SUPPLEMENT TO CHAPTER 5
DECISION THEORY
Answers to Discussion and Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
78
The chief role of the operations manager is that of decision maker.
Decision making consists of the following steps:
(1) Specify objectives and criteria for making a decision.
(2) Develop alternatives.
(3) Analyze and compare alternatives.
(4) Select the best alternative.
(5) Implement the chosen alternative.
(6) Monitor the results.
Bounded rationality is a term that refers to the limits imposed on decision making because of
costs, human abilities, time, technology, and availability of information.
Suboptimization occurs as a result of different departments, each attempting to reach a solution
that is optimum for that department but may not be optimum for the organization as a whole.
Poor decisions can be due to bounded rationality, which refers to limits on decision making in
terms of cost, technology, human abilities and availability of information; managerial style
(including quick decisions, unwillingness to admit a mistake, and procrastination); and
organization difficulties, which can be related to departmentalizing decisions (which often leads
to interdepartment conflicts and suboptimization).
A payoff table shows the expected payoffs for each alternative in every possible state of nature.
Sensitivity analysis refers to examining how sensitive a given solution is to a change in one or
more parameters of a problem. High sensitivity indicates to decision makers that a parameter
must be carefully estimated or determined; low sensitivity would not require such careful
estimation. Also, decision makers can use sensitivity analysis to explore how a change in the
value of a parameter would effect a solution.
Maximax is an optimistic approach that calls for selection of the alternative which has the best
possible payoff; maximin is a conservative approach which seeks the alternative with the best
"worst” payoff. Thus, if the mood is "go for it" maximax would be used, while if the mood is
cautious, maximin would be more appropriate.
The expected monetary value approach implies a linear utility for payoff (e.g., a payoff of $2 has
twice the utility of a payoff of $1). When multiple decisions are to be made, the expected payoff
approach is often used; similarly, when a series of on-going decisions on projects, new
equipment, etc. are to be made, expected value can be useful. Conversely, for unique, one-time
type decision, and/or where utilities are nonlinear, expected value would not be appropriate. Of
course, the expected value approach requires probabilities for states of nature. If these are not
available, the approach cannot be used.
a. The Laplace criterion is an approach to decision making under uncertainty. It treats the states
of nature as equally likely.
b. Minimax regret is an approach to decision making under uncertainty. It seeks to minimize the
opportunity loss, or regret, associated with choosing a decision.
c. Expected value is an approach to decision making under risk: the probabilities of states of
nature are assumed to be known.
Operations Management, 7/e
11.
d. Expected value of perfect information is the maximum amount a decision maker should be
willing to pay to move from a position of decision making under risk to a position of decision
making under certainty.
In order to use an expected value approach to decision making, a decision maker must have state
of nature probabilities (in addition to a list of alternatives, a list of states of nature, and a set of
payoffs). If probabilities are unknown, either additional resources can be used to attempt to
ascertain the probabilities, or adopting an approach which does not call for probabilities.
Sensitivity analysis could be used to show the range of probability for which a particular
alternative would be optimal. Thus, it may not be necessary for a decision maker to pinpoint a
probability; rather, it may only be necessary to decide if a probability is in this "ballpark."
Instructor’s Manual, Chapter 5 Supplement
79
Solutions
1.
a. Maximax:
b. Maximin:
c. Laplace:
Best payoffs:
Do Nothing:
Expand:
Subcontract:
60
80
70
Worst payoffs:
Do Nothing:
Expand:
Subcontract:
50
20
40
[best of the worst payoffs]
Average Payoff
Do Nothing
Expand
Subcontract
55
50
55
[Indifferent between Do Nothing
And Subcontract]
[best of the best payoffs]
d. Minimax Regret
Do Nothing
Expand
Subcontract
2.
Low
0
30
10
a. Expected profit
Do Nothing
$57
Expand
62
Subcontract
61
High
20
0
10
.7
.3
$62
Expand
.7
$61
Subcontr.
[best of worst]
[Best]
.3
$57
Do Nothing
Worst
20
30
10
.3
.7
$50
$60
$20
$80
$40
$70
c. EPC: .30(50) + .70(80) = $71
Exp. Profit:
62
EVPI:
$9
Low
Payoff
3.
Equations:
Do Nothing: 50 + 10P
Expand:
20 + 60P
Subcontract: 40 + 30P
Optimal ranges:
Do nothing:
80
70
60
Do Nothing
50
40
Subcontract
20
Expand
0 to < 50
0
80
High
Payoff
.50 .67
1.0
Operations Management, 7/e
Expand:
> .67 to 1.00
Subcontract: > .50 to < .67
Instructor’s Manual, Chapter 5 Supplement
81
Solutions (continued)
4.
a. (1) Draw the tree diagram:
$400,000 (1)
Demand Low (.4)
Maintain
$50,000 (2)
Build Small
Demand High (.6)
2
Expand
1
Build Large
$450,000 (3)
Demand Low (.4)
$-10,000 (4)
Demand High (.6)
$800,000 (5)
(2) 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).
