Chapter 18
Decision Making Under Uncertainty
CHAPTER OVERVIEW AND OBJECTIVES
This chapter provides an introduction to decision strategies
including structuring a decision problem into various actions under
consideration and states of nature describing the uncertain future.
Defining a decision table by using utility values, rather than dollars,
is presented. The concept of decision trees and using posterior
(revised) probabilities in a decision strategy is discussed and illustrated using several examples. Having completed this material, the
student should be able to:
1. Discuss and implement various decision strategies including
the minimax, maximax and expected payoff procedures.
2. Evaluate the risk associated with a given strategy.
3. Conduct a sensitivity analysis for a decision problem.
4. Construct and apply utility curves.
5. Use a decision tree to structure a decision application and
revise prior probabilities for each state of nature.
6. Discuss and determine the expected value of perfect
information (EVPI) and the expected value of sample
information (EVSI).
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Chapter 18 Glossary
actions (alternatives).
The events in a decision problem over which the
decision maker does have control, such as "purchase" or "lease".
admissable action.
An action for which no other action under
consideration dominates it.
chance node.
A point in a decision tree that is not under the control
of the decision maker.
Leading away from a chance node are states
of nature, not possible actions.
decision node.
A point in a decision tree that is under the control of
the decision maker.
Leading away from a decision node are
possible actions, not states of nature.
decision tree.
A method of representing a multistage decision problem,
containing decision nodes and/or chance nodes connected by
straight lines.
dominated action.
Action Aj is dominated by action Ai if the payoff for
Aj is less than or equal to that for Ai under each state of
nature.
efficiency of sample information.
The ratio EVSI/EVPI times 100.
expected value of perfect information (EVPI).
The amount a decision
maker would be willing to pay for a perfect predictor.
expected value of sample information (EVSI).
The expected payoff when
obtaining sample information minus the expected payoff without
obtaining sample information.
maximax strategy.
A decision strategy that selects the action having
the largest possible payoff.
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minimax strategy.
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A decision strategy (usually conservative) that
minimizes the maximum possible opportunity loss.
opportunity loss (Lij).
The difference between the payoff for action i
and the payoff for the action that would have the largest payoff
under state of nature j.
payoff.
The profit associated with taking a particular action given
that a specific state of nature has occurred.
posterior probability.
A revised probability based upon additional
information.
prior probability.
risk.
An initial estimate for the probability of an event.
The variance of the possible payoffs for a particular action.
risk avoider.
A decision maker who prefers a smaller expected payoff
with a small risk over a larger expected payoff with a large risk.
risk neutral.
A decision maker who maximizes expected utility by
maximizing the expected payoff.
risk taker.
A decision maker who prefers an action with a possible
large payoff even if that action has a large risk.
states of nature.
The uncertain events over which the decision maker
has no control.
utility value.
For a particular outcome, a measure of the
attractiveness and the risk associated with the corresponding
dollar amount.
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18.1
States of Nature
S1
S2
S3
S4
A1
500
500
500
500
A2
400
500
10000
20000
A3
-500
0
10000
30000
Under state S1, action A1 is the best.
or A2 is the best.
Under state S2, action A1
Under state S3, action A2 or A3 is the best.
Under state S4, action A3 is the best.
18.2
a)
Uncertainty
b)
Certainty
c)
Certainty
d)
Uncertainty
18.3
States of Nature
Action
7
8
9
10
11
12
5
5
=
5
=
5
=
5
=
5
=
5
=
x 25
145
-20
x 25
155
-30
x 25
165
-40
x 25
175
-50
x 25
185
-60
x 25
195
-70
6
6
=
6
=
6
=
6
=
6
=
6
=
x 25
145
5
x 25
155
-5
x 25
165
-15
x 25
175
-25
x 25
185
-35
x 25
195
-45
7
8
9
10
11
12
13
30
30
30
30
30
30
30
20
45
45
45
45
45
45
10
35
60
60
60
60
60
0
25
50
75
75
75
75
-10
15
40
65
90
90
90
-20
5
30
55
80
105
105
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13
5 x 25
- 205
= -80
6 x 25
- 205
= -55
-30
-5
20
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Chapter 18 713
70
95
120
18.4
States of Nature
Action
A1
A2
A3
S1
88,000
- 700
= 87,300
85,000
- 700
= 84,300
82,000
- 700
= 81,300
S2
S3
88,000
- 1,400
= 86,600
85,000
- 1,400
= 83,600
82,000
- 1,400
= 80,600
S4
88,000
- 2,100
= 85,900
85,000
- 2,100
= 82,900
82,000
- 2,100
= 79,900
S5
88,000
- 2,800
= 85,200
85,000
- 2,800
= 82,200
82,000
- 2,800
= 79,200
88,000
- 3,500
= 84,500
85,000
- 3,500
= 81,500
82,000
- 3,500
= 78,500
S6
88,000
- 4,200
= 83,800
85,000
- 4,200
= 80,800
82,000
- 4,200
= 77,800
18.5
States of Nature
Action
S1(100)
S2(125)
S3(150)
A1(100)
.90 x 100 = 90
.90 x 100 = 90
.90 x 100 = 90
A2(125)
.90 x 100
1.10 x 25
.90 x 100
1.10 x 50
.90 x 125
= 112.5
.90 x 125 1.10 x 25 = 85
.90 x 125
= 112.5
.90 x 150
= 135
A3(150)
18.6
= 62.5
= 35
Opportunity loss table:
States of Nature
S1
S2
S3
S4
A1
0
0
2
6
A2
1
1
1
2
A3
1
2
0
1
A4
3
3
1
0
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Action
Maximum Opportunity Loss
A1
6
A2
2
A3
2
A4
3
The minimax decision is either action A2 or A3.
