{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "## Welcome to the Second Lab - Week 1, Day 3\n", "\n", "Today we will work with lots of models! This is a way to get comfortable with APIs." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", " \n", " \n", " \n", " \n", "
\n", " \n", " \n", "

Important point - please read

\n", " The way I collaborate with you may be different to other courses you've taken. I prefer not to type code while you watch. Rather, I execute Jupyter Labs, like this, and give you an intuition for what's going on. My suggestion is that you carefully execute this yourself, after watching the lecture. Add print statements to understand what's going on, and then come up with your own variations.

If you have time, I'd love it if you submit a PR for changes in the community_contributions folder - instructions in the resources. Also, if you have a Github account, use this to showcase your variations. Not only is this essential practice, but it demonstrates your skills to others, including perhaps future clients or employers...\n", " \n", "
" ] }, { "cell_type": "code", "execution_count": 13, "metadata": {}, "outputs": [], "source": [ "# Start with imports - ask ChatGPT to explain any package that you don't know\n", "\n", "import os\n", "import json\n", "from dotenv import load_dotenv\n", "from openai import OpenAI\n", "from anthropic import Anthropic\n", "from IPython.display import Markdown, display" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "True" ] }, "execution_count": 14, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# Always remember to do this!\n", "load_dotenv(override=True)" ] }, { "cell_type": "code", "execution_count": 15, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "OpenAI API Key exists and begins sk-proj-\n", "Anthropic API Key not set (and this is optional)\n", "Google API Key exists and begins AI\n", "DeepSeek API Key exists and begins sk-\n", "Groq API Key exists and begins gsk_\n" ] } ], "source": [ "# Print the key prefixes to help with any debugging\n", "\n", "openai_api_key = os.getenv('OPENAI_API_KEY')\n", "anthropic_api_key = os.getenv('ANTHROPIC_API_KEY')\n", "google_api_key = os.getenv('GOOGLE_API_KEY')\n", "deepseek_api_key = os.getenv('DEEPSEEK_API_KEY')\n", "groq_api_key = os.getenv('GROQ_API_KEY')\n", "\n", "if openai_api_key:\n", " print(f\"OpenAI API Key exists and begins {openai_api_key[:8]}\")\n", "else:\n", " print(\"OpenAI API Key not set\")\n", " \n", "if anthropic_api_key:\n", " print(f\"Anthropic API Key exists and begins {anthropic_api_key[:7]}\")\n", "else:\n", " print(\"Anthropic API Key not set (and this is optional)\")\n", "\n", "if google_api_key:\n", " print(f\"Google API Key exists and begins {google_api_key[:2]}\")\n", "else:\n", " print(\"Google API Key not set (and this is optional)\")\n", "\n", "if deepseek_api_key:\n", " print(f\"DeepSeek API Key exists and begins {deepseek_api_key[:3]}\")\n", "else:\n", " print(\"DeepSeek API Key not set (and this is optional)\")\n", "\n", "if groq_api_key:\n", " print(f\"Groq API Key exists and begins {groq_api_key[:4]}\")\n", "else:\n", " print(\"Groq API Key not set (and this is optional)\")" ] }, { "cell_type": "code", "execution_count": 18, "metadata": {}, "outputs": [], "source": [ "# from openai import OpenAI\n", "\n", "# # 1. Provide your Gemini API key\n", "# # 2. Update the base_url to Google's OpenAI-compatible endpoint\n", "# client = OpenAI(\n", "# api_key=google_api_key,\n", "# base_url=\"https://generativelanguage.googleapis.com/v1beta/openai/\"\n", "# )\n", "\n", "# # 3. Use a Gemini model name (e.g., gemini-2.5-flash)\n", "# response = client.chat.completions.create(\n", "# model=\"gemini-2.5-flash\",\n", "# messages=[{\"role\": \"user\", \"content\": \"Hello!\"}],\n", "# )\n", "\n", "# question = response.choices[0].message.content\n", "# print(question)" ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [], "source": [ "request = \"Please come up with a challenging, nuanced question that I can ask a number of LLMs to evaluate their intelligence. \"\n", "request += \"Answer only with the question, no explanation.\"\n", "messages = [{\"role\": \"user\", \"content\": request}]" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "messages" ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "A city's public transit system uses a dynamic pricing model where the base fare increases by 10% during \"peak hours,\" but a frequent rider pass, which costs a flat fee, provides a 20% discount on all rides. If a rider's average monthly cost without the pass is exactly the same as the cost with the pass when 60% of their rides are taken during peak hours, what percentage of their total rides must shift from peak to off-peak hours to make the pass 10% more expensive than not having it, assuming the total number of rides stays constant?\n" ] } ], "source": [ "# openai = OpenAI()\n", "deepseek = OpenAI(api_key=deepseek_api_key, base_url=\"https://api.deepseek.com\")\n", "response = deepseek.chat.completions.create(\n", " # model=\"gpt-5-mini\",\n", " model=\"deepseek-chat\",\n", " messages=messages,\n", ")\n", "question = response.choices[0].message.content\n", "print(question)\n" ] }, { "cell_type": "code", "execution_count": 22, "metadata": {}, "outputs": [], "source": [ "competitors = []\n", "answers = []\n", "messages = [{\"role\": \"user\", \"content\": question}]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Note - update since the videos\n", "\n", "I've updated the model names to use the latest models below, like GPT 5 and Claude Sonnet 4.5. It's worth noting that these models can be quite slow - like 1-2 minutes - but they do a great job! Feel free to switch them for faster models if you'd prefer, like the ones I use in the video." ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [ { "data": { "text/markdown": [ "Let P be the base off-peak fare. Peak fare = 1.10P. With the pass, rides cost 20% less, so per-ride prices are:\n", "- Off-peak with pass: 0.80P\n", "- Peak with pass: 0.80 × 1.10P = 0.88P\n", "\n", "Let N be total rides in a month. The pass price is F.\n", "\n", "Given: when 60% of rides are during peak (0.60N peak, 0.40N off-peak), the monthly cost without the pass equals the monthly cost with the pass.\n", "- Without pass: C0 = N[(0.60)(1.10P) + (0.40)(P)] = NP(0.66 + 0.40) = 1.06 N P\n", "- With pass: Cpass = F + N[(0.60)(0.88P) + (0.40)(0.80P)] = F + NP(0.528 + 0.32) = F + 0.848 N P\n", "\n", "Setting C0 = Cpass gives F = 0.212 N P.\n", "\n", "Now suppose a fraction s of total rides shift from peak to off-peak. Then:\n", "- New peak fraction p' = 0.60 − s\n", "- New off-peak fraction o' = 0.40 + s\n", "\n", "Costs become:\n", "- Without pass: C_without(s) = N P [1.10 p' + o'] = N P [1.10(0.60 − s) + (0.40 + s)] = N P [1.06 − 0.10 s]\n", "- With pass: C_with(s) = F + N P [0.88 p' + 0.80 o'] = F + N P [0.88(0.60 − s) + 0.80(0.40 + s)]\n", " = F + N P [0.848 − 0.08 s] = 0.212 N P + N P [0.848 − 0.08 s] = N P [1.060 − 0.08 s]\n", "\n", "We want C_with(s) = 1.10 × C_without(s). That is:\n", "1.060 − 0.08 s = 1.10(1.060 − 0.10 s)\n", "1.060 − 0.08 s = 1.166 − 0.11 s\n", "0.03 s = 0.106\n", "s ≈ 3.53\n", "\n", "But s must be in [0, 0.60] (you can shift at most 60% of rides). Therefore there is no feasible solution: you cannot reach a 10% higher cost with the pass by shifting rides.\n", "\n", "What is the maximum possible increase? At the extreme s = 0.60:\n", "- C_without(0.60) = N P (1.06 − 0.06) = 1.00 N P\n", "- C_with(0.60) = N P (1.06 − 0.048) = 1.012 N P\n", "\n", "Thus the pass can be at most about 1.2% more expensive than not having it (when 60% of rides shift from peak to off-peak). Conclusion: no feasible shift percentage can make the pass 10% more expensive; the maximum is ≈1.2%." ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# The API we know well\n", "# I've updated this with the latest model, but it can take some time because it likes to think!\n", "# Replace the model with gpt-4.1-mini if you'd prefer not to wait 1-2 mins\n", "\n", "model_name = \"gpt-5-nano\"\n", "openai = OpenAI()\n", "response = openai.chat.completions.create(model=model_name, messages=messages)\n", "answer = response.choices[0].message.content\n", "\n", "display(Markdown(answer))\n", "competitors.append(model_name)\n", "answers.append(answer)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# # Anthropic has a slightly different API, and Max Tokens is required\n", "\n", "# model_name = \"claude-sonnet-4-5\"\n", "\n", "# claude = Anthropic()\n", "# response = claude.messages.create(model=model_name, messages=messages, max_tokens=1000)\n", "# answer = response.content[0].text\n", "\n", "# display(Markdown(answer))\n", "# competitors.append(model_name)\n", "# answers.append(answer)" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/markdown": [ "Let F be the base fare for a single ride during off-peak hours.\n", "Let N be the total number of rides in a month.\n", "\n", "1. **Fares:**\n", " * Off-peak fare (F_o) = F\n", " * Peak hour fare (F_p) = F * (1 + 0.10) = 1.1F\n", "\n", "2. **Cost without the pass:**\n", " Let P be the percentage of rides taken during peak hours.\n", " Number of peak rides = N * P\n", " Number of off-peak rides = N * (1 - P)\n", " Cost_no_pass = (N * P * F_p) + (N * (1 - P) * F_o)\n", " Cost_no_pass = (N * P * 1.1F) + (N * (1 - P) * F)\n", " Cost_no_pass = N * F * (1.1P + 1 - P)\n", " Cost_no_pass = N * F * (0.1P + 1)\n", "\n", "3. **Cost with the frequent rider pass:**\n", " Let C be the flat fee for the pass.\n", " The pass provides a 20% discount on all rides.\n", " * Off-peak fare with pass (F_o_pass) = F * (1 - 0.20) = 0.8F\n", " * Peak hour fare with pass (F_p_pass) = 1.1F * (1 - 0.20) = 0.88F\n", " Cost_with_pass = C + (N * P * F_p_pass) + (N * (1 - P) * F_o_pass)\n", " Cost_with_pass = C + (N * P * 0.88F) + (N * (1 - P) * 0.8F)\n", " Cost_with_pass = C + N * F * (0.88P + 0.8 - 0.8P)\n", " Cost_with_pass = C + N * F * (0.08P + 0.8)\n", "\n", "4. **Determine the pass fee (C) using the initial condition:**\n", " The problem states that when 60% of rides are taken during peak hours (P = 0.6), the cost without the pass is exactly the same as the cost with the pass.\n", " Cost_no_pass (P=0.6) = Cost_with_pass (P=0.6)\n", " N * F * (0.1 * 0.6 + 1) = C + N * F * (0.08 * 0.6 + 0.8)\n", " N * F * (0.06 + 1) = C + N * F * (0.048 + 0.8)\n", " 1.06 NF = C + 0.848 NF\n", " C = (1.06 - 0.848) NF\n", " C = 0.212 NF\n", "\n", "5. **New condition for cost comparison:**\n", " We need to find a new percentage of peak rides, P_new, such that \"the pass is 10% more expensive than not having it.\" This means:\n", " Cost_with_pass_new = 1.10 * Cost_no_pass_new\n", "\n", " Substitute the expressions for costs and the value of C:\n", " C + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", " 0.212 NF + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", "\n", " Divide both sides by NF (assuming N and F are non-zero):\n", " 0.212 + 0.08 * P_new + 0.8 = 1.10 * (0.1 * P_new + 1)\n", " 1.012 + 0.08 * P_new = 0.11 * P_new + 1.10\n", "\n", " Solve for P_new:\n", " 1.012 - 1.10 = 0.11 * P_new - 0.08 * P_new\n", " -0.088 = 0.03 * P_new\n", " P_new = -0.088 / 0.03\n", " P_new = -88 / 30 = -44 / 15\n", " P_new ≈ -2.933\n", "\n", "6. **Interpret the result:**\n", " The percentage of rides taken during peak hours (P_new) must be between 0 and 1 (inclusive). A negative value for P_new is not physically possible.\n", "\n", " Let's analyze the ratio of costs as a function of P:\n", " f(P) = Cost_with_pass / Cost_no_pass = [C + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " Substitute C = 0.212 NF:\n", " f(P) = [0.212 NF + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " f(P) = (1.012 + 0.08P) / (0.1P + 1)\n", "\n", " The initial condition stated that for P = 0.6, f(0.6) = 1.