Availability Analysis Reference Example
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Availability Analysis Reference Example |
This example validates the results for availability analysis BlockSim.
Reference Case
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.
Data
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.
| Components of the System | Number of Failures, [math]\displaystyle{ \lambda_{f} }[/math] (per hour) |
Number of Repairs, [math]\displaystyle{ \lambda_{r} }[/math] (per hour) |
|---|---|---|
| Component 1 | 0.1 | 0.2 |
| Component 2 | 0.1 | 0.2 |
Result
The point availability is formulated in Equation 11.12 on page 257 in the reference book as:
- [math]\displaystyle{ P_{1}(t) = \frac{r}{\lambda + r} + \frac{\lambda}{\lambda + r}e^{-(\lambda+r)1}\,\! }[/math]
The interval availability is formulated in Equation 11.13 on page 258 in the reference book as:
- [math]\displaystyle{ 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 ]\,\! }[/math]
The system availability for n independent components in series, each having a component availability of [math]\displaystyle{ A_{s}(t)\,\! }[/math], is given Equation 11.15 on page 259 in the reference book as:
- [math]\displaystyle{ A_{s}(t) = \prod _{i-1}^{n}A_{i}(t)\,\! }[/math]
The system availability for n independent components in parallel, each having a component availability of [math]\displaystyle{ A_{s}(t)\,\! }[/math], is given Equation 11.16 on page 259 in the reference book as:
- [math]\displaystyle{ A_{s}(t) = 1- \prod _{i-1}^{n}(1-A_{i}(t))\,\! }[/math]
Plugging in the numbers to the given equations, point and interval availability for a 10 hour mission and steady-state availability are calculated as
For series configuration:
- [math]\displaystyle{ \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}\,\! }[/math]
And for parallel configuration:
- [math]\displaystyle{ \begin{align} A_{s}(10) &= (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}\,\! }[/math]
Simulation Proof
Results in BlockSim
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