited 2 n years from today. Tina’s deposits earn an annual eﬀective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits. Calculate n. Problem 6.18 Complete the following table. Simple interest Compound interest a(t) = a(0) = a(1) = Period during which a(t) is greater in = Problem 6.19 On January 1, 2000, you invested $1,000. Your investment grows to $1,400 by December 31, 2007. What was the compound interest rate at which you invested? Problem 6.20 A certiﬁcate of deposit or CD is a ﬁnancial product commonly oﬀered to consumers by banks and credit unions. A CD has a speciﬁc, ﬁxed term (often three months, six months, or one to ﬁve years), and, usually, a ﬁxed interest rate. It is intended that the CD be held until maturity, at which time the money may be withdrawn together with the accrued interest. However, a type a penalty is imposed for early withdrawal. 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. 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. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 49 Problem 6.21 You invest $10,000 at time 0 into each of two accounts. Account A earns interest at an annual simple interest rate of 8%; Account B earns interest at an eﬀective annual compound interest rate of 6%. What is the diﬀerence in the amount of interest earned during the 5th year in these two accounts? Problem 6.22 On April 1, 2013, you will need $10,000. Assuming a 6% annual compound eﬀective rate of interest, what would you have to invest on October 1, 2008, in order for you to fulﬁll that need? Problem 6.23 On January 1, 2009, you invest $500 into an account earning an 8% annual compound eﬀective interest rate. On January 1, 2011, you deposit an additional $1,000 into the account. What is the accumulated value of your account on January 1, 2014? Problem 6.24 Let i be a compound eﬀective interest rate. Show that In = iA(n−1). Problem 6.25 Suppose that A(1) = $110,A(2) = $121, and A(10) = $259.37. (a) Find the eﬀective rate of interest. (b) Is the interest a simple interest or a compound interest? (c) Find I5. Problem 6.26 If A(5) = $1,500 and A(n) = $2,010.14, what is n if i= 0.05 assuming compound interest? Problem 6.27 How much interest is earned in the 9th year by $200 invested today under compound interest with i= 0.07? 50 THE BASICS OF INTEREST THEORY 7 Present Value and Discount Functions In this section we discuss the question of determining the today’s or present value of some amount in the past or in the future. Unless stated otherwise, we assume in this section that we are in a compound interest situation, where a(t) = (1 + i)t. Why would you be interested in present values? Suppose that in your personal ﬁnancial life, you need to save or invest money that will grow to a speciﬁed amount in a speciﬁed number of years. For example, you may want to start saving now to buy a car for $20,000 in two years, or to make a down payment of $25,000 on a home in 3 years, or to pay a year’s college expenses of $30,000 in 8 years. All of these questions involve ﬁnding the present value (PV), i.e., the amount invested today that will grow to the desired amount in the desired time. To start with, suppose we want to know what $1 a period ago, invested at compound interest rate i per period, worths today. If X is the accumulated value then we must have 1(1 + i) = X. Thus, $1 a period ago worths 1 + i dollars now. We call 1 + i the accumulation factor. Similarly, $1 a period from now invested at the rate i worths ν = 1 1+i today. We call ν the discount factor since it discounts the value of an investment at the end of a period to its value at the beginning of the period. The above discussion can be generalized to more than one period. For example, $1 investedtperiods ago worths (1 + i)t today and $1 invested t periods from now is worth 1 (1+i)t today. We call 1 (1+i)t a discount function. It represents the amount that needs to be invested today at interest rate i per period to yield an amount of $1 at the end of t time periods. This function can be expressed in terms of the accumulation function a(t). Indeed, since [ a(t)]−1a(t) = 1, the discount function is [a(t)]−1 = 1 (1+i)t = νt. In a sense, accumulation and discounting are opposite processes. The term (1 + i)t is said to be the accumulated value of $1 at the end of ttime periods. The term νt is said to be the present value or discounted(back) value of $1 to be paid at the end of t periods. Example 7.1 What is the present value of $8,000 to be paid at the end of three years if the interest rate is 11% compounded annually? Solution. Let FV stands for the future value and PV for the present value. We want to ﬁnd PV. We have FV = PV (1 + i)3 or PV = FV (1 + i)−3. Substituting into this equation we ﬁnd PV = 8000(1.11)−3 ≈$5,849.53 Example 7.2 Show that the current value of a payment of 1 made n >1 periods ago and a payment of 1 to be made n periods in the future is greater than 2, if i> 0. 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 51 Solution. We have (1 + i)n + (1 + i)−n = ( (1 + i) n 2 −(1 + i)−n 2 )2 + 2 ≥2. Equality holds if and only if (1 + i) n 2 = (1 + i)−n 2 which is equivalent to (1 + i)n = 1. This happens when either i= 0 or n= 0. Hence, (1 + i)n + (1 +i)−n >2 since by assumption n> 1 and i> 0 Example 7.3 Find an expression for the discount factor during the nth period from the date of investment, i.e. (1 + in)−1, in terms of the amount function. Solution. Recall that in = A(n)−A(n−1) A(n−1) . Thus, 1 1 + in = 1 1 + A(n)−A(n−1) A(n−1) = A(n−1) A(n) Remark 7.1 One should observe that νt extend the deﬁnition of a(t) = (1 + i)t to negative values of t. Thus, a graph of a(t) is given in Figure 7.1. Figure 7.1 What happens to the calculation of present values if simple interest is assumed instead of compound interest? The accumulation function is now a(t) = 1 + it. Hence, the present value of 1, t years in the future, is (a(t))−1 = 1 1 + it, t≥0. Example 7.4 Find the present value of $3,000 to be paid at the end of 5 years with a rate of simple interest of 7% per annum. 52 THE BASICS OF INTEREST THEORY Solution. The answer is 3000[a(5)]−1 = 3000 1+0.07(5) ≈$2,222.22 Example 7.5 Rework Example 7.4 using compound interest instead of simple interest. Solution. The answer is 3000 1.075 = $2,138.96 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 53 Practice Problems Problem 7.1 At an eﬀective rate of interest of 8% per annum, the present value of $100,000 due in X years is $65,322. Determine X. Problem 7.2 What deposit made today will provide for a payment of $1,000 in 1 year and $2,000 in 3 years, if the eﬀective rate of interest is 7.5%? Problem 7.3 The total amount of a loan to which interest has been added is $20,000. The term of the loan was four and one-half years. (a) If money accumulated at simple interest at a rate of 6%, what was the amount of the loan? (b) If the annual rate of interest was 6% and interest was compounded annually, what was the amount of the loan? Problem 7.4 It is known that an investment of $500 will increase to $4,000 at the end of 30 years. Find the sum of the present values of three payments of $10,000 each which will occur at the end of 20, 40, and 60 years. Problem 7.5 The sum of the present value of 1 paid at the end of nperiods and 1 paid at the end of 2 nperiods is 1. Find (1 + i)2n. Problem 7.6 ‡ Sally has two IRA’s. IRA #1 earns interest at 8% eﬀective annually and IRA #2 earns interest at 10% eﬀective annually. She has not made any contributions since January 1, 1985, when the amount in IRA #1 was twice the amount in IRA #2. The sum of the two accounts on January 1, 1993 was $75,000. Determine how much was in IRA #2 on January 1, 1985. Problem 7.7 ‡ At an annual eﬀective interest rate of i,i> 0, the following are all equal: (i) the present value of 10,000 at the end of 6 years; (ii) the sum of the present values of 6,000 at the end of year tand 56,000 at the end of year 2 t; and (iii) 5,000 immediately. Calculate the present value of a payment of 8,000 at the end of year t+ 3 using the same annual eﬀective interest rate. 54 THE BASICS OF INTEREST THEORY Problem 7.8 The Kelly family buys a new house for $93,500 on May 1, 1996. How much was this house worth on May 1, 1992 if real estate prices have risen at a compound rate of 8% per year during that period? Problem 7.9 Find the present (discounted) value of $3,000 to be paid at the end of 5 years if the accumulation function is a(t) = 1 + t2 25 . Problem 7.10 Suppose a(t) = αt2 + 10β. If $X invested at time 0 accumulates to $1,000 at time 10, and to $2,000 at time 20, ﬁnd the original amount of the investment X. Problem 7.11 You want to buy a sailboat when you retire 5 years from now. The sailboat you wish to buy costs $50,000 today and inﬂation is at 4% per year. What do you need to put into an account today to have enough funds to be able to purchase the boat 5 years from now if the account earns 7% per year? Problem 7.12 ‡ A store is running a promotion during which customers have two options for payment. Option one is to pay 90% of the purchase price two months after the date of sale. Option two is to deduct X% oﬀ the purchase price and pay cash on the date of sale. A customer wishes to determine X such that he is indiﬀerent between the two options when valuing them using an eﬀective annual interest rate of 8%. Which of the following equations would the customer need to solve? (A) (X 100 )( 1 + 0.08 6 ) = 0.90 (B) ( 1 − X 100 )( 1 + 0.08 6 ) = 0.90 (C) (X 100 ) (1.08) 1 6 = 0.90 (D) (X 100 )(1.08 1.06 ) = 0.90 (E) ( 1 − X 100 ) (1.08) 1 6 = 0.90 Problem 7.13 Derive a formula for i in terms of ν where ν = 1 1+i. Problem 7.14 In this section the present value is assumed to be the value of an investment at time 0. We can determine the present value of an investment at any time n not necessarily at time 0. A payment of $10 is to be made at time 7 years. (a) Determine the present value of this payment at time 4 years if the annual compound interest is 6%. (b) Determine the present value of this payment at time 4 years if the annual simple interest is 6%. 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 55 Problem 7.15 A treasury bill or T −Bill with face value $100 is a security which is exchangeable for $100 on the maturity date. Suppose that a T −Bill with face value $100 is issued on 08/09/2005 and matures on 03/09/2006 (there are 182 days between the dates). Assuming simple interest, given that the annual rate of interest is 2.810%, ﬁnd the price of the security. Problem 7.16 If we are promised $1,500 5 years from now, what is its present value today at: (a) 7% simple interest? (b) 7% compound interest? Problem 7.17 Show that the discount function of a simple interest rate i is larger than the discount function for a compound interest rate i for t≥1. Problem 7.18 If i 1+i = 0.06, ﬁnd i and ν. Problem 7.19 What is the value 8 years from now of $500 today if ν = 0.96? Problem 7.20 ‡ At an eﬀective annual interest rate of i,i> 0,each of the following two sets of payments has present value K : (i) A payment of 121 immediately and another payment of 121 at the end of one year. (ii) A payment of 144 at the end of two years and another payment of 144 at the end of three years. Calculate K. 56 THE BASICS OF INTEREST THEORY 8 Interest in Advance: Eﬀective Rate of Discount The eﬀective rate of interest is deﬁned as a measure of the interest paid at the end of the period. In this section, we introduce the eﬀective rate of discount, denoted by d, which is a measure of interest where the interest is paid at the beginning of the period. What do we mean by a statement such as “A loan of $1,200 is made for one year at an eﬀective rate of discount of 5%”? This means that the borrower will pay the interest of 1200 ×0.05 = $60 (called the amount of discount)at the beginning of the year and repays $1,200 at the end of the year. So basically, the lender is getting the interest in advance from the borrower. In general, when $kis borrowed at a discount rate of d,the borrower will have to pay $kdin order to receive the use of $ k. Therefore, instead of the borrower having the use of $ k at the beginning of a period he will only have the use of $( k−kd). In our example above, we note that the eﬀective rate of discount 5% is just the ratio 5% = 1200 −1140 1200 . From this we can formulate the deﬁnition of eﬀective rate of discount: The eﬀective rate of discount is the ratio of the amount of discount during the period to the amount at the end of the period. Example 8.1 What is the diﬀerence between the following two situations? (1) A loan of $100 is made for one year at an eﬀective rate of interest of 5%. (2) A loan of $100 is made for one year at an eﬀective rate of discount of 5%. Solution. In both cases the fee for the use of the money is the same which is $5. That is, the amount of discount is the same as the amount of interest. However, in the ﬁrst case the interest is paid at the end of the period so the borrower was able to use the full $100 for the year. He can for example invest this money at a higher rate of interest say 7% and make a proﬁt of $2 at the end of the transaction. In the second case, the interest is paid at the beginning of the period so the borrower had access to only $95 for the year. So, if this amount is invested at 7% like the previous case then the borrower will make a proﬁt of $1.65. Also, note that the eﬀective rate of interest is taken as a percentage of the balance at the beginning of the year whereas the eﬀective rate of discount is taken as a percentage of the balance at the end of the year Keep in mind the diﬀerences between the interest model and the discount model: •Under the interest model, the payment for the use of the money is made at the end of the period, 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 57 based on the balance at the beginning of the period. •Under the discount model, the payment for the use of the money is deducted at the beginning of the period from the ﬁnal amount which will be present unchanged at the end of the period. Even though an eﬀective rate of interest is not the same as an eﬀective rate of discount, there is a relationship between the two. Assume that $1 is invested for one year at an eﬀective rate of discountd.Then the original principal is $(1 −d). The eﬀective rate of interest i for the year is deﬁned to be the ratio of the amount of interest divided by the balance at the beginning of the year. That is, i= d 1 −d (8.1) Solving this last equation for d we ﬁnd d= i 1 + i. (8.2) There is a verbal interpretation of this result: d is the ratio of the amount of interest that 1 will earn during the year to the balance at the end of the year. Example 8.2 The amount of interest earned for one year when X is invested is $108. The amount of discount earned when an investment grows to value X at the end of one year is $100. Find X,i, and d. Solution. We have iX = 108, i 1+iX = 100. Thus, 108 1+i = 100. Solving for i we ﬁnd i = 0.08 = 8% . Hence, X = 108 0.08 = 1,350 and d= i 1+i = 2 27 ≈7.41% Observe that several identities can be derived from Equations (8.1)-(8.2). For example, sinceν = 1 1+i we have d= iν (8.3) that is, discounting i from the end of the period to the beginning of the period with the discount factor ν, we obtain d. Next, we have d= i 1 + i = 1 + i 1 + i − 1 1 + i = 1 −ν. (8.4) This is equivalent to ν = 1 −d. Both sides of the equation represent the present value of 1 to be paid at the end of the period. Now, from i= d 1−d we have i(1 −d) = d or i−id= d. Adding id−d to both sides we ﬁnd i−d= id (8.5) 58 THE BASICS OF INTEREST THEORY that is, the diﬀerence of interest in the two schemes is the same as the interest earned on amount d invested at the rate i for one period. Eﬀective rates of discount can be calculated over any particular measurement period: “The eﬀective rate of discount dn in the nth period is deﬁned to be the ratio of the amount of discount and the accumulated value at the end of the period” . That is dn = a(n) −a(n−1) a(n) . Since A(n) = A(0)a(n), we can write dn = A(n) −A(n−1) A(n) = In A(n). Example 8.3 If a(t) = 1 + t2 25 , ﬁnd d3. Solution. We have d3 = a(3) −a(2) a(3) = 1 + 9 25 −1 −4 25 1 + 9 25 = 5 34 ≈14.71% Analogous to the eﬀective rate of interest in, dn may vary from period to period. Recall that compound interest implies a constant rate of eﬀective interest. A concept parallel to compound interest is compound discount. We say that dis a compound discount if it discounts $1 according to the model shown in Figure 8.1, where t is the number of periods. Figure 8.1 Let ac(t) be the accumulation function for compound discount d> 0 per period. From Figure 8.1, the original principal which will produce an accumulated value of 1 at the end of tperiods is given by [ac(t)]−1 = (1 −d)t,t ≥0. Thus, the accumulation function for a compound discount d is ac(t) = 1 (1 −d)t,t ≥0. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 59 Example 8.4 An investor would like to have $5,000 at the end of 20 years. The annual compound rate of discount is 5%. How much should the investor deposit today to reach that goal? Solution. The investor should set aside 5000(1 −0.05)20 ≈$1,792.43 In the following theorem we prove that compound discount implies a constant rate of eﬀective discount. Theorem 8.1 Assuming compound discount d> 0 per period, we have dn = d for all n≥1. Proof. Compound interest and compound discount have equal accumulation functions. Indeed, (1 + i)t = ( 1 + d 1 −d )t = 1 (1 −d)t, for all t ≥0. Hence, we see that dn = a(n) −a(n−1) a(n) = 1 (1−d)n − 1 (1−d)n−1 1 (1−d)n = d As for compound interest and compound discount, it is possible to deﬁne simple discount in a manner analogous to the deﬁnition of simple interest. We say that d is a simple discount if it discounts $1 according to the model shown in Figure 8.2. Figure 8.2 If as(t) is the accumulation function for simple discountd,then from Figure 8.2 the original principal which will produce an accumulated value of 1 at the end of t periods is given by [as(t)]−1 = 1 −dt, 0 ≤t< 1 d. Note that, unlike the situation for simple interest, we must restrict the length of time over which we propose to apply simple discount. This is necessary to keep [ as(t)]−1 >0. 60 THE BASICS OF INTEREST THEORY Thus, the accumulation function for simple discount d is as(t) = 1 1 −dt, 0 ≤t< 1 d. Example 8.5 Calculate the present value of a payment of 10,000 to be made in 17 years assuming a simple rate of discount of 3% per annum. Solution. The answer is PV = 10000[1 −0.03(17)] = $4,900 It should be noted that formulas (15.3) - (8.5) assume eﬀective rates of compound interest and discount (since a(t) = ac(t)) and are not valid for simple rates of interest and discount unless the period of investment happens to be exactly one period. For example, if i and d are equivalent simple rate of interest and simple discount then 1 + it= 1 1−dt for all t≥0.If (15.3) is satisﬁed then (1 + it)(1 −dt) = 1 is only valid for t= 0. Example 8.6 If i and d are equivalent rates of simple interest and simple discount over t periods, show that i−d= idt. Solution. Since iand dare equivalent we must have 1+it= 1 1−dt or (1−dt)(1+it) = 1.Thus, 1+ti−td−t2id= 1. This can be rearranged to obtain ti−td= t2id. Dividing through by t, the result follows Example 8.7 (a) Find d5 if the rate of simple interest is 10%. (b) Find d5 if the rate of simple discount is 10%. Solution. (a) We are given that i= 0.10. Then d5 = a(5) −a(4) a(5) = 1 + 0.10(5) −(1 + 0.10(4)) 1 + 0.10(5) = 1 15. (b) We are given d= 0.10. Then d5 = as(5) −as(4) as(5) = 1 1−0.10(5) − 1 1−0.10(4) 1 1−0.10(5) = 1 6 In the case of simple interest, in is a decreasing function of n.This is reversed for dn with a simple discount rate. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 61 Theorem 8.2 Assuming simple discount at a rate of discount d >0, dn is an increasing function of n for 0 < n−1 < 1 d. Proof. Under simple discount at a rate d of discount, we have as(n) = 1 1−dn. Thus, dn = as(n) −as(n−1) as(n) = 1 1−dn − 1 1−d(n−1) 1 1−dn = d 1 −dn+ d = d 1 + d(1 −n) >0 since 0 <n −1 < 1 d. As n increases the denominator decreases so dn increases We next introduce the following deﬁnition: Two rates of interest and/or discount are said to be equivalent if a given amount of principal invested over the same period of time at each of the rates produces the same accumulated value. We have seen in the process of proving Theorem 8.1 that an eﬀective rate of compound interest i is equivalent to an eﬀective rate of compound discount d since a(t) = ac(t). Example 8.8 With i and d satisfying (8.1) - (8.5), show that d3 (1 −d)2 = (i−d)2 1 −ν . Solution. We have d3 (1 −d)2 =d ( d 1 −d )2 =di2 = (id)2 d =(i−d)2 1 −ν Simple and compound discount produce the same result over one period. Over a longer period, simple discount produces a smaller present value than compound discount, while the opposite is true over a shorter period. 62 THE BASICS OF INTEREST THEORY Theorem 8.3 Assuming 0 <d< 1, we have (a) (1 −d)t <1 −dt if 0 <t< 1 (b) (1 −d)t = 1 −dt if t= 0 or t= 1 (c) (1 −d)t >1 −dt if t> 1 Proof. (a) Writing the power series expansion of (1 −d)t we ﬁnd (1 −d)t = 1 −dt+ 1 2t(t−1)d2 −1 6t(t−1)(t−2)d3 + ··· If t< 1 then all terms after the second are negative. Hence, (1 −d)t <1 −dt for 0 <t< 1. (b) Straight forward by substitution. (c) Suppose t> 1. Let f(d) = 1 −dt−(1 −d)t. Then f(0) = 0 and f′(d) = −t+ t(1 −d)t−1. Since t> 1 we have t−1 >0.Since 0 <d< 1,we have 1 −d< 1.Thus, (1 −d)t−1 <1 and consequently f′(d) <−t+ t= 0. Hence, f(d) <0. This completes a proof of the theorem Figure 8.3 compares the discount function under simple discount and compound discount. Figure 8.3 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 63 Practice Problems Problem 8.1 The amount of interest earned on A for one year is $336, while the equivalent amount of discount is $300. Find A. Problem 8.2 1000 is to be accumulated by January 1, 1995, at a compound discount of 9% per year. (a) Find the present value on January 1, 1992. (b) Find the value of i equivalent to d. Problem 8.3 The amount of interest earned on Afor one year is $562.50, while the equivalent amount of discount is $500. Find A. Problem 8.4 Let i be a compound interest with equivalent compound discount d. Show the following (a) 1 d −1 i = 1. (b) d ( 1 + i 2 ) = i ( 1 −d 2 ) . (c) i √ 1 −d= d√1 + i. Hint: Use i−d= id for all three identities. Problem 8.5 Calculate the present value of $2000 payable in 10 years using an annual eﬀective discount rate of 8%. Problem 8.6 Calculate the accumulated value at the end of 3 years of 15,000 payable now assuming an interest rate equivalent to an annual discount rate of 8%. Problem 8.7 An investor deposits $1,000 today. The annual compound rate of discount is 6%. What is the accumulated value of the investment at the end of 10 years? Problem 8.8 An investor would like to have $10,000 at the end of 5 years. The annual simple rate of discount is 3%. How much should the investor deposit today to reach that goal? 64 THE BASICS OF INTEREST THEORY Problem 8.9 An investor deposits $5,000 today. The annual simple rate of discount is 5%. What is the accumu- lated value of the investment at the end of 7 months? Problem 8.10 You are given that ν = 0.80. Calculate d. Problem 8.11 A fund earns interest at a rate equivalent to the rate of discount of d.Megan invested 10,000 in the fund. Eleven years later, Megan has 30,042. Calculate d. Problem 8.12 In Account A, an investment of $1,000 grows to $1,700 in four years at an eﬀective annual interest rate of x. In Account B, $3,000 is invested for ﬁve years, at an eﬀective rate of discount d = x. What is the ending balance in Account B? Problem 8.13 ‡ Bruce and Robbie each open up new bank accounts at time 0 . Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns an annual eﬀective discount rate of d. The amount of interest earned in Bruce’s account during the 11th year is equal to X. The amount of interest earned in Robbie’s account during the 17th year is also equal to X. Calculate X. Problem 8.14 Suppose that a(t) = 1.12t. What is the eﬀective rate of discount in the 7th year? Problem 8.15 A fund accumulates based on the following formula: a(t) = ( 1 + t 10 ) ·1.2t. Find d3. Problem 8.16 The eﬀective rate of discount is 5%. Linda will receive $500 one year from today, $1,000 two years from today, and $2,000 ﬁve years from today. Find the combined present value of her three future cash ﬂows. Problem 8.17 A business permits its customers to pay with a credit card or to receive a percentage discount of r for paying cash. For credit card purchases, the business receives 97% of the purchase price one-half month later. At an annual eﬀective rate of discount of 22%, the two payments are equivalent. Find r. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 65 Problem 8.18 A deposit of X is made into a fund which pays an annual eﬀective interest rate of 6% for 10 years. At the same time , X 2 is deposited into another fund which pays an annual eﬀective rate of discount of d for 10 years. The amounts of interest earned over the 10 years are equal for both funds. Calculate d. Problem 8.19 A signs a one-year note for $1000 and receives $920 from the bank. At the end of six months, A makes a payment of $288. Assuming simple discount, to what amount does this reduce the face amount of the note? Problem 8.20 A bank oﬀers a 272 −day discounted loan at a simple discount rate of 12%. (a) How much money would a borrower receive if she asked for a $5000 loan? (b) What size loan should the borrower ask for in order to actually receive $5000? (c) What is the equivalent simple interest rate that is being charged on the loan? Problem 8.21 A discounted loan of $3000 at a simple discount rate of 6.5% is oﬀered to Mr. Jones. If the actual amount of money that Mr. Jones receives is $2869.11, when is the $3000 due to be paid back? Problem 8.22 Show that dn <in. Problem 8.23 At time t= 0, Paul deposits $3500 into a fund crediting interest with an annual discount factor of 0.96. Find the fund value at time 2.5. Problem 8.24 Method A assumes simple interest over ﬁnal fractional periods, while Method B assumes simple discount over ﬁnal fractional periods. The annual eﬀective rate of interest is 20%. Find the ratio of the present value of a payment to be made in 1.5 years computed under methodAto that computed under Method B. 66 THE BASICS OF INTEREST THEORY 9 Nominal Rates of Interest and Discount When we speak of either the “eﬀective” rate of interest/discount we mean interest is paid once per measurement period, either at the end of the period( in the case of interest rate) or at the beginning of the period(in the case of discount rate). In this section, we consider situations where interest is paid more than once per measurement period. Rates of interest and discount in these cases are called nominal. In this section we deﬁne nominal rates of interest and discount and determine relationships between nominal rates and eﬀective rates as well as relationships between nominal rates of interest and discount. Compound interest or discount will always be assumed, unless speciﬁed otherwise. When interest is paid (i.e., reinvested) more frequently than once per period, we say it is “payable” (“convertible”, “compounded”) each fraction of a period, and this fractional period is called the interest conversion period. A nominal rate of interest i(m) payable m times per period, where m is a positive integer, represents mtimes the eﬀective rate of compound interest used for each of the mth of a period. In this case, i(m) m is the eﬀective rate of interest for each mth of a period. Thus, for a nominal rate of 12% compounded monthly, the eﬀective rate of interest per month is 1% since there are twelve months in a year. Suppose that 1 is invested at a nominal rate i(m) compounded m times per measurement period. That is, the period is partitioned into mequal fractions of a period. At the end of the ﬁrst fraction of the period the accumulated value is 1 + i(m) m . At the end of the second fraction of the period the accumulated value is ( 1 + i(m) m )2 .Continuing, we ﬁnd that the accumulated value at the end of the mth fraction of a period, which is the same as the end of one period, is ( 1 + i(m) m )m and at the end of t years the accumulated value is a(t) = ( 1 + i(m) m )mt . Figure 9.1 illustrates accumulation at a nominal rate of interest for one measurement period. Figure 9.1 9 NOMINAL RATES OF INTEREST AND DISCOUNT 67 Example 9.1 Find the accumulated value of 1,000 after three years at a rate of interest of 24% per year convertible monthly. Solution. The accumulated value is 1,000 ( 1 + 0.24 12 )12×3 = 2,039.89 Example 9.2 Find the accumulated value of $3,000 to be paid at the end of 8 years with a rate of compound interest of 5% (a) per annum; (b) convertible quarterly; (c) convertible monthly. Solution. (a) The accumulated value is 3 ,000 ( 1 + 0.05 1 )8 ≈$4,432.37. (b) The accumulated value is 3 ,000 ( 1 + 0.05 4 )8×4 ≈$4,464.39. (c) The accumulated value is 3 ,000 ( 1 + 0.05 12 )8×12 ≈$4,471.76 Next we describe the relationship between eﬀective and nominal rates. If i denotes the eﬀective rate of interest per one measurement period equivalent to i(m) then we can write 1 + i= ( 1 + i(m) m )m since each side represents the accumulated value of a principal of 1 invested for one year. Rearrang- ing we have i= ( 1 + i(m) m )m −1 and i(m) = m[(1 + i) 1 m −1]. For any t≥0 we have (1 + i)t = ( 1 + i(m) m )mt . Example 9.3 Given the nominal interest rate of 12%, compounded monthly. Find the equivalent eﬀective annual interest rate. 68 THE BASICS OF INTEREST THEORY Solution. The answer is i= ( 1 + 0.12 12 )12 −1 = 12.7% Example 9.4 (a) Find the annual eﬀective interest rate i which is equivalent to a rate of compound interest of 8% convertible quarterly. (b) Find the compound interest rate i(2) which is equivalent to an annual eﬀective interest rate of 8%. (c) Find the compound interest rate i(4) which is equivalent to a rate of compound interest of 8% payable semi-annually. Solution. (a) We have 1 + i= ( 1 + 0.08 4 )4 ⇒i= ( 1 + 0.08 4 )4 −1 ≈0.08243216 (b) We have 1 + 0.08 = ( 1 + i(2) 2 )2 ⇒i(2) = 2[(1.08) 1 2 −1] ≈0.07846. (c) We have ( 1 + i(4) 4 )4 = ( 1 + i(2) 2 )2 ⇒i(4) = 4[(1.04) 1 2 −1] ≈0.0792 In the same way that we deﬁned a nominal rate of interest, we could also deﬁne a nominal rate of discount, d(m),as meaning an eﬀective rate of discount of d(m) m for each of the mth of a period with interest paid at the beginning of a mth of a period. Figure 9.2 illustrates discounting at a nominal rate of discount for one measurement period. Figure 9.2 The accumulation function with the nominal rate of discount d(m) is a(t) = ( 1 −d(m) m )−mt , t ≥0. 9 NOMINAL RATES OF INTEREST AND DISCOUNT 69 Example 9.5 Find the present value of $8,000 to be paid at the end of 5 years at a an annual rate of compound interest of 7% (a) convertible semiannually. (b) payable in advance and convertible semiannually. Solution. (a) The answer is 8,000( 1 + 0.07 2 )5×2 ≈$5,671.35 (b) The answer is 8000 ( 1 −0.07 2 )5×2 ≈$5,602.26 If d is the eﬀective discount rate equivalent to d(m) then 1 −d= ( 1 −d(m) m )m since each side of the equation gives the present value of 1 to be paid at the end of the measurement period. Rearranging, we have d= 1 − ( 1 −d(m) m )m and solving this last equation for d(m) we ﬁnd d(m) = m[1 −(1 −d) 1 m] = m(1 −ν 1 m). Example 9.6 Find the present value of $1,000 to be paid at the end of six years at 6% per year payable in advance and convertible semiannually. Solution. The answer is 1,000 ( 1 −0.06 2 )12 = $693.84 There is a close relationship between nominal rate of interest and nominal rate of discount. Since 1 −d= 1 1+i, we conclude that ( 1 + i(m) m )m = 1 + i= (1 −d)−1 = ( 1 −d(n) n )−n . (9.1) 70 THE BASICS OF INTEREST THEORY If m= n then the previous formula reduces to ( 1 + i(n) n ) = ( 1 −d(n) n )−1 . Example 9.7 Find the nominal rate of discount convertible semiannually which is equivalent to a nominal rate of interest of 12% per year convertible monthly. Solution. We have ( 1 −d(2) 2 )−2 = ( 1 + 0.12 12 )12 . Solving for d(2) we ﬁnd d(2) = 0.11591 Note that formula (9.1) can be used in general to ﬁnd equivalent rates of interest or discount, either eﬀective or nominal, converted with any desired frequency. Example 9.8 Express d(4) as a function of i(3). Solution. We have ( 1 −d(4) 4 )−4 = ( 1 + i(3) 3 )3 . Thus, 1 −d(4) 4 = ( 1 + i(3) 3 )−3 4 So d(4) = 4 [ 1 − ( 1 + i(3) 3 )−3 4 ] An analogous formula to i−d= idholds for nominal rates of interest and discount as shown in the next example. Example 9.9 Prove that i(m) m −d(m) m = i(m) m ·d(m) m 9 NOMINAL RATES OF INTEREST AND DISCOUNT 71 Solution. We have ( 1 + i(m) m ) = ( 1 −d(m) m )−1 which is equivalent to ( 1 + i(m) m ) · ( 1 −d(m) m ) = 1. Expanding we obtain 1 −d(m) m + i(m) m −i(m) m ·d(m) m = 1. Hence, i(m) m −d(m) m = i(m) m ·d(m) m Example 9.10 Show that i(m) = d(m)(1 + i) 1 m. Solution. We know that i(m) m −d(m) m = i(m) m ·d(m) m Multiplying through by m and rearranging we ﬁnd i(m) = d(m) ( 1 + i(m) m ) = d(m)(1 + i) 1 m Example 9.11 (a) On occasion, interest is convertible less frequently than once a year. Deﬁne i( 1 m) and d( 1 m) to be the nominal annual rates of interest and discount convertible once every m years. Find a formula analogous to formula (9.1) relating i( 1 m) and d( 1 m). (b) Find the accumulated value of $100 at the end of two years if the nominal annual rate of discount is 6%, convertible once every four years. Solution. (a) If i( 1 m) is convertible once every myears, then m·i( 1 m) is the eﬀective interest rate over myears. Thus, by deﬁnition of equivalent rates we can write 1 + m·i( 1 m) = (1 + i)m. 72 THE BASICS OF INTEREST THEORY Likewise, we have 1 −p·d( 1 p) = (1 + i)−p. It follows that (1 + m·i( 1 m)) 1 m = (1 −p·d( 1 p))−1 p. Note that this formula can be obtained from formula (9.1) by replacingmand pby their reciprocals. (b) If i is the annual eﬀective interest rate than the accumulated value at the end of two years is 100(1 + i)2 = 100(1 −4 ×0.06)−2 4 = $114.71 Remark 9.1 Nominal rates of interest or discount are not relevant under simple interest and simple discount. For example, if we let i(m) be the nominal interest rate and i be the equivalent simple interest rate then at the end of one year we ﬁnd 1 + i= 1 + i(m) or i= i(m). Similarly, we have d= d(m). 9 NOMINAL RATES OF INTEREST AND DISCOUNT 73 Practice Problems Problem 9.1 A man borrows $1,000 at an interest rate of 24% per year compounded monthly. How much does he owe after 3 years? Problem 9.2 If i(6) = 0.15, ﬁnd the equivalent nominal rate of interest convertible semiannually. Problem 9.3 Suppose that $100 is deposited into a savings account, earning at a discount rate of 0.15% biweekly, at the beginning of year 2006. (a) Find the nominal annual discount rate. (b) Find the eﬀective annual discount rate. (c) Find the amount of interest generated during the year. (d) Find the equivalent eﬀective annual interest rate. (e) Find the equivalent nominal annual interest rate, convertible monthly. Problem 9.4 Express i(6) as a function of d(2). Problem 9.5 Given that i(m) = 0.1844144 and d(m) = 0.1802608. Find m. Problem 9.6 It is known that 1 + i(n) n = 1 + i(4) 4 1 + i(5) 5 Find n. Problem 9.7 If r= i(4) d(4) , express ν in terms of r. Problem 9.8 You deposit $1,000 into Account A and $750 into Account B. Account A earns an eﬀective annual interest rate of 5%. Account B earns interest at i(4). Ten years later, the two accounts have the same accumulated value. Find i(4). (Express as a percentage, to two decimal places.) 