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REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 17 3 C H A P T E R Decision Analysis TEACHING SUGGESTIONS Teaching Suggestion 3.1: Using the Steps of the Decision-Making Process. The six steps used in decision theory are discussed in this chapter. Students can be asked to describe a decision they made in the last semester, such as buying a car or selecting an apartment, and describe the steps that they took. This will help in getting students involved in decision theory. It will also help them realize how this material can be useful to them in making important personal decisions. Teaching Suggestion 3.2: Importance of Defining the Problem and Listing All Possible Alternatives. Clearly defining the problem and listing the possible alternatives can be difficult. Students can be asked to do this for a typical decisionmaking problem, such as constructing a new manufacturing plant. Role-playing can be used to make this exercise more interesting. Many students get too involved in the mathematical approaches and do not pay enough attention to the importance of carefully defining the problem and considering all possible alternatives. These initial steps are important. Students need to realize that if they do not carefully define the problem and list all alternatives, most likely their analyses will be wrong. Teaching Suggestion 3.6: Decision Theory and Life-Time Decisions. This chapter investigates large and complex decisions. During one’s life, there are a few very important decisions that have a major impact. Some call these “life-time decisions.” Students can be asked to carefully consider these life-time decisions and how decision theory can be used to assist them. Life-time decisions include decisions about what school to attend, marriage, and the first job. Teaching Suggestion 3.7: Popularity of Decision Trees Among Business Executives. Stress that decision trees are not just an academic subject; they are a technique widely used by top-level managers. Everyone appreciates a graphical display of a tough problem. It clarifies issues and makes a great discussion base. Harvard business students regularly use decision trees in case analysis. Teaching Suggestion 3.8: Importance of Accurate Tree Diagrams. Developing accurate decision trees is an important part of this chapter. Students can be asked to diagram several decision situations. The decisions can come from the end-of-chapter problems, the instructor, or from student experiences. Teaching Suggestion 3.3: Categorizing Decision-Making Types. Decision-making types are discussed in this chapter; decision making under certainty, risk, and uncertainty are included. Students can be asked to describe an important decision they had to make in the past year and categorize the decision type. A good example can be a financial investment of $1,000. In-class discussion can help students realize the importance of decision theory and its potential use. Teaching Suggestion 3.9: Diagramming a Large Decision Problem Using Branches. Some students are intimidated by large and complex decision trees. To avoid this situation, students can be shown that a large decision tree is like having a number of smaller trees or decisions that can be solved separately, starting at the end branches of the tree. This can help students use decision-making techniques on larger and more complex problems. Teaching Suggestion 3.4: Starting the EVPI Concept. The material on the expected value of perfect information (EVPI) can be started with a discussion of how to place a value on information and whether or not new information should be acquired. The use of EVPI to place an upper limit on what you should pay for information is a good way to start the section on this topic. Teaching Suggestion 3.10: Using Tables to Perform Bayesian Analysis. Bayesian analysis can be difficult; the formulas can be hard to remember and use. For many, using tables is the most effective way to learn how to revise probability values. Once students understand how the tables are used, they can be shown that the formulas are making exactly the same calculations. Teaching Suggestion 3.5: Starting the Decision-Making Under Uncertainty Material. The section on decision-making under uncertainty can be started with a discussion of optimistic versus pessimistic decision makers. Students can be shown how maximax is an optimistic approach, while maximin is a pessimistic decision technique. While few people use these techniques to solve real problems, the concepts and general approaches are useful. ALTERNATIVE EXAMPLES Alternative Example 3.1: Goleb Transport George Goleb is considering the purchase of two types of industrial robots. The Rob1 (alternative 1) is a large robot capable of performing a variety of tasks, including welding and painting. The Rob2 (alternative 2) is a smaller and slower robot, but it has all the capabilities 17 REVISED M03_REND6289_10_IM_C03.QXD 18 5/7/08 3:48 PM CHAPTER 3 Page 18 DECISION ANALYSIS of Rob1. The robots will be used to perform a variety of repair operations on large industrial equipment. Of course, George can always do nothing and not buy any robots (alternative 3). The market for the repair could be either favorable (event 1) or unfavorable (event 2). George has constructed a payoff matrix showing the expected returns of each alternative and the probability of a favorable or unfavorable market. The data are presented below: Probability Alternative 1 Alternative 2 Alternative 3 EVENT 1 EVENT 2 0.6 0.4 50,000 30,000 0 40,000 20,000 0 This problem can be solved using expected monetary value. The equations are presented below: EMV (alternative 1) ($50,000)(0.6) ($40,000)(0.4) $14,000 EMV (alternative 2) ($30,000)(0.6) ($20,000)(0.4) $10,000 EMV (alternative 3) 0 The best solution is to purchase Rob1, the large robot. Alternative Example 3.2: George Goleb is not confident about the probability of a favorable or unfavorable market. (See Alternative Example 3.1.) He would like to determine the equally likely (Laplace), maximax, maximin, coefficient of realism (Hurwicz), and minimax regret decisions. The Hurwicz coefficient should be 0.7. The problem data are summarized below: EVENT 1 Probability Alternative 1 Alternative 2 Alternative 3 EVENT 2 0.6 0.4 50,000 30,000 0 40,000 20,000 0 The Laplace (equally likely) solution is computed averaging the payoffs for each alternative and choosing the best. The results are shown below. Alternatives 1 and 2 both give the highest average return of $5,000. Average (alternative 1) [$50,000 ($40,000)]/2 $5,000 Average (alternative 2) [$30,000 ($20,000)]/2 $5,000 Average (alternative 3) 0 The maximin decision (pessimistic) maximizes the minimum payoff outcome for every alternative: these are 40,000; 20,000; and 0. Therefore, the decision is to do nothing. The maximax decision (optimistic) maximizes the maximum payoff for any alternative: these maximums are 50,000; 30,000; and 0. Therefore, the decision is to purchase the large robot (alternative 1). The Hurwicz approach uses a coefficient of realism value of 0.7, and a weighted average of the best and the worst payoffs for each alternative is computed. The results are as follows: Weighted average (alternative 1) ($50,000)(0.7) ($40,000)(0.3) $23,000 Weighted average (alternative 2) ($30,000)(0.7) ($20,000)(0.3) $15,000 Weighted average (alternative 3) 0 The decision would be alternative 1. The minimax regret decision minimizes the maximum opportunity loss. The opportunity loss table for Goleb is as follows: Alternatives Rob1 Rob2 Nothing Favorable Market Unfavorable Market Maximum in Row 0 20,000 50,000 40,000 20,000 0 40,000 20,000 50,000 The alternative that minimizes the maximum opportunity loss is the Rob2. This is due to the $20,000 in the last column in the table above. Rob1 has a maximum opportunity loss of $40,000, and doing nothing has a maximum opportunity loss of $50,000. Alternative Example 3.3: George Goleb is considering the possibility of conducting a survey on the market potential for industrial equipment repair using robots. The cost of the survey is $5,000. George has developed a decision tree that shows the overall decision, as in the figure on the next page. This problem can be solved using EMV calculations. We start with the