10) System of m machines (L2007) A factory has m machines. Each machine fails at

Question

10) System of m machines (L2007) A factory has m machines. Each machine fails at an exponential rate when a machine fails, it remains down during a random time Moreover, the machines are independent room one another. Let X(t) be the number of machines that are in working order at time t 20 It can be shown that the stochastic process (x (t),t 2 0) is a birth and death process. a) Find the birth and death rates of the process {x(t), t 0). b) Show that limP[x(t)=n]=PfY=n], where Yis B(mm) binomially distributed.

Explanation / Answer

When population is the number of customers in the system, n and µn indicate that the arrival and service rates depend on the number in the system. Based on the properties of the Poisson process, i.e. when arrivals are in a Poisson process and service times are exponential, we can make the following probability statements for a transition during (t, t + t]. Birth (n 0): P(one birth) = nt + o(t)

P(no birth) = 1 nt + o(t)

P(more than one birth) = o(t)

Death (n > 0): P(one death) = µnt + o(t)

P(no death) = 1 µnt + o(t)

P(more than one dealth = o(t)

where o(t) is such that o(t) t 0 as t 0

In each of the two cases o(t) terms sum to 0 so that the total probability of the three events is equal to 1.

we get Pn,n+1(t) = nt + o(t) n = 0, 1, 2,...

Pn,n1(t) = µnt + o(t) n = 1, 2, 3 ...

Pnn(t)=1 nt µnt + o(t) n = 1, 2, 3 ...

Pnj (t) = o(t), j not equals to n 1, n, n + 1. (1)

In deriving terms on the right hand side of these equations we have made use of the simplification of the type [nt + o(t)][1 µnt + o(t)] = nt + o(t)

[1 nt + o(t)][1 µnt + o(t)] = 1 nt µnt + o(t).

P'0(t) = 0P0(t) + µ1P1(t)

P'n(t) = (n + µn)Pn(t) + n1Pn1(t) +µn+1Pn+1(t) n = 1, 2,... (2)

As a point of digression, we may note that (2) can also be derived directly using (1) without going through the generator matrix as illustrated below. Considering the transitions of the process Q(t) during (t, t + t], we have

P0(t + t) = P0(t)[1 0t + o(t)] + P1(t)[µ, t + o(t)]

Pn(t + t) = Pn(t)[1 nt µnt + o(t)] +Pn1(t)[n1t + o(t)]+Pn+1(t)[n+1t + o(t)] +o(t) n = 1, 2,... (3)

Subtracting Pn(t) (n = 0, 1, 2 ...) from both sides of the appropriate equation in (3) and dividing by t we get (P0(t + t) P0(t))/ t = 0P0(t) + µ1P1(t) + o(t)

(Pn(t + t) Pn(t)) /t = (n + µn)Pn(t) + o(t) /t +n1Pn1(t) + µn+1Pn+1(t) + o(t)/ t

Now (2) follows by letting t 0.

To determine Pn(t) [ Pin(t)], equations (4.1.3) should be solved along with the initial condition Pi(0) = 1, Pn(0) = 0 for n 6= i. Unfortunately even in simple cases such as n = and µn = µ, n = 0, 1, 2, 3 ..., that is when the arrivals are Poisson and service times are exponential (M/M/1 queue) deriving Pn(t) is an arduous process. Furthermore in most of the applications the need for knowing the time dependent behavior is not all that critical. The most widely used result, therefore, is the limiting result, determined from (4.1.3) by letting t .

This theorem essentially confirms what one can think of a state of equilibrium in a stochastic process and how that affects the Kolmogorov equation. In a state of equilibrium, also known as steady state, the behavior of the process is independent of the time parameter and the initial value; i.e. limt Pin(t) = pn n = 0, 1, 2 ...

and therefore P0 n(t) 0 as t

Using these results in (2) we get

0 = 0p0 + µ1p1

0 = (n + µn)pn + n1pn1 + µn+1pn+1 n = 1, 2,... (4)

These equations can be easily solved through recursion. Re-arranging the first equation in (4.1.5) we have

p1 = 0 /µ1 p0 (5)

For n = 1, the second equation gives

(1 + µ1)p1 = 0p0 + µ2p2

Using (4.1.6), this equation reduces to

µ2p2 = 1p1 p2 = (10 /µ2µ1) p0

Continuing such recursion for n = 2, 3,... we get

µnpn = n1pn1 (6)

and therefore,

pn = 01 ...n1 µ1µ2 ...µn p0 (7)

(b) Here we havw seen that it follows poission distribution with parameter (+µ)

and we know that binomial distribution is a limiting form of poission distribution, from that we can prove that Y follows binomial distributin with parameter (m,(/+µ))

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