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{{template:LDABOOK|30|Bayesian Nonparametric Test Design}}
#REDIRECT [[Reliability Test Design]]
 
 
The nonparametric analysis performed in the last section made no assumptions about the underlying distribution of the data and was performed with only the immediate data at hand.  However, if prior information regarding system performance is available, it can be incorporated into a Bayesian nonparametric analysis.  This subsection will demonstrate how to incorporate prior information about system reliability and also how to incorporate prior information from subsystem tests.
 
====Determining system performance====
The values of system reliability ''R'', confidence level ''CL'', number of units tested ''n'', and number of failures ''r'' are related by the equation
 
<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)\tag{1}
</math>
 
where Beta is the incomplete Beta function.  If <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> are known, then any quantity of interest can be calculated using the remaining three.  The next two examples demonstrate how to calculate <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> depending on the type of prior information available.
 
==== With prior information on system reliability ====
 
Prior information on system reliability can be exploited to determine <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math>. To do so, first approximate the expected value and variance of prior system reliability <span class="texhtml">''R''<sub>0</sub></span>. This requires knowledge of the lowest possible reliability, the most likely possible reliability, and the highest possible reliability of the system. These quantities will be referred to as ''a'', ''b'' and ''c'', respectively. The expected value of the prior system reliability is approximately given as
 
\begin{equation} E\left(R_{0}\right)=\frac{a+4b+c}{6} \end{equation}
 
and the variance is approximately given by
 
\begin{equation} Var\left(R_{0}\right)=\frac{c-a}{6} \end{equation}
 
These approximate values of the expected value and variance of the prior system reliability can then be used to estimate the values of <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math>, assuming that the prior reliability is a beta-distributed random variable. The values of <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> are calculated as
 
\begin{equation} \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] \end{equation} \begin{equation} \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] \end{equation}
 
With <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> known, equation (1) can now be used to calculate a quantity of interest.
 
==== Example – Demonstrate with prior information on system reliability  ====
 
Suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:
<center>
Lowest possible reliability: ''a'' = 0.8
 
Most likely reliability: ''b'' = 0.85
 
Highest possible reliability: ''c'' = 0.97
</center>
This information can be used to approximate the expected value and the variance of the prior system reliability.
 
\begin{equation} E\left(R_{0}\right)=\frac{a+4b+c}{6}=\frac{0.8+4(0.85)+0.97}{6}=0.861667 \end{equation}
 
\begin{equation} Var\left(R_{0}\right)=\frac{c-a}{6}=\frac{0.97-0.8}{6}=0.028333 \end{equation}
 
<br> These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math>.
 
\begin{equation} \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=0.861667\left[\frac{0.861667-\left(0.861667\right)^{2}}{0.028333}-1\right]=2.763331 \end{equation}
 
\begin{equation} \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]=\left(1-0.861667\right)\left[\frac{0.861667-\left(0.861667\right)^{2}}{0.028333}-1\right]=0.44363 \end{equation}
 
With <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> known, any single value of the 4 quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other 3.
 
System reliability ''R'' can be found if confidence level ''CL'', number of units ''n'', and number of failures ''r'' are known. Given the following data
<center>
''CL'' = 0.8
 
''n'' = 20
 
''r'' = 1
</center>
the number of successes ''s'' is
 
\begin{equation} s = n – r = 19 \end{equation}
 
and the posterior distribution is calculated as
 
\begin{equation} \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 \end{equation}
 
\begin{equation} \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 \end{equation}
 
\begin{equation} R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 \end{equation}
 
The confidence level ''CL'' can be found if system reliability ''R'', number of units ''n'', and number of failures ''r'' are known. Given the following data
<center>
''R'' = 0.9
 
''n'' = 20
 
''r'' = 1
</center>
the number of successes ''s'' is
 
\begin{equation} s = n – r = 19 \end{equation}
 
and the posterior distribution is calculated as
 
\begin{equation} \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 \end{equation}
 
\begin{equation} \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 \end{equation}
 
\begin{equation} CL=\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.812164 \end{equation}
 
The number of units ''n'' can be found if system reliability ''R'', confidence level ''CL'', and number of failures ''r'' are known. Given the following data
<center>
''R'' = 0.9
 
''CL'' = 0.8
 
''r'' = 1
</center>
the '''Number of Units''' utility in the '''Non-Parametric Binomial''' tab of the '''Design a Reliability Demonstration Test''' window can be used to solve for ''n''.
 