end of the tree and work toward the beginning computing EMV values. The results of the calculations are shown in the tree. The conditional payoff of the solution is $18,802. Alternative Example 3.4: George (in Alternative Example 3.3) would like to determine the expected value of sample information (EVSI). EVSI is equal to the expected value of the best decision with sample information, assuming no cost to gather it, minus the expected value of the best decision without sample information. Because the cost of the survey is $5,000, the expected value of the best decision with sample information, assuming no cost to gather it, is $23,802. The expected value of the best decision without sample information is found on the lower branch of the decision tree to be $14,000. Thus, EVSI is $9,802. Alternative Example 3.5: This example reveals how the conditional probability values for the George Goleb examples (above) have been determined. The probability values about the survey are summarized in the following table: Results of Survey Positive (P) Negative (N) Favorable Market (FM) Unfavorable Market (UM) P(P | FM) 0.9 P(N | FM) 0.1 P(P | UM) 0.2 P(N | UM) 0.8 Using the values above and the fact that P(FM) 0.6 and P(UM) 0.4, we can compute the conditional probability values of a favorable or unfavorable market given a positive or negative REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 19 CHAPTER 3 First Decision Point 19 DECISION ANALYSIS Second Decision Point $ 0 ,39 33 Favorable Market (0.871) 2 b1 Ro Favorable Market (0.871) Rob2 2) Su 1 (0 3 .6 Unfavorable Market (0.129) lts le su rab e R vo Fa –$45,000 $25,000 –$25,000 –$5,000 8) .3 (0 ey rv ts Su sul tive Re ega N ey ct urv u d tS on e C ark M $45,000 r Favorable Market (0.158) 2 80 , 8 $1 y ve Unfavorable Market (0.129) 4 Unfavorable Market (0.842) b1 Ro Favorable Market (0.158) Rob2 5 Unfavorable Market (0.842) $45,000 –$45,000 $25,000 –$25,000 –$5,000 Do $–5,000 ey rv Su ct du on tC No Favorable Market (0.60) $1 4 0 ,00 6 Unfavorable Market (0.40) b1 Ro Rob2 Favorable Market (0.60) 7 Unfavorable Market (0.40) $50,000 –$40,000 $30,000 –$20,000 $0 Figure for Alternative Example 3.3 survey result. The calculations are presented in the following two tables. Probability revision given a positive survey result State of Nature FM UM Total Conditional Probability Prior Prob. Joint Prob. Posterior Probability 0.9 0.2 0.6 0.4 0.54 0.08 0.62 0.54/0.62 0.871 0.08/0.62 0.129 1.00 Probability given a negative survey result State of Nature FM UM Total Alternative Example 3.6: In the section on utility theory, Mark Simkin used utility theory to determine his best decision. What decision would Mark make if he had the following utility values? Is Mark still a risk seeker? U($10,000) 0.8 U($0) 0.9 U($10,000) 1 Using the data above, we can determine the expected utility of each alternative as follows: U(Mark plays the game) 0.45(1) 0.55(0.8) 0.89 Conditional Probability Prior Prob. Joint Prob. Posterior Probability 0.1 0.8 0.6 0.4 0.06 0.32 0.38 0.06/0.38 0.158 0.32/0.38 0.842 1.00 U(Mark doesn’t play the game) 0.9 Thus, the best decision for Mark is not to play the game with an expected utility of 0.9. Given these data, Mark is a risk avoider. REVISED M03_REND6289_10_IM_C03.QXD 20 5/7/08 CHAPTER 3 3:48 PM Page 20 DECISION ANALYSIS SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 3-1. The purpose of this question is to make students use a personal experience to distinguish between good and bad decisions. A good decision is based on logic and all of the available information. A bad decision is one that is not based on logic and the available information. It is possible for an unfortunate or undesirable outcome to occur after a good decision has been made. It is also possible to have a favorable or desirable outcome occur after a bad decision. 3-2. The decision-making process includes the following steps: (1) define the problem, (2) list the alternatives, (3) identify the possible outcomes, (4) evaluate the consequences, (5) select an evaluation criterion, and (6) make the appropriate decision. The first four steps or procedures are common for all decision-making problems. Steps 5 and 6, however, depend on the decision-making model. 3-3. An alternative is a course of action over which we have complete control. A state of nature is an event or occurrence in which we have no control. An example of an alternative is deciding whether or not to take an umbrella to school or work on a particular day. An example of a state of nature is whether or not it will rain on a particular day. 3-4. The basic differences between decision-making models under certainty, risk, and uncertainty depend on the amount of chance or risk that is involved in the decision. A decision-making model under certainty assumes that we know with complete confidence the future outcomes. Decision-making-under-risk models assume that we do not know the outcomes for a particular decision but that we do know the probability of occurrence of those outcomes. With decision making under uncertainty, it is assumed that we do not know the outcomes that will occur, and furthermore, we do not know the probabilities that these outcomes will occur. 3-5. The techniques discussed in this chapter used to solve decision problems under uncertainty include maximax, maximin, equally likely, coefficient of realism, and minimax regret. The maximax decision-making criterion is an optimistic decision-making criterion, while the maximin is a pessimistic decision-making criterion. 3-6. For a given state of nature, opportunity loss is the difference between the payoff for a decision and the best possible payoff for that state of nature. It indicates how much better the payoff could have been for that state of nature. The minimax regret and the minimum expected opportunity loss are the criteria used with this. 3-7. Alternatives, states of nature, probabilities for all states of nature and all monetary outcomes (payoffs) are placed on the decision tree. In addition, intermediate results, such as EMVs for middle branches, can be placed on the decision tree. 3-8. Using the EMV criterion with a decision tree involves starting at the terminal branches of the tree and working toward the origin, computing expected monetary values and selecting the best alternatives. The EMVs are found by multiplying the probabilities of the states of nature times the economic consequences and summing the results for each alternative. At each decision point, the best alternative is selected. 3-9. A prior probability is one that exists before additional information is gathered. A posterior probability is one that can be computed using Bayes Theorem based on prior probabilities and additional information. 3-10. The purpose of Bayesian analysis is to determine posterior probabilities based on prior probabilities and new information. Bayesian analysis can be used in the decision-making process whenever additional information is gathered. This information can then be combined with prior probabilities in arriving at posterior probabilities. Once these posterior probabilities are computed, they can be used in the decision-making process as any other probability value. 3-11. The expected value of sample information (EVSI) is the increase in expected value that results from having sample information. It is computed as follows: EVSI (expected value with sample information) (cost of information) (expected value without sample information) 3-12. The overall purpose of utility theory is to incorporate a decision maker’s preference for risk in the decision-making process. 3-13. A utility function can be assessed in a number of different ways. A common way is to use a standard gamble. With a standard gamble, the best outcome is assigned a utility of 1, and the worst outcome is assigned a utility of 0. Then, intermediate outcomes are selected and the decision maker is given a choice between having the intermediate outcome for sure and a gamble involving the best and worst outcomes. The probability that makes the decision maker indifferent between having the intermediate outcome for sure and a gamble involving the best and worst outcomes is determined. This probability then becomes the utility of the intermediate value. This process is continued until utility values for all economic consequences are determined. These utility values are then placed on a utility curve. 