Template:Alta prob plotting of ed - Revision history

Richard House: Redirected page to Probability Plotting Example

2012-08-16T01:39:25Z

Redirected page to Probability Plotting Example

← Older revision		Revision as of 01:39, 16 August 2012
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	~~==== Probability Plotting ====~~		#REDIRECT [[Probability_Plotting_Example]]

	~~One method of calculating the parameter of the exponential distribution is by using probability plotting. To better illustrate this procedure, consider the following example.<br>~~

	~~===== Example 1 =====~~

	Let's assume six identical units are reliability tested at the same application and operation stress levels. All of these units fail during the test after operating for the following times (in hours), <span class="texhtml">''T''<sub>''i''</sub></span> : 96, 257, 498, 763, 1051 and 1744. <br> The steps for determining the parameters of the exponential ''pdf'' representing the data, using probability plotting, are as follows: <br>

	~~:*Rank the failure times in ascending order, as shown next.~~

	~~<br>~~
	~~<center><math>\begin{matrix}~~
	~~\text{Failure} & \text{Failure Order Number} \\~~
	~~\text{Time (Hr)} & \text{out of a Sample Size of 6} \\~~
	~~\text{96} & \text{1} \\~~
	~~\text{257} & \text{2} \\~~
	~~\text{498} & \text{3} \\~~
	~~\text{763} & \text{4} \\~~
	~~\text{1,051} & \text{5} \\~~
	~~\text{1,744} & \text{6} \\~~
	~~\end{matrix}</math></center>~~
	~~<br>~~

	~~:*Obtain their median rank plotting positions.~~

	~~<br>~~

	~~Median rank positions are used instead of other ranking methods because median ranks are at a specific confidence level (50%).~~

	~~:*The failure times, with their corresponding median ranks, are shown next:~~

	~~<br>~~
	~~<center><math>\begin{matrix}~~
	~~\text{Failure} & \text{Median} \\~~
	~~\text{Time (Hr)} & \text{Rank, }% \\~~
	~~\text{96} & \text{10}\text{.91} \\~~
	~~\text{257} & \text{26}\text{.44} \\~~
	~~\text{498} & \text{42}\text{.14} \\~~
	~~\text{763} & \text{57}\text{.86} \\~~
	~~\text{1,051} & \text{73}\text{.56} \\~~
	~~\text{1,744} & \text{89}\text{.10} \\~~
	~~\end{matrix}</math></center>~~
	~~<br>~~

	:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.

	~~<br>~~ [[~~Image:ALTA4.1.png\|center\|400px~~]]

	~~<br>~~

	~~:*Draw the best possible straight line that goes through the <span class="texhtml">''t'' = 0</span> and <span class="texhtml">(''t'') = 100%</span> point and through the plotted points.~~
	:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)

	<br> [[Image:ALTA4.2.png\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>

	~~==== MLE Parameter Estimation ====~~

	~~The parameter of the exponential distribution can also be estimated using the maximum likelihood estimation (MLE) method. This log-likelihood function is: <br>~~

	::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]</math>

	~~:where:~~

	~~<br>~~

	~~::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>~~

	~~<br>~~

	~~::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math> <br>~~

	~~<br>~~

	~~:and:~~

	~~<br>~~

	~~:*<span class="texhtml">''F''<sub>''e''</sub></span> is the number of groups of times-to-failure data points.~~
	~~:*<span class="texhtml">''N''<sub>''i''</sub></span> is the number of times-to-failure in the <span class="texhtml">''i''<sup>''th''</sup></span> time-to-failure data group.~~
	~~:*<span class="texhtml">λ</span> is the failure rate parameter (unknown a priori, the only parameter to be found).~~
	~~:*<span class="texhtml">''T''<sub>''i''</sub></span> is the time of the <span class="texhtml">''i''<sup>''th''</sup></span> group of time-to-failure data.~~
	~~:*<span class="texhtml">''S''</span> is the number of groups of suspension data points.~~
	~~:*<math>N_{i}^{\prime }</math> is the number of suspensions in the <span class="texhtml">''i''<sup>''th''</sup></span> group of suspension data points.~~
	~~:*<math>T_{i}^{\prime }</math> is the time of the <span class="texhtml">''i''<sup>''t'''h '''''</sup></span> suspension data group.~~
	~~:*<span class="texhtml">''FI''</span> is the number of interval data groups.~~
	~~:*<math>N_{i}^{\prime \prime }</math> is the number of intervals in the <span class="texhtml">''i''<sup>''th''</sup></span> group of data intervals.~~
	~~:*<math>T_{Li}^{\prime \prime }</math> is the beginning of the <span class="texhtml">''i''<sup>''th''</sup></span> interval.~~
	~~:*<math>T_{Ri}^{\prime \prime }</math> is the ending of the <span class="texhtml">''i''<sup>''th''</sup></span> interval.~~

