Availability Analysis Reference Example - Revision history

Kate Racaza at 16:35, 20 August 2015

2015-08-20T16:35:45Z

← Older revision		Revision as of 16:35, 20 August 2015
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	{{Reference_Example_Heading2}}		{{Reference_Example_Heading2}}

	A two-component system’s point and interval availability for a 10 hour mission and the steady-state availability for both series and parallel configurations are calculated. The components share the same failure rate and repair rate distributions. The failure and repair rates both follow exponential distributions with a failure rate of 0.1 failures per hour and a repair rate of 0.2 repairs per hour.		A two-component system’s point and interval availability for a 10 hour mission, and the steady-state availability for both series and parallel configurations are calculated. The components share the same failure rate and repair rate distributions. The failure and repair rates both follow exponential distributions with a failure rate of 0.1 failures per hour and a repair rate of 0.2 repairs per hour.

	{\| {{Table}}		{\| {{Table}}
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	{{Reference_Example_Heading4\|BlockSim}}		{{Reference_Example_Heading4\|BlockSim}}
	~~<!--~~
	In BlockSim, the system RBDs are configured as shown below for series and parallel configurations. Since this is an availability example, we will validate the results via simulation.		In BlockSim, the system RBDs are configured as shown below for series and parallel configurations. Since this is an availability example, we will validate the results via simulation.

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	Each component is modeled using a 1-parameter exponential distribution with the given failure rate values. They also share a corrective task that is modeled using a 1-parameter exponential distribution with the given repair rate values. In this example, we are assuming that the component is fixed upon item failure to as good as new condition. One important point to keep in mind is that the component that is not failed still accumulates time while the corrective task is taking place for the component that failed. The reference book follows this assumption while driving the equations given above.		Each component is modeled using a 1-parameter exponential distribution with the given lambda (failure rate) values. They also share a corrective task that is modeled using a 1-parameter exponential distribution with the given lambda (repair rate) values. In this example, we are assuming that the component is fixed upon item failure to as good as new condition. One important point to keep in mind is that the component that is not failed still accumulates time while the corrective task is taking place for the component that failed. The reference book follows this assumption while driving the equations given above.

	[[Image:availability_properties.png\|center]]		[[Image:availability_properties.png\|center\|650px]]


	The same simulation setup is used for both series and parallel configurations. To estimate the point and interval availability for a 10 hour mission, we use a simulation end time of 10 hours~~. To estimate the steady-state availability, we use a simulation end time of 1,000 hours, since the system is assumed to reach steady state at that time~~. The ~~simulation setup for each case~~ are ~~given as examples below~~.		The same simulation setup is used for both series and parallel configurations. To estimate the point and interval availability for a 10 hour mission, we use a simulation end time of 10 hours. The settings are shown next.

	[[Image:availability_sim.png\|center]]		[[Image:availability_sim.png\|center\|500px]]


			To estimate the steady-state availability, we use a simulation end time of 1,000 hours, since the system is assumed to reach steady state at that time. The settings are shown next.

			[[Image:availability_sim2.png\|center\|500px]]


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	A_{s} &= 0.8893\\		A_{s} &= 0.8893\\
	\end{align}\,\!</math>		\end{align}\,\!</math>

	~~-->~~

Kate Racaza at 18:46, 14 August 2015

2015-08-14T18:46:42Z

← Older revision		Revision as of 18:46, 14 August 2015
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	The system availability for ''n'' independent components in series, each having a component availability of <math> A_{s}(t)\,\!</math>, is given in Equation 11.15 on page 259 in the reference book as:		The system availability for ''n'' independent components in series, each having a component availability of <math> A_{i}(t)\,\!</math>, is given in Equation 11.15 on page 259 in the reference book as:

	::<math>A_{s}(t) = \prod _{i=1}^{n}A_{i}(t)\,\!</math>		::<math>A_{s}(t) = \prod _{i=1}^{n}A_{i}(t)\,\!</math>


	The system availability for ''n'' independent components in parallel, each having a component availability of <math> A_{s}(t)\,\!</math>, is given in Equation 11.16 on page 259 in the reference book as:		The system availability for ''n'' independent components in parallel, each having a component availability of <math> A_{i}(t)\,\!</math>, is given in Equation 11.16 on page 259 in the reference book as:

	::<math>A_{s}(t) = 1- \prod _{i=1}^{n}(1-A_{i}(t))\,\!</math>		::<math>A_{s}(t) = 1- \prod _{i=1}^{n}(1-A_{i}(t))\,\!</math>
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	Each component is modeled using a 1-parameter exponential distribution with the given ~~lambda (~~failure rate) values. They also share a corrective task that is modeled ~~again with~~ 1-parameter exponential distribution with the given ~~lambda (~~repair rate) values. In this example, we are assuming that the component is fixed upon item failure to as good as new condition. One important point to keep in mind is that the component that is not failed still accumulates time while the corrective task is taking place for the component that failed. The reference book follows this assumption while driving the equations given above.		Each component is modeled using a 1-parameter exponential distribution with the given failure rate values. They also share a corrective task that is modeled using a 1-parameter exponential distribution with the given repair rate values. In this example, we are assuming that the component is fixed upon item failure to as good as new condition. One important point to keep in mind is that the component that is not failed still accumulates time while the corrective task is taking place for the component that failed. The reference book follows this assumption while driving the equations given above.

