Template:System failure rate analytical: Difference between revisions
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::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ | ::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ </math> | ||
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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}} \ | ||
\end{align}</math> | \end{align}</math> | ||
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Revision as of 18:07, 7 March 2012
System Failure Rate
Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the [math]\displaystyle{ pdf }[/math] by the reliability function:
- [math]\displaystyle{ {{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ }[/math]
For the system in Figure "Simple two-component system":
- [math]\displaystyle{ \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 "Failure rate function plot of the two component system" shows a plot of the second equation above.
BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as [math]\displaystyle{ 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]\displaystyle{ t }[/math] can get before floating point problems arise. For example, at [math]\displaystyle{ t=5,000,000 }[/math] both numerator and denominator will tend to zero (e.g. [math]\displaystyle{ {{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]\displaystyle{ \tfrac{0}{0} }[/math] or [math]\displaystyle{ \tfrac{X}{0} }[/math] operation. In these cases, BlockSim will return a value of "[math]\displaystyle{ N/A }[/math]" for the result. Obviously, this does not create any practical constraints.