Question

a. 130000 b. 210000 c.195000 d.160000

A computer chip manufacturer is considering expanding production to meet possible increases in demand. The company's alternatives are to construct a new plant, expand the existing plant, or do nothing in the short run. It will cost them $1 million to build a new facility and $600,000 to expand their existing facility. The market for this particular product may expand, remain stable, or contract. The marketing department estimates the probabilities of these market outcomes as 020, 0.50, and 0.30. respectively. The expected revenue for each alternative is presented in the table below. Build new plant Expand plant Do nothing MKT expands $1,650,000 $1,000,000 $0 MKT stable $1,000,000 $850,000 $0 MKT contracts $450,000 $450,000 $0 What course of action is optimal for thecomputer chip manufacturer? What is the expcted r ithat case?

Answer 1

First we convert the revenues given into profits, (by subtracting 1,000,000 from the first row, subtracting 600,000 from the second row and keeping the third row unchanged).

Thus, the expected profit for each alternative is given by:

	MKT expands(0.20)	MKT stable(0.50)	MKT contracts(0.30)
Build new plant	$650,000	$0	-$550,000
Expand plant	$400,000	$250,000	-$150,000
Do nothing	$0	$0	$0

Now, we find the expected profit for each decision (build new plant, expand plant and do nothing):

E(build new plant) = 650,000 * P(MKT expands) + 0*P(MKT stable) - 550000*P(MKT contracts)

= 650000*0.20 + 0*0.50 - 550000*0.3

= 130000 + 0 - 165000

= -$35000

E(expand plant) = 400,000 * P(MKT expands) + 250000*P(MKT stable) - 150000*P(MKT contracts)

=400,000*0.20 + 250000*0.50 -150000*0.30

= 80000 + 125000 - 45000

= $160000

E(do nothing) = $0

Thus, on comparing the expected values of each outcome we find the expanding plant is optimal for the computer chip manufacturer and the expected profit in that case is $160000.

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