# Question

Blue Ridge Power and Light 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 .10 probability that a defective component will occur, requiring emergency maintenance at a cost of $1,200, or there is a .90 probability that no defects will occur and no maintenance will be necessary.

Strategy 2 (take oil samples) costs $20 to implement (i.e., take a sample), and there is a .10 probability that there will be a defective part and .90 probability that there will not be a defect.

If there is actually a defective part, there is a .70 probability that the sample will correctly identify it, resulting in preventive maintenance at a cost of $500. However, there is a .30 probability that the sample will not identify the defect and indicate that everything is okay, resulting in emergency maintenance later at a cost of $1,200. On the other hand, if there are actually no defects, there is a .20 probability that the sample will erroneously indicate that there is a defect, resulting in unnecessary maintenance at a cost of $250. There is an .80 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 $14.80 to implement and has two outcomes: a .04 probability of a defective component, which will require emergency maintenance at a cost of $1,200, and a .96 probability that no defects will occur, resulting in no maintenance and no cost. Strategy 4 (changing the oil and sampling) costs $34.80 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 .70 probability that the sample will detect it and $500 in preventive maintenance costs will be incurred. Alternatively, there is a .30 probability that the sample will not detect the defect, resulting in emergency maintenance at a cost of $1,200. If there is no defect, there is a .20 probability that the sample will indicate that there is a defect, resulting in an unnecessary maintenance cost of $250, and there is an .80 probability that the sample will correctly indicate no defects, resulting in no cost. Develop a decision strategy for Blue Ridge Power and Light and indicate the expected value of this strategy.1

For autos and light trucks, strategy 1 (no preventive maintenance) costs nothing to implement and results in two possible outcomes: There is a .10 probability that a defective component will occur, requiring emergency maintenance at a cost of $1,200, or there is a .90 probability that no defects will occur and no maintenance will be necessary.

Strategy 2 (take oil samples) costs $20 to implement (i.e., take a sample), and there is a .10 probability that there will be a defective part and .90 probability that there will not be a defect.

If there is actually a defective part, there is a .70 probability that the sample will correctly identify it, resulting in preventive maintenance at a cost of $500. However, there is a .30 probability that the sample will not identify the defect and indicate that everything is okay, resulting in emergency maintenance later at a cost of $1,200. On the other hand, if there are actually no defects, there is a .20 probability that the sample will erroneously indicate that there is a defect, resulting in unnecessary maintenance at a cost of $250. There is an .80 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 $14.80 to implement and has two outcomes: a .04 probability of a defective component, which will require emergency maintenance at a cost of $1,200, and a .96 probability that no defects will occur, resulting in no maintenance and no cost. Strategy 4 (changing the oil and sampling) costs $34.80 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 .70 probability that the sample will detect it and $500 in preventive maintenance costs will be incurred. Alternatively, there is a .30 probability that the sample will not detect the defect, resulting in emergency maintenance at a cost of $1,200. If there is no defect, there is a .20 probability that the sample will indicate that there is a defect, resulting in an unnecessary maintenance cost of $250, and there is an .80 probability that the sample will correctly indicate no defects, resulting in no cost. Develop a decision strategy for Blue Ridge Power and Light and indicate the expected value of this strategy.1

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