4 Temporal stationary probabilities

4.1 Splitting a repair in its availability intervals

An idle period of the repairman ($x=0$) is clearly an availability interval ($y=0$). On the other hand, the $n$-th repair breaks in $k+1$ availability intervals, according to the number $k$ of failures that occur during this repair. Let us denote the inter-arrivals of these failures by $t_{0},\cdots , t_{k-1}$ and put $t_{k}=\tau -\left( t_{0}+\cdots + t_{k-1}\right)$. In other words, the system stays successively with $m-j$ available machines (where $0\leq j\leq k$) during $t_{j}$ seconds.

The inter-arrivals and the non-arrival obey to:

$$\mathrm{Pr}\left( \mathbf{t}_{j}\in \left[ t_{j}, t_{j}+\mathrm{d}t_{j}\right] \right) = \lambda \left( m-j\right) \exp \left( -\lambda \left( m-j\right) t_{j}\right) \mathrm{d}t_{j}$$

$$\mathrm{Pr}\left( T_{q}\left( n \right) \leq T_{a}\left( n+M-m+k \right) \right) = \exp \left( -\lambda \left( m-k\right) t_{k}\right)$$

Taking for example $k=3$, we obtain:

$$pdf_{3}=3! \lambda ^{3} \binom{m}{3}\exp \left( -\lambda \left( 3t_{0}+2t_{1}+t_{2}+\left( m-3\right) \tau \right) \right)$$

Integrating that pdf over $t_{0}+t_{1}+t_{2}\leq \tau$, we obtain again the formerly known $P\left( k \vert m , \tau \right)$. But now, the expectation of any function $\varphi$ of the $t_{j}$ (knowing $k=3$ and $\tau$) follows. We obtain:

$$E_{3}\left( \varphi \right) =6\left( \frac{\lambda \mathbf{e}^... ...0}, t_{1}, t_{2}\right) \mathrm{d}t_{0} \mathrm{d}t_{1} \mathrm{d}t_{2}$$

It should be noticed that $m$ cancels out of that expression. This was an a priori knowledge, since the nttf's of the machines are independent from each other: the machines remaining in function have no effect, even by their number, on the considered conditional expectation.

4.2 Splitting a repair in the M/D case

In the M/D case, any repair lasts $\tau =1/\mu$ and a numerical computation of the preceding formula (taking again $M=4$, $\lambda =0.1$ and $\mu =1$) leads to

$$\left[ .24258, .24741, .25241, .25758\right]$$

for the expected splitting of a repair (knowing that $k=3$). These intervals are slightly increasing: when more machines must break, the waiting time is shorter. But it cannot be too shorter either, since that will provide a sufficient remaining time for one more failure. For the general value of $k$, a somewhat tedious computation [3] leads to:

$$E\left( t_{j} \vert k \right) = \frac{1}{\lambda \left( 1-\mathbf{e}^{\rho }\right) ^{k}}\left( \... ...+\sum _{J}\binom{k}{j}\frac{\left( -\mathbf{e}^{\rho }\right) ^{j}}{j-u}\right)$$ (1)

$$where\quad J=\left\{ j\left\vert 0\leq j\leq k\: ;\: j\neq u\right. \right\}$$

4.3 Comparing to the M/M case

These computations can be repeated for the M/M repairman, involving another integral ($\tau$ is now M-distributed!) and leading first to:

$$P\left( k \vert m \right) =\frac{\lambda ^{k}\mu }{m-k}\times \prod _{j=0}^{k}\frac{m-j}{\left( m-j\right) \lambda +\mu }$$

for the probability of $k$ new failures during a repair started with $m$ available machines. The expectation of any function $\varphi$ of the inter-arrivals of these failures (knowing $k=3$) is now given by:

$$\begin{array}{l} \displaystyle E_{3}\left( \varphi \right) =... ...\mathrm{d}t_{1} \mathrm{d}t_{2} \mathrm{d}\tau \end{array}$$

leading to the well-known expectations

$$E\left( t_{j} \vert k \right) =\frac{1}{\lambda \left( m-j\right) +\mu }$$

The result appears now as independent of $k$ (the PASTA property) but its dependence on $m$ reflects the fact that, for a given repair, $k$ depends on both $\tau$ and $m$.

4.4 From the chain of the edges to the expected sojourn times

Let us now consider the discrete chain whose states are the $M\left( M+1\right) /2$ edges of the repairing chain instead of the $M+1$ (ordinary) states. This new chain is in turn a Markov chain, and the probability of the edge's transition $\left[ a ,b \right] \mapsto \left[ c ,d \right]$ is equal to the probability of the state's transition $c\mapsto d$ when $b=c$ and to zero otherwise, defining the transition matrix $\widehat{\mathrm{M}_{r}}$ of this chain (cf. TAB. 3), and the stationary vector $\mathrm{W}_{\mathrm{r}}$ follows.

