Question

Is it better to order three or four copies?

A small market orders copies of a certain magazine for its magazine rack each week. Let X = demand for the magazine, with pmf

 

$$ \begin{array}{l|cccccc} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(x) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{3}{15} & \frac{2}{15} \end{array} $$

Suppose the store owner actually pays $2.00 for each copy of the magarine and the price to customers is $4.00. If magarines left at the end of the week have no salvage value, is it better to order three or four coples of the magazine? [Hsnt For both three and four coples ordered, express net revenue as a functlon of demand  X, and then compute the expected revenue.]

Answer 1

Concepts and reason

Mathematical expectation is the product of the probability of an event occurring and the value corresponding with the actual observed occurrence of the event.

Fundamentals

The expected value of the random variable can be defined as,

$E(Y) = \sum {y{\rm{ }}p\left( y \right)}$

And

$E({Y^2}) = \sum {{y^2}{\rm{ }}p\left( y \right)}$

$E\left( {aY} \right) = aE\left( Y \right)$

Let us assume that $h\left( X \right)$ be the net revenue (sales revenue-order cost) as a function of X. then ${h_3}\left( X \right)$and ${h_4}\left( X \right)$are the net revenue for 3 and 4 copies purchased, respectively.

The revenue function is defined as a function of demand for three copies as follows:

$\begin{array}{c}\\{h_3}\left( x \right) = \left\{ \begin{array}{l}\\4X - 2\left( 3 \right)\, & X \le 3\\\\4\left( 3 \right) - 2\left( 3 \right)\, & X > 3\\\end{array} \right.\\\\ = \left\{ \begin{array}{l}\\4X - 2 & & X \le 3\\\\6 & \, & X > 3\\\end{array} \right.\\\end{array}$

Hence the expected net revenue for 3 copies, $E\left[ {{h_3}\left( X \right)} \right]$ is,

$\begin{array}{c}\\E\left[ {{h_3}\left( X \right)} \right] = \sum\limits_D {h\left( x \right)p\left( x \right)} \\\\ = \left[ \begin{array}{l}\\h\left( 1 \right)p\left( 1 \right) + h\left( 2 \right)p\left( 2 \right) + h\left( 3 \right)p\left( 3 \right) + \\\\h\left( 4 \right)p\left( 4 \right) + h\left( 5 \right)p\left( 5 \right) + h\left( 6 \right)p\left( 6 \right)\\\end{array} \right]\\\\ = \left[ \begin{array}{l}\\\left( {4 \times 1 - 6} \right)\left( {\frac{1}{{15}}} \right) + \left( {4 \times 2 - 6} \right)\left( {\frac{2}{{15}}} \right) + \left( {4 \times 3 - 6} \right)\left( {\frac{3}{{15}}} \right) + \\\\\left( 6 \right)\left( {\frac{4}{{15}}} \right) + \left( 6 \right)\left( {\frac{3}{{15}}} \right) + \left( 6 \right)\left( {\frac{2}{{15}}} \right)\\\end{array} \right]\\\\ = 4.93\\\end{array}$

Let ${h_4}\left( X \right)$ denote the net revenue for four copies ordered. Note that of demand exceeds 4, the number of copies sold is 4. Therefore, define the revenue function as a function of demand for four copies is,

The revenue function is defined as a function of demand for three copies as follows:

$\begin{array}{c}\\{h_4}\left( x \right) = \left\{ \begin{array}{l}\\4X - 2\left( 4 \right)\, & X \le 4\\\\4\left( 4 \right) - 2\left( 4 \right)\, & X > 4\\\end{array} \right.\\\\ = \left\{ \begin{array}{l}\\4X - 8 & & X \le 4\\\\8 & \, & X > 4\\\end{array} \right.\\\end{array}$

Hence the expected net revenue for 4 copies, $E\left[ {{h_4}\left( X \right)} \right]$ is,

$\begin{array}{c}\\E\left[ {{h_4}\left( X \right)} \right] = \sum\limits_D {h\left( x \right)p\left( x \right)} \\\\ = \left[ \begin{array}{l}\\h\left( 1 \right)p\left( 1 \right) + h\left( 2 \right)p\left( 2 \right) + h\left( 3 \right)p\left( 3 \right) + \\\\h\left( 4 \right)p\left( 4 \right) + h\left( 5 \right)p\left( 5 \right) + h\left( 6 \right)p\left( 6 \right)\\\end{array} \right]\\\\ = \left[ \begin{array}{l}\\\left( {4 \times 1 - 8} \right)\left( {\frac{1}{{15}}} \right) + \left( {4 \times 2 - 8} \right)\left( {\frac{2}{{15}}} \right) + \left( {4 \times 3 - 8} \right)\left( {\frac{3}{{15}}} \right) + \\\\\left( {4 \times 4 - 8} \right)\left( {\frac{4}{{15}}} \right) + \left( 8 \right)\left( {\frac{3}{{15}}} \right) + \left( 8 \right)\left( {\frac{2}{{15}}} \right)\\\end{array} \right]\\\\ = 5.33\\\end{array}$

Ans:

The expected revenue for the 3 copies is$E\left[ {{h_3}\left( X \right)} \right] = 4.93$.

The expected revenue for the 4 copies is$E\left[ {{h_4}\left( X \right)} \right] = 5.33$.