The local grocery store consists of two cashiers. The customers arrive at the checkout according to the Poisson distribution and the service times are based on negative exponential distribution. The average customer service time is 4 minutes and the average time between the arrivals of successive customers is 3 minutes. Assume that customers equally divide themselves between the two cashiers.
-On average, how much time will the customer spend in line waiting to be served?
A) 8 minutes
B) 12 minutes
C) 16 minutes
D) 20 minutes

Question

Quizplus · Accepted Answer

The system can be modeled as an M/M/2 queue (Poisson arrivals, exponential service times, 2 servers). The traffic intensity, $ \rho $, for each server is the ratio of the arrival rate ($ \lambda $) to twice the service rate ($ 2\mu $), since customers are equally divided. With $ \lambda = \frac{1}{3} $ customers per minute and $ \mu = \frac{1}{4} $ services per minute, $ \rho = \frac{1/3}{2/4} = \frac{1}{3} \times \frac{4}{2} = \frac{2}{3} $. The average time a customer spends in the system (waiting time plus service time) in an M/M/2 queue is given by $ T = \frac{1}{\mu - \lambda} $ when $ \rho < 1 $. Here, $ T = \frac{1}{\frac{1}{4} - \frac{1}{6}} = \frac{1}{\frac{1}{12}} = 12 $ minutes. Since this includes both waiting and service times, and the service time is 4 minutes, the average waiting time is $ 12 - 4 = 8 $ minutes.

The Local Grocery Store Consists of Two Cashiers