The question:

A facility has three machines and three mechanics. Machines break down at a rate of $1$ per $24$ hours. Breakdown times are exponentially distributed. The time it takes a mechanic to fix a machine is exponentially distributed with mean $6$ hours. Only one mechanic can work on a failed machine at any given time. Let $X_t$ be the number of machines working at time $t$. Find the long-term probability that all machines are working.

My issue is with finding the rate matrix $Q$. Is it as simple as I think? Could I just say it is

$$ Q= \begin{pmatrix} -\frac16&\frac16&0&0\\ \frac1{24}&-\frac5{24}&\frac16&0\\ 0&\frac1{24}&-\frac{5}{24}&\frac16\\ 0&0&\frac1{24}&-\frac{1}{24} \end{pmatrix}\ \ ? $$

My only problem is figuring $Q$ out, please do not provide an answer with the long-term probability. I'd prefer to do that myself. I just want to be sure that this $Q$ is correct, but I've a feeling it isn't considering I never used the fact that they are exponentially distributed. If this isn't the correct approach, would it be better to find the transition probability matrix function $P(t)$ first and then calculate $Q$ from there? If so, any tips on how I'd find this? Thanks.


1 Answer 1


Remember the minimum of $n$ exponentially distributed random variables is in fact another exponentially distributed random variable, with rate the sum of the other rates, i.e. $X_n\sim Exp(\lambda_n) \\ \implies \min\big( X_1,\ldots,X_n \big) \sim Exp\big( \sum_{i=1}^n \lambda_n \big).$

So when the system is in state 0, we have all three machines broken and all three repairmen are working on repairing them, so the first jump to state 1 is the minimum of these three exponential random variables.

  • $\begingroup$ Ohhh. Yes this makes sense. If I understand correctly, the rate of transition from 0 to 1 would be 3/6? Similarly, the rate from 3 to 2 would be 3/24 correct? $\endgroup$ Apr 19, 2020 at 0:31
  • $\begingroup$ That's what I get: $$Q=\pmatrix{-\frac{1}{2}&\frac{1}{2}&0&0\\ \frac{1}{24}&-\frac{3}{8}&\frac{1}{3}&0\\ 0&\frac{1}{12}&-\frac{1}{4}&\frac{1}{6}\\ 0&0&\frac{1}{8}&-\frac{1}{8}}\ . $$ I calculated the third row in detail from first principles. The remaining entries are educated guesses. $\endgroup$ Apr 19, 2020 at 1:08
  • $\begingroup$ Absolutely! State 0: 3 workmen so jump to state 1 with rate $q_{12}= 3/6=1/2 $. Since there is no other state reachable from state 0, the rate of staying in state 0 is $-1/2$, so the row sums to zero. State 1: 2 workmen so jump to state 2 with rate $2/6=1/3$, and one machine that can break at rate $1/24$ to return to state 0 and the diagonal term makes the row sum to zero, ie $-3/8$. State 2: one workman with 2 machines which can break; $\left( 0,1/12,-1/4,1/6\right)$. State 3: all three machines work for a rate of $3/24=1/8$ to drop down, diagonal as before. So you are correct Ionza and John! $\endgroup$
    – foam78
    Apr 21, 2020 at 1:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .