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Unit 8 Infinite Queuing Models
Structure
8.1. Introduction
8.2. Queuing theory
8.3. Analysis of Queuing process
8.4. Constituents of Queuing system
8.4.1 Arrival Pattern
8.4.2 Completely Random Arrivals
8.5. Service facility
8.6. Queuing discipline
8.6.1 Customer Behaviour
8.6.2 Server Behaviour
8.7. Mathematical analysis of Queuing process
8.7.1 Properties of the System
8.7.2 Notations
8.8. Single channel models
8.9. Multiple service channels
8.10. Erlang family of distribution of service times
8.11. Summary
Terminal Questions
Answers to SAQs and TQs
8.1 Introduction
We are all too familiar with queues in our day to day existence and perhaps there is no need to
define what a queue is. On the way to place of work often we have to wait for a bus, for the traffic
lights to turn green and then at the office premises for the lift. Cars waiting at petrol pumps for
service, customers waiting at the bank, telephone subscribers waiting for connections are all a
common sight. Aircrafts get delayed for want of a free runway. No so familiar to all, are the cases
of broken down machines waiting for repairs, workmen waiting for tools, and goods in production
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shops waiting for cranes. Even inventory situations may be regarded as queues not only in that
customers wait in line for goods, but also that goods form a waiting line awaiting consumption.
All the above examples have one common feature. Customers arrive at a service centre and wait
for service. The arrival of customers is not necessarily regular and so is the time taken for service
not uniform. Queues build up during hours of demand and disappear during the lull period.
Personally we do not like to wait.
In the commercial or industrial situations, it may not be economical to have waiting lines. On the
other hand, it may not be feasible or economical to totally avoid queues. An executive dealing
with the system then would like to find the optimal facilities to be provided
Learning Objectives
After studying this unit, you should be able to understand the following1. Understand the Queuing process
2. Analyze the Queuing process mathematically
3. Apply different models to practical problems
8.2 Queuing Theory
Queuing theory based on probability concepts gives an indication of the capability of a given
system and of the possible changes in its performance with modification to the system. All the
constraints of the process are not taken into account in the formulation of a queuing model. There
is no maximization or minimization of an objective function. Therefore the application of queuing
theory cannot be viewed as an optimization process. With the help of queuing theory the
executive can at best make an informed guess of what the balance between customer waiting
time and service capability could be. He first considers several alternatives, evaluates through
queuing models, their effect on the system, and makes his choice. The criteria for evaluation will
be measures of efficiency of the system like the average length of a queue, expected waiting time
of a customer and the average time spent by the customer in the system. In this approach, his
success primarily depends on the alternatives considered and not so much on the queuing
models developed.
It is essential for the executive to have a succinct understanding of the process so that he will
consider the right alternatives and formulate the correct models.
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Self Assessment Questions 1
State True / False
1. Customers arrive at a Bank at regular interval.
2. Queuing identifies the optimal service facilities to be provided.
3. Queuing theory is based on deterministic model.
8.3 Analysis of a Queuing Process
Very often, the management is interested in finding out the quality of service rendered to the
customers. One of the indictors of efficiency of a system involving flow of customers for service is
how fast they are served. Augmenting the physical facilities to provide service will in general
improve service leading to customer satisfaction. However, such facilities cannot be indiscriminately
increased as they could be quite expensive. If the queuing process involves waiting of machines for
service, it may be possible to find the economic level of the maintenance facilities by balancing the
cost of machine downtime against the cost of maintenance facilities. Such evaluation in monetary
terms may not always be possible where the customers are people.
Then some measures of efficiency of the system as mentioned above may prove helpful.
A typical investigation of a queuing system would comprise the following steps:
Step 1: Preliminary study
At this stage, an analysis of the process is made usually with the help of a flow diagram and an
attempt is made to identify the points which restrict service or the characteristics which indicate
scope for improvement.
Step 2: Exploration of the various alternatives
By introducing changes in the constituents of the queueing system, it should be possible to effect
improvement. The arrival pattern may be altered by withdrawing service facilities to certain
categories of customers or by introducing an appointment system. The time taken for providing
service may be improved by increasing the capacity of the system i.e. by increasing the number
of service channels or the working hours. It is also not uncommon to provide additional service
facilities to relieve congestion during peak periods. Modifying the queue discipline may alsochange the characteristics of the system.