(3) Compute the expected value of the ends of the remaining branches (numbered 1 to 5 in
the diagram), and then determine the expected value for the two initial alternatives.
(1) .4 x $400,000 = $160,000
(2) (eliminated)
$430,000 (expected value if
(3) .6 x $450,000 = $270,000
Build small is chosen)
(4) .4 x -$10,000 = $-4,000
(5) .6 x $800,000 = $480,000
$476,000 (expected value if
Build large is chosen)
(4) Since the expected value of building a large plant has the higher expected value, select
the large plant alternative.
b. Expected payoff under certainty:
.4(400,000) + .6 (800,000) = $640,000
Expected payoff under risk:
476,000
Expected value of perfect information:
$164,000
5.
Low
800
High
Payoff
Build Large
450
Build Large
400
Build Small
-10
1.0
82
P (low)
.54
0
Operations Management, 7/e
Solutions (continued)
5.
EVsubcontract= (.4)(1.0) + (.5)(1.3) + (.1)(1.8) = 1.23
EVexpand = (.4)(1.5) + (.5)(1.6) + (.1)(1.7) = 1.57
EVbuild
= (.4)(1.4) + (.5)(1.1) + (.1)(2.4) = 1.35
Since 1.57 > 1.35 > 1.23, build.
6.
7.
Alternative
Renew
Relocate
Decision:
Alternative
a. Renew
Relocate
Decision:
MaxiMax
MaxiMin
Laplace
Minimax
Max. Payoff (b) Min. Payoff (c) Average
(d) Regret
$4,000,000
$500,000*
$2,250,000
$4,500,000
$5,000,000*
$100,000
$2,550,000*
$3,900,000*
Relocate
Renew
Relocate
Relocate
Expected Value
500,000(.35) + 4,000,000(.65) = $2,775,000*
5,000,000(.35) + 100,000(.65) = $1,815,000
Renew lease
(a)
Approve (.35)
$500,000
E.V.
$2,775,000*
Renew
Relocate
Reject (.65)
Approve (.35)
Reject (.65)
$4,000,000
$5,000,000
$1,815,000
$100,000
EVPI = EPC – EMV
= .35(5,000,000) + .65(4,000,000) - 2,775,000 = $1,575,000
Yes, the manager should sign the lease for $24,000 since it is less than the EVPI of
$1,575,000.
a., b. Let P (application is approved) = x. Then P (application rejected) = 1 - x.
From 7(a)
500,000x + 4,000,000(1 - x) = 4,000,000 - 3,500,000x
and 5,000,000x + 100,000(1 - x) = 100,000 + 4,900,000x
The two alternatives are equally good when
4,000,000 - 3,500,000x = 100,000 + 4,900,000x
i.e. when x + 3,900,000
= 0.4643
8,400,000
c.
8.
Instructor’s Manual, Chapter 5 Supplement
83
Solutions (continued)
Exp.
Value
(millions)
5
4
Renewal better than
Relocation
5
4
Relocate
Renew
3
3
4,000,000 – 3,500,000x
2
Relocate
1
100,000 + 4,900,000x
For 8(a) and 8(b) the decision
should be to renew the 10 year
lease.
2
1
.5
x = .4643
P (application approved)
c. D = am't of decrease in $4,000,000 in order that EVrenew = EVreloc.
EVrenew – EVreloc. = $2,775,000 – $1,815,000 = $960,000
$960,000 = .65D
D = $1,476,020
$4,000,000 – $1,476,920 = $2,523,000
Range is $2,523,080 or more.
$ 42
9.
.2(42)
.2 Low
Subcontract
.8 High
2
Expand Greatly
Small
Medium
1
.2 Low
Do Nothing
.8 High
42
+
48
.8(48)
22
.2(22)
46
+
46.8
44.4
2
Expand
Large
.2 Low
50
.8(50)
(20)
.2(-20)
+
.8 High
72
53.6
.8(72)
a. Decision: Build a large facility.
b. Max (42;22;-20) = $42 million.
Decision: Build a small facility.
c. EVPI = EPC – EMV
EPUC = 42(.2) + 72(.8) = 8.4 + 57.6 = $66.0 - 53.6 = 12.4
84
Operations Management, 7/e
Solutions (continued)
d. Let P(high) = x P(low) = 1-x
I. Small:
42(1-x) + 48x => 42 + 6x
II. Medium: 22(1-x) + 50x => 22 + 28x
III. Large:
-20(1-x) + 72x => -20 + 92x
Value of x where expected value for I and III are the same.
42 + 6x = -20 + 92x => 86x = 62 => x = 0.7209
72
50
48
Small
42
Medium
22
Large
0
.721
1.0
P (High)
-20
10.
.30 low
$90
.3(90)
do nothing
90
+
$104
buy 1
2
subcontract
buy 2nd
.30 low
buy 2
.70 high
110
.7(110)
100
.3(75)
75
130
+
$113.5 (optimum)
.7(130)
Decision: Buy 2 machines.
Instructor’s Manual, Chapter 5 Supplement
85
1/3
Solutions (continued)
11.