74 THE BASICS OF INTEREST THEORY Problem 9.9 ‡ Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a diﬀerent savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i. Problem 9.10 ‡ Calculate the nominal rate of discount convertible monthly that is equivalent to a nominal rate of interest of 18.9% per year convertible monthly. Problem 9.11 A bank credits interest on deposits quarterly at rate 2% per quarter. What is the nominal interest rate (per annum) for interest compounded quarterly? Find the annual eﬀective rate of interest. Problem 9.12 Interest on certain deposits is compounded monthly. Find the nominal rate of interest (per annum) for these deposits if the APR is 15%. Obtain the accumulation of an investment of $1,000 over a period of 6 months. How much interest has accrued by that time? Problem 9.13 Suppose that the eﬀective interest rate for 1 5 year is 2%. Determine the equivalent nominal interest rate, compounded every 6 years. Problem 9.14 Fund A earns interest at a nominal rate of 6% compounded monthly. Fund B earns interest at a nominal rate of discount compounded three times per year. The annual eﬀective rates of interest earned by both funds are equivalent. Calculate the nominal rate of discount earned by Fund B. Problem 9.15 Treasury bills, or T −bills, are sold in terms ranging from a few days to 26 weeks. Bills are sold at a discount from their face value. For instance, you might pay $990 for a $1,000 bill. When the bill matures, you would be paid $1,000. The diﬀerence between the purchase price and face value is interest. A bill for 100 is purchased for 96 three months before it is due. Find: (a) The nominal rate of discount convertible quarterly earned by the purchaser. (b) The annual eﬀective rate of interest earned by the purchaser. Problem 9.16 If i(8) = 0.16, calculate d( 1 2 ). 9 NOMINAL RATES OF INTEREST AND DISCOUNT 75 Problem 9.17 A fund earns a nominal rate of interest of 6% compounded every two years. Calculate the amount that must be contributed now to have 1000 at the end of six years. Problem 9.18 A deposit is made on January 1, 2004. The investment earns 6% compounded semi-annually. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.19 A deposit is made on January 1, 2004. The investment earns interest at a rate equivalent to an annual rate of discount of 6%. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.20 A deposit is made on January 1, 2004. The investment earns interest at a rate equivalent to a rate of discount of 6% convertible quarterly. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.21 Calculate the present value of $1,000 payable in 10 years using a discount rate of 5% convertible quarterly. Problem 9.22 Calculate the accumulated value at the end of 3 years of 250 payable now assuming an interest rate equivalent to a discount rate of 12% convertible monthly. Problem 9.23 Thomas pays 94 into a fund. Six months later, the fund pays Thomas 100. Calculate the nominal rate of discount convertible semi-annually. Problem 9.24 Thomas pays 94 into a fund. Six months later, the fund pays Thomas 100. Calculate the annual eﬀective rate of interest earned. Problem 9.25 Find the present value of 5000, to be paid at the end of 25 months, at a rate of discount of 8% convertible quarterly: (a) assuming compound discount throughout; (b) assuming simple discount during the ﬁnal frac- tional period. 76 THE BASICS OF INTEREST THEORY Problem 9.26 ‡ 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 9.27 A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible semi-annually) 1 5% 2 6% 3 7% 4 8% 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. An investor plans to invest 1,000 in CDs. Calculate the maximum amount that can be withdrawn at the end of six years. Problem 9.28 You are given (1) Fund X accumulates at an interest rate of 8% compounded quarterly (2) Fund Y accumulates at an interest rate of 6% compounded semiannually (3) at the end of 10 years, the total amount in the two funds combined is 1000 (4) at the end of 5 years, the amount in Fund X is twice that in Fund Y Calculate the total amount in the two funds at the end of 2 years. 9 NOMINAL RATES OF INTEREST AND DISCOUNT 77 Problem 9.29 A collection agency pays a doctor $5,000 for invoices that the doctor has not been able to collect on. After two years, the collection agency has collected $6,000 on the invoices. At what nominal rate of discount compounded monthly did the collection agency receive on this transaction? Problem 9.30 A trust company oﬀers guaranteed investment certiﬁcates paying i(2) = 8.9% and i(1) = 9%.Which option yields the higher annual eﬀective rate of interest? 78 THE BASICS OF INTEREST THEORY 10 Force of Interest: Continuous Compounding Eﬀective and nominal rates of interest and discount each measures interest over some interval of time. Eﬀective rates of interest and discount measure interest over one full measurement period, while nominal rates of interest and discount measure interest over mths of a period. In this section we want to measure interest at any particular moment of time. This measure of interest is called the force of interest. To start with, consider the case of a nominal compound interest i(m) converted m times a period. We can think of the force of interest, denoted by δ, as the limit of i(m) as the number of times we credit the compounded interest goes to inﬁnity. That is, δ= lim m→∞ i(m). Letting i be the eﬀective interest rate equivalent to i(m) we have i(m) = m[(1 + i) 1 m −1] = (1 + i) 1 m −1 1 m . Hence, δ= lim m→∞ (1 + i) 1 m −1 1 m . The above limit is of the form 0 0 so that we can apply L’Hopital’s rule to obtain δ= lim m→∞ d dm [ (1 + i) 1 m −1 ] d dm (1 m ) = lim m→∞ [(1 + i) 1 m ln (1 +i)] = ln (1 +i) since limm→∞(1 + i) 1 m = 1. Remark 10.1 The eﬀective interest rate i can be written as a series expansion of δ: i= eδ −1 = δ+ δ2 2! + ··· + δn n! + ··· . Similarly, δ= ln (1 +i) = i−i2 2 + ··· + (−1)nin n + ··· . 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 79 Example 10.1 Given the nominal interest rate of 12%, compounded monthly. Find the equivalent force of interest δ. Solution. The eﬀective annual interest rate is i= (1 + 0.01)12 −1 ≈0.1268250. Hence, δ= ln (1 +i) = ln (1.1268250) ≈0.119404. Remark 10.2 Intuitively, δ represents a nominal interest rate which is converted continuously, a notion of more theoretical than practical importance. Indeed, in theory the most important measure of interest is the force of interest. In practice, however, eﬀective and nominal rates of interest tend to be used more frequently because most ﬁnancial transactions involve discrete, not continuous, processes. However, δ can be used in practice as an approximation to interest converted very frequently, such as daily. Having related δ and iimmediately relates δ and the other measures of interested introduced in the previous sections. Indeed, we have the following important set of equalities ( 1 + i(m) m )m = 1 + i= (1 −d)−1 = ( 1 −d(p) p )−p = eδ. Example 10.2 Using a constant force of interest of 4.2%, calculate the present value of a payment of $1,000 to be made in 8 years’ time. Solution. The present value is 1,000(1 + i)−8 = 1,000e−8δ = 1,000e−8×0.042 ≈$714.62 Example 10.3 A loan of $3,000 is taken out on June 23, 1997. If the force of interest is 14%, ﬁnd each of the following (a) The value of the loan on June 23, 2002. (b) The value of i. (c) The value of i(12). 80 THE BASICS OF INTEREST THEORY Solution. (a) 3,000(1 + i)5 = 3,000e5δ = 3,000e0.7 ≈$6,041.26. (b) i= eδ −1 = e0.14 −1 ≈0.15027. (c) We have ( 1 + i(12) 12 )12 = 1 + i= e0.14. Solving for i(12) we ﬁnd i(12) ≈0.14082 Now, using the accumulation function of compound interest a(t) = (1 + i)t we notice that δ= ln (1 +i) = d dta(t) a(t) . The deﬁnition of force of interest in terms of a compound interest accumulation function can be extended to any accumulation function. That is, for an accumulation function a(t) we deﬁne the force of interest at time t by δt = a′(t) a(t) . Let’s look closely at the expression on the right. From the deﬁnition of the derivative we have d dta(t) a(t) = lim n→∞ a(t+ 1 n)−a(t) a(t) 1 n . Now the expression a(t+ 1 n) −a(t) a(t) is just the eﬀective rate of interest over a very small time period 1 n so that a(t+ 1 n)−a(t) a(t) 1 n is the nominal annual interest rate converted n periods a year with each time period of length 1 n corresponding to that eﬀective rate, which agrees with Remark 10.2. Note that, in general, the force of interest can be a function of t.However, in the case of compound interest δt is a constant. Also, we notice that, since A(t) = A(0)a(t), we can write δt = A′(t) A(t) . 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 81 Example 10.4 Show that, for any amount function A(t), we have ∫ n 0 A(t)δtdt= A(n) −A(0) = I1 + I2 + ··· + In Interpret this result verbally. Solution. We have ∫ n 0 A(t)δtdt= ∫ n 0 A′(t)dt= A(n) −A(0) = I1 + I2 + ··· + In where we used Example 2.6. The term A(n) −A(0) is the amount of interest earned over nmeasurement periods. The term δtdt represents the eﬀective rate of interest over the inﬁnitesimal “period of time” dt.Hence, A(t)δtdtis the amount of interest in this period and ∫n 0 A(t)δtdtrepresents the total amount of interest earned over the entire n periods which is A(n) −A(0) Example 10.5 You are given that A(t) = at2 + bt+ c, for 0 ≤t ≤2, and that A(0) = 100 ,A(1) = 110 , and A(2) = 136. Determine the force of interest at time t= 1 2 . Solution. The condition A(0) = 100 implies c = 100. From A(1) = 110 and A(2) = 136 we ﬁnd the linear system of equations a+ b = 10 and 4 a+ 2b = 36. Solving this system we ﬁnd a = 8 and b = 2. Hence, A(t) = 8t2 + 2t+ 100. It follows that δt = A′(t) A(t) = 16t+ 2 8t2 + 2t+ 100. Hence, δ0.5 = 10 103 = 0.09709 Example 10.6 Find δt in the case of simple interest, that is, when a(t) = 1 + it. Solution. From the deﬁnition of δt we have δt = a′(t) a(t) = i 1 + it 82 THE BASICS OF INTEREST THEORY Note that δt is a decreasing function of t We have now a method for ﬁnding δt given a(t). What if we are given δt instead, and we wish to derive a(t) from it? From the deﬁnition of δt we can write d drln a(r) = δr. Integrating both sides from 0 to t we obtain ∫ t 0 d drln a(r)dr= ∫ t 0 δrdr. Hence, ln a(t) = ∫ t 0 δrdr. From this last equation we ﬁnd a(t) = e ∫t 0 δrdr. Also, we can derive an expression for A(t). Since A(t) = A(0)a(t), we can write A(t) = A(0)e ∫t 0 δrdr. Example 10.7 A deposit of $10 is invested at time 2 years. Using a force of interest of δt = 0.2 −0.02t, ﬁnd the accumulated value of this payment at the end of 5 years. Solution. The accumulated value is A(5) = 10a(5) a(2) = 10e ∫5 2 (0.2−0.02t)dt = 10e[0.2t−0.01t2] 5 2 ≈$14.77 Note that a compound interest rate iimplies a constant force of interest (equals to ln (1 + i)).The converse is also true as shown in the next example. Example 10.8 Show that if δt = δ for all t then a(t) = (1 + i)t for some i. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 83 Solution. Since δt = δ for all t, we obtain ∫t 0 δrdr= tδ.Now, for all integer values n≥1 we have in = a(n) −a(n−1) a(n−1) = enδ −e(n−1)δ e(n−1)δ = eδ −1 = i. In this case, a(t) = eδt = (eδ)t = (1 + i)t The above example shows that a constant force of interest implies a constant eﬀective interest rate. The converse of this result is not always true. See Problem 10.32. Replacing the accumulation function by the discount function in the deﬁnition of δt we obtain the force of discount: δ′ t = − d dt[a(t)]−1 [a(t)]−1 . The negative sign is necessary in order to make the force of discount a positive quantity since the discount function is a decreasing function of t. It is possible to dispence with δ′ t and just use δt according to the following theorem. Theorem 10.1 For all t we have δ′ t = δt. Proof. we have δ′ t = −[a−1(t)]′ a−1(t) =a−2(t)[a(t)]′ a−1(t) =a−2(t)a(t)δt a−1(t) =δt Example 10.9 Find the force of discount under a simple discount rate d. 84 THE BASICS OF INTEREST THEORY Solution. Recall that in the case of a simple discount the discount function is given by [ a(t)]−1 = 1 −dt for 0 ≤t< 1 d. Thus, δt =δ′ t = − d dt[a(t)]−1 [a(t)]−1 = − d dt(1 −dt) 1 −dt = d 1 −dt It follows that the force of interest is increasing with simple discount in contrast to simple interest where it is decreasing. See Example 10.6 Example 10.10 Find δ′ t in the case of compound discount. Solution. We have δ′ t = −[(1 −d)t]′ (1 −d)t = −ln (1−d) Example 10.11 Show that limm→∞d(m) = δ Solution. We have ( 1 −d(m) m )m = (1 + i)−1 = e−δ. Solving for d(m) we ﬁnd d(m) = m[1 −e−δ m]. Using power series expansion of e−δ m we obtain d(m) =m [ 1 − ( 1 + ( −δ m ) + 1 2! ( −δ m )2 + 1 3! ( −δ m )3 + ··· )] =m [δ m −1 2! δ2 m2 + 1 3! δ3 m3 −··· ] =δ−1 2! δ2 m + 1 3! δ3 m2 −···→ δ as m→∞ 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 85 Example 10.12 An investor invests $1,000 into an investment account at time 1. The interest is computed based on the following force of interest: δt =    0.02t 0 ≤t< 3 0.025 t≥3 0 otherwise (a) Find the amount of interest generated between time 2.5 and 3.5. (b) Calculate eﬀective discount rate during the third period, i.e. d3. Solution. (a) We know that A(t) = 1000 a(t) a(1) = 1000e ∫t 1 δrdr. Thus, the amount of interest generated between time 2.5 and 3.5 is A(3.5) −A(2.5) = 1000 ( e ∫3.5 1 δtdt −e ∫2.5 1 δtdt ) . But ∫ 3.5 1 δtdt= ∫ 3 1 0.02tdt+ ∫ 3.5 3 0.025dt= 0.0925 and ∫ 2.5 1 δtdt= ∫ 2.5 1 0.02tdt= 0.0525 Hence, A(3.5) −A(2.5) = 1000(e0.0925 −e0.0525) ≈$43.01. (b) d3 = A(3)−A(2) A(3) = e0.08−e0.03 e0.08 ≈4.877% Example 10.13 Find an expression of t in terms of δ so that f(t) = (1 + it) −(1 + i)t is maximum. Solution. Since f′(t) = i−(1 + i)tln (1 +i) = i−δ(1 + i)t we see that f′(t) = 0 when (1 + i)t = i δ. Solving for t we ﬁnd t= ln i−ln δ ln (1+i) = ln i−ln δ δ . Now, f′′(t) = −δ2(1 + i)t <0 so that the critical point of f is a maximum Example 10.14 On March 15, 2003, a student deposits X into a bank account. The account is credited with simple interest where i= 7.5% 86 THE BASICS OF INTEREST THEORY On the same date, the student’s professor deposits X into a diﬀerent bank account where interest is credited at a force of interest δt = 2t t2 + k, t ≥0 From the end of the fourth year until the end of the eighth year, both accounts earn the same dollar amount of interest. Calculate k. Solution. The interest earned by the student is X[1 + 0.075(8)] −X[1 + 0.075(4)] = 0.3X The interest earned by the professor is Xe ∫8 0 δtdt −Xe ∫4 0 δtdt = Xe ∫8 0 2t t2+kdt −Xe ∫4 0 2t t2+kdt = Xe[ln (t2+k)] 8 0 −Xe[ln (t2+k)] 4 0 = X64 + k k −X16 + k k = 48 k X. Thus, 0.3X = 48 k X ⇒k= 48 0.3 = 160 Example 10.15 Show that d<d (m) <δ <i(m) <i, m> 1. Solution. We have d dm(i(m)) = e δ m ( 1 − δ m −e−δ m ) . From the inequality 1 −x≤e−x for x≥0 we conclude that d dm(i(m)) = e δ m ( 1 − δ m −e−δ m ) <0. That is, i(m) is a decreasing function of m. Since i(1) = i and i(∞) = δ we obtain δ <i(m) <i. Similarly, d dm(d(m)) = e−δ m(e δ m −1 − δ m) >0 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 87 so that d(m) is an increasing function of m. Since d(1) = d and d(∞) = δ we can write d<d (m) <δ. Combining these inequalities we ﬁnd d<d (m) <δ <i(m) <i, m> 1 88 THE BASICS OF INTEREST THEORY Practice Problems Problem 10.1 A deposit of $500 is invested at time 5 years. The constant force of interest is 6% per year. Determine the accumulated value of the investment at the end of 10 years. Problem 10.2 If the constant force of interest is 6%, what is the corresponding annual eﬀective rate of interest? Problem 10.3 Assume that the force of interest varies with time and is given by δt = a+ b t. Find the formula for the accumulation of one unit money from time t1 to time t2. Problem 10.4 You need $500 on Jan 1, 2012. To save for this amount, you invest xon Jan 1, 2008 and 2xon July 1, 2008. The force of interest is δt = 0.02t where t is 0 on Jan 1, 2008. Find x. Problem 10.5 At a constant force of interest, $200 accumulates to $240 over the course of 8 years. Find the force of interest δ. Problem 10.6 ‡ Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest of 4% convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ. After 7.25 years, the value of each account is the same. Calculate δ. Problem 10.7 ‡ Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest i convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ.After 7.25 years, the value of each account is 200. Calculate i−δ. Problem 10.8 ‡ At time t= 0, 1 is deposited into each of Fund X and Fund Y. Fund X accumulates at a force of interest δt = t2 k. Fund Y accumulates at a nominal rate of discount of 8% per annum convertible semiannually. At time t= 5, the accumulated value of Fund X equals the accumulated value of Fund Y. Determine k. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 89 Problem 10.9 ‡ At time 0 , Kis deposited into Fund X, which accumulates at a force of interest δt = 0.006t2. At time m,2K is deposited into Fund Y, which accumulates at an annual eﬀective interest rate of 10%. At time n, where n>m, the accumulated value of each fund is 4 K. Determine m. Problem 10.10 Suppose $2,000 is invested for 3 years with a constant force of interest equal to 9%. (a) Find the accumulation function. (b) Find the eﬀective rate of interest. (c) Find the value of the investment after 3 years. Problem 10.11 ‡ Ernie makes deposits of 100 at time 0, and X at time 3 . The fund grows at a force of interest δt = t2 100 , t> 0. The amount of interest earned from time 3 to time 6 is X. Calculate X. Problem 10.12 If the force of interest δ is 0.1, obtain the nominal rate of interest (per annum) on deposits com- pounded (a) monthly; (b) quarterly; (c) annually. Problem 10.13 The force of interest depends upon time, with δ(t) = 1 10(1+t) , 0 ≤t ≤4. Find the accumulated amount after four years for an initial investment of $1,000 at time zero. Problem 10.14 The force of interest δ is constant. The corresponding nominal rate of discount when interest is compounded p-thly is d(p).Express d(p) in terms of p.Show that d(p) is an increasing function of p. Problem 10.15 ‡ Tawny makes a deposit into a bank account which credits interest at a nominal interest rate of 10% per annum, convertible semiannually. At the same time, Fabio deposits 1,000 into a diﬀerent bank account, which is credited with simple interest. At the end of 5 years, the forces of interest on the two accounts are equal, and Fabio’s account has accumulated to Z. Determine Z. Problem 10.16 Fund A earns interest at a nominal rate of interest of 12% compounded quarterly. Fund B earns interest at a force of interest δ.John invested $1,000 in each fund ﬁve years ago. Today, the amount in Fund A is equal to the amount in Fund B. Calculate δ. 90 THE BASICS OF INTEREST THEORY Problem 10.17 A fund earns interest at a force of interest of δ = 0.01t. Calculate the value at the end of 10 years of 8,000 invested at the end of 5 years. Problem 10.18 A fund earns interest at a force of interest of δ = 0.01t. Calculate the eﬀective rate of interest in the 10th year. Problem 10.19 Calculate the eﬀective annual interest rate equivalent to a nominal rate of discount of 6% com- pounded continuously. Problem 10.20 Fund A accumulates at a simple interest rate of 10%. Find B accumulates at a simple discount rate of 5%. Find the point in time at which the force of interest on the two funds are equal. Problem 10.21 An investment is made for one year in a fund with a(t) = a+ bt+ ct2.The nominal rate of interest earned during the ﬁrst half of the year is 5% convertible semi-annually. The eﬀective rate of interest earned for the entire year is 7%. Find δ0.5. Problem 10.22 If a(t) = 1 + .01(t2 + t), calculate δ5. Problem 10.23 You are given that δ= 0.05. Calculate the amount that must be invested at the end of 10 years to have an accumulated value at the end of 30 years of $1,000. Problem 10.24 Calculate k if a deposit of 1 will accumulate to 2.7183 in 10 years at a force of interest given by: δt = { kt 0 <t ≤5 0.04kt2 t> 5 Problem 10.25 A deposit is made on January 1, 2004. The investment earns interest at a constant force of interest of 6%. Calculate the monthly eﬀective interest rate for the month of December 2004. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 91 Problem 10.26 A fund earns interest at a constant force of interest of δ.Kathy invested 10,000 in the fund. Twenty two years later, Kathy has 30,042. Calculate δ. Problem 10.27 A fund earns interest at a force of interest of δt = kt. Ryan invests 1,000 at time t = 0. After 8 years, Ryan has 1,896.50 . Calculate k. Problem 10.28 You are given that i(12) = 0.09. Calculate δ. Problem 10.29 You are given that d(2) = 0.04. Calculate δ. Problem 10.30 ‡ At time 0, deposits of 10,000 are made into each of Fund X and Fund Y. Fund X accumulates at an annual eﬀective interest rate of 5%. Fund Y accumulates at a simple interest rate of 8%. At time t, the forces of interest on the two funds are equal. At time t, the accumulated value of Fund Y is greater than the accumulated value of Fund X by Z. Determine Z. Problem 10.31 Find an expression for the fraction of a period at which the excess of present values computed at simple discount over compound discount is a maximum. Hint: See Example 10.13. Problem 10.32 Consider the following problem: Invest 1 at a compound interest i for integral periods and at a simple interest for fractional periods. (a) Show that ik = i for k= 1,2,··· ,n. (b) Show that δt1 ̸= δt2 for k <t1 <t2 <k + 1, k= 1,2,···n−1. Thus, a constant eﬀective rate of interest i does not necessarily imply a constant force of interest. Problem 10.33 Which of the following are true? (I) d dd(i) = ν−2 (II) d di(i(m)) = ν−( m−1 m ) (III) d dδ(i) = 1 + i. 92 THE BASICS OF INTEREST THEORY Problem 10.34 You are given that δt = 0.2t 1+0.1t2 . Calculate i2. Problem 10.35 Simplify the following expression (d dνδ )(d did ) . Problem 10.36 An investment is made in an account in which the amount function is given by A(t) = t2 + t+ 2.25, 0 ≤t≤4. Compute the ratio i3 δ2 . Problem 10.37 Fund A accumulates at a force of interest 0.05 1 + 0.05t at time t ≥0. Fund B accumulates at a force of interest 0.05. You are given that the amount in Fund A at time zero is 1,000, the amount in Fund B at time zero is 500, and that the amount in Fund C at any time tis equal to the sum of the amount in FundAand Fund B.Fund C accumulates at force of interest δt. Find δ2. Problem 10.38 Show that d dtδt = A′′(t) A(t) −δ2 t. Problem 10.39 Find δt if A(t) = K2t3t2 52t . Problem 10.40 Show that ∫n 0 δtdt= ln a(n). 11 TIME VARYING INTEREST RATES 93 11 Time Varying Interest Rates In this section we consider situations involving varying interest. The ﬁrst involves a continuously varying force of interest δt. In this case, the accumulated value at time t is given by A(t) = A(0)e ∫t 0 δrdr. Example 11.1 Find the accumulated value of $200 invested for 5 years if the force of interest is δt = 1 8+t. Solution. The accumulated value is A(5) = 200a(5) = 200e ∫5 0 dt 8+t = 200e[ln (8+t)]5 0 = 200 ×13 8 = $325 The second situation involves changes in the eﬀective rate of interest over a period of time. Let- ting in denote the eﬀective rate of interest during the nth period from the date of investment, the accumulated value for integral t is given by a(t) = (1 + i1)(1 + i2) ···(1 + it) (11.1) and the present value is given by (a(t))−1 = (1 + i1)−1(1 + i2)−1 ···(1 + in)−1. (11.2) Example 11.2 Find the accumulated value of $500 invested for 9 years if the rate of interest is 5% for the ﬁrst 3 years, 5.5% for the second 3 years, and 6.25% for the third 3 years. Solution. The accumulated value is 500(1 + 0.05)3(1 + 0.055)3(1 + 0.0625)3 = $815.23 Formulas (11.1) and (11.2) can be extended to include nominal rates of interest or discount. We illustrate this in the next example. Example 11.3 A fund will earn a nominal rate of interest of 5% compounded quarterly during the ﬁrst two years, a nominal rate of discount of 4% compounded monthly during years 3 and 4, and a constant force of interest of 3% during the ﬁfth and sixth year. Calculate the amount that must be invested today in order to accumulate 5,000 after 6 years. 94 THE BASICS OF INTEREST THEORY Solution. The amount that must be invested is 5000 ( 1 + 0.05 4 )−8 ( 1 −0.04 12 )24 e−0.03×2 = $3,935.00 A common question involving varying interest is the question of ﬁnding an equivalent level rate to the rate that vary. Moreover, the answer depends on the period of time chosen for the compari- son . We illustrate this point in the next example. Example 11.4 (a) Find the eﬀective interest rate over the ﬁrst four-year period from now if the eﬀective rate of interest is 5% for the ﬁrst three years from now, 4.5% for the next three years, and 4% for the last three years. (b) What about over the six-year period from now? Solution. (a) We have (1 + i)4 = (1.05)3(1.045). Solving this equation for i we ﬁnd i= [(1.05)3(1.045)] 1 4 −1 = 4.87% (b) We have (1 + i)6 = (1.05)3(1.045)3. Solving this equation for i we ﬁnd i= [(1.05)3(1.045)3] 1 6 −1 = 4.75% Example 11.5 Find the accumulated value of 1 at the end of n periods where the eﬀective rate of interest for the kth period, 1 ≤k≤n, is deﬁned by ik = (1 + r)k(1 + i) −1. Solution. We know that a(n) = (1 + i1)(1 +i2) ···(1 +in) = (1 + r)(1 +i)(1 +r)2(1 +i) ···(1 +r)n(1 +i) = (1 + r)1+2+···+n(1 + i)n = (1 + r) n(n+1) 2 (1 + i)n 11 TIME VARYING INTEREST RATES 95 Practice Problems Problem 11.1 Find the eﬀective interest rate over a three-year period which is equivalent to an eﬀective rate of discount of 8% the ﬁrst year, 7% the second year, and 6% the third year. Problem 11.2 Find the accumulated value of 1 at the end of n years if the force of interest is δt = 1 1+t. Problem 11.3 Find the accumulated value of 1 at the end of 19 years if the force of interest is δt = 0.04(1 + t)−2. Problem 11.4 If δt = 0 .01t, 0 ≤t ≤2, ﬁnd the equivalent annual eﬀective rate of interest over the interval 0 ≤t≤2. Problem 11.5 An investment account earns 4% the ﬁrst year, 5% the second year, and X% the third year. Find X if the three year average interest rate is 10%. Problem 11.6 An investment of $500 was made three years ago. For the ﬁrst six months the annual eﬀective rate of interest paid on the account was 5%, but it then increased to 7% and did not change over the following two and a half years. Find the present accumulation of the investment. Problem 11.7 The annual eﬀective interest rate for a given year is determined by the function: in = 0 .02n, where n = 1,2,3,··· . Brad’s initial investment of $400 earns interest every year according to the preceding. Find the accumulated value of his investment after six years. Problem 11.8 ‡ At a force of interest δt = 2 (k+2t) (i) a deposit of 75 at time t= 0 will accumulate to X at time t= 3; and (ii) the present value at time t= 3 of a deposit of 150 at time t= 5 is also equal to X. Calculate X. Problem 11.9 ‡ Jennifer deposits 1000 into a bank account. The bank credits interest at a nominal annual rate of i compounded semi-annually for the ﬁrst 7 years and a nominal annual rate of 2 i convertible quarterly for all years thereafter. The accumulated amount in the account after 5 years is X. The accumulated amount in the account at the end of 10.5 years is 1,980. Calculate X. 96 THE BASICS OF INTEREST THEORY Problem 11.10 Suppose δt = t3 100 . Find 1 a(3) . Problem 11.11 In Fund X money accumulates at a force of interest δt = 0.01t+ 0.1, 0 ≤t≤20. In Fund Y money accumulates at an annual eﬀective interest rate i. An amount of 1 is invested in each fund for 20 years. The value of Fund X at the end of 20 years is equal to the value of Fund Y at the end of 20 years. Calculate the value of Fund Y at the end of 1.5 years. Problem 11.12 If the eﬀective rate of discount in year k is equal to 0 .01k+ 0.06 for k= 1,2,3, ﬁnd the equivalent rate of simple interest over the three-year period. Problem 11.13 A savings and loan association pays 7% eﬀective on deposits at the end of each year. At the end of every 3 years a 2% bonus is paid on the balance at that time. Find the eﬀective rate of interest earned by an investor if the money is left on deposit (a) Two years. (b) Three years. (c) Four years. Problem 11.14 On July 1, 1999 a person invested 1,000 in a fund for which the force of interest at time t is given by δt = .02(3 + 2t) where t is the number of years since January 1, 1999. Determine the accumulated value of the investment on January 1, 2000. Problem 11.15 You are given that δt = t 100 .Calculate the present value at the end of the 10 year of an accumulated value at the end of 15 years of $1,000. Problem 11.16 Calculate k if a deposit of 1 will accumulate to 2.7183 in 10 years at a force of interest given by: δt = { kt 0 <t ≤5 0.04kt2 t> 5 11 TIME VARYING INTEREST RATES 97 Problem 11.17 The annual eﬀective interest rate for year t is it = 1 (10+t) . Calculate the current value at the end of year 2 of a payment of 6,000 at the end of year 7. Problem 11.18 A fund pays a nominal rate of 12%. The nominal rate is compounded once in year 1, twice in year 2, 3 times in year 3, etc. Calculate the amount that must be invested today in order to accumulate 5,000 after 6 years. Problem 11.19 You are given δt = 2 t−1 , for 2 ≤t ≤20. For any one year interval between n and n+ 1, with 2 ≤n≤9, calculate d(2) n+1. Problem 11.20 Investment X for 100,000 is invested at a nominal rate of interest j convertible semi-annually. After four years it accumulates to 214,358.88. Investment Y for 100,000 is invested at a nominal rate of discount k convertible quarterly. After two years, it accumulates to 232,305.73. Investment Z for 100,000 is invested at an annual eﬀective rate of interest equal to j in year one and an annual eﬀective rate of discount equal to k in year two. Calculate the value of investment Z at the end of two years. Problem 11.21 A bank agrees to lend John 10,000 now and X three years later in exchange for a single repayment of 75,000 at the end of 10 years. The bank charges interest at an annual eﬀective rate of 6% for the ﬁrst 5 years and at a force of interest δt = 1 t+1 for t≥5. Determine X. Problem 11.22 John invests 1000 in a fund which earns interest during the ﬁrst year at a nominal rate of K convertible quarterly. During the 2nd year the fund earns interest at a nominal discount rate of K convertible quarterly. At the end of the 2nd year, the fund has accumulated to 1173.54. Calculuate K. Problem 11.23 At time 0, 100 is deposited into Fund X and also into Fund Y. FundX accumulates at a force of interest δt = 0.5(1 + t)−2. Fund Y accumulates at an annual eﬀective interest rate of i. At the end of 9 years, the accumulated value of Fund X equals the accumulated value of Fund Y. Determine i. 98 THE BASICS OF INTEREST THEORY Problem 11.24 ‡ An investor deposits 1000 on January 1 of year xand deposits 1000 on January 1 of year x+ 2 into a fund that matures on January 1 of year x+ 4. The interest rate on the fund diﬀers every year and is equal to the annual eﬀective rate of growth of the gross domestic product (GDP) during the fourth quarter of the previous year. The following are the relevant GDP values for the past 4 years. Year x−1 x x+ 1 x+ 2 Quarter III 800.0 850.0 900.0 930.0 Quarter IV 808.0 858.5 918.0 948.6 What is the internal rate of return earned by the investor over the 4-year period? Problem 11.25 A deposit of 10,000 is made into a fund at time t = 0. The fund pays interest at a nominal rate of discount of d compounded quarterly for the ﬁrst two years. Beginning at time t = 2, interest is credited using a nominal rate of interest of 8% compounded quarterly. At time t = 5 , the accumulated value of the fund is 14,910. Calculate d. Problem 11.26 Amin deposits 10,000 in a bank. During the ﬁrst year the bank credits an annual eﬀective rate of interest i. During the second year the bank credits an annual eﬀective rate of interest ( i−5%). At the end of two years she has 12,093.75 in the bank. What would Amin have in the bank at the end of three years if the annual eﬀective rate of interest were ( i+ 9%) for each of the three years? Problem 11.27 Fund X starts with 1,000 and accumulates with a force of interest δt = 1 15 −t for 0 ≤t <15. Fund Y starts with 1,000 and accumulates with an interest rate of 8% per annum compounded semi-annually for the ﬁrst three years and an eﬀective interest rate of i per annum thereafter. Fund X equals Fund Y at the end of four years. Calculate i. Problem 11.28 Amin puts 100 into a fund that pays an eﬀective annual rate of discount of 20% for the ﬁrst two years and a force of interest of rate δt = 2t t2 + 8 < 2 ≤t≤4 11 TIME VARYING INTEREST RATES 99 for the next two years. At the end of four years, the amount in Amin’s account is the same as what it would have been if he had put 100 into an account paying interest at the nominal rate of i per annum compounded quarterly for four years. Calculate i. Problem 11.29 On January 1, 1980, Jack deposited 1,000 into Bank X to earn interest at the rate of j per annum compounded semi-annually. On January 1, 1985, he transferred his account to Bank Y to earn interest at the rate of k per annum compounded quarterly. On January 1, 1988, the balance at Bank Y is 1,990.76. If Jack could have earned interest at the rate of k per annum compounded quarterly from January 1, 1980 through January 1, 1988, his balance would have been 2,203.76. Calculate the ratio k j. 100 THE BASICS OF INTEREST THEORY 12 Equations of Value and Time Diagrams Interest problems generally involve four quantities: principal(s), investment period length(s), in- terest rate(s), accumulated value(s). If any three of these quantities are known, then the fourth quantity can be determined. In this section we introduce equations that involve all four quantities with three quantities are given and the fourth to be found. In calculations involving interest, the value of an amount of money at any given point in time depends upon the time elapsed since the money was paid in the past or upon time which will elapse in the future before it is paid. This principle is often characterized as the recognition of the time value of money. We assume that this principle reﬂects only the eﬀect of interest and does not include the eﬀect of inﬂation. Inﬂation reduces the the purchasing power of money over time so investors expect a higher rate of return to compensate for inﬂation. As pointed out, we will neglect the eﬀct of inﬂation when applying the above mentioned principle. As a consequence of the above principle, various amounts of money payable at diﬀerent points in time cannot be compared until all the amount are accumulated or discounted to a common date, called the comparison date, is established. The equation which accumulates or discounts each payment to the comparison date is called the equation of value. One device which is often helpful in the solution of equations of value is the time diagram. A time diagram is a one-dimensional diagram where the only variable is time, shown on a single coordinate axis. We may show above or below the coordinate of a point on the time-axis, values of money intended to be associated with diﬀerent funds. A time diagram is not a formal part of a solution, but may be very helpful in visualizing the solution. Usually, they are very helpful in the solution of complex probems. Example 12.1 In return for a payment of $1,200 at the end of 10 years, a lender agrees to pay $200 immediately, $400 at the end of 6 years, and a ﬁnal amount at the end of 15 years. Find the amount of the ﬁnal payment at the end of 15 years if the nominal rate of interest is 9% converted semiannually. Solution. The comparison date is chosen to be t= 0. The time diagram is given in Figure 12.1. Figure 12.1 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 101 The equation of value is 200 + 400(1 + 0.045)−12 + X(1 + 0.045)−30 = 1200(1 + 0.045)−20. Solving this equation for X we ﬁnd X ≈$231.11 With compound interest, an equation of value will produce the same answer for an unknown value regardless of what comparison date is selected. This is due to the fact that multiplying an equation by a power of an expression yields an equivalent equation. We illustrate this in the next example. Example 12.2 In return for a promise to receive $600 at the end of 8 years, a person agrees to pay $100 at once, $200 at the end of 5 years, and to make further payment at the end of 10 years. Find the payment at the end of 10 years if the nominal interest rate is 8% convertible semi-annually. Solution. The equation of value at t= 0 is 100 + 200(1 + 0.04)−10 + X(1 + 0.04)−20 = 600(1 + 0.04)−16. Solving for X we ﬁnd X ≈186.75. The equation of value at t= 10 is 100(1 + 0.04)20 + 200(1 + 0.04)10 + X = 600(1 + 0.04)4. Solving for X we ﬁnd X ≈186.75.Note that by multiplying the last equation by (1 + 0.04)−20 one gets the equation of value at t= 0 Whereas the choice of a comparison date has no eﬀect on the answer obtained with compound interest, the same cannot be said of simple interest or simple discount. Example 12.3 Find the amount to be paid at the end of 10 years which is equivalent to two payments of 100 each, the ﬁrst to be paid immediately, and the second to be paid at the end of 5 years. Assume 5% simple interest is earned from the date each payment is made, and use a comparison date 1. The end of 10 years. 