Failure Discounting - Revision history

Lisa Hacker at 20:39, 18 September 2023

2023-09-18T20:39:04Z

← Older revision		Revision as of 20:39, 18 September 2023
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	{{Template:~~RGA_BOOK_SUB~~\|Appendix A\|Failure Discounting}}		{{Template:RGA_BOOK\|Appendix A\|Failure Discounting}}
	During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References\|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. However, certain questions arise, such as to what extent should the failure value be diminished or discounted, and how should the failure value be defined? One answer would be to use engineering judgment (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore, that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis needs to be selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, logistic, and the standard and modified Gompertz models.		During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References\|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. However, certain questions arise, such as to what extent should the failure value be diminished or discounted, and how should the failure value be defined? One answer would be to use engineering judgment (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore, that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis needs to be selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, logistic, and the standard and modified Gompertz models.

David J. Groebel at 17:14, 20 June 2014

2014-06-20T17:14:54Z

← Older revision		Revision as of 17:14, 20 June 2014
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	The value of the failure, <math>f\,\!</math>, is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:		The value of the failure, <math>f\,\!</math>, is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:

	::<math>f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\!</math>		:<math>f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\!</math>

	where:		where:
Line 10:		Line 10:
	*<math>CL\,\!</math> is the confidence level.		*<math>CL\,\!</math> is the confidence level.
	*<math>{{S}_{n}}\,\!</math> is the number of successful tests after the first success following the corrective action.		*<math>{{S}_{n}}\,\!</math> is the number of successful tests after the first success following the corrective action.


	For example, after one successful test following a corrective action, <math>{{S}_{n}}=1\,\!</math>, the failure is given a value of 0.9 based on a 90% confidence level. After two successful tests, <math>{{S}_{n}}=2\,\!</math>, the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.		For example, after one successful test following a corrective action, <math>{{S}_{n}}=1\,\!</math>, the failure is given a value of 0.9 based on a 90% confidence level. After two successful tests, <math>{{S}_{n}}=2\,\!</math>, the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.


	'''Example'''		'''Example'''
	{{:Failure_Discounting_Example}}		{{:Failure_Discounting_Example}}

Kate Racaza at 20:04, 31 January 2014

2014-01-31T20:04:20Z

Show changes

Athanasios Gerokostopoulos at 20:06, 27 November 2013

2013-11-27T20:06:29Z

← Older revision		Revision as of 20:06, 27 November 2013
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	:The next figure shows the entered data and the estimated parameters.		:The next figure shows the entered data and the estimated parameters.

	[[Image:rgaA.1.png\|thumb\|center\|~~400px~~\|Entered data and the estimated Standard Gompertz parameters.]]		[[Image:rgaA.1.png\|thumb\|center\|450px\|Entered data and the estimated Standard Gompertz parameters.]]


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	:The predicted reliability after launch 22 is calculated using the Quick Calculation Pad (QCP), as shown next.		:The predicted reliability after launch 22 is calculated using the Quick Calculation Pad (QCP), as shown next.

	[[Image:rgaA.2.png\|thumb\|center\|~~300px~~\|Predicted reliability after launch 22.]]		[[Image:rgaA.2.png\|thumb\|center\|450px\|Predicted reliability after launch 22.]]


	3) In the second table, the predicted reliability values are compared with the reliabilities that are calculated from the raw data using failure discounting. It can be seen in the table, and in the following figure, that the Gompertz curve appears to provide a good fit to the actual data.		3) In the second table, the predicted reliability values are compared with the reliabilities that are calculated from the raw data using failure discounting. It can be seen in the table, and in the following figure, that the Gompertz curve appears to provide a good fit to the actual data.

	[[Image:rgaA.3.png\|thumb\|center\|~~300px~~\|Standard Gompertz reliability growth curve.]]		[[Image:rgaA.3.png\|thumb\|center\|450px\|Standard Gompertz reliability growth curve.]]

