imchap005s

SUPPLEMENT TO CHAPTER 5

Chapter 05S - Decision TheoryCHAPTER 05SDECISION THEORY

Solutions1.Given: A contractor must make a decision on capacity for next year. Estimated profits under each of the two possible states of nature are shown below:Next Years Demand

AlternativeLowHigh

Do nothing$50$60

Expand$20$80

Subcontract$40$70

a. Maximax: Determine best possible payoff for each alternative and choose the alternative that has the best.Next Years Demand

AlternativeLowHighBest Payoff

Do nothing$50$60$60

Expand$20$80$80Best of the Best

Subcontract$40$70$70

Conclusion: Select Expand alternative with a payoff of $80.

b. Maximin: Determine the worst possible payoff for each alternative and choose the alternative that has the best worst.Next Years Demand

AlternativeLowHighWorst Payoff

Do nothing$50$60$50Best of the Worst

Expand$20$80$20

Subcontract$40$70$40

Conclusion: Select Do nothing alternative with a payoff of $50.c.Laplace: Determine the average payoff for each alternative and choose the alternative with the best average.Next Years Demand

AlternativeLowHighAverage Payoff

Do nothing$50$60$55Best

Expand$20$80$50

Subcontract$40$70$55Best

Conclusion: Select Do nothing or Subcontract alternative with an average payoff of $55.

d. Minimax Regret: Prepare a table of regrets (opportunity losses)for each column, subtract every payoff from the best payoff in that column. Identify the worst regret for each alternative. Select the alternative with the best worst.Regrets

AlternativeLowHighWorstRegret

Do nothing$0($50-$50)$20($80-$60)$20

Expand$30($50-$20)$0($80-$80)$30

Subcontract$10($50-$40)$10($80-$70)$10Best of the Worst

Conclusion: Select Subcontract alternative because it has the best of the worst regrets of the three alternatives.

2.Given: P(Low Demand) = .3 and P(High Demand) = .7.a. Determine the best expected profit of the alternatives from Problem 1

Expected Profit:Do nothing=.3($50) + .7($60) = $15 + $42 = $57Expand =.3($20) + .7($80) = $6 + $56 = $62Subcontract = .3($40) + .7($70) = $12 + $49 = $61Conclusion: Expand is the best alternative because it has the highest expected value.

b. Decision Tree Analysis to Select an Alternative:

.3.7.3.7.3.7$50$60$20$80$40$70$62Expand$57Do Nothing$61Subcontr.

Expected Value Calculations:Do nothing=.3($50) + .7($60) = $15 + $42 = $57Expand =.3($20) + .7($80) = $6 + $56 = $62Subcontract = .3($40) + .7($70) = $12 + $49 = $61Conclusion: Expand is the best alternative because it has the highest expected value ($62).

c.Expected Value of Perfect Information:Expected value of perfect information (EVPI) = Expected payoff under certainty Expected payoff under risk Find the best payoff under each state of nature:Low Demand: Best Payoff = $50High Demand: Best Payoff = $80Expected payoff under certainty (apply the probabilities of each state of nature) == (Prob. of Low Demand x Best Payoff) + (Prob. of High Demand x Best Payoff)= .3($50) + .7($80) = $15 + $56 = $71

Expected payoff under risk = Expected value of the alternative selected = $62

EVPI = $71 - $62 = $9

3.Plot each alternative relative to P(High Demand). Plot the payoff value for Low Demand on the left side of the graph and the payoff value for High Demand on the right side of the graph.

Payoff values from Problem 1:

Next Years Demand

AlternativeLowHigh

Do nothing$50$60

Expand$20$80

Subcontract$40$70

205040Do NothingExpand0.50.66671.0P807060Low PayoffHigh PayoffSubcontract

P(High)

The graph above shows the range of values of P(High) over which each alternative is optimal.For low values of P(High), Do Nothing is best because it has the highest expected value.For intermediate values of P(High), Subcontract is best.For higher values of P(High), Expand is best.

To find the exact values of the ranges, we must determine where the upper parts of the lines intersect. For each line, b is the slope of the line and x = P(High). The slope of each line = Right-hand value Left-hand value.

