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| '''Weibull Distribution Example - Demonstrate Reliability'''
| | #REDIRECT [[Reliability_Test_Design#Example_1:_Demonstrate_Reliability]] |
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| In this example, we will design a test to demonstrate a reliability of 90% at <math>t=100</math> hours, with a 95% confidence. We will 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> .
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| '''Determining Units for Available Time'''
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| In the above scenario, we know that we have the testing facilities available for <math>t=48</math> hours. We must now determine the number of units to test for this amount of time with no failures in order to have demonstrated our reliability goal.
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| The first step is to determine the Weibull scale parameter, <math>\eta .</math> The Weibull reliability equation is:
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| ::<math>R={{e}^{-{{(t/\eta )}^{\beta }}}}</math>
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| This can be rewritten as:
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| ::<math>\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}</math>
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| Since we know the values of <math>{{t}_{DEMO}}</math> , <math>{{R}_{DEMO}}</math> and <math>\beta </math> , we can substitute these in the equation and solve for <math>\eta </math> :
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| ::<math>\eta =\frac{100}{{{(-\text{ln}(0.9))}^{\tfrac{1}{1.5}}}}=448.3</math>
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| Next, the value of <math>{{R}_{TEST}}</math> is calculated by:
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| ::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%.</math>
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| The last step is to substitute the appropriate values into the cumulative binomial equation, which for the Weibull distribution appears as:
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| ::<math>1-C.L.=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n!}{i!\cdot (n-i)!}\cdot {{(1-{{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{i}}\cdot {{({{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{(n-i)}}</math>
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| 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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.
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| [[Image:RDT Weibull Demonstrate Reliability Samples.png|thumb|center|400px ]]
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| '''Determining Time for Available Units'''
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| In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.
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| [[Image:RDT Weibull Demonstrate Reliability Test Time.png|thumb|center|400px ]] | |