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ABSTRACT: -

This project contains the analysis of Queuing systems for the empirical data of Bank

service. One of the expected gains from studying queuing systems is to review the efficiency of

the models in terms of utilization and waiting length, hence increasing the number of queues so

customers will not have to wait longer when servers are too busy. Trying to estimate the waiting

time and length of queue(s), is the aim of this project. We may use queuing simulation to obtain

a performance result and we are more interested in obtaining estimated solutions for multiple

queuing models. This project describes a queuing simulation for a multiple server process as well

as for single queue models. This study requires an empirical data which may include the

variables like, arrival time in the queue of checkout operating unit (server), departure time,

service time, etc. The model designed for this example is multiple queues multiple-server model.

In any service system, a queue forms whenever current demand exceeds the existing capacity to

serve. This occurs when the checkout operation unit is too busy to serve the arriving costumers,

immediately.

The purpose of this study is to review Queuing Theory and its empirical analysis based

on the observed data of checking out sales service unit of bank. The main idea in the application

of a mathematical model is to measure the expected queue length in each checkout sales service

unit. The service rate provided to the customers while checking out. Another idea is to give

insight view of the steady-state behaviour of queuing processes . Descriptions of events are given

i.e. the arrivals and service rate in each checkout unit and how they can be generated for any

amount of working hour. The other important factor analysed is about the comparison of two

different queuing models: single-queue multiple-server and multiple-queue multiple-server

model.

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INTRODUCTION:

A flow of customers from finite or infinite population towards the service facility forms

a queue(waiting line) an account of lack of capability to serve them all at a time. In the

absence of a perfect balance between the service facilities and the customers, waiting time

is required either for the service facilities or for the customer’s arrival. In general, the

queueing system consists of one or more queues and one or more servers and operates under a

set of procedures. Depending upon the server status, the incoming customer either waits at the

queue or gets the turn to be served. If the server is free at the time of arrival of a customer, the

customer can directly enter into the counter for getting service and then leave the system. In

this process, over a period of time, the system may experience “Customer waiting” or “Server

idle time”.

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TYPES OF QUEUEING MODELS:-

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QUEUEING SYSTEM:

A queueing system can be completely described by

(1) The input (arrival pattern)

(2) The service mechanism (service pattern)

(3) The queue discipline and

(4) Customer’s behaviour

THE INPUT(ARRIVAL PATTERN) :

The input described the way in which the customers arrive and join the system. Generally,

customers arrive in a more or less random manner which is not possible for prediction. Thus

the arrival pattern can be described in terms of probabilities and consequently the probability

distribution for inter- arri val times (the time between two successive arrivals) must be defined.

We deal with those Queueing system in which the customers arrive in Poisson process. The

mean arrival rate is denoted by .

THE SERVICE MECHANISM :-

This means the arrangement of service facility to serve customers. If there is infinite

number of servers, then all the customers are served instantaneously or arrival and there will

be no queue. If the number of servers is finite, then the customers are served according to a

specific order with service time a constant or a random variable. Distribution of service time

follows ‘Exponential distribution’ defined by

f(t)=ʎ e-ʎt

(t>0)

The mean Service rate is E(t) = 1/

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QUEUEING DISCIPLINE :

It is a rule according to which the customers are selected for service when a queue has been

formed. The most common disciplines are

1. First come first served – (FCFS)

2. First in first out – (FIFO)

3. Last in first out – (LIFO)

4. Selection for service in random order (SIRO)

CUSTOMER BEHAVIOUR: -

There are four types of customer behaviour namely

1)Bulk Arrival

2)Jockeying

3)Balking

4)Reneging

1. Generally, it is assumed that the customers arrive into the system one by one. But

in some cases customers may arrive in groups. Such arrival is called Bulk

Arrival.

2. If there is more than one queue, the customers from one queue may be tempted

to join in another queue because of its smaller size. This behaviour of the

customers is known as jockeying.

3. If the queue length appears very large to a customer, he or she may not join the

queue. This property is known as Balking of the customers.

4. Sometimes, a customer who is already in a queue will leave the queue in the

anticipation of longer waiting time. This kind of departure is known as reneging.

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NOTATION FOR QUEUES: -

Any queuing models may be completely specified in the following symbolic form using

kendal & Lee notations

Since all queues are characterised by arrival, service and queue and its discipline, the

queue system is usually described in shorten form by using these characteristics. The general

notation is:

[A/B/s]:{d/e/f}

Where,

A = Probability distribution of the arrivals

B = Probability distribution of the departures

s = Number of servers (channels)

d = The capacity of the queue(s)

e = The size of the calling population

f = Queue ranking rule (Ordering of the queue).

Queuing models are classified into six types based on the above notation.

There are some special notation that has been developed for various probability

distributions describing the arrivals and departures. Some examples are,

M = Arrival or departure distribution that is a Poisson process

E = Erlang distribution

G = General distribution

GI = General independent distribution.

