QUEUEING THEORY

Introduction

Whenever customers line up at a service facility, a queue system exists.

Queues are common place experiences. Some examples that can easily

be recognized are:

People waiting in shops and at service counters;

Cars waiting at traffic lights;

Aircraft waiting to land;

Students waiting to see an academic adviser;

Ships waiting to enter port.

The less obvious examples of queues are:

Machines waiting for repair;

Goods waiting in inventory;

Telephone subscribers waiting for a clear line;

Papers in an in-tray.

A branch of management science that concerns itself with the general

phenomenon of “customers lining up for service” is queue theory. A

theory of queues is possible because certain regularities about the

patterns of arrival and service of customers exist. These regularities can

be described statistically and analytical results (i.e. formulae)

sometimes follow. Queuing theory is about delay whether or not an

actual queue is observed.

Why does delay occur at all?

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The reason for delay is that it is not possible or not worthwhile to tailor

the supply of a service exactly to the demand for it. A trade-off situation

arises. While delay can be costly, it is also costly to make provision to

avoid or to reduce delay. However, these two costs do not always fall on

the same parties.

Sometimes a system of serving customers can be developed by

experimentation. For example, changing the provision of cashiers at

supermarket checkouts at different times of the day or banking hall at

different periods of the month is possible. However, in some other

cases, such intervention in the real situation would either be impossible

or prohibitively expensive. This is particularly so where large or

irreversible investments are involved, for example, the number of lanes

on a motorway cannot readily be changed. Thus, very complex systems

have to be simulated.

In principle, queue theory can predict how systems will operate. Certain

important measures of performance (or system parameters) can be

obtained from queue models. Among these are:

average waiting time

expected length of queue

average number of customers in the system

probability of experiencing delay.

The three main means of obtaining these parameters are by:

Analytical results

Real world experimentation

Simulation

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In practice, queue theory in pure analytical form is employed relatively

infrequently. However, where it is used, it tends to be used extensively

and it has often been found to be of powerful effect.

Characteristics of Queueing System

Queueing systems (waiting lines) are characterized by the differing

nature of five factors:

1. The arrival pattern of customers

2. The queue discipline

3. The service mechanism

4. Capacitation

5. Population

Arrivals

The pattern of arrivals may be deterministic (e.g. items on a production

flow line) or more usually random (e.g. vehicles entering filling station,

customers at the post-office, supermarket or bank). ‘Customers’ may

arrive one at a time or en-masse. ‘Customer’ or ‘item’ are the general

terms used to describe the queue elements as they do not have to be

people. Arrivals may or may not depend on the state of the system (e.g.,

a customer turning back on seeing a long queue). The arrival rate may

be constant or may vary over time. Customers may be identical or

different (e.g., civilians and military men arriving at a bank, private and

distressed aircraft arriving at an airport).

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Queue Discipline

The simplest arrangement of queue discipline is FIFO (first in, first out).

There is also LIFO (last in, first out, as in items drawn from stock or

redundancies in work force) and random (as with telephone callers

trying to get a connection). In the case where there are several queues,

customers may wish to change from one queue to another (jockey).

Also, some customers may join a queue and then leave before service

(renege). These two cases are elements or signs of queue indiscipline.

Service Mechanism

There may be one or many servers, who may differ in speed of service.

The speed of service at any service point may be constant or random

and may vary with the time of the day. Servers may be in parallel (as in

a self-service cafeteria) or in series.

Capacitation

Capacitation refers to the capacity or maximum available space in the

queueing system. Some systems have a maximum number of customers

that can be contained in the system. Where such a limit exists the

system is said to be CAPACITATED. The theory of queues only takes

cognizance of customers within the queue system and not those waiting

outside the system.

Population

Population refers to the source from which customers come into the

queue system. The ‘population’ from which customers arrive may be

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infinite or finite. The population is sometimes referred to as the ‘calling

source’.

While a very large number of different queueing models can be

constructed, those of interest and of practical relevant to us have

patterns of arrival and service described by probability distributions

rather than being deterministic. A purely random pattern of arrivals is

described by the ‘Poisson Distribution’ and its continuous analogue ‘the

negative exponential distribution’.

The Poisson Distribution

In any time interval T, the probability that there are n arrivals into the

queueing system is given by:

P (n arrivals in time interval T) =

Queueing Notations

The following notation and steady-state statistics will be used to

describe the queueing systems:

n – Number of customers in the system

Pn – Probability that there are exactly n customers in the system (steady

state).