Equations:Do Nothing:50 + 10P (slope = 60 50)Expand:20 + 60P (slope = 80 20)Subcontract:40 + 30P (slope = 70 40)

Find the intersection between Do Nothing & Subcontract:50 + 10P = 40 + 30P10P 30P = 40 50-20P = -10P = -10/-20P = .5000 (this value is shown on the graph above already)

Find the intersection between Subcontract & Expand:40 + 30P = 20 + 60P30P 60P = 20 40-30P = -20P = -20/-30P = .6667 (this value is shown on the graph above already)

Optimal ranges:Do nothing: P(High) = 0 to < .5000Subcontract: P(High) > .5000 to < .6667Expand:P(High) > .6667 to 1.00

4.a. 1) Draw the tree diagram:

$400,000 (1)$50,000 (2)$450,000 (3)$-10,000 (4)$800,000 (5)Demand Low (.4)Demand Low (.4)Demand High (.6)Demand High (.6)MaintainExpandBuild SmallBuild Large122)Analyze decisions from right to left (i.e., work backwards from the end of the tree towards the root). For instance, begin with decision 2 and choose expansion because it has a higher present value ($450,000 vs. $50,000). Draw a double slash through the Maintain alternative.

3)Determine the product of the chance probabilities and their respective payoffs for the remaining branches.

Build SmallDemand Low:.4($400,000) = $160,000Demand High:.6($450,000) = $270,000

Build LargeDemand Low:.4(-$10,000) = -$4,000Demand High:.6($800,000) = $480,000

4)Determine the expected value of each initial alternative.Build Small = $160,000 + $270,000 = $430,000Build Large = -$4,000 + $480,000 = $476,000

Conclusion: Because the expected value of building a large plant is highest, select the large plant alternative. Draw a double slash through the Build Small alternative.

b.Expected payoff under certainty: .4(400,000) + .6(800,000) = $640,000 -Expected payoff under risk: -476,000Expected value of perfect information:$164,000

c.Determine the range over which each alternative would be best in terms of the value of P(Low).Plot each alternative relative to P(Low). Plot the payoff value for High Demand on the left side of the graph and the payoff value for Low Demand on the right side of the graph.AlternativeHigh DemandLow Demand

Build Small$450,000400,000

Build Large$800,000-$10,000

Low PayoffHigh Payoff

Large800,000

Small450,000

400,000

01.0

P(Low)

-10,000

The graph above shows the range of values of P(Low) over which each alternative is optimal. For low values of P(Low), Build Large is best because it has the highest expected value. For high values of P(Low), Build Small is best because it has the highest expected value.To find the exact values of the ranges, we must determine where the upper parts of the lines intersect. For each line, b is the slope of the line and x = P(Low). The slope of each line = Right-hand value Left-hand value.Equations:Build Small:450,000 50,000P (slope = 400,000 450,000)Build Large:800,000 810,000P (slope = -10,000 800,000)

Find the intersection between the two lines:450,000 50,000P = 800,000 810,000P-50,000P + 810,000P = 800,000 450,000760,000P = 350,000P = 350,000/760,000P = .4605

Optimal ranges:Build Large: P(Low) = 0 to < .4605Build Small:P(Low) > .4605 to 1.005.Analyze the decisions from right to left:

1) Determine which alternative would be selected for each possible second decision.Subcontract with Medium demand: Select either alternative. Payoff = $1.3.Subcontract with Large demand: Select Build. Payoff = $1.8. Place a double slash through Do nothing and Expand.Expand with Small demand: Select Other use #1. Payoff = $1.5. Place a double slash through Do nothing and Other use #2.Expand with Large demand: Select Build. Payoff = $1.7. Place a double slash through Do nothing and Subcontract.Build with Small demand: Select Other use #1. Payoff = $1.4. Place a double slash through Do nothing and Other use #2.Build with Medium Demand: Select Other use #1. Payoff = $1.1. Place a double slash through Do nothing and Other use #2.

2) Determine the product of the chance probabilities and their respective payoffs for the remaining branches.

SubcontractSmall demand:.4($1.0) = $0.40Medium demand:.5($1.3) = $0.65Large demand:.1($1.8) = $0.18

ExpandSmall demand:.4($1.5) = $0.60Medium demand:.5($1.6) = $0.80Large demand: .1($1.7) = $0.17

BuildSmall demand:.4($1.4) = $0.56Medium Demand:.5($1.1) = $0.55Large demand:.1($2.4) = $0.24

3) Determine the expected value of each initial alternative.