18.7
a)
No, because under S2 there is no 0.
18.8
b) No, because there is a negative value in the table.
Payoff table:
States of Nature
Action
Complete Loss
No Loss
Insure
-1,000
-1,000
Not Insure
-200,000
0
Opportunity loss table:
Action
S1
S2
A1
0
1,000
A2
199,000
0
The minimax decision is A1.
18.9
The maximax decision is to choose the action having the largest
payoff.
Therefore, a high inventory level would be the maximax
decision.
Opportunity loss table:
States of Nature
Action
S1
S2
S3
A1
0
3000
7000
A2
3000
0
4000
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A3
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0
4000
Action
Maximum Opportunity Loss
A1
7000
A2
4000
A3
8000
The minimax decision is A2.
18.10 The maximax decision is A3 (order 150 gallons a week).
Opportunity loss table:
States of Nature
Action
S1
S2
S3
A1
0
22.5
45
A2
27.5
A3
55
0
22.5
27.5
0
Action
Maximum Opportunity Loss
A1
45
A2
27.5
A3
55
The minimax decision is A2 (order 125 gallons a week).
18.11 P(S1) = .2
P(S2) = .4
P(S3) = .4
The expected payoffs for the 3 actions are:
Expected Payoff
A1: (.2)(40) + (.4)(8) + (.4)(0)
=
A2: (.2)(10) + (.4)(60) + (.4)(20) =
A3: (.2)(0) + (.4)(20) + (.4)(80)
11.2
34
= 40.0
Risk
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A1: (.2)(40)2 + (.4)(8)2 + (.4)(0)2 - (11.2)2
=
220.16
A2: (.2)(10)2 + (.4)(60)2 + (.4)(20)2 - (34)2
=
464.00
A3: (.2)(0)2 + (.4)(20)2 + (.4)(80)2 - (40)2
= 1120.00
18.12 Opportunity loss table:
States of Nature
a)
S1
S2
S3
A1
0
530
100
A2
1500
0
250
A3
2500
400
0
A4
2700
700
220
Action
Maximum Opportunity Loss
A1
530*
A2
1500
A3
2500
A4
2700
* The minimax decision is A1
b)
E(A1) = 5500(.25) + 2670(.50) + 1300(.25) = 3035
E(A2) = 4000(.25) + 3200(.50) + 1150(.25) = 2887.5
E(A3) = 3000(.25) + 2800(.50) + 1400(.25) = 2500
E(A4) = 2800(.25) + 2500(.50) + 1180(.25) = 2245
The decision based on maximum expected payoff is A1.
c)
Risk (A1) = 55002(.25) + 26702(.50) + 13002(.25) - 30352
= 2,338,225
Risk (A2) = 40002(.25) + 32002(.50) + 11502(.25) - 2887.52
= 1,112,968.75
Risk (A3) = 30002(.25) + 28002(.50) + 14002(.25) - 25002
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= 410,000
Risk (A4) = 28002(.25) + 25002(.50) + 11802(.25) - 22452
= 393,075
d)
The decision is A4 based on minimum risk.
18.13 Opportunity loss table:
States of Nature
a)
Action
S1
S2
S3
A1
150
60
10
A2
100
20
0
A3
50
0
35
A4
0
40
160
Action
Maximum Opportunity Loss
A1
150
A2
100
A3
50*
A4
160
* The minimax decision is A3.
b)
E(A1) = 100(.35) + 100(.50) + 100(.15) = 100
E(A2) = 150(.35) + 140(.50) + 110(.15) = 139
E(A3) = 200(.35) + 160(.50) + 75(.15) = 161.25
E(A4) = 250(.35) + 120(.50) + (-50)(.15) = 140
The decision is A3 based on the maximum expected payoff.
c)
Risk (A1) = 0
Risk (A2) = 1502(.35) + 1402(.50) + 1102(.15) - 1392 = 169
Risk (A3) = 2002(.35) + 1602(.50) + 752(.15) - 161.252
= 1642.1875
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Risk (A4) = 2502(.35) + 1202(.50) + 502(.15) - 1402 = 9850
18.14 A1 = low price
A2 = high price
S1 = 1 year
P(S1) = .4
S2 = 2 years
P(S2) = .4
S3 = 3 years
P(S3) = .2
The expected payoffs for the 3 actions are:
A1: (.4)(10,000) + (.4)(15,000) + (.2)(20,000) = 14,000
A2: (.4)(3,000) + (.4)(10,000) + (.2)(30,000) = 11,200
A2 is the maximum expected payoff decision.