	[[Image:availability_properties.png\|center]]		[[Image:availability_properties.png\|center]]


	The same simulation setup is used for both series and parallel configurations. ~~While estimating~~ the point and interval availability for a 10 hour mission, ~~the~~ simulation end time ~~is used as~~ 10 hours. ~~1000 hours is used for the simulation end time while estimating~~ the steady-state availability since the system is assumed to reach steady state at that time. ~~Simulation setups of the series configuration~~ for each case are given as examples below.		The same simulation setup is used for both series and parallel configurations. To estimate the point and interval availability for a 10 hour mission, we use a simulation end time of 10 hours. To estimate the steady-state availability, we use a simulation end time of 1,000 hours, since the system is assumed to reach steady state at that time. The simulation setup for each case are given as examples below.

	[[Image:availability_sim.png\|center]]		[[Image:availability_sim.png\|center]]

Kate Racaza at 20:34, 3 August 2015

2015-08-03T20:34:04Z

← Older revision		Revision as of 20:34, 3 August 2015
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	{{Reference_Example_Heading2}}		{{Reference_Example_Heading2}}

	A two component system’s point and interval availability for a 10 hour mission and the steady-state availability for both series and parallel configurations are calculated. The components share the same failure rate and repair rate distributions. The failure and repair rates both follow exponential distributions with a failure rate of 0.1 failures per hour and a repair rate of 0.2 repairs per hour.		A two-component system’s point and interval availability for a 10 hour mission and the steady-state availability for both series and parallel configurations are calculated. The components share the same failure rate and repair rate distributions. The failure and repair rates both follow exponential distributions with a failure rate of 0.1 failures per hour and a repair rate of 0.2 repairs per hour.

	{\| {{Table}}		{\| {{Table}}
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	The point availability is formulated in Equation 11.12 on page 257 in the reference book as:		The point availability is formulated in Equation 11.12 on page 257 in the reference book as:

	::<math>P_{1}(t) = \frac{r}{\lambda + r} + \frac{\lambda}{\lambda + r}e^{-(\lambda+r)1}\,\!</math>		::<math>P_{1}(t) = \frac{r}{\lambda + r} + \frac{\lambda}{\lambda + r}e^{-(\lambda+r)t}\,\!</math>


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	The system availability for n independent components in series, each having a component availability of <math> A_{s}(t)\,\!</math>, is given Equation 11.15 on page 259 in the reference book as:		The system availability for ''n'' independent components in series, each having a component availability of <math> A_{s}(t)\,\!</math>, is given in Equation 11.15 on page 259 in the reference book as:

	::<math>A_{s}(t) = \prod _{i-1}^{n}A_{i}(t)\,\!</math>		::<math>A_{s}(t) = \prod _{i=1}^{n}A_{i}(t)\,\!</math>


	The system availability for ''n'' independent components in parallel, each having a component availability of <math> A_{s}(t)\,\!</math>, is given Equation 11.16 on page 259 in the reference book as:		The system availability for ''n'' independent components in parallel, each having a component availability of <math> A_{s}(t)\,\!</math>, is given in Equation 11.16 on page 259 in the reference book as:

	::<math>A_{s}(t) = 1- \prod _{i-1}^{n}(1-A_{i}(t))\,\!</math>		::<math>A_{s}(t) = 1- \prod _{i=1}^{n}(1-A_{i}(t))\,\!</math>


	Plugging in the numbers to the given equations, point and interval availability for a 10 hour mission and steady-state availability are calculated as		Plugging in the numbers to the given equations, the point and interval availability for a 10 hour mission, and steady-state availability are calculated below.

	For series configuration:		For a series configuration:

	::<math>\begin{align}		::<math>\begin{align}
	A_{s}(10) &= (0.684)^{2} = 0.468\\		A_{s}(10) = (0.684)^{2} = 0.468\\
	A_{s,0-10} &= (0.772)^{2} = 0.596\\		A_{s,0-10} = (0.772)^{2} = 0.596\\
	A_{s} &= (0.667)^{2} = 0.445\\		A_{s} = (0.667)^{2} = 0.445\\
	\end{align}\,\!</math>		\end{align}\,\!</math>


	And for parallel configuration:		And for a parallel configuration:

	::<math>\begin{align}		::<math>\begin{align}
	A_{s}(10) &= (1- 0.684)^{2} = 0.900\\		A_{s}(10) = 1- (1- 0.684)^{2} = 0.900\\
	A_{s,0-10} &= 1- (1- 0.772)^{2} = 0.948\\		A_{s,0-10} = 1- (1- 0.772)^{2} = 0.948\\
	A_{s} &= 1- (1- 0.667)^{2} = 0.889\\		A_{s} = 1- (1- 0.667)^{2} = 0.889\\
	\end{align}\,\!</math>		\end{align}\,\!</math>


	~~'''~~Simulation ~~Proof'''~~		{{Reference_Example_Heading4\|BlockSim}}
			<!--
			In BlockSim, the system RBDs are configured as shown below for series and parallel configurations. Since this is an availability example, we will validate the results via simulation.

