Template:Example: Weibull Distribution Example-Demonstrate MTTF

Weibull Distribution Example - Demonstrate MTTF

In this example, we will design a test to demonstrate $$MTTF=75$$  hours, with a 95% confidence. We will once again assume a Weibull distribution with a shape parameter $$\beta =1.5$$. No failures will be allowed on this test, or $$f=0$$. We want to determine the number of units to test for $${{t}_{TEST}}=60$$  hours to demonstrate this goal.

The first step in this case involves determining the value of the scale parameter $$\eta $$  from the  $$MTTF$$  equation. The equation for the $$MTTF$$  for the Weibull distribution is:


 * $$MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })$$

where $$\Gamma (x)$$  is the gamma function of  $$x$$. This can be rearranged in terms of .. :


 * $$\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}$$

Since $$MTTF$$  and  $$\beta $$  have been specified, it is a relatively simple matter to calculate  $$\eta =83.1$$. From this point on, the procedure is the same as the reliability demonstration example. Next, the value of $${{R}_{TEST}}$$  is calculated as:


 * $${{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%$$

The last step is to substitute the appropriate values into the cumulative binomial equation. The values of $$CL$$,  $${{t}_{TEST}}$$ ,  $$\beta $$ ,  $$f$$  and  $$\eta $$  have already been calculated or specified, so it merely remains to solve Eqn. (relcum) for $$n.$$  The value is calculated as  $$n=4.8811,$$  or  $$n=5$$  units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.



The procedure for determining the required test time proceeds in the same manner, determining $$\eta $$  from the  $$MTTF$$  equation, and following the previously described methodology to determine  $${{t}_{TEST}}$$  from Eqn. (weibcum).