18.15 A1 = low investment
A2 = medium investment
A3 = high investment
The expected payoffs for the 3 actions are:
A1: (.2)(300,000) + (.5)(400,000) + (.3)(500,000) = 410,000
A2: (.2)(-100,000) + (.5)(900,000) + (.3)(1,000,000) = 730,000
A3: (.2)(-400,000) + (.5)(300,000) + (.3)(3,000,000) = 970,000
The maximum expected payoff action is A3.
18.16 E(insure) = .05(-1000) + .95(-1000) = -1000
E(not insure) = .05(-200,000) + .95(0) = -10,000
The decision is to buy the insurance.
Risk (insure) = .05(-1000)2 + .95(-1000)2 - (-1000)2 = 0
Risk (not insure) = .05(-200,000)2 + .95(0) - (-10,000)2
= 1,900,000,000
18.17 Expected payoff using a perfect predictor is:
(.2)(500) + (.4)(400) + (.3)(150) + (.1)(200) = 325
EVPI = 325 - 180 (maximum expected payoff) = 145
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18.18 The expected payoffs for the 3 actions are:
Expected Payoff
A1: (.3)(30) + (.4)(2) + (.3)(2) = 10.4
A2: (.3)(5) + (.4)(10) + (.3)(1) =
5.8
A3: (.3)(1) + (.4)(1) + (.3)(5)
2.2
=
A1 is the maximum expected payoff decision.
States of Nature
Maximum Payoff
P(Si)
S1
30
.3
S2
10
.4
S3
5
.3
The expected payoff using a perfect predictor is:
(.3)(30) + (.4)(10) + (.3)(5) = 14.5
EVPI = 14.5 - 10.4 = 4.1
18.19 Payoff Table:
States of Nature
Action
1500
2000
2500
1500
750
750
750
2000
750 - .3(500)
= 600
750 - .3(1000)
= 450
1000
1000
1000 - .3(500)
= 850
1250
2500
.40 x 750 + .40 x 1000 + .20 x 1250 = 950
E(A1) = .4(750) + .4(750) + .2(750) = 750
E(A2) = .4(600) + .4(1000) + .2(1000) = 840
E(A3) = .4(450) + .4(850) + .2(1250) = 770
EVPI = 950 - 840 = 110
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$110 is the maximum amount that the manager would be willing to
pay for perfect information.
18.20 The maximum payoff using a perfect predictor
= 40,000 + 25,000 = 65,000
18.21 P(S1) = .30
P(S2) = .40
P(S3) = .30
The expected payoffs for the 3 actions are:
Expected Payoff
A1: (.3)(10,500) + (.4)(8,000) + (.3)(5,000)
= 7,850
A2: (.3)(-1,000) + (.4)(15,000) + (.3)(10,000) = 8,700
A3: (.3)(-4,000) + (.4)(3,000) + (.3)(30,000)
= 9,000
The maximum expected payoff decision is given by A3.
States of Nature
Maximum Payoff
P(Si)
S1
10,500
.3
S2
15,000
.4
S3
30,000
.3
The expected payoff using a perfect predictor is:
(.3)(10,500) + (.4)(15,000) + (.3)(30,000) = 18,150
EVPI = 18,150 - 9,000 = 9,150
An upper limit for the price of advice from a consultant would be
the EVPI.
18.22 5000 x .25 + 3200 x .5 + 2000 x .25 = 3350
E(A1) = 3000
E(A2) = 3000
E(A3) = 2550
E(A4) = 2500
EVPI = 3350 - 3000 = 350
18.23 250 x .35 + 160 x .5 + 110 x .15 = 184
E(A1) = 100
E(A2) = 139
E(A3) = 161.25
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Chapter 18
721
EVPI = 184 - 161.25 = 22.75
A1 is inadmissible because A2 dominates A1.
18.24 a)
U(x) is a utility function for a risk taker.
Note the U(X)
lies below the straight line connecting (0, 0) and (1, 100).
b)
U(x) is a utility function for a risk neutral decision maker.
c)
U(x) is not a utility function since it is a decresing function of x.
