Time-Dependent System Reliability (Analytical) - Revision history

Sharon Honecker: /* Conditional Reliability */ changed R(T,t) to R(t|T)

2017-02-08T20:09:12Z

Conditional Reliability: changed R(T,t) to R(t|T)

← Older revision		Revision as of 20:09, 8 February 2017
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	Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:		Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:

	::<math>R(T,t)=\frac{R(T+t)}{R(T)}\,\!</math>		::<math>R(t\|T)=\frac{R(T+t)}{R(T)}\,\!</math>

	Equation above gives the reliability for a new mission of duration <math>t\,\!</math> having already accumulated <math>T\,\!</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.		Equation above gives the reliability for a new mission of duration <math>t\,\!</math> having already accumulated <math>T\,\!</math> hours of operation up to the start of this new mission. The system is evaluated to assure that it will start the next mission successfully.

	For the simple two-component system, the reliability for mission of <math>t~~=1,000~~\,\!</math> hours, having an age of <math>T~~=500~~\,\!</math> hours, is:		For the simple two-component system, the reliability for mission of <math>t\,\!</math> = 1000 hours, having an age of <math>T\,\!</math> = 500 hours, is:

	::<math>\begin{align}		::<math>\begin{align}
	{{R}_{S}}(T=500~~,t=1000~~)= & \frac{R(T+t)}{R(T)} \\		{{R}_{S}}(t=1000\|T=500)= & \frac{R(T+t)}{R(T)} \\
	= & \frac{R(1500)}{R(500)} \\		= & \frac{R(1500)}{R(500)} \\
	= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\		= & \frac{{{e}^{-\tfrac{1500}{10,000}}}\cdot {{e}^{-{{\left( \tfrac{1500}{10,000} \right)}^{6}}}}}{{{e}^{-\tfrac{500}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{500}{10,000} \right)}^{6}}}}} \\

Richard House: /* Reliability of Standby Systems with a Switching Device */

2016-01-05T22:57:01Z

Reliability of Standby Systems with a Switching Device

← Older revision		Revision as of 22:57, 5 January 2016
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	[[Image:Fig 5.27.PNG\|center\|600px\|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>]]		[[Image:Fig 5.27.PNG\|center\|600px\|<div align="center"> Standby container (switch) failure distribution while waiting to switch </div>\|link=]]

	Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}\,\!</math> attempt.		Note that there are additional properties that can be specified in BlockSim for a switch, such as Switch Restart Probability, No. of Restarts and Switch Delay Time. In many applications, the switch is re-tested (or re-cycled) if it fails to switch the first time. In these cases, it might be possible that it switches in the second or third, or <math>{{n}^{th}}\,\!</math> attempt.

Richard House: /* Reliability of Standby Systems with a Switching Device */

2016-01-05T22:52:35Z

Reliability of Standby Systems with a Switching Device

← Older revision		Revision as of 22:52, 5 January 2016
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	This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in the figures below. In BlockSim, the standby container is acting as the switch.		This problem can be solved in BlockSim by including these probabilities in the container's properties, as shown in the figures below. In BlockSim, the standby container is acting as the switch.

	[[Image:Fig 5.28_2.PNG\|center\|~~500px~~\|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div> ]]		[[Image:Fig 5.28_2.PNG\|center\|600px\|<div align="center"> Standby container (switch) failure probabilities while attempting to switch </div>\|link=]]

Richard House: /* Simple Standby Configuration */

2016-01-05T22:45:23Z

Simple Standby Configuration

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	The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. The following figures illustrate these pages.		The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. The following figures illustrate these pages.
	[[Image: Fig 5.25.PNG\|center\|600px\|<div align="center"> Defining the active failure distribution </div>\|link=]]		[[Image: Fig 5.25.PNG\|center\|600px\|<div align="center"> Defining the active failure distribution </div>\|link=]]


	[[Image:Fig 5.26.PNG\|center\|600px\|<div align="center"> Defining the quiescent failure distribution </div>\|link=]]		[[Image:Fig 5.26.PNG\|center\|600px\|<div align="center"> Defining the quiescent failure distribution </div>\|link=]]

Richard House: /* Simple Standby Configuration */

2016-01-05T22:44:54Z

Simple Standby Configuration

← Older revision		Revision as of 22:44, 5 January 2016
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	The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. The following figures illustrate these pages.		The active and standby blocks are within a container, which is used to specify standby redundancy. Since the standby component has two distributions (active and quiescent), the Block Properties window of the standby block has two pages for specifying each one. The following figures illustrate these pages.
	[[Image: Fig 5.25.PNG\|center\|~~500px~~\|<div align="center"> Defining the active failure distribution </div>]]		[[Image: Fig 5.25.PNG\|center\|600px\|<div align="center"> Defining the active failure distribution </div>\|link=]]

	[[Image:Fig 5.26.PNG\|center\|~~500px~~\|<div align="center"> Defining the quiescent failure distribution </div>]]		[[Image:Fig 5.26.PNG\|center\|600px\|<div align="center"> Defining the quiescent failure distribution </div>\|link=]]

	The system reliability results for 1000 hours are given in the following table:		The system reliability results for 1000 hours are given in the following table:

