Question

The average time to serve a customer at a fast-food restaurant is 5 minutes. The standard deviation of the service time is 4 minutes. What is the coefficient of variation of the service time

Answer 1

=13.62

Explanation

- We are given the interarrival time (a = 15 min), service time (p = 20 min), number of servers (m = 3 people), standard deviation of interarrival time (15 min) and standard deviation of service time (60 min). - Therefore, the coefficient of variation of arrival times is 15 / 15 = 1 and the coefficient of variation of service times is 60 / 20 = 3. Moreover, the utilization is 20 / (15 x 3) = 0.4444. Therefore, the average time in the queue is 6.6667 x 0.4086 x 5.0 = 13.6211 minutes, or 13.62 minutes rounded to two decimals

- We are given the interarrival time (a = 15 min), service time (p = 20 min), number of servers (m = 3 people), standard deviation of interarrival time (15 min) and standard deviation of service time (60 min). - Therefore, the coefficient of variation of arrival times is 15 / 15 = 1 and the coefficient of variation of service times is 60 / 20 = 3. Moreover, the utilization is 20 / (15 x 3) = 0.4444. Therefore, the average time in the queue is 6.6667 x 0.4086 x 5.0 = 13.6211 minutes, or 13.62 minutes rounded to two decim