18.25 a)
The expected payoffs for the 3 actions are:
Expected Payoff
A1: (.25)(4 + 100 + 49 + 9)
= 40.5
A2: (.25)(81 + 25 + 36 + 25) = 41.75
A3: (.25)(100 + 16 + 25 + 9) = 37.5
A2 is the decision based on the maximum expected payoff.
b)
Table of the utility of the payoff:
States of Nature
Action
S1
S2
S3
S4
A1
20
100
70
30
A2
90
50
60
50
A3
100
40
50
30
The expected utility of the payoff for the 3 actions are:
Expected Utility of Payoff
A1: (.25)(20 + 100 + 70 + 30) = 55
A2: (.25)(90 + 50 + 60 + 50)
= 62.5
A3: (.25)(100 + 40 + 50 + 30) = 55
A2 is the decision based on the maximum expected utility.
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18.26 E(A1) = (.2)(50) + (.4)(10) + (.3)(30) + (.1)(10) = 24
E(A2) = (.2)(20) + (.4)(20) + (.3)(30) + (.1)(60) = 27
E(A3) = (.2)(10) + (.4)(50) + (.3)(10) + (.1)(20) = 27
The decision is A2 or A3 based on the maximum expected payoff.
Table of utility values:
States of Nature
S1
S2
S3
S4
A1
85
15
45
15
A2
3
3
45
0
A3
15
85
15
3
E(A1) = (.2)(75) + (.4)(15) + (.3)(45) + (.1)(15) = 36
E(A2) = (.2)(3) + (.4)(3) + (.3)(45) + (.1)(9)
= 40.5
E(A3) = (.2)(15) + (.4)(75) + (.3)(15) + (.1)(3)
= 40.5
The decision is A2 or A3 based on the maximum expected utility of
the payoff.
The decision based on the maximum expected payoff is equivalent to
the decision based on the maximum expected utility of the payoff.
18.27 Table of utility values:
States of Nature
Action
S1
S2
S3
S4
A1
60
12
36
12
A2
24
24
36
72
A3
12
60
12
24
E(A1) = (.2)(60) + (.4)(12) + (.3)(36) + (.1)(12) = 28.8
E(A2) = (.2)(24) + (.4)(24) + (.3)(36) + (.1)(12) = 32.4
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723
E(A3) = (.2)(12) + (.4)(60) + (.3)(12) + (.1)(24) = 32.4
The decision is A2 or A3 based on the maximum expected utility of
the payoff.
The decision based on the maximum expected utility does not differ
from that obtained in Exercise 18.26.
18.28 (1 - P) • U(0) + P • U(500,000) = U(50,000)
(1 - P) • 0 + P • (100) = 75
P = .75
18.29 Table of utility values:
States of Nature
S1
S2
S3
A1
3.699
3.477
3
A2
3.643
3.505
3.079
A3
3.505
3.447
3.146
A4
3.477
3.398
3.301
E(A1) = .25(3.699) + .50(3.477) + .25(3) = 3.413
E(A2) = .25(3.643) + .50(3.505) + .25(3.079) = 3.433
E(A3) = .25(3.505) + .50(3.447) + .25(3.146) = 3.386
E(A4) = .25(3.477) + .50(3.398) + .25(3.30) = 3.394
a)
The decision is A2 based on the maximum expected utility of
the payoff.
b)
The manager is a risk avoider.
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18.30 a)
The expected payoffs for the 3 actions are:
Expected Payoff
A1: (.25)(151) + (.30)(33) + (.40)(95) + (.05)(40) = 87.65
A2: (.25)(75) + (.30)(75) + (.40)(97) + (.05)(180) = 89.05
A3: (.25)(29) + (.30)(162) + (.40)(30) + (.05)(50) = 70.35
b)
The decision based on the maximum expected payoff is A2.
Table of the utility of the payoff:
States of Nature
Action
S1
S2
S3
S4
A1
54.98
05.62
27.44
07.50
A2
19.25
19.25
28.31
71.55
A3
04.63
61.09
04.87
10.48
The expected utility payoff for the 3 actions are:
Expected Utility Payoff
A1: (.25)(54.98) + (.30)(05.62) + (.40)(27.44) + (.05)(07.50) = 26.78
A2: (.25)(19.25) + (.30)(19.25) + (.40)(28.31) + (.05)(71.55) = 25.49
A3: (.25)(04.63) + (.30)(61.09) + (.40)(04.87) + (.05)(10.48) = 21.96
The decision based on the maximum expected utility of the
payoff is A1.