Subcontract = $0.40 + $0.65 + $0.18 = $1.23Expand = $0.60 + $0.80 + $0.17 = $1.57Build = $0.56 + $0.55 + $0.24 = $1.35

Conclusion: Because the expected value of the Expand alternative is greatest, select it. Draw a double slash through Subcontract and Build.

6.Given:Management must decide whether to renew the lease for another 10 years or to relocate near the site of a proposed motel. The net present values of the two alternatives under each state of nature are given in the table below.States of Nature

AlternativeMotel ApprovedMotel Rejected

Renew$500,000$4,000,000

Relocate$5,000,000$100,000

a. Maximax: Determine best possible payoff for each alternative and choose the alternative that has the best.States of Nature

AlternativeMotel ApprovedMotel RejectedBest Payoff

Renew$500,000$4,000,000$4,000,000

Relocate$5,000,000$100,000$5,000,000Best of the Best

Conclusion: Select Relocate alternative with a payoff of $5,000,000.

b. Maximin: Determine the worst possible payoff for each alternative and choose the alternative that has the best worst.States of Nature

AlternativeMotel ApprovedMotel RejectedWorst Payoff

Renew$500,000$4,000,000$500,000Best of the Worst

Relocate$5,000,000$100,000$100,000

Conclusion: Select Renew alternative with a payoff of $500,000.

c. Laplace: Determine the average payoff for each alternative and choose the alternative with the best average.States of Nature

AlternativeMotel ApprovedMotel RejectedAverage Payoff

Renew$500,000$4,000,000$2,250,000

Relocate$5,000,000$100,000$2,550,000Best

Conclusion: Select Relocate alternative with an average payoff of $2,550,000.d. Minimax Regret: Prepare a table of regrets (opportunity losses)for each column, subtract every payoff from the best payoff in that column. Identify the worst regret for each alternative. Select the alternative with the best worst.Regrets

AlternativeMotel ApprovedMotel RejectedWorst Regret

Renew$4,500,000($5,000,000-$500,000)$0($4,000,000-$4,000,000)$4,500,000

Relocate$0($5,000,000-$5,000,000)$3,900,000($4,000,000-$100,000)$3,900,000

Best of the Worst

Conclusion: Select Relocate alternative because it has the best of the worst regrets of the two alternatives.

7.Given: Probability that the motels application will be approved = .35.

The probability that the motel will be rejected = 1.00 - .35 = .65.

AlternativeExpected Value

a.Renew(.35)500,000 + (.65)4,000,000 = $2,775,000

Relocate(.35)5,000,000 + (.65)100,000 = $1,815,000

Conclusion:Renew lease.

b.

$500,000$4,000,000Approve (.35)Reject (.65)Approve (.35)Reject (.65)$5,000,000$100,000$2,775,000$1,815,000E.V.RenewRelocate

Conclusion:Renew lease.

c. Expected value of perfect information (EVPI) = Expected payoff under certainty Expected payoff under risk

Find the best payoff under each state of nature:Motel Approved: Best Payoff = $5,000,000Motel Rejected: Best Payoff = $4,000,000Expected payoff under certainty (apply the probabilities of each state of nature) == (Prob. of Motel Approved x Best Payoff) + (Prob. of Motel Rejected x Best Payoff)= .35($5,000,000) + .65($4,000,000) = $1,750,000 + $2,600,000 = $4,350,000

Expected payoff under risk = Expected value of the alternative selected = $2,775,000

EVPI = $4,350,000 - $2,775,000 = $1,575,000

Conclusion: Yes, the manager should sign the lease for $24,000 because this cost is less than the EVPI of $1,575,000.

8.a. Determine the range over which each alternative would be best in terms of the value of P(Approved).Plot each alternative relative to P(Approved). Plot the payoff value for Rejected on the left side of the graph and the payoff value for Approved on the right side of the graph.AlternativeRejectedApproved

Renew$4,000,000$500,000

Relocate$100,000$5,000,000

RejectedApproved

Renewal better than(millions)RelocationFor 8(a) and 8(b) the decision should be to renew the 10-year lease because a. P(Approved) of .35 < .4643 & b. P(Approved) of .45 < .4643Exp. ValueRenewRelocate100,000 + 4,900,000x54321P (Approved)Relocatex = .46434,000,000 3,500,000x.554321

The graph above shows the range of values of P(Approved) over which each alternative is optimal. For low values of P(Approved), Renewal is best because it has the highest expected value. For high values of P(Approved), Relocate is best because it has the highest expected value.

