Template:System failure rate analytical - Revision history

Lisa Hacker: Replaced content with 'Category: For Deletion'

2012-08-01T07:21:40Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 07:21, 1 August 2012
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	~~====System Failure Rate====~~		[[Category: For Deletion]]
	~~Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the <math>pdf</math> by the reliability function~~:

	~~<br>~~

	~~::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ </math>~~
	~~<br>~~


	~~For the system in Figure "Simple two-component system":~~
	~~<br>~~

	~~::<math>\begin{align}~~
	{{\lambda }_{s}}\left( t \right)= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\
	= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\
	~~= & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\~~
	~~= & {{\lambda }_{1}}+{{\lambda }_{2}} \~~
	~~\end{align}</math>~~
	~~<br>~~

	~~Figure "Failure rate function plot of the two component system" shows a plot of the second equation above.~~
	~~<br>~~

	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.

Dingzhou Cao: /* System Failure Rate */

2012-03-07T18:07:16Z

System Failure Rate

← Older revision		Revision as of 18:07, 7 March 2012
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	<br>		<br>

	::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ ~~(eqn 8)~~</math>		::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ </math>
	<br>		<br>

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	= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\		= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\
	= & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\		= & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\
	= & {{\lambda }_{1}}+{{\lambda }_{2}} \ ~~(eqn 9)~~		= & {{\lambda }_{1}}+{{\lambda }_{2}} \
	\end{align}</math>		\end{align}</math>
	<br>		<br>

Dingzhou Cao: /* System Failure Rate */

2012-03-07T17:01:11Z

System Failure Rate

← Older revision		Revision as of 17:01, 7 March 2012
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	<br>		<br>

	Figure "Failure rate function plot of the two component system" shows a plot of ~~Eqn. (9)~~.		Figure "Failure rate function plot of the two component system" shows a plot of the second equation above.
	<br>		<br>

	BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty </math> , numerical evaluation of ~~Eqn.8~~ 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.

Dingzhou Cao: /* System Failure Rate */

2012-03-06T00:49:17Z

System Failure Rate

← Older revision		Revision as of 00:49, 6 March 2012
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	For the system in Figure ~~Ch5fig2~~:		For the system in Figure "Simple two-component system":
	<br>		<br>

Dingzhou Cao: /* System Failure Rate */

2012-03-06T00:47:35Z

System Failure Rate

← Older revision		Revision as of 00:47, 6 March 2012
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	<br>		<br>

	Figure ~~5.4~~ shows a plot of Eqn. (9).		Figure "Failure rate function plot of the two component system" shows a plot of Eqn. (9).
	<br>		<br>

	BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty </math> , numerical evaluation of Eqn.8 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 Eqn.8 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.

Pengying niu: /* System Failure Rate */

2012-02-09T18:46:28Z

System Failure Rate

← Older revision		Revision as of 18:46, 9 February 2012
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	= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\		= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\
	= & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\		= & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\
	= & {{\lambda }_{1}}+{{\lambda }_{2}}		= & {{\lambda }_{1}}+{{\lambda }_{2}} \ (eqn 9)
	\end{align}</math>		\end{align}</math>
	<br>		<br>
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	<br>		<br>

	BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty </math> , numerical evaluation of Eqn. ~~(system failure rate)~~ 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 Eqn.8 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.

Pengying niu: /* System Failure Rate */

2012-02-09T18:45:24Z

System Failure Rate

← Older revision		Revision as of 18:45, 9 February 2012
Line 4:		Line 4:
	<br>		<br>

	::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)}</math>		::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ (eqn 8)</math>
	<br>		<br>

Pengying niu: /* System Failure Rate */

2012-02-03T22:54:18Z

System Failure Rate

← Older revision		Revision as of 22:54, 3 February 2012
Line 22:		Line 22:
	<br>		<br>

	BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty </math> , numerical evaluation of Eqn. (system failure rate) 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 Eqn. (system failure rate) 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.

Nicolette Young: Created page with '====System Failure Rate==== Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the $pdf$ by the reliability f…'

2012-01-19T18:49:45Z

Created page with '====System Failure Rate==== Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the <math>pdf</math> by the reliability f…'

New page

====System Failure Rate====
Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the <math>pdf</math> by the reliability function:

 

::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)}</math>
 

For the system in Figure Ch5fig2:
 

::<math>\begin{align}
{{\lambda }_{s}}\left( t \right)= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}\cdot {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\
= & \frac{-\tfrac{d}{dt}\left( {{e}^{-\tfrac{1}{10,000}t}} \right)}{{{e}^{-\tfrac{1}{10,000}t}}}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)}{{{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}}} \\
= & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\
= & {{\lambda }_{1}}+{{\lambda }_{2}}
\end{align}</math>
 

Figure 5.4 shows a plot of Eqn. (9).
 

BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as <math>t\to \infty </math> , numerical evaluation of Eqn. (system failure rate) 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.

Template:System failure rate analytical - Revision history

Lisa Hacker: Replaced content with 'Category: For Deletion'

Dingzhou Cao: /* System Failure Rate */

Dingzhou Cao: /* System Failure Rate */

Dingzhou Cao: /* System Failure Rate */

Dingzhou Cao: /* System Failure Rate */

Pengying niu: /* System Failure Rate */

Pengying niu: /* System Failure Rate */

Pengying niu: /* System Failure Rate */

Nicolette Young: Created page with '====System Failure Rate==== Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the pdf by the reliability f…'

Nicolette Young: Created page with '====System Failure Rate==== Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the $pdf$ by the reliability f…'