Question

City Cab, Inc., uses two dispatchers to handle requests for service and to dispatch the cabs. The telephone calls that are made to City Cab use a common telephone number. When both dispatchers are busy, the caller hears a busy signal; no waiting is allowed. Callers who receive a busy signal can call back later or call another cab company for service. Assume that the arrival of calls follows a Poisson distribution, with a mean of 40 calls per hour, and that the call handling time follows an exponential probability distribution with a mean service time of 2 minutes. Based on this information, answer the following questions.

Suppose City Cab added call waiting for the two dispatchers so that at most two calls could be in the hold queue.

      (i) What percentage of the time are both dispatchers idle?

      (ii) What percent of the time are both dispatchers busy?

      (iii)                           What is the probability a caller will receive a busy signal?

      (iv) What is the average number of customers in the system?

      (v) On average, how long does a customer wait before talking to a dispatcher?

Answer 1

Arrival rate lambda = 40 customers per hour

Service rate mu = 60/2 = 30 customers per hour

rho = 40/30 = 1.33

Utilisation of system = 40/30x2 = 0.66

(a) Probability that there are no customer in the system ( both callers are idle)

P(0) = 1/ 1+rho+ [ (rho)² / 2-rho]

P(0) = 1/ 1+1.33+(1.7689)/0.67) = 0.201

(c) Probability that both are busy = 1-0.201 =0.799

(b) Percentage of time the callers are busy = 79.9%.

(d) Avg number of customer is system = lambda x mu x (rho)² xP(0) / (2xmu-lambda)² + rho

Ls = 40x30x 1.7689 x0.201 / ( 2x30-40)² +1.33

= 2.3966

Lq = Avg number of customers in queue = 1.0666

(e) On avg, the waiting time in queue = 1.0666 / lambda = 1.0666 / 40 = 0.0266 hours

= 1.6 min.