Parametric Binomial Example - Demonstrate MTTF

This example appears in the Life Data Analysis Reference book.

In this example, we will use the parametric binomial method to design a test that will demonstrate $$MTTF=75\,\!$$  hours with a 95% confidence if no failure occur during the test $$f=0\,\!$$. We will assume a Weibull distribution with a shape parameter $$\beta =1.5\,\!$$. 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 $$\eta\,\!$$:


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

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 the binomial equation 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 the binomial equation with Weibull distribution.