Kate Racaza at 17:31, 4 October 2012

2012-10-04T17:31:22Z

← Older revision		Revision as of 17:31, 4 October 2012
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	::<math>\begin{align}		::<math>\begin{align}
	c =& \left ( \frac{S_{3}-S_{2}}{S_{2}-S_{1}} \right )^\frac{1}{n\cdot I} \\		c &= \left ( \frac{S_{3}-S_{2}}{S_{2}-S_{1}} \right )^\frac{1}{n\cdot I} \\
	=& \left [ \frac{26.946-26.776}{26.776-22.218} \right ]^\frac{1}{6} \\		&= \left [ \frac{26.946-26.776}{26.776-22.218} \right ]^\frac{1}{6} \\
	=& 0.578 \\		&= 0.578 \\
	\end{align}\,\!</math>		\end{align}\,\!</math>


	::<math>\begin{align}		::<math>\begin{align}
	a =& e^\left [\frac{1}{n}\left (S_{1} + \frac {S_{2}-S_{1}}{1-e^{n\cdot I}} \right )\right ] \\		a &= e^\left [\frac{1}{n}\left (S_{1} + \frac {S_{2}-S_{1}}{1-e^{n\cdot I}} \right )\right ] \\
	=& e^\left [\frac{1}{6}\left (22.218 + \frac{26.776 - 22.218}{1-0.578^{6}}\right ) \right ] \\		&= e^\left [\frac{1}{6}\left (22.218 + \frac{26.776 - 22.218}{1-0.578^{6}}\right ) \right ] \\
	=& 89.31% \\		&= 89.31% \\
	\end{align}\,\!</math>		\end{align}\,\!</math>


	::<math>\begin{align}		::<math>\begin{align}
	b =& e^\left [\frac{(S_{2}-S_{1})(c-1)}{(1-c^{n})^{2}} \right ] \\		b &= e^\left [\frac{(S_{2}-S_{1})(c-1)}{(1-c^{n})^{2}} \right ] \\
	=& e^\left [\frac{(26.776-22.218)(0.578-1)}{(1-0.578^{6})^{2}} \right ] \\		&= e^\left [\frac{(26.776-22.218)(0.578-1)}{(1-0.578^{6})^{2}} \right ] \\
	=& 0.127 \\		&= 0.127 \\
	\end{align}\,\!</math>		\end{align}\,\!</math>

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	::<math>\begin{align}		::<math>\begin{align}
	\widehat{a}= & 0.9299 \\		\widehat{a}&= 0.9299 \\
	\widehat{b}= & 0.0943 \\		\widehat{b} &= 0.0943 \\
	\widehat{c}= & 0.7170		\widehat{c} &= 0.7170
	\end{align}\,\!</math>		\end{align}\,\!</math>

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	::<math>\begin{align}		::<math>\begin{align}
	R= & 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\		R &= 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\
	= & 0.9260		&= 0.9260
	\end{align}\,\!</math>		\end{align}\,\!</math>

Kate Racaza at 17:28, 4 October 2012

2012-10-04T17:28:44Z

@@ Line 195: / Line 195: @@
 :Calculate the initial values of the Gompertz parameters using the second table above. Based on the equations from the [[Gompertz Models]] chapter, the initial values are:
-<INSERT EQUATION>
+::<math>\begin{align}
+c =& \left ( \frac{S_{3}-S_{2}}{S_{2}-S_{1}} \right )^\frac{1}{n\cdot I} \\
 :Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=89.31,\,\!</math> <math>g_{2}^{(0)}=0.127,\,\!</math> <math>g_{3}^{(0)}=0.578\,\!</math>. The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:

Kate Racaza at 17:10, 4 October 2012

2012-10-04T17:10:24Z

Show changes

Kate Racaza at 17:04, 25 September 2012

2012-09-25T17:04:15Z

← Older revision		Revision as of 17:04, 25 September 2012
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	{{~~template~~:~~RGA BOOK~~\|A\|Failure ~~Discounting~~}}		{{Template:RGA_BOOK_SUB\|Appendix A\|Failure Discouting}}
	During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References\|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. But certain questions arise, such as, To what extent should the failure value be diminished or discounted? and How should the failure value be defined? One answer would be to use engineering judgment, (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis is selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, Logistic and the Standard and Modified Gompertz models.		During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References\|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. But certain questions arise, such as, To what extent should the failure value be diminished or discounted? and How should the failure value be defined? One answer would be to use engineering judgment, (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis is selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, Logistic and the Standard and Modified Gompertz models.