	~~<br> <br> The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where: <br>~~

	::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime }-\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}</math>

	~~<br>~~

	~~===== Example 2 =====~~

	~~Using the same data as in the probability plotting example (Example 1), and assuming an exponential distribution, estimate the parameter using the MLE method. <br>~~

	~~===== Solution =====~~

	~~<br> In this example we have non-grouped data without suspensions. Therefore: <br>~~

	::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=0</math>

	~~<br> Substituting the values for <span class="texhtml">''T''</span> we get: <br>~~

	~~::<math>\begin{align}~~
	~~\frac{6}{\lambda }&= 4409,\text{ or:} \\~~
	~~\lambda& = 0.00136\text{ failure/hr}~~
	~~\end{align}</math>~~

	~~<br>~~

Chris Kahn: /* Example 1 */

2012-06-25T10:06:19Z

Example 1

← Older revision		Revision as of 10:06, 25 June 2012
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	:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.		:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.

	<br> [[Image:ALTA4.1.png\|center\|~~200px~~]]		<br> [[Image:ALTA4.1.png\|center\|400px]]

	<br>		<br>
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	:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)		:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)

	<br> [[Image:ALTA4.2.png\|center\|~~200px~~]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.		<br> [[Image:ALTA4.2.png\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>		To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>

Chuck Smith: /* Example 1 */

2012-04-26T16:02:26Z

Example 1

← Older revision		Revision as of 16:02, 26 April 2012
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	:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.		:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.

	<br> [[Image:ALTA4.1.png\|center\|~~400px~~]]		<br> [[Image:ALTA4.1.png\|center\|200px]]

	<br>		<br>
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	:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)		:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)

	<br> [[Image:ALTA4.2.png\|center\|~~400px~~]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.		<br> [[Image:ALTA4.2.png\|center\|200px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>		To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>

Nicolette Young: /* Example 1 */

2012-03-20T16:45:35Z

Example 1

← Older revision		Revision as of 16:45, 20 March 2012
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	:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.		:*On an exponential probability paper, plot the times on the x-axis and their corresponding rank value on the y-axis. The next figure displays an example of an exponential probability paper. The paper is simply a log-linear paper.

	<br> [[Image:ALTA4.1.~~gif\|thumb~~\|center\|400px]]		<br> [[Image:ALTA4.1.png\|center\|400px]]

	<br>		<br>
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	:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)		:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)

	<br> [[Image:ALTA4.2.~~gif\|thumb~~\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.		<br> [[Image:ALTA4.2.png\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>		To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>

Chris Kahn: /* Example 1 */

2012-03-07T17:36:16Z

Example 1

← Older revision		Revision as of 17:36, 7 March 2012
Line 50:		Line 50:

	:*Draw the best possible straight line that goes through the <span class="texhtml">''t'' = 0</span> and <span class="texhtml">(''t'') = 100%</span> point and through the plotted points.		:*Draw the best possible straight line that goes through the <span class="texhtml">''t'' = 0</span> and <span class="texhtml">(''t'') = 100%</span> point and through the plotted points.
	:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)		:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hours which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)

	<br> [[Image:ALTA4.2.gif\|thumb\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.		<br> [[Image:ALTA4.2.gif\|thumb\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>		To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>

	==== MLE Parameter Estimation ====		==== MLE Parameter Estimation ====

Chris Kahn at 23:12, 6 March 2012

2012-03-06T23:12:54Z

← Older revision		Revision as of 23:12, 6 March 2012
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	::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]</math>		::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]</math>

	:where: <br>		:where:

			<br>

	::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>		::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>
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	::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math> <br>		::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math> <br>

			<br>

	:and:		:and:

			<br>

	:*<span class="texhtml">''F''<sub>''e''</sub></span> is the number of groups of times-to-failure data points.		:*<span class="texhtml">''F''<sub>''e''</sub></span> is the number of groups of times-to-failure data points.