\n", " A \"shift from peak to off-peak hours\" means that P_new must be less than the initial percentage of peak rides (0.6). So, P_new must be in the range [0, 0.6).\n", "\n", " Let's evaluate f(P) at the boundaries of this range:\n", " * At P = 0.6 (initial state, no shift): f(0.6) = (1.012 + 0.08 * 0.6) / (0.1 * 0.6 + 1) = (1.012 + 0.048) / (0.06 + 1) = 1.06 / 1.06 = 1.\n", " This means the costs are equal.\n", "\n", " * At P = 0 (all rides are off-peak, which is the maximum possible shift from peak to off-peak, corresponding to 60% of total rides shifting):\n", " f(0) = (1.012 + 0.08 * 0) / (0.1 * 0 + 1) = 1.012 / 1 = 1.012\n", " This means that when all rides are off-peak, the cost with the pass is 1.012 times the cost without the pass. In other words, the pass is 1.2% more expensive than not having it.\n", "\n", " The function f(P) is decreasing for P in [0, 1] (its derivative is f'(P) = -0.0212 / (0.1P+1)^2, which is always negative).\n", " Therefore, as P decreases from 0.6 to 0, f(P) increases from 1 to 1.012.\n", " The maximum value f(P) can reach when shifting rides from peak to off-peak is 1.012 (when P=0).\n", "\n", " The problem requires f(P_new) = 1.10 (i.e., the pass is 10% more expensive).\n", " Since the maximum achievable value for f(P) is 1.012, it is impossible for f(P) to reach 1.10 within the valid range of P_new.\n", "\n", "Conclusion:\n", "Based on the standard interpretation of the problem statement and the mathematical model, it is impossible to make the pass 10% more expensive than not having it by shifting rides from peak to off-peak hours, as the maximum possible cost difference achieved by shifting all peak rides to off-peak rides only makes the pass 1.2% more expensive. Therefore, there is no such percentage of rides that must shift to meet the condition.\n", "\n", "The final answer is $\\boxed{\\text{No such shift is possible}}$." ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "gemini = OpenAI(api_key=google_api_key, base_url=\"https://generativelanguage.googleapis.com/v1beta/openai/\")\n", "model_name = \"gemini-2.5-flash\"\n", "\n", "response = gemini.chat.completions.create(model=model_name, messages=messages)\n", "answer = response.choices[0].message.content\n", "\n", "display(Markdown(answer))\n", "competitors.append(model_name)\n", "answers.append(answer)" ] }, { "cell_type": "code", "execution_count": 28, "metadata": {}, "outputs": [ { "data": { "text/markdown": [ "Alright, let's break this down step-by-step.\n", "\n", "---\n", "\n", "## **Step 1: Define variables**\n", "\n", "Let:\n", "- \\( P \\) = base fare (off-peak fare)\n", "- Peak fare = \\( 1.1P \\) (10% increase)\n", "- Total number of rides per month = \\( N \\)\n", "- Fraction of rides during peak hours = \\( f \\)\n", "- Without pass: \n", " Cost = \\( N \\cdot [ f \\cdot 1.1P + (1-f) \\cdot P ] \\) \n", " = \\( N P [ 1.1 f + 1 - f ] \\) \n", " = \\( N P [ 1 + 0.1 f ] \\)\n", "\n", "- With pass: \n", " Flat fee for pass = \\( F \\) (monthly cost of pass) \n", " Then each ride costs 20% less than the **current** fare (peak or off-peak). \n", " So off-peak fare with pass = \\( 0.8 P \\) \n", " Peak fare with pass = \\( 0.8 \\times 1.1 P = 0.88 P \\) \n", "\n", " Total cost with pass = \\( F + N [ f \\cdot 0.88P + (1-f) \\cdot 0.8P ] \\) \n", " = \\( F + N P [ 0.88 f + 0.8 - 0.8 f ] \\) \n", " = \\( F + N P [ 0.8 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 2: Use given condition to relate \\( F \\) and \\( P \\)**\n", "\n", "Given: Without pass, average monthly cost = cost with pass when \\( f = 0.6 \\).\n", "\n", "Without pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{no pass}} = N P [ 1 + 0.1 \\times 0.6 ] = N P [ 1.06 ] \\)\n", "\n", "With pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}} = F + N P [ 0.8 + 0.08 \\times 0.6 ] \\) \n", "= \\( F + N P [ 0.8 + 0.048 ] \\) \n", "= \\( F + N P [ 0.848 ] \\)\n", "\n", "Set them equal: \n", "\\( N P \\cdot 1.06 = F + N P \\cdot 0.848 \\) \n", "\\( F = N P ( 1.06 - 0.848 ) \\) \n", "\\( F = N P \\cdot 0.212 \\)\n", "\n", "So the pass costs \\( 0.212 N P \\) per month.\n", "\n", "---\n", "\n", "## **Step 3: General cost expressions**\n", "\n", "Without pass: \n", "\\( C_{\\text{no pass}}(f) = N P ( 1 + 0.1 f ) \\)\n", "\n", "With pass: \n", "\\( C_{\\text{pass}}(f) = 0.212 N P + N P ( 0.8 + 0.08 f ) \\) \n", "= \\( N P [ 0.212 + 0.8 + 0.08 f ] \\) \n", "= \\( N P [ 1.012 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 4: Find \\( f \\) such that pass is 10% more expensive than no pass**\n", "\n", "We want: \n", "\\( C_{\\text{pass}}(f) = 1.1 \\times C_{\\text{no pass}}(f) \\)\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f ] = 1.1 \\times N P [ 1 + 0.1 f ] \\) \n", "Cancel \\( N P \\): \n", "\\( 1.012 + 0.08 f = 1.1 + 0.11 f \\) \n", "\\( 1.012 - 1.1 = 0.11 f - 0.08 f \\) \n", "\\( -0.088 = 0.03 f \\) \n", "\\( f = -\\frac{0.088}{0.03} \\approx -2.933 \\)\n", "\n", "This is negative, which is impossible (fraction of peak rides can't be negative). \n", "This means: For pass to be **more expensive** than no pass by 10%, we would need \\( f \\) negative if we keep total rides constant — but maybe they mean **pass 10% more expensive** than no pass **at original \\( f = 0.6 \\)**? Wait, no — they say: \"what percentage of their total rides must shift from peak to off-peak hours to make the pass 10% more expensive than not having it\".\n", "\n", "That means: Start from \\( f = 0.6 \\), shift some rides from peak to off-peak, get new \\( f' \\), such that pass cost is 10% higher than no-pass cost **at new \\( f' \\)**.\n", "\n", "But we already computed the equation for general \\( f \\) for that condition: \n", "\\( 1.012 + 0.08 f = 1.1 (1 + 0.1 f) \\) \n", "We solved: \\( f \\approx -2.933 \\), impossible.\n", "\n", "So maybe they mean: pass is 10% more expensive **than the no-pass cost at the original \\( f = 0.6 \\)**? Let's check.\n", "\n", "---\n", "\n", "## **Step 5: Alternative interpretation**\n", "\n", "Original no-pass cost at \\( f = 0.6 \\): \\( C_0 = N P \\cdot 1.06 \\). \n", "We want \\( C_{\\text{pass}}(f') = 1.1 \\times C_0 \\).\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f' ] = 1.1 \\times N P \\times 1.06 \\) \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — also impossible (>1).\n", "\n", "So that’s not it.\n", "\n", "---\n", "\n", "## **Step 6: Maybe they mean: pass is 10% more expensive than no-pass cost at the NEW \\( f' \\), but starting from \\( f = 0.6 \\) where they were equal.**\n", "\n", "We already tried that in Step 4, got impossible \\( f' \\). \n", "So maybe they mean: pass is 10% more expensive **than the pass cost at \\( f = 0.6 \\)**? That is, increase in cost of pass relative to its original cost.\n", "\n", "---\n", "\n", "Original pass cost at \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}}(0.6) = N P [ 1.012 + 0.08 \\times 0.6 ] \\) \n", "= \\( N P [ 1.012 + 0.048 ] = N P \\times 1.06 \\) (same as no-pass originally, by design).\n", "\n", "We want new \\( f' \\) such that: \n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "= \\( 1.1 \\times 1.06 N P = 1.166 N P \\).\n", "\n", "So: \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — still impossible (>1).\n", "\n", "---\n", "\n", "## **Step 7: Check if maybe they mean: no-pass cost is 10% less than pass cost? That’s same as pass 10% more than no-pass, which we did.**\n", "\n", "Given the impossibility, maybe they mean: **make no-pass 10% cheaper than pass** — but that’s the same ratio.\n", "\n", "Let’s check if maybe the 10% is on base fare difference? Or maybe they mean: **pass becomes 10% more expensive than no-pass for the new pattern** — but we saw that requires \\( f \\) negative, meaning we must shift **so many** rides from peak to off-peak that \\( f \\) drops below zero, so minimum \\( f \\) is 0, let's see cost difference at \\( f = 0 \\):\n", "\n", "At \\( f = 0 \\): \n", "No-pass cost = \\( N P (1 + 0) = N P \\) \n", "Pass cost = \\( N P (1.012 + 0) = 1.012 N P \\) \n", "Ratio = \\( 1.012 / 1 = 1.012 \\) → pass is only 1.2% more expensive.\n", "\n", "So even shifting all peak rides to off-peak, pass is never 10% more expensive than no-pass — it’s always less than or equal to 1.2% more.\n", "\n", "Thus, **it’s impossible** for pass to be 10% more expensive than no-pass for any \\( f \\ge 0 \\) given the formulas.\n", "\n", "---\n", "\n", "## **Step 8: But maybe they mean: pass 10% more expensive than no-pass at the NEW \\( f' \\), but starting from equality at \\( f=0.6 \\), shifting some rides changes both costs.**\n", "\n", "Let’s check difference: \n", "\\( D(f) = C_{\\text{pass}}(f) - C_{\\text{no pass}}(f) \\) \n", "= \\( N P [ 1.012 + 0.08 f - (1 + 0.1 f) ] \\) \n", "= \\( N P [ 0.012 - 0.02 f ] \\)\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(f') \\) for \"pass 10% more expensive than no-pass\":\n", "\n", "\\( 0.012 - 0.02 f' = 0.1 (1 + 0.1 f') \\) \n", "\\( 0.012 - 0.02 f' = 0.1 + 0.01 f' \\) \n", "\\( 0.012 - 0.1 = 0.01 f' + 0.02 f' \\) \n", "\\( -0.088 = 0.03 f' \\) \n", "\\( f' \\approx -2.933 \\) — same impossible.\n", "\n", "So indeed, to get pass 10% more expensive than no-pass, \\( f' \\) must be negative. \n", "The only way to approach that from \\( f=0.6 \\) is to shift **all** peak rides to off-peak and more — impossible since only 60% are peak initially.\n", "\n", "---\n", "\n", "## **Step 9: Conclusion from math**\n", "\n", "The condition “pass 10% more expensive than no-pass” is impossible with given pricing formulas for \\( f \\ge 0 \\). \n", "But if we interpret the question as: shift enough rides so that **the extra cost of the pass relative to no-pass** is 10% **of the original no-pass cost at \\( f=0.6 \\)**? That is:\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(0.6) \\) \n", "= \\( 0.1 \\times 1.06 N P = 0.106 N P \\).\n", "\n", "So: \n", "\\( N P (0.012 - 0.02 f') = 0.106 N P \\) \n", "\\( 0.012 - 0.02 f' = 0.106 \\) \n", "\\( -0.02 f' = 0.094 \\) \n", "\\( f' = -4.7 \\) — also impossible.\n", "\n", "---\n", "\n", "Given the problem constraints, maybe they meant: make pass **10% more expensive than it was originally**? That would be:\n", "\n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "We computed earlier: \\( 1.012 + 0.08 f' = 1.166 \\) → \\( f' = 1.925 \\) impossible.\n", "\n", "So all interpretations give impossible \\( f' \\) except one: maybe they mean **no-pass cost 10% less than pass cost** — which is same equation as before.\n", "\n", "Given the likely intent: They want us to find \\( f' \\) from the equation \\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{no pass}}(f') \\), solve, get \\( f' \\) negative, so minimal \\( f' \\) is 0. \n", "That means shift **all** 60% peak rides to off-peak. \n", "So percentage of total rides shifted = \\( 60\\% \\).\n", "\n", "---\n", "\n", "**Final answer** (assuming they accept the shift to \\( f' = 0 \\) as the point where pass is most expensive relative to no-pass, though not quite 10% more):\n", "\n", "\\[\n", "\\boxed{60}\n", "\\]\n", "\n", "That is, shift all 60% of rides that were peak to off-peak." ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "deepseek = OpenAI(api_key=deepseek_api_key, base_url=\"https://api.deepseek.com/v1\")\n", "model_name = \"deepseek-chat\"\n", "\n", "response = deepseek.chat.completions.create(model=model_name, messages=messages)\n", "answer = response.choices[0].message.content\n", "\n", "display(Markdown(answer))\n", "competitors.append(model_name)\n", "answers.append(answer)" ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/markdown": [ "**Step 1 – Set up the variables**\n", "\n", "| Symbol | Meaning |\n", "|--------|----------|\n", "| \\(F\\) | Base fare for an off‑peak ride (dollar amount) |\n", "| \\(N\\) | Total number of rides the rider takes in a month (kept constant) |\n", "| \\(p\\) | Fraction of rides that occur during peak hours (so off‑peak fraction = \\(1-p\\)) |\n", "| \\(P\\) | Monthly price of the frequent‑rider pass (a flat fee) |\n", "\n", "- Off‑peak fare (no pass) = \\(F\\) \n", "- Peak fare (no pass) = \\(1.10F\\) (10 % increase) \n", "- Discount with the pass = 20 % → you pay 80 % of the listed fare.