Question

A product, sold seasonly, yields a net profit of $100 for each unit sold and a net loss of $60 for each unit left unsold. The number of units for the product ordered at a department store during a season is the random variable X with probability density function P(X = i) = 1/il if 10 i 20 and P(X = i) = 1/10 otherwise. If the store must stock this product in advance, how many should it stores to maximize the expected profit?

Answer 1

Solution

The solution is based on EMV concept.

Back-up Theory

Let Q be the stock D = quantity ordered. Then,

Pay-off = 100Q, if Q ≤ D

             = 100D – 60(Q – D), if Q > D

EMV (Expected Monetary Value) for Q = Σ(pay-off) x P(D), sum over all possible values of D.

Now to work out the solution,

For all possible combinations of Q and D, the pay-off and the EMV are calculated in the table below:

Q	D											EMV
	10	11	12	13	14	15	16	17	18	19	20
10	1000	1000	1000	1000	1000	1000	1000	1000	1000	1000	1000	999.999
11	940	1100	1100	1100	1100	1100	1100	1100	1100	1100	1100	1085.453
12	880	1040	1200	1200	1200	1200	1200	1200	1200	1200	1200	1156.362
13	820	980	1140	1300	1300	1300	1300	1300	1300	1300	1300	1212.726
14	760	920	1080	1240	1400	1400	1400	1400	1400	1400	1400	1254.544
15	700	860	1020	1180	1340	1500	1500	1500	1500	1500	1500	1281.817
16	640	800	960	1120	1280	1440	1600	1600	1600	1600	1600	1294.544
17	580	740	900	1060	1220	1380	1540	1700	1700	1700	1700	1292.726
18	520	680	840	1000	1160	1320	1480	1640	1800	1800	1800	1276.362
19	460	620	780	940	1100	1260	1420	1580	1740	1900	1900	1245.453
20	400	560	720	880	1040	1200	1360	1520	1680	1840	2000	1199.999
P(D) = 1/11 for all values of D.

Since at Q = 16, the EMV is maximum, the quantity to be stocked is 16 Answer

DONE