A repairman fixes broken televisions. The repair time is exponentially distributed with a mean of 20 minutes. Broken television sets arrive at his shop according to a Poisson process with arrival rate 12 sets per working day. (8 hours).
(i) What is the fraction of time that the repairman has work to do?
(ii) What is the mean put-through time (waiting time plus repair time) of a television?
(iii) How many television sets, on average, in his shop?
2. The attempt at a solution
Mean arrival rate = 12/8 hours = 12/480 minutes = 0.025 minutes
Mean servicing time = 20 minutes
[tex]\lambda = 3/2[/tex] = Arrival Rate
[tex]\mu = 1/(1/3) = 3[/tex] = Service Rate
[tex]\rho = (3/2)/(3) = 1/2[/tex] = Traffic Density
(i) - Fraction of time that the repairman has work to do.
I'm basing all this on my assumption that it's an M\M\1 queuing system. So the fraction of time which repairman has no work to do is 1 - the average time per customer, T?
[tex]T = N/\lambda = 1/(\mu - \lambda)[/tex] so, 1/(3 - 3/2) = 2/3? So 1/3 of his time he has no work to do?
(ii) - Mean throughput time:
Average waiting time:
[tex]\rho/(\mu(1-\rho)) + 1/\mu[/tex]
(iii) - [tex]\rho/1-\rho[/tex] = [tex](0.5)/(0.5)[/tex] Which is 1...