Richard House at 21:13, 12 September 2012

2012-09-12T21:13:09Z

← Older revision		Revision as of 21:13, 12 September 2012
Line 2:		Line 2:
	During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References\|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. But certain questions arise, such as, To what extent should the failure value be diminished or discounted? and How should the failure value be defined? One answer would be to use engineering judgment, (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis is selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, Logistic and the Standard and Modified Gompertz models.		During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References\|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. But certain questions arise, such as, To what extent should the failure value be diminished or discounted? and How should the failure value be defined? One answer would be to use engineering judgment, (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis is selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, Logistic and the Standard and Modified Gompertz models.

	The value of the failure, <math>f\,\!</math> , is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:		The value of the failure, <math>f\,\!</math>, is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:

	::<math>f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}</math>		::<math>f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\!</math>

	:where:		:where:
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	::*<math>{{S}_{n}}\,\!</math> is the number of successful tests after the first success following the corrective action.		::*<math>{{S}_{n}}\,\!</math> is the number of successful tests after the first success following the corrective action.

	For example, after one successful test following a corrective action, <math>{{S}_{n}}=1\,\!</math> , the failure is given a value of 0.9 based on a 90% confidence level. After two successful tests, <math>{{S}_{n}}=2\,\!</math> , the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.		For example, after one successful test following a corrective action, <math>{{S}_{n}}=1\,\!</math>, the failure is given a value of 0.9 based on a 90% confidence level. After two successful tests, <math>{{S}_{n}}=2\,\!</math>, the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.

	'''Example'''		'''Example'''
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	\|9\|\| 69.783\|\| 4.245\|\| 66.496		\|9\|\| 69.783\|\| 4.245\|\| 66.496
	\|-		\|-
	\| \|\| \|\| <math>{{S}_{1}}\,\!</math> = 22.218 \|\|		\| \|\| \|\| <math>{{S}_{1}}\,\!</math> = 22.218 \|\|
	\|-		\|-
	\|10\|\| 78.029 \|\|4.357 \|\|73.044		\|10\|\| 78.029 \|\|4.357 \|\|73.044
Line 124:		Line 124:
	\|15\|\| 92.573\|\| 4.528\|\| 88.754		\|15\|\| 92.573\|\| 4.528\|\| 88.754
	\|-		\|-
	\| \|\| \|\| <math>{{S}_{2}}\,\!</math> = 26.776\|\|		\| \|\| \|\| <math>{{S}_{2}}\,\!</math> = 26.776\|\|
	\|-		\|-
	\|16\|\| 93.660\|\| 4.540\|\| 89.923		\|16\|\| 93.660\|\| 4.540\|\| 89.923
Line 148:		Line 148:

	:*Each failure is represented by a single column and named Failure 1, Failure 2 etc. The failure mode for each failure is shown in the Result/Mode column where is also indicated in the launch is a success.		:*Each failure is represented by a single column and named Failure 1, Failure 2 etc. The failure mode for each failure is shown in the Result/Mode column where is also indicated in the launch is a success.
	:*The values of failure are calculated from the equation above and are based on <math>CL=0.90\,\!</math> .		:*The values of failure are calculated from the equation above and are based on <math>CL=0.90\,\!</math>.
	:*These values are summed and the reliability is calculated from:		:*These values are summed and the reliability is calculated from:

	::<math>R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%</math>		::<math>R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%\,\!</math>

	where <math>N\,\!</math> is the number of failures and <math>n\,\!</math> is the number of events, tests, runs or launches.		where <math>N\,\!</math> is the number of failures and <math>n\,\!</math> is the number of events, tests, runs or launches.

	:*Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process.		:*Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process.

	So at launch 3, <math>{{S}_{n}}=1\,\!</math> , and so on.		So at launch 3, <math>{{S}_{n}}=1\,\!</math>, and so on.

