Question

The school of business at abc university has a information desk manned by one student worker. It takes him an average of 3 minutes to answer questions raised by guests who visit the school building. Guests arrive at the desk at the rate of 15 per hour. Suppose that arrivals are Poisson and the service times are exponentially distributed.

A. What is the average waiting time in queue?

B. Suppose the worker earns $7.25 per hour. The cost of waiting time, in terms of customer unhappiness with the service received, is $5 per hour of time spent waiting in line. Find the total expected costs over an 8 hour day.

C. Should the school hire another worker? Justify.

Answer 1

Arrivals = Lambda = 15 per hour

Service rate =mu =60/3 =20 per hour

Utilisation rho = lambda /mu =15/20 =0.75

Avg waiting time in queue = lambda / mu ( mu-lambda) = 15 /20x(20-15) =15/100 =0.15 hours =9 min.

Avg customers waiting in the queue = (lambda) x Avg waiting time in queue = 15 x0.15 = 2.25

Cost of waiting time per hour = 2.25 x0.15 x 5 = 1.687

Total expected cost per day = 1.687x8 = 13.5

If a second person is hired he will have to be paid 7.25x8 =58 per day.

Hence a second worker is not necessary, as the cost of waiting does not warrant this.