– Average number of customers arriving into the system per unit time.λ

µ – Average number of customers a server can service per unit time.

– The traffic intensity or proportion of busy time of the server.ρ

L – The expected (or average) number of customers in the system.

Lq – The expected (or average) number of customers waiting in the

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Queue.

W – The expected time a customer will spend in the system (both

waiting on the queue and being served).

Wq – The expected time a customer will spend waiting in the queue.

A single – Channel Queue Model

The simplest queueing model is referred to as the ‘single – channel

exponential queueing model’ and it is written in shorthand notation as

the “(M/M/1) system”. The assumptions of this model are as follows:

1.The input population is infinite (that is, there are an infinite number of

potential customers (arrivals).

2.The arrival (or input) distribution is Poisson distributed with average

arrival rate . This means that the number of customers (arrivals)λ

arriving in one unit of time is a random variable that has a Poisson

probability density. The average arrival rate per unit time is λ

customers. This is equivalent to saying that the time between

successive customers arriving is a random variable having an

exponential probability density with mean 1/ . In the notationλ

“(M/M/1)”, the first M means that the arrival pattern is Poisson

distributed..

3.The service discipline is first come, first served (FCFS).

4.The service facility consists of a single channel. The one (1) in

(M/M/1) means “one channel” (or one server).

5.The service distribution is also Poisson. The mean number of

customers served per unit of time by a busy server is µ. Equivalently,

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we can say that the service time per customer is a random variable

having an exponential probability density with a mean of . In the

notation (M/M/1), the second M means that the service rate is

Poisson.

6.The average number of customers arriving (arrivals) per unit of time

is assumed to be less than the average number of customers who

could be served by the system per unit time, that is, < µ.λ

7.System capacity is infinite.

8.Balking and reneging are not allowed.

Steady –state Equations for the (M/M/1) System

The following equations can be shown to hold for the (M/M/1) queue

system:

Traffic density, = ρ (1)

Pn = ρn (1- )ρ (2)

where Pn is the probability of there being n customers in the system.

The expected (or average) number of customers in the system, L, is

given by:

L = = [ /(1 – )]ρ ρ (3)

The expected (or average) number of customers waiting in the queue,

Lq is:

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Lq = = [ρ2/(1 – )]ρ (4)

The expected time a customer will spend in the system (both waiting

and being served), W, is given by:

W = (5)

The expected time a customer will spend waiting in the queue, Wq, is

Wq = (6)

Traffic Intensity

The traffic intensity is denoted by and is given by equation (1).ρ

Traffic intensity is the proportion of time that the server will be busy

serving customers who arrive. Sometimes, it is referred to as the

‘utilization factor’ of the queueing system. Since < λ , then < 1.ρ

Idle time

For what proportion of the time will the server be idle?

If equals the busy time, then 1 – is the idle time. This can be seenρ ρ

from Eq. (2). The server is idle when there are no customers in the

system. The probability of there being no customers in the system is ρo,

which from Eq.(2) yields:

Ρ0 = ρo(1- ) = 1(1- ) = 1 - .ρ ρ ρ

Mean Service Time

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If the average number of customers served per unit time is µ, then the

mean (or average) service time per customer is time units. For

example, if = 10 customer per hour, then mean service time is =

= 0.10 hour (per customer) or 6 minutes per customer.

Example 1

Joe’s service station operates a single gas pump. Cars arrive according to

Poisson distribution at an average rate of fifteen cars per hour. Joe can

service cars at the rate of twenty cars per hour on the average, with

service time per car following an exponential probability distribution.

Assuming we model Joe’s service station as an (M/M/1) queueing

system, compute the steady-state statistics for the system.

Solution

We have = 15 cars per hour, and λ = 20 cars per hour. Thus,

1) Traffic intensity, = ρ = = 0.75

This implies that Joe will be busy servicing cars 75 percent of the time.

This leaves an idle time of 1- = 0.25, or 25 percent, for Joe to doρ

paperwork, maintenance, and so on.

2) Expected (average) number of cars is the system, L.

L = = = 3 cars

The implication of this is that on the average Joe can find three

cars at his station.

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3) Expected (or average) number of customers waiting in the

queue:

Lq = = = 2.25

So on the average 2.25 cars will be waiting in line for service. The

practical interpretation of this is that while two cars are still on

queue, it remains ¼ of the tank of the car currently receiving

service to be filled.