\n", "\n", "Hence \n", "\n", "- Off‑peak ride cost **with** the pass = \\(0.8F\\) \n", "- Peak ride cost **with** the pass = \\(0.8\\times1.10F = 0.88F\\).\n", "\n", "---\n", "\n", "**Step 2 – Find the pass fee \\(P\\) from the “break‑even” condition**\n", "\n", "The problem tells us that when 60 % of rides are taken during peak hours the rider spends exactly the same amount whether they buy the pass or not.\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost without pass} &= N\\Big[(0.4)F + (0.6)(1.10F)\\Big] \\\\\n", " &= NF\\big[0.4 + 0.66\\big] = 1.06\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost with pass} &= P + N\\Big[(0.4)(0.8F) + (0.6)(0.88F)\\Big] \\\\\n", " &= P + NF\\big[0.32 + 0.528\\big] = P + 0.848\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "Setting the two equal:\n", "\n", "\\[\n", "1.06\\,NF = P + 0.848\\,NF \\quad\\Longrightarrow\\quad\n", "P = (1.06-0.848)NF = 0.212\\,NF .\n", "\\]\n", "\n", "So the pass costs **21.2 % of the rider’s original monthly expense**.\n", "\n", "---\n", "\n", "**Step 3 – Express the two monthly costs after a shift in the peak‑off‑peak mix**\n", "\n", "If the new peak‑hour proportion is \\(p\\) (the off‑peak proportion is \\(1-p\\)):\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{No‑pass cost} &: \\; C_{\\text{no}}(p)=NF\\Big[(1-p) + 1.10p\\Big] = NF\\big[1+0.10p\\big] ,\\\\[4pt]\n", "\\text{Pass cost} &: \\; C_{\\text{yes}}(p)=P + NF\\Big[0.8(1-p)+0.88p\\Big] \\\\\n", " &: \\; =0.212\\,NF + NF\\big[0.8 + 0.08p\\big] = NF\\big[1.012 + 0.08p\\big] .\n", "\\end{aligned}\n", "\\]\n", "\n", "---\n", "\n", "**Step 4 – Impose the “10 % more expensive” requirement**\n", "\n", "We want the pass to cost **10 % more** than the no‑pass option:\n", "\n", "\\[\n", "C_{\\text{yes}}(p) = 1.10\\,C_{\\text{no}}(p).\n", "\\]\n", "\n", "Plugging the expressions:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10\\big(1 + 0.10p\\big).\n", "\\]\n", "\n", "Simplify:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10 + 0.11p \\\\\n", "\\Longrightarrow\\; 0.08p - 0.11p = 1.10 - 1.012 \\\\\n", "\\Longrightarrow\\; -0.03p = 0.088 \\\\\n", "\\Longrightarrow\\; p = -\\frac{0.088}{0.03} \\approx -2.93 .\n", "\\]\n", "\n", "---\n", "\n", "**Step 5 – Interpret the result**\n", "\n", "A peak‑hour proportion of \\(p\\approx -2.93\\) is impossible (the fraction of rides that are “peak” cannot be negative, nor can it be less than zero). \n", "\n", "What does this tell us?\n", "\n", "* Even if **all** rides are shifted to off‑peak (\\(p = 0\\)), the pass is only\n", "\n", "\\[\n", "\\frac{C_{\\text{yes}}(0)}{C_{\\text{no}}(0)} =\n", "\\frac{1.012}{1.00} = 1.012\n", "\\]\n", "\n", "i.e. **1.2 % more expensive** than going without the pass.\n", "\n", "* The maximum possible increase in cost caused by moving rides away from peak hours is therefore 1.2 %. \n", "\n", "Reaching a 10 % increase would require a negative peak‑hour proportion (or, equivalently, “more off‑peak rides than the total number of rides”), which cannot happen.\n", "\n", "---\n", "\n", "## **Answer**\n", "\n", "It is **not possible**. \n", "Even if the rider moved **all** of his/her rides from peak to off‑peak (a 100 % shift), the monthly cost with the pass would be only about **1.2 % higher** than the cost without the pass. To make the pass 10 % more expensive would require a peak‑hour proportion of \\(p \\approx -2.9\\) (i.e., a negative number of peak rides), which is mathematically impossible. Consequently, no feasible percentage of rides shifted from peak to off‑peak can produce a 10 % cost disadvantage for the pass." ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# Updated with the latest Open Source model from OpenAI\n", "\n", "groq = OpenAI(api_key=groq_api_key, base_url=\"https://api.groq.com/openai/v1\")\n", "model_name = \"openai/gpt-oss-120b\"\n", "# model_name = \"gpt-5-mini\"\n", "\n", "response = groq.chat.completions.create(model=model_name, messages=messages)\n", "answer = response.choices[0].message.content\n", "\n", "display(Markdown(answer))\n", "competitors.append(model_name)\n", "answers.append(answer)\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## For the next cell, we will use Ollama\n", "\n", "Ollama runs a local web service that gives an OpenAI compatible endpoint, \n", "and runs models locally using high performance C++ code.\n", "\n", "If you don't have Ollama, install it here by visiting https://ollama.com then pressing Download and following the instructions.\n", "\n", "After it's installed, you should be able to visit here: http://localhost:11434 and see the message \"Ollama is running\"\n", "\n", "You might need to restart Cursor (and maybe reboot). Then open a Terminal (control+\\`) and run `ollama serve`\n", "\n", "Useful Ollama commands (run these in the terminal, or with an exclamation mark in this notebook):\n", "\n", "`ollama pull ` downloads a model locally \n", "`ollama ls` lists all the models you've downloaded \n", "`ollama rm ` deletes the specified model from your downloads" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", " \n", " \n", " \n", " \n", "
\n", " \n", " \n", "

Super important - ignore me at your peril!

\n", " The model called llama3.3 is FAR too large for home computers - it's not intended for personal computing and will consume all your resources! Stick with the nicely sized llama3.2 or llama3.2:1b and if you want larger, try llama3.1 or smaller variants of Qwen, Gemma, Phi or DeepSeek. See the the Ollama models page for a full list of models and sizes.\n", " \n", "
" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "!ollama pull llama3.2" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "ollama = OpenAI(base_url='http://localhost:11434/v1', api_key='ollama')\n", "model_name = \"llama3.2\"\n", "\n", "response = ollama.chat.completions.create(model=model_name, messages=messages)\n", "answer = response.choices[0].message.content\n", "\n", "display(Markdown(answer))\n", "competitors.append(model_name)\n", "answers.append(answer)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "# So where are we?\n", "\n", "print(competitors)\n", "print(answers)\n" ] }, { "cell_type": "code", "execution_count": 30, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Competitor: gpt-5-nano\n", "\n", "Let P be the base off-peak fare. Peak fare = 1.10P. With the pass, rides cost 20% less, so per-ride prices are:\n", "- Off-peak with pass: 0.80P\n", "- Peak with pass: 0.80 × 1.10P = 0.88P\n", "\n", "Let N be total rides in a month. The pass price is F.\n", "\n", "Given: when 60% of rides are during peak (0.60N peak, 0.40N off-peak), the monthly cost without the pass equals the monthly cost with the pass.\n", "- Without pass: C0 = N[(0.60)(1.10P) + (0.40)(P)] = NP(0.66 + 0.40) = 1.06 N P\n", "- With pass: Cpass = F + N[(0.60)(0.88P) + (0.40)(0.80P)] = F + NP(0.528 + 0.32) = F + 0.848 N P\n", "\n", "Setting C0 = Cpass gives F = 0.212 N P.\n", "\n", "Now suppose a fraction s of total rides shift from peak to off-peak. Then:\n", "- New peak fraction p' = 0.60 − s\n", "- New off-peak fraction o' = 0.40 + s\n", "\n", "Costs become:\n", "- Without pass: C_without(s) = N P [1.10 p' + o'] = N P [1.10(0.60 − s) + (0.40 + s)] = N P [1.06 − 0.10 s]\n", "- With pass: C_with(s) = F + N P [0.88 p' + 0.80 o'] = F + N P [0.88(0.60 − s) + 0.80(0.40 + s)]\n", " = F + N P [0.848 − 0.08 s] = 0.212 N P + N P [0.848 − 0.08 s] = N P [1.060 − 0.08 s]\n", "\n", "We want C_with(s) = 1.10 × C_without(s). That is:\n", "1.060 − 0.08 s = 1.10(1.060 − 0.10 s)\n", "1.060 − 0.08 s = 1.166 − 0.11 s\n", "0.03 s = 0.106\n", "s ≈ 3.53\n", "\n", "But s must be in [0, 0.60] (you can shift at most 60% of rides). Therefore there is no feasible solution: you cannot reach a 10% higher cost with the pass by shifting rides.\n", "\n", "What is the maximum possible increase? At the extreme s = 0.60:\n", "- C_without(0.60) = N P (1.06 − 0.06) = 1.00 N P\n", "- C_with(0.60) = N P (1.06 − 0.048) = 1.012 N P\n", "\n", "Thus the pass can be at most about 1.2% more expensive than not having it (when 60% of rides shift from peak to off-peak). Conclusion: no feasible shift percentage can make the pass 10% more expensive; the maximum is ≈1.2%.\n", "Competitor: gemini-2.5-flash\n", "\n", "Let F be the base fare for a single ride during off-peak hours.\n", "Let N be the total number of rides in a month.\n", "\n", "1. **Fares:**\n", " * Off-peak fare (F_o) = F\n", " * Peak hour fare (F_p) = F * (1 + 0.10) = 1.1F\n", "\n", "2. **Cost without the pass:**\n", " Let P be the percentage of rides taken during peak hours.\n", " Number of peak rides = N * P\n", " Number of off-peak rides = N * (1 - P)\n", " Cost_no_pass = (N * P * F_p) + (N * (1 - P) * F_o)\n", " Cost_no_pass = (N * P * 1.1F) + (N * (1 - P) * F)\n", " Cost_no_pass = N * F * (1.1P + 1 - P)\n", " Cost_no_pass = N * F * (0.1P + 1)\n", "\n", "3. **Cost with the frequent rider pass:**\n", " Let C be the flat fee for the pass.\n", " The pass provides a 20% discount on all rides.\n", " * Off-peak fare with pass (F_o_pass) = F * (1 - 0.20) = 0.8F\n", " * Peak hour fare with pass (F_p_pass) = 1.1F * (1 - 0.20) = 0.88F\n", " Cost_with_pass = C + (N * P * F_p_pass) + (N * (1 - P) * F_o_pass)\n", " Cost_with_pass = C + (N * P * 0.88F) + (N * (1 - P) * 0.8F)\n", " Cost_with_pass = C + N * F * (0.88P + 0.8 - 0.8P)\n", " Cost_with_pass = C + N * F * (0.08P + 0.8)\n", "\n", "4. **Determine the pass fee (C) using the initial condition:**\n", " The problem states that when 60% of rides are taken during peak hours (P = 0.6), the cost without the pass is exactly the same as the cost with the pass.\n", " Cost_no_pass (P=0.6) = Cost_with_pass (P=0.6)\n", " N * F * (0.1 * 0.6 + 1) = C + N * F * (0.08 * 0.6 + 0.8)\n", " N * F * (0.06 + 1) = C + N * F * (0.048 + 0.8)\n", " 1.06 NF = C + 0.848 NF\n", " C = (1.06 - 0.848) NF\n", " C = 0.212 NF\n", "\n", "5. **New condition for cost comparison:**\n", " We need to find a new percentage of peak rides, P_new, such that \"the pass is 10% more expensive than not having it.\" This means:\n", " Cost_with_pass_new = 1.10 * Cost_no_pass_new\n", "\n", " Substitute the expressions for costs and the value of C:\n", " C + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", " 0.212 NF + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", "\n", " Divide both sides by NF (assuming N and F are non-zero):\n", " 0.212 + 0.08 * P_new + 0.8 = 1.10 * (0.1 * P_new + 1)\n", " 1.012 + 0.08 * P_new = 0.11 * P_new + 1.10\n", "\n", " Solve for P_new:\n", " 1.012 - 1.10 = 0.11 * P_new - 0.08 * P_new\n", " -0.088 = 0.03 * P_new\n", " P_new = -0.088 / 0.03\n", " P_new = -88 / 30 = -44 / 15\n", " P_new ≈ -2.933\n", "\n", "6. **Interpret the result:**\n", " The percentage of rides taken during peak hours (P_new) must be between 0 and 1 (inclusive). A negative value for P_new is not physically possible.\n", "\n", " Let's analyze the ratio of costs as a function of P:\n", " f(P) = Cost_with_pass / Cost_no_pass = [C + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " Substitute C = 0.212 NF:\n", " f(P) = [0.212 NF + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " f(P) = (1.012 + 0.08P) / (0.1P + 1)\n", "\n", " The initial condition stated that for P = 0.6, f(0.6) = 1.\n", " A \"shift from peak to off-peak hours\" means that P_new must be less than the initial percentage of peak rides (0.6). So, P_new must be in the range [0, 0.6).