	:*Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore, <math>{{S}_{n}}=1\,\!</math> at launch 4 and at launch 7, and so on.		:*Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore, <math>{{S}_{n}}=1\,\!</math> at launch 4 and at launch 7, and so on.
	:*Failure 3 is Mode 3; it occurs at launch 3 and it recurs at launch 6. Therefore, <math>{{S}_{n}}=1\,\!</math> at launch 5 and at launch 8, and so on.		:*Failure 3 is Mode 3; it occurs at launch 3 and it recurs at launch 6. Therefore, <math>{{S}_{n}}=1\,\!</math> at launch 5 and at launch 8, and so on.
	:*Failure 6 is Mode 4; it occurs at launch 17 and it does not recur throughout the process. So at launch 19, <math>{{S}_{n}}=1\,\!</math> , and so on.		:*Failure 6 is Mode 4; it occurs at launch 17 and it does not recur throughout the process. So at launch 19, <math>{{S}_{n}}=1\,\!</math>, and so on.
	:*Failure 7 is Mode 5; it occurs at launch 19 and it does not recur throughout the process. So at launch 21, <math>{{S}_{n}}=1\,\!</math> , and so on.		:*Failure 7 is Mode 5; it occurs at launch 19 and it does not recur throughout the process. So at launch 21, <math>{{S}_{n}}=1\,\!</math>, and so on.

	For launch 3 and failure 1, <math>{{S}_{n}}=1\,\!</math>.		For launch 3 and failure 1, <math>{{S}_{n}}=1\,\!</math>.

	::<math>\begin{align}		::<math>\begin{align}
	{{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900		{{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900
	\end{align}</math>		\end{align}\,\!</math>

	For launch 4 and failure 1, <math>{{S}_{n}}=2\,\!</math>.		For launch 4 and failure 1, <math>{{S}_{n}}=2\,\!</math>.

	::<math>\begin{align}		::<math>\begin{align}
	{{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684		{{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684
	\end{align}</math>		\end{align}\,\!</math>

	And so on.		And so on.
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	<INSERT EQUATION>		<INSERT EQUATION>

	Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=89.31,\,\!</math> <math>g_{2}^{(0)}=0.127,\,\!</math> <math>g_{3}^{(0)}=0.578\,\!</math> . The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:		Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=89.31,\,\!</math> <math>g_{2}^{(0)}=0.127,\,\!</math> <math>g_{3}^{(0)}=0.578\,\!</math>. The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:

	::<math>\begin{align}		::<math>\begin{align}
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	\widehat{b}= & 0.0943 \\		\widehat{b}= & 0.0943 \\
	\widehat{c}= & 0.7170		\widehat{c}= & 0.7170
	\end{align}</math>		\end{align}\,\!</math>

	The next figure shows the entered data and the estimated parameters.		The next figure shows the entered data and the estimated parameters.
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	[[Image:rgaA.1.png\|thumb\|center\|400px\|Entered data and the estimated Standard Gompertz parameters.]]		[[Image:rgaA.1.png\|thumb\|center\|400px\|Entered data and the estimated Standard Gompertz parameters.]]

	The Gompertz reliability growth curve may now be written as follows where <math>{{L}_{G}}\,\!</math> is the number of launches with the first successful launch being counted as <math>{{L}_{G}}=1\,\!</math> . Therefore, <math>{{L}_{G}}\,\!</math> is equal to 19, since reliability growth starts with launch 4.		The Gompertz reliability growth curve may now be written as follows where <math>{{L}_{G}}\,\!</math> is the number of launches with the first successful launch being counted as <math>{{L}_{G}}=1\,\!</math>. Therefore, <math>{{L}_{G}}\,\!</math> is equal to 19, since reliability growth starts with launch 4.

	::<math>R=0.9299{{(0.0943)}^{{{0.7170}^{{{L}_{G}}}}}}</math>		::<math>R=0.9299{{(0.0943)}^{{{0.7170}^{{{L}_{G}}}}}}\,\!</math>

	:2) The predicted reliability after launch 22 is:		:2) The predicted reliability after launch 22 is:
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	R= & 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\		R= & 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\
	= & 0.9260		= & 0.9260
	\end{align}</math>		\end{align}\,\!</math>

	The predicted reliability after launch 22 is calculated using the Quick Calculation Pad and is shown next.		The predicted reliability after launch 22 is calculated using the Quick Calculation Pad and is shown next.

