Question

A single server waiting line system has an arrival pattern characterized by a Poisson distribution with 3 customers per hour. The average service times is 12 minutes. the service times are distributed according to the negative exponential distribution.

The expected number of customers in the system is :

A) 3.0 B) 1.5 C) 1.0 D) .90

The expected number of customers waiting in line is:

A) 6 B) 7 C) 8 D) 9

Please show work

Answer 1

Answer:

• Lamba = arrival rate = 3 customer per hour
• Mu = mean service rate = 12 mins per customer = 5 customers per hour

Utilization:

• Utilization = rho = lambda / mu = 3/5 = 60%

Average no. of Customers in waiting in line:

• Average no of customers in the waiting in line =L_q = (rho)² / (1-rho)
• Average no of customers in the waiting in line =L_q = (0.6)^2 /(1-0.6) = 0.9 customers (Answer part 2)

Average time spent in waiting:

• W_q = waiting time in the queue for service = L_q / lambda = 0.9/3 = 0.3 hours

Average time spent in the system:

• W = waiting time in the system = W_q + serving time = 0.3+ 1/5 =0.5 hours

Average no. of customers in system.

• L = no. of customers in the system = W *Lambda = 0.5 *3 = 1.5 customers (Answer Part 1)