\n", "\n", " Let's evaluate f(P) at the boundaries of this range:\n", " * At P = 0.6 (initial state, no shift): f(0.6) = (1.012 + 0.08 * 0.6) / (0.1 * 0.6 + 1) = (1.012 + 0.048) / (0.06 + 1) = 1.06 / 1.06 = 1.\n", " This means the costs are equal.\n", "\n", " * At P = 0 (all rides are off-peak, which is the maximum possible shift from peak to off-peak, corresponding to 60% of total rides shifting):\n", " f(0) = (1.012 + 0.08 * 0) / (0.1 * 0 + 1) = 1.012 / 1 = 1.012\n", " This means that when all rides are off-peak, the cost with the pass is 1.012 times the cost without the pass. In other words, the pass is 1.2% more expensive than not having it.\n", "\n", " The function f(P) is decreasing for P in [0, 1] (its derivative is f'(P) = -0.0212 / (0.1P+1)^2, which is always negative).\n", " Therefore, as P decreases from 0.6 to 0, f(P) increases from 1 to 1.012.\n", " The maximum value f(P) can reach when shifting rides from peak to off-peak is 1.012 (when P=0).\n", "\n", " The problem requires f(P_new) = 1.10 (i.e., the pass is 10% more expensive).\n", " Since the maximum achievable value for f(P) is 1.012, it is impossible for f(P) to reach 1.10 within the valid range of P_new.\n", "\n", "Conclusion:\n", "Based on the standard interpretation of the problem statement and the mathematical model, it is impossible to make the pass 10% more expensive than not having it by shifting rides from peak to off-peak hours, as the maximum possible cost difference achieved by shifting all peak rides to off-peak rides only makes the pass 1.2% more expensive. Therefore, there is no such percentage of rides that must shift to meet the condition.\n", "\n", "The final answer is $\\boxed{\\text{No such shift is possible}}$.\n", "Competitor: deepseek-chat\n", "\n", "Alright, let's break this down step-by-step.\n", "\n", "---\n", "\n", "## **Step 1: Define variables**\n", "\n", "Let:\n", "- \\( P \\) = base fare (off-peak fare)\n", "- Peak fare = \\( 1.1P \\) (10% increase)\n", "- Total number of rides per month = \\( N \\)\n", "- Fraction of rides during peak hours = \\( f \\)\n", "- Without pass: \n", " Cost = \\( N \\cdot [ f \\cdot 1.1P + (1-f) \\cdot P ] \\) \n", " = \\( N P [ 1.1 f + 1 - f ] \\) \n", " = \\( N P [ 1 + 0.1 f ] \\)\n", "\n", "- With pass: \n", " Flat fee for pass = \\( F \\) (monthly cost of pass) \n", " Then each ride costs 20% less than the **current** fare (peak or off-peak). \n", " So off-peak fare with pass = \\( 0.8 P \\) \n", " Peak fare with pass = \\( 0.8 \\times 1.1 P = 0.88 P \\) \n", "\n", " Total cost with pass = \\( F + N [ f \\cdot 0.88P + (1-f) \\cdot 0.8P ] \\) \n", " = \\( F + N P [ 0.88 f + 0.8 - 0.8 f ] \\) \n", " = \\( F + N P [ 0.8 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 2: Use given condition to relate \\( F \\) and \\( P \\)**\n", "\n", "Given: Without pass, average monthly cost = cost with pass when \\( f = 0.6 \\).\n", "\n", "Without pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{no pass}} = N P [ 1 + 0.1 \\times 0.6 ] = N P [ 1.06 ] \\)\n", "\n", "With pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}} = F + N P [ 0.8 + 0.08 \\times 0.6 ] \\) \n", "= \\( F + N P [ 0.8 + 0.048 ] \\) \n", "= \\( F + N P [ 0.848 ] \\)\n", "\n", "Set them equal: \n", "\\( N P \\cdot 1.06 = F + N P \\cdot 0.848 \\) \n", "\\( F = N P ( 1.06 - 0.848 ) \\) \n", "\\( F = N P \\cdot 0.212 \\)\n", "\n", "So the pass costs \\( 0.212 N P \\) per month.\n", "\n", "---\n", "\n", "## **Step 3: General cost expressions**\n", "\n", "Without pass: \n", "\\( C_{\\text{no pass}}(f) = N P ( 1 + 0.1 f ) \\)\n", "\n", "With pass: \n", "\\( C_{\\text{pass}}(f) = 0.212 N P + N P ( 0.8 + 0.08 f ) \\) \n", "= \\( N P [ 0.212 + 0.8 + 0.08 f ] \\) \n", "= \\( N P [ 1.012 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 4: Find \\( f \\) such that pass is 10% more expensive than no pass**\n", "\n", "We want: \n", "\\( C_{\\text{pass}}(f) = 1.1 \\times C_{\\text{no pass}}(f) \\)\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f ] = 1.1 \\times N P [ 1 + 0.1 f ] \\) \n", "Cancel \\( N P \\): \n", "\\( 1.012 + 0.08 f = 1.1 + 0.11 f \\) \n", "\\( 1.012 - 1.1 = 0.11 f - 0.08 f \\) \n", "\\( -0.088 = 0.03 f \\) \n", "\\( f = -\\frac{0.088}{0.03} \\approx -2.933 \\)\n", "\n", "This is negative, which is impossible (fraction of peak rides can't be negative). \n", "This means: For pass to be **more expensive** than no pass by 10%, we would need \\( f \\) negative if we keep total rides constant — but maybe they mean **pass 10% more expensive** than no pass **at original \\( f = 0.6 \\)**? Wait, no — they say: \"what percentage of their total rides must shift from peak to off-peak hours to make the pass 10% more expensive than not having it\".\n", "\n", "That means: Start from \\( f = 0.6 \\), shift some rides from peak to off-peak, get new \\( f' \\), such that pass cost is 10% higher than no-pass cost **at new \\( f' \\)**.\n", "\n", "But we already computed the equation for general \\( f \\) for that condition: \n", "\\( 1.012 + 0.08 f = 1.1 (1 + 0.1 f) \\) \n", "We solved: \\( f \\approx -2.933 \\), impossible.\n", "\n", "So maybe they mean: pass is 10% more expensive **than the no-pass cost at the original \\( f = 0.6 \\)**? Let's check.\n", "\n", "---\n", "\n", "## **Step 5: Alternative interpretation**\n", "\n", "Original no-pass cost at \\( f = 0.6 \\): \\( C_0 = N P \\cdot 1.06 \\). \n", "We want \\( C_{\\text{pass}}(f') = 1.1 \\times C_0 \\).\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f' ] = 1.1 \\times N P \\times 1.06 \\) \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — also impossible (>1).\n", "\n", "So that’s not it.\n", "\n", "---\n", "\n", "## **Step 6: Maybe they mean: pass is 10% more expensive than no-pass cost at the NEW \\( f' \\), but starting from \\( f = 0.6 \\) where they were equal.**\n", "\n", "We already tried that in Step 4, got impossible \\( f' \\). \n", "So maybe they mean: pass is 10% more expensive **than the pass cost at \\( f = 0.6 \\)**? That is, increase in cost of pass relative to its original cost.\n", "\n", "---\n", "\n", "Original pass cost at \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}}(0.6) = N P [ 1.012 + 0.08 \\times 0.6 ] \\) \n", "= \\( N P [ 1.012 + 0.048 ] = N P \\times 1.06 \\) (same as no-pass originally, by design).\n", "\n", "We want new \\( f' \\) such that: \n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "= \\( 1.1 \\times 1.06 N P = 1.166 N P \\).\n", "\n", "So: \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — still impossible (>1).\n", "\n", "---\n", "\n", "## **Step 7: Check if maybe they mean: no-pass cost is 10% less than pass cost? That’s same as pass 10% more than no-pass, which we did.**\n", "\n", "Given the impossibility, maybe they mean: **make no-pass 10% cheaper than pass** — but that’s the same ratio.\n", "\n", "Let’s check if maybe the 10% is on base fare difference? Or maybe they mean: **pass becomes 10% more expensive than no-pass for the new pattern** — but we saw that requires \\( f \\) negative, meaning we must shift **so many** rides from peak to off-peak that \\( f \\) drops below zero, so minimum \\( f \\) is 0, let's see cost difference at \\( f = 0 \\):\n", "\n", "At \\( f = 0 \\): \n", "No-pass cost = \\( N P (1 + 0) = N P \\) \n", "Pass cost = \\( N P (1.012 + 0) = 1.012 N P \\) \n", "Ratio = \\( 1.012 / 1 = 1.012 \\) → pass is only 1.2% more expensive.\n", "\n", "So even shifting all peak rides to off-peak, pass is never 10% more expensive than no-pass — it’s always less than or equal to 1.2% more.\n", "\n", "Thus, **it’s impossible** for pass to be 10% more expensive than no-pass for any \\( f \\ge 0 \\) given the formulas.\n", "\n", "---\n", "\n", "## **Step 8: But maybe they mean: pass 10% more expensive than no-pass at the NEW \\( f' \\), but starting from equality at \\( f=0.6 \\), shifting some rides changes both costs.**\n", "\n", "Let’s check difference: \n", "\\( D(f) = C_{\\text{pass}}(f) - C_{\\text{no pass}}(f) \\) \n", "= \\( N P [ 1.012 + 0.08 f - (1 + 0.1 f) ] \\) \n", "= \\( N P [ 0.012 - 0.02 f ] \\)\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(f') \\) for \"pass 10% more expensive than no-pass\":\n", "\n", "\\( 0.012 - 0.02 f' = 0.1 (1 + 0.1 f') \\) \n", "\\( 0.012 - 0.02 f' = 0.1 + 0.01 f' \\) \n", "\\( 0.012 - 0.1 = 0.01 f' + 0.02 f' \\) \n", "\\( -0.088 = 0.03 f' \\) \n", "\\( f' \\approx -2.933 \\) — same impossible.\n", "\n", "So indeed, to get pass 10% more expensive than no-pass, \\( f' \\) must be negative. \n", "The only way to approach that from \\( f=0.6 \\) is to shift **all** peak rides to off-peak and more — impossible since only 60% are peak initially.\n", "\n", "---\n", "\n", "## **Step 9: Conclusion from math**\n", "\n", "The condition “pass 10% more expensive than no-pass” is impossible with given pricing formulas for \\( f \\ge 0 \\). \n", "But if we interpret the question as: shift enough rides so that **the extra cost of the pass relative to no-pass** is 10% **of the original no-pass cost at \\( f=0.6 \\)**? That is:\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(0.6) \\) \n", "= \\( 0.1 \\times 1.06 N P = 0.106 N P \\).\n", "\n", "So: \n", "\\( N P (0.012 - 0.02 f') = 0.106 N P \\) \n", "\\( 0.012 - 0.02 f' = 0.106 \\) \n", "\\( -0.02 f' = 0.094 \\) \n", "\\( f' = -4.7 \\) — also impossible.\n", "\n", "---\n", "\n", "Given the problem constraints, maybe they meant: make pass **10% more expensive than it was originally**? That would be:\n", "\n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "We computed earlier: \\( 1.012 + 0.08 f' = 1.166 \\) → \\( f' = 1.925 \\) impossible.\n", "\n", "So all interpretations give impossible \\( f' \\) except one: maybe they mean **no-pass cost 10% less than pass cost** — which is same equation as before.\n", "\n", "Given the likely intent: They want us to find \\( f' \\) from the equation \\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{no pass}}(f') \\), solve, get \\( f' \\) negative, so minimal \\( f' \\) is 0. \n", "That means shift **all** 60% peak rides to off-peak. \n", "So percentage of total rides shifted = \\( 60\\% \\).\n", "\n", "---\n", "\n", "**Final answer** (assuming they accept the shift to \\( f' = 0 \\) as the point where pass is most expensive relative to no-pass, though not quite 10% more):\n", "\n", "\\[\n", "\\boxed{60}\n", "\\]\n", "\n", "That is, shift all 60% of rides that were peak to off-peak.\n", "Competitor: openai/gpt-oss-120b\n", "\n", "**Step 1 – Set up the variables**\n", "\n", "| Symbol | Meaning |\n", "|--------|----------|\n", "| \\(F\\) | Base fare for an off‑peak ride (dollar amount) |\n", "| \\(N\\) | Total number of rides the rider takes in a month (kept constant) |\n", "| \\(p\\) | Fraction of rides that occur during peak hours (so off‑peak fraction = \\(1-p\\)) |\n", "| \\(P\\) | Monthly price of the frequent‑rider pass (a flat fee) |\n", "\n", "- Off‑peak fare (no pass) = \\(F\\) \n", "- Peak fare (no pass) = \\(1.10F\\) (10 % increase) \n", "- Discount with the pass = 20 % → you pay 80 % of the listed fare.\n", "\n", "Hence \n", "\n", "- Off‑peak ride cost **with** the pass = \\(0.8F\\) \n", "- Peak ride cost **with** the pass = \\(0.8\\times1.10F = 0.88F\\).\n", "\n", "---\n", "\n", "**Step 2 – Find the pass fee \\(P\\) from the “break‑even” condition**\n", "\n", "The problem tells us that when 60 % of rides are taken during peak hours the rider spends exactly the same amount whether they buy the pass or not.