To find the exact values of the ranges, we must determine where the upper parts of the lines intersect. For each line, b is the slope of the line and x = P(Approved). The slope of each line = Right-hand value Left-hand value.Equations:Renew:4,000,000 3,500,000P (slope = 500,000 4,000,000)Relocate:100,000 + 4,900,000P (slope =5,000,000 100,000)

Find the intersection between the two lines:4,000,000 3,500,000P = 100,000 + 4,900,000P-3,500,000P 4,900,000P = 100,000 4,000,000-8,400,000P = -3,900,000P = -3,900,000/-8,400,000P = .4643

Optimal ranges:Renew: P(Approved) = 0 to < .4643Relocate:P(Approved) > .4643 to 1.00

Conclusion: Still should renew because P = .35 is less than P = .4643.

b.Conclusion: Still should renew because P = .45 is less than P = .4643.c.

Given: Renew = .35(500,000) + .65(4,000,000)Relocate = .35(5,000,000) + .65(100,000) = $1,815,000

Plug in x for 4,000,000 in Renew equation:Renew = .35(500,000) + .65xRelocate = .35(5,000,000) + .65(100,000) = $1,815,000

Set the two equations equal and solve for x:.35(500,000) + .65x = $1,815,000175,000 + .65x = 1,815,000.65x = 1,815,000 175,000.65x = 1,640,000x = 1,640,000/.65x = 2,523,077

Original decision at the payoff of $4,000,000 would have to decrease to less than $2,523,077 for the decision to change from Renew to Relocate.

Conclusion: The decision to Renew remains the same for the range of x > $2,523,077.At x = $2,523,077, we would be indifferent between Renew and Relocate.At x < $2,523,077, the decision would change to Relocate.

.2 Low.8 High.2 Low.8 HighSubcontractExpand Greatly2.2 LowDo NothingExpand2.8 High$ 424248224650(20)7246.844.453.6.2(42)+.8(48).2(22)+.8(50).2(-20)+.8(72)SmallMediumLarge19. a.53.646.84844.45053.6

Analyze the decisions from right to left:

1) Determine which alternative would be selected for each possible second decision (marked as 2).Small with High demand: Select Expand Greatly. Payoff = $48. Place a double dash through Subcontract.Medium with High demand: Select Expand. Payoff = $50. Place a double slash through Do Nothing.


SmallLow demand:.2($42) = $8.4High demand:.8($48) = $38.4

MediumLow demand:.2($22) = $4.4Medium demand:.8($50) = $40

LargeLow demand:.2(-$20) = -$4Medium Demand:.8($72) = $57.6


Small = $8.4 + $38.4 = $46.8Medium = $4.4 + $40 = $44.4Large = -$4 + $57.6 = $53.6

Conclusion: Because the expected value of the Large alternative is greatest, select it. Draw a double slash through Small and Medium.

b. Maximin Alternative: Determine the worst possible payoff for each alternative and choose the alternative that has the best worst.

Next Years Demand

AlternativeLowHighWorst Payoff

Small$42$48$42Best of the Worst

Medium$22$50$22

Large-$20$72-$20

Conclusion: Build Small.

c. Expected value of perfect information (EVPI) = Expected payoff under certainty Expected payoff under risk

Find the best payoff under each state of nature:Low Demand: Best Payoff = $42High Demand: Best Payoff = $72Expected payoff under certainty (apply the probabilities of each state of nature) == (Prob. of Low Demand x Best Payoff) + (Prob. of High Demand x Best Payoff)= .2($42) + .8($72) = $8.4 + $57.6 = $66

Expected payoff under risk = Expected value of the alternative selected in part a = $53.6

EVPI = $66 - $53.6 = $12.4

d. Determine the range over which each alternative would be best in terms of the value of P(High).

Plot each alternative relative to P(High). Plot the payoff value for Low on the left side of the graph and the payoff value for High on the right side of the graph.

