Template:Example: Weibull Distribution Example-Demonstrate MTTF

Weibull Distribution Example - Demonstrate MTTF

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

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

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

where [math]\displaystyle{ \Gamma (x) }[/math] is the gamma function of [math]\displaystyle{ x }[/math] . This can be rearranged in terms of .. :

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

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

[math]\displaystyle{ {{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]\displaystyle{ CL }[/math] , [math]\displaystyle{ {{t}_{TEST}} }[/math] , [math]\displaystyle{ \beta }[/math] , [math]\displaystyle{ f }[/math] and [math]\displaystyle{ \eta }[/math] have already been calculated or specified, so it merely remains to solve Eqn. (relcum) for [math]\displaystyle{ n. }[/math] The value is calculated as [math]\displaystyle{ n=4.8811, }[/math] or [math]\displaystyle{ 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]\displaystyle{ \eta }[/math] from the [math]\displaystyle{ MTTF }[/math] equation, and following the previously described methodology to determine [math]\displaystyle{ {{t}_{TEST}} }[/math] from Eqn. (weibcum).