A01_REND6289_10_IM_FM.QXD 5/7/08 12:57 PM Page iwww.elsolucionario.net REVISED Quantitative Analysis for Management www.elsolucionario.net Instructor’s Solutions Manual 5/7/08 12:57 PM www.elsolucionario.net Page ii REVISED www.elsolucionario.net A01_REND6289_10_IM_FM.QXD A01_REND6289_10_IM_FM.QXD 5/7/08 12:57 PM www.elsolucionario.net Page iii REVISED Quantitative Analysis for Management Tenth Edition Barry Render Ralph M Stair Jr Michael E Hanna Michael E Hanna University of Houston-Clear Lake Upper Saddle River, New Jersey 07458 www.elsolucionario.net Instructor’s Solutions Manual 5/7/08 12:57 PM REVISED www.elsolucionario.net Page iv VP/Editorial Director: Sally Yagan Executive Editor: Mark Pfaltzgraff Assistant Editor: Susie Abraham Editorial Assistant: Vanessa Bain Production Editor: Judy Leale Printer/Binder: Bind-Rite Graphics Copyright © 2009 by Pearson Education, Inc., Upper Saddle River, New Jersey, 07458 Pearson Prentice Hall.All rights reserved Printed in the United States of America.This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials Pearson Prentice HallTM is a trademark of Pearson Education, Inc 10 ISBN 0-13-603628-7 978-0-13-603629-8 www.elsolucionario.net A01_REND6289_10_IM_FM.QXD M01_REND6289_10_IM_C01.QXD 5/7/08 11:43 AM Page www.elsolucionario.net REVISED C H A P T E R TEACHING SUGGESTIONS Teaching Suggestion 1.1: Importance of Qualitative Factors Section 1.2 gives students an overview of quantitative analysis In this section, a number of qualitative factors, including federal legislation and new technology, are discussed Students can be asked to discuss other qualitative factors that could have an impact on quantitative analysis Waiting lines and project planning can be used as examples Teaching Suggestion 1.2: Discussing Other Quantitative Analysis Problems Section 1.2 covers an application of the quantitative analysis approach Students can be asked to describe other problems or areas that could benefit from quantitative analysis Teaching Suggestion 1.3: Discussing Conflicting Viewpoints Possible problems in the QA approach are presented in this chapter A discussion of conflicting viewpoints within the organization can help students understand this problem For example, how many people should staff a registration desk at a university? Students will want more staff to reduce waiting time, while university administrators will want less staff to save money A discussion of these types of conflicting viewpoints will help students understand some of the problems of using quantitative analysis Teaching Suggestion 1.4: Difficulty of Getting Input Data A major problem in quantitative analysis is getting proper input data Students can be asked to explain how they would get the information they need to determine inventory ordering or carrying costs Role-playing with students assuming the parts of the analyst who needs inventory costs and the instructor playing the part of a veteran inventory manager can be fun and interesting Students quickly learn that getting good data can be the most difficult part of using quantitative analysis Teaching Suggestion 1.5: Dealing with Resistance to Change Resistance to change is discussed in this chapter Students can be asked to explain how they would introduce a new system or change within the organization People resisting new approaches can be a major stumbling block to the successful implementation of quantitative analysis Students can be asked why some people may be afraid of a new inventory control or forecasting system SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 1-1 Quantitative analysis involves the use of mathematical equations or relationships in analyzing a particular problem In most cases, the results of quantitative analysis will be one or more numbers that can be used by managers and decision makers in making better decisions Calculating rates of return, financial ratios from a balance sheet and profit and loss statement, determining the number of units that must be produced in order to break even, and many similar techniques are examples of quantitative analysis Qualitative analysis involves the investigation of factors in a decision-making problem that cannot be quantified or stated in mathematical terms The state of the economy, current or pending legislation, perceptions about a potential client, and similar situations reveal the use of qualitative analysis In most decisionmaking problems, both quantitative and qualitative analysis are used In this book, however, we emphasize the techniques and approaches of quantitative analysis 1-2 Quantitative analysis is the scientific approach to managerial decision making This type of analysis is a logical and rational approach to making decisions Emotions, guesswork, and whim are not part of the quantitative analysis approach A number of organizations support the use of the scientific approach: the Institute for Operation Research and Management Science (INFORMS), Decision Sciences Institute, and Academy of Management 1-3 Quantitative analysis is a step-by-step process that allows decision makers to investigate problems using quantitative techniques The steps of the quantitative analysis process include defining the problem, developing a model, acquiring input data, developing a solution, testing the solution, analyzing the results, and implementing the results In every case, the analysis begins with defining the problem The problem could be too many stockouts, too many bad debts, or determining the products to produce that will result in the maximum profit for the organization After the problems have been defined, the next step is to develop one or more models These models could be inventory control models, models that describe the debt situation in the organization, and so on Once the models have been developed, the next step is to acquire input data In the inventory problem, for example, such factors as the annual demand, the ordering cost, and the carrying cost would be input data that are used by the model developed in the preceding step In determining the products to produce in order to maximize profits, the input data could be such things as the profitability for all the different products, the amount of time that is available at the various production departments that produce the products, and the amount of time it takes for each product to be produced in each production department The next step is developing the solution This requires manipulation of the model in order to determine the best solution Next, the results are tested, analyzed, and implemented In the inventory control www.elsolucionario.net Introduction to Quantitative Analysis 5/7/08 CHAPTER 11:43 AM Page www.elsolucionario.net INTRODUCTION TO REVISED QUANTITATIVE ANALYSIS problem, this might result in determining and implementing a policy to order a certain