Parametric Binomial Example - Demonstrate Reliability - Revision history

Lisa Hacker at 18:30, 18 September 2023

2023-09-18T18:30:23Z

← Older revision		Revision as of 18:30, 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 17:11, 9 December 2015

2015-12-09T17:11:36Z

← Older revision		Revision as of 17:11, 9 December 2015
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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\|center\|~~650px~~ ]]
			[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|center\|800px]]


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

	[[Image:RDT Weibull Demonstrate Reliability ~~Test Time~~.png\|center\|~~650px~~ ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|center\|800px]]

Richard House at 19:44, 26 September 2012

2012-09-26T19:44:40Z

Richard House at 22:49, 26 August 2012

2012-08-26T22:49:08Z

Chris Kahn at 05:54, 22 August 2012

2012-08-22T05:54:13Z

← Older revision		Revision as of 05:54, 22 August 2012
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	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100</math> hours with 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.		In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100</math> hours with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.

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

Chris Kahn at 05:53, 22 August 2012

2012-08-22T05:53:21Z

← Older revision		Revision as of 05:53, 22 August 2012
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	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100</math> hours ~~and~~ with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.		In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100</math> hours with 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.

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

Chris Kahn at 05:12, 22 August 2012

2012-08-22T05:12:13Z

← Older revision		Revision as of 05:12, 22 August 2012
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	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>t=100</math> hours and with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.		In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100</math> hours and with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.

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

Chris Kahn at 05:09, 22 August 2012

2012-08-22T05:09:50Z

← Older revision		Revision as of 05:09, 22 August 2012
Line 3:		Line 3:

	</noinclude>		</noinclude>

	In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>t=100</math> hours and with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.		In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>t=100</math> hours and with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.

Chris Kahn at 05:09, 22 August 2012

2012-08-22T05:09:42Z

← Older revision		Revision as of 05:09, 22 August 2012
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	</noinclude>		</noinclude>
	~~'''Example: Using Parametric Binomial for Test to 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 use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>t=100</math> hours and with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.

	'''Determining Units for Available Test Time'''		'''Determining Units for Available Test 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.		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:

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

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

2012-08-15T03:00:15Z

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

New page

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

</noinclude>
'''Example: Using Parametric Binomial for Test to 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>.

'''Determining Units for Available Test 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:

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

This can be rewritten as:

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

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

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

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

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

::<math>1-CL=\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.

[[Image:RDT Weibull Demonstrate Reliability Samples.png|center|650px ]]

'''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|center|650px ]]

← Older revision		Revision as of 19:44, 26 September 2012
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	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100\,\!</math> hours with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</math>.		In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100\,\!</math> hours with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</math>.

	'''Determining Units for Available Test Time'''		'''Determining Units for Available Test 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.		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:

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

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

	::<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>

	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>:

	::<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>

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

	::<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>

	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:

	::<math>1-CL=\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-CL=\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.		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\|center\|650px ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|center\|650px ]]
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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 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.		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\|center\|650px ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|center\|650px ]]

← Older revision		Revision as of 22:49, 26 August 2012
Line 3:		Line 3:

	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100</math> hours with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</math>.		In this example, we will use the parametric binomial method to design a test to demonstrate a reliability of 90% at <math>{{t}_{DEMO}}=100\,\!</math> hours with a 95% confidence if no failure occur during the test. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</math>.

	'''Determining Units for Available Test Time'''		'''Determining Units for Available Test 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.		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:

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

	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>:

	::<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>

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

	::<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>
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	::<math>1-CL=\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-CL=\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.		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\|center\|650px ]]		[[Image:RDT Weibull Demonstrate Reliability Samples.png\|center\|650px ]]
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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 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.		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\|center\|650px ]]		[[Image:RDT Weibull Demonstrate Reliability Test Time.png\|center\|650px ]]