Parametric Binomial Example - Demonstrate MTTF - Revision history

Lisa Hacker at 18:55, 18 September 2023

2023-09-18T18:55:17Z

← 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 17:13, 9 December 2015

2015-12-09T17:13:16Z

← Older revision		Revision as of 17:13, 9 December 2015
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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 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.

	[[Image:RDT Weibull Demonstrate MTTF.png\|center\|~~650px~~\| ]]		[[Image:RDT Weibull Demonstrate MTTF.png\|center\|800px\| ]]

	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.

Richard House at 19:46, 26 September 2012

2012-09-26T19:46:42Z

← Older revision		Revision as of 19:46, 26 September 2012
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	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test that will demonstrate <math>MTTF=75\,\!</math> hours with a 95% confidence if no failure occur during the test <math>f=0\,\!</math>. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</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 use the parametric binomial method to design a test that will demonstrate <math>MTTF=75\,\!</math> hours with a 95% confidence if no failure occur during the test <math>f=0\,\!</math>. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</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:

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

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

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

	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:

	::<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>

	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.

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

	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.

Richard House at 22:55, 26 August 2012

2012-08-26T22:55:41Z

← Older revision		Revision as of 22:55, 26 August 2012
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	</noinclude>		</noinclude>
	In this example, we will use the parametric binomial method to design a test that will demonstrate <math>MTTF=75</math> hours with a 95% confidence if no failure occur during the test <math>f=0</math>. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</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 use the parametric binomial method to design a test that will demonstrate <math>MTTF=75\,\!</math> hours with a 95% confidence if no failure occur during the test <math>f=0\,\!</math>. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5\,\!</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:

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

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

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

	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:

	::<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>

	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.

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

	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.

Chris Kahn at 05:49, 22 August 2012

2012-08-22T05:49:00Z

← Older revision		Revision as of 05:49, 22 August 2012
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	</noinclude>		</noinclude>
	~~'''Example: Using Parametric Binomial for Test to Demonstrate MTTF'''~~		In this example, we will use the parametric binomial method to design a test that will demonstrate <math>MTTF=75</math> hours with a 95% confidence if no failure occur during the test <math>f=0</math>. We will assume a Weibull distribution with a shape parameter <math>\beta =1.5</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:

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

2012-08-15T03:14:29Z

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 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.

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:

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

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

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

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:

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

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|center|650px| ]]

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.