Thus for example, the [M/M/1]:{infinity/infinity/FCFS} system is one where the arrivals and

departures are a Poisson distribution with a single server, infinite queue length, calling

population infinite and the queue discipline is FCFS. This is the simplest queue system that

can be studied mathematically. This queue system is also simply referred to as the M/M/1

queue.

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LIST OF VARIABLES :

The list of variable used in queueing models is give below:

n - No of customers in the system

C - No of servers in the system

Pn (t) – Probability of having n customers in the system at time t.

Pn - Steady state probability of having customers in the system

P0 - Probability of having zero customer in the system

Lq - Average number of customers waiting in the queue.

Ls - Average number of customers waiting in the system (in the queue and in the service

counters)

Wq - Average waiting time of customers in the queue.

Ws - Average waiting time of customers in the system (in the queue and in the service counters)

ʎ - Arrival rate of customers

µ - Service rate of server

Ǿ - Utilization factor of the server

δ eff - Effective rate of arrival of customers

M - Poisson distribution

N - Maximum numbers of customers permitted in the system. Also, it denotes the size of the

calling source of the customers.

GD - General discipline for service. This may be first in first – serve (FIFS), last-in-first serve

(LIFS) random order (Ro) etc.

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AN ELEMENTARY QUEUEING PROCESS :-

A single waiting line forms in the front of a single service facility, within which are

stationed one or more servers. Each customer is serviced by one of the servers, perhaps after

some waiting in the queue.

The prototype example is of this type

We usually label a queueing model as ----/----/----

The first spot is for distribution of interarrival times. s

The second spot is for distribution of service times.

The third one is for number of servers.

M = exponential distribution (Markovian), which is the most widely used.

D = degenerate distribution (constant time). ¾

Ek = Erlang distribution.

G = general distribution (any arbitrary distribution allowed)

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TERMINOLOGY AND NOTATION :

State of system = number of customers in queueing system.

Queue length = number of customers waiting for service to begin = state of system minus

number of customers being served.

N(t) = number of customers in queueing system at time t.

Pn(t) = probability of exactly n customers in queueing system at time t.

s = number of servers (parallel service channels) in queueing system.

λ= mean arrival rate (expected number of arrival per unit time) of new customers when n

customers are in system.

o When λn is a constant for all n, this constant is denoted by λ.

o 1/ λ is the expected interarrival time.

μ= mean service rate for overall system (expected number of customers completing

service per unit time) when n customers are in system.

ρ = utilization factor for the system= λ /µ

Ls = average number of units (customers) in the system (waiting and being served)

Ws = average time a unit spends in the system (waiting time plus service time) =

Lq = average number of units waiting in the queue

Wq = average time a unit spends waiting in the queue

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SINGLE SERVER CASE(M/M/s) s>1:-

SYSTEM WITH INFINITE POPULATION :-

The following assumptions are to be made when we model this system:

1. The customers are patient (no balking, reneging or jockeying) and come from a population

that can be considered infinite.

2. Customer arrivals are described by a Poisson distribution with a mean arrival rate of

(lambda). This means that the time between successive customer arrivals follows an exponential

distribution with an average of 1/ʎ.

3. The customer service rate is described by a Poisson distribution with a mean service rate of

(mu). This means that the service time for one customer follows an exponential distribution with

an average of 1/ .

4. The waiting line priority rule used is first-come,first-served.

Ls = average number of units (customers) in the system (waiting and being served)

Ws = average time a unit spends in the system (waiting time plus service time) =

Lq = average number of units waiting in the queue

Wq = average time a unit spends waiting in the queue

ρ = utilization factor for the system=

ʎ/µ

P0 = probability of 0 units in the system (that is, the service unit is idle)

1-(ʎ/µ)

Pn>k = probability of more than k units in the system, where n is the number of units in

the system

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SYSTEM WITH FINITE POPULATION :-

Pn = ((1-Ǿ) x (Ǿ)n)/(1-Ǿ

n+1) (if Ǿ ≠1 )

Pn = 1/(N+1) (if Ǿ=1)

Ls= (Ǿ[1-(N+1)Ǿn+N Ǿ

n+1])/[(1-Ǿ)(1-Ǿ

N+1)] (if Ǿ≠1)

LS =N/2 (if Ǿ =1)

Lq = Ls -ʎEFF/µ

ʎEFF = ʎ(1-PN)

Wq= Lq/ʎEFF

Ws = Ls/ʎEFF

MULTIPLE SERVER CASE :

(k>s)

Similarly

(k>s)

( k<s)

Length of the queue=

Waiting time in the queue WQ =LQ/ʎ

Waiting time in the system Ws= WQ + 1/µ

Length of the system Ls

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SINGLE SERVER INFINITE QUEUE LENGTH MODEL :

PROBLEM 1:

In a bank there is only on window. A solitary employee performs all the service required and the

window remains continuously open from 7am to 1pm. It has discovered that an average number

of clients is 54 during the day and the average service time is 5mins / person. Find

a) Average number of clients in the system

b) Average waiting time

c) The probability that a client has to spend more than 10mins in a system.