amount of inventory at specified intervals For the problem of determining the best products to produce, this might mean testing, analyzing, and implementing a decision to produce a certain quantity of given products 1-4 Although the formal study of quantitative analysis and the refinement of the tools and techniques of the scientific method have occurred only in the recent past, quantitative approaches to decision making have been in existence since the beginning of time In the early 1900s, Frederick W Taylor developed the principles of the scientific approach During World War II, quantitative analysis was intensified and used by the military Because of the success of these techniques during World War II, interest continued after the war 1-5 Model types include the scale model, physical model, and schematic model (which is a picture or drawing of reality) In this book, mathematical models are used to describe mathematical relationships in solving quantitative problems In this question, the student is asked to develop two mathematical models The student might develop a number of models that relate to finance, marketing, accounting, statistics, or other fields The purpose of this part of the question is to have the student develop a mathematical relationship between variables with which the student is familiar 1-6 Input data can come from company reports and documents, interviews with employees and other personnel, direct measurement, and sampling procedures For many problems, a number of different sources are required to obtain data, and in some cases it is necessary to obtain the same data from different sources in order to check the accuracy and consistency of the input data If the input data are not accurate, the results can be misleading and very costly to the organization This concept is called “garbage in, garbage out” 1-7 Implementation is the process of taking the solution and incorporating it into the company or organization This is the final step in the quantitative analysis approach, and if a good job is not done with implementation, all of the effort expended on the previous steps can be wasted 1-8 Sensitivity analysis and postoptimality analysis allow the decision maker to determine how the final solution to the problem will change when the input data or the model change This type of analysis is very important when the input data or model has not been specified properly A sensitive solution is one in which the results of the solution to the problem will change drastically or by a large amount with small changes in the data or in the model When the model is not sensitive, the results or solutions to the model will not change significantly with changes in the input data or in the model Models that are very sensitive require that the input data and the model itself be thoroughly tested to make sure that both are very accurate and consistent with the problem statement 1-9 There are a large number of quantitative terms that may not be understood by managers Examples include PERT, CPM, simulation, the Monte Carlo method, mathematical programming, EOQ, and so on The student should explain each of the four terms selected in his or her own words 1-10 Many quantitative analysts enjoy building mathematical models and solving them to find the optimal solution to a problem Others enjoy dealing with other technical aspects, for example, data analysis and collection, computer programming, or computations The implementation process can involve political aspects, convincing people to trust the new approach or solutions, or the frustrations of getting a simple answer to work in a complex environment Some people with strong analytical skills have weak interpersonal skills; since implementation challenges these “people” skills, it will not appeal to everyone If analysts become involved with users and with the implementation environment and can understand “where managers are coming from,” they can better appreciate the difficulties of implementing what they have solved using QA 1-11 Users need not become involved in technical aspects of the QA technique, but they should have an understanding of what the limitations of the model are, how it works (in a general sense), the jargon involved, and the ability to question the validity and sensitivity of an answer handed to them by an analyst 1-12 Churchman meant that sophisticated mathematical solutions and proofs can be dangerous because people may be afraid to question them Many people not want to appear ignorant and question an elaborate mathematical model; yet the entire model, its assumptions and its approach, may be incorrect 1-13 The breakeven point is the number of units that must be sold to make zero profits To compute this, we must know the selling price, the fixed cost, and the variable cost per unit 1-14 f ϭ 350 s ϭ 15 v ϭ a) Total revenue ϭ 20(15) ϭ $300 Total variable cost ϭ 20(8) ϭ $160 b) BEP ϭ f/(s Ϫ v) ϭ 350/(15 Ϫ 8) ϭ 50 units Total revenue ϭ 50(15) ϭ $750 1-15 f ϭ 150 s ϭ 50 v ϭ 20 BEP ϭ f/(s Ϫ v) ϭ 150/(50 Ϫ 20) ϭ units 1-16 f ϭ 150 s ϭ 50 v ϭ 15 BEP ϭ f/(s Ϫ v) ϭ 150/(50 Ϫ 15) ϭ 4.2 units 1-17 f ϭ 400 ϩ 1,000 ϭ 1,400 sϭ5 vϭ3 BEP ϭ f/(s Ϫ v) ϭ 1400/(5 Ϫ 3) ϭ 700 units 1-18 BEP ϭ f/(s Ϫ v) 500 ϭ 1400/(s Ϫ 3) 500(s Ϫ 3) ϭ 1400 s Ϫ ϭ 1400/500 s ϭ 2.8 ϩ s ϭ $5.80 1-19 f ϭ 2400 s ϭ 40 v ϭ 25 BEP ϭ f/(s Ϫ v) ϭ 2400/(40 Ϫ 25) ϭ 160 per week Total revenue ϭ 40(160) ϭ 6400 1-20 f ϭ 2400 s ϭ 50 v ϭ 25 BEP ϭ f/(s Ϫ v) ϭ 2400/(50 Ϫ 25) ϭ 96 per week Total revenue ϭ 50(96) ϭ 4800 1-21 f ϭ 2400 s ϭ ? v ϭ 25 BEP ϭ f/(s Ϫ v) 120 ϭ 2400/(s Ϫ 25) 120(s Ϫ 25) ϭ 2400 s ϭ 45 1-22 f ϭ 11000 s ϭ 250 v ϭ 60 BEP ϭ f/(s Ϫ v) ϭ 11000/(250 Ϫ 60) ϭ 57.9 www.elsolucionario.net M01_REND6289_10_IM_C01.QXD M01_REND6289_10_IM_C01.QXD 5/7/08 11:43 AM REVISED Page www.elsolucionario.net CHAPTER INTRODUCTION TO QUANTITATIVE ANALYSIS SOLUTION TO FOOD AND BEVERAGES AT SOUTHWESTERN UNIVERSITY FOOTBALL GAMES The total fixed cost per games includes salaries, rental fees, and cost of the workers in the six booths These are: Salaries ϭ $20,000 Rental fees ϭ 2,400 ϫ $2 ϭ $4,800 Booth worker wages ϭ ϫ ϫ ϫ $7 ϭ $1,260 Total fixed cost per game ϭ $20,000 ϩ $4,800 ϩ $1,260 ϭ $26,060 Item Soft drink Coffee Hot dogs Hamburgers Misc snacks Percent revenue Allocated fixed cost 25% 25% 20% 20% 10% $6,515 $6,515 $5,212 $5,212 $2,606 The break-even points for each of these items are found by computing the contribution to profit (profit margin) for each item and dividing this into the allocated fixed cost These are shown in the next table: Item Selling price Soft drink Coffee Hot dogs Hamburgers Misc snacks $1.50 $2.00 $2.00 $2.50 $1.00 Var cost Profit margin Percent revenue Allocated fixed cost $0.75 $0.50 $0.80 $1.00 $0.40 $0.75 $1.50 $1.20 $1.50 $0.60 25% 25% 20% 20% 10% 6515 6515 5212 5212 2606 To determine the total sales for each item that is