Chapter 5: Q. 5.9 (page 212)

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
$F (s^{}) = \frac{b}{b + l}$
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.

Short Answer

Expert verified

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

$w h e r e, F (s) = \int_{0}^{s} f (x) d x$ is the cumulative distribution of demand.

Step by step solution

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

$P (s) = \{\begin{cases} b X - (s - X l) i f X \leq s \\ X b i f X > s \end{cases}$

The expected profit $E [P (s)] = \int_{0}^{s} \{b x - (s - x) l\} f (x) f x + \int_{s}^{\infty} s b f (x) d x$

$= (b + l) \int_{0}^{s} x f (x) d x - s l \int_{0}^{s} f (x) d x + s b [1 - \int_{0}^{s} f (x) d x] = s b + (b + l) \int_{0}^{s} (x - s) f (x) d x$

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

$\frac{d}{d s} E [P (s)] = 0 \Rightarrow b + (b + l) \frac{d}{d s} [\int_{0}^{s} x f (x) d x - s \int_{0}^{s} f (x) d x] = 0 \Rightarrow b + (b + l) [S f (s) - S f (s) - \int_{0}^{s} f (x) d x] = 0 \Rightarrow b - (b + l) [\int_{0}^{s} f (x) d x] = 0 ∴ F (s) = \frac{b}{b + l}$

Where, $F (s) = \int_{0}^{s} f (x) d x$ is the cumulative distribution of demand.

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Short Answer

Step by step solution

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

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

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