4) Expected time a customer will spend in the system (i.e. waiting

and being served):

W = = = 0.20 hour

Or 12 minutes

5) Expected time a customer will spend waiting in the queue, Wq

is:

Wq = = = 0.15 hour

Both (4) and (5), respectively, imply that on the average, a car

spends 0.20 hour in the system (or 12 minutes) and 0.15 hour

waiting for service (9 minutes).

Suppose we wish to find the probability of having four cars or less in the

system, this probability equals the sum of the various probabilities from

zero cars being in the system to four cars being in the system, that is

Ρ0 + P1 + P2 + P3 + P4.

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Note that from Equation (2):

Pn = ρn(1- ) = (0.75)ρ n (1-0.75)

= (0.75)n (0.25)

Thus, we have

-----------------------------------------------

n Pn = (0.75)n (0.25)

----------------------------------------------

0 (0.75)0(0.25) = 0.250

1 (0.75)1(0.25) = 0.188

2 (0.75)2 (0.25) = 0.141

3 (0.75)3(0.25) = 0.105

4 (0.75)4 (0.25) = 0.079

∑ = 0.763

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The above result is interpreted to mean “there is a 76.3 percent chance

of finding four or fewer cars line up at the service station.”

The above example can be represented pictorially as shown below:

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Cars in queque waiting for service

Service facility: a single gas pump serving 20 cars per hour

Departing

Car

Key relationships of the steady-state statistics

In every Poisson queueing system, certain relationship will always hold

between L, Lq, W, and Wq. These are as follows:

L = Wλ (7)

Lq = Wqλ (8)

W = Wq + (9)

Each of the above relationships holds for the (M/M/1) system.

Example 2

Customers arrive in a shop at the average rate of 42 per hour. The

shopkeeper takes on average one minute to serve each customer.

Simple queue conditions thus prevail.

(i) What proportion of his time does the shopkeeper spend

serving customers?

(ii) What is the average queueing time for customers?

(iii) What is the probability that:

a)There are three customers in the shop?

b)There is one customer queueing?

c)There are four or more customers in the ship?

d)There are three or less customers in the shop?

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Arriving cars(15 per hour)

Solution

i)This refers to the proportion of the time when there is at least one

customers in the system. This is given by

Ρ1 + P2 + P3 + ….. + P .

More conveniently, this is the same as 1- P0. In fact it is itself, theρ

traffic intensity or utilization factor. Thus,

= ρ =

= 0.71minute to serve

Since it takes 1 minute to serve each customer, then the number of

customers served in 1 hour is 60.

(ii) Average queueing time is the same as average time a customer will

spend in the queue, Wq.

Wq = =

= hours or minutes

(iii)(a) We shall assume here that ‘being in the shop’ means being in the

queue system. Then the probability of three (3) customers being in the

shop is

Ρ3 = ρ3(1- )ρ

= (0.7)3 (1-0.7)

= (0.7)3 (0.3) = 0.1029

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(b) If one customer is queueing this must mean that one is being served

so that there are two customers in the system. Thus,

Ρ2 = ρ2(1- )ρ

= (0.7)2(1-0.7)

= (0.7)2 (0.3) = 0.147

(c ) Here we have simply:

P (4 or more) = ρ4 = (0.7)4 = 0.2401

(d ) To simplify workings we note that ‘three or less’ and ‘four or more’

cover all possibilities. Thus,

Ρ (3 or less) = 1- (4 or more)Ρ

= 1-0.2401 = 0.7599

Example 3

A telephone is used by eight people per hour on average and the mean

call length is three minutes. A further instrument has been requested

but this will only be installed if average queueing time is at least three

minutes. Determine the percentage increase in demand necessary to

produce this result.

Solution

We note that = 8 callers per hour. Clearly, the ‘service time’ is theλ

length of call so that with an average call length of three minutes, there

will be 20 callers per hour. Thus µ = 20 callers per hour.

Expected queueing time is given by , thus

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Queueing time = = (1)

If average queueing time is 3 minutes = hours = hours (2)

Thus, we require the R.H.S of (1) and (2) to be equal. Hence

=

This (by cross multiplication) implies that

20 = 20 (20 - )λ λ

or 20 = 400 - 20 λ λ

That is,

20 + 20 = 400λ λ

40 = 400λ

= 10 callers per hourλ

Now the initial average arrivals was = 8λ

Increase in the average number of callers per hour is 10 – 8 = 2

Percentage increase Increase in

in the number of callers = callers per hour

per hour Initial average arrivals

= = 25 percent.

Thus, a 25 percent increase in demand (measured by rate of arrivals)

would be needed to install a further instrument.

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by transferring the 20 on the λR.H.S to the L.H.S