Template:Example: Weibull Distribution Example-Demonstrate MTTF: Difference between revisions

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(Created page with '====Weibull Distribution Example - Demonstrate MTTF==== In this example, we will design a test to demonstrate <math>MTTF=75</math> hours, with a 95% confidence. We will once a…')
 
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====Weibull Distribution Example - Demonstrate MTTF====
#REDIRECT [[Reliability_Test_Design]]
In this example, we will design a test to demonstrate  <math>MTTF=75</math>  hours, with a 95% confidence.  We will once again assume a Weibull distribution with a shape parameter  <math>\beta =1.5</math> .  No failures will be allowed on this test, or  <math>f=0</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 =\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 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:lda21.3.gif|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).

Latest revision as of 08:58, 9 August 2012

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