Next Years Demand

AlternativeLowHigh

Small$42$48

Medium$22$50

Large-$20$72

HighLow

7250484222SmallMediumLarge0.72091.0P (High)-20

The graph above shows the range of values of P(High) over which each alternative is optimal.For low values of P(High), Small is best because it has the highest expected value.For interemediate and higher values of P(High), Large is best.

To find the exact values of the ranges, we must determine where the upper parts of the lines intersect. For each line, b is the slope of the line and x = P(High). The slope of each line = Right-hand value Left-hand value.

Equations:Small:42 + 6P (slope = 48 42)Large:-20 + 92P (slope = 72 (-20))

Find the intersection between Small & Large:42 + 6P = -20 + 92P6P 92P = -20 42-86P = -62P = -62/-86P = .7209 (this value is shown on the graph above already)

Optimal ranges:Small: P(High) = 0 to < .7209Large:P(High) > .7209 to 1.00

10. Decision Tree

buy 1.70 high.30 lowdo nothingbuy 2ndsubcontract.30 low1307510011090$90buy 2

.7 high

Analyze the decisions from right to left:

1) Determine which alternative would be selected for each possible second decision.Buy 1 with High Demand: Select Subcontract. Payoff = $110. Place a double dash through Do Nothing and Buy 2nd.


Buy 1Low Demand:.30($90) = $27High demand:.70($110) = $77

Buy 2Low demand:.30($75) = $22.5Medium demand:.70($130) = $91


Buy 1 = $27+ $77 = $104Buy 2 = $22.5 + $91 = $113.5

Conclusion: Because the expected value of the Buy 2 alternative is greatest, select it. Draw a double slash through Buy 1.

EV1 = (.3) (90) + (.7) (110) = 104EV2 = (.3) (75) + (.7) (130) = 113.5 Since 113.5 > 104Decision: Buy 2 machines.

EV=50

(45)99.30.30.506040Alternative AAlternative BD1.20D2a09060 1D2b 21/31/3.5050 51/344D2c401/3.20 3501/2 41/31/345301/24011.Decision Tree

EV=49

EV=33

EV=46

EV=45

1) Determine the product of the chance probabilities and their respective payoffs for the branches on the right hand side. Because this is a complex problem, we have added labels to the circles:EV1 = (1/3)(0) + (1/3)(60) + (1/3)(90) = 50EV2 = (1/3)(45) + (1/3)(45) + (1/3)(99) = 33EV3 = (1/2)(40) + (1/2)(50) = 45

2) Determine which alternative would be selected for each possible second decision. We have labeled these D2a, D2b, and D2c.

D2a: Select upper branch with payoff of 50. Draw a double slash through the lower branch.D2b: Select lower branch with payoff of 40. Draw a double slash through the upper branch.D2c: Select lower branch with payoff of 45. Draw a double slash through the upper branch.

3) Determine the product of the chance probabilities and their respective payoffs for the branches on the left hand side. Because this is a complex problem, we have added labels to the circles:

EV4 = (.30)(50) + (.50)(44) + (.20)(60) = 49EV5 = (.30)(40) + (.50)(50) + (.20)(45) = 46


Alternative A = $49 (work is shown in previous step)Alternative B = $46 (work is shown in previous step)

Conclusion: Because the expected value of Alternative A is greatest, select it for Decision 1 (D1). Draw a double slash through Alternative B.

12.a.1) Draw the tree diagram. Because the probabilities are unknown, we would assume that each state of nature has an equal probability of occurring.

$700$100$500$40$2,000 Demand Low (.50)Demand Low (.50)Demand High (.50)Demand High (.50) LeaseExpandBuild SmallBuild Large122)We have to make a choice for the possible second decision before proceeding.Expand has a higher payoff than Lease ($500 > $100). Select Expand. Draw a double slash through the Lease branch.

b.Use the tree diagram to identify the choice that you would make using each of the four approaches for decision making under uncertainty.States of Nature

AlternativeDemand LowDemand High

Build Small$700$500

Build Large$40$2,000

1) Maximax: Determine best possible payoff for each alternative and choose the alternative that has the best.States of Nature

AlternativeDemand LowDemand HighBest Payoff

Build Small$700$500$700

Build Large$40$2,000$2,000Best of the Best

Conclusion: Select Build Large with a payoff of $2,000.2) Maximin: Determine the worst possible payoff for each alternative and choose the alternative that has the best worst.States of Nature

AlternativeDemand LowDemand HighWorst Payoff

Build Small$700$500$500Best of the Worst

Build Large$40$2,000$40

Conclusion: Select Build Small alternative with a payoff of $500.