[[Image:drt8.jpg|thumb|center|600px| ]]
 
The figure above shows that, in this case, ''n'' = 28.925085. The posterior distribution can now be calculated as \begin{equation} s=n-r=27.925085 \end{equation} \begin{equation} \alpha\,\!=\alpha\,\!_{0}+s=2.763331+27.925085=30.688416 \end{equation} \begin{equation} \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 \end{equation}
 
which results in a confidence level of \begin{equation} CL=1-1-\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.91775 \end{equation} Since the confidence level (0.91775) is greater than that which is required (0.8), we can reduce the number of units ''n'' until the calculated confidence level is close to the required value of 0.8. This results in the number of units ''n'' = 19.31.
 
====With prior information from subsystem tests ====
 
Prior information from subsystem tests can also be used to determine values of alpha and beta. Information from subsystem tests can be used to calculate the expected value and variance of the reliability of individual components, which can then be used to calculate the expected value and variance of the reliability of the entire system. <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> are then calculated as before.
 
\begin{equation}
\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]
\end{equation}
\begin{equation}
\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]
\end{equation}
 
For each subsystem ''i'', calculate the expected value and the variance of the subsystem’s reliability <math>R_{i}</math> as
 
\begin{equation}
E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}
\end{equation}
\begin{equation}
Var\left(R_{i}\right)=\frac{s_{i}\left(n_{i}+1-s_{i}\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)}
\end{equation}
 
The expected value and variance of the entire system’s reliability R can then be calculated as
\begin{equation}
E\left(R_{0}\right)=\prod_{i=1}^{k}E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\times\ldots\times E\left(R_{k}\right)\tag{2}
\end{equation}                                                   
\begin{equation}
Var\left(R_{0}\right)=\prod_{i=1}^{k}\left[E^{2}\left(R_{i}\right)+Var\left(R_{i}\right)\right]-\prod_{i=1}^{k}\left[E^{2}\left(R_{i}\right)\right]\tag{3}
\end{equation}                                                   
 
With the expected value and variance of the system reliability known, and can now be calculated as before.
 
==== Example – Demonstrate with prior information from subsystem tests  ====
 
Given prior information from subsystem tests, the system reliability can be estimated. Assume a system of interest is composed of 3 subsystems A, B and C. Prior information from tests of these subsystems is given in the table below.
<center>
{| cellspacing="1" cellpadding="1" border="1" style="width: 438px; height: 135px;"
|-
! scope="col" | Subsystem
! scope="col" | Number of Units (''n'')
! scope="col" | Number of Failures (''r'')
|-
| A
| 20
| 0
|-
| B
| 30
| 1
|-
| C
| 100
| 4
|}
</center>
This data can be used to calculate expected value and variance of the subsystems’ reliability.
\begin{equation}
E\left(R_{i}\right)=\frac{n-r}{n+1}
\end{equation}
\begin{equation}
Var\left(R_{i}\right)=\frac{\left(n-r\right)\left(r+1\right)}{\left(n+1\right)^{2}\left(n+2\right)}
\end{equation}
 
The results of these calculations are given in the table below.
<center>
{| cellspacing="1" cellpadding="1" border="1" style="width: 438px; height: 135px;"
|-
! scope="col" | Subsystem
! scope="col" | Mean
! scope="col" | Variance
|-
| A
| 0.952380952
| 0.002061
|-
| B
| 0.935483871
| 0.001886
|-
| C
| 0.95049505
| 0.000461
|}
</center>
These values can then be substituted back into equations (2) and (3) to find the prior system reliability.
\begin{equation}
E\left(R_{0}\right)=\left(0.95238095\right)\left(0.935483871\right)\left(0.95049505\right)=0.846831227
\end{equation}
\begin{align}
Var\left(R_{0}\right)=&\left(0.952380952^{2}+0.002061\right)\left(0.935483871^{2}+0.001886\right)\left(0.95049505^{2}+0.000461\right)\\
&-\left(0.952380952^{2}\right)\left(0.935483871^{2}\right)\left(0.95049505^{2}\right)=0.003546663
\end{align}
Calculating and as before yields
 
\begin{equation}
\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=30.12337003
\end{equation}
\begin{equation}
\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
\end{equation}
 
which can then be used to calculate system reliability ''R'', confidence level ''CL'', number of units ''n'', and number of failures ''r'' as discussed in the previous subsection, '''Prior Information on System Reliability'''.

Latest revision as of 08:04, 29 June 2012

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