A company has one repair technician to keep a large group of machines in running order. Treating this group as an infinite calling population, individual breakdowns occur according to a Poisson process at a mean rate of 1 per hour. For each breakdown, the probability is 0.9 that only a minor repair is needed, in which case the repair time has an exponential distribution with a mean of 1

2 hour. Otherwise, a major repair is needed, in which case the repair time has an exponential distribution with a mean of 5 hours. Because both of these conditional distributions are exponential, the unconditional (combined) distribution of repair times is hyperexponential.

(a) Compute the mean and standard deviation of this hyperexponential distribution. [Hint: Use the general relationships from probability theory that, for any random variable X and any pair of mutually exclusive events E1 and E2, E(X) E(XE1)P(E1) 

E(XE2)P(E2) and var(X) E(X2

)  E(X)

2

.] Compare this standard deviation with that for an exponential distribution having this mean.

(b) What are P0, Lq, L, Wq, and W for this queueing system?

(c) What is the conditional value of W, given that the machine involved requires major repair? A minor repair? What is the division of L between machines requiring the two types of repairs? (Hint: Little’s formula still applies for the individual categories of machines.)

(d) How should the states of the system be defined in order to formulate this queueing system as a continuous time Markov chain? (Hint: Consider what additional information must be given, besides the number of machines down, for the conditional distribution of the time remaining until the next event of each kind to be exponential.)

(e) Construct the corresponding rate diagram.