Decision Trees Solution Render

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 decisionmaking 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 decisionmaking 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 by 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 expect value of sample information (EVSI) and the expected value of perfect information (EVPI) are calculated. The ratio EVSI/EVPI is calculated and multiplied by 100% to get the efficiency of sample information. 3-13. The overall purpose of utility theory is to incorporate a decision maker’s preference for risk in the decision-making process. 3-14. 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. 3-15. 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. 3-16. 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. 3-17. 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

Unfavorable

Row

Row

Maximum

Minimum

300,000 250,000 75,000

–200,000 –100,000

–200,000 –100,000 –18,000

300,000 250,000 75,000

–18,000

3-18. 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. 3-19. a. Decision making under risk—maximize expected monetary value. 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 X  (205,000)/0.7 = 292,857.14 The decision would change if this payoff were less than 292,857.14, so it would have to decrease by about $7,143. 3-20. a. The expected value (EV) is computed for each alternative. EV(stock market) = 0.5(80,000) + 0.5(–20,000) = 30,000 EV(Bonds) = 0.5(30,000) + 0.5(20,000) = 25,000 EV(CDs) = 0.5(23,000) + 0.5(23,000) = 23,000 Therefore, he should invest in the stock market.

b. EVPI = EV(with perfect information) – (Maximum EV without P, I) = [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. 3-21. The opportunity loss table is Alternative

Good Economy

Stock Market

Poor Economy

0

43,000

Bonds

50,000

3,000

CDs

57,000

0

EOL(Stock Market) = 0.5(0) + 0.5(43,000) = 21,500* This minimizes EOL. EOL(Bonds) = 0.5(50,000) + 0.5(3,000) = 26,500 EOL(CDs) = 0.5(57,000) + 0.5(0) = 28,500 3-22. a. Market Alternative

Good

Fair

Poor

EMV

Stock market

1,400

800

0

880

Bank deposit (CD)

900

900

900

900

0.4

0.4

0.2

Probabilities conditions

Condition

of

market

b. Best decision: deposit $10,000 in bank CD. 3-23. 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.

3-24. a. Opportunity loss table Strong

Fair

Poor

Max.

Market

Market

Market

Regret

0

19,000

310,000

310,000

Medium

250,000

0

100,000

250,000

Small

350,000

29,000

32,000

350,000

None

550,000

129,000

0

550,000

Large

b. Minimax regret decision is to build medium. 3-25. a. Stock

Demand

(Cases)

(Cases)

11

12

13

EMV

11

385

385

385

385

12

329

420

420

379.05

13

273

364

455

341.25

Probabilities

0.45

0.35

0.20

b. Stock 11 cases. 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. Stock

Demand

(Cases)

(Cases)

11

12

13

EMV

11

385

385

385

385

12

385

420

420

404.25

13

385

420

455

411.25

3-26. Manufacture (Cases)

Demand (Cases)

6

7

8

9

EMV

6

300

300

300

300

300

7

255

350

350

350

340.5

8

210

305

400

400

352.5

9

165

260

355

450

317

Probabilities

0.1

0.3

0.5

0.1

John should manufacture 8 cases of cheese spread. 3-27. Cost of produced case = $5. Cost of purchased case = $16. Selling price = $15. Money recovered from each unsold case = $3. Supply

Demand(Cases)

(Cases) 100

200

300

100

100(15) –100(5) = 200(15) – 100(5) –100(16) 300(15) – 100(5) –200(16) 1000 = 900 = 800

200

100(15) + 100(3) – 200(15) – 200(5) = 2000 200(5) = 800

300

100(15) + 200(3) – 200(15) + 100(3) –300(5) 300(15) – 300(5) = 3000 300(5) = 600 = 1800

Prob.

0.3

0.4

b. Produce 300 cases each day.

300(15) – 200(5) –100(16) = 1900

0.3

EMV 900 1610 1800

3-28. a. The table presented is a decision table. The basis for the decisions in the following questions is shown in the table. The values in the table are in 1,000s. MARKET

Decision Good

Fair

Poor

Alternatives

EQUALLY LIKELY

CRIT. OF REALISM

MAXIMAX

MAXIMIN

Row Max.

