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gavmur [86]
3 years ago
7

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)?
Business
1 answer:
Reptile [31]3 years ago
4 0

Answer:

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).

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A company produces and sells a consumer product and is able to control the demand for the product by varying the selling price. The approximate relationship between price and demand is 50 units.

p = 38 + (2,700 / D) - (5,000 / D2)

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(a) Profit is maximized by equality of Marginal revenue (MR) and MC.

Total revenue (TR) = p x D = 38D + 2,700 - (5,000 / D)

MR = dTR / dD = 38 + (5,000 / D2)

Equating MR with MC,

38 + (5,000 / D2) = 40

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Second order condition: d2\pi/dD2 = d/dD(d\pi/dD) = - 2 x (5,000 / D3) = - 10,000 / D3

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