3-14. When a utility curve is to be used in the decision-making process, utility values from the utility curve replace all monetary values at the terminal branches in a decision tree or in the body of a decision table. Then, expected utilities are determined in the same way as expected monetary values. The alternative with the highest expected utility is selected as the best decision. 3-15. A risk seeker is a decision maker who enjoys and seeks out risk. A risk avoider is a decision maker who avoids risk even if the potential economic payoff is higher. The utility curve for a risk seeker increases at an increasing rate. The utility curve for a risk avoider increases at a decreasing rate. 3-16. a. Decision making under uncertainty. b. Maximax criterion. c. Sub 100 because the maximum payoff for this is $300,000. Equipment Sub 100 Oiler J Texan Favorable 300,000 250,000 75,000 Row Unfavorable Maximum 200,000 100,000 18,000 300,000 250,000 75,000 Row Minimum 200,000 100,000 18,000 3-17. Using the maximin criterion, the best alternative is the Texan (see table above) because the worst payoff for this ($18,000) is better than the worst payoffs for the other decisions. 3-18. a. Decision making under risk—maximize expected monetary value. REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 21 CHAPTER 3 b. EMV (Sub 100) 0.7(300,000) 0.3(–200,000) 150,000 EMV (Oiler J) 0.7(250,000) 0.3(–100,000) 145,000 EMV (Texan) 0.7(75,000) 0.3(–18,000) 47,100 Optimal decision: Sub 100. c. Ken would change decision if EMV(Sub 100) is less than the next best EMV, which is $145,000. Let X payoff for Sub 100 in favorable market. (0.7)(X) (0.3)(200,000) 145,000 0.7X 145,000 60,000 205,000 3-22. a. Expected value with perfect information is 1,400(0.4) 900(0.4) 900(0.2) 1,100; the maximum EMV without the information is 900. Therefore, Allen should pay at most EVPI 1,100 – 900 $200. b. Yes, Allen should pay [1,100(0.4) 900(0.4) 900(0.2)] 900 $80. 3-23. X (205,000)/0.7 292,857.14 3-19. a. The expected value (EV) is computed for each alternative. Stock (Cases) EV(CDs) 0.5(23,000) 0.5(23,000) 23,000 Strong Market Fair Market Poor Market Max. Regret 0 250,000 350,000 550,000 19,000 0 29,000 129,000 310,000 100,000 32,000 0 310,000 250,000 350,000 550,000 b. Minimax regret decision is to build medium. 3-24. EV(Bonds) 0.5(30,000) 0.5(20,000) 25,000 a. Opportunity loss table Large Medium Small None The decision would change if this payoff were less than 292,857.14, so it would have to decrease by about $7,143. EV(stock market) 0.5(80,000) 0.5(20,000) 30,000 21 DECISION ANALYSIS a. Demand (Cases) 11 12 13 EMV 11 12 13 385 329 273 385 420 364 385 420 455 38512. 379.05 341.25 Probabilities 0.45 0.35 0.20 Therefore, he should invest in the stock market. b. Stock 11 cases. b. EVPI EV(with perfect information) (Maximum EV without P, I) c. If no loss is involved in excess stock, the recommended course of action is to stock 13 cases and to replenish stock to this level each week. This follows from the following decision table. [0.5(80,000) 0.5(23,000)] 30,000 51,500 30,000 21,500 Thus, the most that should be paid is $21,500. 3-20. Stock (Cases) The opportunity loss table is Alternative Stock Market Bonds CDs Good Economy Poor Economy 0 50,000 57,000 43,000 3,000 0 11 12 13 EOL(Bonds) 0.5(50,000) 0.5(3,000) 26,500 EOL(CDs) 0.5(57,000) 0.5(0) 28,500 a. Alternative Market Condition Good Fair Stock market 1,400 Bank deposit 900 Probabilities of market conditions 0.4 Poor EMV 800 0 880 900 900 900 0.4 0.2 b. Best decision: deposit $10,000 in bank. 11 12 13 EMV 385 385 385 385 420 420 385 420 455 385 404.25 411.25 3-25. EOL(Stock Market) 0.5(0) 0.5(43,000) 21,500* This minimizes EOL. 3-21. Demand (Cases) Manufacture (Cases) Demand (Cases) 6 7 8 9 Probabilities 6 7 8 9 EMV 300 255 210 165 300 350 305 260 300 350 400 355 300 350 400 450 300 340.5 352.5 317 0.1 0.3 0.5 0.1 John should manufacture 8 cases of cheese spread. 3-26. Cost of produced case $5. Cost of purchased case $16. Selling price $15. REVISED M03_REND6289_10_IM_C03.QXD 22 5/7/08 3:48 PM CHAPTER 3 Page 22 DECISION ANALYSIS Money recovered from each unsold case $3. Supply (Cases) Demand (Cases) 100 200 300 EMV 100 100(15) 100(5) 1000 100(15) 100(3) 200(5) 800 100(15) 200(3) 300(5) 600 300(15) 100(5) 200(16) 800 300(15) 200(5) 100(16) 1900 300(15) 300(5) 3000 900 200 200(15) 100(5) 100(16) 900 200(15) 200(5) 2000 300 Probabilities 200(15) 100(3) 300(5) 1800 0.3 0.4 0.3 b. Produce 300 cases each day. 3-27. a. The table presented is a decision table. The basis for the decisions in the following questions is shown in the table below. MARKET Decision Alternatives Small Medium Large Very Large Good Fair 50,000 80,000 100,000 300,000 20,000 30,000 30,000 25,000 Poor MAXIMIN Row Maximum Row Minimum Row Average 50,000 80,000 100,000 300,000 10,000 20,000 40,000 160,000 20,000 30,000 30,000 55,000 10,000 20,000 40,000 160,000 b. Maximax decision: Very large station. c. Maximin decision: Small station. d. Equally likely decision: Very large station. e. Criterion of realism decision: Very large station. f. Opportunity loss table: MARKET MINIMAX Decision Alternatives Good Market Fair Market Poor Market Row Maximum Small Medium Large Very Large 250,000 220,000 200,000 0 10,000 0 0 5,000 0 10,000 30,000 150,000 250,000 220,000 200,000 150,000 Minimax regret decision: Very large station. 3-28. EMV for node 1 0.5(100,000) 0.5(40,000) $30,000. Choose the highest EMV, therefore construct the clinic. Payoff Favorable Market (0.5) ct tru ns nic o C Cli $100,000 1 Unfavorable Market (0.5) –$40,000 $30,000 Do N oth i ng EMV for no clinic is $0 EQUALLY LIKELY MAXIMAX $0 CRIT. OF REALISM Weighted Average 38,000 60,000 72,000 208,000 1610 1800 REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 23 CHAPTER 3 3-29. DECISION ANALYSIS a. Payoff Favorable Market (0.82) CONSTRUCT 2 Unfavorable Market (0.18) 5) ey .5 rv e (0 $69,800 u S bl a r vo Fa DO NOT CONSTRUCT 1 t y uc ve $36,140 nd Sur o C et k ar M –$45,000 –$5,000 Favorable Market (0.11) CONSTRUCT $95,000 5) ey .4 rv (0 Su tive ga Ne 3 Unfavorable Market (0.89) $95,000 –$45,000 –$5,000 $36,140 –$5,000 DO NOT CONSTRUCT Do tC No on du ct Favorable Market (0.5) Su rv CONSTRUCT CLINIC ey 4 Unfavorable Market (0.5) $100,000 –$40,000 $30,000 DO NOT CONSTRUCT b. EMV(node 2) (0.82)($95,000) (0.18)(–$45,000) 77,900 8,100 $69,800 EMV(node 3) (0.11)($95,000) (0.89)(–$45,000) 10,450 $40,050 –$29,600 EMV(node 4) $30,000 EMV(node 1) (0.55)($69,800) (0.45)(–$5,000) 38,390 2,250 $36,140 The EMV for using the survey $36,140. EMV(no survey) (0.5)($100,000) (0.5)(–$40,000) $30,000 The survey should be used. c. EVSI ($36,140 $5,000) $30,000 $11,140. Thus, the physicians would pay up to $11,140 for the survey. $0 23 REVISED M03_REND6289_10_IM_C03.QXD 24 5/7/08 CHAPTER 3 3:48 PM Page 24 DECISION ANALYSIS 3-30. Favorable Market Large Shop 2 Unfavorable Market No Shop e bl ra o y v e Fa urv S Favorable Market Small Shop 3 Unfavorable Market 1 et k ar ey M urv S N U Su nfa rv vo ey ra b Favorable Market Large Shop 4 le Unfavorable Market No Shop Favorable Market o Small Shop Su 5 rv ey Unfavorable Market Favorable Market Large Shop 6 Unfavorable Market No Shop Favorable Market Small Shop 7 3-31. a. EMV(node 2) (0.9)(55,000) (0.1)(–$45,000) 49,500 4,500 $45,000 EMV(node 3) (0.9)(25,000) (0.1)(–15,000) 22,500 1,500 $21,000 EMV(node 4) (0.12)(55,000) (0.88)(–45,000) 6,600 39,600 –$33,000 EMV(node 5) (0.12)(25,000) (0.88)(–15,000) 3,000 13,200 –$10,200 EMV(node 6) (0.5)(60,000) (0.5)(–40,000) 30,000 20,000 $10,000 EMV(node 7) (0.5)(30,000) (0.5)(–10,000) 15,000 5,000 $10,000 EMV(node 1) (0.6)(45,000) (0.4)(–5,000) 27,000 2,000 $25,000 Since EMV(market survey) > EMV(no survey), Jerry should conduct the survey. Since EMV(large shop | favorable survey) is larger than both EMV(small shop | favorable survey) and EMV(no shop | favorable survey), Jerry should build a large shop if the survey is favorable. If the survey is unfavorable, Jerry should build nothing since EMV(no shop | unfavorable survey) is larger than both EMV(large shop | unfavorable survey) and EMV(small shop | unfavorable survey). Unfavorable Market REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 25 CHAPTER 