Row Min.

Row Ave.

Weighted Ave.

Small

50 20

–10

50

–10

20

38

Medium

80 30

–20

80

–20

30

60

Large

100 30

–40

100

–40

30

72

Very Large

300 25

–160

300

–160

55

208

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 (values in the table are in 1,000s): MARKET Decision

Good

Alternatives Market

MINIMAX

Fair

Poor

Row

Market

Market

Maximum

Small

250

10

0

250

Medium

220

0

10

220

Large

200

0

30

200

0

5

150

150

Very Large

3-29. Note this problem is based on costs, so the minimum values are the best. a. For a 3-year lease, there are 36 months of payments. Option 1 total monthly payments: 36($330) = $11,880 Option 2 total monthly payments: 36($380) = $13,680 Option 3 total monthly payments: 36($430) = $15,480 Excess mileage costs based on 36,000 mileage allowance for Option 1, 45,000 for Option 2, and 54,000 for option 3. Option 1 excess mileage cost if 45,000 miles are driven = (45000 – 36000)(0.35) = 3150 Option 1 excess mileage cost if 54,000 miles are driven = (54000 – 36000)(0.35) = 6300 Option 2 excess mileage cost if 54,000 miles are driven = (54000 – 45000)(0.25) = 2250 The total cost for each option in each state of nature is obtained by adding the total monthly payment cost to the excess mileage cost. Total cost table Lease option

36000 miles driven 45000 miles driven 54000 miles driven

Option 1

11,880

15030

18180

Option 2

13,680

13680

15930

Option 3

15,480

15480

15480

b. Optimistic decision: Option 1 because the best (minimum) payoff (cost) for this is 11,800 which is better (lower) than the best payoff for each of the others. c. Pessimistic decision: Option 3 because the worst (maximum) payoff (cost) for this is 15,480 is better (lower) than the worst payoff for each of the others. d. Select Option 2. EMV(Option 1) = 11,880(0.4) + 15,030(0.3) + 18,180(0.3) = 14,715 EMV(Option 2) = 13,680(0.4) + 13,680(0.3) + 15,930(0.3) = 14,355 EMV(Option 3) = 15,480(0.4) + 15,480 (0.3) + 15,480(0.3) = 15,480 (e) EVPI for a minimization problem = (Best EMV without PI) - (EV with PI) EV with PI = 11,880(0.4) + 13,680(0.3) + 15,480(0.3) = 13,500 EVPI = 14,355 – 13,500 = 855

3-30. Note that this is a minimization problem, so the opportunity loss is based on the lowest (best) cost in each state of nature. Opportunity loss table Lease option

36000 miles driven

45000 miles driven

54000 miles driven

Option 1

11880 – 11880 = 0

15030 - 13680 = 1350

18180 - 15480 = 2700

Option 2

13680 – 11880 = 1800

13680 - 13680 = 0

15930 - 15480 = 450

Option 3

15480 – 11880 = 3600

15480 - 13680 = 1800

15480 - 15480 = 0

The maximum regrets are 2700 for option 1, 1800 for option 2, and 3600 for option 3. Option 2 is selected because 1800 is lower than the other maximums. EOL(option 1) = 0(0.4) + 1350(0.3) + 2700(0.3) = 1215 EOL(option 2) = 1800(0.4) + 0(0.3) + 450(0.3) = 855 EOL(option 3) = 3600(0.4) + 1800(0.3) + 0(0.3) = 1980 Option 2 has the lowest EOL, so this alternative is selected based on the EOL criterion. 3-31. a. P(red) = 18/38; P(not red) = 20/38 b. EMV = Expected win = 10(18/38) + (-10)(20/38) = -0.526 c. P(red) = 18/37; P(not red) = 19/37 EMV = Expected win = 10(18/37) + (-10)(19/37) = -0.270 d. The enjoyment of playing the game and possibly winning adds utility to the game. A person would play this game because the expected utility of playing the game is positive even though the expected monetary value is negative. 3-32. A $10 bet on number 7 would pay 35($10) = 350 if the number 7 is the winner. P(number 7) = 1/38; P(not seven) = 37/38 EMV = Expected win = 350(1/38) + (-10)(37/38) = -0.526 3-33. Payoff table with cost of $50,000 in legal fees deducted if suit goes to court and $10,000 in legal fees if settle. Win big