Chris Kahn at 23:11, 6 March 2012

2012-03-06T23:11:55Z

← Older revision		Revision as of 23:11, 6 March 2012
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	::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math> <br>		::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math> <br>


	:and:		:and:

Chris Kahn at 23:11, 6 March 2012

2012-03-06T23:11:21Z

← Older revision		Revision as of 23:11, 6 March 2012
Line 73:		Line 73:

	:*<span class="texhtml">''F''<sub>''e''</sub></span> is the number of groups of times-to-failure data points.		:*<span class="texhtml">''F''<sub>''e''</sub></span> is the number of groups of times-to-failure data points.
	:*<span class="texhtml">''N''<sub>''i''</sub></span> is the number of times-to-failure in the <span class="texhtml">''i''<sup>''~~t'''h '''~~''</sup></span>~~'''~~time-to-failure data group. ~~'''~~		:*<span class="texhtml">''N''<sub>''i''</sub></span> is the number of times-to-failure in the <span class="texhtml">''i''<sup>''th''</sup></span> time-to-failure data group.
	:*<span class="texhtml">λ</span> is the failure rate parameter (unknown a priori, the only parameter to be found).		:*<span class="texhtml">λ</span> is the failure rate parameter (unknown a priori, the only parameter to be found).
	:*<span class="texhtml">''T''<sub>''i''</sub></span> is the time of the <span class="texhtml">''i''<sup>''~~t'''h'''~~''</sup></span>~~'''''~~group of time-to-failure data. ~~''' ''~~		:*<span class="texhtml">''T''<sub>''i''</sub></span> is the time of the <span class="texhtml">''i''<sup>''th''</sup></span> group of time-to-failure data.
	:*<span class="texhtml">''S''</span> is the number of groups of suspension data points.		:*<span class="texhtml">''S''</span> is the number of groups of suspension data points.
	:*<math>N_{i}^{\prime }</math> is the number of suspensions in the <span class="texhtml">''i''<sup>''~~t'''h'''~~''</sup></span>~~'''~~group of suspension data points. ~~'''~~		:*<math>N_{i}^{\prime }</math> is the number of suspensions in the <span class="texhtml">''i''<sup>''th''</sup></span> group of suspension data points.
	:*<math>T_{i}^{\prime }</math> is the time of the <span class="texhtml">''i''<sup>''t'''h '''''</sup></span>~~'''''~~suspension data group. ~~''' ''~~		:*<math>T_{i}^{\prime }</math> is the time of the <span class="texhtml">''i''<sup>''t'''h '''''</sup></span> suspension data group.
	:*<span class="texhtml">''FI''</span> is the number of interval data groups.		:*<span class="texhtml">''FI''</span> is the number of interval data groups.
	:*<math>N_{i}^{\prime \prime }</math> is the number of intervals in the <span class="texhtml">''i''<sup>''~~t'''h'''~~''</sup></span>~~'''~~group of data intervals. ~~'''~~		:*<math>N_{i}^{\prime \prime }</math> is the number of intervals in the <span class="texhtml">''i''<sup>''th''</sup></span> group of data intervals.
	:*<math>T_{Li}^{\prime \prime }</math> is the beginning of the <span class="texhtml">''i''<sup>''~~t'''h'''~~''</sup></span>~~'''~~interval. ~~'''~~		:*<math>T_{Li}^{\prime \prime }</math> is the beginning of the <span class="texhtml">''i''<sup>''th''</sup></span> interval.
	:*<math>T_{Ri}^{\prime \prime }</math> is the ending of the <span class="texhtml">''i''<sup>''~~t'''h'''~~''</sup></span>~~'''~~interval.~~'''~~		:*<math>T_{Ri}^{\prime \prime }</math> is the ending of the <span class="texhtml">''i''<sup>''th''</sup></span> interval.

