Weibull Demand Distribution¶

Definition¶

The Weibull demand distribution in the exercise has shape $k>0$ $0$ , scale $\lambda>0$ $0$ , and nonnegative support. Its CDF and density are given by

\begin{gathered} F(x)=1-\mathrm{e}^{-(x/\lambda)^k},\qquad x\ge0,\\ f(x)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}\mathrm{e}^{-(x/\lambda)^k},\qquad x>0. \end{gathered}

The CDF is $0$ for $x<0$ $0$ . The density formula is stated for $x>0$ $0$ : as $x\downarrow0$ $0$ , it tends to $+\infty$ if $0<k<1$ $1$ , to $1/\lambda$ if $k=1$ $1$ , and to $0$ if $k>1$ $1$ . A density's assigned value at the single point $0$ does not change the distribution.

For the newsvendor critical ratio $0<\beta<1$ $1$ , inversion gives

y^*=\lambda[-\ln(1-\beta)]^{1/k}.

Interpretation¶

With Weibull demand, the newsvendor answer is still the demand level at which the CDF equals the critical ratio. Because this CDF has an explicit formula, the equation $1-\mathrm{e}^{-(y/\lambda)^k}=\beta$ can be solved by hand with a logarithm, without a table: it gives $(y/\lambda)^k=-\ln(1-\beta)$ $β)$ , which is positive because $0<1-\beta<1$ $1$ .

A Weibull demand density with the area up to about 70.97 shaded, equal to the critical ratio 13/15.

For bananas, $k=2$ and $\lambda=50$ kilograms, and $\beta=1.3/(1.3+0.2)=13/15$ $13/15$ . The density starts at $0$ , peaks before $\lambda$ and is skewed to the right. The shaded area to the left of $y^*\approx70.97$ is $13/15\approx0.867$ $0.867$ : here $-\ln(2/15)\approx2.015$ $2.015$ , its square root is about $1.419$ , and $50\cdot1.419\approx70.97$ $70.97$ .

The scale $\lambda$ has the units of demand and stretches the distribution: doubling $\lambda$ doubles every quantile, including $y^*$ . The shape $k$ changes the form of the curve, but $F(\lambda)=1-\mathrm{e}^{-1}\approx0.632$ for every $k$ , so $\lambda$ is always the demand level that is not exceeded with probability about $0.632$ . As in the continuous newsvendor rule, the CDF is strictly increasing on $[0,\infty)$ , so the order is unique.

A common slip is to write $\ln\beta$ instead of $\ln(1-\beta)$ $β)$ . For the bananas, $50\sqrt{-\ln(13/15)}\approx18.9$ $18.9$ ; this is the level where $F$ equals $2/15$ , not $13/15$ .

Key Results¶

Use the given shape and scale parametrization.
The CDF is zero below zero.
The scale parameter has the same units as demand.

Examples¶

Bananas have $k=2$ $2$ , $\lambda=50$ $50$ , $b=1.3$ $1.3$ , $h=0.2$ $0.2$ , so $\beta=13/15$ $13/15$ .
The optimum is $50\sqrt{-\ln(2/15)}\approx70.97$ kilograms.

Prerequisites¶

Continuous Newsvendor Rule
Logarithms and cumulative distribution functions

Normal Demand Distribution

Sources¶

OR Modelling · Week 6

Symbols: beta, infinity, lambda, square root

Last updated: 27 October 2026 11:52