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

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{{Reference Example|Image=RGA_Reference_Examples_Banner.png|Link=RGA_Reference_Examples}}
{{Reference Example|{{Banner RGA Reference_Examples}}|Test-Fix-Test Data}}
 
This example validates the results for test-fix-test data (time terminated) in RGA.  
This example compares the results for test-fix-test data (time terminated).  




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Crow, L.H., ''Confidence Interval Procedures for the Weibull Process with Applications to Reliability Growth'', U.S. Army Material Systems Analysis Activity, 1981.
Crow, L.H., ''Confidence Interval Procedures for the Weibull Process with Applications to Reliability Growth'', U.S. Army Material Systems Analysis Activity, 1981.
For this example, the following will be calculated:
*Parameters of the Crow-AMSAA (NHPP) model
*Demonstrated MTBF (DMTBF)
*80% two-sided confidence bounds on the DMTBF




{{Reference_Example_Heading2}}
{{Reference_Example_Heading2}}
Sample content
 
The following table shows the data.
 
{| {{table|12%}}
|-
| 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'''
|}
 


{{Reference_Example_Heading3}}
{{Reference_Example_Heading3}}
Sample content


{{Reference_Example_Heading4|Software=RGA}}
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)
 
 
{{Reference_Example_Heading4|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>\beta\,\!</math> and <math>\lambda\,\!</math>.
 
::<math>\begin{align}
\hat{\beta }=&\frac{N}{N\ln T^{*}-\underset{i=1}{\overset{N}{\mathop \sum }}\,T_{i}}\\
\\
=&\frac{23}{{23\cdot \ln \left ( 500 \right )}-87.2106}\\
\\
=&0.4127
\end{align}\,\!</math>
 
 
::<math>\begin{align}
\hat{\lambda }=&\frac{N}{T^{*\beta }}\\
\\
=&\frac{23}{500^{0.4127}}\\
\\
=&1.7691
\end{align}\,\!</math>
 
 
*The model parameters are:
 
[[image:CB for Weibull_Results.png|center]]
 
 
*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]]

Latest revision as of 18:26, 28 September 2015

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Test-Fix-Test Data

This example validates the results for test-fix-test data (time terminated) in RGA.


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.

For this example, the following will be calculated:

  • Parameters of the Crow-AMSAA (NHPP) model
  • Demonstrated MTBF (DMTBF)
  • 80% two-sided confidence bounds on the DMTBF


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^{*}-\underset{i=1}{\overset{N}{\mathop \sum }}\,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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