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| =&1.7691 | | =&1.7691 |
| \end{align}\,\!</math> | | \end{align}\,\!</math> |
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| *The model parameters are: | | *The model parameters are: |
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Test-Fix-Test Data Reference Example
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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
- 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:
- The instantaneous MTBF and its two-sided 80% confidence bounds are:
