Imperfect Repairs - Revision history

Lisa Hacker at 21:33, 18 September 2023

2023-09-18T21:33:57Z

← Older revision		Revision as of 21:33, 18 September 2023
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	<noinclude>{{Banner BlockSim Articles}}{{Navigation box}}		<noinclude>{{Banner BlockSim Articles}}{{Navigation box}}
	''This article also appears in the [~~[Repairable_Systems_Analysis_Through_Simulation#Imperfect_Repairs\|~~System ~~Analysis Reference~~]~~] book~~.''		''This article also appears in the [https://help.reliasoft.com/reference/system_analysis System analysis reference].''
	</noinclude>		</noinclude>
	===Restoration Factors (RF)===		===Restoration Factors (RF)===

Sharon Honecker: /* Illustrating Type I RF Through an Example */

2017-02-08T19:44:41Z

Illustrating Type I RF Through an Example

← Older revision		Revision as of 19:44, 8 February 2017
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	#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>		#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>
	#Then, the first failure event will be at 500 hours. <br>		#Then, the first failure event will be at 500 hours.<br>
	#After instantaneous repair, the component will begin life with an age after repair of 375 hours ~~<math>~~((1-0.25) ~~\times~~ 500)~~\,\!</math>~~.<br>		#After instantaneous repair, the component will begin life with an age after repair of 375 hours ((1 - 0.25) x 500).<br>
	#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>		#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>
	#The next failure event is now determined using the conditional reliability equation, or: <br>		#The next failure event is now determined using the conditional reliability equation, or: <br>

Sharon Honecker: /* Illustrating Type II RF Through an Example */

2017-02-08T19:43:35Z

Illustrating Type II RF Through an Example

← Older revision		Revision as of 19:43, 8 February 2017
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	:Thus, the next failure event will be at 500 + 126.024 = 626.024 hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.		:Thus, the next failure event will be at 500 + 126.024 = 626.024 hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 hours is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.

	:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or 375 + 126.024 = 501.024 hours.		:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or 375 + 126.024 = 501.024 hours.

Sharon Honecker: /* Illustrating Type II RF Through an Example */

2017-02-08T19:43:17Z

Illustrating Type II RF Through an Example

← Older revision		Revision as of 19:43, 8 February 2017
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	#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>		#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>
	#Then, the first failure event will be at 500 hours. <br>		#Then, the first failure event will be at 500 hours.<br>
	#After instantaneous repair, the component will begin life with an age after repair of 375 ~~hrs <math>~~((1-0.25) ~~\times~~ 500)~~\,\!</math>~~.<br>		#After instantaneous repair, the component will begin life with an age after repair of 375 hours ((1 - 0.25) x 500).<br>
	#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>		#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>
	:#The next failure event is now determined using the conditional reliability equation, or:		:#The next failure event is now determined using the conditional reliability equation, or:

Sharon Honecker: /* Illustrating Type II RF Through an Example */

2017-02-08T19:41:40Z

Illustrating Type II RF Through an Example

← Older revision		Revision as of 19:41, 8 February 2017
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	===Illustrating Type II RF Through an Example===		===Illustrating Type II RF Through an Example===
	Assume that you have a component with a Weibull failure distribution ( <math>\beta ~~=1.5~~\,\!</math>, <math>\eta ~~=1000~~\,\!</math> ~~<math>~~hours~~\,\!</math>~~ ), RF type II = 0.25 and the component undergoes instant repair. Furthermore, assume that the component starts life new (i.e., with a start age of zero). The simulation steps are as follows:		Assume that you have a component with a Weibull failure distribution (<math>\beta\,\!</math> = 1.5, <math>\eta\,\!</math> = 1000 hours), RF type II = 0.25 and the component undergoes instant repair. Furthermore, assume that the component starts life new (i.e., with a start age of zero). The simulation steps are as follows:

	#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>		#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>
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	:Thus, the next failure event will be at ~~<math>~~500+126.024=626.024~~\,\!</math>~~ hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 ~~hrs~~ (500 + 250), where 250 is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.		:Thus, the next failure event will be at 500 + 126.024 = 626.024 hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.

	:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or ~~<math>~~375+126.024=501.024~~\,\!</math>~~ hours.		:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or 375 + 126.024 = 501.024 hours.
	:7. Thus, the age after the second repair will be the restoration factor times the age of the component at failure, or ~~<math>~~(1-0.25) ~~\times~~ (375+126.024)=375.768~~\,\!</math>~~ hours.		:7. Thus, the age after the second repair will be the restoration factor times the age of the component at failure, or (1 - 0.25) x (375 + 126.024) = 375.768 hours.
	:8. Go to Step 4 and repeat the process.		:8. Go to Step 4 and repeat the process.

Sharon Honecker: /* Illustrating Type I RF Through an Example */

2017-02-08T19:38:38Z

Illustrating Type I RF Through an Example

← Older revision		Revision as of 19:38, 8 February 2017
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	===Illustrating Type I RF Through an Example===		===Illustrating Type I RF Through an Example===

	Assume that you have a component with a Weibull failure distribution ( <math>\beta ~~=1.5~~\,\!</math>, <math>\eta ~~=1000~~\,\!</math> hours ), RF type I = 0.25 and the component undergoes instant repair. Furthermore, assume that the component starts life new (i.e., with a start age of zero). The simulation steps are as follows:		Assume that you have a component with a Weibull failure distribution (<math>\beta\,\!</math> = 1.5, <math>\eta\,\!</math> = 1000 hours), RF type I = 0.25 and the component undergoes instant repair. Furthermore, assume that the component starts life new (i.e., with a start age of zero). The simulation steps are as follows:

	#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>		#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>
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	:Thus, the next failure event will be at ~~<math>~~500+126.024=626.024~~\,\!</math>~~ hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 hours is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.		:Thus, the next failure event will be at 500 + 126.024 = 626.024 hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 hours is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.

	:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or ~~<math>~~375+126.024=501.024~~\,\!</math>~~ hours.		:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or 375 + 126.024 = 501.024 hours.
	:7. Thus, the age after the second repair will be the sum of the previous age and the restoration factor times the age of the component since the last failure, or ~~<math>~~375+((1-0.25) ~~\times~~ 126.024)=469.518~~\,\!</math>~~ hours.		:7. Thus, the age after the second repair will be the sum of the previous age and the restoration factor times the age of the component since the last failure, or 375 + ((1-0.25) x 126.024) = 469.518 hours.
	:8. Go to Step 4 and repeat the process.		:8. Go to Step 4 and repeat the process.

Sharon Honecker: /* Discussion of Type I and Type II RFs */ edit for consistency in equations

2016-03-10T21:56:52Z

Discussion of Type I and Type II RFs: edit for consistency in equations

← Older revision		Revision as of 21:56, 10 March 2016
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	As an application example, consider an automotive engine that fails after six years of operation. The engine is rebuilt. The rebuild has the effect of rejuvenating the engine to a condition as if it were three years old (i.e., a 50% RF). Assume that the rebuild affects all of the damage on the engine (i.e., a Type II restoration). The engine fails again after three years (when it again reaches an age of six) and another rebuild is required. This rebuild will also rejuvenate the engine by 50%, thus making it three years old again.		As an application example, consider an automotive engine that fails after six years of operation. The engine is rebuilt. The rebuild has the effect of rejuvenating the engine to a condition as if it were three years old (i.e., a 50% RF). Assume that the rebuild affects all of the damage on the engine (i.e., a Type II restoration). The engine fails again after three years (when it again reaches an age of six) and another rebuild is required. This rebuild will also rejuvenate the engine by 50%, thus making it three years old again.