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18.31
18.32
Sum of paths = (.2)(.5) + (.1)(.3) + (.1)(.5) + (.3)(.2)
+ (.3)(.1) = .27
P(AiB)
P(A1B)
P(A2B)
P(A3B)
P(A4B)
P(A5B)
=
=
=
=
=
=
ith path/sum
(.2)(.5)/.27
(.1)(.3)/.27
(.1)(.5)/.27
(.3)(.2)/.27
(.5)(.1)/.27
of paths
= .371
= .111
= .185
= .222
= .111
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18.33
18.34
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18.35 Expected payoffs for the 4 actions are:
Expected Payoff
A1: (.15)(1451) + (.25)(1840) + (.25)(2050) + (.35)(2300)
= 1995.15
A2: (.15)(-1091) + (.25)(1685) + (.25)(2430) + (.35)(2900)
= 1880.1
A3: (.15)(-2015) + (.25)(1100) + (.25)(3060) + (.35)(3561)
= 1984.1
A4: (.15)(-3460) + (.25)(-1350) + (.25)(3340) + (.35)(4300) = 1483.5
18.36 a)
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P(I1) = sum of the branches
P(I2) = sum of the branches
= .23
= .31
P(S1I1) = .12/.23 = .522
P(S1I2) = .015/.31 = .048
P(S2I1) = .025/.23 = .109
P(S2I2) = .175/.31 = .565
P(S3I1) = .05/.23 = .217
P(S3I2) = .05/.31 = .161
P(S4I1) = .035/.23 = .152
P(S4I2) = .07/.31 = .226
P(I3) = sum of the branches
= .2625
P(I4) = sum of the branches
= .1975
P(S1I3) = .0075/.2625 = .029
P(S1I4) = .0075/.1975 = .038
P(S2I3) = .025/.2625 = .095
P(S2I4) = .025/.1975 = .127
P(S3I3) = .125/.2625 = .476
P(S4I3) = .105/.2625 = .4
P(S3I4) = .025/.1975 = .127
P(S4I4) = .14/.1975 = .709
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Consultant's fee = $400
EVSI = 2445.27 - 400 - 1995.15 = 50.12
b)
States of Nature
Maximum Payoff
P(Si)
S1
1451
.15
S2
1840
.25
S3
3340
.25
S4
4300
.35
Average payoff using a perfect predictor:
(1451)(.15) + (1840)(.25) + (3340)(.25) + (4300)(.35) = 3017.65
EVPI = 3017.65 - 1995.15 = 1022.5
The efficiency of the sample information:
(EVSI/EVPI)(100) = (50.12/1022.5)(100) = 4.9%
c)
The consultant's service would be slightly better than using
the action which gives the largest expected payoff without
sample information.
The low efficiency indicates that perhaps another consultant
would be more profitable.
18.37 P(B) = (.3)(.1) + (.1)(.3) + (.2)(.3) + (.5)(.1) + (.5)(.2)
= .03 + .03 + .06 + .05 + .10 = .27
P(A1B) = P(BA1) P(A1) / P(B) = .03/.27 = .111
P(A2B) = P(BA2) P(A2) / P(B) = .03/.27 = .111
P(A3B) = P(BA3) P(A3) / P(B) = .06/.27 = .222
P(A4B) = P(BA4) P(A4) / P(B) = .05/.27 = .185
P(A5B) = P(BA5) P(A5) / P(B) = .10/.27 = .370
730
Chapter 18
18.38 P[ProfitMkt] = .6(.7)/[.6(.7) + .3(.3)] = .42/.51 = .824
The probability that the stores market a particular fashion is
.51.
18.39 P(life) = .4(.05) + .33(.08) + .27(.10) = .0734
P(Hlife) = .33(.08)/.0734 = .0264/.0734 = .3597
18.40 States of Nature
Maximum Payoff
P(Si)
S1
44
.1
S2
27
.3
S3
40
.4
S4
45
.2
The expected payoff using a perfect predictor is:
(.1)(44) + (.3)(27) + (.4)(40) + (.2)(45) = 37.5
EVPI = (expected payoff using a perfect predictor) - (expected
payoff) = 37.5 - 32.4 = 5.1
EVSI = 1.5
Efficiency of the sample information = (EVSI/EVPI)100
= (1.5/5.1)100 = 29.4%
18.41 P(B) = .1(.6) + .3(.2) + .4(.3) + .2(.1) = .26
P(S1B) = (.1)(.6)/.26 = .2308
P(S2B) = (.3)(.2)/.26 = .2308
P(S3B) = (.4)(.3)/.26 = .4615
P(S4B) = (.2)(.1)/.26 = .0769
731
731
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Instructor's Manual
Choose action A2.
18.42 P(G) = (.10)(.04) + (.20)(.06) + (.20)(.08) + (.40)(.03)
+ (.10)(.09)
=
.004 + .012 + .016 + .012 + .009 = .053
where G = the event that an error is found on a legal document.
P(AG) = .004/.053 = .075
P(BG) = .012/.053 = .226
P(CG) = .016/.053 = .302
P(DG) = .012/.053 = .226
P(EG) = .009/.053 = .170
Secretary C would have the highest probability of having typed the
error.
18.43 (1/3)(15000 + 5000 + 1000) = 7000
EVPI = 7000 - 4000 = 3000
Since 2000 < 3000, it is worthwhile to conduct the experiment.