3) Laplace: Determine the average payoff for each alternative and choose the alternative with the best average.States of Nature

AlternativeDemand LowDemand HighAverage Payoff

Build Small$700$500$600

Build Large$40$2,000$1,020Best

Conclusion: Select Build Large alternative with an average payoff of $1,020.4) Minimax Regret: Prepare a table of regrets (opportunity losses)for each column, subtract every payoff from the best payoff in that column. Identify the worst regret for each alternative. Select the alternative with the best worst.Regrets

AlternativeDemand LowDemand HighWorst Regret

Build Small$0($700-$700)$1,500($2,000-$500)$1,500

Build Large$660($700-$40)$0($2,000-$2,000)$660

Best of the Worst

Conclusion: Select Build Large alternative because it has the best of the worst regrets of the two alternatives.

13.Given: We have the estimated costs for various alternatives and caseloads shown below.

Caseload

AlternativeModerateHighVery High

Reassign staff506085

New staff606060

Redesign collection405090

Note: Because this problem uses costs, best is associated with lowest cost.

a. Maximin: Determine the worst possible payoff for each alternative and choose the alternative that has the best worst.Caseload

AlternativeModerateHighVery HighWorst


New staff60606060Best of the Worst


Conclusion: Select New staff alternative.

b. Maximax: Determine best possible payoff for each alternative and choose the alternative that has the best.Caseload

AlternativeModerateHighVery HighBest


New staff60606060

Redesign collection40509040Best of the Best

Conclusion: Select Redesign collection alternative.

c. Minimax Regret: Prepare a table of regrets (opportunity losses)for each column, subtract every payoff from the best payoff in that column. Identify the worst regret for each alternative. Select the alternative with the best worst.Regret

AlternativeModerateHighVery HighWorst Regret

Reassign staff10(50-40)10(60-50)25(85-60)25

New staff20(60-40)10(60-50)0(60-60)20Best of the Worst

Redesign collection0(40-40)0(50-50)30(90-60)30

Conclusion: Select New Staff alternative.

d. Laplace: Determine the average payoff for each alternative and choose the alternative with the best average.Caseload

AlternativeModerateHighVery HighAverage


New staff60606060Best (tie)

Redesign collection40509060Best (tie)

Conclusion: Select either New Staff or Redesign collection alternative.

14.Given: Probabilities for states of nature are now given as follows: .10 for moderate, .30 for high, and .60 for very high.

a. Minimum expected cost:Reassign:.10(50) + .30(60) + .60(85) = $74

New Staff:.10(60) + .30(60) + .60(60) = 60

Redesign:.10(40) + .30(50) + .60(90) = 73

Conclusion: New Staff alternative will yield the minimum expected cost.

New StaffReassignRedesign506085606060405090.30 High.60 Very High.10 Moderate.30 High.60 Very High.10 Moderate.30 High.60 Very High.10 Moderate74607360b.

Conclusion: New Staff alternative will yield the minimum expected cost. Draw a double slash through the Reassign and Redesign branches.

c.Opportunity loss table

Regret

AlternativeModerateHighVery High


New staff20100


Expected regret for each alternative:Reassign:.10(10) +.30(10) + .60(25) = 19New staff:.10(20) +.30(10) + .60(0) = 5Redesign:.10(0) +.30(0) + .60(30) = 18

Conclusion: The lowest expected regret is 5. Hence, the EVPI = 5.

15.a. Given: Payoffs (profits) are provided in the table below.Plot each alternative relative to P(1). Plot the payoff value for #2 on the left side of the graph and the payoff value for #1 on the right side of the graph.

State of Nature

Alternative#2#1

A20120

B4060

C11010

D9090

#1#2

120A

110C

D9090

B

60

40

2010

01.0

P(#1)

Alternative C is best for the lowest range of P(#1), followed by Alternative D for the intermediate range, and then Alternative A for the highest range.

