Template:Bayesian Nonparametric Test Design - Revision history

Kate Racaza: Redirected page to Reliability Test Design

2012-06-29T08:04:22Z

Redirected page to Reliability Test Design

← Older revision		Revision as of 08:04, 29 June 2012
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	~~{{template:LDABOOK\|30\|Bayesian Nonparametric~~ Test Design}}		#REDIRECT [[Reliability Test Design]]

	~~{{bayesian test design}}~~

Harry Guo: moved Bayesian Nonparametric Test Design to Template:Bayesian Nonparametric Test Design: To include it in the reliability test design chapter

2012-02-08T16:13:51Z

moved Bayesian Nonparametric Test Design to Template:Bayesian Nonparametric Test Design: To include it in the reliability test design chapter

← Older revision	Revision as of 16:13, 8 February 2012
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Nicolette Young: Replaced content with '{{template:LDABOOK|30|Bayesian Nonparametric Test Design}} {{bayesian test design}}'

2012-01-05T00:31:53Z

Replaced content with '{{template:LDABOOK|30|Bayesian Nonparametric Test Design}} {{bayesian test design}}'

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Nicolette Young: /* Example – Demonstrate with prior information on system reliability */

2011-08-16T18:49:39Z

Example – Demonstrate with prior information on system reliability

← Older revision		Revision as of 18:49, 16 August 2011
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	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 '''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\|~~400px~~\| ]]		[[Image:drt8.jpg\|thumb\|center\|500px\| ]]

	The figure above shows that, in this case, ''n'' = 28.925085. The posterior distribution can now be calculated as <math> s=n-r=27.925085 </math> <math> \alpha\,\!=\alpha\,\!_{0}+s=2.763331+27.925085=30.688416 </math> <math> \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 </math>		The figure above shows that, in this case, ''n'' = 28.925085. The posterior distribution can now be calculated as <math> s=n-r=27.925085 </math> <math> \alpha\,\!=\alpha\,\!_{0}+s=2.763331+27.925085=30.688416 </math> <math> \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 </math>

Nicolette Young: /* Example – Demonstrate with prior information from subsystem tests */

2011-08-16T18:49:06Z

Example – Demonstrate with prior information from subsystem tests

← Older revision		Revision as of 18:49, 16 August 2011
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	Calculating and as before yields		Calculating and as before yields

	<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>

	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'''.		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'''.

Nicolette Young: /* With prior information from subsystem tests */

2011-08-16T18:48:29Z

With prior information from subsystem tests

← Older revision		Revision as of 18:48, 16 August 2011
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	The expected value and variance of the entire system’s reliability R can then be calculated as		The expected value and variance of the entire system’s reliability R can then be calculated as

	<math>		::<math>
	E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)		E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)
	</math>		</math>
	<math>		::<math>
	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]		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]
	</math>		</math>

Nicolette Young: /* With prior information from subsystem tests */

2011-08-16T18:47:14Z

With prior information from subsystem tests

← Older revision		Revision as of 18:47, 16 August 2011
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	</math>		</math>
	The expected value and variance of the entire system’s reliability R can then be calculated as		The expected value and variance of the entire system’s reliability R can then be calculated as

	<math>		<math>
	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}~~		E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)
	</math>		</math>
	<math>		<math>
	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}~~		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]
	</math>		</math>

Nicolette Young: /* Example – Demonstrate with prior information from subsystem tests */

2011-08-16T18:44:54Z

Example – Demonstrate with prior information from subsystem tests

← Older revision		Revision as of 18:44, 16 August 2011
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	</center>		</center>
	This data can be used to calculate expected value and variance of the subsystems’ reliability.		This data can be used to calculate expected value and variance of the subsystems’ reliability.
	<math>		::<math>
	E\left(R_{i}\right)=\frac{n-r}{n+1}		E\left(R_{i}\right)=\frac{n-r}{n+1}
	</math>		</math>
	<math>		::<math>
	Var\left(R_{i}\right)=\frac{\left(n-r\right)\left(r+1\right)}{\left(n+1\right)^{2}\left(n+2\right)}		Var\left(R_{i}\right)=\frac{\left(n-r\right)\left(r+1\right)}{\left(n+1\right)^{2}\left(n+2\right)}
	</math>		</math>

Nicolette Young: /* Example – Demonstrate with prior information from subsystem tests */

2011-08-16T18:44:24Z

Example – Demonstrate with prior information from subsystem tests

← Older revision		Revision as of 18:44, 16 August 2011
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	\|}		\|}
	</center>		</center>

	These values can then be substituted back into equations (2) and (3) to find the prior system reliability.		These values can then be substituted back into equations (2) and (3) to find the prior system reliability.
	<math>		::<math>E\left(R_{0}\right)=\left(0.95238095\right)\left(0.935483871\right)\left(0.95049505\right)=0.846831227</math>
	E\left(R_{0}\right)=\left(0.95238095\right)\left(0.935483871\right)\left(0.95049505\right)=0.846831227		::<math>\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>
	</math>
	<math>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		Calculating and as before yields

	<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>
	\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		<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>
	<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>

	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'''.		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'''.

Nicolette Young: /* With prior information from subsystem tests */

2011-08-16T18:38:52Z

With prior information from subsystem tests

← Older revision		Revision as of 18:38, 16 August 2011
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	\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]		\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]
	</math>		</math>
	<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]		\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]
	</math>		</math>
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	E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}		E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}
	</math>		</math>
	<math>		::<math>
	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)}		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)}
	</math>		</math>