required to break even, multiply the selling price by the break even volume The results are shown: Item Soft drink Coffee Hot dogs Hamburgers Misc snacks Total Selling price Break even volume Dollar volume of sales $1.50 $2.00 $2.00 $2.50 $1.00 8686.67 4343.33 4343.33 3474.67 4343.33 $13,030.00 $8,686.67 $8,686.67 $8,686.67 $4,343.33 $43,433.33 Thus, to break even, the total sales must be $43,433.33 If the attendance is 35,000 people, then each person would have to spend $43,433.33/35,000 ϭ $1.24 If the attendance is 60,000, then each person would have to spend $43,433.33/60,000 ϭ $0.72 Both of these are very low values, so we should be confident that this food and beverage operation will at least break even Note: While this process provides information about break-even points based on the current percent revenues for each product, there is one difficulty The total revenue using the break-even points will not result in the same percentages (dollar volume of product/total revenue) as originally stated in the problem A more complex model is available to this (see p 284 Jay Heizer and Barry Render, Operations Management, 7th ed., Upper Saddle River, NJ: Prentice Hall, 2004) Break even volume 8686.67 4343.33 4343.33 3474.67 4343.33 www.elsolucionario.net The cost of this allocated to each food item is shown in the table: M02_REND6289_10_IM_C02.QXD 5/7/08 2:31 PM REVISED Page 4www.elsolucionario.net C H A P T E R Probability Concepts and Applications Teaching Suggestion 2.1: Concept of Probabilities Ranging From to People often misuse probabilities by such statements as, “I’m 110% sure we’re going to win the big game.” The two basic rules of probability should be stressed Teaching Suggestion 2.2: Where Do Probabilities Come From? Students need to understand where probabilities come from Sometimes they are subjective and based on personal experiences Other times they are objectively based on logical observations such as the roll of a die Often, probabilities are derived from historical data—if we can assume the future will be about the same as the past Teaching Suggestion 2.3: Confusion Over Mutually Exclusive and Collectively Exhaustive Events This concept is often foggy to even the best of students—even if they just completed a course in statistics Use practical examples and drills to force the point home The table at the end of Example is especially useful Teaching Suggestion 2.4: Addition of Events That Are Not Mutually Exclusive The formula for adding events that are not mutually exclusive is P(A or B) ϭ P(A) ϩ P(B) Ϫ P(A and B) Students must understand why we subtract P(A and B) Explain that the intersect has been counted twice Teaching Suggestion 2.5: Statistical Dependence with Visual Examples Figure 2.3 indicates that an urn contains 10 balls This example works well to explain conditional probability of dependent events An even better idea is to bring 10 golf balls to class Six should be white and orange (yellow) Mark a big letter or number on each to correspond to Figure 2.3 and draw the balls from a clear bowl to make the point You can also use the props to stress how random sampling expects previous draws to be replaced Teaching Suggestion 2.6: Concept of Random Variables Students often have problems understanding the concept of random variables Instructors need to take this abstract idea and provide several examples to drive home the point Table 2.2 has some useful examples of both discrete and continuous random variables Teaching Suggestion 2.7: Expected Value of a Probability Distribution A probability distribution is often described by its mean and variance These important terms should be discussed with such practical examples as heights or weights of students But students need to be reminded that even if most of the men in class (or the United States) have heights between feet inches and feet inches, there is still some small probability of outliers Teaching Suggestion 2.8: Bell-Shaped Curve Stress how important the normal distribution is to a large number of processes in our lives (for example, filling boxes of cereal with 32 ounces of cornflakes) Each normal distribution depends on the mean and standard deviation Discuss Figures 2.8 and 2.9 to show how these relate to the shape and position of a normal distribution Teaching Suggestion 2.9: Three Symmetrical Areas Under the Normal Curve Figure 2.10 is very important, and students should be encouraged to truly comprehend the meanings of Ϯ1, 2, and standard deviation symmetrical areas They should especially know that managers often speak of 95% and 99% confidence intervals, which roughly refer to Ϯ2 and standard deviation graphs Clarify that 95% confidence is actually Ϯ1.96 standard deviations, while Ϯ3 standard deviations is actually a 99.7% spread Teaching Suggestion 2.10: Using the Normal Table to Answer Probability Questions The IQ example in Figure 2.11 is a particularly good way to treat the subject since everyone can relate to it Students are typically curious about the chances of reaching certain scores Go through at least a half-dozen examples until it’s clear that everyone can use the table Students get especially confused answering questions such as P(X Ͻ 85) since the standard normal table shows only right-hand-side Z values The symmetry requires special care ALTERNATIVE EXAMPLES Alternative Example 2.1: In the past 30 days, Roger’s Rural Roundup has sold either 8, 9, 10, or 11 lottery tickets It never sold fewer than nor more than 11 Assuming that the past is similar to the future, here are the probabilities: Sales No Days Probability 10 11 Total 10 12 30 0.333 0.400 0.200 0.067 1.000 Alternative Example 2.2: Grades received for a course have a probability based on the professor’s grading pattern Here are Professor Ernie Forman’s BA205 grades for the past five years www.elsolucionario.net TEACHING SUGGESTIONS 5/7/08 CHAPTER Outcome PROBABILITY CONCEPTS 0.25 0.30 0.35 0.03 0.02 0.05 1.00 APPLICATIONS P(Ͼ1/2 As and regular class) ϭ P(Ͼ1/2 As ͉ regular ) ϫ P(regular) ϭ (0.25)(0.50) ϭ 0.125 P(Ͼ1/2 As and advanced class) ϭ P(Ͼ1/2 As ͉ advanced) ϫ P(advanced) ϭ (0.50)(0.5) ϭ 0.25 These grades are mutually exclusive and collectively exhaustive So P(Ͼ1/2 As) ϭ 0.125 ϩ 0.25 ϭ 0.375 P (advanced and > / As) P (advanced Η > / As) ϭ P (> / As) 0.25 ϭ ϭ2 / 0.375 So there is a 66% chance the class tested was the advanced Alternative Example 2.3: P(drawing a from a deck of cards) ϭ 4/52 ϭ 1/13 P(drawing a club on the same draw) ϭ 13/52 ϭ 1/4 These are neither mutually exclusive nor collectively exhaustive P(3 or club) ϭ P(3) ϩ P(club) Ϫ P(3 and club) ϭ 4/52 ϩ 13/52 Ϫ 1/52 ϭ 16/52 ϭ 4/13 Alternative Example 2.5: A class contains 30 students Ten are female (F) and U.S citizens (U); 12 are male (M) and U.S citizens; are female and non-U.S citizens (N); are male and nonU.S citizens A name is randomly selected from the class roster and it is female What is the probability that the student is a U.S citizen? U.S Not U.S Total 10 12 22 16 14 30 P(U | F) = 10/16 Alternative Example 2.6: Your professor tells you that if you score an 85 or better on your midterm exam, there is a 90% chance you’ll get an A for the course You think you have only a 50% chance of scoring 85 or better The probability that both your score is 85 or better and you receive an A in the course is P(A and 85) ϭ P(A ͉ 85) ϫ P(85) ϭ (0.90)(0.50) ϭ 0.45 ϭ 45% Alternative Example 2.7: An instructor is teaching two sections (classes) of calculus Each class has 24 students, and on the surface, both classes appear identical One class, however, consists of students who have all taken calculus in high school The instructor has no idea which class is which She knows that the probability of at least half the class getting As on the first exam is only 25% in an average class, but 50% in a class with more math background A section is selected at random and quizzed More than half the class received As Now, what is the revised probability that the class was the advanced one? one Alternative Example 2.8: Students in a statistics class were asked how many “away” football games they expected to attend in the upcoming season The number of students responding to each possibility are shown below: Number of games Number of students 40 30 20 10 100 A probability distribution of the results would be: Number of games Probability P(X) 0.4 ϭ 40/100 0.3 ϭ 30/100 0.2 ϭ 20/100 0.1 ϭ 10/100 0.0 ϭ 0/100 1.0 ϭ 100/100 This discrete probability distribution is computed using the relative frequency approach Probabilities are shown in graph form below 0.4 Probability Alternative Example 2.4: In Alternative Example 2.3 we looked at 3s and clubs Here is the probability for or club: F M Total AND P(Ͼ1/2 As ͉ advanced class) ϭ 0.50 Probability A B C D F Withdraw/drop REVISED Page 5www.elsolucionario.net 2:31 PM 0.3 0.2 0.1 P(regular class chosen) ϭ 0.5 P(advanced class chosen) ϭ 0.5 P(Ͼ1/2 As ͉ regular class) ϭ 0.25 Possible Outcomes, x www.elsolucionario.net M02_REND6289_10_IM_C02.QXD 4/28/08 5:55 PM REVISED Pagewww.elsolucionario.net 275 MODULE 275 DYNAMIC PROGRAMMING M2-9 The distances are summarized in Table M2.1 The stages are the same stages that were used to minimize the distance The longest distance to node is 31 The shortest distance that was found in the chapter was 13 Thus, using the shortest route method can potentially save 31 Ϫ 13 ϭ 18 miles Stage M2-10 The shortest route is 1–2–5–8–9 with a total distance of 19 miles See the graph below Beginning node Longest distance to node 14 Arcs along this path 5–7 6–7 Stage Beginning node 14 Longest distance to node 24 26 18 Arcs along this path 10 4–5 5–7 3–5 5–7 2–5 5–7 19 10 12 11 Stage Beginning node Longest distance to node 31 Arcs along this path Stage 1–3 3–5 5–7 20 Stage 10 Stage Stage 14 M2-11 The solution for this shortest route problem can be seen in the following network: 10 3 3 2 11 M2-12 The optimal decision is to ship units of item 1, unit of item 2, and no units of items and www.elsolucionario.net Z02_REND6289_10_IM_MOD2.QXD Z02_REND6289_10_IM_MOD2.QXD 276 4/28/08 5:55 PM MODULE REVISED Pagewww.elsolucionario.net 276 DYNAMIC PROGRAMMING M2-13 Given the data presented in this problem, the shortest route for Leslie is the following: 1, 3, 5, 7, 10, and 12 3 11 1 3 Other optimal solutions for Problem M2-13 are: 12 a) 1, 4, 6, 9, 11, 12 b) 1, 3, 6, 9, 11, 12 c) 1, 3, 5, 8, 11, 12 10 3 Item Items to Ship Optimal Return Total 1 0 $18 0 $30 $35 M2-15 www.elsolucionario.net M2-14 Given the data presented in this problem, the following number of units should be shipped for each item: With these changes, the new shipping pattern is: Item Items to Ship Optimal Return Total 1 0 $18 0 $32 $37 As you can see, the shipping pattern is slightly different M2-16 The shortest route for this problem is 1, 3, 6, 11, 15, 17, 19, and 20 The optimal solution is shown in the following network 13 16 4 3 11 19 15 10 17 5 3 18 12 14 M2-17 The optimal solution is now 1, 3, 7, 12, 15, 17, 19, and 20 20 4/28/08 5:55 PM REVISED Pagewww.elsolucionario.net 277 MODULE viewed as a five stage process—at stage 1, an amount x1 is invested in “Standard”, at stage 2, an amount x2 is invested in the “Micro” model, and so on through stage where an amount x5 is invested in the “Network” model Even though in actuality the allocation is not made in stages, treating the situation as if it were multi-stage allows the use of the dynamic programming approach Define fn(x) as the maximum increased profit that can be realized over stages n through given that the amount not yet invested at stage n is x (Note that some texts number the stages backwards so that stage n would correspond to n allocations still to be made Since the stages are an artificiality used in this problem, it makes no difference which sequence is used) Define gn(y) as the increased profit at stage n if y is invested at that stage Then, if at the start of stage n the amount not yet invested is x, M2-18 The shortest route is 6, 11, 15, 17, 19, and 20 The total distance is 13 If the road from node to node 11 is not available, the shortest route is 6, 10, 14, 17, 19, and 20 The total distance is 16 M2-19 The optimal solution is to carry 1unit of item A and units of item C The total nutritional value is 5,100 The total weight is 18 pounds SOLUTION TO UNITED TRUCKING CASE The optimal shipping pattern is shown in the following table Item Items to Ship Optimal Return 10 Total 1 1 $20 10 11 50 20 $110 $118 fn(y͉x) ϭ gn(y) ϩ fnϩ1(x Ϫ y) is the increased profit that would be realized over stages n through if y is invested at stage n and the remaining x Ϫ y is invested optimally over stages n ϩ through Then fn(x) ϭ max fn(y͉x) ϭ max [gn(y) ϩ fnϩ1(x Ϫ y)] is the recursive relationship that allows one to start at stage and successively determine the optimum allocation for each stage—note that the maximization in this expression is taken over all y р x For stage 5, the calculations are shown in Table If there is still x5 dollars not yet invested at this last stage, the optimum is clearly to invest as much as possible at that stage; either all of it or $50,000, whichever is less Note that all profit figures in this and subsequent Tables represent thousands of dollars The indicated optimum f values are the profit increases due to investing the indicated x5 in the “Network” model Increasing the total capacity to 20 tons has a dramatic impact on the optimal decision, as can be seen in the following table Item Items to Ship Optimal Return 10 Total 1 1 10 11 $20 10 50 11 30 50 20 $190 $198 277 DYNAMIC PROGRAMMING Table Stage Calculations x5 y f5(x5) 10 20 30 40 50 Ͼ50 10 20 30 40 50 50 27 64 101 199 248 248 Proceeding to stage 4, Table summarizes the