|
|
Line 190: |
Line 190: |
| PMTBF=&\frac{1}{\lambda _{p}}\\ | | PMTBF=&\frac{1}{\lambda _{p}}\\ |
| \\ | | \\ |
| =&135.17 hours | | =&135.17 |
| \end{align}\,\!</math> | | \end{align}\,\!</math> |
|
| |
|
Revision as of 22:31, 11 June 2014
RGA_Reference_Examples_Banner.png
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
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
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]