2. The end of 15 years. 102 THE BASICS OF INTEREST THEORY Solution. 1. With a comparison date at t= 10, the payment at the end of 10 years will be 100(1 + 10×0.05) + 100(1 + 5×0.05) = 275. 2. With a comparison date at t = 15 , the amount P that must be paid at the end of 10 years satisﬁes the equation of value 100(1 + 15×0.05) + 100(1 + 10×0.05) = P(1 + 5×0.05) which implies that P = 260 Example 12.4 Investor A deposits 1,000 into an account paying 4% compounded quarterly. At the end of three years, he deposits an additional 1,000. Investor B deposits X into an account with force of interest δt = 1 6+t. After ﬁve years, investors A and B have the same amount of money. Find X. Solution. Consider investor A’s account ﬁrst. The initial 1,000 accumulates at 4% compounded quarterly for ﬁve years; the accumulated amount of this piece is 1,000 ( 1 + 0.04 4 )4×5 = 1000(1.01)20. The second 1,000 accumulates at 4% compounded quarterly for two years, accumulating to 1,000 ( 1 + 0.04 4 )4×2 = 1000(1.01)8. The value in investor A’s account after ﬁve years is A= 1000(1.01)20 + 1000(1.01)8. The accumulated amount of investor B’s account after ﬁve years is given by B = Xe ∫5 0 dt 6+t = Xeln (11 6 ) = 11 6 X. The equation of value at time t= 5 is 11 6 X = 1000(1.01)20 + 1000(1.01)8. Solving for X we ﬁnd X ≈$1,256.21 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 103 Practice Problems Problem 12.1 In return for payments of $5,000 at the end of 3 years and $4,000 at the end of 9 years, an investor agrees to pay $1500 immediately and to make an additional payment at the end of 2 years. Find the amount of the additional payment if i(4) = 0.08. Problem 12.2 At a certain interest rate the present values of the following two payment patterns are equal: (i) 200 at the end of 5 years plus 500 at the end of 10 years; (ii) 400.94 at the end of 5 years. At the same interest rate 100 invested now plus 120 invested at the end of 5 years will accumulate to P at the end of 10 years. Calculate P. Problem 12.3 An investor makes three deposits into a fund, at the end of 1, 3, and 5 years. The amount of the deposit at time t is 100(1.025)t. Find the size of the fund at the end of 7 years, if the nominal rate of discount convertible quarterly is 4 41 . Problem 12.4 ‡ Brian and Jennifer each take out a loan of X. Jennifer will repay her loan by making one payment of 800 at the end of year 10. Brian will repay his loan by making one payment of 1,120 at the end of year 10. The nominal semi-annual rate being charged to Jennifer is exactly one-half the nominal semi-annual rate being charged to Brian. Calculate X. Problem 12.5 Fund A accumulates at 6% eﬀective, and Fund B accumulates at 8% eﬀective. At the end of 20 years the total of the two funds is 2,000. At the end of 10 years the amount in Fund A is half that in Fund B. What is the total of the two funds at the end of 5 years? Problem 12.6 Louis has an obligation to pay a sum of $3,000 in four years from now and a sum of $5,000 in six years from now. His creditor permits him to discharge these debts by paying $X in two years from now, $1000 in three years from now, and a ﬁnal payment of $2X in nine years from now. Assuming an annual eﬀective rate of interest of 6%, ﬁnd X. Problem 12.7 Every Friday in February (7, 14, 21,28) Vick places a 1,000 bet, on credit, with his oﬀ-track bookmaking service, which charges an eﬀective weekly interest rate of 8% on all credit extended. 104 THE BASICS OF INTEREST THEORY Vick looses each bet and agrees to repay his debt in four installments to be made on March 7, 14, 21, and 28. Vick pays 1,100 on March 7, 14, and 21. How much must Vick pay on March 28 to completely repay his debt? Problem 12.8 A borrower is repaying a loan by making payments of 1,000 at the end of each of the next 3 years. The interest rate on the loan is 5% compounded annually. What payment could the borrower make at the end of the ﬁrst year in order to extinguish the loan? Problem 12.9 An investor purchases an investment which will pay 2,000 at the end of one year and 5,000 at the end of four years. The investor pays 1,000 now and agrees to pay X at the end of the third year. If the investor uses an interest rate of 7% compounded annually, what is X? Problem 12.10 Today is New Year’s Day. In return for payments of 1,500 at the end of January, February, and March, and of 3,000 at the end of May, July, and September, an investor agrees to pay now the total value of the 6 payments, and to either make or receive an additional payment at the end of December. Find the amount of that additional payment if it is known that the nominal annual interest rate is 6%, compounded monthly. Problem 12.11 A debt of 7,000 is due at the end of 5 years. If 2,000 is paid at the end of 1 year, what single payment should be made at the end of the 2nd year to liquidate the debt, assuming interest at the rate of 6.5% per year, compounded quarterly. Problem 12.12 George agrees to buy his brother’s car for 7,000. He makes a down payment of 4,000 now, and agrees to pay two equal payments, one at the end of 6 months, and the other at the end of a year. If interest is being charged at 5% per annum eﬀective, how large should each of the equal payments be? Problem 12.13 ‡ Jeﬀ deposits 10 into a fund today and 20 ﬁfteen years later. Interest is credited at a nominal discount rate of dcompounded quarterly for the ﬁrst 10 years, and at a nominal interest rate of 6% compounded semiannually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 105 Problem 12.14 ‡ The parents of three children, ages 1, 3, and 6, wish to set up a trust fund that will pay X to each child upon attainment of age 18, and Y to each child upon attainment of age 21. They will establish the trust fund with a single investment of Z. Find the equation of value for Z. Problem 12.15 An investment fund is established at time 0 with a deposit of 5000. 1000 is added at the end of 4 months, and an additional 4000 is added at the end of 8 months. No withdrawals are made. The fund value, including interest, is 10560 at the end of 1 year. The force of interest at time t is k 1+(1−t)k, for 0 ≤t≤1. Determine k. Problem 12.16 A loan of 1000 is made at an interest rate of 12% compounded quarterly. The loan is to be repaid with three payments: 400 at the end of the ﬁrst year, 800 at the end of the ﬁfth year, and the balance at the end of the tenth year. Calculate the amount of the ﬁnal payment. Problem 12.17 You are given δt = 0.5 1+t for t> 0. A payment of 400 at the end of 3 years and 800 at the end of 15 years has the same present value as a payment of 300 at the end of 8 years and X at the end of 24 years. Calculate X. Problem 12.18 A loan of 12 is to be repaid with payments of 10 at the end of 3 years and 5 at the end of 6 years. Calculate the simple discount rate that is being charged on the loan. Problem 12.19 A deposit of K3 at time t= 0 accumulates to 8000 after 3nyears using an annual eﬀective discount rate of d.Using the same annual eﬀective discount rate, a deposit of K2 at time t= 0 accumulates to X after 2n years. Calculate X. Problem 12.20 An investor puts 1000 into Fund X and 1000 into Fund Y. Fund Y earns compound interest at the annual rate of j >0,and Fund X earns simple interest at the annual rate of 1 .10j.At the end of 2 years, the amount in Fund Y is 30 more than the amount in Fund X. After 5 years, the amount in Fund Y exceeds the amount in Fund X by E. Determine E. Problem 12.21 You are given two loans, with each loan to be repaid by a single payment in the future. Each payment includes both principal and interest. The ﬁrst loan is repaid by a 3,000 payment at the 106 THE BASICS OF INTEREST THEORY end of four years. The interest is accrued at 10% per annum compounded semi-annually. The second loan is repaid by a 4,000 payment at the end of ﬁve years. The interest is accrued at 8% per annum compounded semi-annually. These two loans are to be consolidated. The consolidated loan is to be repaid by two equal installments of X, with interest at 12% per annum compounded semi-annually. The ﬁrst payment is due immediately and the second payment is due one year from now. Calculate X. Problem 12.22 An investment fund accrues interest with force of interest δt = K 1 + (1−t)K for 0 ≤t≤1. At time zero, there is 100,000 in the fund. At time one there is 110,000 in the fund. The only two transactions during the year are a deposit of 15,000 at time 0.25 and a withdrawal of 20,000 at time 0.75. Calculate K. Problem 12.23 Paul borrows 1000 at an annual interest rate of 12%. He repays the loan in full by making the following payments: (1) 500 at the end of 4 months (2) 200 at the end of 14 months (3) R at the end of 18 months Calculate R. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 107 13 Solving for the Unknown Interest Rate In this section, it is the rate of interest i that is unknown in an equation of value. We discuss four ways for detrmining i. Method 1: Direct Method When a single payment is involved, the method that works best is to solve for i directly in the equation of value (using exponential and logarithmic functions). In this situation, the equation of value takes either the form A = P(1 + i)n or A = Peδn. In the ﬁrst case, the interest is found by using the exponential function obtaining i= (A P )1 n −1. In the second case, the interest is found using the logarithmic function obtaining δ= 1 nln (A P ) . Example 13.1 At what interest rate convertible semiannually would $500 accumulate to $800 in 4 years? Solution. Let j = i(2) 2 be the eﬀective rate per six months. We are given that 500(1 + j)8 = 800. Solving for j and using a calculator we ﬁnd j = (8 5 )1 8 −1 ≈0.06051. Thus, i(2) = 2j = 0.121 = 12.1% Method 2: Analytical Method When multiple payments are involved, the equation of value takes the formf(i) = 0.In this case, the problem is the classical problem of determining the nonnegative solutions to the equation f(i) = 0 with f(i) being a diﬀerentiable function. If f(i) is a polynomial then the problem reduces to solving polynomial equations. Some known algebraic methods can be used such as the rational zero test, or the quadratic formula, etc. Example 13.2 At what eﬀective interest rate would the present value of $1000 at the end of 3 years plus $2,000 at the end of 6 years be equal to $2,700? Solution. The equation of value at time t = 0 is 2000 ν6 + 1000ν3 = 2700. Dividing through by 100 we ﬁnd 108 THE BASICS OF INTEREST THEORY 20ν6 + 10ν3 = 27.Using the quadratic formula and the fact that ν3 >0 we obtain ν3 = 0.9385 from which we obtain ν ≈0.9791. Hence, i= 1 0.9791 −1 ≈0.0213 = 2.13% Method 3: Linear Interpolation If the equation f(i) = 0 can not be solved known algebraic methods, then approximation methods can be used such as linear interpolation. We next introduce this concept. Suppose that we know the values of a function f(x) at distinct points x1 and at x2, and that f(x1) ̸= f(x2). If |x1 −x2|is small, it may be reasonable to assume that the graph of f is approxi- mately linear between x1 and x2. That is equivalent to assuming that f(x) = f(x1) + f(x2) −f(x1) x2 −x1 (x−x1) for x1 <x<x 2. If now we wish to determine an approximate value of x where f(x) has a speciﬁc value y0, we may use this equation for that purpose and ﬁnd x= x1 + (x2 −x1) y0 −f(x1) f(x2) −f(x1). In particular, if we wish to ﬁnd a zero of f near the given points (i.e. y0 = 0),and if f(x1) ̸= f(x2), a good approximation could be x≈x1 −f(x1) x2 −x1 f(x2) −f(x1). When we apply these formulas for points between two points x1 and x2,we speak of linear inter- polation. The following example illustrates the use of linear interpolation in determing the unknown rate. Example 13.3 At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Solution. The equation of value at time t= 10 years is 1000(1 + j)20 + 2000(1 +j)14 = 5000 where j = i(2) 2 . We will use linear interpolation to solve for j. For that purpose, we deﬁne f(j) = 1000(1 + j)20 + 2000(1 +j)14 −5000. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 109 We want to ﬁnd j such that f(j) = 0 . By trial and error we see that f(0.03) = −168.71 and f(0.035) = 227.17. Since f is continuous, j is between these two values. Using linear interpolation we ﬁnd j ≈0.03 + 168.71 × 0.005 227.17 + 168.71 ≈0.0321 and f(0.0321) = −6.11. Hence, i(2) = 2j = 0.0642 = 6.42% Method 4: Successive Iterations Methods In order to achieve a higher level of accuracy than the one provided with linear interpolation, it- eration methods come to mind. We will discuss two iteration methods: the bisection method and the Newton-Raphson method. The bisection method is based on the fact that a diﬀerentiable functionf that satisﬁes f(α)f(β) < 0 must satisfy the equation f(x0) = 0 for some x0 between α and β. The ﬁrst step in the methods consists of ﬁnding two starting values x0 <x1 such that f(x0)f(x1) < 0.Usually, these values are found by trial and error. Next, bisect the interval by means of the mid- point x2 = x0+x1 2 . If f(x0)f(x2) < 0 then apply the bisection process to the interval x0 ≤x ≤x2. If f(x1)f(x2) < 0 then apply the bisection process to the interval x2 ≤x ≤x1. We continue the bisection process as many times as necessary to achieve the desired level of accuracy. We illustrate this method in the next example. Example 13.4 At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Use the bisection method. Solution. The equation of value at time t= 10 years is 1000(1 + j)20 + 2000(1 +j)14 = 5000 where j = i(2) 2 . Deﬁne f(j) = 1000(1 + j)20 + 2000(1 +j)14 −5000. Descartes’ rule of signs says that the maximum number of positive roots to a polynomial equation is equal to the number of sign changes in the coeﬃcients of the polynomial. In our case, Descartes’ rule asserts the existence of one positive root to the equation f(j) = 0. Now, by trial and error we see that f(0) = −2000 and f(0.1) = 9322.50. Thus, j is between these two values. Let j2 = 0+0.1 2 = 0.05. Then f(0.05) = 1613.16 so that f(0)f(0.05) <0. Now bisect the interval [0,0.05] by means of the point j3 = 0.5(0 + 0.05) = 0.025 and note that f(j3) = −535.44. Continue this process to obtain the table 110 THE BASICS OF INTEREST THEORY n jn f(jn) 0 0 −2000 1 0.1 9322.50 2 0.05 1613.16 3 0.025 −535.44 4 0.0375 436.75 5 0.03125 −72.55 6 0.034375 176.02 7 0.0328125 50.25 8 0.03203125 −11.52 9 0.032421875 19.27 10 0.0322265625 3.852 11 0.0321289063 −3.84 12 0.0321777344 0.005 Thus, j ≈0.032178 accurate to six decimal places compared to 0 .0321 found in Example 13.3. Hence, i2 = 2j = 0.06436 = 6.436% The problem with the bisection method is that the rate of convergence is slow. An iteration method with a faster rate of convergence is the Newton-Raphson method given by the iteration formula jn+1 = jn −f(jn) f′(jn). This method is discussed in more details in Section 20. Example 13.5 Rework Example 13.3 using the Newton-Raphson method. Solution. We have f(j) = 1000(1 + j)20 + 2000(1 +j)14 −5000. and f′(j) = 20000(1 + j)19 + 28000(1 +j)13. Thus, jn+1 = jn −(1 + jn)20 + 2(1 +jn)14 −5 20(1 + jn)19 + 28(1 +jn)13 . 13 SOLVING FOR THE UNKNOWN INTEREST RATE 111 Let j0 = 0. Then j1 =0 −1 + 2−5 20 + 28 = 0.0416666667 j2 =0.0416666667 − 1.041666666720 + 2(1.0416666667)14 −5 20(1.0416666667)19 + 28(1.0416666667)13 = 0.0328322051 j3 =0.0328322051 − 1.032832205120 + 2(1.0328322051)14 −5 20(1.0328322051)19 + 28(1.0328322051)13 = 0.0321809345 j4 =0.0321809345 − 1.032180934520 + 2(1.0321809345)14 −5 20(1.0321809345)19 + 28(0.0321809345)13 = 0.032177671 Thus, in four iterations we ﬁnd j = 0.032178 and hence i(2) = 6.436% 112 THE BASICS OF INTEREST THEORY Practice Problems Problem 13.1 Find the nominal rate of interest convertible semiannually at which the accumulated value of 1,000 at the end of 15 years is 3,000. Problem 13.2 Find the exact eﬀective rate of interest at which payments of 300 at the present, 200 at the end of one year, and 100 at the end of two years will accumulate to 700 at the end of two years. Problem 13.3 The sum of the accumulated value of 1 at the end of three years at a certain eﬀective rate of interest i, and the present value of 1 to be paid at the end of three years at an eﬀective rate of discount numerically equal to i is 2.0096. Find the rate i. Problem 13.4 ‡ David can receive one of the following two payment streams: (i) 100 at time 0, 200 at time n, and 300 at time 2 n (ii) 600 at time 10 At an annual eﬀective interest rate of i, the present values of the two streams are equal. Given νn = 0.75941, determine i. Problem 13.5 ‡ Mike receives cash ﬂows of 100 today, 200 in one year, and 100 in two years. The present value of these cash ﬂows is 364.46 at an annual eﬀective rate of interest i. Calculate i. Problem 13.6 ‡ At a nominal interest rate of i convertible semi-annually, an investment of 1,000 immediately and 1,500 at the end of the ﬁrst year will accumulate to 2,600 at the end of the second year. Calculate i. Problem 13.7 ‡ Eric deposits 100 into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 200 into a diﬀerent savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 113 Problem 13.8 ‡ A deposit of 100 is made into a fund at time t = 0. The fund pays interest at a nominal rate of discount of d compounded quarterly for the ﬁrst two years. Beginning at time t = 2, interest is credited at a force of interest δt = 1 t+1 . At time t = 5, the accumulated value of the fund is 260. Calculate d. Problem 13.9 ‡ A customer is oﬀered an investment where interest is calculated according to the following force of interest: δt = { 0.02t 0 ≤t≤3 0.045 3 <t. The customer invests 1,000 at time t= 0.What nominal rate of interest, compounded quarterly, is earned over the ﬁrst four-year period? Problem 13.10 A manufacturer sells a product to a retailer who has the option of paying 30% below the retail price immediately, or 25% below the retail price in six months. Find the annual eﬀective rate of interest at which the retailer would be indiﬀerent between the two options. Problem 13.11 An investor deposits 10,000 in a bank. During the ﬁrst year, the bank credits an annual eﬀective rate of interest i. During the second year the bank credits an annual eﬀective interest i−0.05. At the end of two years the account balance is 12,093.75. What would the account balance have been at the end of three years if the annual eﬀective rate of interest were i+ 0.09 for each of the three years? Problem 13.12 You are given δt = 2 t−1 for any t ∈[2,10]. For any one year interval between n and n+ 1; with 2 ≤n≤9; calculate the equivalent nominal rate of discount d(2). Problem 13.13 It is known that an investment of $1,000 will accumulate to $1,825 at the end of 10 years. If it is assumed that the investment earns simple interest at rate i during the 1st year, 2 i during the second year, ··· , 10i during the 10th year, ﬁnd i. Problem 13.14 It is known that an amount of money will double itself in 10 years at a varying force of interest δt = kt. Find k. 114 THE BASICS OF INTEREST THEORY Problem 13.15 A fund pays 1 at time t = 0 , 2 at time t = 2 n and 1 at time t = 4 n. The present value of the payments is 3.61. Calculate (1 + i)n. Problem 13.16 The present value of 300 in 3 years plus 600 in 6 years is equal to 800. Calculate i. Problem 13.17 A savings and loan association pays 7% eﬀective on deposits at the end of each year. At the end of every three years a 2% bonus is paid on the balance at that time. Find the eﬀective rate of interest earned by an investor if the money is left on deposit: (a) Two years. (b) Three years. (c) Four years. Problem 13.18 Consider the following equation of value 25 j = 1 −(1 + j)−40. Use linear interpolation to estimate j. Problem 13.19 Consider the following equation of value 0.0725[1 −(1 + 0.01j)−50] = 0.01j.Use linear interpolation to estimate j. Problem 13.20 Two funds, X and Y, start with the same amount. You are given (1) Fund X accumulates at a force of interest 5% (2) Fund Y accumulates at a rate of interest j compounded semi-annually (3) At the end of eight years, Fund X is 1.05 times as large as Fund Y. Calculate j. Problem 13.21 You are given a loan on which interest is charged over a 4 year period, as follows (1) an eﬀective rate of discount of 6% for the ﬁrst year (2) a nominal rate of discount of 5% compounded every 2 years for the second year (3) a nominal rate of interest of 5% compounded semiannually for the third year (4) a force of interest of 5% for the fourth year. Calculate the annual eﬀective rate of interest over the 4 year period. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 115 Problem 13.22 An investor deposits 10 into a fund today and 20 ﬁfteen years later. Interest is credited at a nominal discount rate of dcompounded quarterly for the ﬁrst 10 years, and at a nominal interest rate of 6% compounded semi-annually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. 116 THE BASICS OF INTEREST THEORY 14 Solving for Unknown Time As pointed out in Section 12, if any three of the four basic quantities: principal(s), investment period length(s), interest rate(s), accumulated value(s) are given, then the fourth can be determined. In this section, we consider the situation in which the length of the investment period is the unknown. Example 14.1 A single payment of $3,938.31 will pay oﬀ a debt whose original repayment plan was $1,000 due on January 1 of each of the next four years, beginning January 1, 2006. If the eﬀective annual rate is 8%, on what date must the payment of $3,938.31 be paid? Solution. The equation of value with comparison date January 1, 2006 is 1,000[1 + (1 + 0.08)−1 + (1 + 0.08)−2 + (1 + 0.08)−3] = 3938.31(1 + 0.08)−t. Solving this equation for t we ﬁnd t= ln {1000(3938.31)−1[1 + (1 + 0.08)−1 + (1 + 0.08)−2 + (1 + 0.08)−3]} ln 1.08 ≈1.25. Therefore, the single payment of 3938.31 must be paid 1.25 years after January 1, 2006, or on March 31, 2007 In the above example, the answer was obtained by using a pocket calculator with exponential and logarithmic functions. An alternative approach to using calculator is the use of linear interpolation in the interest tables as illustrated in the next example. Example 14.2 Find the length of time necessary for $1000 to accumulate to $1500 if invested at 6% per year compounded semiannually: (a) by a direct method, i.e. using a calculator; (b) by interpolating in the interest tables Solution. Let t be the comparison date. Then we must have 1000(1 .03)2t = 1000 or (1.03)2t = 1.5. (a) Using a pocket calculator we ﬁnd t= 1 2 ln 1.5 ln 1.03 = 6.859 years. 14 SOLVING FOR UNKNOWN TIME 117 (b) From interest tables, (1.03)13 = 1.46853 and (1.03)14 = 1.51259.Thus, 13 <2t< 14.Performing a linear interpolation we ﬁnd 2t= 13 + 1.5 −1.46853 1.51259 −1.46853 = 13.714. Thus, t= 6.857 years The Rule of 72 Next, we consider the following problem: Given a particular rate of compound interest, how long will it take an investment to double? Starting with 1 invested at an annually compounded interest i, let n be the time needed for the accumulated value to become 2. That is, (1 + i)n = 2. Taking the natural logarithm of both sides we ﬁnd nln (1 +i) = ln 2 and solving for n we obtain n= ln 2 ln (1 +i). We would like to ﬁnd an approximation that is pretty good for an interest rate of i= 8%.For that purpose, we start by writing the Taylor series approximation of x ln (1+x) around x= 0.08 to obtain x ln (1 +x) =1.039486977 + 0.4874162023(x−0.08) −0.07425872903(x−0.08)2 + ··· ≈1.039486977 for x very “close” to 0.08. Thus, the time necessary for the original principal to double is given by t= ln 2 ln (1 +i) = ln 2 i · i ln (1 +i) ≈ln 2 i (1.039486977) ≈0.72 i . This is called the rule of 72, since t can be written in the form t= 72 100i. Here’s a table that shows the actual number of years required to double your money based on diﬀerent interest rates, along with the number that the rule of 72 gives you. 118 THE BASICS OF INTEREST THEORY % Rate Actual Rule 72 1 69.66 72 2 35.00 36 3 23.45 24 4 17.67 18 5 14.21 14.4 6 11.90 12 7 10.24 10.29 8 9.01 9 9 8.04 8 10 7.27 7.2 ··· ··· ··· 15 4.96 4.8 20 3.80 3.6 25 3.11 2.88 30 2.64 2.4 40 2.06 1.8 50 1.71 1.44 75 1.24 0.96 100 1.00 0.72 Example 14.3 Suppose $2,000 is invested at a rate of 7% compounded semiannually. (a) Find the length of time for the investment to double by using the exact formula. (b) Find the length of time for the investment to double by using the rule of 72. Solution. (a) By using the exact formula for compound interest we ﬁnd 2000(1 + 0.035)2t = 4000. Solving for t we ﬁnd t≈10.074 years. (b) Using the Rule of 72 we ﬁnd t= 72 7 ≈10.286 The Method of Equated Time We consider the following problem: Suppose amounts s1,s2,··· ,sn are to be paid at respective times t1,t2,··· ,tn with annual interest rate i. What would be the time t when one single payment 14 SOLVING FOR UNKNOWN TIME 119 of s1 + s2 + ··· + sn would be equivalent to the individual payments made separately? A time diagram for this situation is shown in Figure 14.1 Figure 14.1 In this case, the equation of value at t= 0 is (s1 + s2 + ··· + sn)νt∗ = s1νt1 + s2νt2 + ··· + snνtn. (14.1) Taking the natural logarithm of both sides we ﬁnd t∗ln ν+ ln (s1 + s2 + ··· + sn) = ln (s1νt1 + s2νt2 + ··· + snνtn). Solving for t∗ and recalling that ln ν = −δ we ﬁnd t∗= −1 δln (s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn ) =1 δ [ ln ( n∑ k=1 sk ) −ln ( n∑ k=1 skνtk )] In order to provide an approximation to t∗ we recall the binomial series expansion (1 + x)α = 1 + α 1!x+ α(α−1) 2! x2 + ··· + α(α−1) ···(α−n+ 1) n! xn + ··· where α∈R and −1 <x< 1. Using this series expansion we can write νt∗ = (1 + i)−t∗ = 1 + (−t∗) 1! i+ (−t∗)(−t∗−1) 2! i2 + ···≈ 1 −it∗. Similarly, we have νtk = (1 + i)−tk = 1 + (−tk) 1! i+ (−tk)(−tk −1) 2! i2 + ···≈ 1 −itk. 120 THE BASICS OF INTEREST THEORY Substituting these approximations in the equation νt∗ = s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn we ﬁnd 1 −it∗≈s1(1 −it1) + s2(1 −it2) + ··· + sn(1 −itn) s1 + s2 + ··· + sn which can be written in the form 1 −it∗≈1 −is1t1 + s2t2 + ··· + sntn s1 + s2 + ··· + sn . Solving for t∗ we obtain the approximation t∗≈s1t1 + s2t2 + ··· + sntn s1 + s2 + ··· + sn = t. Estimating t∗ with t is known as the method of equated time. Note that the formula for t is similar to the center of mass that is covered in a calculus course where the weights are replaced by the various amounts paid. Example 14.4 Payments of $100, $200, and $500 are due at the ends of years 2,3, and 8, respectively. Assuming an eﬀective rate of interest of 5% per year, ﬁnd the point in time at which a payment of $800 would be equivalent using the method of equated time. Solution. Using the method of equating time we ﬁnd t= 100 ·2 + 200·3 + 500·8 100 + 200 + 500 = 6 years With the approximation t, the estimated present value for the equivalent single payment is PV = (s1 + s2 + ··· + sn)νt. The following result shows that tis an overestimate of t.Also, the result shows that the true present value exceeds the present value given by the method of equated time. Theorem 14.1 With s1,s2,··· ,sn,t1,t2,··· ,tn,t, and t as deﬁned above we have s1νt1 + s2νt2 + ··· + snνtn >(s1 + s2 + ··· + sn)νt. Hence, t>t. 14 SOLVING FOR UNKNOWN TIME 121 Proof. Consider s1 payments each equal to νt1,s2 payments each equal to νt2,··· ,sn payments each equal to νtn. The arithmetic mean of these payments is s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn and the geometric mean of these payments is ( νs1t1νs2t2 ···νsntn ) 1 s1+s2+···+sn = ν s1t1+s2t2+···+sntn s1+s2+···+sn = νt. In Problem 14.16 we show that the arithmetic mean is always greater than the geometric mean. Hence, s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn >ν t or s1νt1 + s2νt2 + ··· + snνtn >(s1 + s2 + ··· + sn)νt. The left-hand side is the true present value and the right-hand side is the estimated present value. Note that this last equation implies that νt >ν t by (14.1). Thus, νt−t >1 and since ν <1 we must have t>t Remark 14.1 As seen above, the solution by the method of equated time is relatively easy to compute, and is generally fairly accurate. Also, an interesting thing to notice is that the interest rate plays no role in the computations when the method of equated time is used. Example 14.5 A loan is negotiated with the lender agreeing to accept $1,000 after 10 years, $2,000 after 20 years, and $3,000 after 30 years in full repayments. The borrower wishes to liquidate the loan with a single $6,000 payment. Let T1 represent the time of the $6,000 payment calculated by an equation of value. Let T2 represent the time determined by the method of equated time. If i = 0.01, ﬁnd T2 −T1 to the nearest penny. Solution. The equation of value at time t= 0 is 1,000ν10 + 2,000ν20 + 3,000ν30 = 6,000νT1. Solving we ﬁnd T1 = ln (ν10 + 2ν20 + 3ν30) −ln 6 −ln 1.01 = 23.05. 122 THE BASICS OF INTEREST THEORY Now, using the method of equated time we ﬁnd T2 = 1,000 ×10 + 2,000 ×20 + 3,000 ×30 6,000 = 23.33. Thus, T2 −T1 = 23.33 −23.05 = 0.28 14 SOLVING FOR UNKNOWN TIME 123 Practice Problems Problem 14.1 The present value of a payment of $5,000 to be made in t years is equal to the present value of a payment of $7,100 to be made in 2 t years. If i= 7.5% ﬁnd t. Problem 14.2 The present value of two payments of 100 each to be made at the end of n years and 2 n years is 100. If i= 0.08, ﬁnd n. Problem 14.3 You are asked to develop a rule of n to approximate how long it takes money to triple. Find n. Problem 14.4 ‡ Joe deposits 10 today and another 30 in ﬁve years into a fund paying simple interest of 11% per year. Tina will make the same two deposits, but the 10 will be deposited n years from today and the 30 will be deposited 2 n years from today. Tina’s deposits earn an annual eﬀective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits. Calculate n. Problem 14.5 Find how long $1,000 should be left to accumulate at 6% eﬀective in order that it will amount to twice the accumulated value of another 1,000 deposited at the same time at 4% eﬀective. Problem 14.6 If an investment will be doubled in 8 years at a force of interest δ, in how many years will an investment be tripled at a nominal rate of interest numerically equal to δ and convertible once every three years? Problem 14.7 Amounts $500, $800, and $1,000 are to be paid at respective times 3 years from today, 5 years from today, and 11 years from today. The eﬀective rate of interest is 4.5% per annum. (a) Use the method of equated time to ﬁnd the number of years t from today when one single payment of $500 + $800 + $1000 = $2,300 would be equivalent to the individual payments made separately. (b) Find the exact number of years tfrom today when one single payment of $500 + $800 + $1000 = $2,300 would be equivalent to the individual payments made separately. 124 THE BASICS OF INTEREST THEORY Problem 14.8 A payment of n is made at the end of n years, 2n at the end of 2 n years, ··· , n2 at the end of n2 years. Find the value of t by the method of equated time. Problem 14.9 Fund A accumulates at a rate of 12% convertible monthly. Fund B accumulates with a force of interest δt = t 6 . At time t = 0 equal deposits are made in each fund. Find the next time that the two funds are equal. Problem 14.10 A loan requires the borrower to repay $1,000 after 1 year, 2000 after 2 years, $3,000 after 3 years, and $4,000 after 4 years. At what time could the borrower make a single payment of $10,000 according to the method of equated time? Problem 14.11 A series of payments of 100 are made with the ﬁrst payment made immediately. The second payment is made at the end of year 2. The third payment is made at the end of year 4, etc with the last payment being made at the end of year 12. Calculate t using the method of equated time. Problem 14.12 A series of payments are made at the end of each year for 50 years. The amount of the payment at the end of year n is n. Calculate t using the method of equated time. Problem 14.13 A fund earns interest at a nominal rate of interest of 12% compounded monthly. If $1,000 is invested in the fund, it will grow to be $5,320.97 after n years. Calculate n. Problem 14.14 A fund earns interest at a force of interest of δt = 0.05t. Lauren invests $2,000 at time t= 0. After n years, Lauren has $4,919.20 . Calculate n. Problem 14.15 A fund earns interest at a force of interest of δt = 0.01t2. Calculate the amount of time until the fund triples. Problem 14.16 Consider the positive numbers a1,a2,··· ,an where n is a positive integer. The arithmetic mean is deﬁned by a1 + a2 + ··· + an n 14 SOLVING FOR UNKNOWN TIME 125 whereas the geometric mean is deﬁned by [a1 ·a2 ···an] 1 n. Show that a1 + a2 + ··· + an n ≥[a1 ·a2 ···an] 1 n. Hint: Use the fact that ex ≥1 + x for all x. This implies e ai µ −1 ≥ ai µ,1 ≤i ≤n where µ is the arithmetic mean. Problem 14.17 A fund starts with a zero balance at time zero. The fund accumulates with a varying force of interest δt = 2t t2+1 ,for t≥0.A deposit of 100,000 is made at time 2. Calculate the number of years from the time of deposit for the fund to double. Problem 14.18 At time t= 0 Billy puts 625 into an account paying 6% simple interest. At the end of year 2, George puts 400 into an account paying interest at a force of interest δt = 1 6+t, for t≥2. If both accounts continue to earn interest indeﬁnitely at the levels given above, the amounts in both accounts will be equal at the end of year n. Calculate n. Problem 14.19 Payment of 300, 500, and 700 are made at the end of years ﬁve, six, and eight, respectively. Interest is accumulated at an annual eﬀective interest rate of 4%. You are to ﬁnd the point in time at which a single payment of 1500 is equivalent to the above series of payments. You are given: (i) X is the point in time calculated using the method of equated time. (ii) Y is the exact point in time. Calculate X+ Y. Problem 14.20 The present value of a payment of 1004 at the end of T months is equal to the present value of 314 after 1 month, 271 after 18 months, and 419 after 24 months. The eﬀective annual interest rate is 5%. Calculate T to the nearest integer. Problem 14.21 Using the method of equated time, a payment of 400 at time t = 2 plus a payment of X at time t= 5 is equivalent to a payment of 400 + X at time t= 3.125. At an annual eﬀective interest rate of 10%, the above two payments are equivalent to a payment of 400 + X at time t= k using the exact method. Calculate k. 126 THE BASICS OF INTEREST THEORY Problem 14.22 Investor A deposits 100 into an account that earns simple interest at a rate of 10% per annum. Investor Bmakes two deposits of 50 each into an account that earns compound interest at an annual eﬀective rate of 10%. B′s ﬁrst deposit occurs n years after A′s deposit. B′s second deposit occurs 2 n years after A′s deposit. The balances in the two accounts are equal 10 years after Investor A′s deposit. What is the value of n? 14 SOLVING FOR UNKNOWN TIME 127

