Question

PowerE is an electric utility company with a large fleet of vehicles, including automobiles, light trucks, and construction equipment. The company is evaluating four alternative strategies for maintaining its vehicles at the lowest cost: (1) do no preventive maintenance at all and repair vehicle components when they fail; (2) take oil samples at regular intervals and perform whatever preventive maintenance is indicated by the oil analysis; (3) change the vehicle oil on a regular basis and perform repairs when needed; (4) change the oil at regular intervals, take oil samples regularly, and perform maintenance repairs as indicated by the sample analysis.

For autos and light trucks, strategy 1 (no preventive maintenance) costs nothing to implement and results in two possible outcomes: There is a 0.12 probability that a defective component will occur, requiring emergency maintenance at a cost of $1,000  or there is a 0.88 probability that no defects will occur and no maintenance will be necessary. If defective component is found, the company should consider whether to change some parts at a cost of $400. There is a 0.7 probability that the defect problem is solved after changing some parts. However, the company needs to pay $750 as further maintenance fee if defect is still found.

Strategy 2 (take oil samples) costs $20 to implement (i.e. take a sample), and there is a 0.1 probability that there will be a defective part and 0.9 probability that will not be a defect. If there is actually a defective part, there is a 0.65 probability that the sample will correctly identify it, resulting in preventive maintenance at a cost of $800. However, there is a 0.35 probability that the sample will not identify the defect and indicate that everything is okay, resulting in emergency maintenance later at a cost of $1,650. On the other hand, if there are actually no defects, there is a 0.2 probability that the sample will erroneously indicate that there is a defect, resulting in unnecessary maintenance at a cost of $450. There is an 0.8 probability that the sample will correctly indicate that there are no defects, resulting in no maintenance and no costs.

Strategy 3 (changing the oil regularly) costs $26.8 to implement and has two outcomes: a 0.08 probability of a defective component, which will require emergency maintenance at a cost of $2,250, and the remaining probability that no defects will occur, resulting in no maintenance and no cost.

Strategy 4 (changing the oil and sampling) costs $54.8 to implement and results in the same probabilities of defects and no defects as strategy 3. If there is a defective component, there is a 0.7 probability that the sample will defect it and $700 in preventive maintenance costs will be incurred. Alternatively, there is a 0.3 probability that the sample will not detect the defect, resulting in emergency maintenance at a cost of $1,500, if there is no defect, there is a 0.25 probability that the sample will indicate that there is a defect, resulting in an unnecessary maintenance cost of $350, and there is an 0.75 probability that the sample will correctly indicate no defects, resulting in no cost.

1. If the company conducts another decision analysis for its heavy construction equipment. 5 strategies are identified and four states of nature to predict the future market revenue ($millions).

	States of Nature
	1	2	3	4
Strategy 1	42	68	71	105
Strategy 2	10	45	79	135
Strategy 3	45	60	74	89
Strategy 4	-19	32	70	150
Strategy 5	-28	-5	100	125

1. Determine which strategy should be used under minimax regret approach. Explain your answer by creating an opportunity loss table.                                           

2. If the probability of states of nature 1, 2 ,3 and 4 are 0.26, 0.18, 0.32 and 0.24 respectively, which strategy should be chosen? (you need to show the details of your calculation as evidence)                                                           

3. How much should the company to get the perfect information (EVPI)? (you need to show the details of your calculation)                                              

Answer 1

1. Regret matrix

Strategy /SON	1	2	3	4
1	3	0	29	45
2	35	23	21	15
3	0	8	26	61
4	64	36	30	0
5	73	73	0	25

Minimax regret = Min [ Max regret]

= Min [ 45,35,61,64,73] =35 which corresponds to strategy 2.

2. EMV for a strategy = 0.26xpayoff I+0.18xpayoff 2 +0.32x payoff 3+0.24x payoff 4

Table of EMV

Prob.	0.26	0.18	0.32	0.24
Strategy /SON	1	2	3	4	EMV
1	42	68	71	105	71.08
2	10	45	79	135	68.38
3	45	60	74	89	67.54
4	-19	32	70	150	59.22
5	-28	-5	100	125	53.82

Strategy 1 is the best strategy as it has highest value of EMV = 71.08

3.

EVPI = EV with PI - EV without PI

= 0.26x45+0.18x 68+0.32x100+0.24x150 = 11.7+12.24+32+36 =91.94

EVPI = 91.94-71.08 =20.86