Equations:A: 20 + 100P (slope = 120 20)B: 40 + 20P (slope = 60 40)C: 110 100P (slope = 10 110)D: 90 + 0P (slope = 90 90)

Find the intersection between C & D:110 100P = 90 + 0P-100P = 90 - 110-100P = -20P = -20/-100P = .2000

Find the intersection between D & A:90 + 0P = 20 + 100P90 20 = 100P70 = 100PP = 70/100P =.7000

Optimal ranges:C: P(#1) = 0 to < .2000D:P(#1) > .2000 to < .7000A:P(#1) > .7000 to 1.00

b. Treat the payoffs as costs.

#2#1

120A

110C

D9090

B

60

40

1020

01.0

P(#1)

Alternative A is best for the lowest range of P(#1), followed by Alternative B for the intermediate range, and then Alternative C for the highest range.

Equations:A: 20 + 100P (slope = 120 20)B: 40 + 20P (slope = 60 40)C: 110 100P (slope = 10 110)D: 90 + 0P (slope = 90 90)

Find the intersection between A & B:20 + 100P = 40 + 20P100P 20P = 40 2080P = 20P = 20/80P = 0.2500

Find the intersection between B & C:40 + 20P = 110 100P20P + 100P = 110 40120P = 70P = 70/120P = .5833

Optimal ranges:A: P(#1) = 0 to < .2500B:P(#1) > .2500 to < .5833C:P(#1) > .5833 to 1.00

16.a.Determine the range over which each alternative would be best in terms of the value of P(High).

Plot each alternative relative to P(#2). Plot the payoff value for #1 on the left side of the graph and the payoff value for #2 on the right side of the graph.

State of Nature

Alternative#1#2

A$20$140

B$120$80

C$100$40

Payoff#214080401.00120100Payoff#1P(#2)BAC

20

b.Alternative C is lower than Alternative B for all values of P(#2), so it would never be appropriate in terms of maximizing profits.

c.For low and intermediate values of P(#2), Alternative B is best because it has the highest expected value. For higher values of P(#2), Alternative A is best.

To find the exact values of the ranges, we must determine where the upper parts of the lines intersect. For each line, b is the slope of the line and x = P(#2). The slope of each line = Right-hand value Left-hand value.

Equations:A: 20 +120P (slope = 140 20)B: 120 40P (slope = 80 120)

Find the intersection between A & B:20 +120P = 120 40P120P + 40P = 120 20160P = 100P = 100/160P = .6250Conclusion: Select Alternative A if P(#2) is greater than .6250.

d.Conclusion: For P(#1), choose Alternative A if P(#1) is less than .3750 (i.e., 1.00 .6250).

17.a.[Refer to the diagram in the solution for Problem 16]

b.Alternative B is higher than Alternative C for all values of P(#2), so Alternative B would never be appropriate in terms of minimizing costs.

c.For low values of P(#2), Alternative A is best. For intermediate and higher values of P(#2), Alternative C is best.

Equations:A: 20 +120P (slope = 140 20)C: 100 60P (slope = 40 100)

Find the intersection between A & C:20 +120P = 100 60P120P + 60P = 100 20180P = 80P = 80/180P = .4444

Conclusion: Select Alternative A for P(#2) less than .4444 and choose Alternative C for P(#2) greater than .4444.

d.Conclusion: In terms of P(#1), choose Alternative A for P(#1) greater than .5556 (1.00 - .4444) and choose Alternative C for P(#1) less than .5556.

18.Given: Payoffs are provided in the table below.Plot each alternative relative to P(No Contract). Plot the payoff value for Receive Contract on the left side of the graph and the payoff value for No Contract on the right side of the graph.

State of Nature

AlternativeReceive ContractNo Contract

#110-2

#283

#355

#407

108#3Payoff for No Contract75350-21.0#1#2#4Payoff for Contract

P(No Contract)

Alternative 1 is best for the lowest range of P(No Contract), followed by Alternative 2 for the next range, then Alternative 3 for the range after that, and then Alternative 4 for the highest range.