Human: Which of the following may ﬁt the deﬁnition of interest?

Actuarial AI: A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2 Marcel B. Finan Arkansas Tech University c⃝All Rights Reserved Preliminary Draft Last updated February 27, 2011 2 In memory of my parents August 1, 2008 January 7, 2009 Preface This manuscripts is designed for an introductory course in the theory of interest and annuity. This manuscript is suitablefor a junior level course in the mathematics of ﬁnance. A calculator, such as TI BA II Plus, either the solar or battery version, will be useful in solving many of the problems in this book. A recommended resource link for the use of this calculator can be found at http://www.scribd.com/doc/517593/TI-BA-II-PLUS-MANUAL. The recommended approach for using this book is to read each section, work on the embedded examples, and then try the problems. Answer keys are provided so that you check your numerical answers against the correct ones. Problems taken from previous exams will be indicated by the symbol ‡. This manuscript can be used for personal use or class use, but not for commercial purposes. If you ﬁnd any errors, I would appreciate hearing from you: mﬁnan@atu.edu This project has been supported by a research grant from Arkansas Tech University. Marcel B. Finan Russellville, Arkansas March 2009 3 4 PREFACE Contents Preface 3 The Basics of Interest Theory 9 1 The Meaning of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2 Accumulation and Amount Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3 Eﬀective Interest Rate (EIR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4 Linear Accumulation Functions: Simple Interest . . . . . . . . . . . . . . . . . . . . . . 28 5 Date Conventions Under Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . 35 6 Exponential Accumulation Functions: Compound Interest . . . . . . . . . . . . . . . . 41 7 Present Value and Discount Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Interest in Advance: Eﬀective Rate of Discount . . . . . . . . . . . . . . . . . . . . . . 56 9 Nominal Rates of Interest and Discount . . . . . . . . . . . . . . . . . . . . . . . . . . 66 10 Force of Interest: Continuous Compounding . . . . . . . . . . . . . . . . . . . . . . . 78 11 Time Varying Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 12 Equations of Value and Time Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . 100 13 Solving for the Unknown Interest Rate . . . . . . . . . . . . . . . . . . . . . . . . . . 107 14 Solving for Unknown Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 The Basics of Annuity Theory 143 15 Present and Accumulated Values of an Annuity-Immediate . . . . . . . . . . . . . . . 144 16 Annuity in Advance: Annuity Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 17 Annuity Values on Any Date: Deferred Annuity . . . . . . . . . . . . . . . . . . . . . 168 18 Annuities with Inﬁnite Payments: Perpetuities . . . . . . . . . . . . . . . . . . . . . . 176 19 Solving for the Unknown Number of Payments of an Annuity . . . . . . . . . . . . . . 184 20 Solving for the Unknown Rate of Interest of an Annuity . . . . . . . . . . . . . . . . . 192 21 Varying Interest of an Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 22 Annuities Payable at a Diﬀerent Frequency than Interest is Convertible . . . . . . . . 206 23 Analysis of Annuities Payable Less Frequently than Interest is Convertible . . . . . . . 210 5 6 CONTENTS 24 Analysis of Annuities Payable More Frequently than Interest is Convertible . . . . . . 218 25 Continuous Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 26 Varying Annuity-Immediate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 27 Varying Annuity-Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 28 Varying Annuities with Payments at a Diﬀerent Frequency than Interest is Convertible 257 29 Continuous Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Rate of Return of an Investment 275 30 Discounted Cash Flow Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 31 Uniqueness of IRR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 32 Interest Reinvested at a Diﬀerent Rate . . . . . . . . . . . . . . . . . . . . . . . . . . 292 33 Interest Measurement of a Fund: Dollar-Weighted Interest Rate . . . . . . . . . . . . 302 34 Interest Measurement of a Fund: Time-Weighted Rate of Interest . . . . . . . . . . . . 311 35 Allocating Investment Income: Portfolio and Investment Year Methods . . . . . . . . 320 36 Yield Rates in Capital Budgeting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 Loan Repayment Methods 333 37 Finding the Loan Balance Using Prospective and Retrospective Methods. . . . . . . . 334 38 Amortization Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 39 Sinking Fund Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 40 Loans Payable at a Diﬀerent Frequency than Interest is Convertible . . . . . . . . . . 365 41 Amortization with Varying Series of Payments . . . . . . . . . . . . . . . . . . . . . . 371 Bonds and Related Topics 379 42 Types of Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 43 The Various Pricing Formulas of a Bond . . . . . . . . . . . . . . . . . . . . . . . . . 384 44 Amortization of Premium or Discount . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 45 Valuation of Bonds Between Coupons Payment Dates . . . . . . . . . . . . . . . . . . 405 46 Approximation Methods of Bonds’ Yield Rates . . . . . . . . . . . . . . . . . . . . . . 413 47 Callable Bonds and Serial Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 Stocks and Money Market Instruments 427 48 Preferred and Common Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 49 Buying Stocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 50 Short Sales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 51 Money Market Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 CONTENTS 7 Measures of Interest Rate Sensitivity 453 52 The Eﬀect of Inﬂation on Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . 454 53 The Term Structure of Interest Rates and Yield Curves . . . . . . . . . . . . . . . . . 459 54 Macaulay and Modiﬁed Durations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 55 Redington Immunization and Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . 479 56 Full Immunization and Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 An Introduction to the Mathematics of Financial Derivatives 493 57 Financial Derivatives and Related Issues . . . . . . . . . . . . . . . . . . . . . . . . . 494 58 Derivatives Markets and Risk Sharing . . . . . . . . . . . . . . . . . . . . . . . . . . . 500 59 Forward and Futures Contracts: Payoﬀ and Proﬁt Diagrams . . . . . . . . . . . . . . 504 60 Call Options: Payoﬀ and Proﬁt Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 514 61 Put Options: Payoﬀ and Proﬁt Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 523 62 Stock Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 63 Options Strategies: Floors and Caps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540 64 Covered Writings: Covered Calls and Covered Puts . . . . . . . . . . . . . . . . . . . 547 65 Synthetic Forward and Put-Call Parity . . . . . . . . . . . . . . . . . . . . . . . . . . 553 66 Spread Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 67 Collars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 68 Volatility Speculation: Straddles, Strangles, and Butterﬂy Spreads . . . . . . . . . . . 574 69 Equity Linked CDs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585 70 Prepaid Forward Contracts On Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 71 Forward Contracts on Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598 72 Futures Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 73 Understanding the Economy of Swaps: A Simple Commodity Swap . . . . . . . . . . 618 74 Interest Rate Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 75 Risk Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638 Answer Key 645 BIBLIOGRAPHY 646 8 CONTENTS The Basics of Interest Theory A component that is common to all ﬁnancial transactions is the investment of money at interest. When a bank lends money to you, it charges rent for the money. When you lend money to a bank (also known as making a deposit in a savings account), the bank pays rent to you for the money. In either case, the rent is called “interest”. In Sections 1 through 14, we present the basic theory concerning the study of interest. Our goal here is to give a mathematical background for this area, and to develop the basic formulas which will be needed in the rest of the book. 9 10 THE BASICS OF INTEREST THEORY 1 The Meaning of Interest To analyze ﬁnancial transactions, a clear understanding of the concept of interest is required. Inter- est can be deﬁned in a variety of contexts, such as the ones found in dictionaries and encyclopedias. In the most common context, interest is an amount charged to a borrower for the use of the lender’s money over a period of time. For example, if you have borrowed $100 and you promised to pay back $105 after one year then the lender in this case is making a proﬁt of $5, which is the fee for borrowing his money. Looking at this from the lender’s perspective, the money the lender is investing is changing value with time due to the interest being added. For that reason, interest is sometimes referred to as the time value of money. Interest problems generally involve four quantities: principal(s), investment period length(s), inter- est rate(s), amount value(s). The money invested in ﬁnancial transactions will be referred to as theprincipal, denoted by P.The amount it has grown to will be called the amount value and will be denoted by A.The diﬀerence I = A−P is the amount of interest earned during the period of investment. Interest expressed as a percent of the principal will be referred to as an interest rate. Interest takes into account the risk of default (risk that the borrower can’t pay back the loan). The risk of default can be reduced if the borrowers promise to release an asset of theirs in the event of their default (the asset is called collateral). The unit in which time of investment is measured is called the measurement period. The most measurement period is one year but may be longer or shorter (could be days, months, years, decades, etc.). Example 1.1 (a) The amount I owe on my credit card. (b) The amount of credit remaining on my credit card. (c) The cost of borrowing money for some period of time. (d) A fee charged on the money you’ve earned by the Federal government. Solution. The answer is (c) Example 1.2 Let A(t) denote the amount value of an investment at time t years. (a) Write an expression giving the amount of interest earned from time t to time t+ s in terms of A only. (b) Use (a) to ﬁnd the annual interest rate, i.e., the interest rate from time t years to time t+ 1 years. 1 THE MEANING OF INTEREST 11 Solution. (a) The interest earned during the time t years and t+ s years is A(t+ s) −A(t). (b) The annual interest rate is A(t+ 1) −A(t) A(t) Example 1.3 You deposit $1,000 into a savings account. One year later, the account has accumulated to $1,050. (a) What is the principal in this investment? (b) What is the interest earned? (c) What is the annual interest rate? Solution. (a) The principal is $1,000. (b) The interest earned is $1,050 - $1,000 = $50. (c) The annual interest rate is 50 1000 = 5% Interest rates are most often computed on an annual basis, but they can be determined for non- annual time periods as well. For example, a bank oﬀers you for your deposits an annual interest rate of 10% “compounded” semi-annually. What this means is that if you deposit $1000 now, then after six months, the bank will pay you 5% ×1000 = $50 so that your account balance is $1050. Six months later, your balance will be 5% ×1050 + 1050 = $1102.50.So in a period of one year you have earned $102.50. The annual interest rate is then 10.25% which is higher than the quoted 10% that pays interest semi-annually. In the next several sections, various quantitative measures of interest are analyzed. Also, the most basic principles involved in the measurement of interest are discussed. 12 THE BASICS OF INTEREST THEORY Practice Problems Problem 1.1 You invest $3,200 in a savings account on January 1, 2004. On December 31, 2004, the account has accumulated to $3,294.08. What is the annual interest rate? Problem 1.2 You borrow $12,000 from a bank. The loan is to be repaid in full in one year’s time with a payment due of $12,780. (a) What is the interest amount paid on the loan? (b) What is the annual interest rate? Problem 1.3 The current interest rate quoted by a bank on its savings accounts is 9% per year. You open an account with a deposit of $1,000. Assuming there are no transactions on the account such as depositing or withdrawing during one full year, what will be the amount value in the account at the end of the year? Problem 1.4 The simplest example of interest is a loan agreement two children might make:“I will lend you a dollar, but every day you keep it, you owe me one more penny.” Write down a formula expressing the amount value after t days. Problem 1.5 When interest is calculated on the original principal only it is calledsimple interest. Accumulated interest from prior periods is not used in calculations for the following periods. In this case, the amount value A, the principal P, the period of investment t, and the annual interest rate i are related by the formula A= P(1 + it). At what rate will $500 accumulate to $615 in 2.5 years? Problem 1.6 Using the formula of the previous problem, in how many years will 500 accumulate to 630 if the annual interest rate is 7.8%? Problem 1.7 Compounding is the process of adding accumulated interest back to the principal, so that interest is earned on interest from that moment on. In this case, we have the formula A= P(1 +i)t and we call i a yearly compound interest . You can think of compound interest as a series of back-to- back simple interest contracts. The interest earned in each period is added to the principal of the previous period to become the principal for the next period. You borrow $10,000 for three years at 5% annual interest compounded annually. What is the amount value at the end of three years? 1 THE MEANING OF INTEREST 13 Problem 1.8 Using compound interest formula, what principal does Andrew need to invest at 15% compounding annually so that he ends up with $10,000 at the end of ﬁve years? Problem 1.9 Using compound interest formula, what annual interest rate would cause an investment of $5,000 to increase to $7,000 in 5 years? Problem 1.10 Using compound interest formula, how long would it take for an investment of $15,000 to increase to $45,000 if the annual compound interest rate is 2%? Problem 1.11 You have $10,000 to invest now and are being oﬀered $22,500 after ten years as the return from the investment. The market rate is 10% compound interest. Ignoring complications such as the eﬀect of taxation, the reliability of the company oﬀering the contract, etc., do you accept the investment? Problem 1.12 Suppose that annual interest rate changes from one year to the next. Let i1 be the interest rate for the ﬁrst year, i2 the interest rate for the second year, ··· , in the interest rate for the nth year. What will be the amount value of an investment of P at the end of the nth year? Problem 1.13 Discounting is the process of ﬁnding the present value of an amount of cash at some future date. By the present value we mean the principal that must be invested now in order to achieve a desired accumulated value over a speciﬁed period of time. Find the present value of $100 in ﬁve years time if the annual compound interest is 12%. Problem 1.14 Suppose you deposit $1000 into a savings account that pays annual interest rate of 0.4% compounded quarterly (see the discussion at the end of page 11.) (a) What is the balance in the account at the end of year. (b) What is the interest earned over the year period? (c) What is the eﬀective interest rate? Problem 1.15 The process of ﬁnding the present value P of an amount A, due at the end of t years, is called discounting A.The diﬀerence A−P is called the discount on A. Notice that the discount on A is also the interest on P. For example, if $1150 is the discounted value of $1250, due at the end of 7 months, the discount on the $1250 is $100. What is the interest on $1150 for the same period of time? 14 THE BASICS OF INTEREST THEORY 2 Accumulation and Amount Functions Imagine a fund growing at interest. It would be very convenient to have a function representing the accumulated value, i.e. principal plus interest, of an invested principal at any time. Unless stated otherwise, we will assume that the change in the fund is due to interest only, that is, no deposits or withdrawals occur during the period of investment. If t is the length of time, measured in years, for which the principal has been invested, then the amount of money at that time will be denoted by A(t).This is called the amount function. Note that A(0) is just the principal P. Now, in order to compare various amount functions, it is convenient to deﬁne the function a(t) = A(t) A(0). This is called the accumulation function. It represents the accumulated value of a principal of 1 invested at time t≥0. Note that A(t) is just a constant multiple of a(t), namely A(t) = A(0)a(t). That is, A(t) is the accumulated value of an original investment of A(0). Example 2.1 Suppose that A(t) = αt2 + 10β. If X invested at time 0 accumulates to $500 at time 4, and to $1,000 at time 10, ﬁnd the amount of the original investment, X. Solution. We have A(0) = X = 10β; A(4) = 500 = 16 α+ 10β; and A(10) = 1000 = 100 α+ 10β. Using the ﬁrst equation in the second and third we obtain the following system of linear equations 16α+ X =500 100α+ X =1000. Multiply the ﬁrst equation by 100 and the second equation by 16 and subtract to obtain 1600 α+ 100X−1600α−16X = 50000 −16000 or 84X = 34000. Hence, X = 34000 84 = $404.76 What functions are possible accumulation functions? Ideally, we expect a(t) to represent the way in which money accumulates with the passage of time. Hence, accumulation functions are assumed to possess the following properties: (P1) a(0) = 1. (P2) a(t) is increasing,i.e., if t1 <t2 then a(t1) ≤a(t2).(A decreasing accumulation function implies a negative interest. For example, negative interest occurs when you start an investment with $100 and at the end of the year your investment value drops to $90. A constant accumulation function 2 ACCUMULATION AND AMOUNT FUNCTIONS 15 implies zero interest.) (P3) If interest accrues for non-integer values oft, i.e. for any fractional part of a year, then a(t) is a continuous function. If interest does not accrue between interest payment dates then a(t) possesses discontinuities. That is, the function a(t) stays constant for a period of time, but will take a jump whenever the interest is added to the account, usually at the end of the period. The graph of such an a(t) will be a step function. Example 2.2 Show that a(t) = t2 + 2t+ 1, where t ≥0 is a real number, satisﬁes the three properties of an accumulation function. Solution. (a) a(0) = 02 + 2(0) + 1 = 1. (b) a′(t) = 2t+ 2 >0 for t≥0. Thus, a(t) is increasing. (c) a(t) is continuous being a quadratic function Example 2.3 Figure 2.1 shows graphs of diﬀerent accumulation functions. Describe real-life situations where these functions can be encountered. Figure 2.1 Solution. (1) An investment that is not earning any interest. (2) The accumulation function is linear. As we shall see in Section 4, this is referred to as “simple interest”, where interest is calculated on the original principal only. Accumulated interest from prior periods is not used in calculations for the following periods. (3) The accumulation function is exponential. As we shall see in Section 6, this is referred to as “compound interest”, where the fund earns interest on the interest. (4) The graph is a step function, whose graph is horizontal line segments of unit length (the period). A situation like this can arise whenever interest is paid out at ﬁxed periods of time. If the amount of interest paid is constant per time period, the steps will all be of the same height. However, if the amount of interest increases as the accumulated value increases, then we would expect the steps to get larger and larger as time goes 16 THE BASICS OF INTEREST THEORY Remark 2.1 Properties (P2) and (P3) clearly hold for the amount function A(t). For example, since A(t) is a positive multiple of a(t) and a(t) is increasing, we conclude that A(t) is also increasing. The amount function gives the accumulated value of k invested/deposited at time 0. Then it is natural to ask what if k is not deposited at time 0, say time s> 0,then what will the accumulated value be at time t > s? For example, $100 is deposited into an account at time 2, how much does the $100 grow by time 4? Consider that a deposit of $ k is made at time 0 such that the $ k grows to $100 at time 2 (the same as a deposit of $100 made at time 2). Then A(2) = ka(2) = 100 so that k = 100 a(2) . Hence, the accumulated value of $k at time 4 (which is the same as the accumulated value at time 4 of an investment of $100 at time 2) is given by A(4) = 100 a(4) a(2) . This says that $100 invested at time 2 grows to 100 a(4) a(2) at time 4. In general, if $ k is deposited at time s, then the accumulated value of $ k at time t>s is k×a(t) a(s) , and a(t) a(s) is called the accumulation factor or growth factor. In other words, the accumulation factor a(t) a(s) gives the dollar value at time t of $1 deposited at time s. Example 2.4 It is known that the accumulation function a(t) is of the form a(t) = b(1.1)t + ct2, where b and c are constants to be determined. (a) If $100 invested at time t= 0 accumulates to $170 at time t= 3,ﬁnd the accumulated value at time t= 12 of $100 invested at time t= 1. (b) Show that a(t) is increasing. Solution. (a) By (P1), we must have a(0) = 1. Thus, b(1.1)0 + c(0)2 = 1 and this implies that b= 1. On the other hand, we have A(3) = 100a(3) which implies 170 = 100a(3) = 100[(1.1)3 + c·32] Solving for c we ﬁnd c= 0.041. Hence, a(t) = A(t) A(0) = (1.1)t + 0.041t2. It follows that a(1) = 1.141 and a(12) = 9.042428377. Now, 100 a(t) a(1) is the accumulated value of $100 investment from time t= 1 to t> 1. Hence, 100a(12) a(1) = 100 ×9.042428377 1.141 = 100(7.925002959) = 792.5002959 2 ACCUMULATION AND AMOUNT FUNCTIONS 17 so $100 at time t= 1 grows to $792.50 at time t= 12. (b) Since a(t) = (1.1)t+ 0.041t2,we have a′(t) = (1.1)tln 1.1 + 0.082t> 0 for t≥0.This shows that a(t) is increasing for t≥0 Now, let n be a positive integer. The nthperiod of time is deﬁned to be the period of time between t= n−1 and t= n. More precisely, the period normally will consist of the time interval n−1 ≤t≤n. We deﬁne the interest earned during the nth period of time by In = A(n) −A(n−1). This is illustrated in Figure 2.2. Figure 2.2 This says that interest earned during a period of time is the diﬀerence between the amount value at the end of the period and the amount value at the beginning of the period. It should be noted that In involves the eﬀect of interest over an interval of time, whereas A(n) is an amount at a speciﬁc point in time. In general, the amount of interest earned on an original investment of $ k between time s and t is I[s,t] = A(t) −A(s) = k(a(t) −a(s)). Example 2.5 Consider the amount function A(t) = t2 + 2t+ 1. Find In in terms of n. Solution. We have In = A(n) −A(n−1) = n2 + 2n+ 1 −(n−1)2 −2(n−1) −1 = 2n+ 1 Example 2.6 Show that A(n) −A(0) = I1 + I2 + ··· + In. Interpret this result verbally. Solution. We haveA(n)−A(0) = [A(1)−A(0)]+[A(2)−A(1)]+···+[A(n−1)−A(n−2)]+[A(n)−A(n−1)] = I1 + I2 + ··· + In.Hence, A(n) = A(0) + (I1 + I2 + ··· + In) so that I1 + I2 + ··· + In is the interest earned on the capital A(0). That is, the interest earned over the concatenation of n periods is the 18 THE BASICS OF INTEREST THEORY sum of the interest earned in each of the periods separately. Note that for any 0 ≤t < nwe have A(n) −A(t) = [ A(n) −A(0)] −[A(t) −A(0)] = ∑n j=1 Ij −∑t j=1 Ij = ∑n j=t+1 Ij. That is, the interest earned between time t and time n will be the total interest from time 0 to time n diminished by the total interest earned from time 0 to time t. Example 2.7 Find the amount of interest earned between time t and time n, where t < n,if Ir = r for some positive integer r. Solution. We have A(n) −A(t) = n∑ i=t+1 Ii = n∑ i=t+1 i = n∑ i=1 i− t∑ i=1 i =n(n+ 1) 2 −t(t+ 1) 2 = 1 2(n2 + n−t2 −t) where we apply the following sum from calculus 1 + 2 +··· + n= n(n+ 1) 2 2 ACCUMULATION AND AMOUNT FUNCTIONS 19 Practice Problems Problem 2.1 An investment of $1,000 grows by a constant amount of $250 each year for ﬁve years. (a) What does the graph of A(t) look like if interest is only paid at the end of each year? (b) What does the graph of A(t) look like if interest is paid continuously and the amount function grows linearly? Problem 2.2 It is known that a(t) is of the form at2 + b. If $100 invested at time 0 accumulates to $172 at time 3, ﬁnd the accumulated value at time 10 of $100 invested at time 5. Problem 2.3 Consider the amount function A(t) = t2 + 2t+ 3. (a) Find the the corresponding accumulation function. (b) Find In in terms of n. Problem 2.4 Find the amount of interest earned between time t and time n, where t < n,if Ir = 2r for some positive integer r.Hint: Recall the following sum from Calculus: ∑n i=0 ari = a1−rn+1 1−r , r̸= 1. Problem 2.5 $100 is deposited at time t= 0 into an account whose accumulation function is a(t) = 1 + 0.03 √ t. (a) Find the amount of interest generated at time 4, i.e., between t= 0 and t= 4. (b) Find the amount of interest generated between time 1 and time 4. Problem 2.6 Suppose that the accumulation function for an account is a(t) = (1 + 0.5it).You invest $500 in this account today. Find i if the account’s value 12 years from now is $1,250. Problem 2.7 Suppose that a(t) = 0.10t2 + 1.The only investment made is $300 at time 1. Find the accumulated value of the investment at time 10. Problem 2.8 Suppose a(t) = at2 + 10b.If $X invested at time 0 accumulates to $1,000 at time 10, and to $2,000 at time 20, ﬁnd the original amount of the investment X. Problem 2.9 Show that the function f(t) = 225 −(t−10)2 cannot be used as an amount function for t> 10. 20 THE BASICS OF INTEREST THEORY Problem 2.10 For the interval 0 ≤t≤10, determine the accumulation function a(t) that corresponds to A(t) = 225 −(t−10)2. Problem 2.11 Suppose that you invest $4,000 at time 0 into an investment account with an accumulation function of a(t) = αt2 + 4β. At time 4, your investment has accumulated to $5,000. Find the accumulated value of your investment at time 10. Problem 2.12 Suppose that an accumulation function a(t) is diﬀerentiable and satisﬁes the property a(s+ t) = a(s) + a(t) −a(0) for all non-negative real numbers s and t. (a) Using the deﬁnition of derivative as a limit of a diﬀerence quotient, show that a′(t) = a′(0). (b) Show that a(t) = 1 + it where i= a(1) −a(0) = a(1) −1. Problem 2.13 Suppose that an accumulation function a(t) is diﬀerentiable and satisﬁes the property a(s+ t) = a(s) ·a(t) for all non-negative real numbers s and t. (a) Using the deﬁnition of derivative as a limit of a diﬀerence quotient, show that a′(t) = a′(0)a(t). (b) Show that a(t) = (1 + i)t where i= a(1) −a(0) = a(1) −1. Problem 2.14 Consider the accumulation functions as(t) = 1 + it and ac(t) = (1 + i)t where i> 0. Show that for 0 <t< 1 we have ac(t) ≈a1(t). That is (1 + i)t ≈1 + it. Hint: Expand (1 + i)t as a power series. Problem 2.15 Consider the amount function A(t) = A(0)(1 + i)t. Suppose that a deposit 1 at time t = 0 will increase to 2 in ayears, 2 at time 0 will increase to 3 in byears, and 3 at time 0 will increase to 15 in c years. If 6 will increase to 10 in n years, ﬁnd an expression for n in terms of a,b, and c. 2 ACCUMULATION AND AMOUNT FUNCTIONS 21 Problem 2.16 For non-negative integer n, deﬁne in = A(n) −A(n−1) A(n−1) . Show that (1 + in)−1 = A(n−1) A(n) . Problem 2.17 (a) For the accumulation function a(t) = (1 + i)t, show that a′(t) a(t) = ln (1 +i). (b) For the accumulation function a(t) = 1 + it, show that a′(t) a(t) = i 1+it. Problem 2.18 Deﬁne δt = a′(t) a(t) . Show that a(t) = e ∫t 0 δrdr. Hint: Notice that d dr(ln a(r)) = δr. Problem 2.19 Show that, for any amount function A(t), we have A(n) −A(0) = ∫ n 0 A(t)δtdt. Problem 2.20 You are given that A(t) = at2 + bt+ c, for 0 ≤t ≤2, and that A(0) = 100 ,A(1) = 110 , and A(2) = 136. Determine δ1 2 . Problem 2.21 Show that if δt = δfor all tthen in = a(n)−a(n−1) a(n−1) = eδ−1.Letting i= eδ−1,show that a(t) = (1+i)t. Problem 2.22 Suppose that a(t) = 0 .1t2 + 1. At time 0, $1,000 is invested. An additional investment of $X is made at time 6. If the total accumulated value of these two investments at time 8 is $18,000, ﬁnd X. 22 THE BASICS OF INTEREST THEORY 3 Eﬀective Interest Rate (EIR) Thus far, interest has been deﬁned by Interest = Accumulated value −Principal. This deﬁnition is not very helpful in practical situations, since we are generally interested in compar- ing diﬀerent ﬁnancial situations to ﬁgure out the most proﬁtable one. In this section we introduce the ﬁrst measure of interest which is developed using the accumulation function. Such a measure is referred to as the eﬀective rate of interest: The eﬀective rate of interest is the amount of money that one unit invested at the beginning of a period will earn during the period, with interest being paid at the end of the period. If i is the eﬀective rate of interest for the ﬁrst time period then we can write i= a(1) −a(0) = a(1) −1. where a(t) is the accumulation function. Remark 3.1 We assume that the principal remains constant during the period; that is, there is no contribution to the principal or no part of the principal is withdrawn during the period. Also, the eﬀective rate of interest is a measure in which interest is paid at the end of the period compared to discount interest rate (to be discussed in Section 8) where interest is paid at the beginning of the period. We can write i in terms of the amount function: i= a(1) −a(0) = a(1) −a(0) a(0) = A(1) −A(0) A(0) = I1 A(0). Thus, we have the following alternate deﬁnition: The eﬀective rate of interest for a period is the amount of interest earned in one period divided by the principal at the beginning of the period. One can deﬁne the eﬀective rate of interest for any period: The eﬀective rate of interest in the nth period is deﬁned by in = A(n) −A(n−1) A(n−1) = In A(n−1) where In = A(n) −A(n−1).Note that In represents the amount of growth of the function A(t) in the nth period whereas in is the rate of growth (based on the amount in the fund at the beginning of the period). Thus, the eﬀective rate of interest in is the ratio of the amount of interest earned during the period to the amount of principal invested at the beginning of the period. Note that i1 = i= a(1) −1 and for any accumulation function, it must be true that a(1) = 1 + i. 3 EFFECTIVE INTEREST RATE (EIR) 23 Example 3.1 Assume that A(t) = 100(1.1)t. Find i5. Solution. We have i5 = A(5) −A(4) A(4) = 100(1.1)5 −100(1.1)4 100(1.1)4 = 0.1 Now, using the deﬁnition of in and solving for A(n) we ﬁnd A(n) = A(n−1) + inA(n−1) = (1 + in)A(n−1). Thus, the fund at the end of the nth period is equal to the fund at the beginning of the period plus the interest earned during the period. Note that the last equation leads to A(n) = (1 + i1)(1 + i2) ···(1 + in)A(0). Example 3.2 If A(4) = 1000 and in = 0.01n, ﬁnd A(7). Solution. We have A(7) =(1 + i7)A(6) =(1 + i7)(1 + i6)A(5) =(1 + i7)(1 + i6)(1 + i5)A(4) = (1.07)(1.06)(1.05)(1000) = 1,190.91 Note that in can be expressed in terms of a(t) : in = A(n) −A(n−1) A(n−1) = A(0)a(n) −A(0)a(n−1) A(0)a(n−1) = a(n) −a(n−1) a(n−1) . Example 3.3 Suppose that a(n) = 1 + in, n≥1. Show that in is decreasing as a function of n. Solution. We have in = a(n) −a(n−1) a(n−1) = [1 + in−(1 + i(n−1))] 1 + i(n−1) = i 1 + i(n−1). Since in+1 −in = i 1 + in − i 1 + i(n−1) = − i2 (1 + in)(1 + i(n−1)) <0 we conclude that as n increases in decreases 24 THE BASICS OF INTEREST THEORY Example 3.4 Show that if in = i for all n≥1 then a(n) = (1 + i)n. Solution. We have a(n) = A(n) A(0) = (1 + i1)(1 + i2) ···(1 + in) = (1 + i)n Remark 3.2 In all of the above discussion the interest rate is associated with one complete period; this will be contrasted later with rates −called “nominal”−that are stated for one period, but need to be applied to fractional parts of the period. Most loans and ﬁnancial products are stated with nominal rates such as a nominal rate that is compounded daily, or monthly, or semiannually, etc. To compare these loans, one compare their equivalent eﬀective interest rates. Nominal rates will be discussed in more details in Section 9. We pointed out in the previous section that a decreasing accumulated function leads to negative interest rate. We illustrate this in the next example. Example 3.5 You buy a house for $100,000. A year later you sell it for $80,000. What is the eﬀective rate of return on your investment? Solution. The eﬀective rate of return is i= 80,000 −100,000 100,000 = −20% which indicates a 20% loss of the original value of the house 3 EFFECTIVE INTEREST RATE (EIR) 25 Practice Problems Problem 3.1 Consider the accumulation function a(t) = t2 + t+ 1. (a) Find the eﬀective interest rate i. (b) Find in. (c) Show that in is decreasing. Problem 3.2 If $100 is deposited into an account, whose accumulation function is a(t) = 1 + 0.03 √ t, at time 0, ﬁnd the eﬀective rate for the ﬁrst period(between time 0 and time 1) and second period (between time 1 and time 2). Problem 3.3 Assume that A(t) = 100 + 5t. (a) Find i5. (b) Find i10. Problem 3.4 Assume that A(t) = 225 −(t−10)2, 0 ≤t≤10. Find i6. Problem 3.5 An initial deposit of 500 accumulates to 520 at the end of one year and 550 at the end of the second year. Find i1 and i2. Problem 3.6 A fund is earning 5% simple interest (See Problem 1.5). Calculate the eﬀective interest rate in the 6th year. Problem 3.7 Given A(5) = 2500 and i= 0.05. (a) What is A(7) assuming simple interest (See Problem 1.5)? (b) What is a(10)? Answer: (a) 2700 (b) 1.5 Problem 3.8 If A(4) = 1200,A(n) = 1800, and i= 0.06. (a) What is A(0) assuming simple interest? (b) What is n? 26 THE BASICS OF INTEREST THEORY Problem 3.9 John wants to have $800. He may obtain it by promising to pay $900 at the end of one year; or he may borrow $1,000 and repay $1,120 at the end of the year. If he can invest any balance over $800 at 10% for the year, which should he choose? Problem 3.10 Given A(0) = $1,500 and A(15) = $2,700. What is i assuming simple interest? Problem 3.11 You invest $1,000 now, at an annual simple interest rate of 6%. What is the eﬀective rate of interest in the ﬁfth year of your investment? Problem 3.12 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 3.13 Suppose a(t) = 1.12t −0.05 √ t. (a) How much interest will be earned during the 5th year on an initial investment of $12? (b) What is the eﬀective annual interest rate during the 5th year? Problem 3.14 Assume that A(t) = 100 + 5t, where t is in years. (a) Find the principal. (b) How much is the investment worth after 5 years? (c) How much is earned on this investment during the 5th year? Problem 3.15 If $64 grows to $128 in four years at a constant eﬀective annual interest rate, how much will $10,000 grow to in three years at the same rate of interest? Problem 3.16 Suppose that in = 5% for all n≥1. How long will it take an investment to triple in value? 