732
Chapter 18
18.44 P(I1) = .4(.8) + .4(.3) + .2(.3) = .32 + .12 + .06 = .50
P(I2) = .4(.1) + .4(.5) + .2(.1) = .04 + .2 + .02 = .26
P(I3) = .4(.1) + .4(.2) + .2(.6) = .04 + .08 + .12 = .24
P(S1I1) = .32/.50 = .64
P(S2I1) = .12/.50 = .24
P(S3I1) = .06/.50 = .12
P(S1I2) = .04/.26 = .15
P(S2I2) = .2/.26 = .77
P(S3I2) = .02/.26 = .08
P(S1I3) = .04/.24 = .17
P(S2I3) = .08/.24 = .33
P(S3I3) = .12/.24 = .50
EVPI is 64.651 - 62.8 = 1.851 thousand.
cost of the survey, is too expensive.
733
Therefore, $2000, the
733
734
Instructor's Manual
734
Chapter 18
18.45 a)
735
The expected payoffs for the 4 actions are:
Expected Payoff
A1: (.3)(25) + (.2)(25) + (.5)(25)
= 25
A2: (.3)(12.5) + (.2)(32.5) + (.5)(32.5) = 26.5
A3: (.3)(0) + (.2)(40) + (.5)(40)
= 28
A4: (.3)(-25) + (.2)(15) + (.5)(55)
= 23
The maximum expected payoff decision is given by A3.
The expected payoffs when the consultant predicts I1 are:
Expected Payoff
A1: (.54)(25) + (.2)(25) + (.26)(25)
= 25
A2: (.54)(12.5) + (.2)(32.5) + (.26)(32.5) = 21.7
A3: (.54)(0) + (.2)(40) + (.26)(40)
= 18.4
A4: (.54)(-25) + (.2)(15) + (.26)(55)
= 3.8
The maximum expexcted payoff decision when the consultant predicts
I1 is given by A1.
The expected payoffs when the consultant predicts I2 are:
Expected Payoff
A1: (.231)(25) + (.385)(25) + (.385)(25)
= 25.025
A2: (.231)(12.5) + (.385)(32.5) + (.385)(32.5) = 27.913
A3: (.231)(0) + (.385)(40) + (.385)(40)
= 30.8
A4: (.231)(-25) + (.385)(15) + (.385)(55)
= 21.175
The maximum expected payoff decision when the consultant predicts
I2 is given by A3.
The expected payoffs when the consultant predicts I3 are:
Expected Payoff
A1: (.086)(25) + (.057)(25) + (.857)(25)
735
= 25
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Instructor's Manual
A2: (.086)(12.5) + (.057)(32.5) + (.857)(32.5) = 30.78
A3: (.086)(0) + (.057)(40) + (.857)(40)
= 36.56
A4: (.086)(-25) + (.057)(15) + (.857)(55)
= 45.84
The maximum expected payoff decision when the consultant predicts
I3 is given by A4.
Net expected gain of hiring the consultant (EVSI) is:
(.39)(25) + (.26)(30.8) + (.35)(46.2) - 2.5 - 28 = 3.428
b)
Given a perfect predictor, the following payoffs are possible:
Si
Maximum Payoff
P(Si)
S1
25
.3
S2
40
.2
S3
55
.5
The expected payoff using a perfect predictor is:
(.3)(25) + (.2)(40) + (.5)(55) = 43
EVPI = 43 - 28 = 15
Efficiency of the sample information = (EVSI/EPVI)(100)
= (3.428/15)(100) = 22.85%
18.46 The expected payoffs for the 3 actions are:
A1:
A2:
A3:
The
Expected Payoff
(.4)(10,000) + (.3)(13,000) + (.3)(16,000) = 12,700
(.4)(8,000) + (.3)(23,000) + (.3)(25,000) = 17,600
(.4)(8,000) + (.3)(20,000) + (.3)(40,000) = 21,200
decision based on the maxium e expected payoff is A3.
736
Chapter 18
P(I1) = sum of the branches = .5
P(S1I1) = .32/.5 = .64
P(S1I2) = .04/.22 = .182
P(S2I1) = .12/.5 = .24
P(S2I2) = .12/.22 = .545
P(S3I1) = .06/.5 = .12
P(S3I2) = .06/.22 = .273
737
737
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Instructor's Manual
P(I2) = sum of the branches = .22
P(S1I2) = .04/.22 = .182
P(S2I2) = .12/.22 = .545
P(S3I2) = .06/.22 = .273
738
Chapter 18
P(I3) = sum of the branches = .28
P(S1I3) = .04/.28 = .143
P(S2I3) = .06/.28 = .214
P(S3I3) = .18/.28 = .643
739
739
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Instructor's Manual
I2 .22
I3 .28
740
Chapter 18
741
Note that A3 gives the largest expected payoff for each of the
predictions I1, I2, and I3 of the consultant.
The decision based on the maximum expected payoff is A3.
Therefore, it would not be worth paying $2,000 for the financial
planner's advice.
