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The executives of the General Products Company (GPC) have to decide which of three products to introduce A, B or C. Product C is essentially

The executives of the General Products Company (GPC) have to decide which of three products to introduce A, B or C. Product C is essentially a risk-free proposition, from which the company will obtain a net profit of $1 million. Product B is considerably more risky. Sales may be high, with resulting net profit of $8 million, medium with net profit of $ 4million, or low, in which case the company just breaks even. The probabilities for these outcomes are:

P(Sales High for B) = 0.25

P(Sales Medium for B) = 0.25

P(Sales Low for B) = 0.50

Product A poses something of a difficulty, a problem with the production system has not yet been solved. The engineering division has indicated its confidence in solving the problem, but there is a slight (5%) chance that devising a workable solution may take a long time. In this event, there will be a delay in introducing the product, and that delay will result in lower sales and profits. Another issue is the price for Product A. The options are to introduce it at either high or low price, the price would not be set until just before the product is to be introduced. Both of these issues have an impact on the ultimate net profit. Finally, once the product is introduced, sales can be either high or low. If the company decides to set a low price, then low price are just as likely as high sales. If the company sets a high price, the likelihood of the low sales depends on whether the product was delayed by the production problem. If there was no delay and the company sets a high price, the probability is 0.4 that sales will be high. However, if there is a delay and the price is set high, the probability is only 0.3 that sales will be high. The following table shows the possible net profit figures (in millions) for Product A:

Price High Sales($ Million) Low Sales($ Million)

Time delay High 5.0 ( -0.5)

Low 3.5 1.0

No delay High 8.0 0.0

Low 4.5 1.5

Solve the decision tree using EMV?

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