Question

Consider the machine-repair model involving two repairment and for machines. Assume that the breakdown rate is λ = 1 machine per hour, and that the service rate is µ = 2 machines per hour per serviceman. Let X(t) be the number of machines broken at time t.

Suppose that you observe at 11A.M. that two of the machines are operating and two of the machines are being repaired.

a) What is the probability that, in the next ¼ hour there will be no change in the state of the system?

b) What is the probability that both of the machines being repaired at 11:00 A.M. will be fixed before any more breakdowns occur?

Answer 1

Answer:

Consider the machine-repair model involving two repairment and for machines. Assume that the breakdown rate is λ = 1 machine per hour and that the service rate is µ = 2 machines per hour per serviceman.

Let X(t) be the number of machines broken at time t.

Suppose that you observe at 11 A.M. that two of the machines are operating and two of the machines are being repaired.

(a).

What is the probability that, in the next ¼ hour there will be no change in the state of the system:

It is given that at 11:00 A.M our random variable X (the number of machines broken) is given by X=2.

Then by the memoryless property of the exponential random variables that make up the "births" and "deaths" associated with this machine repair model, the fact that we have two broken machines makes no difference on T, the random variable representing the sojourn time from this state.

It is given that,

The breakdown rate of a machine is =1 per hour

The service rate is = 2 per hour

When X=2 the sojourn time is given by an exponential random variable with a rate,

=2+4

=6

If F(t) is the cumulative distribution function for such a random variable we have,

=1-F(0.25)

=0.1353

Therefore, the probability that during the next 1/4 hour, there will be no change in the state of the system is 0.1353.

(b).

What is the probability that both of the machines being repaired at 11:00 A.M. will be fixed before any more breakdowns occur:

Since we assume that each machine is currently being worked on by repairman the repair time Ri for each machine i=1,2 is an exponential random variable with a rate of 2.

While the time till the next breakdown, Bi, is an exponential random variable with a rate 1. Now since there are two independent machines that could breakdown, the time to any breakdown, B, is an exponential RV with a rate 2.

Thus, we want to evaluate the probability of the combined event R1 B and R2B. By the independence of these two-component events this joint probability is equal to the product of two individual probabilities.

From previous work on the probability that one exponential random variable will be less than, or happen before another one, we see that this joined event has a probability of

=(2/4)(2/4)

=4/16

=0.25

Therefore, the probability that both machines being repaired at 11 am will be fixed before another one breaks 0.25.