Richard House at 21:18, 11 September 2012

2012-09-11T21:18:57Z

← Older revision		Revision as of 21:18, 11 September 2012
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	!Gompertz Reliability (%)		!Gompertz Reliability (%)
	\|-		\|-
	\|1\|\| 0.000		\|1\|\| 0.000\|\| \|\|
	\|-		\|-
	\|2\|\| 0.000		\|2\|\| 0.000\|\| \|\|
	\|-		\|-
	\|3\|\| 3.333\|\| 1.204		\|3\|\| 3.333\|\| 1.204\|\|
	\|-		\|-
	\|4\|\| 35.406\|\| 3.567\|\| 16.426		\|4\|\| 35.406\|\| 3.567\|\| 16.426
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	\|9\|\| 69.783\|\| 4.245\|\| 66.496		\|9\|\| 69.783\|\| 4.245\|\| 66.496
	\|-		\|-
	\| \|\| \|\| <math>{{S}_{1}}</math> = 22.218		\| \|\| \|\| <math>{{S}_{1}}\,\!</math> = 22.218 \|\|
	\|-		\|-
	\|10\|\| 78.029 \|\|4.357 \|\|73.044		\|10\|\| 78.029 \|\|4.357 \|\|73.044
Line 124:		Line 124:
	\|15\|\| 92.573\|\| 4.528\|\| 88.754		\|15\|\| 92.573\|\| 4.528\|\| 88.754
	\|-		\|-
	\| \|\| \|\| <math>{{S}_{2}}</math> = 26.776		\| \|\| \|\| <math>{{S}_{2}}\,\!</math> = 26.776\|\|
	\|-		\|-
	\|16\|\| 93.660\|\| 4.540\|\| 89.923		\|16\|\| 93.660\|\| 4.540\|\| 89.923
Line 138:		Line 138:
	\|21\|\| 89.820\|\| 4.498\|\| 92.396		\|21\|\| 89.820\|\| 4.498\|\| 92.396
	\|-		\|-
	\| \|\| \|\| <math>{{S}_{3}}</math> = 26.946		\| \|\| \|\| <math>{{S}_{3}}\,\!</math> = 26.946\|\|
	\|-		\|-
	\|22\|\| 91.896 4.521 92.565		\|22\|\| 91.896 4.521 92.565\|\| \|\|
	\|}		\|}

	'''Solution'''		'''Solution'''
	~~<br>~~

	:1) The first table above is organized as follows:		:1) The first table above is organized as follows:

	:*Each failure is represented by a single column and named Failure 1, Failure 2 etc. The failure mode for each failure is shown in the Result/Mode column where is also indicated in the launch is a success.		:*Each failure is represented by a single column and named Failure 1, Failure 2 etc. The failure mode for each failure is shown in the Result/Mode column where is also indicated in the launch is a success.
	:*The values of failure are calculated from the equation above and are based on <math>CL=0.90</math> .		:*The values of failure are calculated from the equation above and are based on <math>CL=0.90\,\!</math> .
	:*These values are summed and the reliability is calculated from:		:*These values are summed and the reliability is calculated from:

	::<math>R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%</math>		::<math>R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%</math>

	where <math>N</math> is the number of failures and <math>n</math> is the number of events, tests, runs or launches.		where <math>N\,\!</math> is the number of failures and <math>n\,\!</math> is the number of events, tests, runs or launches.

	:*Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process.		:*Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process.

	So at launch 3, <math>{{S}_{n}}=1</math> , and so on.		So at launch 3, <math>{{S}_{n}}=1\,\!</math> , and so on.

