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

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



Revision as of 16:32, 13 June 2014

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Test-Find-Test Data Reference Example

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

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

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


IEC 61164 Example 4 Plot.png