A manufacturer buys an item for 1600 dollar and sells it for 2000 dollar. The probabilities for a demand of 0, 1, 2, 3, 4, “5 or more” items are 0.05, 0.15, 0.30, 0.25, 0.15, 0.10 respectively. How many items he must stock to maximize his expected profit?
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There is often no magic formula to solve a problem. So we need to do some preliminary exploration to see what's going on.
We also need to make some assumptions. In the real world, unsold items can probably be returned to the supplier, probably with some not very large "restocking" fee. Or else we can keep the unsold item, and perhaps sell it next month. Again, there will be a cost associated with this (warehousing, interest).
Absolutely no information has been supplied about these matters. So we are forced to make the extremely unrealistic assumption that any unsold item has to be junked, like a fish that goes bad. Under that assumption, we start to solve the problem.
Start with the easy stuff. If we stock $0$ items, our expected profit is $0$.
Suppose that we stock $1$ item. With probability $0.05$, we won't sell it. Then our "profit" is $-1600$. With probability $0.95$, we will sell it. Our profit is then $400$. So our expected profit is $(-1600)(0.05)+(400)(0.95)$. Compute. We get $300$.
Suppose that we stock $2$ items. With probability $0.05$, we sell none, profit $-3200$. With probability $0.15$, we sell one, profit $400-1600=-1200$. And with probability $0.8$ we sell both, profit $800$. The expected profit is therefore $(-3200)(0.05)+(-1200)(0.15)+(800)(0.8)$. Compute.
Suppose that we stock $3$ items. In the same way, we can find the expected profit. Probably we should also do the calculation for $4$ items.
For stocking $5$ or more items, we will need to do some mathematics. If we stock $n$ items, where $n \ge 5$, our profit is $400n$ if we sell them all (probability $\le 0.10$). But if we are unlucky and the demand is $0$, we "gain" $-1600n$. So our expected profit is $\le (-1600)(0.05)+(400n)(0.10)$, which is negative.
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