Template:System failure rate analytical: Difference between revisions

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====System Failure Rate====
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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:
 
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::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)}</math>
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For the system in Figure Ch5fig2:
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::<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>
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Figure 5.4 shows a plot of Eqn. (9).
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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.

Latest revision as of 07:21, 1 August 2012

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