# expected value for this question

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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What did you try? And so on. – Did Jun 28 '12 at 15:25
His profit is 400 dollars for each item he sells. Multiply 400 by the expected demand to get the answer. However you cannot calculate the expected demand because all you know is that 5 or more has probability .1 and you do not know how this distributes among the integers above 5. To maximize his expected profit he should buy just enough to meet his expected demand. So if you could calculate the expected demand you would have the answer. All you actually know though is that the expected demand is greater than 2.6 ( the value you would get if 0.1 is the demand for exact 5 and no more). – Michael Chernick Jun 28 '12 at 16:52

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.

There are not many details for you to fill in.

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