3 The repairing chain

3.1 Arrivals during a repairing interval of length $\tau$

Each machine has, independently, a probability $\mathbf{e}^{-\lambda \tau }$ of being still available at the end of a time interval of length $\tau$ (knowing that it was available at the beginning of that interval). Therefore, the probability $P\left( k \vert m , \tau \right)$ that $k$ failures occur in a time interval of length $\tau$, knowing that $m$ machines were available at the beginning of the interval and no repair has been completed during that same interval, is given by the binomial law. In other words:

$$P\left( k \vert m , \tau \right) =\binom{m}{k}\left( 1-\ma... ...lambda \tau }\right) ^{k}\left( \mathbf{e}^{-\lambda \tau }\right) ^{m-k}$$

Thus, when the nttr's are D distributed, the probability $P\left( k \vert m \right)$ that $k$ failures occur during a repair that began when $m$ machines were still available is $P\left( k \vert m , 1/\mu \right)$ . In the general case, we have the following formula (where $b$ denotes the pdf of the nttr's):

$$P\left( k \vert m \right) =\int _{0}^{\infty }P\left( k \vert m , \tau \right) b\left( \tau \right) \mathrm{d}\tau$$

3.2 Reverse conditional probabilities (M/D case)

If we consider the generating series :

$$GG_{s}\left( s , z \right) =\sum _{m\in N}\left( z^{m}\sum _{k\leq m}s^{k} P\left( k \vert m \right) \right)$$

and put $s=1$ in the successive derivatives of that series, we obtain the $z$-generating series of the expectations, for a given $m$, of, respectively, $1$, $k$ and $k\left( k-1\right)$. The results, shown in TAB. 1, are nothing but the usual values for a sum of independent variables.

Table 1: Conditionning the number $k$ of failures by the number $m$ of available machines.

A less obvious result is the following. Renormalizing $GG_{s}\left( s , z \right)$ conducts to the generating series  :

$$GG_{z}\left( s , z \right) =\sum _{k\in N}\left( s^{k} \sum _{m\geq k}z^{m} P\left( m \vert k \right) \right)$$

of the distribution of the number $m$ of available machines at the beginning of a repair knowing the number $k$ of breakdowns that occur during that repair. Expectations and variances follow easily, as shown TAB. 2.

Table 2: Conditionning the number of failures ($k$) by the number of available machines ($m$).

3.3 The M/G repairing chain

Let us now break the continuous time in repairing periods (i.e., a repair or an idle period of the repairman). In other words, the states of that embedded discrete chain are beginning just after each repair or idle period, i.e. at each of the $T_{s}\left( n \right)$ and $T_{q}\left( n \right)$ (some of these events being the same). Let us call it the repairing chain (to avoid confusion with the former defined availability chain).

The key point is as follows: this repairing chain remains a Markov chain even in the M/G case. Its states will be labeled by the number $X$ of machines that were broken at the beginning of the state (including the machine being in repair, if any). It should be noticed that $X\in \left[ 0, M-1\right]$, since just after any repair the yet repaired machine is available!

The $X$-transition probabilities are related to the $P\left( k \vert m \right)$. We have :

$$\mathrm{Pr}\left( X=v \vert X=u \right) = P\left( v-u+1 \vert M-u \right) \qquad when\quad 1\leq u\leq v+1$$

$$\mathrm{Pr}\left( X=1 \vert X=0 \right) = 1$$

$$\mathrm{Pr}\left( X=v \vert X=u \right) = 0\qquad otherwise$$

FIG. 3 illustrates the M/D case with $M=4$, $\lambda =0.1$ and $\mu =1$.

FIG. 3: The M/D repairing chain.

Since the state $X=1$ is directly connected to any other state, the matrix $\mathrm{M}_{r}$ of that chain is convergent and its stationary vector $\mathrm{V}_{r}$ can be obtained as any column of the iterated squares of $\mathrm{M}_{r}$, leading to

$$\mathrm{V}_{r}=\left[ \begin{array}{ccccc} x & 0 & 1 & 2 & 3\\ iPr\left( x \right) & .41779 & .56396 & .01752 & .00071 \end{array}\right]$$

FIG. 4 gives the stationary probabilities of the edges of the $\mathrm{M}_{r}$ chain, obtained again as: $iPr\left( a\mapsto b \right) =iPr\left( a \right) \times \mathrm{Pr}\left( a \vert b \right)$. Here again, these probabilities are conditioned by an uniform sorting over the discrete chain.

FIG. 4: Stationary probabilities of the M/D availability edges.