Question

1. Consider two machines both of which have an exponential lifetime with rate λ. There is a single repairman that can service machines at an exponential rate μ.

   • – Set up the Kolmogorov backward equations in the matrix format P′(t) = RP(t). You do not need to solve the system.

   • – Find the proportion of time that 0, 1, or 2 machines are down

Answer 1

Machine i operates for an exponential time with rate λi and then fails; its repair time is exponential with rate µi, i = 1,2. The machines act independently of each other.

Consider the state space S = {(0,0),(1,0),(0,1),(1,1)}and denote by Pk (i,j) the probability that machine k = 1,2 makes a transition from i to j, where i,j ∈{0,1} (fail/function). Then by independence

P(i,j),(k,l) = P(X(t) = (k,l)|X(t) = (i,j)) = P1 (i,k)(t)P2 j,l(t)

for all i,j,k,l ∈{0,1}.

To compute the probabilities, we compute for instance P1 00(t) and P1 10.