A small market orders copies of a certain magazine for its magazine rack each week. Let X = demand for the magazine, with the following pmf.
x 1 2 3 4 5 6
p(x) 1/14 1/14 4/14 3/14 3/14 2/14
Suppose the store owner actually pays $2.00 for each copy of the magazine and the price to customers is $4.00. If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand X, and then compute the expected revenue.]
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suppose he buys \(n\) magazines, costing him \(2n\) dollars, and the quantity demanded (i.e. the # of magazines people are willing to buy), is denoted \(X\). if \(n\le X\), then he can can sell all \(n\) of the magazines he bought, so the revenue is \(4n\), giving a profit/net-revenue of \(4n-2n=2n\). on the other hand, if \(n> X\), i.e. he bought more than were demanded, then he only sells \(X\) many, so the revenue is \(4X\) and the profit or net-revenue is \(4X-2n\). so we have two cases:
$$X\ge n\implies \text{the net revenue is }2n\\X< n\implies\text{the net revenue is }4X-2n$$
so if he buys \(n=3\), then we see that: $$\begin{array}{c|c}
X & \text{Net Revenue }(\$)\\
\hline
1 & -2\\
2 & 2\\
3 & 6\\
4 & 6\\
5 & 6\\
6 & 6
\end{array}$$
whereas if he buys \(n=4\) we see $$\begin{array}{c|c}
X & \text{Net Revenue }(\$)\\
\hline
1 & -4\\
2 & 0\\
3 & 4\\
4 & 8\\
5 & 8\\
6 & 8
\end{array}$$

- anonymous

now, compute the expected net revenue for \(n=3\) and \(n=4\) using the given probabilities for \(X=1,2,3,\dots,6\): $$\begin{array}{c||cc}X&1&2&3&4&5&6\\\hline p &\frac1{14}&\frac1{14}&\frac4{14}&\frac3{14}&\frac3{14}&\frac2{14}\end{array}$$
so the expected net revenue for \(n=3\) is given by $$\frac1{14}\cdot(-2)+\frac1{14}\cdot2+\frac4{14}\cdot6+\frac3{14}\cdot6+\frac3{14}\cdot6+\frac2{14}\cdot6=\frac{36}7\approx5.143$$whereas for \(n=4\) we get
$$\frac1{14}\cdot(-4)+\frac1{14}\cdot0+\frac4{14}\cdot4+\frac3{14}\cdot8+\frac3{14}\cdot8+\frac2{14}\cdot8=\frac{38}7\approx5.43$$
so on average he can expect more profit from buying \(n=4\) magazines rather than only \(3\)

- anonymous

Ahh, thank you!! That makes sense

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