Priority may be given to arrivals involving high cost of waiting time. In multi channel queues,
separate channels of service may be provided for different types of customers. The fixed change
counter provided at the Church gate Railway Station by the Western Railway is said to have cut
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down the waiting time for a majority of customers.
Step 3: Collection of data and analysis
Direct observation of the system in terms of arrival time of customers, service rate, length of queue
and waiting time is done at this stage. The data is analyzed in sufficient detail to determine the
statistical pattern of the variables. The measures of efficiency of the system are computed on the
basis of queuing theory. These are counterchecked with the results obtained through direct
observation to confirm the validity of the formulae applied. It may sometimes be necessary to collect
additional data regarding the process.
Step 4: Evaluation of alternatives
Effect of modifying the constituents of the system on the basis of the selected alternatives is
evaluated through application of queuing theory. Simulation technique can also be used for the
purpose. Based on the results obtained, the best combination of the changes to be made in theexisting system are selected.
Step 5: Implementation
The proposals formulated are implemented on a small scale and tested. If necessary, further
changes are made and rechecked before implementing them on full scale. It is advisable to
observe the functioning of the system periodically to ensure that the results desired are
maintained.
Self Assessment Questions 2
Fill in the blanks
1. One of the indicators of efficiency of a system is __________ factor.
2. Analysis of Queuing system explores _________ _________.
3. __________ technique can also be used for analysis.
8.4 Constituents Of A Queuing System
In specifying a queuing process one should know the following:
(a) Arrival Pattern: The average rate at which customers arrive as well as the statistical patternof arrivals
(b) Service facility: When service is available, number of customers that can be served at a
time and the statistical pattern of time taken for service
(c) Queue discipline: Method of choosing a customer for service from amongst those
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waiting for service. (First come First served is a common basis but there are also other
methods.)
We shall now examine each of these constituents.
8.4.1 Arrival Pattern
The arrival of customers can be regular as in the case of an appointment system of a dentist which is
closely adhered to, or like the flow of components from a conveyor. The regular pattern of arrivals is
not very common nor is very easy to deal with mathematically. We shall be primarily concerned with a
pattern of completely random arrivals, as it has useful applications.
8.4.2 Completely Random Arrivals
If the number of potential customers is infinitely large, the probability of an arrival in the next
interval of time will not depend upon the number of customers already in the system. (Theassumption is valid by and large, except for queues involving a small finite number of customers.)
Where the arrivals are completely random, they follow Poisson distribution with mean equal to the
average number of arrivals per unit time.
When dealing with arrivals it might sometimes be necessary to distinguish between groups of
customers like male and female callers, or large and small aircraft.
There are several other types of arrival patterns which we shall not deal with due to their
restricted applications.
Self Assessment Questions 3
State Yes / No
1. Queuing process has arrival pattern, service facility and Queue discipline as its constituents.
2. If the arrivals are completely random, then it follows poisson distribution.
8.5 Service Facility
i) Availability of service
Apart from specifying the time span over which service is available it is also necessary to
examine if there are any constraints which reduce the number of customers that can be
served at a time. For instance, in a waiting line for a suburban train, the probability
distribution of the number of passengers that can be accommodated in a train that arrives is
relevant apart from the timings of the train services.
ii) Number of Service Centers
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If there is only one service centre referred to as a service channel, obviously only one
customer can be served at a time. Often there will be more than one service centre and the
behaviour of the queues will vary with the number of channels available for service.
Multiple service channels may be arranged in series or in parallel. If the customer has to gothrough several counters one after the other each providing a different part of the service
required, the arrangement is said to be in series.
Normally for withdrawals in a bank, the customer has to go to at least two counters in series.
Provision of several counters offering the same type of service is a common approach to
reduce congestion. Ticket booths in a railway station have this type of multiple channels with
parallel arrangement.
iii) Duration of service
This is the length of time taken to serve a customer. This can be constant or varying.