Table 3: The chain of the edges of the M/D repairman.

As previously seen, the stationary vector $\mathrm{W}_{\mathrm{r}}$ of this edges's chain, i.e. the vector of the $iPr\left( tr \right)$'s can be directly obtained from $\mathrm{M}_{r}$ (see FIG. 4) and the effective consideration of $\widehat{\mathrm{M}_{r}}$ is not very usefull. On the contrary, the consideration of $\mathrm{W}_{\mathrm{r}}$ leads to the temporal stationary probabilities. Given an edge of the repairing chain, e.g. $\left[ 1 ,2 \right]$ i.e. the transition from $X=1$ to $X=2$, we know that this edge induces sojourns in states $Y=1$, $Y=2$ and $Y=3$ of the availability chain before entering in the $Y=2$ state when the new repair begins. Therefore, we dispose the $E\left( t_{j} \mid k \right)$ obtained in (1) to form a $\left( M+1\right) \times \left( M\left( M+1\right) /2\right)$-sized matrix $\mathrm{U}_{\mathrm{r}}$, the column relative to transition $\left[ a ,b \right]$being built (for $a\neq 0$) with $k=b-a+1$ and starting at line $y=a$ (see TAB. 4).

Table 4: Conversion of edges into sojourn time in the advailability states.

The vector of the expected sojourn times is nothing but $\mathrm{U}_{\mathrm{r}} . \mathrm{W}_{\mathrm{r}}$. The sum of the elements of this vector is obviously $\left( 1-\rho \right) \left( 1/M\lambda \right) +\rho \left( 1/\mu \right) \approx 1.6267$. A renormalization leads to the requested temporal stationary probabilities for the states of the (non Markovian) availability chain. We obtain:

$$\left[ \begin{array}{cccccc} y & 0 & 1 & 2 & 3 & 4\\ tPr\left( y \right) & .64209 & .29952 & .05387 & .00438 & .00013 \end{array}\right]$$

4.5 Back to the availability chain

From the preceding results, we can extract the average transition probabilities for the (non Markovian) availability chain. The relation

$$\mathrm{Pr}\left( Y=j+1 \mid Y=j \right) +\mathrm{Pr}\left( Y=j-1 \mid Y=j \right) =1$$

is obvious and it remains a problem of conditional probabilities.

Let us examplify this problem by considering a sojourn in state $Y=3$ of the availability chain associated to the repairing chain of FIG. 3. That sojourn can occur from the $X$-transitions $\left[ 1 ,2 \right]$, $\left[ 2 ,2 \right]$, $\left[ 3 ,2 \right]$ or from the $X$-transitions $\left[ 1 ,3 \right]$, $\left[ 2 ,3 \right]$, $\left[ 3 ,3 \right]$. In the first case, the next $Y$ state will be $2$ and, in the second, it will be $4$. Therefore the odds are $.01386+.00301+.00064$ to go downwards and $.00048+.00015+.00006$ to go upwards when starting from state $Y=3$. The resulting matrix $\mathrm{M}_{a}$ is pictured in FIG. 5Figure 9 (upper part).

FIG. 5: The M/D availability chain : $M=4$ (top) and $M=5$ (bottom).

Here again, the stationary state-vector $\mathrm{V}_{a}$ and edge-vector $\mathrm{W}_{a}$ can be obtained through the Cesaro's limit of $\left( \mathrm{M}_{a}\right) ^{n}$, disregarding that the availability chain is not Markov (e.g. $\mathrm{M}_{a}^{2}$ does not contain the probabilities for a two-step transition in that chain). But this computation is nevertheless sound, being based upon the general additive properties of the expectations, without any independence hypothesis, and not upon the Chapmann-Kolmogorov formula.

One obtains:

$$\mathrm{V}_{a}=\left[ \begin{array}{cccccc} y & 0 & 1 & 2 & 3 & 4... ...eft( y \right) & .35880 & .48433 & .14058 & .01566 & .00061 \end{array}\right]$$

For checking purposes, we can construct (from matrix $\mathrm{U}_{\mathrm{r}}$) the matrix $\mathrm{U}_{\mathrm{a}}$ of the expected sojourn time in a given $Y$-state knowing it's upward or downward conclusion. It can be checked that normalizing $\mathrm{U}_{\mathrm{a}} . \mathrm{W}_{a}$ leads again to the temporal probabilities for the availability chain.