Question

For the following problems compute (a) utilization, (b) average time a customer waits in the queue, (c) average number of customers waiting in the queue, (d) average number of customers in service, (e) the average time a customer spends in the system.

Problem 1. An average of 10 cars per hour (with variance 4) arrives at a single-server drive-in teller. Assume that the average service time for each customer is 5.5 minutes (with variance 5).

Problem 2. Customers arrive to a bank based on a Poisson process with hourly rate 5. There is one server in the system and the service time is exponentially distributed with mean 11 minutes.
Hint 1: The variance of a Poisson process with rate is .

Hint 2: The standard deviation of an exponential distribution with mean is .

Answer 1

1-

It is a G/G/1 queueing system. In order to solve this, we often use an approximation method.

We have the following information

Arrival rate (lambda) = 10 or inter-arrival time = 6 minutes

Arrival variance = 4

Coefficient of variance for arrival (Ca^2) = 4/36 = 0.11

Service time (1/mu) = 5.5 minutes

Service variance = 5

Coefficient of variance for service (Cs^2) = 5/5.5^2 = 0.165

a)

Utilization (rho) = lambda/mu = 10 / (60/5.5) = 0.916 or 91.6%

b)

Here we can use the Kingman approximation where

Wq for G/G/c = Wq for M/M/s * (Ca^2 + Cs^2)/2

Wq for M/M/s = lambda/(mu*(mu-lambda)) = 10/(10.90*(10.90-10)) = 1.01 hours

Wq for G/G/c = 1.01*(0.11 + 0.165)/2 = 0.138 hours

The waiting time in queue (Wq) = 0.138 hours or 8.3 minutes.

c)

Average number of customers waiting in queue (Lq) = Wq*lambda = 0.138*10 = 1.38

d)

Average number of customers in system (L) = Wq*mu = 0.138*(60/5.5) = 1.50

This means the average number of customers in service = 1.50 – 1.38 = 0.12

e)

Average time spent in the system (W) = Wq/rho = 0.138/0.916 = 0.150 hours or 9.03 minutes

2-

Lambda = arrival rate = 5 per hour

Mu = service rate = 60/11 = 5.4545 per hour

utilisation = lambda / mu = 5/5.4545 = 0.9166

avg number of customers waiting Lq= (lambda)² / mu ( mu-lambda)

= 25 / (5.4545 x 0.4545) = 10.084

avg waiting time in line Wq = Lq / lambda = 10.084/5 = 2.016 hour

avg num,ber in system = lambda / mu-lambda = 5/0.4545 =11

avg waititng time in system = 1/mu-lambda = 1/0.4545 = = 2.2 hours