(a) Constant service time
This is not strictly realized in practice, but an assumption that service time is constant
holds true if the pattern of arrivals is very irregular
(b) Completely random service time
The service time can be considered completely random where the server does not
distinguish between the various arrivals, and he does not change deliberately the duration
of service on the basis of the time taken to serve the previous arrival. The server forgets
the time for which he has been serving a customer. Under these conditions, the service
time follows exponential distribution with mean equal to reciprocal of the average rate ofservice.
(c) Service time following Erlang Distribution
In cases where the assumption of an exponential distribution for service time is not
valid, Erlang family of service time distributions is used.
S.
No
.
Situation
Arriving
Customers Service Facility
1 Passage of customers through
supermarket checkout
Shoppers Checkout
counters
2. Flow of automobile traffic through
a road network.
Automobiles Road Network
3. Transfer of electronic messages Electronic Transmission
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messages lines
4. Banking transactions Bank patrons Bank tellers
5. Flow of computer programmers
through a computer system.
Computer
Programmers
Central
processing unit
6. Sale of theatre tickets Theatre-goers Ticket booking
windows
7. Arrival of trucks to carry fruits and
vegetables from a central market
Trucks Loading crews
and facilities
8. Registration of unemployed at
employment exchange
Unemployed
personnel
Registration
assistants
9. Occurrences of fires Fires Firemen and
equipment
10. Flow of ships to the seashore ships Harbor and
docking facilities
11 Calls at police control room Service calls policemen
Self Assessment Questions 4
Fill in the blanks
1. Multiple service channels may be arranged in ___________ or in _________.
2. The service time can be __________ or __________.
8.6 Queue Discipline
The pattern of selection for service from the pool of customers is of two types. The common
pattern is to select in the order in which the customers arrive. First come first served is a
common example. In issuing materials from a stores inventory sometimes the storekeeper follows
the Last In First Out principle because of the convenience it offers for removal from stocks and
handling.
There can also be queues which accord priority to certain types of customers. Here again there
can be two approaches. In case of non pre-emptive priority the customer getting service is
allowed to continue with service till completion, even if a priority customer arrives midway during
his service. This is a common form of priority. Pre-emptive priority involves stopping the service of
the non priority customers as soon as the priority customer arrives. Priority given to repairs of a
production holding machine over an auxiliary unit for allocation of maintenance labour force is a
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typical example. Preference is given to larger ships over the smaller ones irrespective of the order
in which they arrive for allocation of berths.
8.6.1 Customer Behaviour
a) Balking: Arriving customers are said to balk if they do not join a queue because of their
reluctance to wait.
b) Collusion: Customers may be in collusion in the sense that only one person would join the
queue but would demand service on behalf of several customers.
c) Reneging: Impatient customers who would not wait beyond a certain time and leave the
queue are said to renege.
d) Jockeying: Some customers keep on switching over from one queue to another in a multiple
service centre. This is called jockeying.
8.6.2 Server Behaviour
Although the timings for service might have been specified, the server may not be available
through the entire span of time. For instance, after every hour, he may leave the service
centre for 5 minutes, to attend to his personal needs.
Self Assessment Questions 4
Fill in the blanks
1. When customers keep in switching over from one queue to another then it is called ________
2. _______ ________ ________ ________ are the types of customer behaviour.
8.7. Mathematical Analysis Of Queuing Process
Statistical Equilibrium: In analyzing the queuing process, we are interested in developing
mathematical model which represent the system for a major part of the time. This implies that
changes occurring in the characteristics of the system are not to be considered if they are of short
durations. When specifying the statistical distributions of the arrivals or service times, we are
interested in an equilibrium state.
In a queuing process, at each point of time, there is a probability distribution for the length of the
queue. The number of customers will be very much different 15 minutes after opening the counter
in a post office from that after one hour. After the initial rush, one might reasonably expect to find
the system with the same type of probabilities of arrivals.
The probability distribution of the arrivals will then be different from that of the initial state. The
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state during which the probability distribution remains the same, is called the steady state and the
system is said to have acquired a state of statistical equilibrium.
In the steady state, there will be variations in the queue from time to time but the probability
distributions representing the queueing process will remain the same and are independent of the
time at which the system is examined.
8.7.1 Properties of the System
In developing queuing models, we shall confine our discussion to queuing systems with the
following properties.
