SUPPLEMENT TO CHAPTER 5
DECISION THEORY
Teaching Notes
The presentation of decision theory is standard except for the addition of material on sensitivity analysis.
Decision theory can be omitted if it does not suit your purposes, without loss of continuity.
The presentation should emphasize that decision tables are developed for one-phase decision-making,
whereas decision trees are developed for multi-phase decision-making where several interrelated
decisions and state of nature are considered, and the decisions are dependent on each other and the states
of nature. Decision analysis forces the decision maker to study the states of nature (conditions) carefully
because probabilities must be assigned to each state of nature. The decision analysis provides the
decision maker with reasonably objective method of evaluating the relative value of each decision
alternative
Answers to Discussion and Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
66
The chief role of the operations manager is as a decision maker.
Decision making consists of the following steps:
(1) Identify possible future conditions (states of nature).
(2) Develop alternatives.
(3) Determine the payoffs.
(4) Estimate the probability of each future condition.
(5) Evaluate alternatives according to decision criteria, and select the best alternative.
Bounded rationality is a term that refers to rational decision-making made within limits imposed
by costs, human abilities, time, technology, and availability of information. Another name for it is
satisficing.
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 departmentalized decisions (which often lead 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.
Operations Management, 2/ce
9.
10.
11.
12.
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) and risk-neutral attitude towards risk. 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, where utilities are nonlinear, and for people with
risk aversion and risk lovers, 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
maximum opportunity loss, or regret, associated with decisions.
c. Expected monetary value is an approach to decision making under risk: the probabilities of
states of nature are assumed to be known. It involves choosing that decision that has the
highest expected payoff value.
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 monetary 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 an approach can be adopted 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."
First using sensitivity analysis, the manager can find out which decision is more robust, i.e., will
result in highest EMV under varied probabilities for future conditions. The manager can also look
at other criteria such as maximin. Finally, other non-monetary operations factors such as space
availability can be used to break the tie.
Instructor’s Manual, Chapter 5 Supplement
67
Solutions
1.
a. Maximax:
Expand [$80 is the highest payoff]
b. Maximin:
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]
c. Laplace:
d. Minimax Regret
Do Nothing
Expand
Subcontract
2.
Low
0
30
10
High
20
0
10
Worst = maximum regret
20
30
10
[best of worst]
a. Expected profit
Do Nothing .3 (50) + .7 (60) = $57
Expand
.3 (20) + .7 (80) = $62
Subcontract .3 (40) + .7 (70) = $61
b.
.3
$50
$57
Do Nothing
$62
Expand
$61
Subcontr.
.7
.3
.7
.3
.7
[Best]
$60
$20
$80
$40
$70
c. Expected profit under certainty: .30(50) + .70(80) = $71
Expected profit under risk:
62
EVPI:
$9
Any market research is at most worth $9000 to the contractor.
80
3.
Let P = P (high)
Equations:
Do Nothing: 50 (1 - P) + 60P = 50 + 10P
Expand:
20 (1 - P) + 80P = 20 + 60P
Subcontract: 40 (1 - P) + 70P = 40 + 30P
Optimal ranges:
___P___
Do nothing:
0 to < .50
Expand:
.67 to 1.00
Subcontract: .50 to .67
68
70
60
Do Nothing
50
40
Subcontract
20
Expand
0
.50 .67
1.0
P
Operations Management, 2/ce
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
Build small = 400P + 450 (1 – P)
= 450 – 50P
800
Build large = -10P + 800 (1 – P)
= 800 – 810P
450
Build Small
450 – 50P = 800 – 810 P
760 P = 350
P = .46
400
Build Large
Build small if P(low) ≥ .46
Build large if P(low) < .46
0
.46
1
P (low)
-10
Instructor’s Manual, Chapter 5 Supplement
69
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 is the maximum build.
6.
Alternative
Renew
Relocate
Decision:
7.
Alternative
a. Renew
Relocate
Decision:
(a)
MaxiMax
Max. Payoff
$4,000,000
$5,000,000*
Relocate
MaxiMin
Laplace
Minimax
(b) Min. Payoff (c) Average
(d) Max. Regret
$500,000*
$2,250,000
$4,500,000
$100,000
$2,550,000*
$3,900,000*
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
Motel Approved (.35)
$500,000
E.V.
$2,775,000*
Renew
Relocate
Rejected (.65)
Approved (.35)
Rejected (.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.
Let P (application is approved) = x. Then P (application rejected) = 1 - x.
From Problem 6.
Renew: 500,000x + 4,000,000(1 - x) = 4,000,000 - 3,500,000x
Relocate: 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.
70
Operations Management, 2/ce
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
2
1
.5
x = .4643
0.1
a.
b.
100,000 + 4,900,000x
P (application approved)=x
Not sensitive (i.e., p (approve) can increase to .46 and still the decision remains renew.
Very sensitive because .45 is very close to .464, at which point decision should be
changed.
c. D = amount 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 = $1,476,923
$4,000,000 – $1,476,923 = $2,523,077
Range is $2,523,077 or more.
EMV
Demand
9.
$ 42
.2(42)
.2 Low
Subcontract
.8 High
2
Expand
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
max.