			[[Image:availability_rbd.png\|center]]


			Each component is modeled using a 1-parameter exponential distribution with the given lambda (failure rate) values. They also share a corrective task that is modeled again with 1-parameter exponential distribution with the given lambda (repair rate) values. In this example, we are assuming that the component is fixed upon item failure to as good as new condition. One important point to keep in mind is that the component that is not failed still accumulates time while the corrective task is taking place for the component that failed. The reference book follows this assumption while driving the equations given above.

			[[Image:availability_properties.png\|center]]


			The same simulation setup is used for both series and parallel configurations. While estimating the point and interval availability for a 10 hour mission, the simulation end time is used as 10 hours. 1000 hours is used for the simulation end time while estimating the steady-state availability since the system is assumed to reach steady state at that time. Simulation setups of the series configuration for each case are given as examples below.

			[[Image:availability_sim.png\|center]]


			The results of the BlockSim simulations for each configuration are given below. The numbers match with the results calculated in the reference book.

			For series configuration:

			::<math>\begin{align}
			A_{s}(10) &= 0.468\\
			A_{s,0-10} &= 0.6019\\
			A_{s} &= 0.4459\\
			\end{align}\,\!</math>


	{{~~Reference_Example_Heading4\|BlockSim~~}}		And for parallel configuration:

			::<math>\begin{align}
			A_{s}(10) &= 0.9000\\
			A_{s,0-10} &= 0.9419\\
			A_{s} &= 0.8893\\
			\end{align}\,\!</math>

	~~Your text here.~~		-->

Kate Racaza: Created page with '{{Reference Example|{{Banner BlockSim Reference Examples}}}} This example validates the results for availability analysis BlockSim. {{Reference_Example_Heading1}} The data se…'

2015-08-03T18:20:26Z

Created page with '{{Reference Example|{{Banner BlockSim Reference Examples}}}} This example validates the results for availability analysis BlockSim. {{Reference_Example_Heading1}} The data se…'

New page

{{Reference Example|{{Banner BlockSim Reference Examples}}}}
This example validates the results for availability analysis BlockSim.

{{Reference_Example_Heading1}}

The data set is from example 11.3 on page 259 in the book, ''An Introduction to Reliability and Maintainability Engineering'', by Dr. Charles E. Ebeling, McGraw-Hill, 1997.

{{Reference_Example_Heading2}}

A two component system’s point and interval availability for a 10 hour mission and the steady-state availability for both series and parallel configurations are calculated. The components share the same failure rate and repair rate distributions. The failure and repair rates both follow exponential distributions with a failure rate of 0.1 failures per hour and a repair rate of 0.2 repairs per hour.

{| {{Table}}
!Components of the System
!Number of Failures, <math>\lambda_{f}</math> <br>(per hour)
!Number of Repairs, <math>\lambda_{r}</math> <br>(per hour)
|-
|Component 1||0.1||0.2
|-
|Component 2||0.1||0.2
|}

{{Reference_Example_Heading3}}

The point availability is formulated in Equation 11.12 on page 257 in the reference book as:

::<math>P_{1}(t) = \frac{r}{\lambda + r} + \frac{\lambda}{\lambda + r}e^{-(\lambda+r)1}\,\!</math>

The interval availability is formulated in Equation 11.13 on page 258 in the reference book as:

::<math>A_{t2-t1} = \frac{r}{\lambda+r} + \frac{\lambda}{(\lambda+r)^{2}(t_{2}-t_{1})}\left [e^{-(\lambda+r)t_{1}} - e^{-(\lambda+r)t_{2}} \right ]\,\!</math>

The system availability for n independent components in series, each having a component availability of <math> A_{s}(t)\,\!</math>, is given Equation 11.15 on page 259 in the reference book as:

::<math>A_{s}(t) = \prod _{i-1}^{n}A_{i}(t)\,\!</math>

The system availability for ''n'' independent components in parallel, each having a component availability of <math> A_{s}(t)\,\!</math>, is given Equation 11.16 on page 259 in the reference book as:

::<math>A_{s}(t) = 1- \prod _{i-1}^{n}(1-A_{i}(t))\,\!</math>

Plugging in the numbers to the given equations, point and interval availability for a 10 hour mission and steady-state availability are calculated as

For series configuration:

::<math>\begin{align}
A_{s}(10) &= (0.684)^{2} = 0.468\\
A_{s,0-10} &= (0.772)^{2} = 0.596\\
A_{s} &= (0.667)^{2} = 0.445\\
\end{align}\,\!</math>

And for parallel configuration:

::<math>\begin{align}
A_{s}(10) &= (1- 0.684)^{2} = 0.900\\
A_{s,0-10} &= 1- (1- 0.772)^{2} = 0.948\\
A_{s} &= 1- (1- 0.667)^{2} = 0.889\\
\end{align}\,\!</math>

'''Simulation Proof'''

{{Reference_Example_Heading4|BlockSim}}

Your text here.