Using this generator get a weibull variate for α=8, β=0.75, v=0, and R = 0.612.

Weibull Distribution

$f(x)=\frac{\beta}{\alpha}\left(\frac{x-v}{\alpha}\right)^{\beta-1} e^{-\left(\frac{x-v}{\alpha}\right)^{\beta}}; x \ge v$

Algorithm:

Step 1: Compute c.d.f. for given random variable X for uniform or Rectangular Distribution it is given by

$f(x)=1-e^{-\left(\frac{x-1}{a}\right)^{\beta}}$

Step 2: Set $f(x) = R $ where R represents a random number uniformly Distributed over [0,1]

$\therefore R=1-e^{-\left(\frac{x-V}{\alpha}\right)^{\beta}}$

*Step 3: * Solve the equation in step 2 to find x

$\begin{aligned} e^{-\left(\frac{x-v}{\alpha}\right)^{\beta}} &=1-R \\ \therefore e^{\left(\frac{x-v}{\alpha}\right)^{\beta}} &=\frac{1}{1-R} \\ \left(\frac{x-v}{\alpha}\right)^{\beta} &=\text{ln}\left(\frac{1}{1-R}\right) \\ \frac{x-v}{\alpha} &=\left[\ln \left(\frac{1}{1-R}\right)\right]^{\frac{1}{\beta}} \\ \therefore x-v &=\alpha\left[\ln \left(\frac{1}{1-R}\right)\right]^{\frac{1}{\beta}} \\ x &=v+\alpha\left[\ln \left(\frac{1}{1-R}\right)\right]^{\frac{1}{\beta}} \\ \end{aligned}$

Given:

$\alpha=8$

$B=0 \cdot 75$

$y=0$ and $R=0.612$

$\begin{aligned} x&=v+\alpha\left[\ln \left(\frac{1}{1-R}\right)\right]^{\beta} \\ &=0+8\left[\ln \left(\frac{1}{1-0 \cdot 612}\right)\right]^{\frac{1}{0.75}} \\ &= 0.929 \times 8 \\ &= 7.437 \end{aligned}$