.8(72)
a. Decision: Build a large facility (see the tree), EMV = $53.6m
Instructor’s Manual, Chapter 5 Supplement
71
b. Max (42;22;-20) = $42 million.
Decision: Build a small facility.
c. EVPI = EPC – EMV
EPC = 42(.2) + 72(.8) = $66.0
EVPI = $66.0 – 53.6 = $12.4
Would be willing to pay up to $12.4 million to find out the demand.
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 = 0.721
72
50
48
Small
42
Medium
22
Large
0
.721
1.0
P (High)
-20
Choose Small if p(high) <.721, and Large if p(high) >.721
10.
.30 low demand
$90
.3(90)
do nothing
90
buy 1
$104
+
.70 high demand
2
subcontract
buy 2nd
.30 low
buy 2
.70 high
110
.7(110)
100
75
130
.3(75)
+
.7(130)
$113.5 (optimum)
Decision: Buy 2 machines (initially).
72
Operations Management, 2/ce
1/3
Solutions (continued)
11.
1
1/3
1/3
50
.30
Alternative A
44
60
.20
49
1/3
2
1/3
1/3
40
Alternative B
.30
5
90
40
.50
4
0
60
(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
EV5 = (.3)(40) + (.5)(50) + (.2)(45) = 46
1/2
40
50
Since 49 > 46, choose alternative A.
12.
Moderate
High
Very High
Reassign
New Staff
50
60
60
60
85
60
Worst
(max)
85
60*
Redesign
40
50
90
90
Best
(min)
50
60
Average
40*
65
60 (tie)
60
a. Minimax: New staff
b. Minimin: Redesign
c. Regret table:
Moderate
Reassign
10
New Staff
20
Redesign
0
High
10
10
0
Very High
25
0
30
Worst (max)
25
20*
30
d. New Staff or Redesign (tie).
Instructor’s Manual, Chapter 5 Supplement
73
Solutions (continued)
13.
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
.30 High
74
60
.60 Very High
.10 Moderate
60
New Staff
.30 High
60
.10 Moderate
.30 High
73
Reassign
New Staff
Redesign
EVPI=$5(000)
60
60
40
50
.60 Very High
c. Opportunity loss table:
Moderate
85
60
.60 Very High
Redesign
50
90
High
Very High
10
10
0
(.3)
25
0
30
(.6)
10
20
0
(.1)
14.
Expected
Regret
19
5* (min)
18
.10 Moderate
50
Reassign
74
.60 Very High
.10 Moderate
Hire 2
Initially
60
New Staff
60
.30 High
.30 High
.60 Very High
60
85
60
60
60
60
.10 Moderate
Hire 1
Initially
74.5
.30 High (Hire 1 more)
40
75
.60 V. High (Hire 1 more)
80
.10 Moderate
40
Redesign
73
.30 High
.60 Very High
50
90
Hire 2 initially.
74
Operations Management, 2/ce
Solutions (continued)
15.
a.
Let P = p (#2)
Payoff
#1
Payoff
#2
A: 20 (1-P) + 140 P = 20 + 120 P
B: 120 (1-P) + 80P = 120 – 40 P
C: 100 (1-P) + 40P = 100 – 60P
140
120
A
B
100
B
C
80
C
A
0
40
.625
1.0
P
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).
16.
[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 = 1 - .444
17.
Let P = P (not receiving contract)
EV1 = 10 (1– P) + (-2) P = 10 – 12P
EV2 = 8 (1– P) + 3P = 8 – 5P
10
8
#1
#2
7
EV3 = 5 (1– P) + 5P = 5 – 0P
EV4 = 0 (1– P) + 7P = 0 + 7P
#4
#2
#3
5
5
2
EV1 = EV2 => 10-12P = 8-5P => P = = .285
7
EV2 = EV3 => 8-5P = 5 => 5P = 3 => P = .6
3
#4
EV3 = EV4 => 5 = 7P => P = .714
If P < .286 => Choose #1; if .286 ≤ P < .60 => Choose
0
#2; if .6 ≤ P < .714 => Choose #3; if P ≥ .714 => Choose #4
.286
.60 .714
1.0
-2
Instructor’s Manual, Chapter 5 Supplement
75
Solutions (continued)
18.
EV = 120 (1-P) + 20P = 120 – 100P
A
EV = 60 (1-P) + 40P = 60 – 20p
B
.30
a. A:
P(2) < .30
B: never
C:
P(2) ≥ .8
110
D: .3 ≤ P(2) < .8
.80
120
A
Profits
C
EV = 10 (1-P) + 110P = 10 + 100p
C
EV
D
D
90
90
90 (1-P) + 90P = 90
60
b. A: never
B: .417 ≤ P(2) < .75
C:
P(2) < .417
D:
P(2) ≥ .75
B
C
Costs
A
40
20
10
0
EV = EV => 120 – 100P = 90 => P =.3
A
D
.417
.75
1.0
P (#2)
EV = EV => 90 = 10 + 100P => P = 80
D
C
EV = EV => 10 + 100P = 60 – 20P => P = .417
C
B
EV = EV => 60 – 20P = 120 – 100P => P =.75
B
A
76
Operations Management, 2/ce