Non-Parametric Bayesian - Subsystem Tests
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This example appears in the Life Data Analysis Reference book.
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] E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1} \,\![/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)} \,\![/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]E\left(R_{0}\right)=0.846831227\,\![/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:
- [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]
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.