	:*Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore, <math>{{S}_{n}}=1</math> at launch 4 and at launch 7, and so on.		:*Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore, <math>{{S}_{n}}=1\,\!</math> at launch 4 and at launch 7, and so on.
	:*Failure 3 is Mode 3; it occurs at launch 3 and it recurs at launch 6. Therefore, <math>{{S}_{n}}=1</math> at launch 5 and at launch 8, and so on.		:*Failure 3 is Mode 3; it occurs at launch 3 and it recurs at launch 6. Therefore, <math>{{S}_{n}}=1\,\!</math> at launch 5 and at launch 8, and so on.
	:*Failure 6 is Mode 4; it occurs at launch 17 and it does not recur throughout the process. So at launch 19, <math>{{S}_{n}}=1</math> , and so on.		:*Failure 6 is Mode 4; it occurs at launch 17 and it does not recur throughout the process. So at launch 19, <math>{{S}_{n}}=1\,\!</math> , and so on.
	:*Failure 7 is Mode 5; it occurs at launch 19 and it does not recur throughout the process. So at launch 21, <math>{{S}_{n}}=1</math> , and so on.		:*Failure 7 is Mode 5; it occurs at launch 19 and it does not recur throughout the process. So at launch 21, <math>{{S}_{n}}=1\,\!</math> , and so on.

	For launch 3 and failure 1, <math>{{S}_{n}}=1</math>.		For launch 3 and failure 1, <math>{{S}_{n}}=1\,\!</math>.

	::<math>{{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900</math>		::<math>\begin{align}
			{{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900
			\end{align}</math>

	For launch 4 and failure 1, <math>{{S}_{n}}=2</math>.		For launch 4 and failure 1, <math>{{S}_{n}}=2\,\!</math>.

	::<math>{{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684</math>		::<math>\begin{align}
			{{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684
			\end{align}</math>

	And so on.		And so on.
Line 179:		Line 182:
	<INSERT EQUATION>		<INSERT EQUATION>

	Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting <math>{{Y}_{i}}={{R}_{i}},</math> <math>g_{1}^{(0)}=89.31,</math> <math>g_{2}^{(0)}=0.127,</math> <math>g_{3}^{(0)}=0.578</math> . The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:		Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting <math>{{Y}_{i}}={{R}_{i}},\,\!</math> <math>g_{1}^{(0)}=89.31,\,\!</math> <math>g_{2}^{(0)}=0.127,\,\!</math> <math>g_{3}^{(0)}=0.578\,\!</math> . The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:

	::<math>\begin{align}		::<math>\begin{align}
	& \widehat{a}= & 0.9299 \\		\widehat{a}= & 0.9299 \\
	& \widehat{b}= & 0.0943 \\		\widehat{b}= & 0.0943 \\
	& \widehat{c}= & 0.7170		\widehat{c}= & 0.7170
	\end{align}</math>		\end{align}</math>

Line 191:		Line 194:
	[[Image:rgaA.1.png\|thumb\|center\|400px\|Entered data and the estimated Standard Gompertz parameters.]]		[[Image:rgaA.1.png\|thumb\|center\|400px\|Entered data and the estimated Standard Gompertz parameters.]]

	The Gompertz reliability growth curve may now be written as follows where <math>{{L}_{G}}</math> is the number of launches with the first successful launch being counted as <math>{{L}_{G}}=1</math> . Therefore, <math>{{L}_{G}}</math> is equal to 19, since reliability growth starts with launch 4.		The Gompertz reliability growth curve may now be written as follows where <math>{{L}_{G}}\,\!</math> is the number of launches with the first successful launch being counted as <math>{{L}_{G}}=1\,\!</math> . Therefore, <math>{{L}_{G}}\,\!</math> is equal to 19, since reliability growth starts with launch 4.

	::<math>R=0.9299{{(0.0943)}^{{{0.7170}^{{{L}_{G}}}}}}</math>		::<math>R=0.9299{{(0.0943)}^{{{0.7170}^{{{L}_{G}}}}}}</math>


	:2) The predicted reliability after launch 22 is:		:2) The predicted reliability after launch 22 is:

	::<math>\begin{align}		::<math>\begin{align}
	& R= & 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\		R= & 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\
	& = & 0.9260		= & 0.9260
	\end{align}</math>		\end{align}</math>