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

Lisa Hacker: Redirected page to Reliability Test Design#Example 1: Demonstrate Reliability

2012-08-09T09:55:41Z

Redirected page to Reliability Test Design#Example 1: Demonstrate Reliability

← Older revision		Revision as of 09:55, 9 August 2012
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	[[~~Category~~: ~~For Deletion~~]]		#REDIRECT [[Reliability_Test_Design#Example_1:_Demonstrate_Reliability]]

Kate Racaza: Merged content with page: Reliability_Test_Design

2012-08-01T05:33:24Z

Merged content with page: Reliability_Test_Design

← Older revision		Revision as of 05:33, 1 August 2012
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	~~'''Weibull Distribution Example - Demonstrate Reliability'''~~		[[Category: For Deletion]]

	In this example, we will design a test to demonstrate a reliability of 90% at <math>t=100</math> hours, with a 95% confidence. We will 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>.


	~~'''Determining Units for Available Time'''~~

	In the above scenario, we know that we have the testing facilities available for <math>t=48</math> hours. We must now determine the number of units to test for this amount of time with no failures in order to have demonstrated our reliability goal.
	~~The first step is to determine the Weibull scale parameter, <math>\eta </math>. The Weibull reliability equation is:~~

	~~<br>~~

	~~<center><math>R={{e}^{-{{(t/\eta )}^{\beta }}}}</math></center>~~

	~~<br>~~

	~~This can be rewritten as:~~

	~~<br>~~

	~~<center><math>\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}</math></center>~~

	~~<br>~~

	~~Since we know the values of <math>{{t}_{DEMO}}</math>, <math>{{R}_{DEMO}}</math> and <math>\beta </math>, we can substitute these in the equation and solve for <math>\eta </math>:~~

	~~<br>~~

	~~<center><math>\eta =\frac{100}{{{(-\text{ln}(0.9))}^{\tfrac{1}{1.5}}}}=448.3</math></center>~~

	~~<br>~~

	~~Next, the value of <math>{{R}_{TEST}}</math> is calculated by:~~

	~~<br>~~

	~~<center><math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%</math></center>~~

	~~<br>~~

	~~The last step is to substitute the appropriate values into the cumulative binomial equation, which for the Weibull distribution appears as:~~

	~~<br>~~

	<center><math>1-C.L.=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n!}{i!\cdot (n-i)!}\cdot {{(1-{{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{i}}\cdot {{({{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{(n-i)}}</math></center>

	~~<br>~~

	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 equation for <math>n</math>. This value is <math>n=85.4994</math>, or <math>n=86</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 Reliability Samples.png\|thumb\|center\|650px ]]~~

	~~<br>~~

	~~'''Determining Time for Available Units'''~~

	In this case, we will assume that we have 20 units to test, <math>n=20</math>, and must determine the test time, <math>{{t}_{TEST}}</math>. We have already determined the value of the scale parameter, <math>\eta </math>, in the previous example. Since we know the values of <math>n</math>, <math>CL</math>, <math>f</math>, <math>\eta </math> and <math>\beta </math>, it remains to solve the binomial equation with the Weibull distribution for <math>{{t}_{TEST}}</math>. This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

	~~<br>~~

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

Chris Kahn at 10:25, 18 July 2012

2012-07-18T10:25:19Z

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

	[[Image:RDT Weibull Demonstrate Reliability Samples.png\|thumb\|center\|~~400px~~ ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|thumb\|center\|650px ]]

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

	[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|~~400px~~ ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|650px ]]

Lisa Hacker at 23:34, 29 March 2012

2012-03-29T23:34:38Z

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

2012-03-29T23:32:15Z

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

← Older revision	Revision as of 23:32, 29 March 2012
(No difference)

Chris Kahn at 00:40, 9 March 2012

2012-03-09T00:40:40Z

Harry Guo at 22:15, 2 March 2012

2012-03-02T22:15:13Z

← Older revision		Revision as of 22:15, 2 March 2012
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	'''Determining Time for Available Units'''		'''Determining Time for Available Units'''

	In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve ~~Eqn. (weibcum)~~ for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.		In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve the binomial equation with Weibull distribution for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

	[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]