18.47 a) Opportunity loss table:
States of Nature
b)
S1
S2
S3
S4
A1
0
0
10
20
A2
4
2
0
10
A3
4
4
0
0
A4
6
6
14
5
Action
Maximum Opportunity Loss
A1
20
A2
10
A3
4
A4
14
The minimax decision is action A3.
c)
Action
Maximum Payoff
A1
40
A2
50
A3
60
A4
55
The maximax decision is A3.
741
742 Instructor's Manual
18.48 Payoff table:
S1
(5,000)
-100,000
S2
(10,000)
0
S3
(15,000)
100,000
S4
(20,000)
200,000
S5
(25,000)
300,000
200,000
200,000
200,000
200,000
200,000
S1
S2
S3
S4
S5
A1
300,000
200,000
100,000
0
0
A2
0
0
0
0
100,000
A1
(introduce
software)
A2 (don’t
introduce
software)
Opportunity loss table:
Action
Maximum Opportunity Loss
A1
300,000
A2
100,000
The minimax decision is A2.
18.49
i) With the minimax strategy, the change of probabilities will
not change the decision.
ii) Decision based on a maximum expected payoff.
a)
A1: (.1)(250) + (.5)(320) + (.2)(350) + (.2)(400) = 335
A2: (.1)(150) + (.5)(260) + (.2)(300) + (.2)(370) = 279
A3: (.1)(120) + (.5)(290) + (.2)(380) + (.2)(450) = 323
A4: (.1)(80) + (.5)(280) + (.2)(410) + (.2)(550)
b)
= 340
A1: (.1)(250) + (.5)(320) + (.1)(350) + (.3)(400) = 340
A2: (.1)(150) + (.5)(260) + (.1)(300) + (.3)(370) = 286
A3: (.1)(120) + (.5)(290) + (.1)(380) + (.3)(450) = 330
A4: (.1)(80) + (.5)(280) + (.1)(410) + (.3)(550)
742
= 354
Chapter 18
c)
743
A1: (.1)(250) + (.5)(320) + (.3)(350) + (.1)(400) = 330
A2: (.1)(150) + (.5)(260) + (.3)(300) + (.1)(370) = 272
A3: (.1)(120) + (.5)(290) + (.3)(380) + (.1)(450) = 316
A4: (.1)(80) + (.5)(280) + (.3)(410) + (.1)(550)
d)
= 326
A1: (.1)(250) + (.5)(320) + (.2)(350) + (.2)(400) = 335
A2: (.1)(150) + (.5)(260) + (.2)(300) + (.2)(370) = 279
A3: (.1)(120) + (.5)(290) + (.2)(380) + (.2)(450) = 323
A4: (.1)(80) + (.5)(280) + (.2)(410) + (.2)(550)
e)
= 342
A1: (.2)(250) + (.5)(320) + (.2)(350) + (.1)(400) = 320
A2: (.2)(150) + (.5)(260) + (.2)(300) + (.1)(370) = 257
A3: (.2)(120) + (.5)(290) + (.2)(380) + (.1)(450) = 290
A4: (.2)(80) + (.5)(280) + (.2)(410) + (.1)(550)
= 295
Since three out of five sets of probabilities make the same decision, it is not extremely sensitive to changing the probabilities.
Action A4 is the optimal action for sets 1, 2, and 4.
Action A1
is the optimal action for sets 3 and 5.
Summary of Sensitivity Analysis
Expected Payoff
P(S1)
P(S2)
P(S3)
P(S4)
A1
A2
A3
A4
.1
.5
.2
.2
335
279
323
340*
.1
.5
.1
.3
340
286
330
354*
.1
.5
.3
.1
330*
272
316
326
.1
.5
.2
.2
335
279
323
342*
.1
.5
.2
.1
320*
257
290
295
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Instructor's Manual
18.50 a)
b)
All the actions are admissible.
Opportunity loss table:
States of Nature
Action
S1
S2
A1
1020
375
0
A2
620
0
90
A3
0
215
380
Action
S3
Maximum Opportunity Loss
A1
1020
A2
620
A3
380*
* The minimax decision is A3.
c)
E(A1) = .3(830) + .55(750) + .15(710) = 768
E(A2) = .3(1230) + .55(1125) + .15(620) = 1080.75
E(A3) = .3(1850) + .55(910) + .15(330) = 1105
The decision is A3 based on the maximum expected payoff.
d)
Risk (A1) = .3(830)2 + .55(750)2 + .15(710)2 - 7682 = 1836
Risk (A2) = .3(1230)2 + .55(1125)2 + .15(620)2 - 1080.752
= 39603.1875
Risk (A3) = .3(1850)2 + .55(910)2 + .15(330)2 - 11052 = 277515
e)
The decision is A1 based on minimum risk.
f)
.3(1850) + .55(1125) + .15(710) = 1280.25
EVPI = 1280.25 - 1105 = 175.25
744
Chapter 18
745
18.51 One example is:
0 for x  0


U(x)  .036x for 0  x  1000

36  64(.001x  1)2 for 1000  x  2000

This utility function is scaled so that its values are between 0
and 100.
