Parametric Binomial Example - Demonstrate MTTF: Difference between revisions
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<noinclude>{{Banner Weibull Examples}} | <noinclude>{{Banner Weibull Examples}} | ||
''This example appears in the [ | ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''. | ||
</noinclude> | </noinclude> | ||
In this example, we will use the parametric binomial method to design a test that will demonstrate | 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 | 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> | ::<math>MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })\,\!</math> | ||
where | 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> | ::<math>\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}\,\!</math> | ||
Since | 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> | ::<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 | 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. | ||
[[Image:RDT Weibull Demonstrate MTTF.png|center| | [[Image:RDT Weibull Demonstrate MTTF.png|center|800px| ]] | ||
The procedure for determining the required test time proceeds in the same manner, determining | 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. | ||
Latest revision as of 18:55, 18 September 2023
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This example appears in the Life data analysis reference.
In this example, we will use the parametric binomial method to design a test that will demonstrate [math]\displaystyle{ MTTF=75\,\! }[/math] hours with a 95% confidence if no failure occur during the test [math]\displaystyle{ f=0\,\! }[/math]. We will assume a Weibull distribution with a shape parameter [math]\displaystyle{ \beta =1.5\,\! }[/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\,\! }[/math]:
- [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 the binomial equation 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 the binomial equation with Weibull distribution.