Arrivals
Customers are discrete entities
Population - Finite/InfiniteNo simultaneous arrivals
Pattern of arrivals in a time period t0 follows Poisson distribution with average arrival rate
[ ] ( )!x
texarrivalsofnumberProb
x0
0t-
==x = 0, 1, 2 ..
Service
Single serve channel/Multiple channels
Single queue/Infinite capacity
Pattern disciplineFirst come First served
It may be noted that when the number of arrivals follows Poisson distribution (discrete), the inter-
arrival time i.e., the time between arrivals follows exponential distribution (continuous).
8.7.2 Notations
The Queuing Systems with which we are concerned are denoted by M/M/1 and M/M/c where Ms
stand for exponential inter arrival and exponential service time distributions, and the third figure
indicates the number of channels (or servers) available (1 or c).
l = Average number of arrivals per unit time
m = Average number of customers served per unit time
r = Traffic intensity =
c = Number of service channels
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m = Number of customers in the queue or the length of the queue
n = number of customers in the system
= number of customers in the queue + number of customers being served
Pn = Steady state probability of finding n people in the system
E(m) = Average length of queue
E(m/m > 0) = Average length of a nonempty queue
E(n) = Average number of customers in the system
W = Steady state waiting time of a customer
E(w) = Average waiting time of a customer
P(w = 0) = Probability of not having to wait in the queue
E(w/w > 0) = Average waiting time of a customer given that there is a queue
V = Time spent by a customer in the system in steady state= Waiting time in the queue + Service time
F(v) = Probability density function of the time spent by a customer in the system
E(v) = Average time spent by a customer in the system.
Self Assessment Questions 6
Fill in the blanks
1. E (m) refers to ________ _________ ________ _________.
2. Probability density function of the time spent by a customer in the system is denoted by_____________.
3. ___________ Arrivals are allowed.
8.8 Single Channel Models
The formulae are listed in Tables 1 and 2. The examples which follow not only illustrate the
computation of various measures of efficiency of a queuing system but give an idea of the areas
of application as well.Table 1
Formulae for Poisson Arrivals, Exponential Service, Single Channel Queuing Models Infinite
Population
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)(ve)()v(f
)(we)1()w(f
1)v(E
1)ow(P
1)0w/w(E
)()w(E
)n(E
)0m/m(E
)()m(E
1P
)1(pnP
2
n
n
lmlm
lmrl
lm
r
lm
lmm
l
lm
l
lm
m
lmm
l
r
r
---=
---=
-=
-==
-=>
-=
-=
-=>
-=
-=
-=
Table 2
Formulae for Poisson Arrivals, exponential Service, Single Channel
Queuing Models Number of customers limited to N
pnp1
p1P
p1
p1P
1nn
1no
-
-=
-
-=
+
+
n = 0, 1, 2 . N
1nop1
p1P)0w(P
+-
-===
Example 1:
Patrons arrive at a small post office at the rate of 30 per hour. Service by the clerk on duty takes
an average of 1 minute per customer
a) Calculate the mean customer time
(i) spent waiting in line
(ii) spent receiving or waiting for service
b) Find the mean number of persons
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(i) in line
(ii) receiving or waiting for service
Solution:
Mean arrival rate l = 30 customers per hour
= 2
1
customer per minute
Mean service rate l = 1 per minute
Traffic intensity2
1p ==
m
l
(a) (i) Mean customer time spent waiting in line minute
utemin1
)()w(E
=
-=
lmm
l
(ii) Mean customer time receiving or waiting for service
)(
1)v(E
lm-= = 2 minutes
(b) (i) Mean number of persons in line
2
1
)()m(E
2=
-=
lmm
lcustomer
(ii) Mean number of persons receiving or waiting for service
1)n(E =-= lml customer
Example 2:
The Tool Companys Quality Control Department is manned by a single clerk, who takes an
average of 5 minutes in checking parts of each of the machines coming for inspection. The
machines arrive once in every 8 minutes on the average. One hour of machine is valued at Rs. 15
and a clerks time is valued at Rs. 4 per hour. What are the average hourly queuing system costs
associated with the Quality Control Department?
