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 [math]MTTF=75\,\![/math] hours with a 95% confidence if no failure occur during the test [math]f=0\,\![/math]. We will assume a Weibull distribution with a shape parameter [math]\beta =1.5\,\![/math]. We want to determine the number of units to test for [math]{{t}_{TEST}}=60\,\![/math] hours to demonstrate this goal.

The first step in this case involves determining the value of the scale parameter [math]\eta \,\![/math] from the [math]MTTF\,\![/math] equation. The equation for the [math]MTTF\,\![/math] for the Weibull distribution is:

[math]MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })\,\![/math]

where [math]\Gamma (x)\,\![/math] is the gamma function of [math]x\,\![/math]. This can be rearranged in terms of [math]\eta\,\![/math]:

[math]\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}\,\![/math]

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

[math]{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%\,\![/math]

The last step is to substitute the appropriate values into the cumulative binomial equation. The values of [math]CL\,\![/math], [math]{{t}_{TEST}}\,\![/math], [math]\beta \,\![/math], [math]f\,\![/math] and [math]\eta \,\![/math] have already been calculated or specified, so it merely remains to solve the binomial equation for [math]n\,\![/math]. The value is calculated as [math]n=4.8811,\,\![/math] or [math]n=5\,\![/math] 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 [math]\eta \,\![/math] from the [math]MTTF\,\![/math] equation, and following the previously described methodology to determine [math]{{t}_{TEST}}\,\![/math] from the binomial equation with Weibull distribution.