3 EFFECTIVE INTEREST RATE (EIR) 27 Problem 3.17 You have $1000 that you want to deposit in a savings account. Bank Acomputes the amount value of your investment using the amount function A1(t) = 1 + 0.049twhereas Bank B uses the amount function A2(t) = (1.4)12t. Where should you put your money? Problem 3.18 Consider the amount function A(t) = 12(1.01)4t. (a) Find the principal. (b) Find the eﬀective annual interest rate. Problem 3.19 Given i5 = 0.1 and A(4) = 146.41. Find A(5). Problem 3.20 Given i5 = 0.1 and I5 = 14.641. Find A(4). Problem 3.21 Suppose that in = 0.01n for n≥1. Show that In = 0.01n(1.01)(1.02) ···[1 + 0.01(n−1)]A(0). Problem 3.22 If A(3) = 100 and in = 0.02n, ﬁnd A(6). 28 THE BASICS OF INTEREST THEORY 4 Linear Accumulation Functions: Simple Interest Accumulation functions of two common types of interest are discussed next. The accumulation func- tion of “simple” interest is covered in this section and the accumulation function of “compound” interest is discussed in Section 6. Consider an investment of 1 such that the interest earned in each period is constant and equals to i. Then, at the end of the ﬁrst period, the accumulated value is a(1) = 1 +i,at the end of the second period it is a(2) = 1 + 2i and at the end of the nth period it is a(n) = 1 + in, n ≥0. Thus, the accumulation function is a linear function. The accruing of interest according to this function is called simple interest. Note that the eﬀective rate of interest i= a(1)−1 is also called the simple interest rate. We next show that for a simple interest rate i, the eﬀective interest rate in is decreasing . Indeed, in = a(n) −a(n−1) a(n−1) = [1 + in−(1 + i(n−1))] 1 + i(n−1) = i 1 + i(n−1), n≥1 and in+1 −in = i 1 + in − i 1 + i(n−1) = − i2 (1 + in)(1 + i(n−1)) <0. Thus, even though the rate of simple interest is constant over each period of time, the eﬀective rate of interest per period is not constant −it is decreasing from each period to the next and converges to 0 in the long run. Because of this fact, simple interest is less favorable to the investor as the number of periods increases. Example 4.1 A fund is earning 5% simple interest. Calculate the eﬀective interest rate in the 6th year. Solution. The eﬀective interest rate in the 6th year is i6 which is given by i6 = i 1 + i(n−1) = 0.05 1 + 0.05(5) = 4% Remark 4.1 For simple interest, the absolute amount of interest earned in each time interval, i.e., In = a(n) − a(n−1) is constant whereas in is decreasing in value as n increases; in Section 6 we will see that under compound interest, it is the relative amount of interest that is constant, i.e. in = a(n)−a(n−1) a(n−1) . 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 29 The accumulation function for simple interest has been deﬁned for integral values of n ≥0. In order for this function to have the graph shown in Figure 2.1(2), we need to extend a(n) for nonintegral values of n. This is equivalent to crediting interest proportionally over any fraction of a period. If interest accrued only for completed periods with no credit for fractional periods, then the accumulation function becomes a step function as illustrated in Figure 2.1(4). Unless stated otherwise, it will be assumed that interest is allowed to accrue over fractional periods under simple interest. In order to deﬁne a(t) for real numbers t≥0 we will redeﬁne the rate of simple interest in such a way that the previous deﬁnition is a consequence of this general assumption. The general assumption states the following: Under simple interest, the interest earned by an initial investment of $1 in all time periods of length t+ s is equal to the sum of the interest earned for periods of lengths t and s. Symbolically, a(t+ s) −a(0) = [a(t) −a(0)] + [a(s) −a(0)] or a(t+ s) = a(t) + a(s) −a(0) (4.1) for all non-negative real numbers t and s. Note that the deﬁnition assumes the rule is to hold for periods of any non-negative length, not just of integer length. Are simple interest accumulation functions the only ones which preserve property (4.1)? Suppose that a(t) is a diﬀerentiable function satisfying property (4.1). Then a′(t) = lim s→0 a(t+ s) −a(t) s = lim s→0 a(t) + a(s) −a(0) −a(t) s = lim s→0 a(s) −a(0) s =a′(0), a constant Thus the time derivative of a(t) is shown to be constant. We know from elementary calculus that a(t) must have the form a(t) = a′(0)t+ C where C is a constant; and we can determine that constant by assigning to t the particular value 0, so that C = a(0) = 1. 30 THE BASICS OF INTEREST THEORY Thus, a(t) = 1 + a′(0)t. Letting t= 1 and deﬁning i1 = i= a(1) −a(0) we can write a(t) = 1 + it, t ≥0. Consequently, simple interest accumulation functions are the only ones which preserve property (4.1). It is important to notice that the above derivation does not depend ontbeing a nonnegative integer, and is valid for all nonnegative real numbers t. Example 4.2 You invest $100 at time 0, at an annual simple interest rate of 10%. Find the accumulated value after 6 months. Solution. The accumulated function for simple interest is a continuous function. Thus, A(0.5) = 100[1 + 0.1(0.5)] = $105 Remark 4.2 Simple interest is in general inconvenient for use by banks. For if such interest is paid by a bank, then at the end of each period, depositors will withdraw the interest earned and the original deposit and immediately redeposit the sum into a new account with a larger deposit. This leads to a higher interest earning for the next investment year. We illustrate this in the next example. Example 4.3 Consider the following investments by John and Peter. John deposits $100 into a savings account paying 6% simple interest for 2 years. Peter deposits $100 now with the same bank and at the same simple interest rate. At the end of the year, he withdraws his balance and closes his account. He then reinvests the total money in a new savings account oﬀering the same rate. Who has the greater accumulated value at the end of two years? Solution. John’s accumulated value at the end of two years is 100(1 + 0.06 ×2) = $112. Peter’s accumulated value at the end of two years is 100(1 + 0.06)2 = $112.36. 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 31 Thus, Peter has a greater accumulated value at the end of two years Simple interest is very useful for approximating compound interest, a concept to be discussed in Section 6, for a short time period such as a fraction of a year. To be more speciﬁc, we will see that the accumulation function for compound interest i is given by the formula a(t) = (1 + i)t. Using the binomial theorem we can write the series expansion of a(t) obtaining (1 + i)t = 1 + it+ t(t−1) 2! i2 + t(t−1)(t−2) 3! i3 + ··· . Thus, for 0 <t< 1 we can write the approximation (1 + i)t ≈1 + it. Example 4.4 $10,000 is invested for four months at 12.6% compounded annually, that is A(t) = 10000(1 + 0.126)t, where interest is computed using a quadratic to approximate an exact calculation. Find the accumulated value. Solution. We want to estimate A(1/3) = 10000(1 + 0.126) 1 3 using the ﬁrst three terms of the series expansion of (1 + i)t discussed above. That is, A(1/3) =10000(1.126) 1 3 ≈10000 ( 1 + 1 3 ×0.126 + (1 3 )(−2 3 ) 2! (0.126)2 ) = $10,402.36 Remark 4.3 Under simple interest, what is the accumulated value at time t > sof 1 deposited at time s? The SOA/CAS approach is diﬀerent than the approach discussed in Section 2 right after Remark 2.1. According to the SOA/CAS, “simple interest” is generally understood to mean that the linear function starts all over again from the date of each deposit or withdrawal. We illustrate this approach in the next example Example 4.5 Suppose you make a deposit of $100 at time t = 0. A year later, you make a withdrawal of $50. Assume annual simple interest rate of 10%, what is the accumulated value at time t= 2 years? Solution. With the SOA/CAS recommended approach for simple interest, the answer is 100(1 + 0.1 ×2) −50(1 + 0.1 ×1) = $65 32 THE BASICS OF INTEREST THEORY Practice Problems Problem 4.1 You invest $100 at time 0, at an annual simple interest rate of 9%. (a) Find the accumulated value at the end of the ﬁfth year. (b) How much interest do you earn in the ﬁfth year? Problem 4.2 At what annual rate of simple interest will $500 accumulate to $615 in 2 1 2 years? Problem 4.3 In how many years will $500 accumulate to $630 at 7.8% annual simple interest? Problem 4.4 What principal will earn interest of 100 in 7 years at a simple interest rate of 6%? Problem 4.5 What simple interest rate is necessary for $10,000 to earn $100 interest in 15 months? Problem 4.6 At a certain rate of simple interest $1,000 will accumulate to $1110 after a certain period of time. Find the accumulated value of $500 at a rate of simple interest three fourths as great over twice as long a period of time. Problem 4.7 At time 0, you invest some money into an account earning 5.75% simple interest. How many years will it take to double your money? Problem 4.8 You invest $1,000 now, at an annual simple interest rate of 6%. What is the eﬀective rate of interest in the ﬁfth year of your investment? Problem 4.9 Suppose that the accumulation function for an account is a(t) = 1 + 3it.At time 0, you invest $100 in this account. If the value in the account at time 10 is $420, what is i? Problem 4.10 You have $260 in a bank savings account that earns simple interest. You make no subsequent deposits in the account for the next four years, after which you plan to withdraw the entire account balance and buy the latest version of the iPod at a cost of $299. Find the minimum rate of simple interest that the bank must oﬀer so that you will be sure to have enough money to make the purchase in four years. 4 LINEAR ACCUMULATION FUNCTIONS: SIMPLE INTEREST 33 Problem 4.11 The total amount of a loan to which interest has been added is $20,000. The term of the loan was four and one-half years. If money accumulated at simple interest at a rate of 6%, what was the amount of the loan? Problem 4.12 If ik is the rate of simple interest for period k, where k = 1 ,2,··· ,n, show that a(n) −a(0) = i1 + i2 + ··· + in. Be aware that in is not the eﬀective interest rate of the nth period as deﬁned in the section! Problem 4.13 A fund is earning 5% simple interest. The amount in the fund at the end of the 5th year is $10,000. Calculate the amount in the fund at the end of 7 years. Problem 4.14 Simple interest of i = 4% is being credited to a fund. The accumulated value at t = n−1 is a(n−1).The accumulated value at t= nis a(n) = 1 + 0.04n.Find nso that the accumulated value of investing a(n−1) for one period with an eﬀective interest rate of 2.5% is the same as a(n). Problem 4.15 A deposit is made on January 1, 2004. The investment earns 6% simple interest. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 4.16 Consider an investment with nonzero interest rate i. If i5 is equal to i10, show that interest is not computed using simple interest. Problem 4.17 Smith has just ﬁled his income tax return and is expecting to receive, in 60 days, a refund check of 1000. (a) The tax service that helped him ﬁll out his return oﬀers to buy Smith’s refund check for 850. What annual simple interest rate is implied? (b) Smith negotiates and sells his refund check for 900. What annual simple interest rate does this correspond to? (c) Smith deposits the 900 in an account which earns simple interest at annual rate 9%. How many days would it take from the time of his initial deposit of 900 for the account to reach 1000? Problem 4.18 Let i be a simple interest rate and suppose that i6 = 0.04. Calculate i. 34 THE BASICS OF INTEREST THEORY Problem 4.19 A fund is earning 5% simple interest. If in = 0.04, calculate n. Problem 4.20 Suppose that an account earns simple interest with annual interest rate of i. If an investment of k is made at time s years, what is the accumulated value at time t>s years? Note that the answer is diﬀerent from ka(t) a(s) . Problem 4.21 Suppose A(5) = $2,500 and i= 0.05. (a) What is A(7) assuming simple interest? (b) What is a(10) assuming simple interest? Problem 4.22 If A(4) = $1,200 and A(n) = $1,800, (a) what is A(0), assuming a simple interest of 6%? (b) what is n if i= 0.06 assuming simple interest? Problem 4.23 What is A(15), if A(0) = $1,100, simple interest is assumed and in = 0.01n? 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 35 5 Date Conventions Under Simple Interest In the simple interest problems encountered thus far, the length of the investment has been an integral number of years. What happens if the time is given in days. In this section we discuss three techniques for counting the number of days in a period of investment or between two dates. In all three methods, time= # of days between two dates # of days in a year . In what follows, it is assumed, unless stated otherwise, that in counting days interest is not credited for both the starting date and the ending date, but for only one of these dates. Exact Simple Interest: The “actual/actual” method is to use the exact number of days for the period of investment and to use 365 days in a nonleap year and 366 for a leap year (a year divisible by 4). Simple interest computed with this method is called exact simple interest. For this method, it is important to know the number of days in each month. In counting days between two dates, the last, but not the ﬁrst, date is included. Example 5.1 Suppose that $2,500 is deposited on March 8 and withdrawn on October 3 of the same year, and that the interest rate is 5%. Find the amount of interest earned, if it is computed using exact simple interest. Assume non leap year. Solution. From March 8 (not included) to October 3 (included) there are 23+30+31+30+31+31+30+3 = 209 days. Thus, the amount of interest earned using exact simple interest is 2500(0 .05) ·209 365 = $71.58 Ordinary Simple Interest: This method is also known as “30/360”. The 30/360 day counting scheme was invented in the days before computers to make the computations easier. The premise is that for the purposes of computation, all months have 30 days, and all years have 12 ×30 = 360 days. Simple interest computed with this method is called ordinary simple interest. The Public Securities Association (PSA) publishes the following rules for calculating the number of days between any two dates from M1/D1/Y1 to M2/D2/Y2: •If D1 (resp. D2) is 31, change D1 (resp. D2)to 30. •If M1 (resp. M2) is 2, and D1 (resp. D2) is 28 (in a non-leap year) or 29, then change D1 (resp. D2) to 30. Then the number of days, N is: N = 360(Y2 −Y1) + 30(M2 −M1) + (D2 −D1). 36 THE BASICS OF INTEREST THEORY For exampe, the number of days from February 25 to March 5 of the same year is 10 days. Like the exact simple interest, the ending date is counted and not the starting date. Example 5.2 Jack borrows 1,000 from the bank on January 28, 1996 at a rate of 15% simple interest per year. How much does he owe on March 5, 1996? Use ordinary simple interest. Solution. The amount owed at time t is A(t) = 1000(1 + 0 .15t). Using ordinary simple interest with Y1 = Y2 = 1996,M1 = 1,M2 = 3, and D1 = 28 and D2 = 5 we ﬁnd t = 37 360 and the amount owed on March 5, 1996 is 1,000 ( 1 + 0.15 × 37 360 ) = $1,015.42 Banker’s Rule: This method is also known as “actual/360”. This method uses the exact number of days for the period of investment and that the calendar year has 360 days. Simple interest computed with this method is called Banker’s rule. The number of days between two dates is found in the same way as for exact simple interest. In this method, we also count the last day but not the ﬁrst day. Example 5.3 Jack borrows 1,000 from the bank on January 1, 1996 at a rate of 15% simple interest per year. How much does he owe on January 17, 1996?Use Banker’s rule. Solution. Jack owes 1000 ( 1 + 0.15 × 16 360 ) = $1,006.67 Example 5.4 If an investment was made on the date the United States entered World War II, i.e., December 7, 1941, and was terminated at the end of the war on August 8, 1945, for how many days was the money invested under: 1. the “actual/actual” basis? 2. the “30/360” basis? Solution. 1. From December 7, 1941(not included) to December 31, 1941 (included) there were 24 days. From January 1, 1942 to December 31, 1944 ( including a leap year) there were 3(365) + 1 = 1096 days. 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 37 From January 1, 1945 to August 8, 1945(included) the numbers of days is 31 + 28 + 31 + 30 + 31 + 30 + 31 + 8 = 220 days. The total number of days is 24 + 1096 + 220 = 1340. 2. We have 360(1945 −1941) + 30(8−12) + (8−7) = 1321 . Remark 5.1 If the time is given in months, reduce it to a fraction of a year on the basis of 12 months to the year, without changing to days. Example 5.5 A merchant is oﬀered $50 discount for cash payment of a $1200 bill due in two months. If he pays cash, at what rate may he consider his money to be earning interest in the next two months? Solution. The merchant would pay now $1150 in place of $1200 in two months. To ﬁnd the interest rate under which $1150 is the present value of $1200, due in two months, use the formula I = Pit which by substitution becomes 50 = 1150i 6 . Solving for i we ﬁnd i= 26.087% We end this section by pointing out that the methods discussed above do not only apply for simple interest rate problems but also to compound interest rate problems. Compound interest rates are introduced in the next section. Unless otherwise stated, in later sections we will assume always the actual/actual method is in use. 38 THE BASICS OF INTEREST THEORY Practice Problems Problem 5.1 Find the amount of interest that $2,000 deposited on June 17 will earn, if the money is withdrawn on September 10 in the same year and if the simple rate of interest is 8% using (a) exact simple interest, (b) ordinary simple interest, and (c) Banker’s rule. Assume non-leap year. Problem 5.2 A sum of 10,000 is invested for the months of July and August at 6% simple interest. Find the amount of interest earned: 1. Assuming exact simple interest (Assume non-leap year). 2. Assuming ordinary simple interest. 3. Assuming the Banker’s Rule. Problem 5.3 Show that the Banker’s Rule is always more favorable to the lender than is exact simple interest. Problem 5.4 (a) Show that the Banker’s Rule is usually more favorable to the lender than is ordinary simple interest. (b) Give an example in (a) for which the opposite relationship holds. Problem 5.5 Suppose that $2,500 is deposited on March 8 and withdrawn on October 3 of the same year, and that the interest rate is 5%. Find the amount of interest earned, if it is computed using (a) exact simple interest (Assume non-leap year), (b) ordinary simple interest, (c) the Banker’s Rule. Problem 5.6 The sum of $ 5,000 is invested for the months of April, May, and June at 7% simple interest. Find the amount of interest earned (a) assuming exact simple interest in a non-leap year; (b) assuming exact simple interest in a leap year (with 366 days); (c) assuming ordinary simple interest; (d) assuming the Banker’s Rule. Problem 5.7 Fund A calculates interest using exact simple interest (actual/actual). Fund B calculates interest 5 DATE CONVENTIONS UNDER SIMPLE INTEREST 39 using ordinary simple interest (30/360). Fund C calculates interest using the Banker’s Rule (ac- tual/360). All Funds earn 5% simple interest and have the same amount of money deposited on January 1, 2005. Order the Funds based on the amount in the funds on March 1, 2005 from smallest to largest. Problem 5.8 Suppose you lend $60 to your sister on Sept 14, at an annual rate of simple interest of 10%, to be repaid on Dec 25. How much does she have to pay you back? Use actual/360 time measurement. Problem 5.9 John borrows $60 from Eddie. If he repays Eddie $63 after 5 weeks, what simple interest has John paid? Use actual/actual for days counting. Assume non-leap year. Problem 5.10 Henry invests 1000 on January 15 in an account earning simple interest at an annual eﬀective rate of 10%. On November 25 of the same year, Henry withdraws all his money. How much money will Henry withdraw if the bank counts days: (a) Using exact simple interest (ignoring February 29th) (b) Using ordinary simple interest (c) Using Banker’s Rule Problem 5.11 Let Ie denote the interest earned using the exact simple interest method and Ib the interest earned using the Banker’s rule. Find the ratio Ie Ib . Problem 5.12 Suppose that the interest earned using the Banker’s rule is $40.58. What is the interest earned using the exact simple interest method? Problem 5.13 Show that Ib = Ie + Ie 72 and Ie = Ib −Ib 73 . Thus, the Banker’s rule is more favorable to the investor than the actual/actual rule. Problem 5.14 On January 1, 2000, you invested $1,000. Your investment grows to $1,400 by December 31, 2007. What was the exact simple interest rate at which you invested? Problem 5.15 A 3% discount is oﬀered for cash payment of a $2500 bill, due at the end of 90 days. At what rate of simple interest earned over the 90 days is cash payment is made? Use the Banker’s rule. 40 THE BASICS OF INTEREST THEORY Problem 5.16 The terms of payment of a certain debt are: net cash in 90 days or 2% discount for cash in 30 days. At what rate of simple interest is earned if the discount is taken advantage of? 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 41 6 Exponential Accumulation Functions: Compound Interest Simple interest has the property that the interest earned is not invested to earn additional interest. In contrast, compound interest has the property that the interest earned at the end of one period is automatically invested in the next period to earn additional interest. We next ﬁnd the accumulation function for compound interest. Starting with an investment of 1 and with compound interest rate iper period. At the end of the ﬁrst period, the accumulated value is 1 + i. At the end of the second period, the accumulated value is (1 + i) + i(1 + i) = (1 + i)2. Continuing this way, we ﬁnd that the accumulated value after tperiods is given by the exponential function a(t) = (1 + i)t, for integral t ≥0. Interest accruing according to this function is called compound interest. We call i the rate of compound interest. From the above accumulation function, we can write in = a(n) −a(n−1) a(n−1) = (1 + i)n −(1 + i)n−1 (1 + i)n−1 = i. Thus, the eﬀective rate of interest for compound interest is constant. The accumulation function for compound interest has been deﬁned for nonnegative integers. In order to extend the domain to fractional periods we note that the compound interest accumulation function a(t) = (1 + i)t satisﬁes the property (1 + i)t+s = (1 + i)t ·(1 + i)s. Thus, under compound interest we will require the accumulation function to satisfy the property a(t+ s) = a(t) ·a(s), t,s ≥0. (6.1) This formula says that under compound interest the amount of interest earned by an initial invest- ment of 1 over t+speriods is equal to the amount of interest earned if the investment is terminated at the end of t periods and the accumulated value at that point is immediately reinvested for an additional s periods. Are compound interest accumulation functions the only ones which preserve property (6.1)? Assum- ing a(t) is diﬀerentiable and satisfying the above property, then from the deﬁnition of the derivative 42 THE BASICS OF INTEREST THEORY we have a′(t) = lim s→0 a(t+ s) −a(t) s = lim s→0 a(t)a(s) −a(t) s =a(t) lim s→0 a(s) −a(0) s =a(t)a′(0) Hence, a′(t) a(t) = (ln a(t))′= a′(0) a constant for all t≥0. Hence, ln a(t) = a′(0)t+ C Since a(0) = 1, we obtain C = 0 so ln a(t) = a′(0)t. Setting t= 1 and recalling that a(1) −a(0) = i1 = i yields a′(0) = ln (1 +i) so a(t) = (1 + i)t and this is valid for all t≥0. We thus see the graphical distinction between simple and compound interest: the graph of an accumulation function under simple interest is a straight line −a linear function; the graph of an accumulation function under compound interest is an exponential function. We next present a comparison result between simple interest and compound interest. Theorem 6.1 Let 0 <i< 1. We have (a) (1 + i)t <1 + it for 0 <t< 1, (b) (1 + i)t = 1 + it for t= 0 or t= 1, (c) (1 + i)t >1 + it for t> 1. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 43 Proof. (a) Suppose 0 < t <1. Let f(i) = (1 + i)t −1 −it. Then f(0) = 0 and f′(i) = t(1 + i)t−1 −t. Since i> 0, it follows that 1 + i> 1, and since t< 1 we have t−1 <0. Hence, (1 + i)t−1 <1 and therefore f′(i) = t(1 + i)t−1 −t< 0 for 0 <t< 1. It follows that f(i) <0 for 0 <t< 1. (b) Follows by substitution. (c) Suppose t >1. Let g(i) = 1 + it−(1 + i)t. Then g(0) = 0 and g′(i) = t−t(1 + i)t−1. Since i >0, we have 1 + i >1. Since t >1 we have t−1 > 0. Hence (1 + i)t−1 > 1 and therefore g′(i) = t−t(1 + i)t−1 <t −t= 0. We conclude that g(i) <0. This establishes part (c) Thus, simple and compound interest produce the same result over one measurement period. Com- pound interest produces a larger return than simple interest for periods greater than 1 and smaller return for periods smaller than 1. See Figure 6.1. Figure 6.1 It is worth observing the following (1) With simple interest, the absolute amount of growth is constant, that is, for a ﬁxed s the diﬀerence a(t+ s) −a(t) = a(s) −1 does not depend on t. (2) With compound interest, the relative rate of growth is constant, that is, for a ﬁxed s the ratio [a(t+s)−a(t)] a(t) = a(s) −1 does not depend on t. Example 6.1 It is known that $600 invested for two years will earn $264 in interest. Find the accumulated value of $2,000 invested at the same rate of compound interest for three years. Solution. We are told that 600(1 + i)2 = 600 + 264 = 864 . Thus, (1 + i)2 = 1.44 and solving for i we ﬁnd i = 0 .2. Thus, the accumulated value of investing $2,000 for three years at the rate i = 20% is 2,000(1 + 0.2)3 = $3,456 44 THE BASICS OF INTEREST THEORY Example 6.2 At an annual compound interest rate of 5%, how long will it take you to triple your money? (Provide an answer in years, to three decimal places.) Solution. We must solve the equation (1 + 0.05)t = 3. Thus, t= ln 3 ln 1.05 ≈22.517 Example 6.3 At a certain rate of compound interest, 1 will increase to 2 in ayears, 2 will increase to 3 in byears, and 3 will increase to 15 in c years. If 6 will increase to 10 in n years, ﬁnd an expression for n in terms of a,b, and c. Solution. If the common rate is i, the hypotheses are that 1(1 + i)a =2 →ln 2 = aln (1 +i) 2(1 + i)b =3 →ln 3 2 = bln (1 +i) 3(1 + i)c =15 →ln 5 = cln (1 +i) 6(1 + i)n =10 →ln 5 3 = nln (1 +i) But ln 5 3 = ln 5−ln 3 = ln 5−(ln 2 + ln 1.5). Hence, nln (1 +i) = cln (1 +i) −aln (1 +i) −bln (1 +i) = (c−a−b) ln (1 +i) and this implies n= c−a−b We conclude this section by noting that the three counting days techniques discussed in Section 5 for simple interest applies as well for compound interest. Example 6.4 Christina invests 1000 on April 1 in an account earning compound interest at an annual eﬀective rate of 6%. On June 15 of the same year, Christina withdraws all her money. Assume non-leap year, how much money will Christina withdraw if the bank counts days: (a) Using actual/actual method. (b) Using 30/360 method. (c) Using actual/360 method. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 45 Solution. (a) The number of days is 29 + 31 + 15 = 75. Thus, the amount of money withdrawn is 1000(1 + 0.06) 75 365 = $1,012.05. (b) Using the 30/360, we ﬁnd 1000(1 + 0.06) 74 360 = $1,012.05. (c) Using the actual/360 method, we ﬁnd 1000(1 + 0.06) 75 360 = $1,012.21 46 THE BASICS OF INTEREST THEORY Practice Problems Problem 6.1 If $4,000 is invested at an annual rate of 6.0% compounded annually, what will be the ﬁnal value of the investment after 10 years? Problem 6.2 Jack has deposited $1,000 into a savings account. He wants to withdraw it when it has grown to $2,000. If the interest rate is 4% annual interest compounded annually, how long will he have to wait? Problem 6.3 At a certain rate of compound interest, $250 deposited on July 1, 2005 has to accumulate to $275 on January 1, 2006. Assuming the interest rate does not change and there are no subsequent deposits, ﬁnd the account balance on January 1, 2008. Problem 6.4 You want to triple your money in 25 years. What is the annual compound interest rate necessary to achieve this? Problem 6.5 You invest some money in an account earning 6% annual compound interest. How long will it take to quadruple your account balance? (Express your answer in years to two decimal places.) Problem 6.6 An amount of money is invested for one year at a rate of interest of 3% per quarter. Let D(k) be the diﬀerence between the amount of interest earned on a compound interest basis, and on a simple interest basis for quarter k, where k= 1,2,3,4. Find the ratio of D(4) to D(3). Problem 6.7 Show that the ratio of the accumulated value of 1 invested at rateifor nperiods, to the accumulated value of 1 invested at ratejfor nperiods, where i>j, is equal to the accumulated value of 1 invested for n periods at rate r.Find an expression for r as a function of i and j. Problem 6.8 At a certain rate of compound interest an investment of $1,000 will grow to $1,500 at the end of 12 years. Determine its value at the end of 5 years. Problem 6.9 At a certain rate of compound interest an investment of $1,000 will grow to $1,500 at the end of 12 years. Determine precisely when its value is exactly $1,200. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 47 Problem 6.10 ‡ Bruce and Robbie each open up new bank accounts at time 0. Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns the same annual eﬀective interest rate. The amount of interest earned in Bruce’s account during the 11th year is equal to X. The amount of interest earned in Robbie’s account during the 17th year is also equal to X. Calculate X. Problem 6.11 Given A(5) = $7,500 and A(11) = $9,000. What is A(0) assuming compound interest? Problem 6.12 If $200 grows to $500 overnyears, what will $700 grow to over 3nyears? Assuming same compound annual interest rate. Problem 6.13 Suppose that your sister repays you $62 two months after she borrows $60 from you. What eﬀective annual rate of interest have you earned on the loan? Assume compound interest. Problem 6.14 An investor puts 100 into Fund X and 100 into Fund Y. Fund Y earns compound interest at the annual rate of j >0,and Fund X earns simple interest at the annual rate of 1 .05j.At the end of 2 years, the amount in Fund Y is equal to the amount in Fund X. Calculate the amount in Fund Y at the end of 5 years. Problem 6.15 Fund A is invested at an eﬀective annual interest rate of 3%. Fund B is invested at an eﬀective annual interest rate of 2.5%. At the end of 20 years, the total in the two funds is 10,000. At the end of 31 years, the amount in Fund A is twice the amount in Fund B. Calculate the total in the two funds at the end of 10 years. Problem 6.16 Carl puts 10,000 into a bank account that pays an annual eﬀective interest rate of 4% for ten years. If a withdrawal is made during the ﬁrst ﬁve and one-half years, a penalty of 5% of the withdrawal amount is made. Carl withdraws K at the end of each of years 4, 5, 6, 7. The balance in the account at the end of year 10 is 10,000. Calculate K. Problem 6.17 ‡ Joe deposits 10 today and another 30 in ﬁve years into a fund paying simple interest of 11% per 48 THE BASICS OF INTEREST THEORY year. Tina will make the same two deposits, but the 10 will be deposited nyears from today and the 30 will be deposited 2 n years from today. Tina’s deposits earn an annual eﬀective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits. Calculate n. Problem 6.18 Complete the following table. Simple interest Compound interest a(t) = a(0) = a(1) = Period during which a(t) is greater in = Problem 6.19 On January 1, 2000, you invested $1,000. Your investment grows to $1,400 by December 31, 2007. What was the compound interest rate at which you invested? Problem 6.20 A certiﬁcate of deposit or CD is a ﬁnancial product commonly oﬀered to consumers by banks and credit unions. A CD has a speciﬁc, ﬁxed term (often three months, six months, or one to ﬁve years), and, usually, a ﬁxed interest rate. It is intended that the CD be held until maturity, at which time the money may be withdrawn together with the accrued interest. However, a type a penalty is imposed for early withdrawal. 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. 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. 6 EXPONENTIAL ACCUMULATION FUNCTIONS: COMPOUND INTEREST 49 Problem 6.21 You invest $10,000 at time 0 into each of two accounts. Account A earns interest at an annual simple interest rate of 8%; Account B earns interest at an eﬀective annual compound interest rate of 6%. What is the diﬀerence in the amount of interest earned during the 5th year in these two accounts? Problem 6.22 On April 1, 2013, you will need $10,000. Assuming a 6% annual compound eﬀective rate of interest, what would you have to invest on October 1, 2008, in order for you to fulﬁll that need? Problem 6.23 On January 1, 2009, you invest $500 into an account earning an 8% annual compound eﬀective interest rate. On January 1, 2011, you deposit an additional $1,000 into the account. What is the accumulated value of your account on January 1, 2014? Problem 6.24 Let i be a compound eﬀective interest rate. Show that In = iA(n−1). Problem 6.25 Suppose that A(1) = $110,A(2) = $121, and A(10) = $259.37. (a) Find the eﬀective rate of interest. (b) Is the interest a simple interest or a compound interest? (c) Find I5. Problem 6.26 If A(5) = $1,500 and A(n) = $2,010.14, what is n if i= 0.05 assuming compound interest? Problem 6.27 How much interest is earned in the 9th year by $200 invested today under compound interest with i= 0.07? 