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| {{Reference_Example_Heading4|RGA}} | | {{Reference_Example_Heading4|RGA}} |
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| | In RGA, the following equations are used to calculate <math>\beta\,\!</math> and <math>\lambda\,\!</math>. |
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| ::<math>\begin{align} | | ::<math>\begin{align} |
| \widehat{\beta }=&\frac{N}{{N\ln T^{*}}-\sum_{i=1}^{N}T_{i}}\\ | | \hat{\beta }=&\frac{N}{{N\ln T^{*}}-\sum_{i=1}^{N}T_{i}}\\ |
| \\ | | \\ |
| =&\frac{23}{{23\cdot \ln \left ( 500 \right )}-87.2106}\\ | | =&\frac{23}{{23\cdot \ln \left ( 500 \right )}-87.2106}\\ |
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| ::<math>\begin{align} | | ::<math>\begin{align} |
| \widehat{\lambda }=&\frac{N}{T^{*\beta }}\\ | | \hat{\lambda }=&\frac{N}{T^{*\beta }}\\ |
| \\ | | \\ |
| =&\frac{23}{500^{0.4127}}\\ | | =&\frac{23}{500^{0.4127}}\\ |
Revision as of 16:17, 13 June 2014
RGA_Reference_Examples_Banner.png
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 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]