Solution:
Mean arrival rate l = 1/8 per minute
=8
60per hour
Mean service rate 5
1=m per minute
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= 12 per hour
Average time spent by a machine in the system E(v) = )(
1
lm-
= 92hours
Average queuing cost per machine is3
10.Rs
9
215 =
An average arrival of8
60machines per hour costs
8
60
3
10
= Rs. 25 per hour
Average hourly queuing cost = Rs. 25
Average hourly cost for the clerk = Rs. 4
Hence total hourly cost for the department = Rs. 29Example 3:
A hospital emergency room can accommodate at most M = 5 patients. The patients arrive at a
rate of 4 per hour. The single staff physician can only treat 5 patients per hour. Any patient
overflow is directed to another hospital.
(a) Determine the probability distribution for the number of patients either waiting for or receiving
treatment at any given time.
(b) Determine the mean values for the number of patients in the emergency room , and the
number of patients waiting to see the doctor.
Solution:
M = Maximum number of customers in the system = 5
l = 4 per hour and m = 5 per hour
8.==m
lr
The probability distribution for the number of patients in the system
271.
1
1
P
Mn0forPP
1xo
o
n
n
=
-
-=
=
+
m
l
ml
m
l
The probability distribution for the number of patients either waiting for treatment or receiving
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treatment is as follows.
n Pn
0 0.271
1 0.217
2 0.173
3 0.139
4 0.111
5 0.089
(b) (i) The mean value for the number of patients in the emergency room = 0 (0.271) + 1
(0.217) + 2 (0.173) + 3 (0.139) + 4 (0.111) + 5 (0.089) = 1.869
(ii) Average number of patients waiting to see the doctor E(m) = 0 0.271 + 0 0.217 + 1
0.173 + 2 0.139 + 3 0.111 + 4 0.089 = 1.140.
Example 4:
Cars arrive at a toll gate on a frequency according to Poisson distribution with mean 90 per hour.
Average time for passing through the gate is 38 seconds. Drivers complain of long waiting time.
Authorities are willing to decrease the passing time through the gate to 30 seconds by introducing
new automatic devices. This can be justified only if under the old system, the number of waiting
cars exceeds 5. In addition the percentage of gates idle time under the new system should not
exceed 10. Can the new device be justified?
Solution:
In the old system,
Mean arrival rate = 90 per hour
= 1.5 per minute
Mean service rate =38
60 per minute
Traffic intensity =20
19
60
385.1 =
Expected number of waiting cars
181
2=
-=
r
r
Hence the new system is justified on the basis of the expected number of waiting cars.
Under the new system, the probability of device being idle
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= P0 = 1 r
= 1 .75
= .25 which is greater than 10% required for the new device to be introduced.
As only one of the stipulated conditions is fulfilled the new automatic device is not justified.
Example 5:
Customers arrive at a one-window drive-in bank according to a Poisson distribution with mean 10
per hour. Service time per customer is exponential with mean 5 minutes. The space in front of the
window, including that for the serviced car can accommodate a maximum of 3 cars. Other cars
can wait outside this space.
(a) What is the probability that an arriving customer can drive directly to the space in front of the
window?
(b) What is the probability that an arriving customer will have to wait outside the indicatedspace?
(c) How long is an arriving customer expected to wait before starting service?
(d) How many spaces should be provided in front of the window so that all the arriving
customers can wait in front of the window at least 20% of the time?
Solution:
(a) l = 10 m = 12 r = 10 /12 = 5/6 .
The probability that an arriving customer can drive directly to the space in front of the window
= P0 + P1 + P2
= Po + pPo + p2Po
= Po (1+p + p2)
= (1 p) (1+p+p2)
Hence required probability = .42
(b) The probability that an arriving customer has to wait outside the indicated space = Probability
that there are at least 3 customers in the space in front of the window
= 1 (P0 +P1 + P2 + P3)
= 1 (1 p) (1+p+p2+ p
8)
= .48
(c) The average waiting time of a customer in queue.
=417.0
12
5
)(==
- lmml
(d) The existing space in front of the window can accommodate upto three cars. Probability that a
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customer can wait in front of the window must be at least 0.20. This is possible if the number
of customers in the system is 0, 1 or2for at least 20% of the time.
i.e. P0 + P1 + P20.20
It seen from (a) that P0 + P1 + P2 = 0.42
As even with one space in front of the window 42% of the times an arrival an wait in the space,
the number of spaces required is one or more.