analysis To illustrate, suppose x4 ϭ 20, there is $20,000 remaining after allocations at stages 1, 2, If none is invested at stage (x4 ϭ 0) the increased profit at stage is zero and the surviving $20,000 carries over to stage where it generates $64,000 On the other hand, if $10,000 is invested at stage 4, it will increase profit by $37,000 and SOLUTION TO INTERNET CASE Briarcliff Electronics The apportionment of the $100,000 among the various models can be accomplished by means of dynamic programming This can be Table Stage Calculations x4͉y 10 20 30 40 50 f4(y͉x4) Optimal y 10 20 30 40 50 60 70 80 90 100 27 64 101 199 248 248 248 248 248 248 — 37 64 101 138 236 285 285 285 285 285 — — 59 86 123 160 258 307 307 307 307 — — — 96 123 160 197 295 344 344 344 — — — — 156 183 220 257 355 404 404 — — — — — 287 314 351 388 486 535 37 64 101 199 287 314 351 388 486 535 10 or 10 or 10 50 50 50 50 50 50 www.elsolucionario.net Z02_REND6289_10_IM_MOD2.QXD Z02_REND6289_10_IM_MOD2.QXD 278 4/28/08 MODULE 5:55 PM REVISED Pagewww.elsolucionario.net 278 DYNAMIC PROGRAMMING available The calculations for the “Standard” model are shown in Table This reveals that zero should be invested in the Standard model leaving $100,000 for stage Table shows that $10,000 should be invested in the Micro model leaving $90,000 for stage Table shows that zero should be invested in the Major model leaving $90,000 for stage Table shows that $50,000 should be invested in the Extended model leaving $40,000 for stage investment in the Network model The $10,000 investment in the Micro returns $71,000, the $50,000 investment in the Extended returns $287,000, and the $40,000 investment in the Network returns $199,000, a total return of $577,000 the remaining $10,000 will increase the stage profit by $27,000— a total of $64,000 again If all $20,000 is invested at stage 4, the increased profit is $59,000 at that stage with zero to invest at stage yielding no profit at that stage Since no other investment would be possible at stage 4, it is clear that if there is $20,000 available, either or $10,000 should be invested on the “Extended” model since y ϭ or y ϭ 10 yields the maximum f4(y͉20) The entries in the other rows are obtained in a similar manner Table 3, on the previous page, shows the equivalent stage calculations for the “Major” model and Table shows the stage calculations for the “Micro” model At stage 1, all $100,000 is x3͉y 10 20 30 40 50 f3(y͉͉x3) Optimal y 10 20 30 40 50 60 70 80 90 100 37 64 101 199 287 314 351 388 486 535 — 60 97 124 161 259 347 374 411 448 546 — — 119 156 183 220 318 406 433 470 507 — — — 151 188 215 252 350 438 465 502 — — — — 183 220 247 284 382 470 497 — — — — — 243 280 307 344 442 530 60 119 156 199 287 347 406 438 486 546 10 20 20 0 10 20 30 10 Table Stage Calculations x2͉y 10 20 30 40 50 f2(y͉͉x2) Optimal y 10 20 30 40 50 60 70 80 90 100 60 119 156 199 287 347 406 438 486 546 — 71 131 190 227 270 358 418 477 509 557 — — 92 152 211 248 291 379 439 498 530 — — — 112 172 231 268 311 399 459 518 — — — — 134 194 253 290 333 421 481 — — — — — 188 248 307 344 387 475 71 131 190 227 287 358 418 477 509 557 10 10 10 10 10 10 10 10 10 Table Stage Calculations x1͉y 10 20 30 40 50 f1(y͉͉x1) Optimal y 100 557 525 533 498 552 530 557 www.elsolucionario.net Table Stage Calculations Z03_REND6289_10_IM_MOD3.QXD 5/15/08 8:57 PM REVISED Pagewww.elsolucionario.net 279 M O D U L E Decision Theory and the Normal Distribution Teaching Suggestion M3.1: Reviewing the Normal Curve Most of the material in this supplement requires the use of the normal curve A review of the basic principles of the normal curve found in the probability chapter (Chapter 2) would be helpful before this module is started information to the number you obtained in step two EVPI will always be equal to EOL M3-5 VC/U = $16 P/U = $24 FC = $160,000 Teaching Suggestion M3.2: Covering Break-Even Analysis First Covering break-even calculations first helps students get into decision theory and normal curve analysis This material will also help students get back into the fundamental principles of normal curve theory Once break-even analysis has been mastered by students, they should be ready for the rest of the material in this module Teaching Suggestion M3.3: Spending More Time on EVPI and the Normal Distribution EVPI and the normal distribution concepts are difficult for many students You may need to spend more time on this topic and reinforce the basic steps involved Some instructors reduce coverage or eliminate this topic M = 60,000 σ = 10,000 a SOLUTIONS TO QUESTIONS AND PROBLEMS M3-1 The purpose of break-even analysis is to help a manager determine at what point overall revenue will equal overall cost It can also help the manager to determine at a certain sales volume what revenues will be generated This knowledge can assist the manager in making decisions as to whether or not to introduce a new product to the market M3-2 The normal distribution can be used in break-even analysis when sales are symmetrical around the mean expected demand and follow a bell-shaped distribution (when demand is normally distributed), and when there is only one random variable Usually, the normal distribution represents the demand for a new product M3-3 The relationship between EMV and the state of nature must be linear when you use the computations presented in Equation M3-5 in determining EMV from the mean and the standard deviation When this relationship is not linear, the approach used in computing EMV cannot be used M3-4 When EVPI is to be computed using a state of nature that follows a normal distribution, three steps are required The first step is to determine the opportunity loss function The second step is to determine the opportunity loss using the unit normal loss integral The third step is to equate the expected value of perfect BE = FC 160, 000 = = 20, 000 P / U − VC / U 24 − 16 books b EMV ϭ (P/U Ϫ VC/U)(M) Ϫ FC ϭ (24 Ϫ 16)(60,000) Ϫ 160,000 ϭ $480,000 Ϫ $160,000 ϭ $320,000 M3-6 a OLF ϭ K(BE Ϫ X) for X Յ BE OLF ϭ for X ജ BE where K ϭ (P/U Ϫ VC/U) ϭ Thus, EOL ϭ $8 (20,000 Ϫ X) for X Յ 20,000 $0 for X ജ 20,000 where X is sales in units b EOL ϭ KN(D) with K ϭ $8 ϭ 10,000 60, 000 − 20, 000 D= =4 10, 000 and N(D) ϭ 0.000007145 279 www.elsolucionario.net TEACHING SUGGESTIONS Z03_REND6289_10_IM_MOD3.QXD 280 5/15/08 MODULE 8:57 PM DECISION THEORY AND THE NORMAL DISTRIBUTION Thus, d EVPI ϭ EOL EOL ϭ (8)(10,000)(0.000007145) ϭ $0.57160 EOL ϭ K N(D) c EVPI ϭ $0.57160, since EOL ϭ EVPI 20, 000 − 60, 000 = −4 d Z = 10, 000 Kϭ6 ϭ 3,571 9, 000 − 6, 000 D= = 0.8401 3, 571 standard deviations from A Z value for ϩ4 is not found in the table, but we used 0.99997 Thus, P(profit) ϭ 0.99997 ϭ 99.99% P(loss) REVISED Pagewww.elsolucionario.net 280 ϭ 0.00003 ϭ 0.003% e The firm should print the book N(.84) ϭ 1120 Thus, EOL ϭ (6)(3571)(0.1120) ϭ $2,399.71, and Rudy should be willing to pay up to $2,399.71 for a marketing research study M3-8 FC ϭ $24,000 VC/U ϭ $8 M3-7 P/U ϭ $24 FC 24, 000 = = 1, 500 sets P / U − VC / U 24 − b If D ϭ 2,000, True Lens should produce the lenses The expected profit would be BE = Revenue ϭ (2,000 ϫ $24/set) 20% a Z= D− Thus, 0.84 = Less expenses Fixed cost 24,000 Variable cost (2,000 ϫ $8/set) 16,000 Total expenses (40,000) Profit ϭ $8,000) 20% 6,000 = 9,000 12,000 (area to the left of 12,000 ϭ 0.80; from Appendix A, Z value for 0.80 ϭ 0.84) M3-9 EMV ϭ ($28 Ϫ $20)(35,000) Ϫ $16,000 ϭ $264,000 No effect M3-10 12, 000 − 9, 000 $10(30 Ϫ x) for X Ͻ 30 where X is actual sales a OLF ϭ otherwise { b D= 0.84 ϭ 3,000 ϭ 3,571 6, 000 − 9, 000 −3, 000 = = −0.84 b Z = 3, 571 3, 571 Using Appendix A gives ϭ 0.5∴N(D) ϭ N(0.5) ϭ 0.1978 c EVPI ϭ EOL ϭ $59.34 M3-11 EOL ϭ K N(D) Z(0.84) ϭ 0.79955 K ϭ $8, $10, or $15 ϭ 30 Thus, D= P(loss) ϭ 0.20045 ϭ 20.045% P(profit) ϭ 0.79955 ϭ 79.955% ϭ ($10 Ϫ $4)(9,000) Ϫ $36,000 ϭ $54,000 Ϫ $36,000 ϭ $18,000 −X 45 − 30 = 30 EOL ϭ KN(D) ϭ $10 ϫ 30 ϫ 0.1978 ϭ $59.34 Ϫ 0.79955 ϭ 0.20045 c EMV ϭ (P/U Ϫ VC/U)() Ϫ FC ϭ $48,000 45 − 30 = 0.5 30 N(D) ϭ 0.1978 Thus, EOL if K ϭ $8 ϭ (8)(30)(0.1978) ϭ $47.47 EOL if K ϭ $10 ϭ (10)(30)(0.1978) ϭ $59.34 EOL if K ϭ $15 ϭ (15)(30)(0.1978) ϭ $89.01 Thus, as the loss per lamp increases, the expected opportunity loss increases www.elsolucionario.net a 5/15/08 8:57 PM MODULE DECISION THEORY M3-12 a New EMV ϭ ($28 Ϫ $19)(35,000) Ϫ $32,000 ϭ $283,000 AND THE b New EMV ϭ ($32 Ϫ $20)(26,000) Ϫ $16,000 ϭ $296,000 EVPI ϭ EOL ϭ KN(D) ϭ $80 ϫ 150 ϫ 0.0833 ϭ $999.60 The most Joe would be willing to pay is $999.60 M3-14 EVPI ϭ EOL ϭ K N(D) ϭ $15 ϫ 200 ϫ 0.08332 ϭ $249.96 M3-16 ϭ 750; still ϭ 200 D= EOL ϭ $15 ϫ 0.05059 ϫ 200 ϭ $151.77 M3-17 Fixed cost ϭ $4,000, Profit per job ϭ $40 Break even point ϭ 4,000/40 ϭ 100 jobs M3-18 The EVPI is equal to the minimum EOL We use the formula EOL ϭ K N(D); K ϭ 80; ϭ 15; BEP ϭ 100 N(D) ϭ 0.1; from OL tables, D ϭ 0.9 − X b 5, 000 − X b D= = Xb ϭ 4,955 50 D= EVPI ϭ EOL ϭ K N(D) ϭ $80 ϫ 15 ϫ 0.0427 ϭ 51.24 M3-15 ϭ 700 D= 750 − 700 = 200 0.25 − X b 700 − 500 = = 1.00 200 − BEP 120 − 100 = = 1.33 15 N(1.33) ϭ 0.0427 Break-even point is 4,955 pumps ∴ = 750 − 500 200 ϭ 1.25; N(D) decreased to 0.05059 $500 ϭ $100 ϫ 50 ϫ N(D) 750 − 700 ⎞ ⎛ P ( X < 750) = P ⎜ Z < ⎟⎠ = 0.60 ⎝ 281 EVPI ϭ EOL ϭ K N(D) Increase selling price − Xb 350 − 200 D= = =1 150 N(D) ϭ 0.0833 NORMAL DISTRIBUTION N(D) ϭ 0.08332 Go ahead with new process M3-13 REVISED Pagewww.elsolucionario.net 281 M3-19 If the selling price increases to $150, the profit per job would increase to $70 and the break even point would be 4,000/70 ϭ 57.14 units www.elsolucionario.net Z03_REND6289_10_IM_MOD3.QXD Z04_REND6289_10_IM_MOD4.QXD 5/2/08 3:01 PM Page www.elsolucionario.net 282 REVISED M O D U L E Game Theory Teaching Suggestion M4.1: Game Theory and Conflict This module covers zero-sum, two-person games Conflict is a part of our world Students can be asked to discuss the use of conflict analysis and game theory in corporate and political settings Teaching Suggestion M4.2: Use of Pure Strategy Games and Dominance The use of pure strategy games and dominance shows students that some strategies or alternatives can be eliminated from consideration by carefully analyzing the situation This is an important concept that can be applied to many problems in addition to game theory ALTERNATIVE EXAMPLE Alternative Example M4.1: Melinda (person A) and Stanley (person B) are involved in a competitive situation Both have two strategies (1 and 2) that they can play A table showing the winnings is presented below STANLEY (B) Melinda (A) Strategy Strategy Strategy Strategy 10 To solve this game, we determine the strategies for both players We begin with Melinda (player A) The equations are as follows: For player A—Melinda: 10Q ϩ 2(1 Ϫ Q) ϭ 1Q ϩ 7(1 Ϫ Q) 10Q ϩ Ϫ 2Q ϭ 1Q ϩ Ϫ 7Q 14Q ϭ 5; Q ϭ 5/14 ϭ 0.357 Ϫ strategy for player A—Melinda Ϫ Q ϭ 9/14; Q ϭ 0.643 Ϫ strategy for player A—Melinda For player B—Stanley: 10P ϩ 1(1 Ϫ P) ϭ 2P ϩ 7(1 Ϫ P) 10P ϩ Ϫ P ϭ 2P ϩ Ϫ 7P 14P ϭ 6; P ϭ 6/14 ϭ 0.429 Ϫ strategy for player B—Stanley Ϫ P ϭ 8/14 ϭ 0.571 Ϫ strategy for player B—Stanley organizations, governments, and so on A zero-sum game means that when one person wins the other person must lose Therefore, the sum of gains and losses for both players will always be equal to zero because when one player wins, the other player loses M4-2 The value of the game can be computed by multiplying the percentage that each player plays a given strategy times the game outcomes embodied in the table of the game Since the optimal strategies for each player are obtained by equating the expected gains of both strategies for each player, there is a shortcut method for determining the value of the game This shortcut method involves multiplying game outcomes times their probabilities of occurrence for any row or any column M4-3 A pure strategy is one in which a player will always play one strategy in the game Dominance can be used in game theory to reduce the size of the game This is done by eliminating strategies that would never be played by one of the players of the game M4-4 A strategy is dominated if there is another strategy that has outcomes at least as good as the outcomes of this first strategy Whenever a strategy is dominated, it can be eliminated from consideration because it will never be selected M4-5 A saddle point is found by finding the largest number in each column and the smallest number in each row If a number in the table has both of these characteristics, it is a saddle point M4-6 If a game has a saddle point, it will be a pure strategy game If there is no saddle point, the game is a mixed strategy game M4-7 A mixed game is one in which each player would play every strategy a given percent of the time In other words, there is no pure strategy in a mixed game A mixed game can be solved by equating a player’s expected winnings for one of the strategies with his or her expected winnings for the other opponent’s strategy M4-8 Strategy for X ϭ X2 Strategy for Y ϭ Y1 Value of game ϭ M4-9 A’s strategy ϭ A1 B’s strategy ϭ B1 Value of game ϭ 19 M4-10 X’s strategy: SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS M4-1 A two-person game is one in which only two players can participate These players could be people, companies, other 282 86Q ϩ 36(1 Ϫ Q) ϭ 42Q ϩ 106(1 Ϫ Q), 35 Qϭ 57 22 1ϪQϭ 57 www.elsolucionario.net TEACHING SUGGESTIONS 5/2/08 3:01 PM REVISED Page www.elsolucionario.net 283 MODULE Y’s strategy: 6P Ϫ 25(1 Ϫ P) ϭ Ϫ11P ϩ 30(1 Ϫ P), 55 Pϭ 72 17 1ϪPϭ 72 So strategies remain identical M4-14 The game can be reduced to a ϫ game, since X would never play X1 or X4 since X stands to lose in every eventuality under those two strategies Thus, the game is Y2 Y1 A1: A selects $5 bill