Reliability Demonstration Test Design for Repairable Systems - Revision history

Richard House at 15:09, 14 February 2019

2019-02-14T15:09:50Z

← Older revision		Revision as of 15:09, 14 February 2019
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	:<math>{{\lambda }_{c}}\left( t \right)=\lambda {{t}^{\beta -1}}=\frac{{{\lambda }_{i}}\left( t \right)}{\beta }\,\!</math>		:<math>{{\lambda }_{c}}\left( t \right)=\lambda {{t}^{\beta -1}}=\frac{{{\lambda }_{i}}\left( t \right)}{\beta }\,\!</math>

	The instantaneous ~~MTTF~~ is:		The instantaneous MTBF is:

	:<math>\begin{align}		:<math>\begin{align}
	~~MTT~~{{F}_{i}}\left( t \right)= & \frac{1}{{{\lambda }_{i\left( t \right)}}} \\		MTB{{F}_{i}}\left( t \right)= & \frac{1}{{{\lambda }_{i\left( t \right)}}} \\
	= & \frac{1}{\lambda \beta }{{t}^{1-\beta }} \\		= & \frac{1}{\lambda \beta }{{t}^{1-\beta }} \\
	= & \frac{~~MTT~~{{F}_{c}}\left( t \right)}{\beta }		= & \frac{MTB{{F}_{c}}\left( t \right)}{\beta }
	\end{align}\,\!</math>		\end{align}\,\!</math>

	The cumulative MTBF at time <math>t\,\!</math> is:		The cumulative MTBF at time <math>t\,\!</math> is:

	:<math>MTB{{F}_{c}}\left( ~~\beta~~ \right)=\frac{1}{{{\lambda }_{c}}\left( t \right)}=\frac{1}{\lambda }{{t}^{1-\beta }}=~~MTT~~{{F}_{i}}\left( t \right)\beta \,\!</math>		:<math>MTB{{F}_{c}}\left( t \right)=\frac{1}{{{\lambda }_{c}}\left( t \right)}=\frac{1}{\lambda }{{t}^{1-\beta }}=MTB{{F}_{i}}\left( t \right)\beta \,\!</math>

	The relation between the confidence level, required test time, number of systems under test and allowed total number of failures in the test is:		The relation between the confidence level, required test time, number of systems under test and allowed total number of failures in the test is:

David J. Groebel at 17:15, 20 June 2014

2014-06-20T17:15:40Z

Kate Racaza: /* Example: Solve for Number of Samples */

2014-01-29T19:28:41Z

Example: Solve for Number of Samples

← Older revision		Revision as of 19:28, 29 January 2014
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	==Example: Solve for Number of Samples==		==Example: Solve for Number of Samples==
	At the end of a reliability growth testing program, a manufacturer wants to demonstrate that a new product has achieved an MTBF of 10,000 hours with 80% confidence. The available time for the demonstration test is 4,000 hours for each test unit. Assuming zero failures are allowed, what is the required number of units to be tested in order to demonstrate the desired MTBF?		{{:Repairable_System_Test_Design_Example_-_Solve_for_Sample_Size}}

	~~'''Solution'''~~

	We can obtain the required number of units by using the Design of Reliability Tests (DRT) tool in the RGA software. Since this is a demonstration test then <math>\beta =1\,\!</math>, and no growth will be achieved. The results are shown next. It can be seen that in order to demonstrate a 10,000 hours MTBF with 80% confidence, 5 test units will be required.

	~~[[Image~~:~~rga14.16.png\|thumb\|center\|600px\|Calculated number of units for the demonstration test.]]~~


	The next figure shows the different combinations of test time per unit, and the number of units in the test for different numbers of allowable failures. It helps to visually examine other possible test scenarios.

	~~[[Image:rga14.17.png\|thumb\|center\|600px\|Combinations of test time and number of units for different numbers of allowable failures.]]~~

Kate Racaza: /* Example: Solve for Time */

2014-01-29T19:27:28Z

Example: Solve for Time

@@ Line 42: / Line 42: @@
 ==Example: Solve for Time==
+{{:Repairable_System_Test_Design_Example_-_Solve_for_Time}}
 ==Example: Solve for Number of Samples==

David J. Groebel at 00:58, 16 January 2014

2014-01-16T00:58:08Z

← Older revision		Revision as of 00:58, 16 January 2014
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	{{template:RGA BOOK\|8}}		{{template:RGA BOOK\|8\|Reliability Demonstration Test Design for Repairable Systems}}
	The RGA software provides a utility that allows you to design reliability demonstration tests for repairable systems. For example, you may want to design a test to demonstrate a specific cumulative MTBF for a specific operating period at a specific confidence level for a repairable system. The same applies for demonstrating an instantaneous MTBF or cumulative and instantaneous failure intensity at a given time <math>t.\,\!</math>		The RGA software provides a utility that allows you to design reliability demonstration tests for repairable systems. For example, you may want to design a test to demonstrate a specific cumulative MTBF for a specific operating period at a specific confidence level for a repairable system. The same applies for demonstrating an instantaneous MTBF or cumulative and instantaneous failure intensity at a given time <math>t.\,\!</math>