\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost without pass} &= N\\Big[(0.4)F + (0.6)(1.10F)\\Big] \\\\\n", " &= NF\\big[0.4 + 0.66\\big] = 1.06\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost with pass} &= P + N\\Big[(0.4)(0.8F) + (0.6)(0.88F)\\Big] \\\\\n", " &= P + NF\\big[0.32 + 0.528\\big] = P + 0.848\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "Setting the two equal:\n", "\n", "\\[\n", "1.06\\,NF = P + 0.848\\,NF \\quad\\Longrightarrow\\quad\n", "P = (1.06-0.848)NF = 0.212\\,NF .\n", "\\]\n", "\n", "So the pass costs **21.2 % of the rider’s original monthly expense**.\n", "\n", "---\n", "\n", "**Step 3 – Express the two monthly costs after a shift in the peak‑off‑peak mix**\n", "\n", "If the new peak‑hour proportion is \\(p\\) (the off‑peak proportion is \\(1-p\\)):\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{No‑pass cost} &: \\; C_{\\text{no}}(p)=NF\\Big[(1-p) + 1.10p\\Big] = NF\\big[1+0.10p\\big] ,\\\\[4pt]\n", "\\text{Pass cost} &: \\; C_{\\text{yes}}(p)=P + NF\\Big[0.8(1-p)+0.88p\\Big] \\\\\n", " &: \\; =0.212\\,NF + NF\\big[0.8 + 0.08p\\big] = NF\\big[1.012 + 0.08p\\big] .\n", "\\end{aligned}\n", "\\]\n", "\n", "---\n", "\n", "**Step 4 – Impose the “10 % more expensive” requirement**\n", "\n", "We want the pass to cost **10 % more** than the no‑pass option:\n", "\n", "\\[\n", "C_{\\text{yes}}(p) = 1.10\\,C_{\\text{no}}(p).\n", "\\]\n", "\n", "Plugging the expressions:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10\\big(1 + 0.10p\\big).\n", "\\]\n", "\n", "Simplify:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10 + 0.11p \\\\\n", "\\Longrightarrow\\; 0.08p - 0.11p = 1.10 - 1.012 \\\\\n", "\\Longrightarrow\\; -0.03p = 0.088 \\\\\n", "\\Longrightarrow\\; p = -\\frac{0.088}{0.03} \\approx -2.93 .\n", "\\]\n", "\n", "---\n", "\n", "**Step 5 – Interpret the result**\n", "\n", "A peak‑hour proportion of \\(p\\approx -2.93\\) is impossible (the fraction of rides that are “peak” cannot be negative, nor can it be less than zero). \n", "\n", "What does this tell us?\n", "\n", "* Even if **all** rides are shifted to off‑peak (\\(p = 0\\)), the pass is only\n", "\n", "\\[\n", "\\frac{C_{\\text{yes}}(0)}{C_{\\text{no}}(0)} =\n", "\\frac{1.012}{1.00} = 1.012\n", "\\]\n", "\n", "i.e. **1.2 % more expensive** than going without the pass.\n", "\n", "* The maximum possible increase in cost caused by moving rides away from peak hours is therefore 1.2 %. \n", "\n", "Reaching a 10 % increase would require a negative peak‑hour proportion (or, equivalently, “more off‑peak rides than the total number of rides”), which cannot happen.\n", "\n", "---\n", "\n", "## **Answer**\n", "\n", "It is **not possible**. \n", "Even if the rider moved **all** of his/her rides from peak to off‑peak (a 100 % shift), the monthly cost with the pass would be only about **1.2 % higher** than the cost without the pass. To make the pass 10 % more expensive would require a peak‑hour proportion of \\(p \\approx -2.9\\) (i.e., a negative number of peak rides), which is mathematically impossible. Consequently, no feasible percentage of rides shifted from peak to off‑peak can produce a 10 % cost disadvantage for the pass.\n" ] } ], "source": [ "# It's nice to know how to use \"zip\"\n", "for competitor, answer in zip(competitors, answers):\n", " print(f\"Competitor: {competitor}\\n\\n{answer}\")\n" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [], "source": [ "# Let's bring this together - note the use of \"enumerate\"\n", "\n", "together = \"\"\n", "for index, answer in enumerate(answers):\n", " together += f\"# Response from competitor {index+1}\\n\\n\"\n", " together += answer + \"\\n\\n\"" ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "# Response from competitor 1\n", "\n", "Let P be the base off-peak fare. Peak fare = 1.10P. With the pass, rides cost 20% less, so per-ride prices are:\n", "- Off-peak with pass: 0.80P\n", "- Peak with pass: 0.80 × 1.10P = 0.88P\n", "\n", "Let N be total rides in a month. The pass price is F.\n", "\n", "Given: when 60% of rides are during peak (0.60N peak, 0.40N off-peak), the monthly cost without the pass equals the monthly cost with the pass.\n", "- Without pass: C0 = N[(0.60)(1.10P) + (0.40)(P)] = NP(0.66 + 0.40) = 1.06 N P\n", "- With pass: Cpass = F + N[(0.60)(0.88P) + (0.40)(0.80P)] = F + NP(0.528 + 0.32) = F + 0.848 N P\n", "\n", "Setting C0 = Cpass gives F = 0.212 N P.\n", "\n", "Now suppose a fraction s of total rides shift from peak to off-peak. Then:\n", "- New peak fraction p' = 0.60 − s\n", "- New off-peak fraction o' = 0.40 + s\n", "\n", "Costs become:\n", "- Without pass: C_without(s) = N P [1.10 p' + o'] = N P [1.10(0.60 − s) + (0.40 + s)] = N P [1.06 − 0.10 s]\n", "- With pass: C_with(s) = F + N P [0.88 p' + 0.80 o'] = F + N P [0.88(0.60 − s) + 0.80(0.40 + s)]\n", " = F + N P [0.848 − 0.08 s] = 0.212 N P + N P [0.848 − 0.08 s] = N P [1.060 − 0.08 s]\n", "\n", "We want C_with(s) = 1.10 × C_without(s). That is:\n", "1.060 − 0.08 s = 1.10(1.060 − 0.10 s)\n", "1.060 − 0.08 s = 1.166 − 0.11 s\n", "0.03 s = 0.106\n", "s ≈ 3.53\n", "\n", "But s must be in [0, 0.60] (you can shift at most 60% of rides). Therefore there is no feasible solution: you cannot reach a 10% higher cost with the pass by shifting rides.\n", "\n", "What is the maximum possible increase? At the extreme s = 0.60:\n", "- C_without(0.60) = N P (1.06 − 0.06) = 1.00 N P\n", "- C_with(0.60) = N P (1.06 − 0.048) = 1.012 N P\n", "\n", "Thus the pass can be at most about 1.2% more expensive than not having it (when 60% of rides shift from peak to off-peak). Conclusion: no feasible shift percentage can make the pass 10% more expensive; the maximum is ≈1.2%.\n", "\n", "# Response from competitor 2\n", "\n", "Let F be the base fare for a single ride during off-peak hours.\n", "Let N be the total number of rides in a month.\n", "\n", "1. **Fares:**\n", " * Off-peak fare (F_o) = F\n", " * Peak hour fare (F_p) = F * (1 + 0.10) = 1.1F\n", "\n", "2. **Cost without the pass:**\n", " Let P be the percentage of rides taken during peak hours.\n", " Number of peak rides = N * P\n", " Number of off-peak rides = N * (1 - P)\n", " Cost_no_pass = (N * P * F_p) + (N * (1 - P) * F_o)\n", " Cost_no_pass = (N * P * 1.1F) + (N * (1 - P) * F)\n", " Cost_no_pass = N * F * (1.1P + 1 - P)\n", " Cost_no_pass = N * F * (0.1P + 1)\n", "\n", "3. **Cost with the frequent rider pass:**\n", " Let C be the flat fee for the pass.\n", " The pass provides a 20% discount on all rides.\n", " * Off-peak fare with pass (F_o_pass) = F * (1 - 0.20) = 0.8F\n", " * Peak hour fare with pass (F_p_pass) = 1.1F * (1 - 0.20) = 0.88F\n", " Cost_with_pass = C + (N * P * F_p_pass) + (N * (1 - P) * F_o_pass)\n", " Cost_with_pass = C + (N * P * 0.88F) + (N * (1 - P) * 0.8F)\n", " Cost_with_pass = C + N * F * (0.88P + 0.8 - 0.8P)\n", " Cost_with_pass = C + N * F * (0.08P + 0.8)\n", "\n", "4. **Determine the pass fee (C) using the initial condition:**\n", " The problem states that when 60% of rides are taken during peak hours (P = 0.6), the cost without the pass is exactly the same as the cost with the pass.\n", " Cost_no_pass (P=0.6) = Cost_with_pass (P=0.6)\n", " N * F * (0.1 * 0.6 + 1) = C + N * F * (0.08 * 0.6 + 0.8)\n", " N * F * (0.06 + 1) = C + N * F * (0.048 + 0.8)\n", " 1.06 NF = C + 0.848 NF\n", " C = (1.06 - 0.848) NF\n", " C = 0.212 NF\n", "\n", "5. **New condition for cost comparison:**\n", " We need to find a new percentage of peak rides, P_new, such that \"the pass is 10% more expensive than not having it.\" This means:\n", " Cost_with_pass_new = 1.10 * Cost_no_pass_new\n", "\n", " Substitute the expressions for costs and the value of C:\n", " C + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", " 0.212 NF + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", "\n", " Divide both sides by NF (assuming N and F are non-zero):\n", " 0.212 + 0.08 * P_new + 0.8 = 1.10 * (0.1 * P_new + 1)\n", " 1.012 + 0.08 * P_new = 0.11 * P_new + 1.10\n", "\n", " Solve for P_new:\n", " 1.012 - 1.10 = 0.11 * P_new - 0.08 * P_new\n", " -0.088 = 0.03 * P_new\n", " P_new = -0.088 / 0.03\n", " P_new = -88 / 30 = -44 / 15\n", " P_new ≈ -2.933\n", "\n", "6. **Interpret the result:**\n", " The percentage of rides taken during peak hours (P_new) must be between 0 and 1 (inclusive). A negative value for P_new is not physically possible.\n", "\n", " Let's analyze the ratio of costs as a function of P:\n", " f(P) = Cost_with_pass / Cost_no_pass = [C + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " Substitute C = 0.212 NF:\n", " f(P) = [0.212 NF + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " f(P) = (1.012 + 0.08P) / (0.1P + 1)\n", "\n", " The initial condition stated that for P = 0.6, f(0.6) = 1.\n", " A \"shift from peak to off-peak hours\" means that P_new must be less than the initial percentage of peak rides (0.6). So, P_new must be in the range [0, 0.6).\n", "\n", " Let's evaluate f(P) at the boundaries of this range:\n", " * At P = 0.6 (initial state, no shift): f(0.6) = (1.012 + 0.08 * 0.6) / (0.1 * 0.6 + 1) = (1.012 + 0.048) / (0.06 + 1) = 1.06 / 1.06 = 1.\n", " This means the costs are equal.\n", "\n", " * At P = 0 (all rides are off-peak, which is the maximum possible shift from peak to off-peak, corresponding to 60% of total rides shifting):\n", " f(0) = (1.012 + 0.08 * 0) / (0.1 * 0 + 1) = 1.012 / 1 = 1.012\n", " This means that when all rides are off-peak, the cost with the pass is 1.012 times the cost without the pass. In other words, the pass is 1.2% more expensive than not having it.\n", "\n", " The function f(P) is decreasing for P in [0, 1] (its derivative is f'(P) = -0.0212 / (0.1P+1)^2, which is always negative).\n", " Therefore, as P decreases from 0.6 to 0, f(P) increases from 1 to 1.012.\n", " The maximum value f(P) can reach when shifting rides from peak to off-peak is 1.012 (when P=0).\n", "\n", " The problem requires f(P_new) = 1.10 (i.e., the pass is 10% more expensive).\n", " Since the maximum achievable value for f(P) is 1.012, it is impossible for f(P) to reach 1.10 within the valid range of P_new.\n", "\n", "Conclusion:\n", "Based on the standard interpretation of the problem statement and the mathematical model, it is impossible to make the pass 10% more expensive than not having it by shifting rides from peak to off-peak hours, as the maximum possible cost difference achieved by shifting all peak rides to off-peak rides only makes the pass 1.2% more expensive. Therefore, there is no such percentage of rides that must shift to meet the condition.\n", "\n", "The final answer is $\\boxed{\\text{No such shift is possible}}$.\n", "\n", "# Response from competitor 3\n", "\n", "Alright, let's break this down step-by-step.\n", "\n", "---\n", "\n", "## **Step 1: Define variables**\n", "\n", "Let:\n", "- \\( P \\) = base fare (off-peak fare)\n", "- Peak fare = \\( 1.1P \\) (10% increase)\n", "- Total number of rides per month = \\( N \\)\n", "- Fraction of rides during peak hours = \\( f \\)\n", "- Without pass: \n", " Cost = \\( N \\cdot [ f \\cdot 1.1P + (1-f) \\cdot P ] \\) \n", " = \\( N P [ 1.1 f + 1 - f ] \\) \n", " = \\( N P [ 1 + 0.1 f ] \\)\n", "\n", "- With pass: \n", " Flat fee for pass = \\( F \\) (monthly cost of pass) \n", " Then each ride costs 20% less than the **current** fare (peak or off-peak). \n", " So off-peak fare with pass = \\( 0.8 P \\) \n", " Peak fare with pass = \\( 0.8 \\times 1.1 P = 0.88 P \\) \n", "\n", " Total cost with pass = \\( F + N [ f \\cdot 0.88P + (1-f) \\cdot 0.8P ] \\) \n", " = \\( F + N P [ 0.88 f + 0.8 - 0.8 f ] \\) \n", " = \\( F + N P [ 0.8 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 2: Use given condition to relate \\( F \\) and \\( P \\)**\n", "\n", "Given: Without pass, average monthly cost = cost with pass when \\( f = 0.6 \\).