= Solution
A <renewal-reward process> consists of i.i.d. regenerative cycles $(\xi_n,R_n)$, where $\xi_n>0$ is the $n$th cycle length and $R_n$ is the reward accumulated during that cycle. If $0<\mathbb E\xi_1<\infty$ and $\mathbb E|R_1|<\infty$, the <renewal-reward theorem> states that the cumulative reward $Z(t)$ satisfies
$$
\frac{Z(t)}t\longrightarrow\frac{\mathbb E R_1}{\mathbb E\xi_1}
$$
almost surely, with the usual negligible-remainder condition for an unfinished cycle.
For the machine, let $X\sim\operatorname{Exp}(\lambda)$ be its working lifetime after a repair. The next repair occurs at the first inspection following failure, so one repair-to-repair cycle has length
$$
\xi=m\left\lceil\frac Xm\right\rceil,
$$
while its working-time reward is $R=X$. Since an <exponential distribution> of rate $\lambda$ has mean $1/\lambda$,
$$
\mathbb ER=\frac1\lambda.
$$
By the <tail-sum formula> for the positive integer-valued variable $\lceil X/m\rceil$,
$$
\mathbb E\left\lceil\frac Xm\right\rceil
=\sum_{k=0}^{\infty}\mathbb P(X>km)
=\sum_{k=0}^{\infty}e^{-\lambda km}
=\frac1{1-e^{-\lambda m}}.
$$
Therefore
$$
\mathbb E\xi=\frac{m}{1-e^{-\lambda m}},
$$
and the <long-run availability under periodic inspection>[long-run proportion of time for which the machine is working] is
$$
\boxed{\frac{\mathbb ER}{\mathbb E\xi}
=\frac{1-e^{-\lambda m}}{\lambda m}}.
$$
Back to article page
