Test-Find-Test Data Reference Example
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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 |
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]
- [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]