1
1/3
1/3
50
.30
Alternative A
90
40
.50
4
0
60
44
60
.20
49
1/3
2
1/3
1/3
40
Alternative B
.30
5
(45)
45
99
40
.50
50
.20
30
1/2
45
3
EV1 = (1/3)(0) + (1/3)(60) + (1/3)(90) = 50
EV2 = (1/3)(-45) + (1/3)(45) + (1/3)(99) = 33
EV3 = (1/2)(40) + (1/2)(50) = 45
EV4 = (.3)(50) + (.5)(44) + (.2)(60) = 49
EV1 = (.3)(40) + (.5)(50) + (.2)(45) = 46
1/2
40
50
Since 49 > 46, choose alternative A.
12.
(1) Draw the tree diagram:
$400,000 (1)
Demand Low (.4)
Maintain
$50,000 (2)
Build Small
Demand High (.6)
1
Build Large
2
Expand
$450,000 (3)
Demand Low (.4)
$-10,000 (4)
Demand High (.6)
$800,000 (5)
(2) 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).
86
Operations Management, 7/e
Solutions (continued)
(3) Compute the expected value of the ends of the remaining branches (numbered 1 to 5 in the
diagram), and then determine the expected value for the two initial alternatives.
(1) .4 x $400,000 = $160,000
(2) (eliminated)
$430,000 (expected value if
(3) .6 x $450,000 = $270,000
Build small is chosen)
(4) .4 x -$10,000 = $-4,000
(5) .6 x $800,000 = $480,000
$476,000 (expected value if
Build large is chosen)
(4) Since the expected value of building a large plant has the higher expected value, select the
large plant alternative.
13.
Reassign
New Staff
Redesign
Moderate
50
60
40
High
60
60
50
Very High
85
60
90
Worst
85
60*
90
Best
50
60
40*
Average
65
60 (tie)
60
a. Maximin: New staff
b. Maximax: Redesign
c. Regret table:
Moderate
Reassign
10
New Staff
20
Redesign
0
High
10
10
0
Very High
25
0
30
Worst
25
20*
30
d. Insufficient reason: (tie) New Staff or Redesign
14.
a. 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
b.
.10 Moderate
Reassign
74
.30 High
.60 Very High
.10 Moderate
60
New Staff
60
.30 High
.60 Very High
.10 Moderate
Redesign
73
.30 High
.60 Very High
Instructor’s Manual, Chapter 5 Supplement
50
60
85
60
60
60
40
50
90
87
Solutions (continued)
c. Opportunity loss table:
Moderate
Reassign
10
New Staff
20
Redesign
0
(.1)
High
10
10
0
(.3)
Very High
25
0
30
(.6)
.10 Moderate
15.
Reassign
.30 High
.60 Very High
74
.10 Moderate
Hire 2
Initially
.30 High
60
.60 Very High
New Staff
60
60
.10 Moderate
Hire 1
Initially
Redesign
EOL
19
5*
18
50
60
85
60
60
60
40
.30 High (Hire 1)
75
.60 V. High (Hire 1) 80
.10 Moderate
40
.30 High
50
.60 Very High
90
74.5
73
Solution continued
16.
Payoff
#1
Payoff
#2
140
120
B
100
A
B
C
80
C
A
0
40
P(#2)
1.0
b. Alternative C is lower than Alternative B for all values of P(#2), so it would
never be appropriate.
c. EVB = 120 - 40P; EVA = 20 + 120P. Solving, P = .625.
Therefore, choose Alternative A if P(#2) is greater than .625.
d. For P(#1), choose Alternative A, if P(#1) is less than .375 (i.e., 1.000 - .625).
88
Operations Management, 7/e
Solutions (continued)
17.
[Refer to the diagram in the previous solution]
b. Alternative B is now the one that is never appropriate.
c. EVA = 20 + 120P; EVC = 100 - 60P. Solving, P = .444.
Therefore, choose Alternative A for P(#2) less than .444, and choose
Alternative C for P(#2) greater than .444.
d. In terms of P(#1), choose A for P(#1) greater than .556 and C for P(#1) less
than .556.
18.
EV1 = 10 – 12P
Payoff for
no contract
Payoff for
contract 10
EV2 = 8 – 5P
8
#1
#2
7
EV3 = 5 – 0P
#4
#2
#3
5
EV4 = 0 + 7P
5
3
#4
0
.286
1.0
.60 .714
-2
19.
EVA = 120 – 100P
EVB = 60 – 20p
Payoff
#1
120
EVC = 10 + 100p
.30
Payoff
#2
.80
A
Profits
110
C
D
90
90
EVD = 90
Note that in terms of P(#1), .30 60
becomes .70, .80 becomes .20,
.417 becomes .583, and .75
becomes .25.
B
C
Costs
A
40
20
10
0
.417
.75
1.0
P (#2)
Instructor’s Manual, Chapter 5 Supplement
89
2.3 > 1.5, hire no additional employees when the probability of rain is 20%.
90
Operations Management, 7/e