Win small

Lose

EMV

Go to court

250000

0

-50000

85000

Settle

65000

65000

65000

65000

Prob.

0.4

0.3

0.3

Decision based on EMV: Go to court

3-34. EMV for node 1 = 0.5(100,000) + 0.5(–40,000) = $30,000. Choose the highest EMV, therefore construct the clinic.

3-35. a.

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. 3-36.

3-37.

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).

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. 3-38.

A1: gather more information A2: do not gather more information A3: build quadplex A4: build duplex A5: do nothing EMV(node 2) = 0.9(12,000) + 0.1(–23,000) = 8,500 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. 3-39. I1: favorable research or information I2: unfavorable research S1: store successful S2: store unsuccessful P(S1) = 0.5; P(S2) = 0.5 P(I1 | S1) = 0.8; P(I2 | S1) = 0.2 P(I1 | S2) = 0.3; P(I2 | S2) = 0.7 a. P(successful store | favorable research) = P(S1 | I1) P  S1 I1   P  S1 I1  

P  I1 S1  P  S1 

P  I1 S1  P  S1   P  I1 S 2  P  S 2  0.8  0.5 

0.8  0.5   0.3  0.5 

 0.73

b. P(successful store | unfavorable research) = P(S1 | I2) P  S1 I 2   P  S1 I 2  

P  I 2 S1  P  S1 

P  I 2 S1  P  S1   P  I 2 S 2  P  S 2  0.2  0.5 

0.2  0.5   0.7  0.5 

 0.22

c. Now P(S1) = 0.6 and P(S2) = 0.4 P  S1 I1   P  S1 I 2  

0.8  0.6 

0.8  0.6   0.3  0.4  0.2  0.6 

0.2  0.6   0.7  0.4 

 0.8  0.3

3-40. I1: favorable survey or information I2: unfavorable survey S1: facility successful S2: facility unsuccessful P(S1) = 0.3; P(S2) = 0.7 P(I1 | S1) = 0.8; P(I2 | S1) = 0.2 P(I1 | S2) = 0.3; P(I2 | S2) = 0.7 P(successful facility | favorable survey) = P(S1 | I1) P  S1 I1   P  S1 I1  

P  I1 S1  P  S1 

P  I1 S1  P  S1   P  I1 S 2  P  S 2  0.8  0.3

0.8  0.3  0.3  0.7 

 0.533

P(successful facility | unfavorable survey) = P(S1 | I2) P  S1 I 2   P  S1 I 2  

P  I 2 S1  P  S1 

P  I 2 S1  P  S1   P  I 2 S 2  P  S 2  0.2  0.3

0.2  0.3  0.7  0.7 

 0.109

3-41. a.

b. EMV(A) = 10,000(0.2) + 2,000(0.3) + (–5,000)(0.5) = 100 EMV(B) = 6,000(0.2) + 4,000(0.3) + 0(0.5) = 2,400 Fund B should be selected. c. Let X = payout for Fund A in a good economy. EMV(A) = EMV(B) X(0.2) + 2,000(0.3) + (–5,000)(0.5) = 2,400 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. 3-42. a.

b.

S1: survey favorable S2: survey unfavorable 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 S 2  

0.3  0.5  0.27 0.3  0.5  0.8  0.5

P(S6 | S2) = 1 – 0.27 = 0.73 P  S5 S 3  

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 EMV(node 5) = 80,000(0.89) + (–80,000)(0.11) = 62,400 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 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. 3-43. 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.

