Question

# A firm has received a production order of 10 injection-molded parts. The parts will sell for...

A firm has received a production order of 10 injection-molded parts. The parts will sell for \$4000 each; it will cost \$2500 to produce an individual part. The probability of a part meeting the final inspection standards is .80. Any good parts in excess of 10 can be salvaged for \$2000; any bad parts produced can be salvaged for \$1000. If an insufficient number of good parts are produced, each shortage must be overcome at a cost of \$3000 in excess of the cost of production. Determine the number of parts to be scheduled for production to maximize expected profit.

The number of parts schedule for production to maximize expected profit should be 15.

EXPLANATION

As the probability of a part meeting the final inspection standards is .80 i.e. 80%

Which means if we produce 15 parts then their will be 12 parts perfect meeting the final standard inspection. And their will be 3 bad parts.

So for the production of 15 parts money required will be = \$2500 × 15 = \$37500

And we will sell the 10 good parts at = \$4000 × 10 = \$40000

Now, their are 2 good parts left extra, that will sold for = \$2000 × 2 = \$4000

And their are 3 bad parts left, that will be sold for = \$1000 × 3 = \$3000

Total cost of selling = \$40000 + \$4000 + \$3000 = \$47000

Maximum profit = seeling price - production price = \$47000 - \$37500 = \$9500

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