Template:Bayesian test design: Difference between revisions
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which results in a confidence level of <math> CL=1-1-\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.91775 </math> 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. | which results in a confidence level of <math> CL=1-1-\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.91775 </math> 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. | ||
{{btd w info from subsystem tests}} | |||
==== Example – Demonstrate with prior information from subsystem tests ==== | ==== Example – Demonstrate with prior information from subsystem tests ==== | ||
Revision as of 00:34, 5 January 2012
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]\displaystyle{ 1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right) }[/math]
where Beta is the incomplete Beta function. If [math]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \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]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \beta\,\!_{0} }[/math] depending on the type of prior information available.
Template:Btd w info on reliability
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:
Lowest possible reliability: a = 0.8
Most likely reliability: b = 0.85
Highest possible reliability: c = 0.97
This information can be used to approximate the expected value and the variance of the prior system reliability.
- [math]\displaystyle{ E\left(R_{0}\right)=\frac{a+4b+c}{6}=\frac{0.8+4(0.85)+0.97}{6}=0.861667 }[/math]
- [math]\displaystyle{ Var\left(R_{0}\right)=\frac{c-a}{6}=\frac{0.97-0.8}{6}=0.028333 }[/math]
These approximations of the expected value and variance of the prior system reliability can then be used to estimate [math]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \beta\,\!_{0} }[/math].
- [math]\displaystyle{ \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 }[/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]=\left(1-0.861667\right)\left[\frac{0.861667-\left(0.861667\right)^{2}}{0.028333}-1\right]=0.44363 }[/math]
With [math]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \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
CL = 0.8
n = 20
r = 1
the number of successes s is
- [math]\displaystyle{ s = n – r = 19 }[/math]
and the posterior distribution is calculated as
- [math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 }[/math]
- [math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 }[/math]
- [math]\displaystyle{ R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 }[/math]
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
R = 0.9
n = 20
r = 1
the number of successes s is
- [math]\displaystyle{ s = n – r = 19 }[/math]
and the posterior distribution is calculated as
- [math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 }[/math]
- [math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 }[/math]
- [math]\displaystyle{ CL=\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.812164 }[/math]
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
R = 0.9
CL = 0.8
r = 1
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.
The figure above shows that, in this case, n = 28.925085. The posterior distribution can now be calculated as [math]\displaystyle{ s=n-r=27.925085 }[/math] [math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=2.763331+27.925085=30.688416 }[/math] [math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 }[/math]
which results in a confidence level of [math]\displaystyle{ CL=1-1-\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.91775 }[/math] 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.
Template:Btd w info from subsystem tests
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.
| 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 expected value and variance of the subsystems’ reliability.
- [math]\displaystyle{ E\left(R_{i}\right)=\frac{n-r}{n+1} }[/math]
- [math]\displaystyle{ Var\left(R_{i}\right)=\frac{\left(n-r\right)\left(r+1\right)}{\left(n+1\right)^{2}\left(n+2\right)} }[/math]
The results of these calculations are given in the table below.
| Subsystem | Mean | Variance |
|---|---|---|
| A | 0.952380952 | 0.002061 |
| B | 0.935483871 | 0.001886 |
| C | 0.95049505 | 0.000461 |
These values can then be substituted back into equations (2) and (3) to find the prior system reliability.
- [math]\displaystyle{ E\left(R_{0}\right)=\left(0.95238095\right)\left(0.935483871\right)\left(0.95049505\right)=0.846831227 }[/math]
- [math]\displaystyle{ \text{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 }[/math]
Calculating and as before yields
- [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]
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.