	Now consider a similar engine subjected to a similar rebuild, but that the rebuild only affects the damage since the last repair (i.e., a Type I restoration of 50%). The first rebuild will rejuvenate the engine to a three-year-old condition. The engine will fail again after three years, but the rebuild this time will only affect the age (of three years) after the first rebuild. Thus the engine will have an age of four and a half years after the second rebuild ( <math>3+~~3\times~~ (1-0.5)=4.5\,\!</math> ). After the second rebuild the engine will fail again after a period of one and a half years and a third rebuild will be required. The age of the engine after the third rebuild will be five years and three months ( <math>4.5+1.5\times (1-0.5)=5.25\,\!</math> ).		Now consider a similar engine subjected to a similar rebuild, but that the rebuild only affects the damage since the last repair (i.e., a Type I restoration of 50%). The first rebuild will rejuvenate the engine to a three-year-old condition. The engine will fail again after three years, but the rebuild this time will only affect the age (of three years) after the first rebuild. Thus the engine will have an age of four and a half years after the second rebuild ( <math>3+(1-0.5) \times 3=4.5\,\!</math> ). After the second rebuild the engine will fail again after a period of one and a half years and a third rebuild will be required. The age of the engine after the third rebuild will be five years and three months ( <math>4.5+(1-0.5) \times 1.5=5.25\,\!</math> ).

	It should be pointed out that when dealing with constant failure rates (i.e., with a distribution such as the exponential), the restoration factor has no effect.		It should be pointed out that when dealing with constant failure rates (i.e., with a distribution such as the exponential), the restoration factor has no effect.

Sharon Honecker: /* Illustrating Type II RF Through an Example */ significant math edits

2016-03-10T21:52:55Z

Illustrating Type II RF Through an Example: significant math edits

← Older revision		Revision as of 21:52, 10 March 2016
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	===Illustrating Type II RF Through an Example===		===Illustrating Type II RF Through an Example===
	Assume that you have a component with a Weibull failure distribution ( <math>\beta =1.5\,\!</math>, <math>\eta =1000\,\!</math> <math>hr\,\!</math> ), RF type II = 0.25 and the component undergoes instant repair. Furthermore, assume that the component starts life new (i.e., with a start age of zero). The simulation steps are as follows:		Assume that you have a component with a Weibull failure distribution ( <math>\beta =1.5\,\!</math>, <math>\eta =1000\,\!</math> <math>hours\,\!</math> ), RF type II = 0.25 and the component undergoes instant repair. Furthermore, assume that the component starts life new (i.e., with a start age of zero). The simulation steps are as follows:

	#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>		#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>
	#~~The~~ first failure event will ~~then~~ be at 500 ~~hrs~~. <br>		#Then, the first failure event will be at 500 hours. <br>
	#After instantaneous repair, the component will begin life with an age after repair of ~~350~~ hrs <math>(~~500\times~~ (1-0.25))\,\!</math>.<br>		#After instantaneous repair, the component will begin life with an age after repair of 375 hrs <math>((1-0.25) \times 500)\,\!</math>.<br>
	#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>		#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>
	:#The next failure event is now determined using the conditional reliability equation, or:		:#The next failure event is now determined using the conditional reliability equation, or:
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	::<math>\begin{align}		::<math>\begin{align}
	R(t+T)= & R(t,T)\cdot R(T) \\		R(x+v)= & R(x,v)\cdot R(v) \\
	R(t+~~350~~)= & 0.8824969\cdot R(~~350~~) \\		R(x+375)= & 0.8824969\cdot R(375) \\
	R(t+~~350~~)= & 0.8824969\cdot 0.~~8129686~~ \\		R(x+375)= & 0.8824969\cdot 0.7948200 \\
	R(t+~~350~~)= & 0.~~71744226~~ \\		R(x+375)= & 0.7014261 \\
	t+~~350~~= & ~~479~~.~~527~~ \\		x+375= & 501.024 \\
	t= & ~~129~~.~~527~~		x= & 126.024
	\end{align}\,\!</math>		\end{align}\,\!</math>


	:Thus, the next failure event will be at <math>500+~~129~~.~~527~~=~~629~~.~~527~~\,\!</math> ~~hrs~~. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hrs (500 + 250), where 250 is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.		:Thus, the next failure event will be at <math>500+126.024=626.024\,\!</math> hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hrs (500 + 250), where 250 is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.

	:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or <math>~~350~~+~~129~~.~~527~~\,\!</math>.		:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or <math>375+126.024=501.024\,\!</math> hours.
	:7. Thus, the age after the second repair will be the restoration factor times the age of the component at failure, or <math>(~~350+129~~.~~527~~)\times (~~1-0~~.25)=~~359~~.~~64525~~\,\!</math> ~~hrs~~.		:7. Thus, the age after the second repair will be the restoration factor times the age of the component at failure, or <math>(1-0.25) \times (375+126.024)=375.768\,\!</math> hours.
	:8. Go to Step 4 and repeat the process.		:8. Go to Step 4 and repeat the process.

Sharon Honecker: /* Illustrating Type I RF Through an Example */ significant math edits

2016-03-10T21:44:56Z

Illustrating Type I RF Through an Example: significant math edits

← Older revision		Revision as of 21:44, 10 March 2016
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	#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>		#Generate a uniform random number, <math>{{U}_{R}}[0,1] = 0.7021885\,\!</math>.<br>
	#~~The~~ first failure event will ~~then~~ be at 500 hours. <br>		#Then, the first failure event will be at 500 hours. <br>
	#After instantaneous repair, the component will begin life with an age after repair of ~~350~~ hours <math>(~~500\times~~ (1-0.25))\,\!</math>.<br>		#After instantaneous repair, the component will begin life with an age after repair of 375 hours <math>((1-0.25) \times 500)\,\!</math>.<br>
	#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>		#Generate another uniform random number, <math>{{U}_{R}}[0,1] = 0.8824969\,\!</math>.<br>
	#The next failure event is now determined using the conditional reliability equation, or: <br>		#The next failure event is now determined using the conditional reliability equation, or: <br>
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	::<math>\begin{align}		::<math>\begin{align}
	R(t+T)= & R(t,T)\cdot R(T) \\		R(x+v)= & R(x,v)\cdot R(v) \\
	R(t+~~350~~)= & 0.8824969\cdot R(~~350~~) \\		R(x+375)= & 0.8824969\cdot R(375) \\
	R(t+~~350~~)= & 0.8824969\cdot 0.~~8129686~~ \\		R(x+375)= & 0.8824969\cdot 0.7948200 \\
	R(t+~~350~~)= & 0.~~71744226~~ \\		R(x+375)= & 0.7014261 \\
	t+~~350~~= & ~~479~~.~~527~~ \\		x+375= & 501.024 \\
	t = & ~~129~~.~~527~~		x = & 126.024
	\end{align}\,\!</math>		\end{align}\,\!</math>


	:Thus, the next failure event will be at <math>500+~~129~~.~~527~~=~~629~~.~~527~~\,\!</math> hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.		:Thus, the next failure event will be at <math>500+126.024=626.024\,\!</math> hours. Note that if the component had been as good as new (i.e., RF = 100%), then the next failure would have been at 750 hours (500 + 250), where 250 hours is the time corresponding to a reliability of 0.8824969, which is the random number that was generated in Step 4.

	:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or <math>~~350~~+~~129~~.~~527~~\,\!</math> hours.		:6. At this failure point, the item's age will now be equal to the initial age, after the first corrective action, plus the additional time it operated, or <math>375+126.024=501.024\,\!</math> hours.
	:7. Thus, the age after the second repair will be the sum of the previous age and the restoration factor times the age of the component since the last failure, or <math>~~350~~+(~~129.527\times~~ (1-0.25))=~~447~~.~~14525~~\,\!</math> hours.		:7. Thus, the age after the second repair will be the sum of the previous age and the restoration factor times the age of the component since the last failure, or <math>375+((1-0.25) \times 126.024)=469.518\,\!</math> hours.
	:8. Go to Step 4 and repeat the process.		:8. Go to Step 4 and repeat the process.