3 $45,000 DECISION ANALYSIS Favorable Market (0.9) Large Shop 2 Unfavorable Market (0.1) $45,000 Payoff $55,000 –$45,000 No Shop –$5,000 le 6) ab 0. or y ( v e Fa urv S $21,000 Favorable Market (0.9) Small Shop 3 $25,000 Unfavorable Market (0.1) $25,000 –$15,000 1 t ke ar ey M urv S U Su nfa rv vo ey ra (0 ble .4 ) –$33,000 Favorable Market (0.12) $55,000 Large Shop 4 Unfavorable Market (0.88) –$5,000 –$45,000 No Shop –$5,000 –$10,200 N o Favorable Market (0.12) $25,000 Small Shop Su 5 rv ey $10,000 Unfavorable Market (0.88) Favorable Market (0.5) Large Shop 6 Unfavorable Market (0.5) $10,000 –$15,000 $60,000 –$40,000 No Shop $0 $10,000 Favorable Market (0.5) $30,000 Small Shop 7 b. If no survey, EMV 0.5(30,000) 0.5(–10,000) $10,000. To keep Jerry from changing decisions, the following must be true: EMV(survey) ≥ EMV(no survey) Let P probability of a favorable survey. Then, P[EMV(favorable survey)] (1 P) [EMV(unfavorable survey)] ≥ EMV(no survey) This becomes: P(45,000) (1 P)(–5,000) ≥ $10,000 Solving gives 45,000P 5,000 5,000P ≥ 10,000 50,000P ≥ 15,000 P ≥ 0.3 Thus, the probability of a favorable survey could be as low as 0.3. Since the marketing professor estimated the probability at 0.6, the value can decrease by 0.3 without causing Jerry to change his decision. Jerry’s decision is not very sensitive to this probability value. Unfavorable Market (0.5) –$10,000 25 REVISED M03_REND6289_10_IM_C03.QXD 26 5/7/08 3:48 PM CHAPTER 3 Page 26 DECISION ANALYSIS 3-32. Payoff $8,500 (0.9) 2 (0.1) $500 (0.9) 3 (0.1) A3 $8,500 $2,750 e or M on r i e t A 1 ath ma G for In A4 ) n 5 i o 0. at le ( rm b fo ra In avo F A5 $12,000 –$23,000 $2,000 –$13,000 –$3,000 1 In U for nf m av at or ion ab le (0 . –$9,000 (0.4) 4 (0.6) –$7,000 (0.4) 5 (0.6) A3 –$3,000 A4 5) A5 r n he io at at G rm ot fo A 2 o N e In D or M A4 –$23,000 $2,000 –$13,000 –$3,000 $4,500 (0.7) 6 (0.3) $500 (0.7) 7 (0.3) A3 $4,500 $12,000 $15,000 –$20,000 $5,000 –$10,000 A5 A1: gather more information $0 P(S1) 0.5; P(S2) 0.5 A2: do not gather more information P(I1 | S1) 0.8; P(I2 | S1) 0.2 A3: build quadplex P(I1 | S2) 0.3; P(I2 | S2) 0.7 A4: build duplex a. P(successful store | favorable research) P(S1 | I1) A5: do nothing EMV(node 2) 0.9(12,000) 0.1(23,000) 8,500 P ( S1 | I1 ) = EMV(node 3) 0.9(2,000) 0.1(13,000) 500 EMV(get information and then do nothing) 3,000 EMV(node 4) 0.4(12,000) 0.6(23,000) 9,000 EMV(node 5) 0.4(2,000) 0.6(13,000) 7,000 EMV(get information and then do nothing) 3,000 EMV(node 1) 0.5(8,500) 0.5(-3,000) 2,750 EMV(build quadplex) 0.7(15,000) 0.3(20,000) 4,500 EMV(build duplex) 0.7(5,000) 0.3(10,000) 500 EMV(do nothing) 0 Decisions: do not gather information; build quadplex. 3-33. I1: favorable research or information P ( I1 | S1 ) P ( S1 ) P ( I1 | S1 ) P ( S1 ) + P ( I1 | S2 ) P ( S2 ) 0.8(0.5) = 0.73 0.8(0.5) + 0.3(0.5) P ( S1 | I1 ) = b. P(successful store | unfavorable research) P(S1 | I2) P ( S1 | I 2 ) = P ( I 2 | S1 ) P ( S1 ) P ( I 2 | S1 ) P ( S1 ) + P ( I 2 | S2 ) P ( S2 ) P ( S1 | I 2 ) = c. Now P(S1) 0.6 and P(S2) 0.4 P ( S1 | I1 ) = 0.8(0.6 ) = 0.8 0.8(0.6 ) + 0.3(0.4 ) P ( S1 | I 2 ) = 0.2(0.6 ) = 0.3 0.2(0.6 ) + 0.7(0.4 ) I2: unfavorable research S1: store successful S2: store unsuccessful 0.2(0.5) = 0.22 0.2(0.5) + 0.7(0.5) REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 27 CHAPTER 3 3-34. I1: favorable survey or information EMV(B) 6,000(0.2) 4,000(0.3) 0(0.5) 2,400 S1: facility successful S2: facility unsuccessful Fund B should be selected. P(S1) 0.3; P(S2) 0.7 c. Let X payout for Fund A in a good economy. P(I1 | S1) 0.8; P(I2 | S1) 0.2 EMV(A) EMV(B) P(I1 | S2) 0.3; P(I2 | S2) 0.7 X(0.2) 2,000(0.3) (–5,000)(0.5) 2,400 P(successful facility | favorable survey) P(S1 | I1) P ( S1 | I1 ) = P ( I1 | S1 ) P ( S1 ) P ( I1 | S1 ) P ( S1 ) + P ( I1 | S2 ) P ( S2 ) P ( S1 | I1 ) = 0.8(0.3) = 0.533 0.8(0.3) + 0.3(0.7) 0.2X 4,300 X 4,300/0.2 21,500 Therefore, the return would have to be $21,500 for Fund A in a good economy for the two alternatives to be equally desirable based on the expected values. P(successful facility | unfavorable survey) P(S1 | I2) P ( S1 | I 2 ) = P ( I 2 | S1 ) P ( S1 ) P ( I 2 | S1 ) P ( S1 ) + P ( I 2 | S2 ) P ( S2 ) P ( S1 | I 2 ) = 0.2(0.3) = 0.109 0.2(0.3) + 0.7(0.7) a. Good economy 0.2 Fair economy 0.3 10,000 2,000 Fund A Poor economy 0.5 Good economy 0.2 Fund B 27 b. EMV(A) 10,000(0.2) 2,000(0.3) (5,000)(0.5) 100 I2: unfavorable survey 3-35. DECISION ANALYSIS Fair economy 0.3 Poor economy 0.5 –5,000 6,000 4,000 0 REVISED M03_REND6289_10_IM_C03.QXD 28 5/7/08 3:48 PM CHAPTER 3 3-36. Page 28 DECISION ANALYSIS a. Payoff Favorable Market Survey Favorable Produce Razor 3 Do Not Produce Razor Favorable Market 1 Survey Unfavorable Produce Razor 4 Unfavorable Market Do Not Produce Razor tS urv ey Unfavorable Market Co nd uc Favorable Market Study Favorable Conduct Pilot Study Produce Razor 5 Unfavorable Market Do Not Produce Razor Favorable Market 2 Ne Unfavorable Do Not Produce Razor 6 Unfavorable Market er ith Study Produce Razor Te st Favorable Market Produce Razor 7 Unfavorable Market Do Not Produce Razor b. $95,000 –$65,000 –$5,000 $95,000 –$65,000 –$5,000 $80,000 –$80,000 –$20,000 $80,000 –$80,000 –$20,000 $100,000 –$60,000 $0 S1: survey favorable EMV(node 5) 80,000(0.89) (80,000)(0.11) 62,400 S2: survey unfavorable EMV(node 6) 80,000(0.18) (80,000)(0.82) 51,200 EMV(node 7) 100,000(0.5) (60,000)(0.5) 20,000 EMV(conduct survey) 59,800(0.45) (–5,000)(0.55) 24,160 EMV(conduct pilot study) 62,400(0.45) (20,000)(0.55) 17,080 EMV(neither) 20,000 S3: study favorable S4: study unfavorable S5: market favorable S6: market unfavorable P ( S5 | S1 ) = 0.7(0.5) = 0.78 0.7(0.5) + 0.2(0.5) P(S6 | S1) 1 – 0.778 0.222 P ( S5 | S2 ) = 0.3(0.5) = 0.27 0.3(0.5) + 0.8(0.5) P(S6 | S2) 1 – 0.27 0.73 P ( S5 | S3 ) = 0.8(0.5) = 0.89 0.8(0.5) + 0.1(0.5) P(S6 | S3) 1 – 0.89 0.11 P ( S5 | S 4 ) = 0.2(0.5) = 0.18 0.2(0.5) + 0.9(0.5) P(S6 | S4) 1 – 0.18 0.82 c. EMV(node 3) 95,000(0.78) (65,000)(0.22) 59,800 EMV(node 4) 95,000(0.27) (65,000)(0.73) 21,800 Therefore, the best decision is to conduct the survey. If it is favorable, produce the razor. If it is unfavorable, do not produce the razor. 3-37. The following computations are for the decision tree that follows. EU(node 3) 0.95(0.78) 0.5(0.22) 0.85 EU(node 4) 0.95(0.27) 0.5(0.73) 0.62 EU(node 5) 0.9(0.89) 0(0.11) 0.80 EU(node 6) 0.9(0.18) 0(0.82) 0.16 EU(node 7) 1(0.5) 0.55(0.5) 0.78 EU(conduct survey) 0.85(0.45) 0.8(0.55) 0.823 EU(conduct pilot study) 0.80(0.45) 0.7(0.55) 0.745 EU(neither test) 0.81 Therefore, the best decision is to conduct the survey. Jim is a risk avoider. REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 29 CHAPTER 3 DECISION ANALYSIS 29 Utility 0.85 Survey 0.82 Favorable (0.45) Produce Razor 0.62 ey Survey urv Unfavorable (0.55) Produce Razor 4 uc nd Co Conduct 0.745 Pilot 2 Study Favorable (0.45) Produce Razor 5 Study Ne er ith Unfavorable (0.55) Market Unfavorable (0.73) Produce Razor 6 Market Favorable (0.89) Market Unfavorable (0.11) Te st 0.78 7 Market Unfavorable (0.82) a. P(good economy | prediction of good economy) 0.8(0.6 ) = 0.923 0.8(0.6 ) + 0.1(0.4 ) P(poor economy | prediction of good economy) 0.1(0.4 ) = 0.077 0.8(0.6 ) + 0.1(0.4 ) P(good economy | prediction of poor economy) 0.2(0.6 ) = 0.25 0.2(0.6 ) + 0.9(0.4 ) P(poor economy | prediction of poor economy) 0.9(0.6 ) = 0.75 0.2(0.6 ) + 0.9(0.4 ) b. P(good economy | prediction of good economy) 0.8(0.7) = 0.949 0.8(0.7) + 0.1(0.3) P(poor economy | prediction of good economy) 0.1(0.3) = 0.051 0.8(0.7) + 0.1(0.3) P(good economy | prediction of poor economy) 0.2(0.7) = 0.341 0.2(0.7) + 0.9(0.3) 0.5 0.9 0 0.9 0 0.7 Market Favorable (0.5) Market Unfavorable (0.5) Do Not Produce Razor 3-38. 