Question 3. [9 Marks]. Recall that the exponential PDF has the form f (t; λ) = λe−λtwhere t, λ > 0 and λ is known as the rate parameter. Consider a job shop that consists of3 identical machines and 2 technicians. Suppose that, the amount of time each machineoperates before breaking down is exponentially distributed with rate parameter 0.1, and theamount of time a technician takes to fix a machine is exponentially distributed with rateparameter 0.4. Suppose that all the times to breakdown and times to repair are independentrandom variables and let X(t) be the number of machines which are operating at time t.(a) Determine the Q-matrix for this Markov process.(b) Write the forward equations involving the P ′0j (t), j = 0, 1, 2, 3, in terms ofpj (t) := P0j (t) = P(X(t) = j|X(0) = 0).(c) Obtain the equilibrium probabilities pj = limt→∞ pj (t).(d) What is the average number of busy technicians in the long-run? [Note: do not expectyour answer to be an integer.]

This is a complex question that involves understanding of Markov processes, exponential distribution, and probability. Here's how you can approach each part:

(a) The Q-matrix for this Markov process can be determined by considering the rates at which transitions occur between states. The states in this case are the number of machines operating, which can be 0, 1, 2, or 3. The rate of transition from state i to state i+1 is the rate at which a machine breaks down, which is 0.1 per machine. The rate of transition from state i to state i-1 is the rate at which a machine is repaired, which is 0.4 per technician. However, when all machines are operating (state 3), a breakdown can occur but no repair can occur, and when no machines are operating (state 0), a repair can occur but no breakdown can occur. The Q-matrix is therefore:

Q = [ -0.4   0.4   0     0
       0.1  -0.5   0.4   0
       0     0.2  -0.6   0.4
       0     0     0.3  -0.3 ]

(b) The forward equations involving the P'0j(t), j = 0, 1, 2, 3, in terms of pj(t) := P0j(t) = P(X(t) = j|X(0) = 0) can be written by differentiating the probabilities pj(t) with respect to time and setting this equal to the sum of the rates of transition into and out of state j, multiplied by the probabilities of being in those states. This gives a system of differential equations:

P'00(t) = -0.4*P00(t) + 0.1*P01(t)
P'01(t) = 0.4*P00(t) - 0.5*P01(t) + 0.2*P02(t)
P'02(t) = 0.1*P01(t) - 0.6*P02(t) + 0.3*P03(t)
P'03(t) = 0.2*P02(t) - 0.3*P03(t)

(c) The equilibrium probabilities pj = limt→∞ pj(t) can be found by setting the derivatives P'j(t) equal to zero and solving the resulting system of linear equations. This gives:

p0 = 0.4/(0.4 + 0.1) = 0.8
p1 = 0.1*0.8/(0.5 + 0.2) = 0.16
p2 = 0.2*0.16/(0.6 + 0.3) = 0.032
p3 = 0.3*0.032/0.3 = 0.032

(d) The average number of busy technicians in the long-run is the sum of the number of technicians times the probability of that number of technicians being busy, summed over all possible numbers of technicians. Since there are 2 technicians, this is:

E[technicians] = 0*p0 + 1*p1 + 2*p2 + 2*p3 = 0*0.8 + 1*0.16 + 2*0.032 + 2*0.032 = 0.32

So, on average, there are 0.32 technicians busy in the long-run.