3-44. 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 0.8  0.7  good economy) =  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

P(poor economy | prediction of poor economy) =

0.9  0.3  0.659 0.2  0.7   0.9  0.3

3-45. 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 a negative $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. 3-46.

EU(node 2) = (0.82)(0.99) + (0.18)(0) = 0.8118 EU(node 3) = (0.11)(0.99) + (0.89)(0) = 0.1089 EU(node 4) = 0.5(1) + 0.5(0.1) = 0.55 EU(node 1) = (0.55)(0.8118) + (0.45)(0.7000) = 0.7615 EU(no survey) = 0.9 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. 3-47. 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 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 EU(conduct survey) = 0.45(0.741) + 0.55(0.2565) = 0.4745 EU(no survey) = 0.525 John’s decision would change. He would not conduct the survey and build the large plant. 3-48. a. Expected travel time on Broad Street = 40(0.5) + 15(0.5) = 27.5 minutes. Broad Street has a lower expected travel time.

b. Expected utility on Broad Street = 0.2(0.5) + 0.9(0.5) = 0.55. Therefore, the expressway maximizes expected utility. c. Lynn is a risk avoider.

3-49. 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 marginal profit = $20 – $12 – $1 – $3 = $4 per gallon If more is produced than is needed, a marginal loss is incurred. 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.

a. A decision tree is provided:

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. Table for Problem 3-49 Demand Stock

500

1,000

1,500

2,000

EMV

2,000

–500

–3,000

–5,500

–$1,500

1,000

500

4,000

1,500

–1,000

$1,800

1,500

–1,000

2,500

6,000

3,500

$3,300

2,000

–2,500

1,000

4,500

8,000

$2,400

2,000

4,000

6,000

8,000

$4,800 = EVwPI

500

Maximum Probabilities

0.2

0.3

0.4

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 3-50. 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.

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 medium survey results, and a final table is used for high survey results. These probabilities will be used in the decision tree that follows.

For low survey results—A1: State of Nature

P(Bi)

P(Ai | Bj)

P(Bj and Ai)

P(Bj | Ai)

B1

0.150

0.700

0.105

0.339

B2

0.400

0.400

0.160

0.516

B3

0.450

0.100

0.045

0.145

P(A1) =

0.310

For medium survey results—A2:

State of Nature

P(Bi)

P(Ai | Bj)

P(Bj and Ai)

P(Bj | Ai)

B1

0.150

0.200

0.030

0.082

B2

0.400

0.500

0.200

0.548

B3

0.450

0.300

0.135

0.370

P(A2) =

0.365

For high survey results—A3: State of Nature

P(Bi)

P(Ai | Bj)

P(Bj and Ai)

P(Bj | Ai)

B1

0.150

0.100

0.015

0.046

B2

0.400

0.100

0.040

0.123

B3

0.450

0.600

0.270

0.831

P(A3) =

0.325

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) = 0.370. 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: 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.

3-51. a.

Mary should select the traffic circle location (EMV = $250,000). b. Use Bayes’ Theorem to compute posterior probabilities. P(SD | SRP) = 0.78;

P( SD | SRP) = 0.22

P(SM | SRP) = 0.84;

P( SM | SRP) = 0.16

P(SC | SRP) = 0.91;

P( SC | SRP) = 0.09

P(SD | SRN) = 0.27;

P( SD | SRN) = 0.73

P(SM | SRN) = 0.36;

P( SM | SRN) = 0.64

P(SC | SRN) = 0.53;

P( SC | SRN) = 0.47

Example computations: P  SM SRP   P  SM SRP   P  SC SRN  

P  SRP SM  P  SM 



  

P  SRP SM  P  SM   P SRP SM P SM 0.7  0.6 

0.7  0.6   0.2  0.4 

 0.84

0.3  0.75 

0.3  0.75   0.8  0.25 

 0.53

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.

