Non-Parametric Bayesian - Subsystem Tests

This example appears in the Life Data Analysis Reference book.

Example: Non-Parametric Bayesian with Prior Information from Subsystem Tests

Assume a system of interest is composed of three subsystems A, B and C. Prior information from tests of these subsystems is given in the table below. This data can be used to calculate the expected value and variance of the reliability for each subsystem.



E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1} $$

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)} $$

The results of these calculations are given in the table below.

These values can then be used to find the prior system reliability and its variance:


 * $$E\left(R_{0}\right)=0.846831227$$
 * $$\text{Var}\left(R_{0}\right)=0.003546663$$

From the above two values, the parameters of the prior distribution of the system reliability can be calculated by:


 * $$\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{\text{Var}\left(R_{0}\right)}-1\right]=30.12337003$$
 * $$\beta_{0}=\left(1-E \left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var \left(R_{0}\right)}-1\right]=5.448499634$$

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.