Self Assessment Questions 7
Write one line answers
1. Expected number of customers in non-empty Queue is given by
2. The probability that an arriving customer has to wait for receiving service.
8.9 Multiple Service Channels
The analysis of systems involving several service channels is more complex. However these
models are of major practical utility as one of the ways of improving service is to provide
additional service facilities. The list of formulae to be used is given in Table below this is following
by a few examples.
Table
Formulae for Poisson Arrivals, Exponential ServiceMulti channel Queuing Models Infinite Population
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mlm
mlm
lm
m
lm
m
l
lm
m
llm
lm
m
llm
r
1P
)c(!)1c()v(E
P)c()!1c(
)n(E
P)c()!1c(
)n(E
p)c()!1c(
)m(E
c
p1!c
pe!npn
1P
cnforP)c(!c
pu
cnforP!n
uP
o2
c
o2
c
o2
c
o2
c
1c
0n
o
ucn
on
+--
=
--
=
+--
=
--
=
-
+
=
=
4
Let us calculate the waiting time when c = 5
)c/p1(1c
o
!n
nP
1c
0n
1o -
+= -
=
- rr
Substituting for n, c and p
231
3Po =
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Average waiting time of a ship
o2P
1
)c()!1c(
1pe)w(E
lr--
+=
= 6.65
Which is greater than 6 hours and inadequate.
When c = 6, Po =899
15
And average waiting time of a ship E(w) = 1.71 hours.
Hence 6 berths should be provided at the port.
Example 7:
A tool crib is operated by M servers, and demands for service arrive randomly at mean rate of 1.4
per minute. The mean service time per server is 1.25 minutes, and times are exponentially
distributed. Finite population effects may be ignored. If the average hourly pay rate of a tool crib
operator is Rs. 2 per hour and the average hourly pay rate of production employee is Rs. 4 per
hour, determine the optimum value of M. (You may assume an 8 hour working day).
l = 1.4 per minute
m = .80 per minute
-
=
-
+
=
==
1c
0n
o
e
p1!c
pe!n
pn
1P
75.1m
lr
Substituting for and c,
P0 will be 0.067, 0.156 and 0.170 for c = 2, 3 and 4 respectively. Average waiting time of an
arrival in the queue.
o2
c
P)c()!1c(
)w(Elm
m
lm
--
=
As the number of arrivals is per minute average waiting time of production employee per day = E
(w) l 60 8
= E (v) 672
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For a c channel system, one server will be idle if n = c 1, two will be idle for n = c 2 and so
on. It can be shown that average idle time of service facilities m
l-= c .
For various values of c, the cost implications are presented in the Table below:Table
c
Average
waiting
time per
arrival
E(w) mts.
Average
waiting time
of all arrivals
per day
(E (w) 672
mts. per day)
Cost of
production
employees
per day
60
4(3)
(Rs.)
Average
idle time
of tool crib
operator
per day
= (cp)
hrs.
Cost of
idle time
of tool crib
operator
per day
(5) 2 8
(Rs.)
Total
cost per
day
(4) + (6)
(Rs.)
(1) (2) (3) (4) (5) (6) (7)
2 4.08 2741.76 182.79 0.25 4 186.79
3 0.33 221.76 14.78 1.25 20 34.78
4 0.07 47.04 3.14 2.25 36 39.14
The optimal number of servers is seen to be three
Example 8:
A bank has two counters for withdrawals. Counter one handles withdrawals of value less than Rs.300 and counter two Rs. 300 and above. Analysis of service time shows a negative exponential
distribution with mean service time of 6 minutes per customer for both the counters. Arrival of
customers follows Poisson distribution with mean eight per hour for counter one and five per hour
for counter two.
(a) What are the average waiting times per customer of each counter?
(b) If each counter could handle all withdrawals irrespective of their value how would the
average waiting time change?