\n", "\n", "Without pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{no pass}} = N P [ 1 + 0.1 \\times 0.6 ] = N P [ 1.06 ] \\)\n", "\n", "With pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}} = F + N P [ 0.8 + 0.08 \\times 0.6 ] \\) \n", "= \\( F + N P [ 0.8 + 0.048 ] \\) \n", "= \\( F + N P [ 0.848 ] \\)\n", "\n", "Set them equal: \n", "\\( N P \\cdot 1.06 = F + N P \\cdot 0.848 \\) \n", "\\( F = N P ( 1.06 - 0.848 ) \\) \n", "\\( F = N P \\cdot 0.212 \\)\n", "\n", "So the pass costs \\( 0.212 N P \\) per month.\n", "\n", "---\n", "\n", "## **Step 3: General cost expressions**\n", "\n", "Without pass: \n", "\\( C_{\\text{no pass}}(f) = N P ( 1 + 0.1 f ) \\)\n", "\n", "With pass: \n", "\\( C_{\\text{pass}}(f) = 0.212 N P + N P ( 0.8 + 0.08 f ) \\) \n", "= \\( N P [ 0.212 + 0.8 + 0.08 f ] \\) \n", "= \\( N P [ 1.012 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 4: Find \\( f \\) such that pass is 10% more expensive than no pass**\n", "\n", "We want: \n", "\\( C_{\\text{pass}}(f) = 1.1 \\times C_{\\text{no pass}}(f) \\)\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f ] = 1.1 \\times N P [ 1 + 0.1 f ] \\) \n", "Cancel \\( N P \\): \n", "\\( 1.012 + 0.08 f = 1.1 + 0.11 f \\) \n", "\\( 1.012 - 1.1 = 0.11 f - 0.08 f \\) \n", "\\( -0.088 = 0.03 f \\) \n", "\\( f = -\\frac{0.088}{0.03} \\approx -2.933 \\)\n", "\n", "This is negative, which is impossible (fraction of peak rides can't be negative). \n", "This means: For pass to be **more expensive** than no pass by 10%, we would need \\( f \\) negative if we keep total rides constant — but maybe they mean **pass 10% more expensive** than no pass **at original \\( f = 0.6 \\)**? Wait, no — they say: \"what percentage of their total rides must shift from peak to off-peak hours to make the pass 10% more expensive than not having it\".\n", "\n", "That means: Start from \\( f = 0.6 \\), shift some rides from peak to off-peak, get new \\( f' \\), such that pass cost is 10% higher than no-pass cost **at new \\( f' \\)**.\n", "\n", "But we already computed the equation for general \\( f \\) for that condition: \n", "\\( 1.012 + 0.08 f = 1.1 (1 + 0.1 f) \\) \n", "We solved: \\( f \\approx -2.933 \\), impossible.\n", "\n", "So maybe they mean: pass is 10% more expensive **than the no-pass cost at the original \\( f = 0.6 \\)**? Let's check.\n", "\n", "---\n", "\n", "## **Step 5: Alternative interpretation**\n", "\n", "Original no-pass cost at \\( f = 0.6 \\): \\( C_0 = N P \\cdot 1.06 \\). \n", "We want \\( C_{\\text{pass}}(f') = 1.1 \\times C_0 \\).\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f' ] = 1.1 \\times N P \\times 1.06 \\) \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — also impossible (>1).\n", "\n", "So that’s not it.\n", "\n", "---\n", "\n", "## **Step 6: Maybe they mean: pass is 10% more expensive than no-pass cost at the NEW \\( f' \\), but starting from \\( f = 0.6 \\) where they were equal.**\n", "\n", "We already tried that in Step 4, got impossible \\( f' \\). \n", "So maybe they mean: pass is 10% more expensive **than the pass cost at \\( f = 0.6 \\)**? That is, increase in cost of pass relative to its original cost.\n", "\n", "---\n", "\n", "Original pass cost at \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}}(0.6) = N P [ 1.012 + 0.08 \\times 0.6 ] \\) \n", "= \\( N P [ 1.012 + 0.048 ] = N P \\times 1.06 \\) (same as no-pass originally, by design).\n", "\n", "We want new \\( f' \\) such that: \n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "= \\( 1.1 \\times 1.06 N P = 1.166 N P \\).\n", "\n", "So: \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — still impossible (>1).\n", "\n", "---\n", "\n", "## **Step 7: Check if maybe they mean: no-pass cost is 10% less than pass cost? That’s same as pass 10% more than no-pass, which we did.**\n", "\n", "Given the impossibility, maybe they mean: **make no-pass 10% cheaper than pass** — but that’s the same ratio.\n", "\n", "Let’s check if maybe the 10% is on base fare difference? Or maybe they mean: **pass becomes 10% more expensive than no-pass for the new pattern** — but we saw that requires \\( f \\) negative, meaning we must shift **so many** rides from peak to off-peak that \\( f \\) drops below zero, so minimum \\( f \\) is 0, let's see cost difference at \\( f = 0 \\):\n", "\n", "At \\( f = 0 \\): \n", "No-pass cost = \\( N P (1 + 0) = N P \\) \n", "Pass cost = \\( N P (1.012 + 0) = 1.012 N P \\) \n", "Ratio = \\( 1.012 / 1 = 1.012 \\) → pass is only 1.2% more expensive.\n", "\n", "So even shifting all peak rides to off-peak, pass is never 10% more expensive than no-pass — it’s always less than or equal to 1.2% more.\n", "\n", "Thus, **it’s impossible** for pass to be 10% more expensive than no-pass for any \\( f \\ge 0 \\) given the formulas.\n", "\n", "---\n", "\n", "## **Step 8: But maybe they mean: pass 10% more expensive than no-pass at the NEW \\( f' \\), but starting from equality at \\( f=0.6 \\), shifting some rides changes both costs.**\n", "\n", "Let’s check difference: \n", "\\( D(f) = C_{\\text{pass}}(f) - C_{\\text{no pass}}(f) \\) \n", "= \\( N P [ 1.012 + 0.08 f - (1 + 0.1 f) ] \\) \n", "= \\( N P [ 0.012 - 0.02 f ] \\)\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(f') \\) for \"pass 10% more expensive than no-pass\":\n", "\n", "\\( 0.012 - 0.02 f' = 0.1 (1 + 0.1 f') \\) \n", "\\( 0.012 - 0.02 f' = 0.1 + 0.01 f' \\) \n", "\\( 0.012 - 0.1 = 0.01 f' + 0.02 f' \\) \n", "\\( -0.088 = 0.03 f' \\) \n", "\\( f' \\approx -2.933 \\) — same impossible.\n", "\n", "So indeed, to get pass 10% more expensive than no-pass, \\( f' \\) must be negative. \n", "The only way to approach that from \\( f=0.6 \\) is to shift **all** peak rides to off-peak and more — impossible since only 60% are peak initially.\n", "\n", "---\n", "\n", "## **Step 9: Conclusion from math**\n", "\n", "The condition “pass 10% more expensive than no-pass” is impossible with given pricing formulas for \\( f \\ge 0 \\). \n", "But if we interpret the question as: shift enough rides so that **the extra cost of the pass relative to no-pass** is 10% **of the original no-pass cost at \\( f=0.6 \\)**? That is:\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(0.6) \\) \n", "= \\( 0.1 \\times 1.06 N P = 0.106 N P \\).\n", "\n", "So: \n", "\\( N P (0.012 - 0.02 f') = 0.106 N P \\) \n", "\\( 0.012 - 0.02 f' = 0.106 \\) \n", "\\( -0.02 f' = 0.094 \\) \n", "\\( f' = -4.7 \\) — also impossible.\n", "\n", "---\n", "\n", "Given the problem constraints, maybe they meant: make pass **10% more expensive than it was originally**? That would be:\n", "\n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "We computed earlier: \\( 1.012 + 0.08 f' = 1.166 \\) → \\( f' = 1.925 \\) impossible.\n", "\n", "So all interpretations give impossible \\( f' \\) except one: maybe they mean **no-pass cost 10% less than pass cost** — which is same equation as before.\n", "\n", "Given the likely intent: They want us to find \\( f' \\) from the equation \\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{no pass}}(f') \\), solve, get \\( f' \\) negative, so minimal \\( f' \\) is 0. \n", "That means shift **all** 60% peak rides to off-peak. \n", "So percentage of total rides shifted = \\( 60\\% \\).\n", "\n", "---\n", "\n", "**Final answer** (assuming they accept the shift to \\( f' = 0 \\) as the point where pass is most expensive relative to no-pass, though not quite 10% more):\n", "\n", "\\[\n", "\\boxed{60}\n", "\\]\n", "\n", "That is, shift all 60% of rides that were peak to off-peak.\n", "\n", "# Response from competitor 4\n", "\n", "**Step 1 – Set up the variables**\n", "\n", "| Symbol | Meaning |\n", "|--------|----------|\n", "| \\(F\\) | Base fare for an off‑peak ride (dollar amount) |\n", "| \\(N\\) | Total number of rides the rider takes in a month (kept constant) |\n", "| \\(p\\) | Fraction of rides that occur during peak hours (so off‑peak fraction = \\(1-p\\)) |\n", "| \\(P\\) | Monthly price of the frequent‑rider pass (a flat fee) |\n", "\n", "- Off‑peak fare (no pass) = \\(F\\) \n", "- Peak fare (no pass) = \\(1.10F\\) (10 % increase) \n", "- Discount with the pass = 20 % → you pay 80 % of the listed fare.\n", "\n", "Hence \n", "\n", "- Off‑peak ride cost **with** the pass = \\(0.8F\\) \n", "- Peak ride cost **with** the pass = \\(0.8\\times1.10F = 0.88F\\).\n", "\n", "---\n", "\n", "**Step 2 – Find the pass fee \\(P\\) from the “break‑even” condition**\n", "\n", "The problem tells us that when 60 % of rides are taken during peak hours the rider spends exactly the same amount whether they buy the pass or not.\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost without pass} &= N\\Big[(0.4)F + (0.6)(1.10F)\\Big] \\\\\n", " &= NF\\big[0.4 + 0.66\\big] = 1.06\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost with pass} &= P + N\\Big[(0.4)(0.8F) + (0.6)(0.88F)\\Big] \\\\\n", " &= P + NF\\big[0.32 + 0.528\\big] = P + 0.848\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "Setting the two equal:\n", "\n", "\\[\n", "1.06\\,NF = P + 0.848\\,NF \\quad\\Longrightarrow\\quad\n", "P = (1.06-0.848)NF = 0.212\\,NF .\n", "\\]\n", "\n", "So the pass costs **21.2 % of the rider’s original monthly expense**.\n", "\n", "---\n", "\n", "**Step 3 – Express the two monthly costs after a shift in the peak‑off‑peak mix**\n", "\n", "If the new peak‑hour proportion is \\(p\\) (the off‑peak proportion is \\(1-p\\)):\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{No‑pass cost} &: \\; C_{\\text{no}}(p)=NF\\Big[(1-p) + 1.10p\\Big] = NF\\big[1+0.10p\\big] ,\\\\[4pt]\n", "\\text{Pass cost} &: \\; C_{\\text{yes}}(p)=P + NF\\Big[0.8(1-p)+0.88p\\Big] \\\\\n", " &: \\; =0.212\\,NF + NF\\big[0.8 + 0.08p\\big] = NF\\big[1.012 + 0.08p\\big] .\n", "\\end{aligned}\n", "\\]\n", "\n", "---\n", "\n", "**Step 4 – Impose the “10 % more expensive” requirement**\n", "\n", "We want the pass to cost **10 % more** than the no‑pass option:\n", "\n", "\\[\n", "C_{\\text{yes}}(p) = 1.10\\,C_{\\text{no}}(p).\n", "\\]\n", "\n", "Plugging the expressions:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10\\big(1 + 0.10p\\big).\n", "\\]\n", "\n", "Simplify:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10 + 0.11p \\\\\n", "\\Longrightarrow\\; 0.08p - 0.11p = 1.10 - 1.012 \\\\\n", "\\Longrightarrow\\; -0.03p = 0.088 \\\\\n", "\\Longrightarrow\\; p = -\\frac{0.088}{0.03} \\approx -2.93 .\n", "\\]\n", "\n", "---\n", "\n", "**Step 5 – Interpret the result**\n", "\n", "A peak‑hour proportion of \\(p\\approx -2.93\\) is impossible (the fraction of rides that are “peak” cannot be negative, nor can it be less than zero). \n", "\n", "What does this tell us?\n", "\n", "* Even if **all** rides are shifted to off‑peak (\\(p = 0\\)), the pass is only\n", "\n", "\\[\n", "\\frac{C_{\\text{yes}}(0)}{C_{\\text{no}}(0)} =\n", "\\frac{1.012}{1.00} = 1.012\n", "\\]\n", "\n", "i.e. **1.2 % more expensive** than going without the pass.\n", "\n", "* The maximum possible increase in cost caused by moving rides away from peak hours is therefore 1.2 %. \n", "\n", "Reaching a 10 % increase would require a negative peak‑hour proportion (or, equivalently, “more off‑peak rides than the total number of rides”), which cannot happen.\n", "\n", "---\n", "\n", "## **Answer**\n", "\n", "It is **not possible**. \n", "Even if the rider moved **all** of his/her rides from peak to off‑peak (a 100 % shift), the monthly cost with the pass would be only about **1.2 % higher** than the cost without the pass. To make the pass 10 % more expensive would require a peak‑hour proportion of \\(p \\approx -2.9\\) (i.e., a negative number of peak rides), which is mathematically impossible. Consequently, no feasible percentage of rides shifted from peak to off‑peak can produce a 10 % cost disadvantage for the pass.\n", "\n", "\n" ] } ], "source": [ "print(together)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [] }, { "cell_type": "code", "execution_count": 33, "metadata": {}, "outputs": [], "source": [ "judge = f\"\"\"You are judging a competition between {len(competitors)} competitors.