	<br> <br> The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where: <br>		<br> <br> The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where: <br>

Chris Kahn at 23:09, 6 March 2012

2012-03-06T23:09:16Z

← Older revision		Revision as of 23:09, 6 March 2012
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	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>		To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>

	====MLE Parameter Estimation====		==== MLE Parameter Estimation ====
	The parameter of the exponential distribution can also be estimated using the maximum likelihood estimation (MLE) method. This log-likelihood function is:
	<br>		The parameter of the exponential distribution can also be estimated using the maximum likelihood estimation (MLE) method. This log-likelihood function is: <br>

	::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]</math>		::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]</math>
	~~<br>~~
	:where:		:where: <br>
	<br>
	::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>		::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>

	<br>		<br>
	::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math>
			::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math> <br>

	~~<br>~~
	:and:		:and:
	:• <~~math~~>{{F~~}_{~~e}}</~~math~~> is the number of groups of times-to-failure data points.
	:• <~~math~~>{{N~~}_{~~i}}</~~math~~> is the number of times-to-failure in the <~~math~~>{{i~~}^{th}}~~</~~math~~> time-to-failure data group.		:*<span class="texhtml">''F''<sub>''e''</sub></span> is the number of groups of times-to-failure data points.
	:• <~~math~~>~~\lambda~~ </~~math~~> is the failure rate parameter (unknown a priori, the only parameter to be found).		:*<span class="texhtml">''N''<sub>''i''</sub></span> is the number of times-to-failure in the <span class="texhtml">''i''<sup>''t'''h '''''</sup></span>'''time-to-failure data group. '''
	:• <~~math~~>{{T~~}_{~~i}}</~~math~~> is the time of the <~~math~~>{{i~~}^{th}}~~</~~math~~> group of time-to-failure data.		:*<span class="texhtml">λ</span> is the failure rate parameter (unknown a priori, the only parameter to be found).
	:• <~~math~~>S</~~math~~> is the number of groups of suspension data points.		:*<span class="texhtml">''T''<sub>''i''</sub></span> is the time of the <span class="texhtml">''i''<sup>''t'''h'''''</sup></span>'''''group of time-to-failure data. ''' ''
	:• <math>N_{i}^{\prime }</math> is the number of suspensions in the <~~math~~>{{i~~}^{th}}~~</~~math~~> group of suspension data points.		:*<span class="texhtml">''S''</span> is the number of groups of suspension data points.
	:• <math>T_{i}^{\prime }</math> is the time of the <~~math~~>{{i~~}^{th}}~~</~~math~~> suspension data group.		:*<math>N_{i}^{\prime }</math> is the number of suspensions in the <span class="texhtml">''i''<sup>''t'''h'''''</sup></span>'''group of suspension data points. '''
	:• <~~math~~>FI</~~math~~> is the number of interval data groups.		:*<math>T_{i}^{\prime }</math> is the time of the <span class="texhtml">''i''<sup>''t'''h '''''</sup></span>'''''suspension data group. ''' ''
	:• <math>N_{i}^{\prime \prime }</math> is the number of intervals in the i <~~math~~>~~^{th}~~</~~math~~> group of data intervals.		:*<span class="texhtml">''FI''</span> is the number of interval data groups.
	:• <math>T_{Li}^{\prime \prime }</math> is the beginning of the i <~~math~~>~~^{th}~~</~~math~~> interval.		:*<math>N_{i}^{\prime \prime }</math> is the number of intervals in the <span class="texhtml">''i''<sup>''t'''h'''''</sup></span>'''group of data intervals. '''
	:• <math>T_{Ri}^{\prime \prime }</math> is the ending of the i <~~math~~>~~^{th}~~</~~math~~> interval.		:*<math>T_{Li}^{\prime \prime }</math> is the beginning of the <span class="texhtml">''i''<sup>''t'''h'''''</sup></span>'''interval. '''
	<br>		:*<math>T_{Ri}^{\prime \prime }</math> is the ending of the <span class="texhtml">''i''<sup>''t'''h'''''</sup></span>'''interval.'''
	<br>
	The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where:		<br> <br> The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where: <br>
	<br>
	::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime }-\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}</math>		::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime }-\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}</math>
	~~<br>~~

