Template:Example: Test Design Using Expected Failure Times Plot - Revision history

Lisa Hacker: Redirected page to Expected Failure Times Plot Example

2012-08-15T03:53:44Z

Redirected page to Expected Failure Times Plot Example

← Older revision		Revision as of 03:53, 15 August 2012
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	[[~~Category:For Deletion~~]]		#REDIRECT [[Expected_Failure_Times_Plot_Example]]

Kate Racaza at 06:45, 1 August 2012

2012-08-01T06:45:57Z

← Older revision		Revision as of 06:45, 1 August 2012
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	~~Category~~[[For Deletion]]		[[Category:For Deletion]]

Kate Racaza: Merged content with page: Reliability_Test_Design

2012-08-01T06:25:05Z

Merged content with page: Reliability_Test_Design

← Older revision		Revision as of 06:25, 1 August 2012
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	~~'''Test Design Using Expected Failure Times Plot'''~~		Category[[For Deletion]]

	Four units were allocated for a reliability test. The test engineers want to know how long the test will last if all the units are tested to failure. Based on previous experiments, they assume the underlying failure distribution is a Weibull distribution with <math>\beta</math> = 2 and <math>\eta</math> = 500.


	~~'''Solution'''~~

	~~The expected failure times with 80% 2-sided confidence bounds are given below.~~

	[[~~Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|500px~~]]
	~~[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|500px]]~~

	~~From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.~~

	As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the second failure is calculated as:

	~~<center>~~
	~~<math>\begin{align}~~
	~~& Q=1-{{e}^{{{\left( \frac{t}{\eta } \right)}^{\beta }}}}\Rightarrow \\~~
	~~& \ln (1-Q)={{\left( \frac{t}{\eta } \right)}^{\beta }} \\~~
	~~& \Rightarrow t=\text{349.04}\\~~
	~~\end{align}</math>~~
	~~</center>~~

	~~Its bounds and other failure times can be calculated in a similar way.~~

Chris Kahn at 10:42, 18 July 2012

2012-07-18T10:42:26Z

← Older revision		Revision as of 10:42, 18 July 2012
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	The expected failure times with 80% 2-sided confidence bounds are given below.		The expected failure times with 80% 2-sided confidence bounds are given below.

	[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|~~300px~~]]		[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|500px]]
	[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|~~250px~~]]		[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|500px]]

	From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.		From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.

Nicolette Young at 23:22, 25 April 2012

2012-04-25T23:22:28Z

← Older revision		Revision as of 23:22, 25 April 2012
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	The expected failure times with 80% 2-sided confidence bounds are given below.		The expected failure times with 80% 2-sided confidence bounds are given below.

	[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|~~350px~~]]		[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|300px]]
	[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|~~350px~~]]		[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|250px]]

	From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.		From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.

Nicolette Young at 23:21, 25 April 2012

2012-04-25T23:21:54Z

← Older revision		Revision as of 23:21, 25 April 2012
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	The expected failure times with 80% 2-sided confidence bounds are given below.		The expected failure times with 80% 2-sided confidence bounds are given below.

	[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|~~400px~~]]		[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|350px]]
	[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|~~400px~~]]		[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|350px]]

	From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.		From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.

Lisa Hacker at 00:18, 30 March 2012

2012-03-30T00:18:40Z

← Older revision		Revision as of 00:18, 30 March 2012
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	'''Solution'''		'''Solution'''

	The expected failure ~~time~~ with 80% 2-sided confidence bounds are given below.		The expected failure times with 80% 2-sided confidence bounds are given below.

	[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|400px]]		[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|400px]]
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	From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.		From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.

	As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the ~~2nd~~ failure is calculated as:		As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the second failure is calculated as:

	<center>		<center>

Chris Kahn at 18:56, 8 March 2012

2012-03-08T18:56:57Z

← Older revision		Revision as of 18:56, 8 March 2012
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	From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.		From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.

	As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the 2nd failure is ~~calcualted~~ as:		As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the 2nd failure is calculated as:

	<center>		<center>
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	</center>		</center>

	Its bounds and other failure times can be ~~calcualted~~ in a similar way.		Its bounds and other failure times can be calculated in a similar way.

Harry Guo at 18:00, 24 February 2012

2012-02-24T18:00:36Z

← Older revision		Revision as of 18:00, 24 February 2012
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	As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the 2nd failure is calcualted as:		As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the 2nd failure is calcualted as:

			<center>
	<math>\begin{align}		<math>\begin{align}
	& Q=1-{{e}^{{{\left( \frac{t}{\eta } \right)}^{\beta }}}}\Rightarrow \\		& Q=1-{{e}^{{{\left( \frac{t}{\eta } \right)}^{\beta }}}}\Rightarrow \\
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	& \Rightarrow t=\text{349.04}\\		& \Rightarrow t=\text{349.04}\\
	\end{align}</math>		\end{align}</math>
			</center>

	Its bounds and other failure times can be calcualted in a similar way.		Its bounds and other failure times can be calcualted in a similar way.

Harry Guo at 18:00, 24 February 2012

2012-02-24T18:00:06Z

← Older revision		Revision as of 18:00, 24 February 2012
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	[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|400px]]		[[Image: Expected Failure Plot example Result Summary.png\|thumb\|center\|400px]]
	[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|400px]]		[[Image: Expected Failure Plot example Result Plot.png\|thumb\|center\|400px]]

			From the above results, we can see the upper bound of the last failure is about 955 hours. Therefore, the test probably will last for around 955 hours.

			As we know, with four samples, the median rank for the second failure is 0.385728. Using this value and the assumed Weibull distribution, the median value of the failure time of the 2nd failure is calcualted as:

			<math>\begin{align}
			& Q=1-{{e}^{{{\left( \frac{t}{\eta } \right)}^{\beta }}}}\Rightarrow \\
			& \ln (1-Q)={{\left( \frac{t}{\eta } \right)}^{\beta }} \\
			& \Rightarrow t=\text{349.04}\\
			\end{align}</math>

			Its bounds and other failure times can be calcualted in a similar way.