Template:Example: Weibull Distribution Example-Demonstrate MTTF - Revision history

Lisa Hacker: Redirected page to Reliability Test Design

2012-08-09T08:58:28Z

Redirected page to Reliability Test Design

← Older revision		Revision as of 08:58, 9 August 2012
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	[[~~Category: For Deletion~~]]		#REDIRECT [[Reliability_Test_Design]]

Kate Racaza: Merged content with page: Reliability_Test_Design

2012-08-01T05:32:57Z

Merged content with page: Reliability_Test_Design

← Older revision		Revision as of 05:32, 1 August 2012
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	~~'''Weibull Distribution Example - Demonstrate MTTF'''~~		[[Category: For Deletion]]

	In this example, we will design a test to demonstrate <math>MTTF=75</math> hours, with a 95% confidence. We will once again assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>. No failures will be allowed on this test, or <math>f=0</math>. We want to determine the number of units to test for <math>{{t}_{TEST}}=60</math> hours to demonstrate this goal.

	The first step in this case involves determining the value of the scale parameter <math>\eta </math> from the <math>MTTF</math> equation. The equation for the <math>MTTF</math> for the Weibull distribution is:

	~~<br>~~

	~~<center><math>MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })</math></center>~~

	~~<br>~~

	~~where <math>\Gamma (x)</math> is the gamma function of <math>x</math>. This can be rearranged in terms of <math>\eta</math>:~~

	~~<br>~~

	~~<center><math>\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}</math></center>~~

	~~<br>~~

	Since <math>MTTF</math> and <math>\beta </math> have been specified, it is a relatively simple matter to calculate <math>\eta =83.1</math>. From this point on, the procedure is the same as the reliability demonstration example. Next, the value of <math>{{R}_{TEST}}</math> is calculated as:

	~~<br>~~

	~~<center><math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%</math></center>~~

	~~<br>~~

	The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math>, <math>{{t}_{TEST}}</math>, <math>\beta </math>, <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve the binomial equation for <math>n</math>. The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.

	~~<br>~~

	[[~~Image~~:~~RDT Weibull Demonstrate MTTF.png\|thumb\|center\|650px\|~~ ]]

	~~<br>~~

	The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from the binomial equation with Weibull distribution.

Chris Kahn at 10:25, 18 July 2012

2012-07-18T10:25:43Z

← Older revision		Revision as of 10:25, 18 July 2012
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	<br>		<br>

	[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|~~300px~~\| ]]		[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|650px\| ]]

	<br>		<br>

	The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from the binomial equation with Weibull distribution.		The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from the binomial equation with Weibull distribution.

Lisa Hacker at 23:39, 29 March 2012

2012-03-29T23:39:23Z

← Older revision		Revision as of 23:39, 29 March 2012
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	<br>		<br>

	::<math>MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })</math>		<center><math>MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })</math></center>

	<br>		<br>
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	<br>		<br>

	::<math>\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}</math>		<center><math>\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}</math></center>

	<br>		<br>
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	<br>		<br>

	::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%</math>		<center><math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%</math></center>

	<br>		<br>

Lisa Hacker: moved Template:Example: Weibull Distribution Example-Demostrate MTTF to Template:Example: Weibull Distribution Example-Demonstrate MTTF: Correct typo

2012-03-29T23:37:44Z

moved Template:Example: Weibull Distribution Example-Demostrate MTTF to Template:Example: Weibull Distribution Example-Demonstrate MTTF: Correct typo

← Older revision	Revision as of 23:37, 29 March 2012
(No difference)

Chris Kahn at 00:44, 9 March 2012

2012-03-09T00:44:13Z

← Older revision		Revision as of 00:44, 9 March 2012
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	'''Weibull Distribution Example - Demonstrate MTTF'''		'''Weibull Distribution Example - Demonstrate MTTF'''

	In this example, we will design a test to demonstrate <math>MTTF=75</math> hours, with a 95% confidence. We will once again assume a Weibull distribution with a shape parameter <math>\beta =1.5</math> . No failures will be allowed on this test, or <math>f=0</math> . We want to determine the number of units to test for <math>{{t}_{TEST}}=60</math> hours to demonstrate this goal.		In this example, we will design a test to demonstrate <math>MTTF=75</math> hours, with a 95% confidence. We will once again assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>. No failures will be allowed on this test, or <math>f=0</math>. We want to determine the number of units to test for <math>{{t}_{TEST}}=60</math> hours to demonstrate this goal.