50 THE BASICS OF INTEREST THEORY 7 Present Value and Discount Functions In this section we discuss the question of determining the today’s or present value of some amount in the past or in the future. Unless stated otherwise, we assume in this section that we are in a compound interest situation, where a(t) = (1 + i)t. Why would you be interested in present values? Suppose that in your personal ﬁnancial life, you need to save or invest money that will grow to a speciﬁed amount in a speciﬁed number of years. For example, you may want to start saving now to buy a car for $20,000 in two years, or to make a down payment of $25,000 on a home in 3 years, or to pay a year’s college expenses of $30,000 in 8 years. All of these questions involve ﬁnding the present value (PV), i.e., the amount invested today that will grow to the desired amount in the desired time. To start with, suppose we want to know what $1 a period ago, invested at compound interest rate i per period, worths today. If X is the accumulated value then we must have 1(1 + i) = X. Thus, $1 a period ago worths 1 + i dollars now. We call 1 + i the accumulation factor. Similarly, $1 a period from now invested at the rate i worths ν = 1 1+i today. We call ν the discount factor since it discounts the value of an investment at the end of a period to its value at the beginning of the period. The above discussion can be generalized to more than one period. For example, $1 investedtperiods ago worths (1 + i)t today and $1 invested t periods from now is worth 1 (1+i)t today. We call 1 (1+i)t a discount function. It represents the amount that needs to be invested today at interest rate i per period to yield an amount of $1 at the end of t time periods. This function can be expressed in terms of the accumulation function a(t). Indeed, since [ a(t)]−1a(t) = 1, the discount function is [a(t)]−1 = 1 (1+i)t = νt. In a sense, accumulation and discounting are opposite processes. The term (1 + i)t is said to be the accumulated value of $1 at the end of ttime periods. The term νt is said to be the present value or discounted(back) value of $1 to be paid at the end of t periods. Example 7.1 What is the present value of $8,000 to be paid at the end of three years if the interest rate is 11% compounded annually? Solution. Let FV stands for the future value and PV for the present value. We want to ﬁnd PV. We have FV = PV (1 + i)3 or PV = FV (1 + i)−3. Substituting into this equation we ﬁnd PV = 8000(1.11)−3 ≈$5,849.53 Example 7.2 Show that the current value of a payment of 1 made n >1 periods ago and a payment of 1 to be made n periods in the future is greater than 2, if i> 0. 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 51 Solution. We have (1 + i)n + (1 + i)−n = ( (1 + i) n 2 −(1 + i)−n 2 )2 + 2 ≥2. Equality holds if and only if (1 + i) n 2 = (1 + i)−n 2 which is equivalent to (1 + i)n = 1. This happens when either i= 0 or n= 0. Hence, (1 + i)n + (1 +i)−n >2 since by assumption n> 1 and i> 0 Example 7.3 Find an expression for the discount factor during the nth period from the date of investment, i.e. (1 + in)−1, in terms of the amount function. Solution. Recall that in = A(n)−A(n−1) A(n−1) . Thus, 1 1 + in = 1 1 + A(n)−A(n−1) A(n−1) = A(n−1) A(n) Remark 7.1 One should observe that νt extend the deﬁnition of a(t) = (1 + i)t to negative values of t. Thus, a graph of a(t) is given in Figure 7.1. Figure 7.1 What happens to the calculation of present values if simple interest is assumed instead of compound interest? The accumulation function is now a(t) = 1 + it. Hence, the present value of 1, t years in the future, is (a(t))−1 = 1 1 + it, t≥0. Example 7.4 Find the present value of $3,000 to be paid at the end of 5 years with a rate of simple interest of 7% per annum. 52 THE BASICS OF INTEREST THEORY Solution. The answer is 3000[a(5)]−1 = 3000 1+0.07(5) ≈$2,222.22 Example 7.5 Rework Example 7.4 using compound interest instead of simple interest. Solution. The answer is 3000 1.075 = $2,138.96 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 53 Practice Problems Problem 7.1 At an eﬀective rate of interest of 8% per annum, the present value of $100,000 due in X years is $65,322. Determine X. Problem 7.2 What deposit made today will provide for a payment of $1,000 in 1 year and $2,000 in 3 years, if the eﬀective rate of interest is 7.5%? Problem 7.3 The total amount of a loan to which interest has been added is $20,000. The term of the loan was four and one-half years. (a) If money accumulated at simple interest at a rate of 6%, what was the amount of the loan? (b) If the annual rate of interest was 6% and interest was compounded annually, what was the amount of the loan? Problem 7.4 It is known that an investment of $500 will increase to $4,000 at the end of 30 years. Find the sum of the present values of three payments of $10,000 each which will occur at the end of 20, 40, and 60 years. Problem 7.5 The sum of the present value of 1 paid at the end of nperiods and 1 paid at the end of 2 nperiods is 1. Find (1 + i)2n. Problem 7.6 ‡ Sally has two IRA’s. IRA #1 earns interest at 8% eﬀective annually and IRA #2 earns interest at 10% eﬀective annually. She has not made any contributions since January 1, 1985, when the amount in IRA #1 was twice the amount in IRA #2. The sum of the two accounts on January 1, 1993 was $75,000. Determine how much was in IRA #2 on January 1, 1985. Problem 7.7 ‡ At an annual eﬀective interest rate of i,i> 0, the following are all equal: (i) the present value of 10,000 at the end of 6 years; (ii) the sum of the present values of 6,000 at the end of year tand 56,000 at the end of year 2 t; and (iii) 5,000 immediately. Calculate the present value of a payment of 8,000 at the end of year t+ 3 using the same annual eﬀective interest rate. 54 THE BASICS OF INTEREST THEORY Problem 7.8 The Kelly family buys a new house for $93,500 on May 1, 1996. How much was this house worth on May 1, 1992 if real estate prices have risen at a compound rate of 8% per year during that period? Problem 7.9 Find the present (discounted) value of $3,000 to be paid at the end of 5 years if the accumulation function is a(t) = 1 + t2 25 . Problem 7.10 Suppose a(t) = αt2 + 10β. If $X invested at time 0 accumulates to $1,000 at time 10, and to $2,000 at time 20, ﬁnd the original amount of the investment X. Problem 7.11 You want to buy a sailboat when you retire 5 years from now. The sailboat you wish to buy costs $50,000 today and inﬂation is at 4% per year. What do you need to put into an account today to have enough funds to be able to purchase the boat 5 years from now if the account earns 7% per year? Problem 7.12 ‡ A store is running a promotion during which customers have two options for payment. Option one is to pay 90% of the purchase price two months after the date of sale. Option two is to deduct X% oﬀ the purchase price and pay cash on the date of sale. A customer wishes to determine X such that he is indiﬀerent between the two options when valuing them using an eﬀective annual interest rate of 8%. Which of the following equations would the customer need to solve? (A) (X 100 )( 1 + 0.08 6 ) = 0.90 (B) ( 1 − X 100 )( 1 + 0.08 6 ) = 0.90 (C) (X 100 ) (1.08) 1 6 = 0.90 (D) (X 100 )(1.08 1.06 ) = 0.90 (E) ( 1 − X 100 ) (1.08) 1 6 = 0.90 Problem 7.13 Derive a formula for i in terms of ν where ν = 1 1+i. Problem 7.14 In this section the present value is assumed to be the value of an investment at time 0. We can determine the present value of an investment at any time n not necessarily at time 0. A payment of $10 is to be made at time 7 years. (a) Determine the present value of this payment at time 4 years if the annual compound interest is 6%. (b) Determine the present value of this payment at time 4 years if the annual simple interest is 6%. 7 PRESENT VALUE AND DISCOUNT FUNCTIONS 55 Problem 7.15 A treasury bill or T −Bill with face value $100 is a security which is exchangeable for $100 on the maturity date. Suppose that a T −Bill with face value $100 is issued on 08/09/2005 and matures on 03/09/2006 (there are 182 days between the dates). Assuming simple interest, given that the annual rate of interest is 2.810%, ﬁnd the price of the security. Problem 7.16 If we are promised $1,500 5 years from now, what is its present value today at: (a) 7% simple interest? (b) 7% compound interest? Problem 7.17 Show that the discount function of a simple interest rate i is larger than the discount function for a compound interest rate i for t≥1. Problem 7.18 If i 1+i = 0.06, ﬁnd i and ν. Problem 7.19 What is the value 8 years from now of $500 today if ν = 0.96? Problem 7.20 ‡ At an eﬀective annual interest rate of i,i> 0,each of the following two sets of payments has present value K : (i) A payment of 121 immediately and another payment of 121 at the end of one year. (ii) A payment of 144 at the end of two years and another payment of 144 at the end of three years. Calculate K. 56 THE BASICS OF INTEREST THEORY 8 Interest in Advance: Eﬀective Rate of Discount The eﬀective rate of interest is deﬁned as a measure of the interest paid at the end of the period. In this section, we introduce the eﬀective rate of discount, denoted by d, which is a measure of interest where the interest is paid at the beginning of the period. What do we mean by a statement such as “A loan of $1,200 is made for one year at an eﬀective rate of discount of 5%”? This means that the borrower will pay the interest of 1200 ×0.05 = $60 (called the amount of discount)at the beginning of the year and repays $1,200 at the end of the year. So basically, the lender is getting the interest in advance from the borrower. In general, when $kis borrowed at a discount rate of d,the borrower will have to pay $kdin order to receive the use of $ k. Therefore, instead of the borrower having the use of $ k at the beginning of a period he will only have the use of $( k−kd). In our example above, we note that the eﬀective rate of discount 5% is just the ratio 5% = 1200 −1140 1200 . From this we can formulate the deﬁnition of eﬀective rate of discount: The eﬀective rate of discount is the ratio of the amount of discount during the period to the amount at the end of the period. Example 8.1 What is the diﬀerence between the following two situations? (1) A loan of $100 is made for one year at an eﬀective rate of interest of 5%. (2) A loan of $100 is made for one year at an eﬀective rate of discount of 5%. Solution. In both cases the fee for the use of the money is the same which is $5. That is, the amount of discount is the same as the amount of interest. However, in the ﬁrst case the interest is paid at the end of the period so the borrower was able to use the full $100 for the year. He can for example invest this money at a higher rate of interest say 7% and make a proﬁt of $2 at the end of the transaction. In the second case, the interest is paid at the beginning of the period so the borrower had access to only $95 for the year. So, if this amount is invested at 7% like the previous case then the borrower will make a proﬁt of $1.65. Also, note that the eﬀective rate of interest is taken as a percentage of the balance at the beginning of the year whereas the eﬀective rate of discount is taken as a percentage of the balance at the end of the year Keep in mind the diﬀerences between the interest model and the discount model: •Under the interest model, the payment for the use of the money is made at the end of the period, 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 57 based on the balance at the beginning of the period. •Under the discount model, the payment for the use of the money is deducted at the beginning of the period from the ﬁnal amount which will be present unchanged at the end of the period. Even though an eﬀective rate of interest is not the same as an eﬀective rate of discount, there is a relationship between the two. Assume that $1 is invested for one year at an eﬀective rate of discountd.Then the original principal is $(1 −d). The eﬀective rate of interest i for the year is deﬁned to be the ratio of the amount of interest divided by the balance at the beginning of the year. That is, i= d 1 −d (8.1) Solving this last equation for d we ﬁnd d= i 1 + i. (8.2) There is a verbal interpretation of this result: d is the ratio of the amount of interest that 1 will earn during the year to the balance at the end of the year. Example 8.2 The amount of interest earned for one year when X is invested is $108. The amount of discount earned when an investment grows to value X at the end of one year is $100. Find X,i, and d. Solution. We have iX = 108, i 1+iX = 100. Thus, 108 1+i = 100. Solving for i we ﬁnd i = 0.08 = 8% . Hence, X = 108 0.08 = 1,350 and d= i 1+i = 2 27 ≈7.41% Observe that several identities can be derived from Equations (8.1)-(8.2). For example, sinceν = 1 1+i we have d= iν (8.3) that is, discounting i from the end of the period to the beginning of the period with the discount factor ν, we obtain d. Next, we have d= i 1 + i = 1 + i 1 + i − 1 1 + i = 1 −ν. (8.4) This is equivalent to ν = 1 −d. Both sides of the equation represent the present value of 1 to be paid at the end of the period. Now, from i= d 1−d we have i(1 −d) = d or i−id= d. Adding id−d to both sides we ﬁnd i−d= id (8.5) 58 THE BASICS OF INTEREST THEORY that is, the diﬀerence of interest in the two schemes is the same as the interest earned on amount d invested at the rate i for one period. Eﬀective rates of discount can be calculated over any particular measurement period: “The eﬀective rate of discount dn in the nth period is deﬁned to be the ratio of the amount of discount and the accumulated value at the end of the period” . That is dn = a(n) −a(n−1) a(n) . Since A(n) = A(0)a(n), we can write dn = A(n) −A(n−1) A(n) = In A(n). Example 8.3 If a(t) = 1 + t2 25 , ﬁnd d3. Solution. We have d3 = a(3) −a(2) a(3) = 1 + 9 25 −1 −4 25 1 + 9 25 = 5 34 ≈14.71% Analogous to the eﬀective rate of interest in, dn may vary from period to period. Recall that compound interest implies a constant rate of eﬀective interest. A concept parallel to compound interest is compound discount. We say that dis a compound discount if it discounts $1 according to the model shown in Figure 8.1, where t is the number of periods. Figure 8.1 Let ac(t) be the accumulation function for compound discount d> 0 per period. From Figure 8.1, the original principal which will produce an accumulated value of 1 at the end of tperiods is given by [ac(t)]−1 = (1 −d)t,t ≥0. Thus, the accumulation function for a compound discount d is ac(t) = 1 (1 −d)t,t ≥0. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 59 Example 8.4 An investor would like to have $5,000 at the end of 20 years. The annual compound rate of discount is 5%. How much should the investor deposit today to reach that goal? Solution. The investor should set aside 5000(1 −0.05)20 ≈$1,792.43 In the following theorem we prove that compound discount implies a constant rate of eﬀective discount. Theorem 8.1 Assuming compound discount d> 0 per period, we have dn = d for all n≥1. Proof. Compound interest and compound discount have equal accumulation functions. Indeed, (1 + i)t = ( 1 + d 1 −d )t = 1 (1 −d)t, for all t ≥0. Hence, we see that dn = a(n) −a(n−1) a(n) = 1 (1−d)n − 1 (1−d)n−1 1 (1−d)n = d As for compound interest and compound discount, it is possible to deﬁne simple discount in a manner analogous to the deﬁnition of simple interest. We say that d is a simple discount if it discounts $1 according to the model shown in Figure 8.2. Figure 8.2 If as(t) is the accumulation function for simple discountd,then from Figure 8.2 the original principal which will produce an accumulated value of 1 at the end of t periods is given by [as(t)]−1 = 1 −dt, 0 ≤t< 1 d. Note that, unlike the situation for simple interest, we must restrict the length of time over which we propose to apply simple discount. This is necessary to keep [ as(t)]−1 >0. 60 THE BASICS OF INTEREST THEORY Thus, the accumulation function for simple discount d is as(t) = 1 1 −dt, 0 ≤t< 1 d. Example 8.5 Calculate the present value of a payment of 10,000 to be made in 17 years assuming a simple rate of discount of 3% per annum. Solution. The answer is PV = 10000[1 −0.03(17)] = $4,900 It should be noted that formulas (15.3) - (8.5) assume eﬀective rates of compound interest and discount (since a(t) = ac(t)) and are not valid for simple rates of interest and discount unless the period of investment happens to be exactly one period. For example, if i and d are equivalent simple rate of interest and simple discount then 1 + it= 1 1−dt for all t≥0.If (15.3) is satisﬁed then (1 + it)(1 −dt) = 1 is only valid for t= 0. Example 8.6 If i and d are equivalent rates of simple interest and simple discount over t periods, show that i−d= idt. Solution. Since iand dare equivalent we must have 1+it= 1 1−dt or (1−dt)(1+it) = 1.Thus, 1+ti−td−t2id= 1. This can be rearranged to obtain ti−td= t2id. Dividing through by t, the result follows Example 8.7 (a) Find d5 if the rate of simple interest is 10%. (b) Find d5 if the rate of simple discount is 10%. Solution. (a) We are given that i= 0.10. Then d5 = a(5) −a(4) a(5) = 1 + 0.10(5) −(1 + 0.10(4)) 1 + 0.10(5) = 1 15. (b) We are given d= 0.10. Then d5 = as(5) −as(4) as(5) = 1 1−0.10(5) − 1 1−0.10(4) 1 1−0.10(5) = 1 6 In the case of simple interest, in is a decreasing function of n.This is reversed for dn with a simple discount rate. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 61 Theorem 8.2 Assuming simple discount at a rate of discount d >0, dn is an increasing function of n for 0 < n−1 < 1 d. Proof. Under simple discount at a rate d of discount, we have as(n) = 1 1−dn. Thus, dn = as(n) −as(n−1) as(n) = 1 1−dn − 1 1−d(n−1) 1 1−dn = d 1 −dn+ d = d 1 + d(1 −n) >0 since 0 <n −1 < 1 d. As n increases the denominator decreases so dn increases We next introduce the following deﬁnition: Two rates of interest and/or discount are said to be equivalent if a given amount of principal invested over the same period of time at each of the rates produces the same accumulated value. We have seen in the process of proving Theorem 8.1 that an eﬀective rate of compound interest i is equivalent to an eﬀective rate of compound discount d since a(t) = ac(t). Example 8.8 With i and d satisfying (8.1) - (8.5), show that d3 (1 −d)2 = (i−d)2 1 −ν . Solution. We have d3 (1 −d)2 =d ( d 1 −d )2 =di2 = (id)2 d =(i−d)2 1 −ν Simple and compound discount produce the same result over one period. Over a longer period, simple discount produces a smaller present value than compound discount, while the opposite is true over a shorter period. 62 THE BASICS OF INTEREST THEORY Theorem 8.3 Assuming 0 <d< 1, we have (a) (1 −d)t <1 −dt if 0 <t< 1 (b) (1 −d)t = 1 −dt if t= 0 or t= 1 (c) (1 −d)t >1 −dt if t> 1 Proof. (a) Writing the power series expansion of (1 −d)t we ﬁnd (1 −d)t = 1 −dt+ 1 2t(t−1)d2 −1 6t(t−1)(t−2)d3 + ··· If t< 1 then all terms after the second are negative. Hence, (1 −d)t <1 −dt for 0 <t< 1. (b) Straight forward by substitution. (c) Suppose t> 1. Let f(d) = 1 −dt−(1 −d)t. Then f(0) = 0 and f′(d) = −t+ t(1 −d)t−1. Since t> 1 we have t−1 >0.Since 0 <d< 1,we have 1 −d< 1.Thus, (1 −d)t−1 <1 and consequently f′(d) <−t+ t= 0. Hence, f(d) <0. This completes a proof of the theorem Figure 8.3 compares the discount function under simple discount and compound discount. Figure 8.3 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 63 Practice Problems Problem 8.1 The amount of interest earned on A for one year is $336, while the equivalent amount of discount is $300. Find A. Problem 8.2 1000 is to be accumulated by January 1, 1995, at a compound discount of 9% per year. (a) Find the present value on January 1, 1992. (b) Find the value of i equivalent to d. Problem 8.3 The amount of interest earned on Afor one year is $562.50, while the equivalent amount of discount is $500. Find A. Problem 8.4 Let i be a compound interest with equivalent compound discount d. Show the following (a) 1 d −1 i = 1. (b) d ( 1 + i 2 ) = i ( 1 −d 2 ) . (c) i √ 1 −d= d√1 + i. Hint: Use i−d= id for all three identities. Problem 8.5 Calculate the present value of $2000 payable in 10 years using an annual eﬀective discount rate of 8%. Problem 8.6 Calculate the accumulated value at the end of 3 years of 15,000 payable now assuming an interest rate equivalent to an annual discount rate of 8%. Problem 8.7 An investor deposits $1,000 today. The annual compound rate of discount is 6%. What is the accumulated value of the investment at the end of 10 years? Problem 8.8 An investor would like to have $10,000 at the end of 5 years. The annual simple rate of discount is 3%. How much should the investor deposit today to reach that goal? 64 THE BASICS OF INTEREST THEORY Problem 8.9 An investor deposits $5,000 today. The annual simple rate of discount is 5%. What is the accumu- lated value of the investment at the end of 7 months? Problem 8.10 You are given that ν = 0.80. Calculate d. Problem 8.11 A fund earns interest at a rate equivalent to the rate of discount of d.Megan invested 10,000 in the fund. Eleven years later, Megan has 30,042. Calculate d. Problem 8.12 In Account A, an investment of $1,000 grows to $1,700 in four years at an eﬀective annual interest rate of x. In Account B, $3,000 is invested for ﬁve years, at an eﬀective rate of discount d = x. What is the ending balance in Account B? Problem 8.13 ‡ Bruce and Robbie each open up new bank accounts at time 0 . Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns an annual eﬀective discount rate of d. The amount of interest earned in Bruce’s account during the 11th year is equal to X. The amount of interest earned in Robbie’s account during the 17th year is also equal to X. Calculate X. Problem 8.14 Suppose that a(t) = 1.12t. What is the eﬀective rate of discount in the 7th year? Problem 8.15 A fund accumulates based on the following formula: a(t) = ( 1 + t 10 ) ·1.2t. Find d3. Problem 8.16 The eﬀective rate of discount is 5%. Linda will receive $500 one year from today, $1,000 two years from today, and $2,000 ﬁve years from today. Find the combined present value of her three future cash ﬂows. Problem 8.17 A business permits its customers to pay with a credit card or to receive a percentage discount of r for paying cash. For credit card purchases, the business receives 97% of the purchase price one-half month later. At an annual eﬀective rate of discount of 22%, the two payments are equivalent. Find r. 8 INTEREST IN ADVANCE: EFFECTIVE RATE OF DISCOUNT 65 Problem 8.18 A deposit of X is made into a fund which pays an annual eﬀective interest rate of 6% for 10 years. At the same time , X 2 is deposited into another fund which pays an annual eﬀective rate of discount of d for 10 years. The amounts of interest earned over the 10 years are equal for both funds. Calculate d. Problem 8.19 A signs a one-year note for $1000 and receives $920 from the bank. At the end of six months, A makes a payment of $288. Assuming simple discount, to what amount does this reduce the face amount of the note? Problem 8.20 A bank oﬀers a 272 −day discounted loan at a simple discount rate of 12%. (a) How much money would a borrower receive if she asked for a $5000 loan? (b) What size loan should the borrower ask for in order to actually receive $5000? (c) What is the equivalent simple interest rate that is being charged on the loan? Problem 8.21 A discounted loan of $3000 at a simple discount rate of 6.5% is oﬀered to Mr. Jones. If the actual amount of money that Mr. Jones receives is $2869.11, when is the $3000 due to be paid back? Problem 8.22 Show that dn <in. Problem 8.23 At time t= 0, Paul deposits $3500 into a fund crediting interest with an annual discount factor of 0.96. Find the fund value at time 2.5. Problem 8.24 Method A assumes simple interest over ﬁnal fractional periods, while Method B assumes simple discount over ﬁnal fractional periods. The annual eﬀective rate of interest is 20%. Find the ratio of the present value of a payment to be made in 1.5 years computed under methodAto that computed under Method B. 66 THE BASICS OF INTEREST THEORY 9 Nominal Rates of Interest and Discount When we speak of either the “eﬀective” rate of interest/discount we mean interest is paid once per measurement period, either at the end of the period( in the case of interest rate) or at the beginning of the period(in the case of discount rate). In this section, we consider situations where interest is paid more than once per measurement period. Rates of interest and discount in these cases are called nominal. In this section we deﬁne nominal rates of interest and discount and determine relationships between nominal rates and eﬀective rates as well as relationships between nominal rates of interest and discount. Compound interest or discount will always be assumed, unless speciﬁed otherwise. When interest is paid (i.e., reinvested) more frequently than once per period, we say it is “payable” (“convertible”, “compounded”) each fraction of a period, and this fractional period is called the interest conversion period. A nominal rate of interest i(m) payable m times per period, where m is a positive integer, represents mtimes the eﬀective rate of compound interest used for each of the mth of a period. In this case, i(m) m is the eﬀective rate of interest for each mth of a period. Thus, for a nominal rate of 12% compounded monthly, the eﬀective rate of interest per month is 1% since there are twelve months in a year. Suppose that 1 is invested at a nominal rate i(m) compounded m times per measurement period. That is, the period is partitioned into mequal fractions of a period. At the end of the ﬁrst fraction of the period the accumulated value is 1 + i(m) m . At the end of the second fraction of the period the accumulated value is ( 1 + i(m) m )2 .Continuing, we ﬁnd that the accumulated value at the end of the mth fraction of a period, which is the same as the end of one period, is ( 1 + i(m) m )m and at the end of t years the accumulated value is a(t) = ( 1 + i(m) m )mt . Figure 9.1 illustrates accumulation at a nominal rate of interest for one measurement period. Figure 9.1 9 NOMINAL RATES OF INTEREST AND DISCOUNT 67 Example 9.1 Find the accumulated value of 1,000 after three years at a rate of interest of 24% per year convertible monthly. Solution. The accumulated value is 1,000 ( 1 + 0.24 12 )12×3 = 2,039.89 Example 9.2 Find the accumulated value of $3,000 to be paid at the end of 8 years with a rate of compound interest of 5% (a) per annum; (b) convertible quarterly; (c) convertible monthly. Solution. (a) The accumulated value is 3 ,000 ( 1 + 0.05 1 )8 ≈$4,432.37. (b) The accumulated value is 3 ,000 ( 1 + 0.05 4 )8×4 ≈$4,464.39. (c) The accumulated value is 3 ,000 ( 1 + 0.05 12 )8×12 ≈$4,471.76 Next we describe the relationship between eﬀective and nominal rates. If i denotes the eﬀective rate of interest per one measurement period equivalent to i(m) then we can write 1 + i= ( 1 + i(m) m )m since each side represents the accumulated value of a principal of 1 invested for one year. Rearrang- ing we have i= ( 1 + i(m) m )m −1 and i(m) = m[(1 + i) 1 m −1]. For any t≥0 we have (1 + i)t = ( 1 + i(m) m )mt . Example 9.3 Given the nominal interest rate of 12%, compounded monthly. Find the equivalent eﬀective annual interest rate. 68 THE BASICS OF INTEREST THEORY Solution. The answer is i= ( 1 + 0.12 12 )12 −1 = 12.7% Example 9.4 (a) Find the annual eﬀective interest rate i which is equivalent to a rate of compound interest of 8% convertible quarterly. (b) Find the compound interest rate i(2) which is equivalent to an annual eﬀective interest rate of 8%. (c) Find the compound interest rate i(4) which is equivalent to a rate of compound interest of 8% payable semi-annually. Solution. (a) We have 1 + i= ( 1 + 0.08 4 )4 ⇒i= ( 1 + 0.08 4 )4 −1 ≈0.08243216 (b) We have 1 + 0.08 = ( 1 + i(2) 2 )2 ⇒i(2) = 2[(1.08) 1 2 −1] ≈0.07846. (c) We have ( 1 + i(4) 4 )4 = ( 1 + i(2) 2 )2 ⇒i(4) = 4[(1.04) 1 2 −1] ≈0.0792 In the same way that we deﬁned a nominal rate of interest, we could also deﬁne a nominal rate of discount, d(m),as meaning an eﬀective rate of discount of d(m) m for each of the mth of a period with interest paid at the beginning of a mth of a period. Figure 9.2 illustrates discounting at a nominal rate of discount for one measurement period. Figure 9.2 The accumulation function with the nominal rate of discount d(m) is a(t) = ( 1 −d(m) m )−mt , t ≥0. 9 NOMINAL RATES OF INTEREST AND DISCOUNT 69 Example 9.5 Find the present value of $8,000 to be paid at the end of 5 years at a an annual rate of compound interest of 7% (a) convertible semiannually. (b) payable in advance and convertible semiannually. Solution. (a) The answer is 8,000( 1 + 0.07 2 )5×2 ≈$5,671.35 (b) The answer is 8000 ( 1 −0.07 2 )5×2 ≈$5,602.26 If d is the eﬀective discount rate equivalent to d(m) then 1 −d= ( 1 −d(m) m )m since each side of the equation gives the present value of 1 to be paid at the end of the measurement period. Rearranging, we have d= 1 − ( 1 −d(m) m )m and solving this last equation for d(m) we ﬁnd d(m) = m[1 −(1 −d) 1 m] = m(1 −ν 1 m). Example 9.6 Find the present value of $1,000 to be paid at the end of six years at 6% per year payable in advance and convertible semiannually. Solution. The answer is 1,000 ( 1 −0.06 2 )12 = $693.84 There is a close relationship between nominal rate of interest and nominal rate of discount. Since 1 −d= 1 1+i, we conclude that ( 1 + i(m) m )m = 1 + i= (1 −d)−1 = ( 1 −d(n) n )−n . (9.1) 70 THE BASICS OF INTEREST THEORY If m= n then the previous formula reduces to ( 1 + i(n) n ) = ( 1 −d(n) n )−1 . Example 9.7 Find the nominal rate of discount convertible semiannually which is equivalent to a nominal rate of interest of 12% per year convertible monthly. Solution. We have ( 1 −d(2) 2 )−2 = ( 1 + 0.12 12 )12 . Solving for d(2) we ﬁnd d(2) = 0.11591 Note that formula (9.1) can be used in general to ﬁnd equivalent rates of interest or discount, either eﬀective or nominal, converted with any desired frequency. Example 9.8 Express d(4) as a function of i(3). Solution. We have ( 1 −d(4) 4 )−4 = ( 1 + i(3) 3 )3 . Thus, 1 −d(4) 4 = ( 1 + i(3) 3 )−3 4 So d(4) = 4 [ 1 − ( 1 + i(3) 3 )−3 4 ] An analogous formula to i−d= idholds for nominal rates of interest and discount as shown in the next example. Example 9.9 Prove that i(m) m −d(m) m = i(m) m ·d(m) m 9 NOMINAL RATES OF INTEREST AND DISCOUNT 71 Solution. We have ( 1 + i(m) m ) = ( 1 −d(m) m )−1 which is equivalent to ( 1 + i(m) m ) · ( 1 −d(m) m ) = 1. Expanding we obtain 1 −d(m) m + i(m) m −i(m) m ·d(m) m = 1. Hence, i(m) m −d(m) m = i(m) m ·d(m) m Example 9.10 Show that i(m) = d(m)(1 + i) 1 m. Solution. We know that i(m) m −d(m) m = i(m) m ·d(m) m Multiplying through by m and rearranging we ﬁnd i(m) = d(m) ( 1 + i(m) m ) = d(m)(1 + i) 1 m Example 9.11 (a) On occasion, interest is convertible less frequently than once a year. Deﬁne i( 1 m) and d( 1 m) to be the nominal annual rates of interest and discount convertible once every m years. Find a formula analogous to formula (9.1) relating i( 1 m) and d( 1 m). (b) Find the accumulated value of $100 at the end of two years if the nominal annual rate of discount is 6%, convertible once every four years. Solution. (a) If i( 1 m) is convertible once every myears, then m·i( 1 m) is the eﬀective interest rate over myears. Thus, by deﬁnition of equivalent rates we can write 1 + m·i( 1 m) = (1 + i)m. 72 THE BASICS OF INTEREST THEORY Likewise, we have 1 −p·d( 1 p) = (1 + i)−p. It follows that (1 + m·i( 1 m)) 1 m = (1 −p·d( 1 p))−1 p. Note that this formula can be obtained from formula (9.1) by replacingmand pby their reciprocals. (b) If i is the annual eﬀective interest rate than the accumulated value at the end of two years is 100(1 + i)2 = 100(1 −4 ×0.06)−2 4 = $114.71 Remark 9.1 Nominal rates of interest or discount are not relevant under simple interest and simple discount. For example, if we let i(m) be the nominal interest rate and i be the equivalent simple interest rate then at the end of one year we ﬁnd 1 + i= 1 + i(m) or i= i(m). Similarly, we have d= d(m). 9 NOMINAL RATES OF INTEREST AND DISCOUNT 73 Practice Problems Problem 9.1 A man borrows $1,000 at an interest rate of 24% per year compounded monthly. How much does he owe after 3 years? Problem 9.2 If i(6) = 0.15, ﬁnd the equivalent nominal rate of interest convertible semiannually. Problem 9.3 Suppose that $100 is deposited into a savings account, earning at a discount rate of 0.15% biweekly, at the beginning of year 2006. (a) Find the nominal annual discount rate. (b) Find the eﬀective annual discount rate. (c) Find the amount of interest generated during the year. (d) Find the equivalent eﬀective annual interest rate. (e) Find the equivalent nominal annual interest rate, convertible monthly. Problem 9.4 Express i(6) as a function of d(2). Problem 9.5 Given that i(m) = 0.1844144 and d(m) = 0.1802608. Find m. Problem 9.6 It is known that 1 + i(n) n = 1 + i(4) 4 1 + i(5) 5 Find n. Problem 9.7 If r= i(4) d(4) , express ν in terms of r. Problem 9.8 You deposit $1,000 into Account A and $750 into Account B. Account A earns an eﬀective annual interest rate of 5%. Account B earns interest at i(4). Ten years later, the two accounts have the same accumulated value. Find i(4). (Express as a percentage, to two decimal places.) 74 THE BASICS OF INTEREST THEORY Problem 9.9 ‡ Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a diﬀerent savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i. Problem 9.10 ‡ Calculate the nominal rate of discount convertible monthly that is equivalent to a nominal rate of interest of 18.9% per year convertible monthly. Problem 9.11 A bank credits interest on deposits quarterly at rate 2% per quarter. What is the nominal interest rate (per annum) for interest compounded quarterly? Find the annual eﬀective rate of interest. Problem 9.12 Interest on certain deposits is compounded monthly. Find the nominal rate of interest (per annum) for these deposits if the APR is 15%. Obtain the accumulation of an investment of $1,000 over a period of 6 months. How much interest has accrued by that time? Problem 9.13 Suppose that the eﬀective interest rate for 1 5 year is 2%. Determine the equivalent nominal interest rate, compounded every 6 years. Problem 9.14 Fund A earns interest at a nominal rate of 6% compounded monthly. Fund B earns interest at a nominal rate of discount compounded three times per year. The annual eﬀective rates of interest earned by both funds are equivalent. Calculate the nominal rate of discount earned by Fund B. Problem 9.15 Treasury bills, or T −bills, are sold in terms ranging from a few days to 26 weeks. Bills are sold at a discount from their face value. For instance, you might pay $990 for a $1,000 bill. When the bill matures, you would be paid $1,000. The diﬀerence between the purchase price and face value is interest. A bill for 100 is purchased for 96 three months before it is due. Find: (a) The nominal rate of discount convertible quarterly earned by the purchaser. (b) The annual eﬀective rate of interest earned by the purchaser. Problem 9.16 If i(8) = 0.16, calculate d( 1 2 ). 