Question

CHAPTER 5 PRODUCT DESIGN 213 DECISION TREE APPLIED TO PRODUCT DESIGN Silicon, Inc, a semcductor manufacturer, is invessigating the possility of producing and marketing a microprocessor. hiring and training several additional engineers. The mark unfavorable. Silicon, Inc., of course, has the option of not developing the new product at all. king this project will require either purchasing a sophisticated CAD system or et for the product could be either favorable or With favorable acceptance by the market, sales would be 25,000 processors selling for S100 each. With unfavorable acceptance, sales would be only 8,000 processors selling for $100 each. The cost of CAD equipment is S500,000, but that of hiring and training three new enginers is only $375,000. However manufacturing costs should drop from $50 each when manufacturing without CAD to $40 each when manufacturing with CAD The probability of favorable acceptance of the new microprocessor is 40: the probability of unfa- vorable acceptance is 60. APPROACH Use of a decision tree seems appropriate as Silicon, Inc., has the basic ingredients: a choice of decisions, probabilities, and payofs. SOLUTIONIn Figure 5.13 we draw a decision tree with a branch for each of the three decisions, as- sign the respective probabilities and payoff for each branch, and then compute the respective EMVs. The expected monctary values (EMVs) have been circled at each step of the decision tree. For the top branch: EMV (Purchase CAD system)(4$1,000,000) +.6-$20,000) $388,000 This figure represents the results that will occur if Silicon, Inc, purchases CAD The expected value of hiring and training engineers is the second series of branches EMV (Hire/train engineers) (4)875,000) + (6yS25,000) $365,000 $2.500,000 Revenue 1-1.000.000 High sales 500,000 CAD cost Purchase CAD $388,000 Mfgoost($40× 25,000) $1,000,000 Net ment $800,000 Revenue C (6)320,000 Mig cost (540 x 8,000) Low sales-500,000 CAD cost $20,000 Net loss Hire and train engineers $2,500,000 -1,250,000 High sales 375.000 Revenue MIg.cost ($50×25.000) Hire and train cost $875,000 Net $800,000 Revenue 400000 (6) Mig, cost (S50 × 8,000) Low sales 1-375,000 Hire and train cost 25,000 Net Do nothing $0 $0 Net DESIGNING OPERATIONS The EMV of doing nothing is S0 Because the top branch has the highest expected monetary value (an EMV of $388,000 vs. S365 s. SO), it represents the best decision. Management should purchase the CAD system. 0 Pro Pro Qua 0 Pro Con INSIGHT Use of the decision tree provides both objectivity and structure to our analysis of the St con, Inc., decision. LEARNING EXERCISEIf Silicon, Inc., thinks the probabilities of high sales and low sales may be equal, at S each, what is the best decision? (Answer: Purchase CAD remains the best decision, but with an EMV of $490,000.] RELATED PROBLEMS5.10-5.17 Rob ACTIVE MODEL 5.1 This example is further ilustrated in Active Model 5.1 at www.pearsonglobaleditions.com/heizer Con Desi How I solve this in a simple way?

Answer 1

Learning Excercise:

The probabilities of high sales and low sales are equal =0.5

EMV of purchasing CAD = Sum of (Probability of demand * Net profit) = 0.5 * (1,000,000 + (-20000)) = 490,000

EMV of Hire/train = 0.5 * (875000 + 25000) = 450,000

It is still better to purchase CAD but with and EMV of 490,000.