David J. Groebel at 00:56, 16 January 2014

2014-01-16T00:56:15Z

← Older revision		Revision as of 00:56, 16 January 2014
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	{{template:RGA BOOK\|15}}		{{template:RGA BOOK\|8}}
	The RGA software provides a utility that allows you to design reliability demonstration tests for repairable systems. For example, you may want to design a test to demonstrate a specific cumulative MTBF for a specific operating period at a specific confidence level for a repairable system. The same applies for demonstrating an instantaneous MTBF or cumulative and instantaneous failure intensity at a given time <math>t.\,\!</math>		The RGA software provides a utility that allows you to design reliability demonstration tests for repairable systems. For example, you may want to design a test to demonstrate a specific cumulative MTBF for a specific operating period at a specific confidence level for a repairable system. The same applies for demonstrating an instantaneous MTBF or cumulative and instantaneous failure intensity at a given time <math>t.\,\!</math>

Athanasios Gerokostopoulos: /* Example: Solve for Number of Samples */

2013-11-27T20:05:40Z

Example: Solve for Number of Samples

← Older revision		Revision as of 20:05, 27 November 2013
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	We can obtain the required number of units by using the Design of Reliability Tests (DRT) tool in the RGA software. Since this is a demonstration test then <math>\beta =1\,\!</math>, and no growth will be achieved. The results are shown next. It can be seen that in order to demonstrate a 10,000 hours MTBF with 80% confidence, 5 test units will be required.		We can obtain the required number of units by using the Design of Reliability Tests (DRT) tool in the RGA software. Since this is a demonstration test then <math>\beta =1\,\!</math>, and no growth will be achieved. The results are shown next. It can be seen that in order to demonstrate a 10,000 hours MTBF with 80% confidence, 5 test units will be required.

	[[Image:rga14.16.png\|thumb\|center\|~~450px~~\|Calculated number of units for the demonstration test.]]		[[Image:rga14.16.png\|thumb\|center\|600px\|Calculated number of units for the demonstration test.]]


	The next figure shows the different combinations of test time per unit, and the number of units in the test for different numbers of allowable failures. It helps to visually examine other possible test scenarios.		The next figure shows the different combinations of test time per unit, and the number of units in the test for different numbers of allowable failures. It helps to visually examine other possible test scenarios.

	[[Image:rga14.17.png\|thumb\|center\|~~450px~~\|Combinations of test time and number of units for different numbers of allowable failures.]]		[[Image:rga14.17.png\|thumb\|center\|600px\|Combinations of test time and number of units for different numbers of allowable failures.]]

Athanasios Gerokostopoulos: /* Example: Solve for Time */

2013-11-27T20:05:13Z

Example: Solve for Time

← Older revision		Revision as of 20:05, 27 November 2013
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	The same result can be obtained in the RGA software, by using the Design of Reliability Tests (DRT) tool. The next figure shows the calculated required test time per system of 0.3566 on the results of the example.		The same result can be obtained in the RGA software, by using the Design of Reliability Tests (DRT) tool. The next figure shows the calculated required test time per system of 0.3566 on the results of the example.

	[[File:DesignReliabilityDemTst.png\|center\|~~450px~~\|]]		[[File:DesignReliabilityDemTst.png\|center\|600px\|]]

	==Example: Solve for Number of Samples==		==Example: Solve for Number of Samples==

Athanasios Gerokostopoulos: /* Example: Solve for Time */

2013-11-27T19:59:56Z

Example: Solve for Time

← Older revision		Revision as of 19:59, 27 November 2013
Line 103:		Line 103:

	The same result can be obtained in the RGA software, by using the Design of Reliability Tests (DRT) tool. The next figure shows the calculated required test time per system of 0.3566 on the results of the example.		The same result can be obtained in the RGA software, by using the Design of Reliability Tests (DRT) tool. The next figure shows the calculated required test time per system of 0.3566 on the results of the example.
	[[File:DesignReliabilityDemTst.png]]
			[[File:DesignReliabilityDemTst.png\|center\|450px\|]]

	==Example: Solve for Number of Samples==		==Example: Solve for Number of Samples==

Athanasios Gerokostopoulos: /* Example: Solve for Number of Samples */

2013-11-27T19:59:03Z

Example: Solve for Number of Samples

← Older revision		Revision as of 19:59, 27 November 2013
Line 112:		Line 112:
	We can obtain the required number of units by using the Design of Reliability Tests (DRT) tool in the RGA software. Since this is a demonstration test then <math>\beta =1\,\!</math>, and no growth will be achieved. The results are shown next. It can be seen that in order to demonstrate a 10,000 hours MTBF with 80% confidence, 5 test units will be required.		We can obtain the required number of units by using the Design of Reliability Tests (DRT) tool in the RGA software. Since this is a demonstration test then <math>\beta =1\,\!</math>, and no growth will be achieved. The results are shown next. It can be seen that in order to demonstrate a 10,000 hours MTBF with 80% confidence, 5 test units will be required.