Harry Guo at 22:11, 2 March 2012

2012-03-02T22:11:12Z

← Older revision		Revision as of 22:11, 2 March 2012
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	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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</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 Reliability Samples.png\|thumb\|center ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|thumb\|center\|400px ]]


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	In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.		In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

	[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]

Harry Guo at 22:10, 2 March 2012

2012-03-02T22:10:45Z

← Older revision		Revision as of 22:10, 2 March 2012
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	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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</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 Reliability Samples.png\|thumb\|center~~\|300px\|~~ ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|thumb\|center ]]


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	In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.		In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

	[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center~~\|300px\|~~ ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center ]]

Harry Guo at 22:09, 2 March 2012

2012-03-02T22:09:43Z

← Older revision		Revision as of 22:09, 2 March 2012
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	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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</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~~.~~1.gif~~\|thumb\|center\|300px\| ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|thumb\|center\|300px\| ]]


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	In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.		In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve Eqn. (weibcum) for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

	[[Image:~~lda21~~.~~2.gif~~\|thumb\|center\|300px\| ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|300px\| ]]

← Older revision		Revision as of 23:34, 29 March 2012
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	<br>		<br>

	::<math>R={{e}^{-{{(t/\eta )}^{\beta }}}}</math>		<center><math>R={{e}^{-{{(t/\eta )}^{\beta }}}}</math></center>

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

	::<math>\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}</math>		<center><math>\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}</math></center>

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

	::<math>\eta =\frac{100}{{{(-\text{ln}(0.9))}^{\tfrac{1}{1.5}}}}=448.3</math>		<center><math>\eta =\frac{100}{{{(-\text{ln}(0.9))}^{\tfrac{1}{1.5}}}}=448.3</math></center>

	<br>		<br>

	Next, the value of <math>{{R}_{TEST}}</math> is calculated by:		Next, the value of <math>{{R}_{TEST}}</math> is calculated by:

	<br>		<br>

	::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%</math>		<center><math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%</math></center>

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

	::<math>1-C.L.=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n!}{i!\cdot (n-i)!}\cdot {{(1-{{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{i}}\cdot {{({{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{(n-i)}}</math>		<center><math>1-C.L.=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n!}{i!\cdot (n-i)!}\cdot {{(1-{{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{i}}\cdot {{({{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{(n-i)}}</math></center>

	<br>		<br>

	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 equation for <math>n</math>. This value is <math>n=85.4994</math>, or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		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 equation for <math>n</math>. This value is <math>n=85.4994</math>, or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.

	<br>		<br>
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	'''Determining Time for Available Units'''		'''Determining Time for Available Units'''

	In this case, we will assume that we have 20 units to test, <math>n=20</math>, and must determine the test time, <math>{{t}_{TEST}}</math>. We have already determined the value of the scale parameter, <math>\eta </math>, in the previous example. Since we know the values of <math>n</math>, <math>CL</math>, <math>f</math>, <math>\eta </math> and <math>\beta </math>, it remains to solve the binomial equation with Weibull distribution for <math>{{t}_{TEST}}</math>. This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.		In this case, we will assume that we have 20 units to test, <math>n=20</math>, and must determine the test time, <math>{{t}_{TEST}}</math>. We have already determined the value of the scale parameter, <math>\eta </math>, in the previous example. Since we know the values of <math>n</math>, <math>CL</math>, <math>f</math>, <math>\eta </math> and <math>\beta </math>, it remains to solve the binomial equation with the Weibull distribution for <math>{{t}_{TEST}}</math>. This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

	<br>		<br>

	[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]

← Older revision		Revision as of 00:40, 9 March 2012
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	'''Weibull Distribution Example - Demonstrate Reliability'''		'''Weibull Distribution Example - Demonstrate Reliability'''

	In this example, we will design a test to demonstrate a reliability of 90% at <math>t=100</math> hours, with a 95% confidence. We will 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> .		In this example, we will design a test to demonstrate a reliability of 90% at <math>t=100</math> hours, with a 95% confidence. We will 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>.


