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

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{{Reference Example|RGA_Reference_Examples_Banner.png|RGA_Reference_Examples}}
{{Reference Example|{{Banner RGA Reference_Examples}}|Test-Find-Test Data}}
This example compares the results for test-find-test data.  
This example validates the results for test-find-test data in RGA.  




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International Standard IEC 61164 (Reliability Growth in Product Design and Test – Statistical Test and Estimation Methods), Example 4, pg. 32.
International Standard IEC 61164 (Reliability Growth in Product Design and Test – Statistical Test and Estimation Methods), Example 4, pg. 32.
For this example, the following will be calculated:
*Parameters of the Crow Extended  model
*Unseen BD mode failure intensity
*Cramér-von Mises (CVM) goodness of fit for BD modes
*Demonstrated MTBF (DMTBF)
*Projected PMTBF (PMTBF)




{{Reference_Example_Heading2}}
{{Reference_Example_Heading2}}
The following table shows the data.


{| {{table|15%}}
{| {{table|15%}}
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{{Reference_Example_Heading3}}
{{Reference_Example_Heading3}}


BetaBD (UnB) = 0.7472, LambdaBD = 0.0326
The book has the following results:


Unseen BD Mode Failure Intensity = 0.0030/hr
*BetaBD (UnB) = 0.7472, LambdaBD = 0.0326


Goodness of fit for BD modes: CVM = 0.085, critical value = 0.171 with significance level = 0.1. Since CVM < critical value can fail to reject hypothesis that the model fits the data.
*Unseen BD Mode Failure Intensity = 0.0030/hr


DMTBF = 88.9 hours
*Goodness of fit for BD modes: CVM = 0.085, critical value = 0.171 with significance level = 0.1. Since CVM < critical value can fail to reject hypothesis that the model fits the data.


PMTBF = 135.1 hours
*DMTBF = 88.9 hours
 
*PMTBF = 135.1 hours




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


Since test-find-test, the assumption is <math>\,\!\beta =1</math>. Therefore:  
In RGA, the Crow Extended model with the maximum likelihood estimation (MLE) method was used to calculate the results.
 
For a test-find-test, the assumption is <math>\,\!\beta =1</math>. Therefore:  


::<math>\begin{align}
::<math>\begin{align}
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=&\frac{4000}{45}\\
=&\frac{4000}{45}\\
\\
\\
=&88.8889
=&88.8889\; \; \mathrm{hours}
\end{align}\,\!</math>
\end{align}\,\!</math>


*The model parameters are:


[[image:IEC 61164 Example 4_Results.png|center]]
[[image:IEC 61164 Example 4_Results.png|center]]
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PMTBF=&\frac{1}{\lambda _{p}}\\
PMTBF=&\frac{1}{\lambda _{p}}\\
\\
\\
=&135.17 hours
=&135.17 \; \; \mathrm{hours}
\end{align}\,\!</math>
\end{align}\,\!</math>




*The growth potential MTBF plot:


[[image:IEC 61164 Example 4_Plot.png|center|550px]]
[[image:IEC 61164 Example 4_Plot.png|center|550px]]

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Test-Find-Test Data

This example validates the results for test-find-test data in RGA.


Reference Case

International Standard IEC 61164 (Reliability Growth in Product Design and Test – Statistical Test and Estimation Methods), Example 4, pg. 32.

For this example, the following will be calculated:

  • Parameters of the Crow Extended model
  • Unseen BD mode failure intensity
  • Cramér-von Mises (CVM) goodness of fit for BD modes
  • Demonstrated MTBF (DMTBF)
  • Projected PMTBF (PMTBF)


Data

The following table shows the data.

Failure Time Classification Mode
150 BD1
253 BD2
475 BD3
540 BD4
564 BD5
636 A
722 BD5
871 A
996 BD6
1003 BD7
1025 A
1120 BD8
1209 BD2
1255 BD9
1334 BD10
1647 BD9
1774 BD10
1927 BD11
2130 A
2214 A
2293 A
2448 A
2490 BD12
2508 A
2601 BD1
2635 BD8
2731 A
2747 BD6
2850 BD13
3040 BD9
3154 BD4
3171 A
3206 A
3245 DB12
3249 BD10
3420 BD5
3502 BD3
3646 BD10
3649 A
3663 BD2
3730 BD8
3794 BD14
3890 BD15
3949 A
3952 BD16
Termination Time = 4000 hours


BD Mode EF
1 0.7
2 0.7
3 0.8
4 0.8
5 0.9
6 0.9
7 0.5
8 0.8
9 0.9
10 0.7
11 0.7
12 0.6
13 0.6
14 0.7
15 0.7
16 0.5


Result

The book has the following results:

  • BetaBD (UnB) = 0.7472, LambdaBD = 0.0326
  • Unseen BD Mode Failure Intensity = 0.0030/hr
  • Goodness of fit for BD modes: CVM = 0.085, critical value = 0.171 with significance level = 0.1. Since CVM < critical value can fail to reject hypothesis that the model fits the data.
  • DMTBF = 88.9 hours
  • PMTBF = 135.1 hours


Results in RGA

In RGA, the Crow Extended model with the maximum likelihood estimation (MLE) method was used to calculate the results.

For a test-find-test, the assumption is [math]\displaystyle{ \,\!\beta =1 }[/math]. Therefore:

[math]\displaystyle{ \begin{align} DMTBF=&\frac{T}{N}\\ \\ =&\frac{4000}{45}\\ \\ =&88.8889\; \; \mathrm{hours} \end{align}\,\! }[/math]


  • The model parameters are:
IEC 61164 Example 4 Results.png


IEC 61164 Example 4 Stat Tests.png


[math]\displaystyle{ \begin{align} \lambda _{p}=&\lambda _{A}+\sum_{i=1}^{N_{BD}}\left ( 1-d_{i} \right )\lambda _{i}+\bar{d}h\left ( T \right )\\ \\ =&\frac{N_{A}}{T}+\sum_{i=1}^{N_{BD}}\left ( 1-d_{i} \right )\frac{N_{i}}{T}+\bar{d}\lambda _{BD}\beta _{BD}T^{\beta -1}\\ \\ =&\frac{13}{4000}+0.002+\left ( 0.7188 \right )\left ( 0.032572 \right )\left ( 0.74715 \right )\left ( 4000 \right )^{0.7472-1}\\ \\ =&0.007398\\ \\ PMTBF=&\frac{1}{\lambda _{p}}\\ \\ =&135.17 \; \; \mathrm{hours} \end{align}\,\! }[/math]


  • The growth potential MTBF plot:
IEC 61164 Example 4 Plot.png
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