Ajob shop consists of three machines and two repairmen. the amount of time a machine works before breaking down is exponentially distributed with mean 10. if the amount of time it takes a single repairman to fix a machine is exponentially distributed with mean 8, then(a) what is the average number of machines not in use? (b) what proportion of time are both repairmen busy?

Step-by-step explanation:

Let X(t) denote the number of machines breakdown at time t.

The givenn problem follows birth-death process with finite space

S={0, 1, 2, 3} with

The birth-death process having balance equations

since, state  rate at which leave = rate at which enter

            0      

             1    

             2  

Since

a)

Average number not in use equals the mean of the stationary distribution

b)

Proportion of time both repairmen are busy

part 1 answer = 8

part 2 answer = 72

step-by-step explanation:

part 1 explanation = add all of the numbers, then divide by how many numbers you have. so, 200/25 = 8.

part 2 explanation = not gonna lie, this is hard to explain, so just google how to do it.