# Time-Dependent System Reliability for Components in Series

*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): [math]{{\lambda }_{1}} = 0.0002\,\![/math], [math]{{\lambda }_{2}} = 0.0005\,\![/math] and [math]{{\lambda }_{3}} = 0.0001\,\![/math].

- 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

- 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:

- [math]\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}\,\![/math]

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

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

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:

- [math]\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}\,\![/math]

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:

- [math]\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}\,\![/math]

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

- [math]\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}\,\![/math]

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:

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

Thus, the warranty period should be 132 hours.