A biotechnology company produces diagnostic test kits in a once per week product

Question

A biotechnology company produces diagnostic test kits in a once per week production run. Assume the demand for a week follows a Poisson distribution according to a mean predicted by a model developed by the marketing department. To meet demand, they can produce any number of kits. Some of the chemical components in the kit requires that for kits not sold in the week of production they must be scrapped. The production cost is $20.00 per kit. For any weekly demand kits can be sold for $100. The cost of scrapping a kit is $5.

If the demand forecast for the coming week is X = 100 kits, what production run size maximizes the expected profit? What if X = 150? X = 200?

What is the standard deviation of this profit?  

Explanation / Answer

Production cost c = $ 20.00 per kit

Salvage cost s = $ 5.00 per kit

Selling price p = $ 100 per kit

Overage cost = c - s = 20 - 5 = $ 15

Underage cost = p - c = 100 - 20 = $ 80

Here demand forecast X = 100 kits

here we have to calculate the fractile F(X) = [Underage cost/ (underage cost + overage cost)] = [80/(15 + 80)] = 80/95 = 16/19

Here if production run size is K

Pr(X < K) = POISSON (X < K; 100) = 16/19

by normal approximation

standard deviation of poisson distribution = sqrt (150) = 12.25

NORM(X < k ; 100 ; 10) = 16/19 = 0.842

Z value for the given p ;

Z = 1.00

(k - 100)/10 = 1

k = 110

if X = 150

then Z = 1

so the prodution run size k = 150 + 1 * sqrt (150) = 10 = 162.25 or 162

for X = 200 ; production run size k = 200 + 1 * sqrt (200) = = 214.14 or 214

Here profit in terms of production run and demand (X) = (100-20)(X) - (5)(K - X) = 80 X - 5K + 5X = 85X - 5K = 5 (17X - K)

Var(Profit) = Var(17X - K) = 172 Var(X) = 289 * 100 = 28900

Std(profit) = sqrt(28900) = $ 170

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