Template:Example: Weibull Distribution Example-Demonstrate MTTF: Difference between revisions
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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 Eqn. (relcum) 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 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 Eqn. (relcum) 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. | ||
[[Image: | [[Image:RDT Weibull Demonstrate MTTF.png|thumb|center|300px| ]] | ||
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 Eqn. (weibcum). | 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 Eqn. (weibcum). | ||
Revision as of 22:11, 2 March 2012
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).