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| ::<math>\begin{align} | | ::<math>\begin{align} |
| \lambda _{p}=\lambda _{A}+\sum_{i=1}^{N_{BD}}\left ( 1-d_{i} \right )\lambda _{i}+\bar{d}h\left ( T \right )\\ | | \lambda _{p}=&\lambda _{A}+\sum_{i=1}^{N_{BD}}\left ( 1-d_{i} \right )\lambda _{i}+\bar{d}h\left ( T \right )\\ |
| =&\frac{4000}{45}\\
| |
| \\ | | \\ |
| =&88.8889 | | =&\\ |
| | \\ |
| | =&=\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 |
| \end{align}\,\!</math> | | \end{align}\,\!</math> |
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| |
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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
| 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 Weibull++
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{13}{4000}+0.002+\left ( 0.7188 \right )\left ( 0.032572 \right )\left ( 0.74715 \right )\left ( 4000 \right )^{0.7472-1}\\
\\
=&0.007398
\end{align}\,\! }[/math]
