queueing problems in banking

i
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.

ii
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”.

iii
TYPES OF QUEUEING MODELS:-

iv
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-arrival 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/


v
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.

vi
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.

vii
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 probabilityof 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.

viii
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

x
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

xi
SYSTEM WITH FINITE POPULATION :-
Pn = ((1-Ǿ) x (Ǿ)n
)/(1-Ǿn+1
) (if Ǿ ≠1 )
Pn = 1/(N+1) (if Ǿ=1)
Ls= (Ǿ[1-(N+1)Ǿn
+N Ǿn+1
])/[(1-Ǿ)(1-ǾN+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

xii
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

xiii
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
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

xiv
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
=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

xv
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 Ǿn+1
])/[(1-Ǿ)(1-ǾN+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

xvi
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

xvii
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%
2Fwps.prenhall.com%2Fwps%2Fmedia%2Fobjects%2F2234%2F2288589%2FModD.pd
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
t=8&ved=0CCoQFjACahUKEwi3lcy8z__IAhXQv44KHSQ6AxU&url=http%3A%2F%
2Fwww.csus.edu%2Findiv%2Fb%2Fblakeh%2Fmgmt%2Fdocuments%2Fopm101supplc
.pdf&usg=AFQjCNGviWaTXBx6KdJ6YLnTdXcsDBNIQA

queueing problems in banking

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