Question

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.

Answer 1

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