quantitative decisions in business This assignment must be in Excel/QM and submi
ID: 411571 • Letter: Q
Question
quantitative decisions in business
This assignment must be in Excel/QM and submitted through Assignment. Also provide a written response for each part of each question.
3.2) The following payoff table provides profits based on various possible decision alternatives and various levels of demand.
States of Nature
Demand
Alternatives
Low
Medium
High
Alternative 1
80
120
140
Alternative 2
90
90
90
Alternative 3
50
70
150
The probability of a low demand is 0.4, while the probability of a medium and high demand is each 0.3.
(a) What decision would an optimist make?
(b) What decision would a pessimist make?
(c) What is the highest possible expected monetary value?
(d) Calculate the expected value of perfect information for this situation
States of Nature
Demand
Alternatives
Low
Medium
High
Alternative 1
80
120
140
Alternative 2
90
90
90
Alternative 3
50
70
150
Explanation / Answer
(a) An optimist would make decision as per MaxiMax criterion. In this criterion, the Maximum out of the Maximum Payoffs of each decision alternative is used as the selection criterion.
Maximum payoff of Alternative 1 = 140
Maximum payoff of Alternative 2 = 90
Maximum payoff of Alternative 3 = 150
The Maximum out of the above is 150. It pertains to alternative 3. Therefore, an optimist would select Alternative 3.
(b) A pessimist would make decision as per MaxiMin criterion. In this criterion, the Maximum out of the Minimum Payoffs of each decision alternative is used as the selection criterion.
Minimum payoff of Alternative 1 = 80
Minimum payoff of Alternative 2 = 90
Minimum payoff of Alternative 3 = 50
The Maximum out of the above is 90. It pertains to alternative 2. Therefore, a perssimist would select Alternative 2.
(c) Expected Monetary Value (EMV) of alternative 1 = 80*.4 + 120*.3 + 140*.3 = 110
Expected Monetary Value (EMV) of alternative 2 = 90*.4 + 90*.3 + 90*.3 = 90
Expected Monetary Value (EMV) of alternative 3 = 50*.4 + 70*.3 + 150*.3 = 86
Highest EMV is 110 pertaining to alternative 1.
Highest possible EMV = 110
(d) Expected Value with Perfect Information (EVwPI) = MAX(80,90,50)*.4 + MAX(120,90,70)*.3 + MAX(140,90,150)*.3 = 117
Expected Value of Perfect Information (EVPI) = EVwPI - EMVmax = 117 - 110 = 7
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