\n", "Each model has been given this question:\n", "\n", "{question}\n", "\n", "Your job is to evaluate each response for clarity and strength of argument, and rank them in order of best to worst.\n", "Respond with JSON, and only JSON, with the following format:\n", "{{\"results\": [\"best competitor number\", \"second best competitor number\", \"third best competitor number\", ...]}}\n", "\n", "Here are the responses from each competitor:\n", "\n", "{together}\n", "\n", "Now respond with the JSON with the ranked order of the competitors, nothing else. Do not include markdown formatting or code blocks.\"\"\"\n" ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "You are judging a competition between 4 competitors.\n", "Each model has been given this question:\n", "\n", "A city's public transit system uses a dynamic pricing model where the base fare increases by 10% during \"peak hours,\" but a frequent rider pass, which costs a flat fee, provides a 20% discount on all rides. If a rider's average monthly cost without the pass is exactly the same as the cost with the pass when 60% of their rides are taken during peak hours, what percentage of their total rides must shift from peak to off-peak hours to make the pass 10% more expensive than not having it, assuming the total number of rides stays constant?\n", "\n", "Your job is to evaluate each response for clarity and strength of argument, and rank them in order of best to worst.\n", "Respond with JSON, and only JSON, with the following format:\n", "{\"results\": [\"best competitor number\", \"second best competitor number\", \"third best competitor number\", ...]}\n", "\n", "Here are the responses from each competitor:\n", "\n", "# Response from competitor 1\n", "\n", "Let P be the base off-peak fare. Peak fare = 1.10P. With the pass, rides cost 20% less, so per-ride prices are:\n", "- Off-peak with pass: 0.80P\n", "- Peak with pass: 0.80 × 1.10P = 0.88P\n", "\n", "Let N be total rides in a month. The pass price is F.\n", "\n", "Given: when 60% of rides are during peak (0.60N peak, 0.40N off-peak), the monthly cost without the pass equals the monthly cost with the pass.\n", "- Without pass: C0 = N[(0.60)(1.10P) + (0.40)(P)] = NP(0.66 + 0.40) = 1.06 N P\n", "- With pass: Cpass = F + N[(0.60)(0.88P) + (0.40)(0.80P)] = F + NP(0.528 + 0.32) = F + 0.848 N P\n", "\n", "Setting C0 = Cpass gives F = 0.212 N P.\n", "\n", "Now suppose a fraction s of total rides shift from peak to off-peak. Then:\n", "- New peak fraction p' = 0.60 − s\n", "- New off-peak fraction o' = 0.40 + s\n", "\n", "Costs become:\n", "- Without pass: C_without(s) = N P [1.10 p' + o'] = N P [1.10(0.60 − s) + (0.40 + s)] = N P [1.06 − 0.10 s]\n", "- With pass: C_with(s) = F + N P [0.88 p' + 0.80 o'] = F + N P [0.88(0.60 − s) + 0.80(0.40 + s)]\n", " = F + N P [0.848 − 0.08 s] = 0.212 N P + N P [0.848 − 0.08 s] = N P [1.060 − 0.08 s]\n", "\n", "We want C_with(s) = 1.10 × C_without(s). That is:\n", "1.060 − 0.08 s = 1.10(1.060 − 0.10 s)\n", "1.060 − 0.08 s = 1.166 − 0.11 s\n", "0.03 s = 0.106\n", "s ≈ 3.53\n", "\n", "But s must be in [0, 0.60] (you can shift at most 60% of rides). Therefore there is no feasible solution: you cannot reach a 10% higher cost with the pass by shifting rides.\n", "\n", "What is the maximum possible increase? At the extreme s = 0.60:\n", "- C_without(0.60) = N P (1.06 − 0.06) = 1.00 N P\n", "- C_with(0.60) = N P (1.06 − 0.048) = 1.012 N P\n", "\n", "Thus the pass can be at most about 1.2% more expensive than not having it (when 60% of rides shift from peak to off-peak). Conclusion: no feasible shift percentage can make the pass 10% more expensive; the maximum is ≈1.2%.\n", "\n", "# Response from competitor 2\n", "\n", "Let F be the base fare for a single ride during off-peak hours.\n", "Let N be the total number of rides in a month.\n", "\n", "1. **Fares:**\n", " * Off-peak fare (F_o) = F\n", " * Peak hour fare (F_p) = F * (1 + 0.10) = 1.1F\n", "\n", "2. **Cost without the pass:**\n", " Let P be the percentage of rides taken during peak hours.\n", " Number of peak rides = N * P\n", " Number of off-peak rides = N * (1 - P)\n", " Cost_no_pass = (N * P * F_p) + (N * (1 - P) * F_o)\n", " Cost_no_pass = (N * P * 1.1F) + (N * (1 - P) * F)\n", " Cost_no_pass = N * F * (1.1P + 1 - P)\n", " Cost_no_pass = N * F * (0.1P + 1)\n", "\n", "3. **Cost with the frequent rider pass:**\n", " Let C be the flat fee for the pass.\n", " The pass provides a 20% discount on all rides.\n", " * Off-peak fare with pass (F_o_pass) = F * (1 - 0.20) = 0.8F\n", " * Peak hour fare with pass (F_p_pass) = 1.1F * (1 - 0.20) = 0.88F\n", " Cost_with_pass = C + (N * P * F_p_pass) + (N * (1 - P) * F_o_pass)\n", " Cost_with_pass = C + (N * P * 0.88F) + (N * (1 - P) * 0.8F)\n", " Cost_with_pass = C + N * F * (0.88P + 0.8 - 0.8P)\n", " Cost_with_pass = C + N * F * (0.08P + 0.8)\n", "\n", "4. **Determine the pass fee (C) using the initial condition:**\n", " The problem states that when 60% of rides are taken during peak hours (P = 0.6), the cost without the pass is exactly the same as the cost with the pass.\n", " Cost_no_pass (P=0.6) = Cost_with_pass (P=0.6)\n", " N * F * (0.1 * 0.6 + 1) = C + N * F * (0.08 * 0.6 + 0.8)\n", " N * F * (0.06 + 1) = C + N * F * (0.048 + 0.8)\n", " 1.06 NF = C + 0.848 NF\n", " C = (1.06 - 0.848) NF\n", " C = 0.212 NF\n", "\n", "5. **New condition for cost comparison:**\n", " We need to find a new percentage of peak rides, P_new, such that \"the pass is 10% more expensive than not having it.\" This means:\n", " Cost_with_pass_new = 1.10 * Cost_no_pass_new\n", "\n", " Substitute the expressions for costs and the value of C:\n", " C + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", " 0.212 NF + N * F * (0.08 * P_new + 0.8) = 1.10 * N * F * (0.1 * P_new + 1)\n", "\n", " Divide both sides by NF (assuming N and F are non-zero):\n", " 0.212 + 0.08 * P_new + 0.8 = 1.10 * (0.1 * P_new + 1)\n", " 1.012 + 0.08 * P_new = 0.11 * P_new + 1.10\n", "\n", " Solve for P_new:\n", " 1.012 - 1.10 = 0.11 * P_new - 0.08 * P_new\n", " -0.088 = 0.03 * P_new\n", " P_new = -0.088 / 0.03\n", " P_new = -88 / 30 = -44 / 15\n", " P_new ≈ -2.933\n", "\n", "6. **Interpret the result:**\n", " The percentage of rides taken during peak hours (P_new) must be between 0 and 1 (inclusive). A negative value for P_new is not physically possible.\n", "\n", " Let's analyze the ratio of costs as a function of P:\n", " f(P) = Cost_with_pass / Cost_no_pass = [C + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " Substitute C = 0.212 NF:\n", " f(P) = [0.212 NF + NF(0.08P + 0.8)] / [NF(0.1P + 1)]\n", " f(P) = (1.012 + 0.08P) / (0.1P + 1)\n", "\n", " The initial condition stated that for P = 0.6, f(0.6) = 1.\n", " A \"shift from peak to off-peak hours\" means that P_new must be less than the initial percentage of peak rides (0.6). So, P_new must be in the range [0, 0.6).\n", "\n", " Let's evaluate f(P) at the boundaries of this range:\n", " * At P = 0.6 (initial state, no shift): f(0.6) = (1.012 + 0.08 * 0.6) / (0.1 * 0.6 + 1) = (1.012 + 0.048) / (0.06 + 1) = 1.06 / 1.06 = 1.\n", " This means the costs are equal.\n", "\n", " * At P = 0 (all rides are off-peak, which is the maximum possible shift from peak to off-peak, corresponding to 60% of total rides shifting):\n", " f(0) = (1.012 + 0.08 * 0) / (0.1 * 0 + 1) = 1.012 / 1 = 1.012\n", " This means that when all rides are off-peak, the cost with the pass is 1.012 times the cost without the pass. In other words, the pass is 1.2% more expensive than not having it.\n", "\n", " The function f(P) is decreasing for P in [0, 1] (its derivative is f'(P) = -0.0212 / (0.1P+1)^2, which is always negative).\n", " Therefore, as P decreases from 0.6 to 0, f(P) increases from 1 to 1.012.\n", " The maximum value f(P) can reach when shifting rides from peak to off-peak is 1.012 (when P=0).\n", "\n", " The problem requires f(P_new) = 1.10 (i.e., the pass is 10% more expensive).\n", " Since the maximum achievable value for f(P) is 1.012, it is impossible for f(P) to reach 1.10 within the valid range of P_new.\n", "\n", "Conclusion:\n", "Based on the standard interpretation of the problem statement and the mathematical model, it is impossible to make the pass 10% more expensive than not having it by shifting rides from peak to off-peak hours, as the maximum possible cost difference achieved by shifting all peak rides to off-peak rides only makes the pass 1.2% more expensive. Therefore, there is no such percentage of rides that must shift to meet the condition.\n", "\n", "The final answer is $\\boxed{\\text{No such shift is possible}}$.\n", "\n", "# Response from competitor 3\n", "\n", "Alright, let's break this down step-by-step.\n", "\n", "---\n", "\n", "## **Step 1: Define variables**\n", "\n", "Let:\n", "- \\( P \\) = base fare (off-peak fare)\n", "- Peak fare = \\( 1.1P \\) (10% increase)\n", "- Total number of rides per month = \\( N \\)\n", "- Fraction of rides during peak hours = \\( f \\)\n", "- Without pass: \n", " Cost = \\( N \\cdot [ f \\cdot 1.1P + (1-f) \\cdot P ] \\) \n", " = \\( N P [ 1.1 f + 1 - f ] \\) \n", " = \\( N P [ 1 + 0.1 f ] \\)\n", "\n", "- With pass: \n", " Flat fee for pass = \\( F \\) (monthly cost of pass) \n", " Then each ride costs 20% less than the **current** fare (peak or off-peak). \n", " So off-peak fare with pass = \\( 0.8 P \\) \n", " Peak fare with pass = \\( 0.8 \\times 1.1 P = 0.88 P \\) \n", "\n", " Total cost with pass = \\( F + N [ f \\cdot 0.88P + (1-f) \\cdot 0.8P ] \\) \n", " = \\( F + N P [ 0.88 f + 0.8 - 0.8 f ] \\) \n", " = \\( F + N P [ 0.8 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 2: Use given condition to relate \\( F \\) and \\( P \\)**\n", "\n", "Given: Without pass, average monthly cost = cost with pass when \\( f = 0.6 \\).\n", "\n", "Without pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{no pass}} = N P [ 1 + 0.1 \\times 0.6 ] = N P [ 1.06 ] \\)\n", "\n", "With pass cost when \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}} = F + N P [ 0.8 + 0.08 \\times 0.6 ] \\) \n", "= \\( F + N P [ 0.8 + 0.048 ] \\) \n", "= \\( F + N P [ 0.848 ] \\)\n", "\n", "Set them equal: \n", "\\( N P \\cdot 1.06 = F + N P \\cdot 0.848 \\) \n", "\\( F = N P ( 1.06 - 0.848 ) \\) \n", "\\( F = N P \\cdot 0.212 \\)\n", "\n", "So the pass costs \\( 0.212 N P \\) per month.\n", "\n", "---\n", "\n", "## **Step 3: General cost expressions**\n", "\n", "Without pass: \n", "\\( C_{\\text{no pass}}(f) = N P ( 1 + 0.1 f ) \\)\n", "\n", "With pass: \n", "\\( C_{\\text{pass}}(f) = 0.212 N P + N P ( 0.8 + 0.08 f ) \\) \n", "= \\( N P [ 0.212 + 0.8 + 0.08 f ] \\) \n", "= \\( N P [ 1.012 + 0.08 f ] \\)\n", "\n", "---\n", "\n", "## **Step 4: Find \\( f \\) such that pass is 10% more expensive than no pass**\n", "\n", "We want: \n", "\\( C_{\\text{pass}}(f) = 1.1 \\times C_{\\text{no pass}}(f) \\)\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f ] = 1.1 \\times N P [ 1 + 0.1 f ] \\) \n", "Cancel \\( N P \\): \n", "\\( 1.012 + 0.08 f = 1.1 + 0.11 f \\) \n", "\\( 1.012 - 1.1 = 0.11 f - 0.08 f \\) \n", "\\( -0.088 = 0.03 f \\) \n", "\\( f = -\\frac{0.088}{0.03} \\approx -2.933 \\)\n", "\n", "This is negative, which is impossible (fraction of peak rides can't be negative). \n", "This means: For pass to be **more expensive** than no pass by 10%, we would need \\( f \\) negative if we keep total rides constant — but maybe they mean **pass 10% more expensive** than no pass **at original \\( f = 0.6 \\)**? Wait, no — they say: \"what percentage of their total rides must shift from peak to off-peak hours to make the pass 10% more expensive than not having it\".\n", "\n", "That means: Start from \\( f = 0.6 \\), shift some rides from peak to off-peak, get new \\( f' \\), such that pass cost is 10% higher than no-pass cost **at new \\( f' \\)**.