18.52 P(S1) = .2
P(S2) = .2
P(S3) = .4
P(S4) = .2
The expected payoffs for the 4 actions are:
A1: (.2)(9) + (.2)(19) + (.4)(29) + (.2)(39) = 25
A2: (.2)(9) + (.2)(24) + (.4)(39) + (.2)(54) = 33
A3: (.2)(5) + (.2)(25) + (.4)(45) + (.2)(65) = 37
Risk
A1: (.2)(9)2 + (.2)(19)2 + (.4)(29)2 + (.2)(39)2 - 252 = 104
A2: (.2)(9)2 + (.2)(24)2 + (.4)(39)2 + (.2)(54)2 - 332 = 234
A3: (.2)(5)2 + (.2)(25)2 + (.4)(45)2 + (.2)(65)2 - 372 = 416
18.53 EVSI = 21 - 18 = 3
EVPI = 30 - 18 = 12
Efficiency of the sample information = (EVSI/EVPI)(100) = 95.45%
18.54 P(I1) = .3(.6) + .55(.2) + .15(.2) = .18 + .11 + .03 = .32
P(S1 I1) = .18/.32 = .56
P(S2I1) = .11/.32 = .34
P(S3I1) = .03/.32 = .09
745
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Instructor's Manual
Under I1:
E(A1) = .56(830) + .34(750) + .10(710) = 790.8
E(A2) = .56(1230) + .34(1125) + .10(620) = 1133.3
E(A3) = .56(1850) + .34(910) + .10(330) = 1378.4
The decision is A3 under the revised probabilities.
18.55 P(I2) = .3 x .1 + .55 x .5 + .15 x .3 = .03 + .275 + .045 = .35
P(I3) = .3 x .3 + .55 x .3 + .15 x .5 = .09 + .165 + .075 = .33
P(S1I2) = .03/.35 = .08
P(S1I3) = .09/.33 = .27
P(S2I2) = .275/.35 = .79
P(S2I3) = .165/.33 = .5
P(S3I2) = .045/.35 = .13
P(S3I3) = .075/.33 = .23
Under I2:
E(A1) = .08(830) + .79(750) + .13(710) = 751.2
E(A2) = .08(1230) + .79(1125) + .13(620) = 1067.75
E(A3) = .08(1850) + .79(910) + .13(330) = 909.8
Under I3:
E(A1) = .27(830) + .5(750) + .23(710) = 762.4
E(A2) = .27(1230) + .5(1125) + .23(620) = 1037.2
E(A3) = .27(1850) + .5(910) + .23(330) = 1030.4
The maximum payoff with consultant
= .32(1378.4) + .35(1067.75) + .33(1037.2) = 1157.0765
The expected payoff without the consultant is 1105 (from A3).
Since 1157.0765 - 1105 = 52.0765 is less than 350, it is not
worthwhile to have the consultant's service.
18.56 Let E represent the event that an electronic component is
defective.
746
Chapter 18
747
P(E) = (.009)(.48) + (.005)(.22) + (.002)(.30)
= .00432 + .0011 + .0006 = .006
P(AE) = .00432/.006 = .72
P(BE) = .0011/.006 = .18
P(CE) = .0006/.006 = .10
18.57 a)
Risk taker
b)
Risk neutral
c)
Risk avoider
18.58 a)
Opportunity loss table:
States of Nature
S1
S2
S3
A1
350
100
0
A2
100
0
150
A3
0
600
650
Action
Maximum Opportunity Loss
A1
350
A2
150
A3
650
The minimax decision is A2.
b)
E(A1) = .40(450) + .35(400) + .25(350) = 407.50
E(A2) = .40(700) + .35(500) + .25(200) = 505.00
E(A3) = .40(800) + .35(-100) + .25(-300) = 210.00
A2 would be the decision based on the maximum expected payoff.
c)
Risk (A1) = (450)2(.40) + (400)2(.35) + (350)2(.25)
- (407.50)2 = 1568.75
747
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Instructor's Manual
Risk (A2) = (700)2(.40) + (500)2(.35) + (200)2(.25)
- (505.00)2 = 38475.00
Risk (A3) = (800)2(.40) + (-100)2(.35) + (-300)2(.25)
- (210.00)2 = 237900.00
18.59 Let A represent drives own car.
Let B represent uses car pool.
Let C represent uses city bus.
Let M represent male.
P[M] = P[MA]•P[A] + P[MB]•P[B] + P[MC]•P[C]
= (.70)(.55) + (.30)(.25) + (.20)(.20) = .50
P[AM] = P[MA] P[A] / P(M) = .385/.50 = .77
748