Template:System failure rate analytical

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 function:


 * $${{\lambda }_{s}}\left( t \right)=\frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)}  \ (eqn 8)$$

For the system in Figure "Simple two-component system":


 * $$\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)}+\frac{-\tfrac{d}{dt}\left( {{e}^{-{{\left( \tfrac{t}{10,000} \right)}^{6}}}} \right)} \\ = & \frac+\frac \\ = & {{\lambda }_{1}}+{{\lambda }_{2}}  \ (eqn 9) \end{align}$$

Figure "Failure rate function plot of the two component system" shows a plot of Eqn. (9).

BlockSim uses numerical methods to estimate the failure rate. It should be pointed out that as $$t\to \infty $$, numerical evaluation of Eqn.8 is constrained by machine numerical precision. That is, there are limits as to how large $$t$$  can get before floating point problems arise. For example, at $$t=5,000,000$$  both numerator and denominator will tend to zero (e.g.  $${{e}^{-\tfrac{5,000,000}{10,000}}}=7.1245\times {{10}^{-218}}$$ ). 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 $$\tfrac{0}{0}$$  or  $$\tfrac{X}{0}$$  operation. In these cases, BlockSim will return a value of "$$N/A$$" for the result. Obviously, this does not create any practical constraints.