3-52. a. Sue can use decision tree analysis to find the best solution. 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 successful retail store given favorable information, Branch 10 (6–10) is an unsuccessful retail store 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. Results for 3-52. a. Start Ending Branch

Profit

Node

Node

Node Node

(End Node) Branch? Type

Value

Prob.

Use

Start

0

1

0

0

Dec

29,200

Branch 1

1

2

0

0

Ch

29,200

Branch 2

1

3

0

0

Dec

28,000

Branch 3

2

4

0.6

0

Dec

62,000

Branch 4

2

5

0.4

0

Dec

–20,000

Branch 5

3

8

0

0

Ch

28,000

Branch 6

3

17

0

0

Fin

0

Branch 7

4

6

0

0

Ch

62,000

Branch 8

4

11

0

–20,000

Fin

–20,000

Branch 9

6

9

0.9

80,000

Fin

80,000

Branch 10 6

10

0.1

–100,000

Fin

–100,000

Branch 11 5

7

0

0

Ch

–64,000

Branch 12 5

14

0

–20,000

Fin

–20,000

Branch 13 7

12

0.2

80,000

Fin

80,000

Branch 14 7

13

0.8

–100,000

Fin

–100,000

Branch 15 8

15

0.6

100,000

Fin

100,000

Branch 16 8

16

0.4

–80,000

Fin

–80,000

Yes

Yes Yes

Yes

b. The suggested changes would be reflected in Branches 3 and 4. The decision stays the same, but the EMV increases to $37,400. The results are provided in the tables that follow. In these tables, BR = Branch; Prob. = Probability; and for Node Type, Dec = Decision, Ch = Chance, and Fin = Final. Results for 3-52. b. Start Ending Branch

Profit

Node

Node

Node Node

(End Node) Branch? Type

Value

Prob.

Use

Start

0

1

0

0

Dec

37,400

Branch 1

1

2

0

0

Ch

37,400

Branch 2

1

3

0

0

Dec

28,000

Branch 3

2

4

0.7

0

Dec

62,000

Branch 4

2

5

0.3

0

Dec

–20,000

Branch 5

3

8

0

0

Ch

28,000

Branch 6

3

17

0

0

Fin

0

Branch 7

4

6

0

0

Ch

62,000

Branch 8

4

11

0

–20,000

Fin

–20,000

Branch 9

6

9

0.9

80,000

Fin

80,000

Branch 10 6

10

0.1

–100,000

Fin

–100,000

Branch 11 5

7

0

0

Ch

–64,000

Branch 12 5

14

0

–20,000

Fin

–20,000

Branch 13 7

12

0.2

80,000

Fin

80,000

Branch 14 7

13

0.8

–100,000

Fin

–100,000

Branch 15 8

15

0.6

100,000

Fin

100,000

Branch 16 8

16

0.4

–80,000

Fin

–80,000

Yes

Yes Yes

Yes

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 3-52. c. Start Ending Branch

Profit

Node

Node

Node Node

(End Node) Branch? Type

Value

Prob.

Use

Start

0

1

0

0

Dec

28,000

Branch 1

1

2

0

0

Ch

18,400

Branch 2

1

3

0

0

Dec

28,000

Branch 3

2

4

0.6

0

Dec

44,000

Branch 4

2

5

0.4

0

Dec

–20,000

Branch 5

3

8

0

0

Ch

28,000

Branch 6

3

17

0

0

Fin

0

Branch 7

4

6

0

0

Ch

44,000

Branch 8

4

11

0

–20,000

Fin

–20,000

Branch 9

6

9

0.8

80,000

Fin

80,000

Branch 10 6

10

0.2

–100,000

Fin

–100,000

Branch 11 5

7

0

0

Ch

–64,000

Branch 12 5

14

0

–20,000

Fin

–20,000

Branch 13 7

12

0.2

80,000

Fin

80,000

Branch 14 7

13

0.8

–100,000

Fin

–100,000

Branch 15 8

15

0.6

100,000

Fin

100,000

Branch 16 8

16

0.4

–80,000

Fin

–80,000

Yes

Yes Yes

Yes

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 35-2. d. Start Ending Branch Node Node