	[[Image:rga14.16.png\|thumb\|center\|~~400px~~\|Calculated number of units for the demonstration test.]]		[[Image:rga14.16.png\|thumb\|center\|450px\|Calculated number of units for the demonstration test.]]


	The next figure shows the different combinations of test time per unit, and the number of units in the test for different numbers of allowable failures. It helps to visually examine other possible test scenarios.		The next figure shows the different combinations of test time per unit, and the number of units in the test for different numbers of allowable failures. It helps to visually examine other possible test scenarios.

	[[Image:rga14.17.png\|thumb\|center\|~~400px~~\|Combinations of test time and number of units for different numbers of allowable failures.]]		[[Image:rga14.17.png\|thumb\|center\|450px\|Combinations of test time and number of units for different numbers of allowable failures.]]

← Older revision		Revision as of 17:15, 20 June 2014
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	The failure process in a repairable system is considered to be a non-homogeneous Poisson process (NHPP) with power law failure intensity. The instantaneous failure intensity at time <math>t\,\!</math> is:		The failure process in a repairable system is considered to be a non-homogeneous Poisson process (NHPP) with power law failure intensity. The instantaneous failure intensity at time <math>t\,\!</math> is:

	::<math>{{\lambda }_{i}}\left( t \right)=\lambda \beta {{t}^{\beta -1}}={{\lambda }_{c}}\left( t \right)\beta \,\!</math>		:<math>{{\lambda }_{i}}\left( t \right)=\lambda \beta {{t}^{\beta -1}}={{\lambda }_{c}}\left( t \right)\beta \,\!</math>


	So, the cumulative failure intensity at time <math>t\,\!</math> is:		So, the cumulative failure intensity at time <math>t\,\!</math> is:

	::<math>{{\lambda }_{c}}\left( t \right)=\lambda {{t}^{\beta -1}}=\frac{{{\lambda }_{i}}\left( t \right)}{\beta }\,\!</math>		:<math>{{\lambda }_{c}}\left( t \right)=\lambda {{t}^{\beta -1}}=\frac{{{\lambda }_{i}}\left( t \right)}{\beta }\,\!</math>


	The instantaneous MTTF is:		The instantaneous MTTF is:

	::<math>\begin{align}		:<math>\begin{align}
	MTT{{F}_{i}}\left( t \right)= & \frac{1}{{{\lambda }_{i\left( t \right)}}} \\		MTT{{F}_{i}}\left( t \right)= & \frac{1}{{{\lambda }_{i\left( t \right)}}} \\
	= & \frac{1}{\lambda \beta }{{t}^{1-\beta }} \\		= & \frac{1}{\lambda \beta }{{t}^{1-\beta }} \\
	= & \frac{MTT{{F}_{c}}\left( t \right)}{\beta }		= & \frac{MTT{{F}_{c}}\left( t \right)}{\beta }
	\end{align}\,\!</math>		\end{align}\,\!</math>


	The cumulative MTBF at time <math>t\,\!</math> is:		The cumulative MTBF at time <math>t\,\!</math> is:

	::<math>MTB{{F}_{c}}\left( \beta \right)=\frac{1}{{{\lambda }_{c}}\left( t \right)}=\frac{1}{\lambda }{{t}^{1-\beta }}=MTT{{F}_{i}}\left( t \right)\beta \,\!</math>		:<math>MTB{{F}_{c}}\left( \beta \right)=\frac{1}{{{\lambda }_{c}}\left( t \right)}=\frac{1}{\lambda }{{t}^{1-\beta }}=MTT{{F}_{i}}\left( t \right)\beta \,\!</math>


	The relation between the confidence level, required test time, number of systems under test and allowed total number of failures in the test is:		The relation between the confidence level, required test time, number of systems under test and allowed total number of failures in the test is:

	::<math>1-CL=\underset{i=0}{\overset{r}{\mathop \sum }}\,\frac{{{\left( m\lambda {{T}^{\beta }} \right)}^{i}}\exp (-m\lambda {{T}^{\beta }})}{i!}\,\!</math>		:<math>1-CL=\underset{i=0}{\overset{r}{\mathop \sum }}\,\frac{{{\left( m\lambda {{T}^{\beta }} \right)}^{i}}\exp (-m\lambda {{T}^{\beta }})}{i!}\,\!</math>

	where:		where:
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	*<math>r\,\!</math> is the number of allowed failures in the test.		*<math>r\,\!</math> is the number of allowed failures in the test.
	*<math>CL\,\!</math> is the confidence level.		*<math>CL\,\!</math> is the confidence level.


	Given any three of the parameters, the equation above can be solved for the fourth unknown parameter. Note that when <math>\beta =1,\,\!</math> the number of failures is a homogeneous Poisson process, and the time between failures is given by the exponential distribution.		Given any three of the parameters, the equation above can be solved for the fourth unknown parameter. Note that when <math>\beta =1,\,\!</math> the number of failures is a homogeneous Poisson process, and the time between failures is given by the exponential distribution.