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	In the above scenario, we know that we have the testing facilities available for <math>t=48</math> hours. We must now determine the number of units to test for this amount of time with no failures in order to have demonstrated our reliability goal.		In the above scenario, we know that we have the testing facilities available for <math>t=48</math> hours. We must now determine the number of units to test for this amount of time with no failures in order to have demonstrated our reliability goal.
	The first step is to determine the Weibull scale parameter, <math>\eta .</math> The Weibull reliability equation is:		The first step is to determine the Weibull scale parameter, <math>\eta </math>. The Weibull reliability equation is:

			<br>

	::<math>R={{e}^{-{{(t/\eta )}^{\beta }}}}</math>		::<math>R={{e}^{-{{(t/\eta )}^{\beta }}}}</math>

			<br>

	This can be rewritten as:		This can be rewritten as:

			<br>

	::<math>\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}</math>		::<math>\eta =\frac{{{t}_{DEMO}}}{{{(-\text{ln}({{R}_{DEMO}}))}^{\tfrac{1}{\beta }}}}</math>

			<br>

	Since we know the values of <math>{{t}_{DEMO}}</math> , <math>{{R}_{DEMO}}</math> and <math>\beta </math> , we can substitute these in the equation and solve for <math>\eta </math> :		Since we know the values of <math>{{t}_{DEMO}}</math>, <math>{{R}_{DEMO}}</math> and <math>\beta </math>, we can substitute these in the equation and solve for <math>\eta </math>:

			<br>

	::<math>\eta =\frac{100}{{{(-\text{ln}(0.9))}^{\tfrac{1}{1.5}}}}=448.3</math>		::<math>\eta =\frac{100}{{{(-\text{ln}(0.9))}^{\tfrac{1}{1.5}}}}=448.3</math>

			<br>

			Next, the value of <math>{{R}_{TEST}}</math> is calculated by:

	~~Next, the value of~~ <~~math~~>~~{{R}_{TEST}}</math> is calculated by:~~		<br>


	::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%.</math>		::<math>{{R}_{TEST}}={{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}}={{e}^{-{{(48/448.3)}^{1.5}}}}=0.966=96.6%</math>

			<br>

	The last step is to substitute the appropriate values into the cumulative binomial equation, which for the Weibull distribution appears as:		The last step is to substitute the appropriate values into the cumulative binomial equation, which for the Weibull distribution appears as:

			<br>

	::<math>1-C.L.=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n!}{i!\cdot (n-i)!}\cdot {{(1-{{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{i}}\cdot {{({{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{(n-i)}}</math>		::<math>1-C.L.=\underset{i=0}{\overset{f}{\mathop \sum }}\,\frac{n!}{i!\cdot (n-i)!}\cdot {{(1-{{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{i}}\cdot {{({{e}^{-{{({{t}_{TEST}}/\eta )}^{\beta }}}})}^{(n-i)}}</math>

	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 equation for <math>n.</math> This value is <math>n=85.4994,</math> or <math>n=86</math> units, since the fractional value must be rounded up to the next integer value. This example solved in Weibull++ is shown next.		<br>

			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 equation for <math>n</math>. This value is <math>n=85.4994</math>, or <math>n=86</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 Reliability Samples.png\|thumb\|center\|400px ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|thumb\|center\|400px ]]

			<br>

	'''Determining Time for Available Units'''		'''Determining Time for Available Units'''

	In this case, we will assume that we have 20 units to test, <math>n=20</math> , and must determine the test time, <math>{{t}_{TEST}}.</math> We have already determined the value of the scale parameter, <math>\eta </math> , in the previous example. Since we know the values of <math>n</math> , <math>CL</math> , <math>f</math> , <math>\eta </math> and <math>\beta </math> , it remains to solve the binomial equation with Weibull distribution for <math>{{t}_{TEST}}</math> . This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.		In this case, we will assume that we have 20 units to test, <math>n=20</math>, and must determine the test time, <math>{{t}_{TEST}}</math>. We have already determined the value of the scale parameter, <math>\eta </math>, in the previous example. Since we know the values of <math>n</math>, <math>CL</math>, <math>f</math>, <math>\eta </math> and <math>\beta </math>, it remains to solve the binomial equation with Weibull distribution for <math>{{t}_{TEST}}</math>. This value is <math>{{t}_{TEST}}=126.4339</math> hours. This example solved in Weibull++ is shown next.

			<br>

	[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|thumb\|center\|400px ]]