\n", "\n", "But we already computed the equation for general \\( f \\) for that condition: \n", "\\( 1.012 + 0.08 f = 1.1 (1 + 0.1 f) \\) \n", "We solved: \\( f \\approx -2.933 \\), impossible.\n", "\n", "So maybe they mean: pass is 10% more expensive **than the no-pass cost at the original \\( f = 0.6 \\)**? Let's check.\n", "\n", "---\n", "\n", "## **Step 5: Alternative interpretation**\n", "\n", "Original no-pass cost at \\( f = 0.6 \\): \\( C_0 = N P \\cdot 1.06 \\). \n", "We want \\( C_{\\text{pass}}(f') = 1.1 \\times C_0 \\).\n", "\n", "So: \n", "\\( N P [ 1.012 + 0.08 f' ] = 1.1 \\times N P \\times 1.06 \\) \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — also impossible (>1).\n", "\n", "So that’s not it.\n", "\n", "---\n", "\n", "## **Step 6: Maybe they mean: pass is 10% more expensive than no-pass cost at the NEW \\( f' \\), but starting from \\( f = 0.6 \\) where they were equal.**\n", "\n", "We already tried that in Step 4, got impossible \\( f' \\). \n", "So maybe they mean: pass is 10% more expensive **than the pass cost at \\( f = 0.6 \\)**? That is, increase in cost of pass relative to its original cost.\n", "\n", "---\n", "\n", "Original pass cost at \\( f = 0.6 \\): \n", "\\( C_{\\text{pass}}(0.6) = N P [ 1.012 + 0.08 \\times 0.6 ] \\) \n", "= \\( N P [ 1.012 + 0.048 ] = N P \\times 1.06 \\) (same as no-pass originally, by design).\n", "\n", "We want new \\( f' \\) such that: \n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "= \\( 1.1 \\times 1.06 N P = 1.166 N P \\).\n", "\n", "So: \n", "\\( 1.012 + 0.08 f' = 1.166 \\) \n", "\\( 0.08 f' = 0.154 \\) \n", "\\( f' = 1.925 \\) — still impossible (>1).\n", "\n", "---\n", "\n", "## **Step 7: Check if maybe they mean: no-pass cost is 10% less than pass cost? That’s same as pass 10% more than no-pass, which we did.**\n", "\n", "Given the impossibility, maybe they mean: **make no-pass 10% cheaper than pass** — but that’s the same ratio.\n", "\n", "Let’s check if maybe the 10% is on base fare difference? Or maybe they mean: **pass becomes 10% more expensive than no-pass for the new pattern** — but we saw that requires \\( f \\) negative, meaning we must shift **so many** rides from peak to off-peak that \\( f \\) drops below zero, so minimum \\( f \\) is 0, let's see cost difference at \\( f = 0 \\):\n", "\n", "At \\( f = 0 \\): \n", "No-pass cost = \\( N P (1 + 0) = N P \\) \n", "Pass cost = \\( N P (1.012 + 0) = 1.012 N P \\) \n", "Ratio = \\( 1.012 / 1 = 1.012 \\) → pass is only 1.2% more expensive.\n", "\n", "So even shifting all peak rides to off-peak, pass is never 10% more expensive than no-pass — it’s always less than or equal to 1.2% more.\n", "\n", "Thus, **it’s impossible** for pass to be 10% more expensive than no-pass for any \\( f \\ge 0 \\) given the formulas.\n", "\n", "---\n", "\n", "## **Step 8: But maybe they mean: pass 10% more expensive than no-pass at the NEW \\( f' \\), but starting from equality at \\( f=0.6 \\), shifting some rides changes both costs.**\n", "\n", "Let’s check difference: \n", "\\( D(f) = C_{\\text{pass}}(f) - C_{\\text{no pass}}(f) \\) \n", "= \\( N P [ 1.012 + 0.08 f - (1 + 0.1 f) ] \\) \n", "= \\( N P [ 0.012 - 0.02 f ] \\)\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(f') \\) for \"pass 10% more expensive than no-pass\":\n", "\n", "\\( 0.012 - 0.02 f' = 0.1 (1 + 0.1 f') \\) \n", "\\( 0.012 - 0.02 f' = 0.1 + 0.01 f' \\) \n", "\\( 0.012 - 0.1 = 0.01 f' + 0.02 f' \\) \n", "\\( -0.088 = 0.03 f' \\) \n", "\\( f' \\approx -2.933 \\) — same impossible.\n", "\n", "So indeed, to get pass 10% more expensive than no-pass, \\( f' \\) must be negative. \n", "The only way to approach that from \\( f=0.6 \\) is to shift **all** peak rides to off-peak and more — impossible since only 60% are peak initially.\n", "\n", "---\n", "\n", "## **Step 9: Conclusion from math**\n", "\n", "The condition “pass 10% more expensive than no-pass” is impossible with given pricing formulas for \\( f \\ge 0 \\). \n", "But if we interpret the question as: shift enough rides so that **the extra cost of the pass relative to no-pass** is 10% **of the original no-pass cost at \\( f=0.6 \\)**? That is:\n", "\n", "We want \\( D(f') = 0.1 \\times C_{\\text{no pass}}(0.6) \\) \n", "= \\( 0.1 \\times 1.06 N P = 0.106 N P \\).\n", "\n", "So: \n", "\\( N P (0.012 - 0.02 f') = 0.106 N P \\) \n", "\\( 0.012 - 0.02 f' = 0.106 \\) \n", "\\( -0.02 f' = 0.094 \\) \n", "\\( f' = -4.7 \\) — also impossible.\n", "\n", "---\n", "\n", "Given the problem constraints, maybe they meant: make pass **10% more expensive than it was originally**? That would be:\n", "\n", "\\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{pass}}(0.6) \\) \n", "We computed earlier: \\( 1.012 + 0.08 f' = 1.166 \\) → \\( f' = 1.925 \\) impossible.\n", "\n", "So all interpretations give impossible \\( f' \\) except one: maybe they mean **no-pass cost 10% less than pass cost** — which is same equation as before.\n", "\n", "Given the likely intent: They want us to find \\( f' \\) from the equation \\( C_{\\text{pass}}(f') = 1.1 \\times C_{\\text{no pass}}(f') \\), solve, get \\( f' \\) negative, so minimal \\( f' \\) is 0. \n", "That means shift **all** 60% peak rides to off-peak. \n", "So percentage of total rides shifted = \\( 60\\% \\).\n", "\n", "---\n", "\n", "**Final answer** (assuming they accept the shift to \\( f' = 0 \\) as the point where pass is most expensive relative to no-pass, though not quite 10% more):\n", "\n", "\\[\n", "\\boxed{60}\n", "\\]\n", "\n", "That is, shift all 60% of rides that were peak to off-peak.\n", "\n", "# Response from competitor 4\n", "\n", "**Step 1 – Set up the variables**\n", "\n", "| Symbol | Meaning |\n", "|--------|----------|\n", "| \\(F\\) | Base fare for an off‑peak ride (dollar amount) |\n", "| \\(N\\) | Total number of rides the rider takes in a month (kept constant) |\n", "| \\(p\\) | Fraction of rides that occur during peak hours (so off‑peak fraction = \\(1-p\\)) |\n", "| \\(P\\) | Monthly price of the frequent‑rider pass (a flat fee) |\n", "\n", "- Off‑peak fare (no pass) = \\(F\\) \n", "- Peak fare (no pass) = \\(1.10F\\) (10 % increase) \n", "- Discount with the pass = 20 % → you pay 80 % of the listed fare.\n", "\n", "Hence \n", "\n", "- Off‑peak ride cost **with** the pass = \\(0.8F\\) \n", "- Peak ride cost **with** the pass = \\(0.8\\times1.10F = 0.88F\\).\n", "\n", "---\n", "\n", "**Step 2 – Find the pass fee \\(P\\) from the “break‑even” condition**\n", "\n", "The problem tells us that when 60 % of rides are taken during peak hours the rider spends exactly the same amount whether they buy the pass or not.\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost without pass} &= N\\Big[(0.4)F + (0.6)(1.10F)\\Big] \\\\\n", " &= NF\\big[0.4 + 0.66\\big] = 1.06\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{Cost with pass} &= P + N\\Big[(0.4)(0.8F) + (0.6)(0.88F)\\Big] \\\\\n", " &= P + NF\\big[0.32 + 0.528\\big] = P + 0.848\\,NF .\n", "\\end{aligned}\n", "\\]\n", "\n", "Setting the two equal:\n", "\n", "\\[\n", "1.06\\,NF = P + 0.848\\,NF \\quad\\Longrightarrow\\quad\n", "P = (1.06-0.848)NF = 0.212\\,NF .\n", "\\]\n", "\n", "So the pass costs **21.2 % of the rider’s original monthly expense**.\n", "\n", "---\n", "\n", "**Step 3 – Express the two monthly costs after a shift in the peak‑off‑peak mix**\n", "\n", "If the new peak‑hour proportion is \\(p\\) (the off‑peak proportion is \\(1-p\\)):\n", "\n", "\\[\n", "\\begin{aligned}\n", "\\text{No‑pass cost} &: \\; C_{\\text{no}}(p)=NF\\Big[(1-p) + 1.10p\\Big] = NF\\big[1+0.10p\\big] ,\\\\[4pt]\n", "\\text{Pass cost} &: \\; C_{\\text{yes}}(p)=P + NF\\Big[0.8(1-p)+0.88p\\Big] \\\\\n", " &: \\; =0.212\\,NF + NF\\big[0.8 + 0.08p\\big] = NF\\big[1.012 + 0.08p\\big] .\n", "\\end{aligned}\n", "\\]\n", "\n", "---\n", "\n", "**Step 4 – Impose the “10 % more expensive” requirement**\n", "\n", "We want the pass to cost **10 % more** than the no‑pass option:\n", "\n", "\\[\n", "C_{\\text{yes}}(p) = 1.10\\,C_{\\text{no}}(p).\n", "\\]\n", "\n", "Plugging the expressions:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10\\big(1 + 0.10p\\big).\n", "\\]\n", "\n", "Simplify:\n", "\n", "\\[\n", "1.012 + 0.08p = 1.10 + 0.11p \\\\\n", "\\Longrightarrow\\; 0.08p - 0.11p = 1.10 - 1.012 \\\\\n", "\\Longrightarrow\\; -0.03p = 0.088 \\\\\n", "\\Longrightarrow\\; p = -\\frac{0.088}{0.03} \\approx -2.93 .\n", "\\]\n", "\n", "---\n", "\n", "**Step 5 – Interpret the result**\n", "\n", "A peak‑hour proportion of \\(p\\approx -2.93\\) is impossible (the fraction of rides that are “peak” cannot be negative, nor can it be less than zero). \n", "\n", "What does this tell us?\n", "\n", "* Even if **all** rides are shifted to off‑peak (\\(p = 0\\)), the pass is only\n", "\n", "\\[\n", "\\frac{C_{\\text{yes}}(0)}{C_{\\text{no}}(0)} =\n", "\\frac{1.012}{1.00} = 1.012\n", "\\]\n", "\n", "i.e. **1.2 % more expensive** than going without the pass.\n", "\n", "* The maximum possible increase in cost caused by moving rides away from peak hours is therefore 1.2 %. \n", "\n", "Reaching a 10 % increase would require a negative peak‑hour proportion (or, equivalently, “more off‑peak rides than the total number of rides”), which cannot happen.\n", "\n", "---\n", "\n", "## **Answer**\n", "\n", "It is **not possible**. \n", "Even if the rider moved **all** of his/her rides from peak to off‑peak (a 100 % shift), the monthly cost with the pass would be only about **1.2 % higher** than the cost without the pass. To make the pass 10 % more expensive would require a peak‑hour proportion of \\(p \\approx -2.9\\) (i.e., a negative number of peak rides), which is mathematically impossible. Consequently, no feasible percentage of rides shifted from peak to off‑peak can produce a 10 % cost disadvantage for the pass.\n", "\n", "\n", "\n", "Now respond with the JSON with the ranked order of the competitors, nothing else. Do not include markdown formatting or code blocks.\n" ] } ], "source": [ "print(judge)" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [], "source": [ "judge_messages = [{\"role\": \"user\", \"content\": judge}]" ] }, { "cell_type": "code", "execution_count": 36, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "{\"results\": [\"4\", \"2\", \"1\", \"3\"]}\n" ] } ], "source": [ "# Judgement time!\n", "\n", "openai = OpenAI()\n", "response = openai.chat.completions.create(\n", " model=\"gpt-5-mini\",\n", " messages=judge_messages,\n", ")\n", "results = response.choices[0].message.content\n", "print(results)\n" ] }, { "cell_type": "code", "execution_count": 37, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Rank 1: openai/gpt-oss-120b\n", "Rank 2: gemini-2.5-flash\n", "Rank 3: gpt-5-nano\n", "Rank 4: deepseek-chat\n" ] } ], "source": [ "# OK let's turn this into results!\n", "\n", "results_dict = json.loads(results)\n", "ranks = results_dict[\"results\"]\n", "for index, result in enumerate(ranks):\n", " competitor = competitors[int(result)-1]\n", " print(f\"Rank {index+1}: {competitor}\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", " \n", " \n", " \n", " \n", "
\n", " \n", " \n", "

Exercise

\n", " Which pattern(s) did this use? Try updating this to add another Agentic design pattern.\n", " \n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", " \n", " \n", " \n", " \n", "
\n", " \n", " \n", "

Commercial implications

\n", " These kinds of patterns - to send a task to multiple models, and evaluate results,\n", " are common where you need to improve the quality of your LLM response. This approach can be universally applied\n", " to business projects where accuracy is critical.\n", " \n", "
" ] } ], "metadata": { "kernelspec": { "display_name": ".venv", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.12.5" } }, "nbformat": 4, "nbformat_minor": 2 }