Richard House: /* Example */

2016-01-05T21:34:23Z

Example

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	The calculated system reliability at time 1,500 is 0.8569, as given below.		The calculated system reliability at time 1,500 is 0.8569, as given below.
	[[Image:loadsharingresults.png\|center\|~~400px~~\|Calculated System Reliability at Time 1,500]]		[[Image:loadsharingresults.png\|center\|450px\|Calculated System Reliability at Time 1,500\|link=]]

	==Standby Components== <!-- THIS SECTION HEADER IS LINKED TO: Standby_Configuration_Example. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK. -->		==Standby Components== <!-- THIS SECTION HEADER IS LINKED TO: Standby_Configuration_Example. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK. -->

Richard House: /* System Failure Rate */

2016-01-05T20:29:35Z

System Failure Rate

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	The following figure shows a plot of the equation.		The following figure shows a plot of the equation.

	[[Image:BS5.4.png\|center\|~~700px~~\|Failure rate function plot of the two component system\|link=]]		[[Image:BS5.4.png\|center\|650px\|Failure rate function plot of the two component system\|link=]]

	BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty \,\!</math>, numerical evaluation of the first equation above is constrained by machine numerical precision. That is, there are limits as to how large <math>t\,\!</math> can get before floating point problems arise. For example, at <math>t=5,000,000\,\!</math> both numerator and denominator will tend to zero (e.g., <math>{{e}^{-\tfrac{5,000,000}{10,000}}}=7.1245\times {{10}^{-218}}\,\!</math> ). As these numbers become very small they will start looking like a zero to the computer, or cause a floating point error, resulting in a <math>\tfrac{0}{0}\,\!</math> or <math>\tfrac{X}{0}\,\!</math> operation. In these cases, BlockSim will return a value of "<math>N/A\,\!</math>" for the result. Obviously, this does not create any practical constraints.		BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty \,\!</math>, numerical evaluation of the first equation above is constrained by machine numerical precision. That is, there are limits as to how large <math>t\,\!</math> can get before floating point problems arise. For example, at <math>t=5,000,000\,\!</math> both numerator and denominator will tend to zero (e.g., <math>{{e}^{-\tfrac{5,000,000}{10,000}}}=7.1245\times {{10}^{-218}}\,\!</math> ). As these numbers become very small they will start looking like a zero to the computer, or cause a floating point error, resulting in a <math>\tfrac{0}{0}\,\!</math> or <math>\tfrac{X}{0}\,\!</math> operation. In these cases, BlockSim will return a value of "<math>N/A\,\!</math>" for the result. Obviously, this does not create any practical constraints.

Richard House: /* System pdf */

2016-01-05T20:25:46Z

System pdf

← Older revision		Revision as of 20:25, 5 January 2016
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	The next figure shows a plot of the ''pdf'' equation.		The next figure shows a plot of the ''pdf'' equation.

	[[Image:BS5.3.png\|center\|~~700px~~\|pdf plot of the two-component system\|link=]]		[[Image:BS5.3.png\|center\|650px\|pdf plot of the two-component system\|link=]]

	===Conditional Reliability===		===Conditional Reliability===

Richard House: /* Obtaining Other Functions of Interest */

2016-01-05T20:21:36Z

Obtaining Other Functions of Interest

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	Consider the following simple system:		Consider the following simple system:

	[[Image:5-2.png\|center\|~~250px~~\|<div align="center"> Simple two-component system. </div>\|link=]]		[[Image:5-2.png\|center\|200px\|<div align="center"> Simple two-component system. </div>\|link=]]

	Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,\,\!</math> <math>\lambda =1/10,000)\,\!</math> and component 2 follows a Weibull distribution with <math>\beta =6\,\!</math> and <math>\eta =10,000\,\!</math>. The reliability equation of this system is:		Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (<math>\mu =10,000,\,\!</math> <math>\lambda =1/10,000)\,\!</math> and component 2 follows a Weibull distribution with <math>\beta =6\,\!</math> and <math>\eta =10,000\,\!</math>. The reliability equation of this system is:

Richard House: /* Reliability of Standby Systems with a Switching Device */

2016-01-04T23:20:09Z

Reliability of Standby Systems with a Switching Device

← Older revision		Revision as of 23:20, 4 January 2016
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	Solving the analytical solution (as given by the above equation), the following results are obtained.		Solving the analytical solution (as given by the above equation), the following results are obtained.

	[[Image:5-30.png\|center\|~~500px\|~~]]		[[Image:5-30.png\|center\|600px]]

	From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g., switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e., first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.		From the table above, it can be seen that the presence of a switching device has a significant effect on the reliability of a standby system. It is therefore important when modeling standby redundancy to incorporate the switching device reliability properties. It should be noted that this methodology is not the same as treating the switching device as another series component with the standby subsystem. This would be valid only if the failure of the switch resulted in the failure of system (e.g., switch failing open). In equation above, the Switch Probability per Request and quiescent probability are present only in the second term of the equation. Treating these two failure modes as a series configuration with the standby subsystem would imply that they are also present when the active component is functioning (i.e., first term of equation above). This is invalid and would result in the underestimation of the reliability of the system. In other words, these two failure modes become significant only when the active component fails.