Sol:- The mean arrival rate =

= 54/6 =9 clients/hr

The mean service rate

= 1/5 *60

=12 clients/hr

a) Average no of clients in the system(L)

=9/(12-9)

L =3 clients

b) Average waiting time(Wq)

=9/(12(12-9))

=0.25

c) Average waiting time in the system

=.25+(1/12)=0.333

d) Average no of clients in the queue

=2.25 clients

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e) Probability of having 10 clients in the system

Utilisation factor

=9/12

=0.75

=(1-0.75)*(0.75)^10

=0.0140783

Problem 2:- In a bank there is only on window. A solitary employee performs all the service

required and the window remains continuously open from 7am to 1pm. It has discovered that an

average number of clients is 8/hr and the average service time is 11 clients / hr . Find

a) Average number of clients in the system

b) Average waiting time

c) The probability that a client has to spend more than 10mins in a system.

The mean arrival rate =

= 8 clients/hr

The mean service rate

= 11 clients/hr

a) Average no of clients in the system(L)

=8/(11-8)

L =2.667 clients

b) Average waiting time(Wq)

=8/(11(11-8))

=0.2424

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c) The probability that a client has to spend more than 10 min in a system

Utilisation factor

=8/11

=0.7272

Pn>10 = (0.7272)^10

=0.041

MULTIPLE SERVER FINITE QUEUE LENGHT MODEL:

PROBLEM 3:

In a bank there is only on window. A solitary employee performs all the service required and the

window remains continuously open from 7am to 1pm. It has discovered that an average number

of clients is 54 during the day and the average service time is 5mins / person. Find

a) Average number of clients in the system

b) Average waiting time

c) The probability that a client has to spend more than 10mins in a system.

=9, =12 N=10 s=2

a) Average no of clients in the system

Ls

LS=0.00951211

b) Average no of clients in the queue

LQ=0.7595

c) Waiting time in the system

Ws= WQ + 1/µ

WS =0.010506

d) Waiting time in the queue

WQ =LQ/ʎ

WQ=0.08438

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SINGLE SERVER FINITE QUEUE LENGHT MODEL :

PROBLEM 4:

A repair person is attending to work on 6 A.T.M machines in a city .Each machine breakdown

according to poisson distribution with a mean of 4/hr .Service time per machine is expected with

a mean of 10 min.

i) Average no.of machines waiting in queue as well as in system ?

ii) Average waiting time of machines in queue as well as in system ?

Given N=6 machines

Arrival rate (ʎ) = 4/hr

Service time for 1 machine = 10 min

1 minute =1/10 machines

Service rate =1/10 machines per minute

=1/10 x60 per hour

µ =6 /hr

Utilisation factor (Ǿ) = 0.66

Ls = (Ǿ[1-(N+1)Ǿn+N Ǿ

n+1])/[(1-Ǿ)(1-Ǿ

N+1)]

= 1.537

=2 machines

Lq =Ls - ʎEFF/µ

ʎEFF =ʎ(1-PN)

PN=0.0297

ʎEFF = 3.8812 /hr

LQ = 0.8901

= 1 machine(approx)

Wq= Lq/ʎEFF =0.2293 hr

Ws = Ls/ʎEFF =0.39601

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PROBLEM 5:

In a bank the arrival rate of customers is 24 customers/hr and the service rate of the customers is

20/hr .The arrival rate and service rate follows the poission distribution .The no.of persons

allowed at a time only 5 .find

i)Avg. number of persons waiting in queue as well as in system ?

ii)Avg. waiting time of persons in queue as well as in system ?

Given N=5

Arrival rate (ʎ) = 24 cars/hr

Service rate (µ) =20 cars/hr

Utilization factor (Ǿ) =1.2

LS= 3.02 cars

LQ = Ls - ʎEFF/µ

ʎEFF =ʎ(1-PN)

PN=0.3343

ʎEFF = 15.97 /hr

LQ = 2.2215

= 2 machine(approx)

Wq= Lq/ʎEFF = 0.1391 hr

Ws = Ls/ʎEFF =0.1891 hr

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RESULTS :

Problem no: LS LQ WS WQ

1 3 2.25 0.333 0.25

2 2.667 1.939 0.333 0.2424

3 0.7595 0.0095 0.010506 0.084

4 1.537 0.8901 0.39601 0.2293

5 3.02 2.2215 0.1391 0.1391

CONCLUSION:-

Among all the queuing models we solved multiple server models give the less waiting

time for the customers in the queue as well as in the system.

No of people served will be more in case of multiple server models because more no of

customers are served at the same time with the help of parallel servers and it is most

efficient.

REFERENCE:-

1. https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uac

t=8&ved=0CCQQFjABahUKEwi3lcy8z__IAhXQv44KHSQ6AxU&url=http%3A%2F%

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f&usg=AFQjCNEY_VNaZIaWH6vXVVmV5ao93oyPMg

2. https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uac

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.pdf&usg=AFQjCNGviWaTXBx6KdJ6YLnTdXcsDBNIQA