Effective failure rate - Simple situation

Assume that time-to-failure is Weibull distributed with ageing parameter α and mean time to failure, MTTF. Further assume that the item is periodically maintained with time between maintenace equal to τ.

The effective failure rate could then be approximated by:

$ \color{blue}{\lambda_{\mathrm{E}}(\tau) = \left ( \frac{\Gamma(1+1/\alpha)}{\mathrm{MTTF}} \right )^\alpha \tau^{\alpha-1}} $

where Γ(x) = the gamma function, and may be found by Γ(x) = Gamma(x) in MS Excel!

To verify this, remember

• The failure rate function is given by: $ \color{blue}{z(t)=\alpha \lambda^{\alpha} t^{\alpha -1}}$
• $ \color{blue}{\mathrm{MTTF}=\frac{1}{\lambda } \Gamma \left( \frac{1}{\alpha }+1 \right) } $
• For small values of $\tau$ we can take the average value of $z(t)$ and replace $\lambda$ with $\frac{1}{\mathrm{MTTF} } \Gamma \left( \frac{1}{\alpha }+1 \right) $