	The first step in this case involves determining the value of the scale parameter <math>\eta </math> from the <math>MTTF</math> equation. The equation for the <math>MTTF</math> for the Weibull distribution is:		The first step in this case involves determining the value of the scale parameter <math>\eta </math> from the <math>MTTF</math> equation. The equation for the <math>MTTF</math> for the Weibull distribution is:

			<br>

	::<math>MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })</math>		::<math>MTTF=\eta \cdot \Gamma (1+\frac{1}{\beta })</math>

			<br>

	where <math>\Gamma (x)</math> is the gamma function of <math>x</math> . This can be rearranged in terms of .. :		where <math>\Gamma (x)</math> is the gamma function of <math>x</math>. This can be rearranged in terms of <math>\eta</math>:

			<br>

	::<math>\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}</math>		::<math>\eta =\frac{MTTF}{\Gamma (1+\tfrac{1}{\beta })}</math>

			<br>

	Since <math>MTTF</math> and <math>\beta </math> have been specified, it is a relatively simple matter to calculate <math>\eta =83.1</math> . From this point on, the procedure is the same as the reliability demonstration example. Next, the value of <math>{{R}_{TEST}}</math> is calculated as:		Since <math>MTTF</math> and <math>\beta </math> have been specified, it is a relatively simple matter to calculate <math>\eta =83.1</math>. From this point on, the procedure is the same as the reliability demonstration example. Next, the value of <math>{{R}_{TEST}}</math> is calculated as:

			<br>

	::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%</math>		::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(60/83.1)}^{1.5}}}}=0.541=54.1%</math>

			<br>

	The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math> , <math>{{t}_{TEST}}</math> , <math>\beta </math> , <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve the binomial equation for <math>n.</math> The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math>, <math>{{t}_{TEST}}</math>, <math>\beta </math>, <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve the binomial equation for <math>n</math>. The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.

			<br>

	[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|300px\| ]]		[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|300px\| ]]

			<br>

	The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from the binomial equation with Weibull distribution.		The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from the binomial equation with Weibull distribution.

Harry Guo at 22:12, 2 March 2012

2012-03-02T22:12:59Z

← Older revision		Revision as of 22:12, 2 March 2012
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	The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math> , <math>{{t}_{TEST}}</math> , <math>\beta </math> , <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve ~~Eqn. (relcum)~~ for <math>n.</math> The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math> , <math>{{t}_{TEST}}</math> , <math>\beta </math> , <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve the binomial equation for <math>n.</math> The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.

	[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|300px\| ]]		[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|300px\| ]]

	The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from ~~Eqn. (weibcum)~~.		The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from the binomial equation with Weibull distribution.

Harry Guo at 22:11, 2 March 2012

2012-03-02T22:11:39Z

← Older revision		Revision as of 22:11, 2 March 2012
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	The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math> , <math>{{t}_{TEST}}</math> , <math>\beta </math> , <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve Eqn. (relcum) for <math>n.</math> The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		The last step is to substitute the appropriate values into the cumulative binomial equation. The values of <math>CL</math> , <math>{{t}_{TEST}}</math> , <math>\beta </math> , <math>f</math> and <math>\eta </math> have already been calculated or specified, so it merely remains to solve Eqn. (relcum) for <math>n.</math> The value is calculated as <math>n=4.8811,</math> or <math>n=5</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.

	[[Image:~~lda21~~.~~3.gif~~\|thumb\|center\|300px\| ]]		[[Image:RDT Weibull Demonstrate MTTF.png\|thumb\|center\|300px\| ]]

	The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from Eqn. (weibcum).		The procedure for determining the required test time proceeds in the same manner, determining <math>\eta </math> from the <math>MTTF</math> equation, and following the previously described methodology to determine <math>{{t}_{TEST}}</math> from Eqn. (weibcum).

Harry Guo at 23:51, 22 February 2012

2012-02-22T23:51:43Z

← Older revision		Revision as of 23:51, 22 February 2012
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	~~====~~Weibull Distribution Example - Demonstrate MTTF~~====~~		'''Weibull Distribution Example - Demonstrate MTTF'''

	In this example, we will design a test to demonstrate <math>MTTF=75</math> hours, with a 95% confidence. We will once again assume a Weibull distribution with a shape parameter <math>\beta =1.5</math> . No failures will be allowed on this test, or <math>f=0</math> . We want to determine the number of units to test for <math>{{t}_{TEST}}=60</math> hours to demonstrate this goal.		In this example, we will design a test to demonstrate <math>MTTF=75</math> hours, with a 95% confidence. We will once again assume a Weibull distribution with a shape parameter <math>\beta =1.5</math> . No failures will be allowed on this test, or <math>f=0</math> . We want to determine the number of units to test for <math>{{t}_{TEST}}=60</math> hours to demonstrate this goal.

Harry Guo: moved Template:Weibull distribution ex-mttf to Template:Example: Weibull Distribution Example-Demostrate MTTF

2012-02-22T23:50:33Z

moved Template:Weibull distribution ex-mttf to Template:Example: Weibull Distribution Example-Demostrate MTTF

← Older revision	Revision as of 23:50, 22 February 2012
(No difference)