9 NOMINAL RATES OF INTEREST AND DISCOUNT 75 Problem 9.17 A fund earns a nominal rate of interest of 6% compounded every two years. Calculate the amount that must be contributed now to have 1000 at the end of six years. Problem 9.18 A deposit is made on January 1, 2004. The investment earns 6% compounded semi-annually. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.19 A deposit is made on January 1, 2004. The investment earns interest at a rate equivalent to an annual rate of discount of 6%. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.20 A deposit is made on January 1, 2004. The investment earns interest at a rate equivalent to a rate of discount of 6% convertible quarterly. Calculate the monthly eﬀective interest rate for the month of December 2004. Problem 9.21 Calculate the present value of $1,000 payable in 10 years using a discount rate of 5% convertible quarterly. Problem 9.22 Calculate the accumulated value at the end of 3 years of 250 payable now assuming an interest rate equivalent to a discount rate of 12% convertible monthly. Problem 9.23 Thomas pays 94 into a fund. Six months later, the fund pays Thomas 100. Calculate the nominal rate of discount convertible semi-annually. Problem 9.24 Thomas pays 94 into a fund. Six months later, the fund pays Thomas 100. Calculate the annual eﬀective rate of interest earned. Problem 9.25 Find the present value of 5000, to be paid at the end of 25 months, at a rate of discount of 8% convertible quarterly: (a) assuming compound discount throughout; (b) assuming simple discount during the ﬁnal frac- tional period. 76 THE BASICS OF INTEREST THEORY Problem 9.26 ‡ 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 9.27 A bank oﬀers the following certiﬁcates of deposit: Nominal annual interest rate Term in years (convertible semi-annually) 1 5% 2 6% 3 7% 4 8% 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. An investor plans to invest 1,000 in CDs. Calculate the maximum amount that can be withdrawn at the end of six years. Problem 9.28 You are given (1) Fund X accumulates at an interest rate of 8% compounded quarterly (2) Fund Y accumulates at an interest rate of 6% compounded semiannually (3) at the end of 10 years, the total amount in the two funds combined is 1000 (4) at the end of 5 years, the amount in Fund X is twice that in Fund Y Calculate the total amount in the two funds at the end of 2 years. 9 NOMINAL RATES OF INTEREST AND DISCOUNT 77 Problem 9.29 A collection agency pays a doctor $5,000 for invoices that the doctor has not been able to collect on. After two years, the collection agency has collected $6,000 on the invoices. At what nominal rate of discount compounded monthly did the collection agency receive on this transaction? Problem 9.30 A trust company oﬀers guaranteed investment certiﬁcates paying i(2) = 8.9% and i(1) = 9%.Which option yields the higher annual eﬀective rate of interest? 78 THE BASICS OF INTEREST THEORY 10 Force of Interest: Continuous Compounding Eﬀective and nominal rates of interest and discount each measures interest over some interval of time. Eﬀective rates of interest and discount measure interest over one full measurement period, while nominal rates of interest and discount measure interest over mths of a period. In this section we want to measure interest at any particular moment of time. This measure of interest is called the force of interest. To start with, consider the case of a nominal compound interest i(m) converted m times a period. We can think of the force of interest, denoted by δ, as the limit of i(m) as the number of times we credit the compounded interest goes to inﬁnity. That is, δ= lim m→∞ i(m). Letting i be the eﬀective interest rate equivalent to i(m) we have i(m) = m[(1 + i) 1 m −1] = (1 + i) 1 m −1 1 m . Hence, δ= lim m→∞ (1 + i) 1 m −1 1 m . The above limit is of the form 0 0 so that we can apply L’Hopital’s rule to obtain δ= lim m→∞ d dm [ (1 + i) 1 m −1 ] d dm (1 m ) = lim m→∞ [(1 + i) 1 m ln (1 +i)] = ln (1 +i) since limm→∞(1 + i) 1 m = 1. Remark 10.1 The eﬀective interest rate i can be written as a series expansion of δ: i= eδ −1 = δ+ δ2 2! + ··· + δn n! + ··· . Similarly, δ= ln (1 +i) = i−i2 2 + ··· + (−1)nin n + ··· . 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 79 Example 10.1 Given the nominal interest rate of 12%, compounded monthly. Find the equivalent force of interest δ. Solution. The eﬀective annual interest rate is i= (1 + 0.01)12 −1 ≈0.1268250. Hence, δ= ln (1 +i) = ln (1.1268250) ≈0.119404. Remark 10.2 Intuitively, δ represents a nominal interest rate which is converted continuously, a notion of more theoretical than practical importance. Indeed, in theory the most important measure of interest is the force of interest. In practice, however, eﬀective and nominal rates of interest tend to be used more frequently because most ﬁnancial transactions involve discrete, not continuous, processes. However, δ can be used in practice as an approximation to interest converted very frequently, such as daily. Having related δ and iimmediately relates δ and the other measures of interested introduced in the previous sections. Indeed, we have the following important set of equalities ( 1 + i(m) m )m = 1 + i= (1 −d)−1 = ( 1 −d(p) p )−p = eδ. Example 10.2 Using a constant force of interest of 4.2%, calculate the present value of a payment of $1,000 to be made in 8 years’ time. Solution. The present value is 1,000(1 + i)−8 = 1,000e−8δ = 1,000e−8×0.042 ≈$714.62 Example 10.3 A loan of $3,000 is taken out on June 23, 1997. If the force of interest is 14%, ﬁnd each of the following (a) The value of the loan on June 23, 2002. (b) The value of i. (c) The value of i(12). 80 THE BASICS OF INTEREST THEORY Solution. (a) 3,000(1 + i)5 = 3,000e5δ = 3,000e0.7 ≈$6,041.26. (b) i= eδ −1 = e0.14 −1 ≈0.15027. (c) We have ( 1 + i(12) 12 )12 = 1 + i= e0.14. Solving for i(12) we ﬁnd i(12) ≈0.14082 Now, using the accumulation function of compound interest a(t) = (1 + i)t we notice that δ= ln (1 +i) = d dta(t) a(t) . The deﬁnition of force of interest in terms of a compound interest accumulation function can be extended to any accumulation function. That is, for an accumulation function a(t) we deﬁne the force of interest at time t by δt = a′(t) a(t) . Let’s look closely at the expression on the right. From the deﬁnition of the derivative we have d dta(t) a(t) = lim n→∞ a(t+ 1 n)−a(t) a(t) 1 n . Now the expression a(t+ 1 n) −a(t) a(t) is just the eﬀective rate of interest over a very small time period 1 n so that a(t+ 1 n)−a(t) a(t) 1 n is the nominal annual interest rate converted n periods a year with each time period of length 1 n corresponding to that eﬀective rate, which agrees with Remark 10.2. Note that, in general, the force of interest can be a function of t.However, in the case of compound interest δt is a constant. Also, we notice that, since A(t) = A(0)a(t), we can write δt = A′(t) A(t) . 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 81 Example 10.4 Show that, for any amount function A(t), we have ∫ n 0 A(t)δtdt= A(n) −A(0) = I1 + I2 + ··· + In Interpret this result verbally. Solution. We have ∫ n 0 A(t)δtdt= ∫ n 0 A′(t)dt= A(n) −A(0) = I1 + I2 + ··· + In where we used Example 2.6. The term A(n) −A(0) is the amount of interest earned over nmeasurement periods. The term δtdt represents the eﬀective rate of interest over the inﬁnitesimal “period of time” dt.Hence, A(t)δtdtis the amount of interest in this period and ∫n 0 A(t)δtdtrepresents the total amount of interest earned over the entire n periods which is A(n) −A(0) Example 10.5 You are given that A(t) = at2 + bt+ c, for 0 ≤t ≤2, and that A(0) = 100 ,A(1) = 110 , and A(2) = 136. Determine the force of interest at time t= 1 2 . Solution. The condition A(0) = 100 implies c = 100. From A(1) = 110 and A(2) = 136 we ﬁnd the linear system of equations a+ b = 10 and 4 a+ 2b = 36. Solving this system we ﬁnd a = 8 and b = 2. Hence, A(t) = 8t2 + 2t+ 100. It follows that δt = A′(t) A(t) = 16t+ 2 8t2 + 2t+ 100. Hence, δ0.5 = 10 103 = 0.09709 Example 10.6 Find δt in the case of simple interest, that is, when a(t) = 1 + it. Solution. From the deﬁnition of δt we have δt = a′(t) a(t) = i 1 + it 82 THE BASICS OF INTEREST THEORY Note that δt is a decreasing function of t We have now a method for ﬁnding δt given a(t). What if we are given δt instead, and we wish to derive a(t) from it? From the deﬁnition of δt we can write d drln a(r) = δr. Integrating both sides from 0 to t we obtain ∫ t 0 d drln a(r)dr= ∫ t 0 δrdr. Hence, ln a(t) = ∫ t 0 δrdr. From this last equation we ﬁnd a(t) = e ∫t 0 δrdr. Also, we can derive an expression for A(t). Since A(t) = A(0)a(t), we can write A(t) = A(0)e ∫t 0 δrdr. Example 10.7 A deposit of $10 is invested at time 2 years. Using a force of interest of δt = 0.2 −0.02t, ﬁnd the accumulated value of this payment at the end of 5 years. Solution. The accumulated value is A(5) = 10a(5) a(2) = 10e ∫5 2 (0.2−0.02t)dt = 10e[0.2t−0.01t2] 5 2 ≈$14.77 Note that a compound interest rate iimplies a constant force of interest (equals to ln (1 + i)).The converse is also true as shown in the next example. Example 10.8 Show that if δt = δ for all t then a(t) = (1 + i)t for some i. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 83 Solution. Since δt = δ for all t, we obtain ∫t 0 δrdr= tδ.Now, for all integer values n≥1 we have in = a(n) −a(n−1) a(n−1) = enδ −e(n−1)δ e(n−1)δ = eδ −1 = i. In this case, a(t) = eδt = (eδ)t = (1 + i)t The above example shows that a constant force of interest implies a constant eﬀective interest rate. The converse of this result is not always true. See Problem 10.32. Replacing the accumulation function by the discount function in the deﬁnition of δt we obtain the force of discount: δ′ t = − d dt[a(t)]−1 [a(t)]−1 . The negative sign is necessary in order to make the force of discount a positive quantity since the discount function is a decreasing function of t. It is possible to dispence with δ′ t and just use δt according to the following theorem. Theorem 10.1 For all t we have δ′ t = δt. Proof. we have δ′ t = −[a−1(t)]′ a−1(t) =a−2(t)[a(t)]′ a−1(t) =a−2(t)a(t)δt a−1(t) =δt Example 10.9 Find the force of discount under a simple discount rate d. 84 THE BASICS OF INTEREST THEORY Solution. Recall that in the case of a simple discount the discount function is given by [ a(t)]−1 = 1 −dt for 0 ≤t< 1 d. Thus, δt =δ′ t = − d dt[a(t)]−1 [a(t)]−1 = − d dt(1 −dt) 1 −dt = d 1 −dt It follows that the force of interest is increasing with simple discount in contrast to simple interest where it is decreasing. See Example 10.6 Example 10.10 Find δ′ t in the case of compound discount. Solution. We have δ′ t = −[(1 −d)t]′ (1 −d)t = −ln (1−d) Example 10.11 Show that limm→∞d(m) = δ Solution. We have ( 1 −d(m) m )m = (1 + i)−1 = e−δ. Solving for d(m) we ﬁnd d(m) = m[1 −e−δ m]. Using power series expansion of e−δ m we obtain d(m) =m [ 1 − ( 1 + ( −δ m ) + 1 2! ( −δ m )2 + 1 3! ( −δ m )3 + ··· )] =m [δ m −1 2! δ2 m2 + 1 3! δ3 m3 −··· ] =δ−1 2! δ2 m + 1 3! δ3 m2 −···→ δ as m→∞ 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 85 Example 10.12 An investor invests $1,000 into an investment account at time 1. The interest is computed based on the following force of interest: δt =    0.02t 0 ≤t< 3 0.025 t≥3 0 otherwise (a) Find the amount of interest generated between time 2.5 and 3.5. (b) Calculate eﬀective discount rate during the third period, i.e. d3. Solution. (a) We know that A(t) = 1000 a(t) a(1) = 1000e ∫t 1 δrdr. Thus, the amount of interest generated between time 2.5 and 3.5 is A(3.5) −A(2.5) = 1000 ( e ∫3.5 1 δtdt −e ∫2.5 1 δtdt ) . But ∫ 3.5 1 δtdt= ∫ 3 1 0.02tdt+ ∫ 3.5 3 0.025dt= 0.0925 and ∫ 2.5 1 δtdt= ∫ 2.5 1 0.02tdt= 0.0525 Hence, A(3.5) −A(2.5) = 1000(e0.0925 −e0.0525) ≈$43.01. (b) d3 = A(3)−A(2) A(3) = e0.08−e0.03 e0.08 ≈4.877% Example 10.13 Find an expression of t in terms of δ so that f(t) = (1 + it) −(1 + i)t is maximum. Solution. Since f′(t) = i−(1 + i)tln (1 +i) = i−δ(1 + i)t we see that f′(t) = 0 when (1 + i)t = i δ. Solving for t we ﬁnd t= ln i−ln δ ln (1+i) = ln i−ln δ δ . Now, f′′(t) = −δ2(1 + i)t <0 so that the critical point of f is a maximum Example 10.14 On March 15, 2003, a student deposits X into a bank account. The account is credited with simple interest where i= 7.5% 86 THE BASICS OF INTEREST THEORY On the same date, the student’s professor deposits X into a diﬀerent bank account where interest is credited at a force of interest δt = 2t t2 + k, t ≥0 From the end of the fourth year until the end of the eighth year, both accounts earn the same dollar amount of interest. Calculate k. Solution. The interest earned by the student is X[1 + 0.075(8)] −X[1 + 0.075(4)] = 0.3X The interest earned by the professor is Xe ∫8 0 δtdt −Xe ∫4 0 δtdt = Xe ∫8 0 2t t2+kdt −Xe ∫4 0 2t t2+kdt = Xe[ln (t2+k)] 8 0 −Xe[ln (t2+k)] 4 0 = X64 + k k −X16 + k k = 48 k X. Thus, 0.3X = 48 k X ⇒k= 48 0.3 = 160 Example 10.15 Show that d<d (m) <δ <i(m) <i, m> 1. Solution. We have d dm(i(m)) = e δ m ( 1 − δ m −e−δ m ) . From the inequality 1 −x≤e−x for x≥0 we conclude that d dm(i(m)) = e δ m ( 1 − δ m −e−δ m ) <0. That is, i(m) is a decreasing function of m. Since i(1) = i and i(∞) = δ we obtain δ <i(m) <i. Similarly, d dm(d(m)) = e−δ m(e δ m −1 − δ m) >0 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 87 so that d(m) is an increasing function of m. Since d(1) = d and d(∞) = δ we can write d<d (m) <δ. Combining these inequalities we ﬁnd d<d (m) <δ <i(m) <i, m> 1 88 THE BASICS OF INTEREST THEORY Practice Problems Problem 10.1 A deposit of $500 is invested at time 5 years. The constant force of interest is 6% per year. Determine the accumulated value of the investment at the end of 10 years. Problem 10.2 If the constant force of interest is 6%, what is the corresponding annual eﬀective rate of interest? Problem 10.3 Assume that the force of interest varies with time and is given by δt = a+ b t. Find the formula for the accumulation of one unit money from time t1 to time t2. Problem 10.4 You need $500 on Jan 1, 2012. To save for this amount, you invest xon Jan 1, 2008 and 2xon July 1, 2008. The force of interest is δt = 0.02t where t is 0 on Jan 1, 2008. Find x. Problem 10.5 At a constant force of interest, $200 accumulates to $240 over the course of 8 years. Find the force of interest δ. Problem 10.6 ‡ Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest of 4% convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ. After 7.25 years, the value of each account is the same. Calculate δ. Problem 10.7 ‡ Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest i convertible semiannually. At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of δ.After 7.25 years, the value of each account is 200. Calculate i−δ. Problem 10.8 ‡ At time t= 0, 1 is deposited into each of Fund X and Fund Y. Fund X accumulates at a force of interest δt = t2 k. Fund Y accumulates at a nominal rate of discount of 8% per annum convertible semiannually. At time t= 5, the accumulated value of Fund X equals the accumulated value of Fund Y. Determine k. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 89 Problem 10.9 ‡ At time 0 , Kis deposited into Fund X, which accumulates at a force of interest δt = 0.006t2. At time m,2K is deposited into Fund Y, which accumulates at an annual eﬀective interest rate of 10%. At time n, where n>m, the accumulated value of each fund is 4 K. Determine m. Problem 10.10 Suppose $2,000 is invested for 3 years with a constant force of interest equal to 9%. (a) Find the accumulation function. (b) Find the eﬀective rate of interest. (c) Find the value of the investment after 3 years. Problem 10.11 ‡ Ernie makes deposits of 100 at time 0, and X at time 3 . The fund grows at a force of interest δt = t2 100 , t> 0. The amount of interest earned from time 3 to time 6 is X. Calculate X. Problem 10.12 If the force of interest δ is 0.1, obtain the nominal rate of interest (per annum) on deposits com- pounded (a) monthly; (b) quarterly; (c) annually. Problem 10.13 The force of interest depends upon time, with δ(t) = 1 10(1+t) , 0 ≤t ≤4. Find the accumulated amount after four years for an initial investment of $1,000 at time zero. Problem 10.14 The force of interest δ is constant. The corresponding nominal rate of discount when interest is compounded p-thly is d(p).Express d(p) in terms of p.Show that d(p) is an increasing function of p. Problem 10.15 ‡ Tawny makes a deposit into a bank account which credits interest at a nominal interest rate of 10% per annum, convertible semiannually. At the same time, Fabio deposits 1,000 into a diﬀerent bank account, which is credited with simple interest. At the end of 5 years, the forces of interest on the two accounts are equal, and Fabio’s account has accumulated to Z. Determine Z. Problem 10.16 Fund A earns interest at a nominal rate of interest of 12% compounded quarterly. Fund B earns interest at a force of interest δ.John invested $1,000 in each fund ﬁve years ago. Today, the amount in Fund A is equal to the amount in Fund B. Calculate δ. 90 THE BASICS OF INTEREST THEORY Problem 10.17 A fund earns interest at a force of interest of δ = 0.01t. Calculate the value at the end of 10 years of 8,000 invested at the end of 5 years. Problem 10.18 A fund earns interest at a force of interest of δ = 0.01t. Calculate the eﬀective rate of interest in the 10th year. Problem 10.19 Calculate the eﬀective annual interest rate equivalent to a nominal rate of discount of 6% com- pounded continuously. Problem 10.20 Fund A accumulates at a simple interest rate of 10%. Find B accumulates at a simple discount rate of 5%. Find the point in time at which the force of interest on the two funds are equal. Problem 10.21 An investment is made for one year in a fund with a(t) = a+ bt+ ct2.The nominal rate of interest earned during the ﬁrst half of the year is 5% convertible semi-annually. The eﬀective rate of interest earned for the entire year is 7%. Find δ0.5. Problem 10.22 If a(t) = 1 + .01(t2 + t), calculate δ5. Problem 10.23 You are given that δ= 0.05. Calculate the amount that must be invested at the end of 10 years to have an accumulated value at the end of 30 years of $1,000. Problem 10.24 Calculate k if a deposit of 1 will accumulate to 2.7183 in 10 years at a force of interest given by: δt = { kt 0 <t ≤5 0.04kt2 t> 5 Problem 10.25 A deposit is made on January 1, 2004. The investment earns interest at a constant force of interest of 6%. Calculate the monthly eﬀective interest rate for the month of December 2004. 10 FORCE OF INTEREST: CONTINUOUS COMPOUNDING 91 Problem 10.26 A fund earns interest at a constant force of interest of δ.Kathy invested 10,000 in the fund. Twenty two years later, Kathy has 30,042. Calculate δ. Problem 10.27 A fund earns interest at a force of interest of δt = kt. Ryan invests 1,000 at time t = 0. After 8 years, Ryan has 1,896.50 . Calculate k. Problem 10.28 You are given that i(12) = 0.09. Calculate δ. Problem 10.29 You are given that d(2) = 0.04. Calculate δ. Problem 10.30 ‡ At time 0, deposits of 10,000 are made into each of Fund X and Fund Y. Fund X accumulates at an annual eﬀective interest rate of 5%. Fund Y accumulates at a simple interest rate of 8%. At time t, the forces of interest on the two funds are equal. At time t, the accumulated value of Fund Y is greater than the accumulated value of Fund X by Z. Determine Z. Problem 10.31 Find an expression for the fraction of a period at which the excess of present values computed at simple discount over compound discount is a maximum. Hint: See Example 10.13. Problem 10.32 Consider the following problem: Invest 1 at a compound interest i for integral periods and at a simple interest for fractional periods. (a) Show that ik = i for k= 1,2,··· ,n. (b) Show that δt1 ̸= δt2 for k <t1 <t2 <k + 1, k= 1,2,···n−1. Thus, a constant eﬀective rate of interest i does not necessarily imply a constant force of interest. Problem 10.33 Which of the following are true? (I) d dd(i) = ν−2 (II) d di(i(m)) = ν−( m−1 m ) (III) d dδ(i) = 1 + i. 92 THE BASICS OF INTEREST THEORY Problem 10.34 You are given that δt = 0.2t 1+0.1t2 . Calculate i2. Problem 10.35 Simplify the following expression (d dνδ )(d did ) . Problem 10.36 An investment is made in an account in which the amount function is given by A(t) = t2 + t+ 2.25, 0 ≤t≤4. Compute the ratio i3 δ2 . Problem 10.37 Fund A accumulates at a force of interest 0.05 1 + 0.05t at time t ≥0. Fund B accumulates at a force of interest 0.05. You are given that the amount in Fund A at time zero is 1,000, the amount in Fund B at time zero is 500, and that the amount in Fund C at any time tis equal to the sum of the amount in FundAand Fund B.Fund C accumulates at force of interest δt. Find δ2. Problem 10.38 Show that d dtδt = A′′(t) A(t) −δ2 t. Problem 10.39 Find δt if A(t) = K2t3t2 52t . Problem 10.40 Show that ∫n 0 δtdt= ln a(n). 11 TIME VARYING INTEREST RATES 93 11 Time Varying Interest Rates In this section we consider situations involving varying interest. The ﬁrst involves a continuously varying force of interest δt. In this case, the accumulated value at time t is given by A(t) = A(0)e ∫t 0 δrdr. Example 11.1 Find the accumulated value of $200 invested for 5 years if the force of interest is δt = 1 8+t. Solution. The accumulated value is A(5) = 200a(5) = 200e ∫5 0 dt 8+t = 200e[ln (8+t)]5 0 = 200 ×13 8 = $325 The second situation involves changes in the eﬀective rate of interest over a period of time. Let- ting in denote the eﬀective rate of interest during the nth period from the date of investment, the accumulated value for integral t is given by a(t) = (1 + i1)(1 + i2) ···(1 + it) (11.1) and the present value is given by (a(t))−1 = (1 + i1)−1(1 + i2)−1 ···(1 + in)−1. (11.2) Example 11.2 Find the accumulated value of $500 invested for 9 years if the rate of interest is 5% for the ﬁrst 3 years, 5.5% for the second 3 years, and 6.25% for the third 3 years. Solution. The accumulated value is 500(1 + 0.05)3(1 + 0.055)3(1 + 0.0625)3 = $815.23 Formulas (11.1) and (11.2) can be extended to include nominal rates of interest or discount. We illustrate this in the next example. Example 11.3 A fund will earn a nominal rate of interest of 5% compounded quarterly during the ﬁrst two years, a nominal rate of discount of 4% compounded monthly during years 3 and 4, and a constant force of interest of 3% during the ﬁfth and sixth year. Calculate the amount that must be invested today in order to accumulate 5,000 after 6 years. 94 THE BASICS OF INTEREST THEORY Solution. The amount that must be invested is 5000 ( 1 + 0.05 4 )−8 ( 1 −0.04 12 )24 e−0.03×2 = $3,935.00 A common question involving varying interest is the question of ﬁnding an equivalent level rate to the rate that vary. Moreover, the answer depends on the period of time chosen for the compari- son . We illustrate this point in the next example. Example 11.4 (a) Find the eﬀective interest rate over the ﬁrst four-year period from now if the eﬀective rate of interest is 5% for the ﬁrst three years from now, 4.5% for the next three years, and 4% for the last three years. (b) What about over the six-year period from now? Solution. (a) We have (1 + i)4 = (1.05)3(1.045). Solving this equation for i we ﬁnd i= [(1.05)3(1.045)] 1 4 −1 = 4.87% (b) We have (1 + i)6 = (1.05)3(1.045)3. Solving this equation for i we ﬁnd i= [(1.05)3(1.045)3] 1 6 −1 = 4.75% Example 11.5 Find the accumulated value of 1 at the end of n periods where the eﬀective rate of interest for the kth period, 1 ≤k≤n, is deﬁned by ik = (1 + r)k(1 + i) −1. Solution. We know that a(n) = (1 + i1)(1 +i2) ···(1 +in) = (1 + r)(1 +i)(1 +r)2(1 +i) ···(1 +r)n(1 +i) = (1 + r)1+2+···+n(1 + i)n = (1 + r) n(n+1) 2 (1 + i)n 11 TIME VARYING INTEREST RATES 95 Practice Problems Problem 11.1 Find the eﬀective interest rate over a three-year period which is equivalent to an eﬀective rate of discount of 8% the ﬁrst year, 7% the second year, and 6% the third year. Problem 11.2 Find the accumulated value of 1 at the end of n years if the force of interest is δt = 1 1+t. Problem 11.3 Find the accumulated value of 1 at the end of 19 years if the force of interest is δt = 0.04(1 + t)−2. Problem 11.4 If δt = 0 .01t, 0 ≤t ≤2, ﬁnd the equivalent annual eﬀective rate of interest over the interval 0 ≤t≤2. Problem 11.5 An investment account earns 4% the ﬁrst year, 5% the second year, and X% the third year. Find X if the three year average interest rate is 10%. Problem 11.6 An investment of $500 was made three years ago. For the ﬁrst six months the annual eﬀective rate of interest paid on the account was 5%, but it then increased to 7% and did not change over the following two and a half years. Find the present accumulation of the investment. Problem 11.7 The annual eﬀective interest rate for a given year is determined by the function: in = 0 .02n, where n = 1,2,3,··· . Brad’s initial investment of $400 earns interest every year according to the preceding. Find the accumulated value of his investment after six years. Problem 11.8 ‡ At a force of interest δt = 2 (k+2t) (i) a deposit of 75 at time t= 0 will accumulate to X at time t= 3; and (ii) the present value at time t= 3 of a deposit of 150 at time t= 5 is also equal to X. Calculate X. Problem 11.9 ‡ Jennifer deposits 1000 into a bank account. The bank credits interest at a nominal annual rate of i compounded semi-annually for the ﬁrst 7 years and a nominal annual rate of 2 i convertible quarterly for all years thereafter. The accumulated amount in the account after 5 years is X. The accumulated amount in the account at the end of 10.5 years is 1,980. Calculate X. 96 THE BASICS OF INTEREST THEORY Problem 11.10 Suppose δt = t3 100 . Find 1 a(3) . Problem 11.11 In Fund X money accumulates at a force of interest δt = 0.01t+ 0.1, 0 ≤t≤20. In Fund Y money accumulates at an annual eﬀective interest rate i. An amount of 1 is invested in each fund for 20 years. The value of Fund X at the end of 20 years is equal to the value of Fund Y at the end of 20 years. Calculate the value of Fund Y at the end of 1.5 years. Problem 11.12 If the eﬀective rate of discount in year k is equal to 0 .01k+ 0.06 for k= 1,2,3, ﬁnd the equivalent rate of simple interest over the three-year period. Problem 11.13 A savings and loan association pays 7% eﬀective on deposits at the end of each year. At the end of every 3 years a 2% bonus is paid on the balance at that time. Find the eﬀective rate of interest earned by an investor if the money is left on deposit (a) Two years. (b) Three years. (c) Four years. Problem 11.14 On July 1, 1999 a person invested 1,000 in a fund for which the force of interest at time t is given by δt = .02(3 + 2t) where t is the number of years since January 1, 1999. Determine the accumulated value of the investment on January 1, 2000. Problem 11.15 You are given that δt = t 100 .Calculate the present value at the end of the 10 year of an accumulated value at the end of 15 years of $1,000. Problem 11.16 Calculate k if a deposit of 1 will accumulate to 2.7183 in 10 years at a force of interest given by: δt = { kt 0 <t ≤5 0.04kt2 t> 5 11 TIME VARYING INTEREST RATES 97 Problem 11.17 The annual eﬀective interest rate for year t is it = 1 (10+t) . Calculate the current value at the end of year 2 of a payment of 6,000 at the end of year 7. Problem 11.18 A fund pays a nominal rate of 12%. The nominal rate is compounded once in year 1, twice in year 2, 3 times in year 3, etc. Calculate the amount that must be invested today in order to accumulate 5,000 after 6 years. Problem 11.19 You are given δt = 2 t−1 , for 2 ≤t ≤20. For any one year interval between n and n+ 1, with 2 ≤n≤9, calculate d(2) n+1. Problem 11.20 Investment X for 100,000 is invested at a nominal rate of interest j convertible semi-annually. After four years it accumulates to 214,358.88. Investment Y for 100,000 is invested at a nominal rate of discount k convertible quarterly. After two years, it accumulates to 232,305.73. Investment Z for 100,000 is invested at an annual eﬀective rate of interest equal to j in year one and an annual eﬀective rate of discount equal to k in year two. Calculate the value of investment Z at the end of two years. Problem 11.21 A bank agrees to lend John 10,000 now and X three years later in exchange for a single repayment of 75,000 at the end of 10 years. The bank charges interest at an annual eﬀective rate of 6% for the ﬁrst 5 years and at a force of interest δt = 1 t+1 for t≥5. Determine X. Problem 11.22 John invests 1000 in a fund which earns interest during the ﬁrst year at a nominal rate of K convertible quarterly. During the 2nd year the fund earns interest at a nominal discount rate of K convertible quarterly. At the end of the 2nd year, the fund has accumulated to 1173.54. Calculuate K. Problem 11.23 At time 0, 100 is deposited into Fund X and also into Fund Y. FundX accumulates at a force of interest δt = 0.5(1 + t)−2. Fund Y accumulates at an annual eﬀective interest rate of i. At the end of 9 years, the accumulated value of Fund X equals the accumulated value of Fund Y. Determine i. 98 THE BASICS OF INTEREST THEORY Problem 11.24 ‡ An investor deposits 1000 on January 1 of year xand deposits 1000 on January 1 of year x+ 2 into a fund that matures on January 1 of year x+ 4. The interest rate on the fund diﬀers every year and is equal to the annual eﬀective rate of growth of the gross domestic product (GDP) during the fourth quarter of the previous year. The following are the relevant GDP values for the past 4 years. Year x−1 x x+ 1 x+ 2 Quarter III 800.0 850.0 900.0 930.0 Quarter IV 808.0 858.5 918.0 948.6 What is the internal rate of return earned by the investor over the 4-year period? Problem 11.25 A deposit of 10,000 is made into a fund at time t = 0. The fund pays interest at a nominal rate of discount of d compounded quarterly for the ﬁrst two years. Beginning at time t = 2, interest is credited using a nominal rate of interest of 8% compounded quarterly. At time t = 5 , the accumulated value of the fund is 14,910. Calculate d. Problem 11.26 Amin deposits 10,000 in a bank. During the ﬁrst year the bank credits an annual eﬀective rate of interest i. During the second year the bank credits an annual eﬀective rate of interest ( i−5%). At the end of two years she has 12,093.75 in the bank. What would Amin have in the bank at the end of three years if the annual eﬀective rate of interest were ( i+ 9%) for each of the three years? Problem 11.27 Fund X starts with 1,000 and accumulates with a force of interest δt = 1 15 −t for 0 ≤t <15. Fund Y starts with 1,000 and accumulates with an interest rate of 8% per annum compounded semi-annually for the ﬁrst three years and an eﬀective interest rate of i per annum thereafter. Fund X equals Fund Y at the end of four years. Calculate i. Problem 11.28 Amin puts 100 into a fund that pays an eﬀective annual rate of discount of 20% for the ﬁrst two years and a force of interest of rate δt = 2t t2 + 8 < 2 ≤t≤4 11 TIME VARYING INTEREST RATES 99 for the next two years. At the end of four years, the amount in Amin’s account is the same as what it would have been if he had put 100 into an account paying interest at the nominal rate of i per annum compounded quarterly for four years. Calculate i. Problem 11.29 On January 1, 1980, Jack deposited 1,000 into Bank X to earn interest at the rate of j per annum compounded semi-annually. On January 1, 1985, he transferred his account to Bank Y to earn interest at the rate of k per annum compounded quarterly. On January 1, 1988, the balance at Bank Y is 1,990.76. If Jack could have earned interest at the rate of k per annum compounded quarterly from January 1, 1980 through January 1, 1988, his balance would have been 2,203.76. Calculate the ratio k j. 100 THE BASICS OF INTEREST THEORY 12 Equations of Value and Time Diagrams Interest problems generally involve four quantities: principal(s), investment period length(s), in- terest rate(s), accumulated value(s). If any three of these quantities are known, then the fourth quantity can be determined. In this section we introduce equations that involve all four quantities with three quantities are given and the fourth to be found. In calculations involving interest, the value of an amount of money at any given point in time depends upon the time elapsed since the money was paid in the past or upon time which will elapse in the future before it is paid. This principle is often characterized as the recognition of the time value of money. We assume that this principle reﬂects only the eﬀect of interest and does not include the eﬀect of inﬂation. Inﬂation reduces the the purchasing power of money over time so investors expect a higher rate of return to compensate for inﬂation. As pointed out, we will neglect the eﬀct of inﬂation when applying the above mentioned principle. As a consequence of the above principle, various amounts of money payable at diﬀerent points in time cannot be compared until all the amount are accumulated or discounted to a common date, called the comparison date, is established. The equation which accumulates or discounts each payment to the comparison date is called the equation of value. One device which is often helpful in the solution of equations of value is the time diagram. A time diagram is a one-dimensional diagram where the only variable is time, shown on a single coordinate axis. We may show above or below the coordinate of a point on the time-axis, values of money intended to be associated with diﬀerent funds. A time diagram is not a formal part of a solution, but may be very helpful in visualizing the solution. Usually, they are very helpful in the solution of complex probems. Example 12.1 In return for a payment of $1,200 at the end of 10 years, a lender agrees to pay $200 immediately, $400 at the end of 6 years, and a ﬁnal amount at the end of 15 years. Find the amount of the ﬁnal payment at the end of 15 years if the nominal rate of interest is 9% converted semiannually. Solution. The comparison date is chosen to be t= 0. The time diagram is given in Figure 12.1. Figure 12.1 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 101 The equation of value is 200 + 400(1 + 0.045)−12 + X(1 + 0.045)−30 = 1200(1 + 0.045)−20. Solving this equation for X we ﬁnd X ≈$231.11 With compound interest, an equation of value will produce the same answer for an unknown value regardless of what comparison date is selected. This is due to the fact that multiplying an equation by a power of an expression yields an equivalent equation. We illustrate this in the next example. Example 12.2 In return for a promise to receive $600 at the end of 8 years, a person agrees to pay $100 at once, $200 at the end of 5 years, and to make further payment at the end of 10 years. Find the payment at the end of 10 years if the nominal interest rate is 8% convertible semi-annually. Solution. The equation of value at t= 0 is 100 + 200(1 + 0.04)−10 + X(1 + 0.04)−20 = 600(1 + 0.04)−16. Solving for X we ﬁnd X ≈186.75. The equation of value at t= 10 is 100(1 + 0.04)20 + 200(1 + 0.04)10 + X = 600(1 + 0.04)4. Solving for X we ﬁnd X ≈186.75.Note that by multiplying the last equation by (1 + 0.04)−20 one gets the equation of value at t= 0 Whereas the choice of a comparison date has no eﬀect on the answer obtained with compound interest, the same cannot be said of simple interest or simple discount. Example 12.3 Find the amount to be paid at the end of 10 years which is equivalent to two payments of 100 each, the ﬁrst to be paid immediately, and the second to be paid at the end of 5 years. Assume 5% simple interest is earned from the date each payment is made, and use a comparison date 1. The end of 10 years. 