How would you go about modelling this as a Markov chain?

There is two machines which break down at different rates, µ$$_A$$ for machine A and µ$$_B$$ for machine B. When they break down, a machine can be fixed by one of two repairmen. Assume that two repairmen, X and Y, have different abilities, and they repair in exponential random times with parameters λ$$_X$$ and λ$$_Y$$ , respectively. (But the rate doesn’t depend on which machine they repair.)

How would you go about modelling this as a Markov chain?.

• If both machines are working, and one fails, what determines which repairman fixes the machine? Nov 23 '20 at 0:08
• It’s random with an equal probability of either repairman fixing the machine.
– user851087
Nov 23 '20 at 0:33

Assume that if both machines are working when one fails, then the repairmen have equal probability of being chosen to fix the machine. The state space is $$S=\{(1,1), (0_X,1), (0_Y,1), (1,0_X), (1,0_Y), (0_X,0_Y), (0_Y,0_X) \}.$$ The transition rates are given by $$q_{(i,j),(i',j')} = \begin{cases} \frac{\mu_A}2,& (i,j)=(1,1)\text{ and } (i',j') \in \{(0_X,1),(0_Y,1)\}\\ \frac{\mu_B}2,& (i,j)=(1,1)\text{ and } (i',j') \in \{(1,0_X),(1,0_Y)\}\\ \lambda_X,& (i,j) \in \{(0_X,1),(1,0_X)\} \text{ and } (i',j') = (1,1)\\ \lambda_Y,& (i,j) \in \{(0_Y,1),(1,0_Y)\} \text{ and } (i',j') = (1,1)\\ \mu_A,& (i,j) = (1,0_X)\text{ and } (i',j') = (0_Y,0_X)\\ \mu_A,& (i,j) = (1,0_Y)\text{ and } (i',j') = (0_X,0_Y)\\ \mu_B,& (i,j) = (0_X,1)\text{ and } (i',j') = (0_X,0_Y)\\ \mu_B,& (i,j) = (0_Y,1)\text{ and } (i',j') = (0_Y,0_X)\\ \lambda_X,& (i,j) = (0_X,0_Y)\text{ and } (i',j') = (1,0_Y)\\ \lambda_X,& (i,j) = (0_Y,0_X)\text{ and } (i',j') = (0_Y,1)\\ \lambda_Y,& (i,j) = (0_X,0_Y)\text{ and } (i',j') = (0_X,1)\\ \lambda_Y,& (i,j) = (0_Y,0_X)\text{ and } (i',j') = (1,0_X)\\ 0,& \text{otherwise.} \end{cases}$$ Let $$Z(t)$$ be the state of the system at time $$t$$, then $$\{Z(t):t\geqslant 0\}$$ is a continuous-time Markov chain with generator matrix $$Q = \small\left( \begin{array}{ccccccc} -\left(\mu _A+\mu _B\right) & \frac{\mu _A}{2} & \frac{\mu _B}{2} & \frac{\mu _A}{2} & \frac{\mu _B}{2} & 0 & 0 \\ \lambda _X & -\left(\mu _B+\lambda _X\right) & 0 & 0 & 0 & \mu _B & 0 \\ \lambda _Y & 0 & -\left(\mu _B+\lambda _Y\right) & 0 & 0 & 0 & \mu _B \\ \lambda _X & 0 & 0 & -\left(\mu _A+\lambda _X\right) & 0 & 0 & \mu _A \\ \lambda _Y & 0 & 0 & 0 & -\left(\mu _B+\lambda _Y\right) & \mu _B & 0 \\ 0 & \lambda _Y & 0 & 0 & \lambda _X & -\left(\lambda _X+\lambda _Y\right) & 0 \\ 0 & 0 & \lambda _X & \lambda _Y & 0 & 0 & -\left(\lambda _X+\lambda _Y\right) \\ \end{array} \right).$$ The process has a unique stationary distribution $$\pi$$ which satisfies $$\pi_{(i,j)} = \lim_{t\to\infty} \mathbb P(Z_t = (i,j))$$ (independent of the distribution of $$Z_0$$). We can find $$\pi$$ by computing the matrix exponential $$e^{Qt}$$ (which is the unique solution to the Kolmogorov backward equation $$P'(t)=QP(t)$$, $$P'(0)=Q$$) and taking any of the rows of $$\lim_{t\to\infty} e^{Qt}$$. More practically, $$\pi$$ satisfies the system of linear equations $$\pi Q=0$$. Note that $$Q$$ is singular (i.e. $$\det Q=0$$) as its rows all sum to zero, so we must replace one of the equations with $$\sum_{(i,j)\in S} \pi_{(i,j)}=1$$. However, due to the size of this matrix and the number of parameters, the closed form solution is a bit unwieldy. For example, I found that $$\pi_{(1,1)} = \tiny\frac{2 \lambda _X \lambda _Y \left(2 \mu _B+\lambda _X+\lambda _Y\right) \left(\mu _A+\mu _B+\lambda _X+\lambda _Y\right)}{\lambda _X^2 \left(\mu _B \left(3 \mu _A+10 \lambda _Y\right)+\left(\mu _A+2 \lambda _Y\right){}^2+6 \mu _B^2\right)+\lambda _X \left(\mu _B \left(7 \mu _A \mu _B+4 \mu _A^2+5 \mu _B^2\right)+\lambda _Y^2 \left(6 \mu _A+8 \mu _B\right)+\lambda _Y \left(\mu _A+3 \mu _B\right) \left(3 \mu _A+4 \mu _B\right)+2 \lambda _Y^3\right)+\left(\mu _B \left(3 \mu _A+4 \lambda _Y\right)+2 \lambda _Y \left(\mu _A+\lambda _Y\right)+\mu _B^2\right) \left(\mu _B \left(\mu _A+\mu _B\right)+\mu _A \lambda _Y\right)+2 \lambda _X^3 \left(\mu _B+\lambda _Y\right)}$$ (the denominator is broken into two lines to prevent page stretching).