Simple calculation of $Q_0(\tau)$

In the following we show an approach for calculating the probability that the inspection regime will not reveal the potential failure in due time. We denote this probability $Q_0(\tau)$. In this presentation we assume that inspections are perfect, i.e., a potential failure will be revealed with 100% certainty by an inspection. In a more general setting there is some probability that a potential failure is not revealed by an inspection. This is discussed in the course compendium.

We make the following assumptions:

• The system is inspected periodically with a time period of $\tau$ between the inspections
• An inspection will reveal a potential failure with 100% certainty
• The PF interval is assumed to be Weibull distributed with shape (aging) parameter $\alpha$, and expected value $E_\mathrm{PF}$
• Potential failures follow a homogeneous Poisson process, i.e., the time between potential failures are exponentially distributed

If a failure occurs in an interval, assume that it occurs $u$ time units after the last inspection. The potential failure will then not be revealed in due time if the PF-interval is shorter than $\tau-u$. Let $F|u$ denote this event. Since the PF-interval is Weibull distributed we have:

$ \color{blue}{\Pr(F|u) = 1-e^{-\left[\lambda (\tau-u\right]^{\alpha}}} $

where

$ \color{blue}{\lambda =\frac{1}{ E_\mathrm{PF}} \Gamma \left( \frac{1}{\alpha }+1 \right) } $

The time of occurence of the potential failure, $u$, is a stochastic variable which we denote $U$. No values are more likely than the otherr values, hence $U$ is is uniformly distributed on the interval $\left[0,\tau\right]$. It follows from the law of total probability that:

$\color{blue}{Q_0(\tau)=\Pr(F) = \int_0^{\tau} {1 \over \tau} \Pr(F|u) du = \int_0^{\tau} {1 \over \tau} \left [1-e^{-\left[\lambda (\tau-u\right]^{\alpha}}\right] du} $