0.95 0.7 Market Favorable (0.18) Do Not Produce Razor Produce Razor 0.5 0.8 Do Not Produce Razor 0.16 0.95 0.8 Market Favorable (0.27) Do Not Produce Razor 0.80 Study Market Unfavorable (0.22) Do Not Produce Razor 1 tS 3 Market Favorable (0.78) 1 0.55 0.81 P(poor economy | prediction of poor economy) 0.9(0.3) = 0.659 0.2(0.7) + 0.9(0.3) 3-39. The expected value of the payout by the insurance company is EV 0(0.999) 100,000(0.001) 100 The expected payout by the insurance company is $100, but the policy costs $200, so the net gain for the individual buying this policy is negative (–$100). Thus, buying the policy does not maximize EMV since not buying this policy would have an EMV of 0, which is better than –$100. However, a person who buys this policy would be maximizing the expected utility. The peace of mind that goes along with the insurance policy has a relatively high utility. A person who buys insurance would be a risk avoider. REVISED M03_REND6289_10_IM_C03.QXD 30 5/7/08 3:48 PM CHAPTER 3 Page 30 DECISION ANALYSIS 3-40. Survey Favorable (0.55) U = 0.76 ct du t 1 n e Co ark M Survey o D Unfavorable (0.45) U = 0.8118 Construct 2 Clinic Favorable Market (0.82) Unfavorable Market (0.18) Do Not Construct Clinic U = 0.1089 Construct 3 Clinic Favorable Market (0.11) Unfavorable Market (0.89) Do Not Construct Clinic c du on C y ot rve Su N U = 0.55 t Construct Clinic 4 Favorable Market (0.5) Unfavorable Market (0.5) Payoff Utility $95,000 0.99 –$45,000 0 –$5,000 0.7 $95,000 0.99 –$45,000 0 –$5,000 0.7 $100,000 1.0 –$40,000 0.1 $0 0.9 Do Not Construct Clinic EU(node 2) (0.82)(0.99) (0.18)(0) 0.8118 b. Expected utility on Broad Street 0.2(0.5) 0.9(0.5) 0.55. Therefore, the expressway maximizes expected utility. EU(node 3) (0.11)(0.99) (0.89)(0) 0.1089 EU(node 4) 0.5(1) 0.5(0.1) 0.55 c. Lynn is a risk avoider. EU(node 1) (0.55)(0.8118) (0.45)(0.7000) 0.7615 EU(no survey) 0.9 1.0 The expected utility with no survey (0.9) is higher than the expected utility with a survey (0.7615), so the survey should be not used. The medical professionals are risk avoiders. EU(large plant | survey favorable) 0.78(0.95) 0.22(0) 0.741 EU(small plant | survey favorable) 0.78(0.5) 0.22(0.1) 0.412 EU(no plant | survey favorable) 0.2 Utility 3-41. 0.8 0.6 0.4 0.2 0 EU(large plant | survey negative) 0.27(0.95) 0.73(0) 0.2565 EU(small plant | survey negative) 0.27(0.5) 0.73(0.10) 0.208 EU(no plant | survey negative) 0.2 EU(large plant | no survey) 0.5(1) 0.5(0.05) 0.525 EU(small plant | no survey) 0.5(0.6) 0.5(0.15) 0.375 EU(no plant | no survey) 0.3 0 10 20 30 40 Time (minutes) 3-43. Selling price $20 per gallon; manufacturing cost $12 per gallon; salvage value $13; handling costs $1 per gallon; and advertising costs $3 per gallon. From this information, we get: marginal profit selling price minus the manufacturing, handling, and advertising costs EU(conduct survey) 0.45(0.741) 0.55(0.2565) 0.4745 marginal profit $20 $12 $1 $3 $4 per gallon EU(no survey) 0.525 If more is produced than is needed, a marginal loss is incurred. John’s decision would change. He would not conduct the survey and build the large plant. 3-42. a. Expected travel time on Broad Street 40(0.5) 15(0.5) 27.5 minutes. Broad Street has a lower expected travel time. Expressway Broad Street 30 Minutes, U = 0.7 1 Congestion (0.5) 40 Minutes, U = 0.2 No Congestion (0.5) 15 Minutes, U = 0.9 marginal loss $13 $12 $1 $3 $3 per gallon In addition, there is also a shortage cost. Coren has agreed to fulfill any demand that cannot be met internally. This requires that Coren purchase chemicals from an outside company. Because the cost of obtaining the chemical from the outside company is $25 and the price charged by Coren is $20, this results in shortage cost $5 per gallon In other words, Coren will lose $5 for every gallon that is sold that has to be purchased from an outside company due to a shortage. REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 31 CHAPTER 3 31 DECISION ANALYSIS a. A decision tree is shown below: Decision Tree (0.2) Demand 500 $2,000 = (500)(4) (0.3) 1,000 –$1,500 (0.4) 1,500 –$3,000 = (500)(4) – (1,000)(5) Stock 500 (0.1) 2,000 –$5,500 = (500)(4) – (1,500)(5) (0.2) 500 (0.3) 1,000 $4,000 = (1,000)(4) (0.4) 1,500 $1,500 = (1,000)(4) – (5)(500) (0.1) 2,000 –$1,000 = (1,000)(4) – (5)(1,000) (0.2) 500 (0.3) 1,000 (0.4) 1,500 $6,000 = (1,500)(4) (0.1) 2,000 $3,500 = (1,500)(4) – (5)(500) (0.2) 500 (0.3) 1,000 (0.4) 1,500 (0.1) 2,000 $1,800 Stock 1,000 $3,300 Stock 1,500 Stock 2,000 $2,400 –$500 = (500)(4) – (500)(5) $500 = (500)(4) – (500)(3) –$1,000 = (500)(4) – (3)(1,000) $2,500 = (1,000)(4) – (3)(500) –$2,500 = (500)(4) – (3)(1,500) $1,000 = (1,000)(4) – (3)(1,000) $4,500 = (1,500)(4) – (3)(500) $8,000 = (2,000)(4) b. The computations are shown in the following table. These numbers are entered into the tree above. The best decision is to stock 1,500 gallons. Decision Tree–No Survey (0.15) (0.40) Table for Problem 3-43 00 0,0 Demand Stock 500 500 2,000 1,000 500 1,500 1,000 2,000 2,500 Maximum 2,000 Probabilities 0.2 1,000 1,500 2,000 500 3,000 5,500 4,000 1,500 1,000 2,500 6,000 3,500 1,000 4,500 8,000 4,000 6,000 8,000 0.3 0.4 EMV $1,500 $1,800 $3,300 $2,400 $4,800 EVwPI 0.1 c. EVwPI (0.2)(2,000) (0.3)(4,000) (0.4)(6,000) (0.1)(8,000) $4,800 EVPI EVwPI EMV $4,800 $3,300 $1,500 3-44. If no survey is to be conducted, the decision tree is fairly straightforward. There are three main decisions, which are building a small, medium, or large facility. Extending from these decision branches are three possible demands, representing the possible states of nature. The demand for this type of facility could be either low (L), medium (M), or high (H). It was given in the problem that the probability for a low demand is 0.15. The probabilities for a medium and a high demand are 0.40 and 0.45, respectively. The problem also gave monetary consequences for building a small, medium, or large facility when the demand could be low, medium, or high for the facility. These data are reflected in the following decision tree. 50 ll $ a Sm (0.45) (0.15) Medium $670,000 (0.40) La (0.45) rge $5 80 ,00 0 (0.15) (0.40) $500,000 $500,000 $500,000 $200,000 $700,000 $800,000 –$200,000 $400,000 (0.45) $1,000,000 With no survey, we have: EMV(Small) 500,000; EMV(Medium) 670,000; and EMV(Large) 580,000. The medium facility, with an expected monetary value of $670,000, is selected because it represents the highest expected monetary value. If the survey is used, we must compute the revised probabilities using Bayes’ theorem. For each alternative facility, three revised probabilities must be computed, representing low, medium, and high demand for a facility. These probabilities can be computed using tables. One table is used to compute the probabilities for low survey results, another table is used for REVISED M03_REND6289_10_IM_C03.QXD 32 5/7/08 CHAPTER 3 3:48 PM Page 32 DECISION ANALYSIS medium survey results, and a final table is used for high survey results. These tables are shown below. These probabilities will be used in the decision tree that follows. Decision Tree–Survey For low survey results—A1: State of Nature B1 B2 B3 P(Bi) 0.150 0.400 0.450 Small P(Ai | Bj) P(Bj and Ai) 0.700 0.400 0.100 P(A1) 0.105 0.160 0.045 0.310 P(Bj | Ai) 0.339 0.516 0.145 0.150 0.400 0.450 P(Ai | Bj) B1 B2 B3 P(Bi) P(Bj and Ai) 0.200 0.500 0.300 P(A2) 0.150 0.400 0.450 P(Ai | Bj) 450,000 Medium M H 0.030 0.200 0.135 0.365 P(Bj and Ai) 0.100 0.100 0.600 P(A3) 0.015 0.040 0.270 0.325 L P(Bj | Ai) 0.082 0.548 0.370 For high survey results—A3: State of Nature M L Large $49 Low 5,000 (0.3 10) B1 B2 B3 P(Bi) 450,000 H For medium survey results—A2: State of Nature L H L Small P(Bj | Ai) 0.046 0.123 0.831 EMV(with Survey) 0.310(495,000) 0.365(646,000) 0.325(821,000) 656,065 Because the expected monetary value for not conducting the survey is greater (670,000), the decision is not to conduct the survey and to build the medium-sized facility. M H L $646,000 Medium Medium (0.365) M H L Large ,000 $821 .325) (0 High When survey results are low, the probabilities are P(L) 0.339; P(M) 0.516; and P(H) 0.145. This results in EMV(Small) 450,000; EMV(Medium) 495,000; and EMV(Large) 233,600. When survey results are medium, the probabilities are P(L) 0.082; P(M) 0.548; and P(H) 0.378. This results in EMV (Small) 450,000; EMV(Medium) 646,000; and EMV(Large) 522,800. When survey results are high, the probabilities are P(L) 0.046; P(M) 0.123; and P(H) 0.831. This results in EMV(Small) 450,000; EMV(Medium) 710,100; and EMV(Large) 821,000. If the survey results are low, the best decision is to build the medium facility with an expected return of $495,000. If the survey results are medium, the best decision is also to build the medium plant with an expected return of $646,000. On the other hand, if the survey results are high, the best decision is to build the large facility with an expected monetary value of $821,000. The expected value of using the survey is computed as follows: M M H L Small M 450,000 150,000 650,000 750,000 –250,000 350,000 950,000 450,000 450,000 450,000 150,000 650,000 750,000 –250,000 350,000 950,000 450,000 450,000 H 450,000 L Medium M H L Large M H 150,000 650,000 750,000 –250,000 350,000 950,000 REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 33 CHAPTER 3 3-45. DECISION ANALYSIS a. Payoff $75,000 Succeed (0.5) $250,000 1 Do w w nto Don’t Succeed (0.5) n $140,000 Mall Succeed (0.6) –$100,000 $300,000 2 Don’t Succeed (0.4) Tra ffic Cir cle No Gr oc er yS to $250,000 re Succeed (0.75) –$100,000 $400,000 3 Don’t Succeed (0.25) –$200,000 $0 Mary should select the traffic circle location (EMV $250,000). b. Use Bayes’ Theorem to compute posterior probabilities. ¯¯¯¯ | SRP) = 0.22 P(SD | SRP) = 0.78; P(SD P(SM | SRP) = 0.84; P(SC | SRP) = 0.91; ¯¯¯¯¯ | SRP) = 0.16 P(SM ¯¯¯¯ | SRP) = 0.09 P(SC P(SM | SRN) = 0.36; ¯¯¯¯ | SRN) = 0.73 P(SD ¯¯¯¯¯ P(SM | SRN) = 0.64 P(SC | SRN) = 0.53; ¯¯¯¯ | SRN) = 0.47 P(SC P(SD | SRN) = 0.27; Example computations: P ( SM | SRP ) = P ( SRP | SM ) P ( SM ) P ( SRP | SM ) P ( SM ) + P ( SRP | SM ) P ( SM ) 0.7(0.6 ) P ( SM | SRP ) = = 0.84 0.7(0.6 ) + 0.2(0.4 ) P ( SC | SRN ) = 0.3(0.75) = 0.53 0.3(0.75) + 0.8(0.25) These calculations are for the tree that follows: EMV(2) $171,600 $28,600 $143,000 EMV(3) $226,800 $20,800 $206,000 EMV(4) $336,700 $20,700 $316,000 EMV(no grocery A) –$30,000 EMV(5) $59,400 $94,900 –$35,500 EMV(6) $97,200 $83,200 $14,000 EMV(7) $196,100 $108,100 $88,000 EMV(no grocery B) –$30,000 EMV(8) $75,000 EMV(9) $140,000 EMV(10) $250,000 EMV(no grocery C) $0 EMV(A) (best of four alternatives) $316,000 EMV(B) (best of four alternatives) $88,000 EMV(C) (best of four alternatives) $250,000 EMV(1) (0.6)($316,000) (0.4)($88,000) $224,800 EMV(D) (best of two alternatives) $250,000 c. EVSI [EMV(1) cost] (best EMV without sample information) $254,800 – $250,000 $4,800. 33 REVISED M03_REND6289_10_IM_C03.QXD 34 5/7/08 3:48 PM CHAPTER 3 Page 34 DECISION ANALYSIS Second Decision Point First Decision Point Payoff SD (0.78) Downtown 2 ts ul es .6) R (0 ey e rv itiv u S os P A 3 SM (0.16) Circle 4 SC (0.09) No Grocery Store lts su ) Re (0.4 ey e rv tiv Su ega N D 5 Mall SD (0.73) SM (0.36) 6 SM (0.64) SC (0.53) No tP Circle ur c ha se 7 SC (0.47) No Grocery Store M ar ke tS ur v 8 Mall C SD (0.5) SM (0.6) 9 SM (0.4) SC (0.75) Circle 10 SC (0.25) No Grocery Store 3-46. a. Sue can use decision tree analysis to find the best solution. The results are presented below. In this case, the best decision is to get information. If the information is favorable, she should build the retail store. If the information is not favorable, she should not build the retail store. The EMV for this decision is $29,200. In the following results (using QM for Windows), Branch 1 (1–2) is to get information, Branch 2 (1–3) is the decision to not get information, Branch 3 (2–4) is favorable information, Branch 4 (2–5) is unfavorable information, Branch 5 (3–8) is the decision to build the retail store and get no information, Branch 6 (3–17) is the decision to not build the retail store and to get no information, Branch 7 (4–6) is the decision to build the retail store given favorable information, Branch 8 (4–11) is the decision to not build given favorable information, Branch 9 (6–9) is a good market given favorable –$130,000 $270,000 –$130,000 $370,000 –$230,000 $220,000 –$130,000 $270,000 –$130,000 $370,000 –$230,000 –$30,000 SD (0.5) Downtown ey $220,000 –$30,000 SD (0.27) Downtown B Do SM (0.84) SC (0.91) 1 t ke ar M se ey a h rv rc Su u P Mall SD (0.22) $250,000 –$100,000 $300,000 –$100,000 $400,000 –$200,000 $0 information, Branch 10 (6–10) is a bad market given favorable information, Branch 11 (5–7) is the decision to build the retail store given unfavorable information, Branch 12 (5–14) is the decision not to build the retail store given unfavorable information, Branch 13 (7–12) is a successful retail store given unfavorable information, Branch 14 (7–13) is an unsuccessful retail store given unfavorable information, Branch 15 (8–15) is a successful retail store given that no information is obtained, and Branch 16 (8–16) is an unsuccessful retail store given no information is obtained. REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 35 CHAPTER 3 DECISION ANALYSIS b. The suggested changes would be reflected in Branches 4 and 5. The decision stays the same, but the EMV increases to $46,000. The results are provided in the tables that follow: Results for 3-46. a. Start Node Ending Node Branch Probability Profit (End Node) 0 1 1 2 2 3 3 4 4 6 6 5 5 7 7 8 8 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 0 0 0 0.6 0.4 0 0 0 0 0.9 0.1 0 0 0.2 0.8 0.6 0.4 0 0 0 0 0 0 0 0 20,000 80,000 100,000 0 20,000 80,000 100,000 100,000 80,000 Start Node Ending Node Branch Probability Profit (End Node) 0 1 1 2 2 3 3 4 4 6 6 5 5 7 7 8 8 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 0 0 0 0.7 0.3 0 0 0 0 0.9 0.1 0 0 0.2 0.8 0.6 0.4 0 0 0 0 0 0 0 0 20,000 80,000 100,000 0 20,000 80,000 100,000 100,000 80,000 Start Branch 1 Branch 2 Branch 3 Branch 4 Branch 5 Branch 6 Branch 7 Branch 8 Branch 9 Branch 10 Branch 11 Branch 12 Branch 13 Branch 14 Branch 15 Branch 16 Use Branch? Yes Yes Yes Yes Node Type Decision Chance Decision Decision Decision Chance Final Chance Final Final Final Chance Final Final Final Final Final Node Value 29,200 29,200 28,000 62,000 20,000 28,000 0 62,000 20,000 80,000 100,000 64,000 20,000 80,000 100,000 100,000 80,000 Results for 3-46. b. Start Branch 1 Branch 2 Branch 3 Branch 4 Branch 5 Branch 6 Branch 7 Branch 8 Branch 9 Branch 10 Branch 11 Branch 12 Branch 13 Branch 14 Branch 15 Branch 16 Use Branch? Yes Yes Yes Yes Node Type Decision Chance Decision Decision Decision Chance Final Chance Final Final Final Chance Final Final Final Final Final Node Value 37,400 37,400 28,000 62,000 20,000 28,000 0 62,000 20,000 80,000 100,000 64,000 20,000 80,000 100,000 100,000 80,000 35 REVISED M03_REND6289_10_IM_C03.QXD 36 5/7/08 3:48 PM CHAPTER 3 Page 36 DECISION ANALYSIS c. Sue can determine the impact of the change by changing the probabilities and recomputing EMVs. This analysis shows the decision changes. Given the new probability values, Sue’s best decision is build the retail store without getting additional information. The EMV for this decision is $28,000. The results are presented below: Results for 3-46. c. Start Node Ending Node Branch Probability Profit (End Node) 0 1 1 2 2 3 3 4 4 6 6 5 5 7 7 8 8 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 0 0 0 0.6 0.4 0 0 0 0 0.8 0.2 0 0 0.2 0.8 0.6 0.4 0 0 0 0 0 0 0 0 20,000 80,000 100,000 0 20,000 80,000 100,000 100,000 80,000 Start Branch 1 Branch 2 Branch 3 Branch 4 Branch 5 Branch 6 Branch 7 Branch 8 Branch 9 Branch 10 Branch 11 Branch 12 Branch 13 Branch 14 Branch 15 Branch 16 Use Branch? Yes Yes Yes Yes Node Type Decision Chance Decision Decision Decision Chance Final Chance Final Final Final Chance Final Final Final Final Final Node Value 28,000 18,400 28,000 44,000 20,000 28,000 0 44,000 20,000 80,000 100,000 64,000 20,000 80,000 100,000 100,000 80,000 d. Yes, Sue’s decision would change from her original decision. With the higher cost of information, Sue’s decision is to not get the information and build the retail store. The EMV of this decision is $28,000. The results are given below: Results for 3-46. d. Start Branch 1 Branch 2 Branch 3 Branch 4 Branch 5 Branch 6 Branch 7 Branch 8 Branch 9 Branch 10 Branch 11 Branch 12 Branch 13 Branch 14 Branch 15 Branch 16 Start Node Ending Node Branch Probability Profit (End Node) 0 1 1 2 2 3 3 4 4 6 6 5 5 7 7 8 8 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 0 0 0 0.6 0.4 0 0 0 0 0.9 0.1 0 0 0.2 0.8 0.6 0.4 0 0 0 0 0 0 0 0 30,000 70,000 110,000 0 30,000 70,000 110,000 100,000 80,000 Use Branch? Yes Yes Yes Yes Node Type Decision Chance Decision Decision Decision Chance Final Chance Final Final Final Chance Final Final Final Final Final Node Value 28,000 19,200 28,000 52,000 30,000 28,000 0 52,000 30,000 70,000 110,000 74,000 30,000 70,000 110,000 100,000 80,000 REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 37 CHAPTER 3 37 DECISION ANALYSIS e. The expected utility can be computed by replacing the monetary values with utility