Equations:1: 10 12P (slope = -2 10)2: 8 5P (slope = 3 8)3: 5 + 0P (slope = 5 5)4: 0 + 7P (slope = 7 0)Find the intersection between 1 & 2:10 12P = 8 5P-12P (-5P) = 8 10-7P = -2P = -2/-7P = .2857

Find the intersection between 2& 3:8 5P = 5 + 0P8 5P = 5-5P = 5 8-5P = -3P = -3/-5P = .6000

Find the intersection between 3 & 4:5 + 0P = 0 + 7P5 = 7PP = 5/7P = .7143

Optimal ranges:#1: P(No Contract) = 0 to < .2857#2:P(No Contract) > .2857 to < .6000#3:P(No Contract) > .6000 to < .7143#4:P(No Contract) > .7143 to 1.00

Payoff#2Payoff#1.30.801209060100.417.751.0204090110ADCBACCostsProfitsP (#2)

Enrichment Model: Advanced Decision Tree ProblemsIn this section two additional decision tree problems are presented1. Space engineers have three alternative designs for the configuration of a component for an unmanned space shuttle. The space vehicle is likely to encounter one of four different conditions, which have probabilities of occurrence as listed in the following payoff table (costs in $00) with the payoffs for each combination of design and state of nature. Additional data from previous flights are available but will require additional expenditures to analyze. However, the project director is confident that analysis of the data will clearly indicate which state of nature will be encountered. What amount would be justified for the data analysis?States of Nature

ABCD

Probability:.3.4.2.1

Design00120*10100

0021510040

00310203030

* Costs in $ hundreds.

2.Demand for movie rentals at a video store on Saturdays during summer months is related to the weather. If it is raining, or if the chance of rain is greater than 50%, demand tends to follow one distribution, whereas if it is not raining and the chance of rain does not exceed 50%, demand follows a different distribution. This is important to the video store because the manager must decide early on Saturday how many employees to schedule for Saturday afternoon and early evening.The two distributions are:P(Rain) > 50%P(Rain) 50%DemandProbabilityDemandProbability

Low.10Low.60

Moderate.20Moderate.30

High.70High.10

The regular staff can handle Low demand. Moderate demand requires two additional employees, and High demand requires another two employees. The payoff table (profits in $000) is shown below:DemandLowModerateHigh

0234

Extra Staff2145

4036

a. Construct a tree diagram showing the payoffs for this situation.b.Determine the number of additional staff needed for a rainy Saturday.c.Determine the number of staff needed when the chance of rain is 20% for a Saturday.

Solution to Problem 11.ABCD

Design.3.4.2.1Expected Value

001201010012

002151004012.5

0031020303020

Design 001 minimizes the expected cost. Expected payoff under risk = $12 ($1,200).

Find the best payoff under each state of nature:A: Best (minimum cost) = 10B: Best (minimum cost) =10C: Best (minimum cost) = 0D: Best (minimum cost) = 0

Develop the opportunity loss table:

DesignABCD

001100100

00250040

0030103030

Expected regret for each alternative:001: .3(10) + .4(0) + .2(10) + .1(0) = $5002: .3(5) + .4(0) + .2(0) + .1(40) = $5.5003: .3(0) + .4(10) + .2(30) + .1(30) = $13

Conclusion: The lowest expected regret = $5 ($500). Hence, the EVPI = $500.

Solution to Problem 2

Two additional staffNo additional staffour additional staff234145036Medium demand (.2)High demand (.7)Low demand (.1)Medium demand (.2)High demand (.7)Low demand (.1)Medium demand (.2)mad(.2)High demand (.7)Low demand (.1)123b.Two additional staffNo additional staffFour additional staff234145036Medium demandHigh demandLow demandMedium demandHigh demandLow demandMedium demandHigh demandLow demand1232.a.

Four additional staff

EV1 = (.1)(2) + (.2)(3) + (.7)(4) = 3.6EV2 = (.1)(1) + (.2)(4) + (.7)(5) = 4.4EV3 = (.1)(0) + (.2)(3) + (.7)(6) = 4.8Because 4.8 > 4.4 > 3.6, hire four additional employees when it is rainy.

Two additional staffNo additional staffFour additional staff234145036Medium demand (.3)High demand (.1)Low demand (.6)Medium demand (.3)High demand (.1)Low demand (.6)Medium demand (.3)High demand (.1)Low demand (.6)123c.

EV1 = (.6)(2) + (.3)(3) + (.1)(4) = 2.5EV2 = (.6)(1) + (.3)(4) + (.1)(5) = 2.3EV3 = (.6)(0) + (.3)(3) + (.1)(6) = 1.5 Because 2.5 > 2.3 > 1.5, hire no additional employees when the probability of rain is 20%.16

116Operations Management, 10/e5S-1Copyright 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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