# markov chain application

Two workers handle three machines(i.e. we can at most repair two machines at a time). The time until the machine breaks down is exponentially distributed with expectation value 1/2 and independent of other repair and break down times. Every repair time is exponentially distributed with expectation value 1/3 and independent of other repair and break down times. Let $X_t$ be the amount of broken down machines at time $t$.

1. Determine the birth and death frequencies in the birth and death chain $(X_t)_{t\geq 0}$.
2. Determine $\lim_{t \to \infty} \mathbb{P}_i (X_t=0)$ for i=0,1,2,3.

I need some explain aboute the question like this for compute the birth and death rate as a markov chain in continues time

The system can be in three states $\{0,1,2,3\}$ which correspond to the number of working machines. A single machine is repaired at rate 3, so when both workers are repairing they repair at rate 6 (but only when two or more machines are broken). Machines break at rate 2 independently so when 2 or 3 are active the failure rates are 4 and 6 respectively.
Therefore the $Q$ matrix is $$Q = \begin{pmatrix} -6 & 6 & 0 & 0 \\ 2 & -8 & 6 & 0 \\ 0 & 4 & -7 & 3 \\ 0 & 0 & 6 & -6 \end{pmatrix}$$
and the stationary distribution (the limiting distribution as $t\rightarrow \infty$) is $$\pi = \begin{pmatrix} \frac{4}{43} & \frac{12}{43} & \frac{18}{43} & \frac{9}{43} \end{pmatrix}.$$
• No, this model isn't an M/M/$c$ queue because the arrival rate isn't constant in all states. For a finite capacity M/M/$c$ queue the arrival rate is $\lambda$ in states $\{0,1,2,\ldots,N-1\}$, but that isn't the case here. The arrival rate (repair rate) is 3 or 6 depending on the state. Please 'accept' my answer if it answers your question. Thanks! – Gareth May 29 '13 at 15:28