The owner of an amusement park has decided to open a second entry booth and hire another employee to service customers entering the park.Customers arrive at the rate of 20 per hour and will wait in a single line until one of the two employees is available to provide service.The average service time of both employees is 2 minutes to provide service.T The arrival rate follows Poisson distribution,and the service time follows a negative exponential distribution.Determine the average time the customer spends in the queue.
A)2.25 minutes
B)3 minutes
C)0.25 minutes
D)1 minute

Question

Quizplus · Accepted Answer

The arrival rate is 20 customers per hour (λ = 20), and the service rate for each employee is 30 customers per hour (μ = 30, since 60/2 = 30). With two employees, the total service rate is 60 customers per hour. Using the formula for average time in the queue for an M/M/2 system, Wq = λ / (2μ(2μ - λ)), we get Wq = 20 / (60(60 - 20)) = 0.25 minutes.

The Owner of an Amusement Park Has Decided to Open