Non-Parametric Bayesian - Subsystem Tests: Difference between revisions
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<noinclude>{{Banner Weibull Examples}} | <noinclude>{{Banner Weibull Examples}} | ||
''This example appears in the [ | ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''. | ||
</noinclude> | </noinclude> | ||
You can use the non-parametric Bayesian method to design a test for a system using information from tests on its subsystems. For example, suppose a system of interest is composed of three subsystems A, B and C -- with prior information from tests of these subsystems given in the table below. | |||
<center> | <center> | ||
{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5" | {| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5" | ||
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::<math> | ::<math> | ||
E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1} | E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1} | ||
</math> | \,\!</math> | ||
:<math> | ::<math> | ||
Var\left(R_{i}\right)=\frac{\left(n_{i}-r_{i}\right)\left(r_{i}+1\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)} | Var\left(R_{i}\right)=\frac{\left(n_{i}-r_{i}\right)\left(r_{i}+1\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)} | ||
</math> | \,\!</math> | ||
The results of these calculations are given in the table below. | The results of these calculations are given in the table below. | ||
| Line 59: | Line 57: | ||
These values can then be used to find the prior system reliability and its variance: | These values can then be used to find the prior system reliability and its variance: | ||
::<math>E\left(R_{0}\right)=0.846831227</math> | ::<math>E\left(R_{0}\right)=0.846831227\,\!</math> | ||
::<math>\text{Var}\left(R_{0}\right)=0.003546663</math> | ::<math>\text{Var}\left(R_{0}\right)=0.003546663\,\!</math> | ||
From the above two values, the parameters of the prior distribution of the system reliability can be calculated by: | From the above two values, the parameters of the prior distribution of the system reliability can be calculated by: | ||
::<math>\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{\text{Var}\left(R_{0}\right)}-1\right]=30.12337003</math> | ::<math>\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{\text{Var}\left(R_{0}\right)}-1\right]=30.12337003\,\!</math> | ||
::<math>\beta_{0}=\left(1-E \left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var \left(R_{0}\right)}-1\right]=5.448499634</math> | ::<math>\beta_{0}=\left(1-E \left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var \left(R_{0}\right)}-1\right]=5.448499634\,\!</math> | ||
With this prior distribution, we now can design a system reliability demonstration test by calculating system reliability ''R'', confidence level ''CL'', number of units ''n'' or number of failures ''r'', as needed. | With this prior distribution, we now can design a system reliability demonstration test by calculating system reliability ''R'', confidence level ''CL'', number of units ''n'' or number of failures ''r'', as needed. | ||
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Using Weibull++, the results are given in the figure below. The result shows that at least 49 test units are needed. | Using Weibull++, the results are given in the figure below. The result shows that at least 49 test units are needed. | ||
[[Image: Bayesian_Test_Subsystem_Piror_Solve_for_n.png|center| | [[Image: Bayesian_Test_Subsystem_Piror_Solve_for_n.png|center|800px]] | ||
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This example appears in the Life data analysis reference.
You can use the non-parametric Bayesian method to design a test for a system using information from tests on its subsystems. For example, suppose a system of interest is composed of three subsystems A, B and C -- with prior information from tests of these subsystems given in the table below.
| Subsystem | Number of Units (n) | Number of Failures (r) |
|---|---|---|
| A | 20 | 0 |
| B | 30 | 1 |
| C | 100 | 4 |
This data can be used to calculate the expected value and variance of the reliability for each subsystem.
- [math]\displaystyle{ E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1} \,\! }[/math]
- [math]\displaystyle{ Var\left(R_{i}\right)=\frac{\left(n_{i}-r_{i}\right)\left(r_{i}+1\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)} \,\! }[/math]
The results of these calculations are given in the table below.
| Subsystem | Mean of Reliability | Variance of Reliability |
|---|---|---|
| A | 0.952380952 | 0.002061 |
| B | 0.935483871 | 0.001886 |
| C | 0.95049505 | 0.000461 |
These values can then be used to find the prior system reliability and its variance:
- [math]\displaystyle{ E\left(R_{0}\right)=0.846831227\,\! }[/math]
- [math]\displaystyle{ \text{Var}\left(R_{0}\right)=0.003546663\,\! }[/math]
From the above two values, the parameters of the prior distribution of the system reliability can be calculated by:
- [math]\displaystyle{ \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{\text{Var}\left(R_{0}\right)}-1\right]=30.12337003\,\! }[/math]
- [math]\displaystyle{ \beta_{0}=\left(1-E \left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var \left(R_{0}\right)}-1\right]=5.448499634\,\! }[/math]
With this prior distribution, we now can design a system reliability demonstration test by calculating system reliability R, confidence level CL, number of units n or number of failures r, as needed.
Solve for Sample Size n
Given the above subsystem test information, in order to demonstrate the system reliability of 0.9 at a confidence level of 0.8, how many samples are needed in the test? Assume the allowed number of failures is 1.
Using Weibull++, the results are given in the figure below. The result shows that at least 49 test units are needed.