	=====Example 2=====		<br>
	Using the same data as in the probability plotting example (Example 1), and assuming an exponential distribution, estimate the parameter using the MLE method.
	<br>		===== Example 2 =====

			Using the same data as in the probability plotting example (Example 1), and assuming an exponential distribution, estimate the parameter using the MLE method. <br>

			===== Solution =====

			<br> In this example we have non-grouped data without suspensions. Therefore: <br>

	~~=====Solution=====~~
	~~<br>~~
	~~In this example we have non-grouped data without suspensions. Therefore:~~
	~~<br>~~
	::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=0</math>		::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=0</math>
	<br>
	Substituting the values for <~~math~~>T</~~math~~> we get:		<br> Substituting the values for <span class="texhtml">''T''</span> we get: <br>
	<br>
	::<math>\begin{align}		::<math>\begin{align}
	\frac{6}{\lambda }&= 4409,\text{ or:} \\		\frac{6}{\lambda }&= 4409,\text{ or:} \\
	\lambda& = 0.00136\text{ failure/hr}		\lambda& = 0.00136\text{ failure/hr}
	\end{align}</math>		\end{align}</math>

	<br>		<br>

Chris Kahn at 23:04, 6 March 2012

2012-03-06T23:04:34Z

← Older revision		Revision as of 23:04, 6 March 2012
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	Let's assume six identical units are reliability tested at the same application and operation stress levels. All of these units fail during the test after operating for the following times (in hours), <span class="texhtml">''T''<sub>''i''</sub></span> : 96, 257, 498, 763, 1051 and 1744. <br> The steps for determining the parameters of the exponential ''pdf'' representing the data, using probability plotting, are as follows: <br>		Let's assume six identical units are reliability tested at the same application and operation stress levels. All of these units fail during the test after operating for the following times (in hours), <span class="texhtml">''T''<sub>''i''</sub></span> : 96, 257, 498, 763, 1051 and 1744. <br> The steps for determining the parameters of the exponential ''pdf'' representing the data, using probability plotting, are as follows: <br>

	:*Rank the times~~-to-failure~~ in ascending order, as shown next.		:*Rank the failure times in ascending order, as shown next.

	<br>		<br>
	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	\text{~~Time-to-~~} & \text{Failure Order Number} \\		\text{Failure} & \text{Failure Order Number} \\
	\text{~~Failure~~ (Hr)} & \text{out of a Sample Size of 6} \\		\text{Time (Hr)} & \text{out of a Sample Size of 6} \\
	\text{96} & \text{1} \\		\text{96} & \text{1} \\
	\text{257} & \text{2} \\		\text{257} & \text{2} \\
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	Median rank positions are used instead of other ranking methods because median ranks are at a specific confidence level (50%).		Median rank positions are used instead of other ranking methods because median ranks are at a specific confidence level (50%).

	:*The times~~-to-failure~~, with their corresponding median ranks, are shown next:		:*The failure times, with their corresponding median ranks, are shown next:

	<br>		<br>
	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	\text{~~Time-to-~~} & \text{Median} \\		\text{Failure} & \text{Median} \\
	\text{~~Failure~~ (Hr)} & \text{Rank, }% \\		\text{Time (Hr)} & \text{Rank, }% \\
	\text{96} & \text{10}\text{.91} \\		\text{96} & \text{10}\text{.91} \\
	\text{257} & \text{26}\text{.44} \\		\text{257} & \text{26}\text{.44} \\
Line 52:		Line 52:
	:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)		:*At the <span class="texhtml">''Q''(''t'') = 63.2%</span> or <span class="texhtml">''R''(''t'') = 36.8%</span> ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%</math>.)

	<br> [[Image:ALTA4.2.gif\|thumb\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.		<br> [[Image:ALTA4.2.gif\|thumb\|center\|400px]] <br> <br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 15 hour (or any other time) can now be obtained either from the plot or analytically.

	To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>		To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 15</span> hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''R''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 15) = 98.15%</span> . This can also be obtained analytically, from the exponential reliability function. <br>