2. The end of 15 years. 102 THE BASICS OF INTEREST THEORY Solution. 1. With a comparison date at t= 10, the payment at the end of 10 years will be 100(1 + 10×0.05) + 100(1 + 5×0.05) = 275. 2. With a comparison date at t = 15 , the amount P that must be paid at the end of 10 years satisﬁes the equation of value 100(1 + 15×0.05) + 100(1 + 10×0.05) = P(1 + 5×0.05) which implies that P = 260 Example 12.4 Investor A deposits 1,000 into an account paying 4% compounded quarterly. At the end of three years, he deposits an additional 1,000. Investor B deposits X into an account with force of interest δt = 1 6+t. After ﬁve years, investors A and B have the same amount of money. Find X. Solution. Consider investor A’s account ﬁrst. The initial 1,000 accumulates at 4% compounded quarterly for ﬁve years; the accumulated amount of this piece is 1,000 ( 1 + 0.04 4 )4×5 = 1000(1.01)20. The second 1,000 accumulates at 4% compounded quarterly for two years, accumulating to 1,000 ( 1 + 0.04 4 )4×2 = 1000(1.01)8. The value in investor A’s account after ﬁve years is A= 1000(1.01)20 + 1000(1.01)8. The accumulated amount of investor B’s account after ﬁve years is given by B = Xe ∫5 0 dt 6+t = Xeln (11 6 ) = 11 6 X. The equation of value at time t= 5 is 11 6 X = 1000(1.01)20 + 1000(1.01)8. Solving for X we ﬁnd X ≈$1,256.21 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 103 Practice Problems Problem 12.1 In return for payments of $5,000 at the end of 3 years and $4,000 at the end of 9 years, an investor agrees to pay $1500 immediately and to make an additional payment at the end of 2 years. Find the amount of the additional payment if i(4) = 0.08. Problem 12.2 At a certain interest rate the present values of the following two payment patterns are equal: (i) 200 at the end of 5 years plus 500 at the end of 10 years; (ii) 400.94 at the end of 5 years. At the same interest rate 100 invested now plus 120 invested at the end of 5 years will accumulate to P at the end of 10 years. Calculate P. Problem 12.3 An investor makes three deposits into a fund, at the end of 1, 3, and 5 years. The amount of the deposit at time t is 100(1.025)t. Find the size of the fund at the end of 7 years, if the nominal rate of discount convertible quarterly is 4 41 . Problem 12.4 ‡ Brian and Jennifer each take out a loan of X. Jennifer will repay her loan by making one payment of 800 at the end of year 10. Brian will repay his loan by making one payment of 1,120 at the end of year 10. The nominal semi-annual rate being charged to Jennifer is exactly one-half the nominal semi-annual rate being charged to Brian. Calculate X. Problem 12.5 Fund A accumulates at 6% eﬀective, and Fund B accumulates at 8% eﬀective. At the end of 20 years the total of the two funds is 2,000. At the end of 10 years the amount in Fund A is half that in Fund B. What is the total of the two funds at the end of 5 years? Problem 12.6 Louis has an obligation to pay a sum of $3,000 in four years from now and a sum of $5,000 in six years from now. His creditor permits him to discharge these debts by paying $X in two years from now, $1000 in three years from now, and a ﬁnal payment of $2X in nine years from now. Assuming an annual eﬀective rate of interest of 6%, ﬁnd X. Problem 12.7 Every Friday in February (7, 14, 21,28) Vick places a 1,000 bet, on credit, with his oﬀ-track bookmaking service, which charges an eﬀective weekly interest rate of 8% on all credit extended. 104 THE BASICS OF INTEREST THEORY Vick looses each bet and agrees to repay his debt in four installments to be made on March 7, 14, 21, and 28. Vick pays 1,100 on March 7, 14, and 21. How much must Vick pay on March 28 to completely repay his debt? Problem 12.8 A borrower is repaying a loan by making payments of 1,000 at the end of each of the next 3 years. The interest rate on the loan is 5% compounded annually. What payment could the borrower make at the end of the ﬁrst year in order to extinguish the loan? Problem 12.9 An investor purchases an investment which will pay 2,000 at the end of one year and 5,000 at the end of four years. The investor pays 1,000 now and agrees to pay X at the end of the third year. If the investor uses an interest rate of 7% compounded annually, what is X? Problem 12.10 Today is New Year’s Day. In return for payments of 1,500 at the end of January, February, and March, and of 3,000 at the end of May, July, and September, an investor agrees to pay now the total value of the 6 payments, and to either make or receive an additional payment at the end of December. Find the amount of that additional payment if it is known that the nominal annual interest rate is 6%, compounded monthly. Problem 12.11 A debt of 7,000 is due at the end of 5 years. If 2,000 is paid at the end of 1 year, what single payment should be made at the end of the 2nd year to liquidate the debt, assuming interest at the rate of 6.5% per year, compounded quarterly. Problem 12.12 George agrees to buy his brother’s car for 7,000. He makes a down payment of 4,000 now, and agrees to pay two equal payments, one at the end of 6 months, and the other at the end of a year. If interest is being charged at 5% per annum eﬀective, how large should each of the equal payments be? Problem 12.13 ‡ Jeﬀ deposits 10 into a fund today and 20 ﬁfteen years later. Interest is credited at a nominal discount rate of dcompounded quarterly for the ﬁrst 10 years, and at a nominal interest rate of 6% compounded semiannually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. 12 EQUATIONS OF VALUE AND TIME DIAGRAMS 105 Problem 12.14 ‡ The parents of three children, ages 1, 3, and 6, wish to set up a trust fund that will pay X to each child upon attainment of age 18, and Y to each child upon attainment of age 21. They will establish the trust fund with a single investment of Z. Find the equation of value for Z. Problem 12.15 An investment fund is established at time 0 with a deposit of 5000. 1000 is added at the end of 4 months, and an additional 4000 is added at the end of 8 months. No withdrawals are made. The fund value, including interest, is 10560 at the end of 1 year. The force of interest at time t is k 1+(1−t)k, for 0 ≤t≤1. Determine k. Problem 12.16 A loan of 1000 is made at an interest rate of 12% compounded quarterly. The loan is to be repaid with three payments: 400 at the end of the ﬁrst year, 800 at the end of the ﬁfth year, and the balance at the end of the tenth year. Calculate the amount of the ﬁnal payment. Problem 12.17 You are given δt = 0.5 1+t for t> 0. A payment of 400 at the end of 3 years and 800 at the end of 15 years has the same present value as a payment of 300 at the end of 8 years and X at the end of 24 years. Calculate X. Problem 12.18 A loan of 12 is to be repaid with payments of 10 at the end of 3 years and 5 at the end of 6 years. Calculate the simple discount rate that is being charged on the loan. Problem 12.19 A deposit of K3 at time t= 0 accumulates to 8000 after 3nyears using an annual eﬀective discount rate of d.Using the same annual eﬀective discount rate, a deposit of K2 at time t= 0 accumulates to X after 2n years. Calculate X. Problem 12.20 An investor puts 1000 into Fund X and 1000 into Fund Y. Fund Y earns compound interest at the annual rate of j >0,and Fund X earns simple interest at the annual rate of 1 .10j.At the end of 2 years, the amount in Fund Y is 30 more than the amount in Fund X. After 5 years, the amount in Fund Y exceeds the amount in Fund X by E. Determine E. Problem 12.21 You are given two loans, with each loan to be repaid by a single payment in the future. Each payment includes both principal and interest. The ﬁrst loan is repaid by a 3,000 payment at the 106 THE BASICS OF INTEREST THEORY end of four years. The interest is accrued at 10% per annum compounded semi-annually. The second loan is repaid by a 4,000 payment at the end of ﬁve years. The interest is accrued at 8% per annum compounded semi-annually. These two loans are to be consolidated. The consolidated loan is to be repaid by two equal installments of X, with interest at 12% per annum compounded semi-annually. The ﬁrst payment is due immediately and the second payment is due one year from now. Calculate X. Problem 12.22 An investment fund accrues interest with force of interest δt = K 1 + (1−t)K for 0 ≤t≤1. At time zero, there is 100,000 in the fund. At time one there is 110,000 in the fund. The only two transactions during the year are a deposit of 15,000 at time 0.25 and a withdrawal of 20,000 at time 0.75. Calculate K. Problem 12.23 Paul borrows 1000 at an annual interest rate of 12%. He repays the loan in full by making the following payments: (1) 500 at the end of 4 months (2) 200 at the end of 14 months (3) R at the end of 18 months Calculate R. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 107 13 Solving for the Unknown Interest Rate In this section, it is the rate of interest i that is unknown in an equation of value. We discuss four ways for detrmining i. Method 1: Direct Method When a single payment is involved, the method that works best is to solve for i directly in the equation of value (using exponential and logarithmic functions). In this situation, the equation of value takes either the form A = P(1 + i)n or A = Peδn. In the ﬁrst case, the interest is found by using the exponential function obtaining i= (A P )1 n −1. In the second case, the interest is found using the logarithmic function obtaining δ= 1 nln (A P ) . Example 13.1 At what interest rate convertible semiannually would $500 accumulate to $800 in 4 years? Solution. Let j = i(2) 2 be the eﬀective rate per six months. We are given that 500(1 + j)8 = 800. Solving for j and using a calculator we ﬁnd j = (8 5 )1 8 −1 ≈0.06051. Thus, i(2) = 2j = 0.121 = 12.1% Method 2: Analytical Method When multiple payments are involved, the equation of value takes the formf(i) = 0.In this case, the problem is the classical problem of determining the nonnegative solutions to the equation f(i) = 0 with f(i) being a diﬀerentiable function. If f(i) is a polynomial then the problem reduces to solving polynomial equations. Some known algebraic methods can be used such as the rational zero test, or the quadratic formula, etc. Example 13.2 At what eﬀective interest rate would the present value of $1000 at the end of 3 years plus $2,000 at the end of 6 years be equal to $2,700? Solution. The equation of value at time t = 0 is 2000 ν6 + 1000ν3 = 2700. Dividing through by 100 we ﬁnd 108 THE BASICS OF INTEREST THEORY 20ν6 + 10ν3 = 27.Using the quadratic formula and the fact that ν3 >0 we obtain ν3 = 0.9385 from which we obtain ν ≈0.9791. Hence, i= 1 0.9791 −1 ≈0.0213 = 2.13% Method 3: Linear Interpolation If the equation f(i) = 0 can not be solved known algebraic methods, then approximation methods can be used such as linear interpolation. We next introduce this concept. Suppose that we know the values of a function f(x) at distinct points x1 and at x2, and that f(x1) ̸= f(x2). If |x1 −x2|is small, it may be reasonable to assume that the graph of f is approxi- mately linear between x1 and x2. That is equivalent to assuming that f(x) = f(x1) + f(x2) −f(x1) x2 −x1 (x−x1) for x1 <x<x 2. If now we wish to determine an approximate value of x where f(x) has a speciﬁc value y0, we may use this equation for that purpose and ﬁnd x= x1 + (x2 −x1) y0 −f(x1) f(x2) −f(x1). In particular, if we wish to ﬁnd a zero of f near the given points (i.e. y0 = 0),and if f(x1) ̸= f(x2), a good approximation could be x≈x1 −f(x1) x2 −x1 f(x2) −f(x1). When we apply these formulas for points between two points x1 and x2,we speak of linear inter- polation. The following example illustrates the use of linear interpolation in determing the unknown rate. Example 13.3 At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Solution. The equation of value at time t= 10 years is 1000(1 + j)20 + 2000(1 +j)14 = 5000 where j = i(2) 2 . We will use linear interpolation to solve for j. For that purpose, we deﬁne f(j) = 1000(1 + j)20 + 2000(1 +j)14 −5000. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 109 We want to ﬁnd j such that f(j) = 0 . By trial and error we see that f(0.03) = −168.71 and f(0.035) = 227.17. Since f is continuous, j is between these two values. Using linear interpolation we ﬁnd j ≈0.03 + 168.71 × 0.005 227.17 + 168.71 ≈0.0321 and f(0.0321) = −6.11. Hence, i(2) = 2j = 0.0642 = 6.42% Method 4: Successive Iterations Methods In order to achieve a higher level of accuracy than the one provided with linear interpolation, it- eration methods come to mind. We will discuss two iteration methods: the bisection method and the Newton-Raphson method. The bisection method is based on the fact that a diﬀerentiable functionf that satisﬁes f(α)f(β) < 0 must satisfy the equation f(x0) = 0 for some x0 between α and β. The ﬁrst step in the methods consists of ﬁnding two starting values x0 <x1 such that f(x0)f(x1) < 0.Usually, these values are found by trial and error. Next, bisect the interval by means of the mid- point x2 = x0+x1 2 . If f(x0)f(x2) < 0 then apply the bisection process to the interval x0 ≤x ≤x2. If f(x1)f(x2) < 0 then apply the bisection process to the interval x2 ≤x ≤x1. We continue the bisection process as many times as necessary to achieve the desired level of accuracy. We illustrate this method in the next example. Example 13.4 At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Use the bisection method. Solution. The equation of value at time t= 10 years is 1000(1 + j)20 + 2000(1 +j)14 = 5000 where j = i(2) 2 . Deﬁne f(j) = 1000(1 + j)20 + 2000(1 +j)14 −5000. Descartes’ rule of signs says that the maximum number of positive roots to a polynomial equation is equal to the number of sign changes in the coeﬃcients of the polynomial. In our case, Descartes’ rule asserts the existence of one positive root to the equation f(j) = 0. Now, by trial and error we see that f(0) = −2000 and f(0.1) = 9322.50. Thus, j is between these two values. Let j2 = 0+0.1 2 = 0.05. Then f(0.05) = 1613.16 so that f(0)f(0.05) <0. Now bisect the interval [0,0.05] by means of the point j3 = 0.5(0 + 0.05) = 0.025 and note that f(j3) = −535.44. Continue this process to obtain the table 110 THE BASICS OF INTEREST THEORY n jn f(jn) 0 0 −2000 1 0.1 9322.50 2 0.05 1613.16 3 0.025 −535.44 4 0.0375 436.75 5 0.03125 −72.55 6 0.034375 176.02 7 0.0328125 50.25 8 0.03203125 −11.52 9 0.032421875 19.27 10 0.0322265625 3.852 11 0.0321289063 −3.84 12 0.0321777344 0.005 Thus, j ≈0.032178 accurate to six decimal places compared to 0 .0321 found in Example 13.3. Hence, i2 = 2j = 0.06436 = 6.436% The problem with the bisection method is that the rate of convergence is slow. An iteration method with a faster rate of convergence is the Newton-Raphson method given by the iteration formula jn+1 = jn −f(jn) f′(jn). This method is discussed in more details in Section 20. Example 13.5 Rework Example 13.3 using the Newton-Raphson method. Solution. We have f(j) = 1000(1 + j)20 + 2000(1 +j)14 −5000. and f′(j) = 20000(1 + j)19 + 28000(1 +j)13. Thus, jn+1 = jn −(1 + jn)20 + 2(1 +jn)14 −5 20(1 + jn)19 + 28(1 +jn)13 . 13 SOLVING FOR THE UNKNOWN INTEREST RATE 111 Let j0 = 0. Then j1 =0 −1 + 2−5 20 + 28 = 0.0416666667 j2 =0.0416666667 − 1.041666666720 + 2(1.0416666667)14 −5 20(1.0416666667)19 + 28(1.0416666667)13 = 0.0328322051 j3 =0.0328322051 − 1.032832205120 + 2(1.0328322051)14 −5 20(1.0328322051)19 + 28(1.0328322051)13 = 0.0321809345 j4 =0.0321809345 − 1.032180934520 + 2(1.0321809345)14 −5 20(1.0321809345)19 + 28(0.0321809345)13 = 0.032177671 Thus, in four iterations we ﬁnd j = 0.032178 and hence i(2) = 6.436% 112 THE BASICS OF INTEREST THEORY Practice Problems Problem 13.1 Find the nominal rate of interest convertible semiannually at which the accumulated value of 1,000 at the end of 15 years is 3,000. Problem 13.2 Find the exact eﬀective rate of interest at which payments of 300 at the present, 200 at the end of one year, and 100 at the end of two years will accumulate to 700 at the end of two years. Problem 13.3 The sum of the accumulated value of 1 at the end of three years at a certain eﬀective rate of interest i, and the present value of 1 to be paid at the end of three years at an eﬀective rate of discount numerically equal to i is 2.0096. Find the rate i. Problem 13.4 ‡ David can receive one of the following two payment streams: (i) 100 at time 0, 200 at time n, and 300 at time 2 n (ii) 600 at time 10 At an annual eﬀective interest rate of i, the present values of the two streams are equal. Given νn = 0.75941, determine i. Problem 13.5 ‡ Mike receives cash ﬂows of 100 today, 200 in one year, and 100 in two years. The present value of these cash ﬂows is 364.46 at an annual eﬀective rate of interest i. Calculate i. Problem 13.6 ‡ At a nominal interest rate of i convertible semi-annually, an investment of 1,000 immediately and 1,500 at the end of the ﬁrst year will accumulate to 2,600 at the end of the second year. Calculate i. Problem 13.7 ‡ Eric deposits 100 into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 200 into a diﬀerent savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 113 Problem 13.8 ‡ A deposit of 100 is made into a fund at time t = 0. The fund pays interest at a nominal rate of discount of d compounded quarterly for the ﬁrst two years. Beginning at time t = 2, interest is credited at a force of interest δt = 1 t+1 . At time t = 5, the accumulated value of the fund is 260. Calculate d. Problem 13.9 ‡ A customer is oﬀered an investment where interest is calculated according to the following force of interest: δt = { 0.02t 0 ≤t≤3 0.045 3 <t. The customer invests 1,000 at time t= 0.What nominal rate of interest, compounded quarterly, is earned over the ﬁrst four-year period? Problem 13.10 A manufacturer sells a product to a retailer who has the option of paying 30% below the retail price immediately, or 25% below the retail price in six months. Find the annual eﬀective rate of interest at which the retailer would be indiﬀerent between the two options. Problem 13.11 An investor deposits 10,000 in a bank. During the ﬁrst year, the bank credits an annual eﬀective rate of interest i. During the second year the bank credits an annual eﬀective interest i−0.05. At the end of two years the account balance is 12,093.75. What would the account balance have been at the end of three years if the annual eﬀective rate of interest were i+ 0.09 for each of the three years? Problem 13.12 You are given δt = 2 t−1 for any t ∈[2,10]. For any one year interval between n and n+ 1; with 2 ≤n≤9; calculate the equivalent nominal rate of discount d(2). Problem 13.13 It is known that an investment of $1,000 will accumulate to $1,825 at the end of 10 years. If it is assumed that the investment earns simple interest at rate i during the 1st year, 2 i during the second year, ··· , 10i during the 10th year, ﬁnd i. Problem 13.14 It is known that an amount of money will double itself in 10 years at a varying force of interest δt = kt. Find k. 114 THE BASICS OF INTEREST THEORY Problem 13.15 A fund pays 1 at time t = 0 , 2 at time t = 2 n and 1 at time t = 4 n. The present value of the payments is 3.61. Calculate (1 + i)n. Problem 13.16 The present value of 300 in 3 years plus 600 in 6 years is equal to 800. Calculate i. Problem 13.17 A savings and loan association pays 7% eﬀective on deposits at the end of each year. At the end of every three years a 2% bonus is paid on the balance at that time. Find the eﬀective rate of interest earned by an investor if the money is left on deposit: (a) Two years. (b) Three years. (c) Four years. Problem 13.18 Consider the following equation of value 25 j = 1 −(1 + j)−40. Use linear interpolation to estimate j. Problem 13.19 Consider the following equation of value 0.0725[1 −(1 + 0.01j)−50] = 0.01j.Use linear interpolation to estimate j. Problem 13.20 Two funds, X and Y, start with the same amount. You are given (1) Fund X accumulates at a force of interest 5% (2) Fund Y accumulates at a rate of interest j compounded semi-annually (3) At the end of eight years, Fund X is 1.05 times as large as Fund Y. Calculate j. Problem 13.21 You are given a loan on which interest is charged over a 4 year period, as follows (1) an eﬀective rate of discount of 6% for the ﬁrst year (2) a nominal rate of discount of 5% compounded every 2 years for the second year (3) a nominal rate of interest of 5% compounded semiannually for the third year (4) a force of interest of 5% for the fourth year. Calculate the annual eﬀective rate of interest over the 4 year period. 13 SOLVING FOR THE UNKNOWN INTEREST RATE 115 Problem 13.22 An investor deposits 10 into a fund today and 20 ﬁfteen years later. Interest is credited at a nominal discount rate of dcompounded quarterly for the ﬁrst 10 years, and at a nominal interest rate of 6% compounded semi-annually thereafter. The accumulated balance in the fund at the end of 30 years is 100. Calculate d. 116 THE BASICS OF INTEREST THEORY 14 Solving for Unknown Time As pointed out in Section 12, if any three of the four basic quantities: principal(s), investment period length(s), interest rate(s), accumulated value(s) are given, then the fourth can be determined. In this section, we consider the situation in which the length of the investment period is the unknown. Example 14.1 A single payment of $3,938.31 will pay oﬀ a debt whose original repayment plan was $1,000 due on January 1 of each of the next four years, beginning January 1, 2006. If the eﬀective annual rate is 8%, on what date must the payment of $3,938.31 be paid? Solution. The equation of value with comparison date January 1, 2006 is 1,000[1 + (1 + 0.08)−1 + (1 + 0.08)−2 + (1 + 0.08)−3] = 3938.31(1 + 0.08)−t. Solving this equation for t we ﬁnd t= ln {1000(3938.31)−1[1 + (1 + 0.08)−1 + (1 + 0.08)−2 + (1 + 0.08)−3]} ln 1.08 ≈1.25. Therefore, the single payment of 3938.31 must be paid 1.25 years after January 1, 2006, or on March 31, 2007 In the above example, the answer was obtained by using a pocket calculator with exponential and logarithmic functions. An alternative approach to using calculator is the use of linear interpolation in the interest tables as illustrated in the next example. Example 14.2 Find the length of time necessary for $1000 to accumulate to $1500 if invested at 6% per year compounded semiannually: (a) by a direct method, i.e. using a calculator; (b) by interpolating in the interest tables Solution. Let t be the comparison date. Then we must have 1000(1 .03)2t = 1000 or (1.03)2t = 1.5. (a) Using a pocket calculator we ﬁnd t= 1 2 ln 1.5 ln 1.03 = 6.859 years. 14 SOLVING FOR UNKNOWN TIME 117 (b) From interest tables, (1.03)13 = 1.46853 and (1.03)14 = 1.51259.Thus, 13 <2t< 14.Performing a linear interpolation we ﬁnd 2t= 13 + 1.5 −1.46853 1.51259 −1.46853 = 13.714. Thus, t= 6.857 years The Rule of 72 Next, we consider the following problem: Given a particular rate of compound interest, how long will it take an investment to double? Starting with 1 invested at an annually compounded interest i, let n be the time needed for the accumulated value to become 2. That is, (1 + i)n = 2. Taking the natural logarithm of both sides we ﬁnd nln (1 +i) = ln 2 and solving for n we obtain n= ln 2 ln (1 +i). We would like to ﬁnd an approximation that is pretty good for an interest rate of i= 8%.For that purpose, we start by writing the Taylor series approximation of x ln (1+x) around x= 0.08 to obtain x ln (1 +x) =1.039486977 + 0.4874162023(x−0.08) −0.07425872903(x−0.08)2 + ··· ≈1.039486977 for x very “close” to 0.08. Thus, the time necessary for the original principal to double is given by t= ln 2 ln (1 +i) = ln 2 i · i ln (1 +i) ≈ln 2 i (1.039486977) ≈0.72 i . This is called the rule of 72, since t can be written in the form t= 72 100i. Here’s a table that shows the actual number of years required to double your money based on diﬀerent interest rates, along with the number that the rule of 72 gives you. 118 THE BASICS OF INTEREST THEORY % Rate Actual Rule 72 1 69.66 72 2 35.00 36 3 23.45 24 4 17.67 18 5 14.21 14.4 6 11.90 12 7 10.24 10.29 8 9.01 9 9 8.04 8 10 7.27 7.2 ··· ··· ··· 15 4.96 4.8 20 3.80 3.6 25 3.11 2.88 30 2.64 2.4 40 2.06 1.8 50 1.71 1.44 75 1.24 0.96 100 1.00 0.72 Example 14.3 Suppose $2,000 is invested at a rate of 7% compounded semiannually. (a) Find the length of time for the investment to double by using the exact formula. (b) Find the length of time for the investment to double by using the rule of 72. Solution. (a) By using the exact formula for compound interest we ﬁnd 2000(1 + 0.035)2t = 4000. Solving for t we ﬁnd t≈10.074 years. (b) Using the Rule of 72 we ﬁnd t= 72 7 ≈10.286 The Method of Equated Time We consider the following problem: Suppose amounts s1,s2,··· ,sn are to be paid at respective times t1,t2,··· ,tn with annual interest rate i. What would be the time t when one single payment 14 SOLVING FOR UNKNOWN TIME 119 of s1 + s2 + ··· + sn would be equivalent to the individual payments made separately? A time diagram for this situation is shown in Figure 14.1 Figure 14.1 In this case, the equation of value at t= 0 is (s1 + s2 + ··· + sn)νt∗ = s1νt1 + s2νt2 + ··· + snνtn. (14.1) Taking the natural logarithm of both sides we ﬁnd t∗ln ν+ ln (s1 + s2 + ··· + sn) = ln (s1νt1 + s2νt2 + ··· + snνtn). Solving for t∗ and recalling that ln ν = −δ we ﬁnd t∗= −1 δln (s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn ) =1 δ [ ln ( n∑ k=1 sk ) −ln ( n∑ k=1 skνtk )] In order to provide an approximation to t∗ we recall the binomial series expansion (1 + x)α = 1 + α 1!x+ α(α−1) 2! x2 + ··· + α(α−1) ···(α−n+ 1) n! xn + ··· where α∈R and −1 <x< 1. Using this series expansion we can write νt∗ = (1 + i)−t∗ = 1 + (−t∗) 1! i+ (−t∗)(−t∗−1) 2! i2 + ···≈ 1 −it∗. Similarly, we have νtk = (1 + i)−tk = 1 + (−tk) 1! i+ (−tk)(−tk −1) 2! i2 + ···≈ 1 −itk. 120 THE BASICS OF INTEREST THEORY Substituting these approximations in the equation νt∗ = s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn we ﬁnd 1 −it∗≈s1(1 −it1) + s2(1 −it2) + ··· + sn(1 −itn) s1 + s2 + ··· + sn which can be written in the form 1 −it∗≈1 −is1t1 + s2t2 + ··· + sntn s1 + s2 + ··· + sn . Solving for t∗ we obtain the approximation t∗≈s1t1 + s2t2 + ··· + sntn s1 + s2 + ··· + sn = t. Estimating t∗ with t is known as the method of equated time. Note that the formula for t is similar to the center of mass that is covered in a calculus course where the weights are replaced by the various amounts paid. Example 14.4 Payments of $100, $200, and $500 are due at the ends of years 2,3, and 8, respectively. Assuming an eﬀective rate of interest of 5% per year, ﬁnd the point in time at which a payment of $800 would be equivalent using the method of equated time. Solution. Using the method of equating time we ﬁnd t= 100 ·2 + 200·3 + 500·8 100 + 200 + 500 = 6 years With the approximation t, the estimated present value for the equivalent single payment is PV = (s1 + s2 + ··· + sn)νt. The following result shows that tis an overestimate of t.Also, the result shows that the true present value exceeds the present value given by the method of equated time. Theorem 14.1 With s1,s2,··· ,sn,t1,t2,··· ,tn,t, and t as deﬁned above we have s1νt1 + s2νt2 + ··· + snνtn >(s1 + s2 + ··· + sn)νt. Hence, t>t. 14 SOLVING FOR UNKNOWN TIME 121 Proof. Consider s1 payments each equal to νt1,s2 payments each equal to νt2,··· ,sn payments each equal to νtn. The arithmetic mean of these payments is s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn and the geometric mean of these payments is ( νs1t1νs2t2 ···νsntn ) 1 s1+s2+···+sn = ν s1t1+s2t2+···+sntn s1+s2+···+sn = νt. In Problem 14.16 we show that the arithmetic mean is always greater than the geometric mean. Hence, s1νt1 + s2νt2 + ··· + snνtn s1 + s2 + ··· + sn >ν t or s1νt1 + s2νt2 + ··· + snνtn >(s1 + s2 + ··· + sn)νt. The left-hand side is the true present value and the right-hand side is the estimated present value. Note that this last equation implies that νt >ν t by (14.1). Thus, νt−t >1 and since ν <1 we must have t>t Remark 14.1 As seen above, the solution by the method of equated time is relatively easy to compute, and is generally fairly accurate. Also, an interesting thing to notice is that the interest rate plays no role in the computations when the method of equated time is used. Example 14.5 A loan is negotiated with the lender agreeing to accept $1,000 after 10 years, $2,000 after 20 years, and $3,000 after 30 years in full repayments. The borrower wishes to liquidate the loan with a single $6,000 payment. Let T1 represent the time of the $6,000 payment calculated by an equation of value. Let T2 represent the time determined by the method of equated time. If i = 0.01, ﬁnd T2 −T1 to the nearest penny. Solution. The equation of value at time t= 0 is 1,000ν10 + 2,000ν20 + 3,000ν30 = 6,000νT1. Solving we ﬁnd T1 = ln (ν10 + 2ν20 + 3ν30) −ln 6 −ln 1.01 = 23.05. 122 THE BASICS OF INTEREST THEORY Now, using the method of equated time we ﬁnd T2 = 1,000 ×10 + 2,000 ×20 + 3,000 ×30 6,000 = 23.33. Thus, T2 −T1 = 23.33 −23.05 = 0.28 14 SOLVING FOR UNKNOWN TIME 123 Practice Problems Problem 14.1 The present value of a payment of $5,000 to be made in t years is equal to the present value of a payment of $7,100 to be made in 2 t years. If i= 7.5% ﬁnd t. Problem 14.2 The present value of two payments of 100 each to be made at the end of n years and 2 n years is 100. If i= 0.08, ﬁnd n. Problem 14.3 You are asked to develop a rule of n to approximate how long it takes money to triple. Find n. Problem 14.4 ‡ Joe deposits 10 today and another 30 in ﬁve years into a fund paying simple interest of 11% per year. Tina will make the same two deposits, but the 10 will be deposited n years from today and the 30 will be deposited 2 n years from today. Tina’s deposits earn an annual eﬀective rate of 9.15% . At the end of 10 years, the accumulated amount of Tina’s deposits equals the accumulated amount of Joe’s deposits. Calculate n. Problem 14.5 Find how long $1,000 should be left to accumulate at 6% eﬀective in order that it will amount to twice the accumulated value of another 1,000 deposited at the same time at 4% eﬀective. Problem 14.6 If an investment will be doubled in 8 years at a force of interest δ, in how many years will an investment be tripled at a nominal rate of interest numerically equal to δ and convertible once every three years? Problem 14.7 Amounts $500, $800, and $1,000 are to be paid at respective times 3 years from today, 5 years from today, and 11 years from today. The eﬀective rate of interest is 4.5% per annum. (a) Use the method of equated time to ﬁnd the number of years t from today when one single payment of $500 + $800 + $1000 = $2,300 would be equivalent to the individual payments made separately. (b) Find the exact number of years tfrom today when one single payment of $500 + $800 + $1000 = $2,300 would be equivalent to the individual payments made separately. 124 THE BASICS OF INTEREST THEORY Problem 14.8 A payment of n is made at the end of n years, 2n at the end of 2 n years, ··· , n2 at the end of n2 years. Find the value of t by the method of equated time. Problem 14.9 Fund A accumulates at a rate of 12% convertible monthly. Fund B accumulates with a force of interest δt = t 6 . At time t = 0 equal deposits are made in each fund. Find the next time that the two funds are equal. Problem 14.10 A loan requires the borrower to repay $1,000 after 1 year, 2000 after 2 years, $3,000 after 3 years, and $4,000 after 4 years. At what time could the borrower make a single payment of $10,000 according to the method of equated time? Problem 14.11 A series of payments of 100 are made with the ﬁrst payment made immediately. The second payment is made at the end of year 2. The third payment is made at the end of year 4, etc with the last payment being made at the end of year 12. Calculate t using the method of equated time. Problem 14.12 A series of payments are made at the end of each year for 50 years. The amount of the payment at the end of year n is n. Calculate t using the method of equated time. Problem 14.13 A fund earns interest at a nominal rate of interest of 12% compounded monthly. If $1,000 is invested in the fund, it will grow to be $5,320.97 after n years. Calculate n. Problem 14.14 A fund earns interest at a force of interest of δt = 0.05t. Lauren invests $2,000 at time t= 0. After n years, Lauren has $4,919.20 . Calculate n. Problem 14.15 A fund earns interest at a force of interest of δt = 0.01t2. Calculate the amount of time until the fund triples. Problem 14.16 Consider the positive numbers a1,a2,··· ,an where n is a positive integer. The arithmetic mean is deﬁned by a1 + a2 + ··· + an n 14 SOLVING FOR UNKNOWN TIME 125 whereas the geometric mean is deﬁned by [a1 ·a2 ···an] 1 n. Show that a1 + a2 + ··· + an n ≥[a1 ·a2 ···an] 1 n. Hint: Use the fact that ex ≥1 + x for all x. This implies e ai µ −1 ≥ ai µ,1 ≤i ≤n where µ is the arithmetic mean. Problem 14.17 A fund starts with a zero balance at time zero. The fund accumulates with a varying force of interest δt = 2t t2+1 ,for t≥0.A deposit of 100,000 is made at time 2. Calculate the number of years from the time of deposit for the fund to double. Problem 14.18 At time t= 0 Billy puts 625 into an account paying 6% simple interest. At the end of year 2, George puts 400 into an account paying interest at a force of interest δt = 1 6+t, for t≥2. If both accounts continue to earn interest indeﬁnitely at the levels given above, the amounts in both accounts will be equal at the end of year n. Calculate n. Problem 14.19 Payment of 300, 500, and 700 are made at the end of years ﬁve, six, and eight, respectively. Interest is accumulated at an annual eﬀective interest rate of 4%. You are to ﬁnd the point in time at which a single payment of 1500 is equivalent to the above series of payments. You are given: (i) X is the point in time calculated using the method of equated time. (ii) Y is the exact point in time. Calculate X+ Y. Problem 14.20 The present value of a payment of 1004 at the end of T months is equal to the present value of 314 after 1 month, 271 after 18 months, and 419 after 24 months. The eﬀective annual interest rate is 5%. Calculate T to the nearest integer. Problem 14.21 Using the method of equated time, a payment of 400 at time t = 2 plus a payment of X at time t= 5 is equivalent to a payment of 400 + X at time t= 3.125. At an annual eﬀective interest rate of 10%, the above two payments are equivalent to a payment of 400 + X at time t= k using the exact method. Calculate k. 126 THE BASICS OF INTEREST THEORY Problem 14.22 Investor A deposits 100 into an account that earns simple interest at a rate of 10% per annum. Investor Bmakes two deposits of 50 each into an account that earns compound interest at an annual eﬀective rate of 10%. B′s ﬁrst deposit occurs n years after A′s deposit. B′s second deposit occurs 2 n years after A′s deposit. The balances in the two accounts are equal 10 years after Investor A′s deposit. What is the value of n? 14 SOLVING FOR UNKNOWN TIME 127