values. Given the utility values in the problem, the expected utility is 0.62. The utility table represents a risk seeker. The results are given below: Results for 3-46. e. Start Node Ending Node Branch Probability 0 1 1 2 2 3 3 4 4 6 6 5 5 7 7 8 8 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 0 0 0 0.6 0.4 0 0 0 0 0.9 0.1 0 0 0.2 0.8 0.6 0.4 Start Branch 1 Branch 2 Branch 3 Branch 4 Branch 5 Branch 6 Branch 7 Branch 8 Branch 9 Branch 10 Branch 11 Branch 12 Branch 13 Branch 14 Branch 15 Branch 16 Profit (End Node) 0 0 0 0 0 0 0.2 0 0.1 0.4 0 0 0.1 0.4 0 1 0.05 Use Branch? Yes Yes Yes Yes Ending Node Node Type Node Value 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 Decision Chance Decision Decision Decision Chance Final Chance Final Final Final Chance Final Final Final Final Final 0.62 0.256 0.62 0.36 0.1 0.62 0.20 0.36 0.1 0.4 0 0.08 0.1 0.4 0 1 0.05 f. This problem can be solved by replacing monetary values with utility values. The expected utility is 0.80. The utility table given in the problem is representative of a risk avoider. The results are presented below: Results for 3-46. f. Start Branch 1 Branch 2 Branch 3 Branch 4 Branch 5 Branch 6 Branch 7 Branch 8 Branch 9 Branch 10 Branch 11 Branch 12 Branch 13 Branch 14 Branch 15 Branch 16 Start Node Ending Node Branch Probability Profit (End Node) 0 1 1 2 2 3 3 4 4 6 6 5 5 7 7 8 8 1 2 3 4 5 8 17 6 11 9 10 7 14 12 13 15 16 0 0 0 0.6 0.4 0 0 0 0 0.9 0.1 0 0 0.2 0.8 0.6 0.4 0 0 0 0 0 0 0.8 0 0.6 0.9 0 0 0.6 0.9 0 1 0.4 3-47. a. The decision table for Chris Dunphy along with the expected profits or expected monetary values (EMVs) for each alternative are shown on the next page. Use Branch? Yes Yes Yes Yes Node Type Decision Chance Decision Decision Decision Chance Final Chance Final Final Final Chance Final Final Final Final Final Node Value 0.80 0.726 0.80 0.81 0.60 0.76 0.80 0.81 0.60 0.90 0.00 0.18 0.60 0.90 0.00 1.00 0.40 REVISED M03_REND6289_10_IM_C03.QXD 38 5/7/08 3:48 PM CHAPTER 3 Page 38 DECISION ANALYSIS Table for Problem 3-47a Return in $1,000: Number of Watches Probability 100,000 150,000 200,000 250,000 300,000 350,000 400,000 450,000 500,000 Alternative 1 Alternative 2 Alternative 3 Alternative 4 Alternative 5 Alternative 6 Alternative 7 Alternative 8 Alternative 9 Event 1 Event 2 Event 3 Event 4 Event 5 0.100 0.200 0.500 0.100 0.100 100,000 90,000 85,000 80,000 65,000 50,000 45,000 30,000 20,000 110,000 120,000 110,000 120,000 100,000 100,000 95,000 90,000 85,000 120,000 140,000 135,000 155,000 155,000 160,000 170,000 165,000 160,000 135,000 155,000 160,000 170,000 180,000 190,000 200,000 230,000 270,000 140,000 170,000 175,000 180,000 195,000 210,000 230,000 245,000 295,000 Expected profit: Alternative Expected Profit 1 2 3 4 5 6 7 8 9 119,500 135,500 131,500 144,500 141,500 145,000 151,500 151,000 155,500 ← best alternative For this decision problem, Alternative 9 gives the highest expected profit of $155,500. b. The expected value with perfect information is $175,500, and the expected value of perfect information (EVPI) is $20,000. c. The new probability estimates will give more emphasis to event 2 and less to event 5. The overall impact is shown below. As you can see, stocking 400,000 watches is now the best decision with an expected value of $140,700. Return in $1,000: Probability Alternative 1 Alternative 2 Alternative 3 Alternative 4 Alternative 5 Alternative 6 Alternative 7 Alternative 8 Alternative 9 EVENT 1 EVENT 2 EVENT 3 EVENT 4 EVENT 5 0.100 0.280 0.500 0.100 0.020 100,000 90,000 85,000 80,000 65,000 50,000 45,000 30,000 20,000 110,000 120,000 110,000 120,000 100,000 100,000 95,000 90,000 85,000 120,000 140,000 135,000 155,000 155,000 160,000 170,000 165,000 160,000 135,000 155,000 160,000 170,000 180,000 190,000 200,000 230,000 270,000 140,000 170,000 175,000 180,000 195,000 210,000 230,000 245,000 295,000 Expected profit: Alternative 1 2 3 4 5 6 7 8 9 Expected Profit 117.100 131,500 126,300 139,700 133,900 136,200 140,700 ← best alternative: stock 400,000 watches 138,600 138,700 REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 39 CHAPTER 3 39 DECISION ANALYSIS d. Stocking 400,000 is still the best alternative. The results are shown below. Return in $1,000: Event 1 Probability Alternative 1 Alternative 2 Alternative 3 Alternative 4 Alternative 5 Alternative 6 Alternative 7 Alternative 8 Alternative 9 Event 2 Event 3 0.100 0.280 0.500 0.100 0.020 110,000 120,000 110,000 120,000 100,000 100,000 95,000 90,000 85,000 120,000 140,000 135,000 155,000 155,000 160,000 170,000 165,000 160,000 135,000 155,000 160,000 170,000 180,000 190,000 200,000 230,000 270,000 140,000 170,000 175,000 180,000 195,000 210,000 230,000 245,000 340,000 b. Back roads (minimum time used). c. Expected time with perfect information: 15 1/2 + 25 1/3 + 30 1/6 = 20.83 minutes Expected Profit 1 2 3 4 5 6 7 8 9 3-48. 117,100 131,500 126,300 139,700 133,900 136,200 140,700 ← best alternative: stock 400,000 watches 138,600 139,600 Time saved is 31⁄3; minutes. 3-51. a. Decision under uncertainty. b. Population Same Large wing Small wing No wing Population Grows 85,000 45,000 0 150,000 60,000 0 Row Average 32,500 7,500 0 c. Best alternative: large wing. 3-49. a. Note: This problem can also be solved using marginal analysis. Large wing Small wing No wing Population Same Population Grows 85,000 45,000 0 150,000 60,000 0 Weighted Average with ␣ = 0.75 91,250 33,750 0 b. Best decision: large wing. c. No. 3-50. a. No Mild Severe Expected Congestion Congestion Congestion Time Tennessee Back roads Expressway Probabilities Event 5 100,000 90,000 85,000 80,000 65,000 50,000 45,000 30,000 20,000 Expected profit Alternative Event 4 15 20 30 30 25 30 45 35 30 (30 days)/ (20 days)/ (60 days) = 1/2 (60 days) = 1/3 25 24.17 30 (10 days)/ (60 days) = 1/6 a. EMV can be used to determine the best strategy to minimize costs. The QM for Windows solution is shown on the next page. The best decision is to go with the partial service (maintenance) agreement. REVISED M03_REND6289_10_IM_C03.QXD 40 5/7/08 3:48 PM CHAPTER 3 Page 40 DECISION ANALYSIS Solution to 3-51a Probabilities 0.2 0.8 Maint. Cost ($) No Maint. Cost ($) 3,000 1,500 500 0 300 500 No Service Agreement Partial Service Agreement Complete Service Agreement Expected Value ($) Row Minimum ($) 600 540 500 0 0 500 3,000 1,500 500 500 0 500 Column best Row Maximum ($) The minimum expected monetary value is 500 given by Complete Service Agreement b. The new probability estimates dramatically change Sim’s decision. The best decision given this new information is to still go with the complete service or maintenance policy with an expected cost of $500. The results are shown below. Solution to 3-51b Needs Repair ($) Probabilities No Service Agreement Partial Service Agreement Complete Service Agreement Does Not Need Repair ($) 0.8 Expected Value ($) 0.2 3,000 0 2,400 1,500 300 1,260 500 500 500 Column best 500 3-52. We can use QM for Windows to solve this decision making under uncertainty problem. We have made up probability values, which will be ignored in the analysis. As you can see, the maximax decision is Option 4, and the maximum decision is Option 1. To compute the equality likely decision, we used equal probability values of 0.25 for each of the four scenarios. As seen below, the equally likely decision, which is the same as the EMV decision in this case, is Option 3. Solution to 3-52 Expected Value ($) Probabilities Option 1 Option 2 Option 3 Option 4 0.25 0.25 0.25 0.25 Judge ($) Trial ($) Court ($) Arbitration ($) 5,000 10,000 20,000 30,000 5,000 5,000 7,000 15,000 5,000 2,000 1,000 10,000 5,000 0 5,000 20,000 Column best The maximum expected monetary value is 5,750 given by Option 3. The maximum is 5,000 given by Option 1. The maximax is 30,000 given by Option 4. Row Minimum ($) Row Miximum ($) 5,000 4,250 5,750 3,750 5,000 0 5,000 20,000 5,000 10,000 20,000 30,000 5,750 5,000 30,000 REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 41 CHAPTER 3 41 DECISION ANALYSIS SOLUTION TO STARTING RIGHT CASE This is a decision-making-under-uncertainty case. There are two events: a favorable market (event 1) and an unfavorable market (event 2). There are four alternatives, which include do nothing (alternative 1), invest in corporate bonds (alternative 2), invest in preferred stock (alternative 3), and invest