Non-Parametric Bayesian - Subsystem Tests - Revision history

Lisa Hacker at 18:55, 18 September 2023

2023-09-18T18:55:48Z

← Older revision		Revision as of 18:55, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the [~~[Reliability_Test_Design\|~~Life ~~Data Analysis Reference book]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''.

	</noinclude>		</noinclude>

Kate Racaza at 20:25, 9 December 2015

2015-12-09T20:25:36Z

← Older revision		Revision as of 20:25, 9 December 2015
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	Using Weibull++, the results are given in the figure below. The result shows that at least 49 test units are needed.		Using Weibull++, the results are given in the figure below. The result shows that at least 49 test units are needed.

	[[Image: Bayesian_Test_Subsystem_Piror_Solve_for_n.png\|center\|~~650px~~]]		[[Image: Bayesian_Test_Subsystem_Piror_Solve_for_n.png\|center\|800px]]

Richard House at 19:48, 26 September 2012

2012-09-26T19:48:34Z

← Older revision		Revision as of 19:48, 26 September 2012
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	::<math>		::<math>
	E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1}		E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1}
	</math>		\,\!</math>
	::<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)}		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>		\,\!</math>

	The results of these calculations are given in the table below.		The results of these calculations are given in the table below.
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	These values can then be used to find the prior system reliability and its variance:		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>E\left(R_{0}\right)=0.846831227\,\!</math>
	::<math>\text{Var}\left(R_{0}\right)=0.003546663</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:		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>\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>

	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.		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.

Chris Kahn at 05:39, 22 August 2012

2012-08-22T05:39:07Z

← Older revision		Revision as of 05:39, 22 August 2012
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	</noinclude>		</noinclude>
	You can use the non-parametric Bayesian method to design a test for ~~this~~ 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.		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.
	<center>		<center>
	{\| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"		{\| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"

Chris Kahn at 05:38, 22 August 2012

2012-08-22T05:38:51Z

← Older revision		Revision as of 05:38, 22 August 2012
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	</noinclude>		</noinclude>
	~~'''Example: Non~~-~~Parametric~~ Bayesian ~~with Prior Information~~ from ~~Subsystem Tests'''~~		You can use the non-parametric Bayesian method to design a test for this 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.

	~~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.
	<center>		<center>
	{\| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"		{\| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"

Chris Kahn at 06:20, 17 August 2012

2012-08-17T06:20:54Z

← Older revision		Revision as of 06:20, 17 August 2012
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	E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1}		E\left(R_{i}\right)=\frac{n_{i}-r_{i}}{n_{i}+1}
	</math>		</math>
	:<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)}		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>		</math>

Lisa Hacker: Created page with '{{Banner Weibull Examples}} ''This example appears in the Life Data Analysis Reference book''. '''Example: Non-Parametric Bay…'

2012-08-15T01:31:16Z

Created page with '<noinclude>{{Banner Weibull Examples}} ''This example appears in the Life Data Analysis Reference book''. </noinclude> '''Example: Non-Parametric Bay…'

New page

<noinclude>{{Banner Weibull Examples}}
''This example appears in the [[Reliability_Test_Design|Life Data Analysis Reference book]]''.

</noinclude>
'''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.
<center>
{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
! 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 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.
<center>
{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
! scope="col" | Subsystem
! scope="col" | Mean of Reliability
! scope="col" | Variance of Reliability
|-
| A
| 0.952380952
| 0.002061
|-
| B
| 0.935483871
| 0.001886
|-
| C
| 0.95049505
| 0.000461
|}
</center>

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.

[[Image: Bayesian_Test_Subsystem_Piror_Solve_for_n.png|center|650px]]