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

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{{InProgress}}{{Banner RGA Reference Examples}}__NOTOC__
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This example compares the results for test-find-test data.  
This example compares the results for test-find-test data.  


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{{Reference_Example_Heading4|Software=RGA}}
{{Reference_Example_Heading4|RGA}}


Since test-find-test, the assumption is <math>\,\!\beta =1</math>. Therefore:  
Since test-find-test, the assumption is <math>\,\!\beta =1</math>. Therefore:  

Revision as of 20:18, 9 June 2014

RGA_Reference_Examples_Banner.png

RGA_Reference_Examples

This example compares the results for test-find-test data.


Reference Case

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


Data

Termination Time = 4000 hours
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


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

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

Since 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 \end{align}\,\! }[/math]


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 hours \end{align}\,\! }[/math]


IEC 61164 Example 4 Plot.png