Question

5. 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)

Answer 1

Formula used

= mean arrival rate

= mean service rate

Expected no. of customers in the queue, Lq = ^2/((-))

Expected no. of customers in the system, L = L_q+/

Expected waiting time in the queue, W_q = Lq/

Expected waiting time in the system, W = W_q+1/

For multiple servers,

k= no. of server

P₀ = Probability that there are no customer in the system

P₀ = 1/[1/n!(/)ⁿ]+1/k!(/)^k(k/k-)

Expected no. of customers in the queue, Lq = (/)^kP₀/(k-1)!(k-)^2

Expected no. of customers in the system, L = L_q+/

Expected waiting time in the queue, W_q = Lq/

Expected waiting time in the system, W = W_q+1/