Availability Analysis Reference Example

This example validates the results for availability analysis BlockSim.

The data set is from example 11.3 on page 259 in the book, An Introduction to Reliability and Maintainability Engineering, by Dr. Charles E. Ebeling, McGraw-Hill, 1997.

A two-component system’s point and interval availability for a 10 hour mission and the steady-state availability for both series and parallel configurations are calculated. The components share the same failure rate and repair rate distributions. The failure and repair rates both follow exponential distributions with a failure rate of 0.1 failures per hour and a repair rate of 0.2 repairs per hour.

The point availability is formulated in Equation 11.12 on page 257 in the reference book as:


 * $$P_{1}(t) = \frac{r}{\lambda + r} + \frac{\lambda}{\lambda + r}e^{-(\lambda+r)t}\,\!$$

The interval availability is formulated in Equation 11.13 on page 258 in the reference book as:


 * $$A_{t2-t1} = \frac{r}{\lambda+r} + \frac{\lambda}{(\lambda+r)^{2}(t_{2}-t_{1})}\left [e^{-(\lambda+r)t_{1}} - e^{-(\lambda+r)t_{2}} \right ]\,\!$$

The system availability for n independent components in series, each having a component availability of $$ A_{i}(t)\,\!$$, is given in Equation 11.15 on page 259 in the reference book as:


 * $$A_{s}(t) = \prod _{i=1}^{n}A_{i}(t)\,\!$$

The system availability for n independent components in parallel, each having a component availability of $$ A_{i}(t)\,\!$$, is given in Equation 11.16 on page 259 in the reference book as:


 * $$A_{s}(t) = 1- \prod _{i=1}^{n}(1-A_{i}(t))\,\!$$

Plugging in the numbers to the given equations, the point and interval availability for a 10 hour mission, and steady-state availability are calculated below.

For a series configuration:


 * $$\begin{align}

A_{s}(10) = (0.684)^{2} = 0.468\\ A_{s,0-10} = (0.772)^{2} = 0.596\\ A_{s} = (0.667)^{2} = 0.445\\ \end{align}\,\!$$

And for a parallel configuration:


 * $$\begin{align}

A_{s}(10) = 1- (1- 0.684)^{2} = 0.900\\ A_{s,0-10} = 1- (1- 0.772)^{2} = 0.948\\ A_{s} = 1- (1- 0.667)^{2} = 0.889\\ \end{align}\,\!$$