in common stock (alternative 4). The decision table is presented below. Note that for alternative 2, the return in a good market is $30,000 (1 0.13)5 $55,273. The return in a good market is $120,000 (4 x $30,000) for alternative 3, and $240,000 (8 x $30,000) for alternative 4. Payoff table Alternative 1 Alternative 2 Alternative 3 Alternative 4 Event 1 Event 2 Laplace Average Value Minimum Maximum Hurwicz Value 0 55,273 120,000 240,000 0 10,000 15,000 30,000 0.0 22,636.5 152,500.0 105,000.0 0 10,000 15,000 30,000 0 55,273 120,000 240,000 0.00 2,819.97 150.00 300.00 Regret table Alternative Event 1 Event 2 Maximum Regret Alternative 1 Alternative 2 Alternative 3 Alternative 4 240,000 184,727 120,000 0 0 10,000 15,000 30,000 240,000 184,727 120,000 30,000 a. Sue Pansky is a risk avoider and should use the maximin decision approach. She should do nothing and not make an investment in Starting Right. b. Ray Cahn should use a coefficient of realism of 0.11. The best decision is to do nothing. c. Lila Battle should eliminate alternative 1 of doing nothing and apply the maximin criterion. The result is to invest in the corporate bonds. d. George Yates should use the equally likely decision criterion. The best decision for George is to invest in common stock. e. Pete Metarko is a risk seeker. He should invest in common stock. f. Julia Day can eliminate the preferred stock alternative and still offer alternatives to risk seekers (common stock) and risk avoiders (doing nothing or investing in corporate bonds). SOLUTIONS TO INTERNET CASES Drink-at-Home, Inc. Case Abbreviations and values used in the following decision trees: Normal—proceed with research and development at a normal pace. 6 Month—Adopt the 6-month program: if a competitor’s product is available at the end of 6 months, then copy; otherwise proceed with research and development. 8 Month—Adopt the 6-month program: proceed for 8 months; if no competition at 8 months, proceed; otherwise stop development. Success or failure of development effort: Ok—Development effort ultimately a success No—Development effort ultimately a failure Column: S— Sales revenue R—Research and development expenditures E—Equipment costs I—Introduction to market costs Market size and Revenues: S—Substantial (P 0.1) M—Moderate (P 0.6) L—Low (P 0.3) Without Competition With Competition $800,000 $600,000 $500,000 $400,000 $300,000 $250,000 Competition: C6—Competition at end of 6 months (P .5) No C6—No competition at end of 6 months (P .5) C8—Competition at end of 8 months (P .6) No C8—No competition at end of 8 months (P .4) C12—Competition at end of 12 months (P .8) No C12—No competition at end of 12 months (P .2) REVISED M03_REND6289_10_IM_C03.QXD 42 5/7/08 3:48 PM CHAPTER 3 Page 42 DECISION ANALYSIS Drink-at-Home. Inc. Case (continued) Mkt S (.1) C12 (.8) L (.3) Ok (.9) S (.1) No C12 (.2) Normal M (.6) M (.6) L (.3) No (.1) (Stop) S (.1) Ok (.9) M (.6) L (.3) No C8 (.4) S (.1) No (.1) M (.6) nth 6 Mo L (.3) S (.1) C6 (.5) M (.6) L (.3) S (.1) C12 (.8) M (.6) L (.3) Ok (.9) S (.1) No C12 (.2) No C6 (.5) R E M (.6) L (.3) No (.1) = –100 – 80 = – 80 400 – 140 – 100 – 150 = 10 300 – 140 – 100 – 150 = –90 250 – 140 – 100 – 150 = –140 800 – 140 – 100 – 150 = 410 600 – 140 – 100 – 150 = 210 500 – 140 – 100 – 150 = 110 400 – 90 – 100 – 150 = 60 300 – 90 – 100 – 150 = –40 250 – 90 – 100 – 150 = –90 400 – 100 – 100 – 150 = 50 300 – 100 – 100 – 150 = –50 250 – 100 – 100 – 150 = –100 800 – 100 – 100 – 150 = 450 600 – 100 – 100 – 150 = 250 500 – 100 – 100 – 150 = 150 – 100 = –100 Mkt S (.1) C12 (.8) S (.1) No C12 (.2) Normal M (.6) L (.3) Ok (.9) M (.6) L (.3) No (.1) (Stop) Ok (.9) No C8 (.4) No (.1) nth 6 Mo S (.1) M (.6) L (.3) S (.1) M (.6) L (.3) S (.1) C6 (.5) M (.6) L (.3) S (.1) C12 (.8) M (.6) L (.3) Ok (.9) S (.1) No C12 (.2) No C6 (.5) M (.6) L (.3) No (.1) 50 –50 –100 450 250 150 –100 C8 (.6) 8 Month I 400 – 100 – 100 – 150 = 50 300 – 100 – 100 – 150 = –50 250 – 100 – 100 – 150 = –100 800 – 100 – 100 – 150 = 450 600 – 100 – 100 – 150 = 250 500 – 100 – 100 – 150 = 150 – 100 C8 (.6) 8 Month S –80 10 –90 –140 410 210 110 60 –40 –90 50 –50 –100 450 250 150 –100 REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 43 CHAPTER 3 DECISION ANALYSIS Drink-at-Home, Inc. Case (continued) Mkt S (.1) Ok (.9) C12 (.8) M (.6) –55 L (.3) 240 M (.6) (4) S (.1) Normal –6.4 L (.3) No C12 (.2) No (.1) (Stop) –100 C8 (.6) S (.1) 8 Month –74.2 Ok (.9) –95 (–74.2) No C8 (.4) M (.6) L (.3) S (.1) M (.6) 200 L (.3) 6 Mo No (.1) nth S (.1) C6 (.5) M (.6) –45 L (.3) S (.1) C 12 (.8) M (.6) –55 Ok (.9) No C6 (.5) L (.3) S (.1) M (.6) 240 (19.3) L (.3) No C12 (.2) No (.1) –80 10 –90 –140 410 210 110 60 –40 –90 50 –50 –100 450 250 150 –100 The optimal program is to adopt the 6-month program Ruth Jones’ Heart By-Pass Operation Case N o Su Byrg pa er ss y 50 –50 –100 450 250 150 Prob. Years Expected Rate One Year .50 1 .50 Two Years .20 2 .40 Five Years .20 5 1.00 Eight Years .10 8 .80 2.7 years 0.0 .05 0 One Year .45 1 .45 Five Years .20 5 1.00 Ten Years .13 10 1.30 Fifteen Years .08 15 1.20 Twenty Years .05 20 1.00 Twenty-five Years .04 25 1.00 ry e rg Su 0 Years 5.95 years 43 REVISED M03_REND6289_10_IM_C03.QXD 44 5/7/08 3:48 PM CHAPTER 3 Page 44 DECISION ANALYSIS Expected survival rate with surgery (5.95 years) exceeds the nonsurgical survival rate of 2.70 years. Surgery is favorable. Ski Right Case a. Bob can solve this case using decision analysis. As you can see, the best decision is to have Leadville Barts make the helmets and have Progressive Products do the rest with an expected value of $2,600. The final option of not using Progressive, however, was very close with an expected value of $2,500. POOR Probabilities 0.1 AVERAGE GOOD EXCELLENT 0.3 0.4 0.2 EXPECTED VALUE Option 1—PP Option 2—LB and PP Option 3—TR and PP Option 4—CC and PP Option 5—LB, CC, and TR 5,000 10,000 15,000 30,000 60,000 2,000 4,000 10,000 20,000 35,000 2,000 6,000 7,000 10,000 20,000 5,000 12,000 13,000 30,000 55,000 700 2,600 900 1,000 2,500 With Perfect Information 5,000 2,000 25,000 55,000 17,900 The maximum expected monetary value is 2,600 given by Option 2 LB and PP. b and c. The opportunity loss and the expected value of perfect information is presented below. The EVPI is $15,300. Expected value with perfect information 17,900 Expected monetary value 2,600 Expected value of perfect information 15,300 Opportunity loss table POOR MARKET Probabilities Option 1—PP Option 2—LB and PP Option 3—TR and PP Option 4—CC and PP Option 5—LB, CC, and TR AVERAGE GOOD EXCELLENT 0.1 0.3 0.4 0.2 0 5,000 10,000 25,000 55,000 0 2,000 8,000 18,000 33,000 18,000 14,000 13,000 10,000 0 50,000 43,000 42,000 25,000 0 d. Bob was logical in approaching this problem. However, there are other alternatives that might be considered. One possibility is to sell the idea and the rights to produce this product to Progressive Products for a fixed amount. STUDY TIME CASE Raquel must decide which of the three cases (1, 2, or 3) to study, and how much time to devote to each. We will assume that it is equally likely (a 1/3 chance) that each case is chosen. If she misses at most 8 points (let’s assume she is correct in thinking that) on the other parts of the exam, scoring 20 points or more on this part will give her an A for the course. Scoring 0 or 12 points on this portion of the exam will REVISED M03_REND6289_10_IM_C03.QXD 5/7/08 3:48 PM Page 45 CHAPTER 3 DECISION ANALYSIS result in a grade of B for the course. The table below gives the different possibilities – points and grade in the course. Study 1, 2, 3 Study 1,2 Study 1,3 Study 2,3 Study 1 Study 2 Study 3 Case 1 on Exam Case 2 on Exam Case 3 on Exam EV 12 B 20 A 20 A 0B 25 A 0B 0B 12 B 20 A 0B 20 A 0B 25 A 0B 12 B 0B 20 A 20 A 0B 0B 25 A 12 40/3 40/3 40/3 25/3 25/3 25/3 Thus, Raquel should study 2 cases since this will give her a 2/3 chance of an A in the course. Notice that this also has the highest expected value. This is a situation in which the values (points) are not always indicative of the importance of the result since 0 or 12 results in a B for the course, and 20 or 25 results in an A for the course. Grade in Course B A 2/3 chance or B 1/3 chance A 2/3 chance or B 1/3 chance A 2/3 chance or B 1/3 chance A 1/3 chance or B 2/3 chance A 1/3 chance or B 2/3 chance A 1/3 chance or B 2/3 chance 45