Q. 5.9 - A First Course in Probability

Question

Consider Example 4b of Chapter 4, but now suppose that the seasonal demand is a continuous random variable having probability density function $f$ . Show that the optimal amount to stock is the value $s*$ that satisfies

$F\left(s^{*}\right)=\frac{b}{b+l}$

where $b$ is net profit per unit sale, $l$ is the net loss per unit

unsold, and $F$ is the cumulative distribution function of the

seasonal demand.

Step-by-Step Solution

Answer

$F\left(s\right)=\frac{b}{b+l}$

$where, F\left(s\right)={\int }_0^{s}f\left(x\right)dx$ is the cumulative distribution of demand.

Step 1. Find expected profit for P ( s ) .

Let $X$ be the number of units demanded and $s$ be the units stocked, then the profit is

$P\left(s\right) = \left\{\begin{array}{l}bX-\left(s-Xl\right) if X\le s\\ Xb if X>s\end{array}\right.$

The expected profit $E\left[P\left(s\right)\right]={\int }_0^s\left\{bx-\left(s-x\right)l\right\}f\left(x\right)fx + {\int }_s^{\infty }sb f\left(x\right) dx$

$$\begin{gathered} =(b+l){\int }_0^{s}x f\left(x\right) dx -sl {\int }_0^{s}f\left(x\right) dx + sb\left[1-{\int }_0^{s}f\left(x\right) dx\right] \\ =sb+\left(b+l\right){\int }_0^s(x-s) f\left(x\right) dx \end{gathered}$$

Step 2. Take differentiation with respect to s , and equate to zero.

$$\begin{gathered} \frac{d}{ds}E\left[P\left(s\right)\right] = 0 \\ \Rightarrow b +\hspace{0.17em}(b+l) \frac{d}{ds}\left[{\int }_0^{s}x f\left(x\right)dx - s{\int }_0^{s}f\left(x\right)dx\right] =0 \\ \Rightarrow b +\hspace{0.17em}(b+l) \left[S f\left(s\right) - S f\left(s\right) - {\int }_0^{s}f\left(x\right) dx\right] = 0 \\ \Rightarrow b -\hspace{0.17em}(b+l) \left[{\int }_0^{s}f\left(x\right) dx\right] = 0 \\ \therefore F\left(s\right)=\frac{b}{b+l} \end{gathered}$$

Where,  $F\left(s\right) = {\int }_0^{s}f\left(x\right) dx$ is the cumulative distribution of demand.