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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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| <br>
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| ::<math>{{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \ (eqn 8)</math>
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| <br>
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| For the system in Figure Ch5fig2:
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| <br>
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| ::<math>\begin{align}
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| {{\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}}}}} \\
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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}}}}} \\
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| = & \frac{{{f}_{1}}}{{{R}_{1}}}+\frac{{{f}_{2}}}{{{R}_{2}}} \\
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| = & {{\lambda }_{1}}+{{\lambda }_{2}} \ (eqn 9)
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| \end{align}</math>
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| <br>
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| Figure 5.4 shows a plot of Eqn. (9).
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| <br>
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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.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.
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