Mary developed the following payoff table based on the plan to build housing units. Mary has 3 decision alternatives to build 30 units, 50 units, and 60 units. The probabilities of strong, fair, and poor housing market are provided.
Housing Units Good Market Fair Market Poor Market
30, d1 $50,000 . $20,000 $10,000
50, d2 80,000 30,000 -10,000
60, d3 150,000 35000 -30,000
Probability .5 .3 .2
Question # 1: If Mary wanted to make a decision based on the expected value (EV) of the decision alternatives, what would be the value ($) of the housing units chosen?
Construct a regret table based on the payoff table above and calculate the expected value of the opportunity loss for each decision alternatives using the regret payoff amounts.
Question #2: If Mary can't make a decision as to what alternative to choose, what is the minimum expected value (S) of the opportunity loss for Mary?
Question # 1
Expected value for 30 units = 0.5 * 50,000 + 0.3 * 20,000 + 0.2 * 10,000 = $33,000
Expected value for 50 units = 0.5 * 80,000 + 0.3 * 30,000 + 0.2 * -10,000 = $47,000
Expected value for 60 units = 0.5 * 150,000 + 0.3 * 35000 + 0.2 * -30,000 = $79,500
Since the highest EMV is for housing units 60, the decision based on the expected value (EV) is 60 units.
Regret = Best Payoff - Payoff Received
Regret table is,
Housing Units | Good Market | Fair Market | Poor Market |
30 | 150,000 - 50,000 = 100,000 | 35000 - 20,000 = 15000 | 10,000 - 10,000 = 0 |
50 | 150,000 - 80,000 = 70,000 | 35000 - 30,000 = 5000 | 10,000 - (-10,000) = 20,000 |
60 | 150,000 - 150,000 = 0 | 35000 - 35000 = 0 | 10,000 - (-30,000) = 40,000 |
Question #2:
For 30 units, the maximum opportunity loss (Regret) is 100,000
For 50 units, the maximum opportunity loss (Regret) is 70,000
For 60 units, the maximum opportunity loss (Regret) is 40,000
The minimum of these opportunity loss is $40,000 which is for house for 60 units.
Mary should choose housing units 60 to minimize the expected value of the opportunity loss.
Mary developed the following payoff table based on the plan to build housing units. Mary has 3 decision alternatives to build 30 units, 50 units, and 60 units. The probabilities of strong, fair, and p...
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