Question

In Example $4b$, suppose that the department store incurs an additional cost of $c$for each unit of unmet demand. (This type of cost is often referred to as a goodwill cost because the store loses the goodwill of those customers whose demands it cannot meet.) Compute the expected profit when the store stocks $s$ units, and determine the value of data-custom-editor="chemistry" $\mathrm{s}$ that maximizes the expected profit.

Short answer

The expected profit when the store stocks $s$units, and determine the value of $s$that maximizes the expected profit is $\sum _{i=0}^{S^{\ast }} P\left(i\right)<\frac{b+c}{b+c+l}$

Step-by-step solution

Given information

suppose that the department store incurs an additional cost of $c$for each unit of unmet demand.

Solution

The calculation is given below,

$\mathrm{P}\left(\mathrm{s}\right)=\mathrm{b}X-(s-X)\times l$

If $X\le \mathrm{s}$,

$=\mathrm{sb}-c(X-s)$

If $X>\mathrm{s}$

$\Rightarrow \mathrm{E}\left[\mathrm{p}\right(\mathrm{s}\left)\right]=\sum _{\mathrm{i}=0}^{\mathrm{s}} [\mathrm{bi}-(s-i\left)l\right]\mathrm{p}\left(\mathrm{i}\right)+\sum _{i=s+1}^{\mathrm{\infty }} [\mathrm{sb}-c(i-s\left)\right]\mathrm{p}\left(\mathrm{i}\right)$

$=(\mathrm{b}+l)\sum _{\mathrm{i}=0}^{\mathrm{s}} i\mathrm{p}\left(\mathrm{i}\right)-sl\sum _{\mathrm{i}=0}^{\mathrm{s}} p\left(i\right)-c\sum _{\mathrm{i}=\mathrm{s}+1}^{\mathrm{\infty }} i\mathrm{p}\left(\mathrm{i}\right)+(sb+cs)\sum _{\mathrm{i}=\mathrm{s}+1}^{\mathrm{\infty }} \mathrm{p}\left(\mathrm{i}\right)$

$=(\mathrm{b}+l)\sum _{\mathrm{i}=0}^5 \mathrm{i}.\mathrm{p}\left(\mathrm{i}\right)-sl\sum _{\mathrm{i}=0}^s p\left(i\right)-c\sum _{\mathrm{i}=\mathrm{s}+1}^{\mathrm{\infty }} ip\left(i\right)+(sb+cs)\left[1-\sum _{\mathrm{i}=0}^S p\left(i\right)\right]$

$=(\mathrm{sb}+\mathrm{cs})+(\mathrm{b}+l)\sum _{\mathrm{i}=0}^s ip\left(i\right)-(sl+sb+sc)\sum _{\mathrm{i}=0}^s p\left(i\right)-c\sum _{\mathrm{i}=\mathrm{s}+1}^{\mathrm{\infty }} ip\left(i\right)$

$=\mathrm{s}(\mathrm{b}+c)-s(b+c+l)\sum _{\mathrm{i}=0}^s p\left(i\right)+(b+l)\sum _{\mathrm{i}=0}^s ip\left(i\right)-c\sum _{\mathrm{i}=\mathrm{s}+1}^{\mathrm{\infty }} ip\left(i\right)$

Final solution

Here we need to simplify the equation,

$\mathrm{E}\left[\mathrm{P}\right(\mathrm{s}+1\left)\right]=(\mathrm{s}+1)(\mathrm{b}+\mathrm{c})-(s+1)(b+c+l)\sum _{i=0}^{S+1} P\left(i\right)+(b+l)\sum _{i=0}^{S+1} ip\left(i\right)-c\sum _{i=\mathrm{s}+2}^{\mathrm{\infty }} ip\left(i\right)$

$\begin{array}{r}\mathrm{E}\left[\mathrm{P}\right(\mathrm{s}+\mathrm{l}\left)\right]-\mathrm{E}\left[\mathrm{P}\right(\mathrm{s}\left)\right]=(\mathrm{b}+\mathrm{c})-(b+c+l)\sum _{\mathrm{i}=0}^S p\left(i\right)-(s+1)(b+c+l)P(s+1)(\mathrm{b}+l)(s+1)P(s+1)\\ +c(s+1)P(s+1)\end{array}$

$=(\mathrm{b}+\mathrm{c})-(b+c+l)\sum _{i=0}^s P\left(i\right)+(s+1)P(s+1)[b+l+c-b-c-l]$

$=(\mathrm{b}+\mathrm{c})-(b+c+l)\sum _{i=0}^s P\left(i\right)+(s+1)P(s+1)\left(0\right)$

$\Rightarrow \mathrm{E}\left[\mathrm{P}\right(\mathrm{s}+1\left)\right]-E\left[P\right(s\left)\right]=(\mathrm{b}+\mathrm{c})-(b+c+l)\sum _{i=0}^s P\left(i\right)$Now,$\mathrm{s}+1$units will be better than $\mathrm{S}$units

$\Rightarrow \sum _{i=0}^S P\left(i\right)<\frac{b+c}{b+c+l}$

As Right hand side is constant, say ${\mathrm{s}}^{*}$is the maximum value of s such that

$\sum _{i=0}^{S^{\ast }} P\left(i\right)<\frac{b+c}{b+c+l}$

Therefore,$s^{*}+1$Items will lead to maximum expected profit

Final answer

The expected profit when the store stocks $s$units, and determine the value of $s$that maximizes the expected profit is $\sum _{i=0}^{S^{\ast }} P\left(i\right)<\frac{b+c}{b+c+l}$