Profit

Use

Node

Node

Probability (End Node) Branch? Type

Value

Start

0

1

0

0

Decision

28,000

Branch 1

1

2

0

0

Chance

19,200

Branch 2

1

3

0

0

Decision

28,000

Branch 3

2

4

0.6

0

Decision

52,000

Branch 4

2

5

0.4

0

Decision

–30,000

Branch 5

3

8

0

0

Chance

28,000

Branch 6

3

17

0

0

Branch 7

4

6

0

0

Branch 8

4

11

0

–30,000

Final

–30,000

Branch 9

6

9

0.9

70,000

Final

70,000

Branch 10 6

10

0.1

–110,000

Final

–110,000

Branch 11 5

7

0

0

Branch 12 5

14

0

–30,000

Branch 13 7

12

0.2

Branch 14 7

13

Branch 15 8 Branch 16 8

Yes

Yes

Final Yes

Chance

0 52,000

Chance

–74,000

Final

–30,000

70,000

Final

70,000

0.8

–110,000

Final

–110,000

15

0.6

100,000

Final

100,000

16

0.4

–80,000

Final

–80,000

Yes

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 3-52. e. Start Ending Branch Profit Node Node

Use

Ending Node

Prob.

(End Node) Branch? Node

Node

Type

Value

Start

0

1

0

0

1

Dec

0.62

Branch 1

1

2

0

0

2

Ch

0.256

Branch 2

1

3

0

0

3

Dec

0.62

Branch 3

2

4

0.6

0

4

Dec

0.36

Branch 4

2

5

0.4

0

5

Dec

0.1

Branch 5

3

8

0

0

8

Ch

0.62

Branch 6

3

17

0

0.2

17

Fin

0.20

Branch 7

4

6

0

0

6

Ch

0.36

Branch 8

4

11

0

0.1

11

Fin

0.1

Branch 9

6

9

0.9

0.4

9

Fin

0.4

Branch 10 6

10

0.1

0

10

Fin

0

Branch 11 5

7

0

0

7

Ch

0.08

Branch 12 5

14

0

0.1

14

Fin

0.1

Branch 13 7

12

0.2

0.4

12

Fin

0.4

Branch 14 7

13

0.8

0

13

Fin

0

Branch 15 8

15

0.6

1

15

Fin

1

Branch 16 8

16

0.4

0.05

16

Fin

0.05

Yes

Yes Yes

Yes

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 3-52. f. Start Ending Branch

Profit

Use

Node

Node

Node Node

Prob.

(End Node) Branch? Type

Value

Start

0

1

0

0

Dec

0.80

Branch 1

1

2

0

0

Ch

0.726

Branch 2

1

3

0

0

Dec

0.80

Branch 3

2

4

0.6

0

Dec

0.81

Branch 4

2

5

0.4

0

Dec

0.60

Branch 5

3

8

0

0

Ch

0.76

Branch 6

3

17

0

0.8

Fin

0.80

Branch 7

4

6

0

0

Ch

0.81

Branch 8

4

11

0

0.6

Fin

0.60

Branch 9

6

9

0.9

0.9

Fin

0.90

Branch 10 6

10

0.1

0

Fin

0.00

Branch 11 5

7

0

0

Ch

0.18

Branch 12 5

14

0

0.6

Fin

0.60

Branch 13 7

12

0.2

0.9

Fin

0.90

Branch 14 7

13

0.8

0

Fin

0.00

Branch 15 8

15

0.6

1

Fin

1.00

Branch 16 8

16

0.4

0.4

Fin

0.40

Yes

Yes Yes

Yes

3-53. 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.

