Question

Suppose that the customers arrive at a hamburger stand at an average rate of 49 per hour, and the arrivals follow a Poisson distribution. Joe, the stand owner, works alone and takes an average of 0.857 minutes to serve one customer. Assume that the service time is exponentially distributed.

a) What is the average number of people waiting in queue and in the system? (2 points)

b) What is the average time that a customer spends waiting in the queue and in the system? (2 points) Recently, Joe has started receiving complaints regarding long waiting time from customers who want faster service. In response, Joe is considering hiring Jim to work with him at the stand. Assume that Jim can also serve customers at the same rate as Joe. Compute the average time that a customer spends waiting in the queue and in the system for the following cases:

c) There is a single queue at the stand for both servers. (2 points)

d) There are two queues at the stand, each operating independently, and customers are served FCFS within each queue. Assume that the average arrival rate to each queue is half of that in the previous case, i.e., 49/2, and follows a Poisson distribution. (2 points)

e) From a customer’s perspective, which system is better in terms of waiting time in the system (a shared queue for the two servers as in Part c or dedicated queues for each server as in Part d; select from the list in the Answersheet)? (1 point)

Answer 1

This is a M/M/1 queue model

Arrival rate, a = 49 per hour

Service rate, s = 1/(0.857/60) = 70 per hour

a) Average number of people waiting in the queue (Lq) = a²/(s*(s-a)) = 49²/(70*(70-49)) = 1.63

Average number of people in system (L) = a/(s-a) = 49/(70-49) = 2.33

b) Average time a customer spends in the queue (Wq) = Lq/a = 1.63/49 = 0.033 hours =  2 minutes

Average time a customer spends in the system (W) = L/a = 2.33/49 = 0.0476 hours = 2.857 minutes

c) With 2 servers, this is M/M/s queue model with s=2

Average time in queue (Wq) = 0.0020 hours = 0.11966 minutes   (refer cell I13 of above excel sheet)

Average time in system (W) = 0.0163 hours = 0.9768 minutes   (refer cell J13 of above excel sheet)

d) Average arrival rate, a = 49/2 = 24.5 per hour

Average time in queue (Wq) = a/(s*(s-a)) = 24.5/(70*(70-24.5)) = 0.0077 hours = 0.4615 minutes

Average time in system (W) = 1/(s-a) = 1/(70-24.5) = 0.022 hours = 1.32 minutes

e) From a customer perspective, shared queue for two servers is better, because it gives lower waiting time (Wq) and time in system (W)