X2 12 X3 12 B1: B selects $1 bill B2: A selects $20 bill B1 B2 A1 Ϫ6 25 A2 11 Ϫ30 b Strategy for A: X’s strategy: 12Q ϩ (1 Ϫ Q) ϭ 8Q ϩ 12(1 Ϫ Q) Q ϭ 2/3, Ϫ Q ϭ 1/3 Y’s strategy: 12P ϩ (1 Ϫ P) ϭ 4P ϩ 12(1 Ϫ P) P ϭ 1/3, Ϫ P ϭ 2/3 Value of game ϭ 12 ϫ 2/3 ϩ ϫ 1/3 ϭ 9.33 Ϫ6P ϩ 25(1 Ϫ P) ϭ 11P Ϫ 30(1 Ϫ P), 55 Pϭ , 72 17 1ϪPϭ 72 31 41 c Value of game ϭ Ϫ6 ϫ ϩ 11 ϫ ϭ 1.32 72 72 Since game value is positive, I’d rather be A a B1 B2 A1 Ϫ25 A2 Ϫ11 30 b A’s strategy: 6Q Ϫ 11(1 Ϫ Q) ϭ Ϫ25Q ϩ 30(1 Ϫ Q), 41 Qϭ , 72 31 1ϪQϭ 72 A1: Shoe Town does no advertising M4-15 Ϫ6Q ϩ 11(1 Ϫ Q) ϭ 25Q Ϫ 30(1 Ϫ Q), 41 Qϭ , 72 31 1ϪQϭ 72 Strategy for B: M4-13 31 41 Ϫ 11 ϫ ϭ Ϫ1.32 72 72 Since game value is negative, I’d rather be B Value of game ϭ ϫ c A2: A selects $10 bill a 283 B’s strategy: 86P ϩ 42(1 Ϫ P) ϭ 36P ϩ 106(1 Ϫ P), 32 Pϭ 57 25 1ϪPϭ 57 35 22 Value of game ϭ 86 ϫ ϩ 36 ϫ ϭ 66.70 57 57 M4-11 21Q ϩ 89(1 Ϫ Q) ϭ 116Q ϩ 3(1 Ϫ Q) 95 86 Qϭ ,1ϪQϭ 181 181 86 95 value of game ϭ ϫ 21 ϩ ϫ 89 181 181 ϭ 56.69 M4-12 GAME THEORY A2: Shoe Town invests $15,000 in advertising B1: Fancy Foot does nothing B2: Fancy Foot invests $10,000 in advertising B3: Fancy Foot invests $20,000 in advertising B1 B2 B3 A1 Ϫ2 Ϫ5 A2 Ϫ1 a This particular problem has a saddle point with strategies A2 and B3 and game value of Ϫ1 M4-16 a B2 B1 B3 A1 Ϫ2,000 Ϫ5,000 A2 3,000 1,000 Ϫ1,000 Once again we have a saddle point at A2 and B3 www.elsolucionario.net Z04_REND6289_10_IM_MOD4.QXD Z04_REND6289_10_IM_MOD4.QXD 284 5/2/08 MODULE 3:01 PM REVISED Page www.elsolucionario.net 284 GAME THEORY M4-17 The value of the game is 3.17 The optimal strategies for A and B can be computed along with the value of the game using QM for Windows The results are presented below M4-20 The best strategy for Petroleum Research is to play strategy 14 all of the time Petroleum Research can expect to get a return of $3 million from this approach These results are summarized below Mixed Strategy Mixed Strategy Probability Probability Probability Probability Probability Probability of of of of of of strategy strategy strategy strategy strategy strategy 0.390 0.244 0.366 0.000 0.000 0.000 Probability of strategy Probability of strategy Probability of strategy 0.190 0.707 0.102 For player B: Value for this game is 3.17 M4-18 Strategy X2 is dominated by both X1 and X3, so we may eliminate X2 When this is eliminated, we find that strategy Y3 is dominated by both Y1 and Y2, so we can eliminate Y3 The value of the game is M4-19 Y1 Y2 P 1ϪP Expected gain X1 Q Ϫ4 2P Ϫ 4(1 Ϫ P) X2 1ϪQ 10 6P ϩ 10(1 Ϫ P) Expected gain 2Q ϩ 6(1 Ϫ Q) Ϫ4Q ϩ 10(1 Ϫ Q) To solve this as a mixed strategy game, the expected gain for each decision should be the same Taking the expected gain for player X, we have 2P Ϫ 4(1 Ϫ P) ϭ 6P ϩ 10(1 Ϫ P) 2P Ϫ ϩ 4P ϭ 6P ϩ 10 Ϫ 10P 6P Ϫ ϭ 10 Ϫ 4P 10P ϭ 14 P ϭ 1.4 Also, Ϫ P ϭ Ϫ 1.4 ϭ Ϫ0.4 Thus, for a mixed strategy game, the probability that X would play strategy must be 1.4 which is greater than (this strategy must be played 140% of the time) This is impossible Therefore, if a person mistakenly solves a pure strategy game with the techniques of a mixed strategy game, the fraction (probability) for one of the players will be greater than one (or less than zero) For player A: Probability Probability Probability Probability Probability Probability Probability Probability Probability Probability Probability Probability Probability Probability Probability of of of of of of of of of of of of of of of strategy strategy strategy strategy strategy strategy strategy strategy strategy strategy strategy strategy strategy strategy strategy 10 11 12 13 14 15 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.000 0.000 strategy strategy strategy strategy strategy 0.000 0.615 0.385 0.000 0.000 For player B: Probability Probability Probability Probability Probability of of of of of Value for this game is 3.00 www.elsolucionario.net For player A: Z05_REND6289_10_IM_MOD5.QXD 5/15/08 10:56 AM REVISED www.elsolucionario.net Page 285 M O D U L E Mathematical Tools: Determinants and Matrices M5-7 b Value ϭ (3)(Ϫ1)(Ϫ2) ϩ (7)(2)(4) Teaching Suggestion M5.1: Why Discuss Matrices and Determinants? These mathematical tools are needed in Markov analysis to compute absorbing states In a more advanced treatment of linear programming, matrix manipulation skills are also desired Teaching Suggestion M5.2: Matrices Can Be Used to Display Data in Tabular Form As seen in Section M5-3 in this module, a matrix is also a useful tool for presenting data SOLUTIONS TO PROBLEMS M5-1 Matrices are used in Markov analysis and game theory They may also be used to represent a system of equations in linear programming M5-2 One way to calculate the determinant of a matrix is to draw the primary and secondary diagonals The numbers in each diagonal are multiplied Each of the products found on a primary diagonal are added, and each of the products found on a secondary diagonal are subtracted The resulting number is the determinant ϩ(Ϫ6)(1) (3) Ϫ (4)(Ϫ1)(Ϫ6) Ϫ (3)(2)(3) Ϫ (Ϫ2)(1)(7) ϭ ϩ 56 Ϫ 18 Ϫ 24 Ϫ 18 ϩ 14 ϭ 16 M5-8 3 ( )( 3)( ) + ( )(1)(( 3) + ( 3)( )(1) − ( 3)( 3)( 3) − (1)(1)( ) − ( )( )( ) X= = (5)( 3)( ) + ( )(1)( 3) + ( 3)( )(1) 3 − ( 3)( 3)( 3) − (1)(1)(5) − ( )( )( 2) = (5)( )( ) + ( )(1)(( 3) + ( 3)( )( 3) − ( 3)( )( 3) − ( 3)(1)(5) − ( )( )( ) Y= = 3 A matrix of cofactors is developed as follows: = Select an element in the original matrix Draw a line through the row and column of the element selected The numbers uncovered represent the cofactor for that element Calculate the value of the determinant of the cofactor −3 2 3 M5-3 Determinants are used in solving a system of equations and in finding the inverse of a matrix M5-4 a Value ϭ (6)(2) Ϫ (Ϫ5)(3) ϭ 27 www.elsolucionario.net TEACHING SUGGESTIONS (5)( 3)( 3) + ( )( )(( 3) + ( )( )(1) − ( 3)( 3)( ) − (1)( )(5) − ( 3)( )( ) Z= = 3 3 Add together the location numbers of the row and column crossed out in step If the sum is even, the sign of the determinant’s value from step does not change If the sum is an odd number, change the sign of the determinant’s value 3 = The number just computed becomes an entry in the matrix of cofactors; it is located in the same position as the element selected in step To verify, we check to be sure that 5X ϩ 2Y ϩ 3Z ϭ 4; namely, Return to step and continue until all elements in the original matrix have been replaced by their cofactor values M5-9 M5-5 The adjoint of a matrix is the transpose of the matrix of cofactors M5-6 The inverse of a matrix may be found by dividing each term in the adjoint by the determinant of the original matrix 5(Ϫ ) ϩ 2( ) ϩ 3( ) ϭ ϭ ⎡ 3⎤ ⎡ X ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢Y ⎥ = ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ Z ⎥⎦ ⎢⎣ ⎥⎦ 285 286 5/15/08 10:56 AM MODULE MATHEMATICAL TOOLS: DETERMINANTS AND M5-10 a Matrix A ϩ matrix B ⎛ 1⎞ ⎛ 5⎞ ⎛ 10 6⎞ =⎜ + = ⎝ 7⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ 9 ⎟⎠ ⎛ 5⎞ ⎛ 1⎞ − b Matrix B Ϫ matrix A = ⎜ ⎝ ⎟⎠ ⎜⎝ 7⎟⎠ Cost of dormitory job is $380, of office building job is $440, and of apartment complex job is $260 M5-14 ⎡6 ⎢ Transpose of matrix R = ⎢ ⎢2 ⎢ ⎢⎣ c Matrix C ϩ matrix D ⎛ ⎞ ⎛ ⎞ ⎛ 15 ⎞ = ⎜ ⎟ + ⎜ 6⎟ = ⎜ 11 ⎟ ⎟ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜⎝ 4⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ 12 9⎟⎠ d Matrix C ϩ matrix A: cannot be added; matrix A is (2 ϫ 3) and matrix C is (3 ϫ 3) Only matrices of the same dimension can be added or subtracted M5-11 a B ϫ C is ϫ C ϫ B is not possible c A ϫ B is not possible d M5-12 a Matrix A Matrix B Matrix C ⎛ 10⎞ ⎛ 2⎞ ⎜⎝ ⎟⎠ × ( 5) = ⎜⎝ ⎟⎠ b Matrix E Matrix F c Matrix R Matrix S Matrix F ⎛ 3⎞ ⎛ 0⎞ ⎛ + 0 + 3⎞ ⎛ 3⎞ ⎜⎝ ⎟⎠ × ⎜⎝ 1⎟⎠ = ⎜⎝ + 0 + 4⎟⎠ = ⎜⎝ 4⎟⎠ d Matrix W ⎛ 5⎞ ⎜ ⎟ × ⎛1 ⎞ ⎜ ⎟ ⎜⎝ 5⎟⎠ ⎝⎜ 4⎠⎟ ⎛ + 10 12 + 15 15 + 30 + 25 ⎞ = ⎜2 + + 10 + + ⎟ ⎜ ⎟ ⎜⎝ + 16 + 12 20 + 24 + 20⎟⎠ ⎛ 13 27 45 28 ⎞ = ⎜ 11 16 ⎟ ⎜ ⎟ ⎜⎝ 12 28 44 24⎟⎠ Job matrix Cost matrix ⎛ 50 100 10 20⎞ ⎜ 70 80 20 30 ⎟ ⎜ ⎟ ⎜⎝ 20 50 30 10 ⎟⎠ ⎛ $1⎞ ⎜ $2⎟ ⎜ ⎟ ⎜ $3⎟ ⎜ ⎟ ⎝ $5⎠ × 3⎤ ⎥ 1⎥ 2⎥ ⎥ ⎥⎦ M5-15 Element Removed Determinant of Cofactors Value of Cofactor Row 1, column = −48 Ϫ48 Row 1, column 2 = −6 ϩ6 Row 1, column = 12 12 Row 2, column = −6 ϩ6 Row 2, column = −12 Ϫ12 Row 2, column = −6 ϩ6 Row 3, column = 32 32 Row 3, column = −6 ϩ6 Row 3, column = −8 Ϫ8 Matrix G ⎛ 4⎞ ⎜ 3⎟ (5 1) × ⎜⎜ 2⎟⎟ = ( 20 + + 12 + 0) = ( 38) ⎜ ⎟ ⎝ 0⎠ ⎛ 5⎞ Transpose of matrix S = ⎜ ⎝ ⎟⎠ C ϫ D is ϫ e D ϫ C is ϫ M5-13 MATRICES ⎛ $50 + 200 + 30 + 100 ⎞ ⎛ $380⎞ = ⎜ $70 + 160 + 60 + 150⎟ = ⎜ $440⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ $20 + 100 + 90 + 50 ⎟⎠ ⎜⎝ $260⎟⎠ ⎛ 4⎞ =⎜ ⎝ −3 −7 −5 ⎟⎠ b REVISED www.elsolucionario.net Page 286 ⎛ −48 12 ⎞ Matrix of cofactors = ⎜ −12 ⎟ ⎜ ⎟ ⎜⎝ 32 −8⎟⎠ ⎛ −48 32⎞ Adjoint of matrix = ⎜ −12 ⎟ ⎜ ⎟ ⎜⎝ 12 −8⎟⎠ www.elsolucionario.net Z05_REND6289_10_IM_MOD5.QXD Z05_REND6289_10_IM_MOD5.QXD 5/15/08 10:56 AM MODULE M5-16 REVISED www.elsolucionario.net Page 287 MATHEMATICAL TOOLS: DETERMINANTS ⎛ 7⎞ Original matrix = ⎜⎜ 8⎟⎟ ⎜⎝ 9⎟⎠ Value of ϭ ϩ 96 ϩ 84 Ϫ Ϫ 48 Ϫ 72 determinant ϭ 60 AND MATRICES 287 To verify, we multiply the original matrix by its inverse An identity matrix indeed results M5-17 5X1 ϩ X2 ϭ 240 4X1 ϩ 2X2 ϭ 320 M5-18 0X1 ϩ 4X2 ϩ 3X3 ϭ 28 1X1 ϩ 2X2 ϩ 2X3 ϭ 16 ⎛ −48 32⎞ Adjoint calculation ⎜ −12 ⎟ = ⎜ ⎟ in Problem M5–7 ⎜⎝ 12 −8⎟⎠ www.elsolucionario.net ⎛ −48 / 60 / 60 32 / 60⎞ = ⎜ / 60 −12 / 60 / 60 ⎟ ⎜ ⎟ ⎜⎝ 12 / 60 / 60 −8 / 60⎟⎠ Z06_REND6289_10_IM_MOD6.QXD 5/15/08 10:57 AM REVISED www.elsolucionario.net Page 288 M O D U L E TEACHING SUGGESTIONS M6-6 Teaching Suggestion M6.1: Why Discuss Calculus? Many of the quantitative models are derived with the help of calculus The economic order quantity (EOQ) model is one example Nonlinear programming often uses calculus in finding the optimal solution to a problem Teaching Suggestion M6.2: Graphical Presentation Aids in Understanding Maxima and Minima When discussing the maximum and minimum values for a function, it is helpful to draw a graph and illustrate the problem Since the first derivative gives the slope of a tangent line, when this is set equal to 0, the result is a point where the tangent line has a slope of This means that the tangent line is a horizontal line Looking at a horizontal tangent line makes it clear that the point must be a minimum, maximum, or point of inflection a YЈЈ ϭ 12X Ϫ b YЈЈ ϭ 80X3 ϩ 12X YЉ = X c d M6-7 YЉ = 500 X6 a YЈϭ 6X5 Ϫ 0.5(2)X ϭ 6X5 Ϫ X b YЈϭ 20X3 ϩ 24X ϩ 10 −6 YЈ = X c d YЈ = −100 2X = −50 X5 a YЈЈ ϭ 30X4 Ϫ SOLUTIONS TO DISCUSSION QUESTIONS b M6-1 The slope of a straight line is found by choosing any two points on the line and dividing the change in Y by the change in X c YЈЈ ϭ 60X2 ϩ 24 24 YЉ = X M6-2 To find the slope of a nonlinear function at a particular point, we find the slope of a line that is tangent at that point To this, we add a small value (⌬X) to X, and find the resulting value of Y The slope is the ratio of the change in Y to the change in X As ⌬X approaches zero, the result is the slope d M6-3 To find the maximum or minimum of a function, we take the first derivative of this function, set it equal to zero, and solve for X If the second derivative is negative at this point, then the point is a maximum If the second derivative is positive, then the point is a minimum If the second derivative is zero, then it is a point of inflection M6-4 zero M6-8 M6-9 YЉ = 250 X6 YЈϭ 12X Ϫ Set this equal to and solve 12X Ϫ ϭ X ϭ 5/12 YЈЈ ϭ 12 which is greater than 0, so this point is a minimum M6-10 YЈϭ X2 Ϫ 10X ϩ 25 Set this equal to and solve X2 Ϫ 10X ϩ 25 ϭ (X Ϫ 5)(X Ϫ 5) ϭ A critical point is a point where the first derivative equals Xϭ5 YЈЈ ϭ 2X Ϫ 10 SOLUTIONS TO PROBLEMS M6-5 a YЈϭ 2(3)X3Ϫ1 Ϫ 3(2)X2Ϫ1 ϩ ϭ 6X2 Ϫ 6X b YЈϭ 4(5)X5Ϫ1 ϩ 2(3)X3Ϫ1 Ϫ 12(1)X1Ϫ1 ϭ 20X4 ϩ 6X2 Ϫ 12 −1(2) −2 YЈ = + = X X c d 288 YЈ = 25(−4) X 4+1 = −100 X5 If X ϭ 5, then YЈЈ ϭ 2(5) Ϫ 10 ϭ Thus, this is a point of inflection M6-11 YЈϭ 3X2 Set this equal to 3X2 ϭ 0, so X ϭ YЈЈ ϭ 6X If X ϭ 0, then YЈЈ ϭ 6(0) ϭ Thus, this is a point of inflection www.elsolucionario.net Calculus Based Optimization 5/15/08 10:57 AM MODULE M6-12 1,200 ϭ 0.5Q Q ϭ 2,400 TRЉ ϭ Ϫ0.5 This second derivative is always negative, so Q ϭ 2,400 is a maximum M6-15 Total revenue ϭ TR ϭ QP TR ϭ 75P Ϫ 2P2 To maximize total revenue, find the first derivative of TR 75 Ϫ 4P ϭ Total Revenue TR ϭ QP ϭ (180 Ϫ 2P2)P TR ϭ 180P Ϫ 2P3 TRЈϭ 180 Ϫ 6P2 Set this equal to and solve Q 2000 25 + 10 + 2000( 40) Q 50000 TC = + 5Q + 80, 000 Q −50000 TCЈ = +5 Q2 TC = = TRЈϭ 75 Ϫ 4P Set this equal to and solve M6-14 289 P = 30 = 5.48 The price that will maximize revenue is 5.48 TR ϭ (75 Ϫ 2P)P P ϭ 75/4 ϭ 18.75 At this price, Q ϭ 75 Ϫ 2(18.75) ϭ 37.5 CALCULUS BASED OPTIMIZATIONS 180 Ϫ 6P2 ϭ TRЈϭ 1,200 Ϫ 0.5Q 1,200 Ϫ 0.5Q ϭ M6-13 REVISED www.elsolucionario.net Page 289 −50000 Q2 +5= Q = 10000 = 100 M6-16 TCЉ = −50000(−2) Q2+1 For Q = 100, TCЉ = this is a minimum = 100, 000 Q3 100, 000 100 = 100, 000 = 0.1 > 0, so 1, 000, 000 www.elsolucionario.net Z06_REND6289_10_IM_MOD6.QXD ... 10 0,0 00 9 0,0 00 8 5,0 00 8 0,0 00 6 5,0 00 5 0,0 00 4 5,0 00 3 0,0 00 2 0,0 00 11 0,0 00 12 0,0 00 11 0,0 00 12 0,0 00 10 0,0 00 10 0,0 00 9 5,0 00 9 0,0 00 8 5,0 00 12 0,0 00 14 0,0 00 13 5,0 00 15 5,0 00 15 5,0 00 16 0,0 00 17 0,0 00 16 5,0 00... 10 0,0 00 9 0,0 00 8 5,0 00 8 0,0 00 6 5,0 00 5 0,0 00 4 5,0 00 3 0,0 00 2 0,0 00 11 0,0 00 12 0,0 00 11 0,0 00 12 0,0 00 10 0,0 00 10 0,0 00 9 5,0 00 9 0,0 00 8 5,0 00 12 0,0 00 14 0,0 00 13 5,0 00 15 5,0 00 15 5,0 00 16 0,0 00 17 0,0 00 16 5,0 00... 11 0,0 00 12 0,0 00 11 0,0 00 12 0,0 00 10 0,0 00 10 0,0 00 9 5,0 00 9 0,0 00 8 5,0 00 12 0,0 00 14 0,0 00 13 5,0 00 15 5,0 00 15 5,0 00 16 0,0 00 17 0,0 00 16 5,0 00 16 0,0 00 13 5,0 00 15 5,0 00 16 0,0 00 17 0,0 00 18 0,0 00 19 0,0 00 20 0,0 00