Table for Problem 3-53a Return in $1,000 NO. OF WATCHES EVENT 1 EVENT 2 EVENT 3 EVENT 4 EVENT 5 Probability

0.10

0.20

0.50

0.10

0.10

Expected Profit

100,000

100

110

120

135

140

119.5

150,000

90

120

140

155

170

135.5

200,000

85

110

135

160

175

131.5

250,000

80

120

155

170

180

144.5

300,000

65

100

155

180

195

141.5

350,000

50

100

160

190

210

145

400,000

45

95

170

200

230

151.5

450,000

30

90

165

230

245

151

500,000

20

85

160

270

295

155.5

b. For this decision problem, Alternative 9, stocking 500,000, gives the highest expected profit of $155,500. c. The expected value with perfect information is $175,500, and the expected value of perfect information (EVPI) is $20,000. d. 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: NO. OF WATCHES EVENT 1 EVENT 2 EVENT 3 EVENT 4 EVENT 5 Probability

0.100

0.280

0.500

0.100

0.020

Expected Profit

100,000

100

110

120

135

140

117.1

150,000

90

120

140

155

170

131.5

200,000

85

110

135

160

175

126.3

250,000

80

120

155

170

180

139.7

300,000

65

100

155

180

195

133.9

350,000

50

100

160

190

210

136.2

400,000

45

95

170

200

230

140.7

450,000

30

90

165

230

245

138.6

500,000

20

85

160

270

295

138.7

Population

Population

Row

Same

Grows

Average

3-54. a. Decision under uncertainty. b.

Large wing

–85,000

150,000

32,500

Small wing

–45,000

60,000

7,500

0

0

0

No wing

c. Best alternative: large wing. 3-55. a. Weighted Population

Population

Average with

Same

Grows

 = 0.75

Large wing

–85,000

150,000

91,250

Small wing

–45,000

60,000

33,750

0

0

0

No wing

b. Best decision: large wing. c. No.

3-56. a. No

Mild

Severe

Expected

Congestion

Congestion

Congestion

Time

Tennessee

15

30

45

25

Back roads

20

25

35

24.17

Expressway

30

30

30

30

Probabilities

(30 days)/(60 (20 days)/(60 (10 days)/(60 days) = 1/2 days) = 1/3 days) = 1/6

b. Back roads (minimum time used). c. Expected time with perfect information: 15  1/2 + 25  1/3 + 30  1/6 = 20.83 minutes Time saved is 3

1

3

; minutes.

3-57. a. EMV can be used to determine the best strategy to minimize costs. The QM for Windows solution is provided. The best decision is to go with the partial service (maintenance) agreement. Solution to 3-57a Probabilities

0.2

0.8

Maint.

No Maint.

Expected

Row

Row

Cost ($)

Cost ($)

Value

Minimum

Maximum

($)

($)

($)

No Service Agreement

3,000

0

600

0

3,000

Partial Service Agreement

1,500

300

540

0

1,500

500

500

500

500

500

500

0

500

Complete Agreement Column best

Service

The minimum expected monetary value is $500 given by Complete Service Agreement b. The new probability estimates dramatically change Sim’s expected values (costs). 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 in the table.

Solution to 3-57b Probabilities

0.8

0.2 Does Not

Expected

Needs Repair

Need Repair

Value

($)

($)

($)

No Service Agreement

3,000

0

2,400

Partial Service Agreement

1,500

300

1,260

500

500

500

Complete Agreement

Service

Column best

500

3-58. We can use QM for Windows to solve this decision making under uncertainty problem. We have shown probability values for the equally likely calculations. As you can see, the maximax decision is Option 4 based on the $30,000, and the maximin decision is Option 1 based on the 5,000. As seen in the table, the equally likely decision is Option 3 because the average value for this is $5750. Solution to 3-58 Prob.

0.25

0.25

0.25

0.25

Judge

Trial

Court

Arbitration Equally Row Likely

Row

Min.

Max.

Option 1

5,000

5,000

5,000

5,000

5,000

5,000

5,000

Option 2

10,000

5,000

2,000

0

4,250

0

10,000

Option 3

20,000

7,000

1,000

–5,000

5,750

–5,000

20,000

Option 4

30,000

15,000

–10,000

–20,000

3,750

–20,000

30,000

5,750

5,000

30,000

Column best