If Ed Lusk, VP for operations, proceeds with the existing prototype (option a), the firm can expect sales to be 110 comma 000 units at $580 each, with a probability of 0.65 and a 0.35 probability of 65 comma 000 at $580. If, however, he uses the value analysis team (option b), the firm expects sales of 85 comma 000 units at $740 , with a probability of 0.62 and a 0.38 probability of 60 comma 000 units at $740. Value engineering, at a cost of $100 comma 000 , is only used in option b. Which option has the highest expected monetary value (EMV) ?

Question

Ulises Carson

Business

30 December, 04:14

If Ed Lusk, VP for operations, proceeds with the existing prototype (option a), the firm can expect sales to be 110 comma 000 units at $580 each, with a probability of 0.65 and a 0.35 probability of 65 comma 000 at $580. If, however, he uses the value analysis team (option b), the firm expects sales of 85 comma 000 units at $740 , with a probability of 0.62 and a 0.38 probability of 60 comma 000 units at $740. Value engineering, at a cost of $100 comma 000 , is only used in option b. Which option has the highest expected monetary value (EMV) ?

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Answers (1)

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Jovani Rasmussen · Answer 1

Option b has the highest expected monetary value (EMV)

Explanation:

For Option a:

At probability 0.65: At probability 0.35

Units=110,000 Units=65,000

Amount=$580 each Amount=$580

EMV/Total Sales=0.65 (110,000*580) + 0.35 (65,000*580)

EMV/Total Sales=$54,665,000

For Option b:

At probability 0.62: At probability 0.38

Units=85,000 Units=60,000

Amount=$740 each Amount=$740

EMV/Total Sales=0.62 (85,000*740) + 0.38 (60,000*740)

EMV/Total Sales=$55,870,000

So Option B has the highest expected monetary value (EMV).