## anonymous one year ago 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.] Please explain what steps you took and why. I really want to learn the material but I'm having trouble interpreting what information the question is giving me, what it means, and how to set it up. Any help would be greatly appreciated!

1. anonymous

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}$$

2. 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 $$3$$

3. anonymous

Ahh, thank you!! That makes sense