Test-Fix-Test Data Reference Example: Difference between revisions

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{{Reference_Example_Heading4|RGA}}
{{Reference_Example_Heading4|RGA}}


In RGA, the following equations are used to calculate <math>\beta\,\!</math> and <math>\lambda\,\!</math>.
In RGA, the Crow-AMSAA (NHPP) model with the maximum likelihood estimation (MLE) method was used to calculate the results. The following equations are used to calculate <math>\beta\,\!</math> and <math>\lambda\,\!</math>.


::<math>\begin{align}
::<math>\begin{align}
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\end{align}\,\!</math>
\end{align}\,\!</math>


*The model parameters are:


[[image:CB for Weibull_Results.png|center]]
[[image:CB for Weibull_Results.png|center]]




The instantaneous MTBF and its two-sided 80% confidence bounds are:
*The instantaneous MTBF and its two-sided 80% confidence bounds are:
 
[[image:CB for Weibull_QPC.png|center]]
[[image:CB for Weibull_QPC.png|center]]

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Test-Fix-Test Data Reference Example

This example compares the results for test-fix-test data (time terminated).


Reference Case

Crow, L.H., Confidence Interval Procedures for the Weibull Process with Applications to Reliability Growth, U.S. Army Material Systems Analysis Activity, 1981.


Data

The following table shows the data.

0.2
4.2
4.5
5
5.4
6.1
7.9
14.8
19.2
48.6
85.8
108.9
127.2
129.8
150.1
159.7
227.4
244.7
262.7
315.3
329.6
404.3
486.2
Termination Time = 500 hours


Result

The book has the following results:

  • Beta = 0.413, Lambda = 1.769
  • DMTBF = 52.7
  • Confidence Bounds on DMTBF (CL = 80%) = (35.6, 82.9)


Results in RGA

In RGA, the Crow-AMSAA (NHPP) model with the maximum likelihood estimation (MLE) method was used to calculate the results. The following equations are used to calculate [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \lambda\,\! }[/math].

[math]\displaystyle{ \begin{align} \hat{\beta }=&\frac{N}{{N\ln T^{*}}-\sum_{i=1}^{N}T_{i}}\\ \\ =&\frac{23}{{23\cdot \ln \left ( 500 \right )}-87.2106}\\ \\ =&0.4127 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} \hat{\lambda }=&\frac{N}{T^{*\beta }}\\ \\ =&\frac{23}{500^{0.4127}}\\ \\ =&1.7691 \end{align}\,\! }[/math]
  • The model parameters are:
CB for Weibull Results.png


  • The instantaneous MTBF and its two-sided 80% confidence bounds are:
CB for Weibull QPC.png
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