Richard House: /* Calculations to Obtain RFs */

2016-01-06T16:59:03Z

Calculations to Obtain RFs

← Older revision		Revision as of 16:59, 6 January 2016
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	The two types of restoration factors discussed in the previous sections can be calculated using the parametric RDA (Recurrent Data Analysis) tool in Weibull++. This tool uses the GRP (General Renewal Process) model to analyze failure data of a repairable item. More information on the Parametric RDA tool and the GRP (General Renewal Process) model can be found in [[Appendix_B:_References \| [25]]]. As an example, consider the times to failure for an air-conditioning unit of an aircraft recorded in the following table. Assume that each time the unit is repaired, the repair can only remove the damage incurred during the last period of operation. This assumption implies a type I RF factor which is specified as an analysis setting in the Weibull++ folio. The type I RF for the air-conditioning unit can be calculated using the results from Weibull++ shown in the figure below.		The two types of restoration factors discussed in the previous sections can be calculated using the parametric RDA (Recurrent Data Analysis) tool in Weibull++. This tool uses the GRP (General Renewal Process) model to analyze failure data of a repairable item. More information on the Parametric RDA tool and the GRP (General Renewal Process) model can be found in [[Appendix_B:_References \| [25]]]. As an example, consider the times to failure for an air-conditioning unit of an aircraft recorded in the following table. Assume that each time the unit is repaired, the repair can only remove the damage incurred during the last period of operation. This assumption implies a type I RF factor which is specified as an analysis setting in the Weibull++ folio. The type I RF for the air-conditioning unit can be calculated using the results from Weibull++ shown in the figure below.

	[[Image:8.14.png\|center\|727px\|thumb\|Using the Parametric RDA tool in Weibull++ to calculate restoration factors.]]		[[Image:8.14.png\|center\|727px\|thumb\|Using the Parametric RDA tool in Weibull++ to calculate restoration factors.\|link=]]

	[[Image:8.14t.png\|center\|300px\|link=]]		[[Image:8.14t.png\|center\|300px\|link=]]

← Older revision		Revision as of 16:59, 6 January 2016
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	The two types of restoration factors discussed in the previous sections can be calculated using the parametric RDA (Recurrent Data Analysis) tool in Weibull++. This tool uses the GRP (General Renewal Process) model to analyze failure data of a repairable item. More information on the Parametric RDA tool and the GRP (General Renewal Process) model can be found in [[Appendix_B:_References \| [25]]]. As an example, consider the times to failure for an air-conditioning unit of an aircraft recorded in the following table. Assume that each time the unit is repaired, the repair can only remove the damage incurred during the last period of operation. This assumption implies a type I RF factor which is specified as an analysis setting in the Weibull++ folio. The type I RF for the air-conditioning unit can be calculated using the results from Weibull++ shown in the figure below.		The two types of restoration factors discussed in the previous sections can be calculated using the parametric RDA (Recurrent Data Analysis) tool in Weibull++. This tool uses the GRP (General Renewal Process) model to analyze failure data of a repairable item. More information on the Parametric RDA tool and the GRP (General Renewal Process) model can be found in [[Appendix_B:_References \| [25]]]. As an example, consider the times to failure for an air-conditioning unit of an aircraft recorded in the following table. Assume that each time the unit is repaired, the repair can only remove the damage incurred during the last period of operation. This assumption implies a type I RF factor which is specified as an analysis setting in the Weibull++ folio. The type I RF for the air-conditioning unit can be calculated using the results from Weibull++ shown in the figure below.

	[[Image:8.14.png\|center\|727px\|thumb\|Using the Parametric RDA tool in Weibull++ to calculate restoration factors.]]		[[Image:8.14.png\|center\|727px\|thumb\|Using the Parametric RDA tool in Weibull++ to calculate restoration factors.\|link=]]

	[[Image:8.14t.png\|center\|300px\|link=]]		[[Image:8.14t.png\|center\|300px\|link=]]