# Time-Dependent System Reliability for Components in Series

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This example appears in the System Analysis Reference book.

Time-Dependent System Reliability for Components with Constant Failure Rates in Series

Consider a system consisting of 3 exponential units, connected in series, with the following failure rates (in failures per hour): ${{\lambda }_{1}} = 0.0002\,\!$, ${{\lambda }_{2}} = 0.0005\,\!$ and ${{\lambda }_{3}} = 0.0001\,\!$.

• Obtain the reliability equation for the system.
• What is the reliability of the system after 150 hours of operation?
• Obtain the system's pdf.
• Obtain the system's failure rate equation.
• What is the MTTF for the system?
• What should the warranty period be for a 90% reliability?

Solution

The analytical expression for the reliability of the system is given by:

\begin{align} {{R}_{s}}(t)= & {{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot {{R}_{3}}(t) \\ = & {{e}^{-{{\lambda }_{1}}t}}\cdot {{e}^{-{{\lambda }_{2}}t}}\cdot {{e}^{-{{\lambda }_{1}}t}} \\ = & {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \end{align}\,\!

At 150 hours of operation, the reliability of the system is:

\begin{align} {{R}_{s}}(t)= & {{e}^{-(0.0002+0.0005+0.0001)150}} \\ = & 0.8869\text{ or }88.69% \end{align}\,\!

In order to obtain the system's pdf, the derivative of the reliability equation given in the first equation above is taken with respect to time, or:

\begin{align} {{f}_{s}}(t)= & -\frac{d[{{R}_{s}}(t)]}{dt} \\ = & -\frac{d\left[ {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \right]}{dt} \\ = & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}} \end{align}\,\!

The system's failure rate can now be obtained simply by dividing the system's pdf given in the equation above by the system's reliability function given in the first equation above, and:

\begin{align} {{\lambda }_{s}}\left( t \right)= & \frac{{{f}_{s}}\left( t \right)}{{{R}_{s}}\left( t \right)} \\ = & \frac{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})\cdot {{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}}{{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}} \\ = & ({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}) \\ = & 0.0008\text{ }fr/hr \end{align}\,\!

Combining $MTTF=\int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \,\!$ and the first equation above, the system's MTTF can be obtained:

\begin{align} MTTF= & \int_{0}^{\infty }{{R}_{s}}\left( t \right)dt \\ = & \int_{0}^{\infty }{{e}^{-({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})t}}dt \\ = & \frac{1}{({{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}})} \\ = & 1250\text{ }hr \end{align}\,\!

Solving the first equation above with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:

\begin{align} t= & -\frac{\ln (R)}{{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}} \\ = & -\frac{\ln (0.9)}{0.0008} \\ = & 131.7\text{ }hr \end{align}\,\!

Thus, the warranty period should be 132 hours.