Solution:
(a) For counter one m1 = 10 per hour, l1 = 8 per hour.
minutes24)(
E111
1)1n( =-
=lmm
l
For counter two, m 2 = 10 per hour, per hour, l2 = 5 per hour
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minutes6)(
)w(E222
22 =-
=lmm
l
(b) When each counter can handle all withdrawals irrespective of their value with an average
arrival rate of 13 per hour, mean service time of 10 per hour at each counter and number of
channels = 2,
21.0P
71.4
)o/p1(
pe
!c
1pn
!n
1P
30.110
13
o
1c
0n
1o
=
=
-+=
==
-
=
-
r
Average waiting time E(w) =02
c
P.)c(!)1c( lm
m
lm
--
= 4.2 minutes
Example 9:
St. Peters Public Library wants to improve its service facilities in terms of reducing the waiting
time of the borrowers. At present there are two counters in operation and the arrival of the
borrowers follows Poisson distribution with one arriving every 4 minutes on an average and the
service time follows Poisson distribution with one arriving every 4 minutes on an average and the
service time follows a negative exponential distribution with a mean of 5 minutes. The library has
relaxed its membership rules and a substantial increase in the number of borrowers is expected .
Find the number of additional counters to be provided if the present arrival rate is expected to
double and the average waiting time of a borrower is to be half of the present value
Solution:
l = 1/4 per minute, m = 1/5 per minute,
r = 5/4
The number of counters at present c = 2
)c/P1(!c
pc
!n
pn
P
1c
0n
1
o -+=
-
=
-
= 4.33
P0 = .23
Average waiting time of a customer
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E (w) =o2
c
P)c(!)1c( lm
m
lm
--
= 3.19 minutes.
As the arrival rate is expected to be doubled with the relaxation of membership rules, l =
2/4 = 1/2 per minute and m = 1/5 per minute.
Hence r = 5/2
We shall examine the average waiting time of a customer for different number of counters
When c = 3
P0 1
= 22.25
P0 = .04
E(w) = 7.02 minutes
This is more than the present average waiting time.
By adding one more counter, c = 4 and E(w) = 1.07 minute.
This is less than half of the present average waiting time. Thus it will be necessary to provide 2
additional counters to fulfill the stipulated service standards.
Self Assessment Questions 8
Write the formula for.
1. Expected number of customer in the system.
2. Average waiting time of customer in the Queue.
8.10 Erlang Family Of Distribution Of Service Times
Queuing processes discussed so far are mathematically simple. When assuming that the service
time follows negative exponential distribution, we were also taking that its standard deviation is
equal to its mean. As there would be situations where the mean and the standard deviation
substantially differ, the models have to be made more general by using a distribution which
conforms closely to the practical problems but yet retains the simplicity of the properties of
negative exponential distribution. A.K. Erlang had first studied such a distribution.
Consider the distribution of a service time involving a fixed number of phases k, each phase
having a negative exponential distribution. If there are k phases, and the average time taken by a
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customer through each phase is mk
1units, the service time distribution f(t) is given by
)!1k(
ej.)k()t(f
kt1kk
-
=-- m
m
The mean of this distribution is m
1and standard deviation is
k
1
m
If k = 1, f (f) = mem1 which is the negative exponential distribution same as the models
considered earlier.
For k = 2, f (f) = 4m 2te 2 m f and so on.
The mode is located at t = k
1k
m-
.
If k = the variance is zero and this corresponds to a case where the service time is constant
and has value.1
m
The Figure below shows the way the density functions varies as k increases.
Fig. 7.1
The measures of efficiency should take into account the number of customers getting service,
number having entered any one or more of the phases, and the number yet to join. For arrivals
following Poisson distribution, with mean l and service time following the k th Erlang distribution
with mean, m
1the formulae applicable are given in Table below
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Table
Formulae for Poisson Arrivals, Service with k phases, each phase having a negative exponential
distribution
Average queue length)(k2
1k)m(E
2
lmm
l-
+=
Average number of units in the systemm
l
lmm
l +-
+=)(k2
1k)n(E
2
Average waiting time )(k2
1k)w(E
lmm
l-
+=
Average time spent in the systemmlmm
l 1
)(k2
1k)v(E +
-+=
For constant service time, equating k to
mlmm
l
lmm
l
m
l
lmm
l
lmm
l
1
)(2)v(E
)(2)w(E
)(2)n(E
)(2)m(E
2
2
+-
=
-=
+-
=
-=
Example 10:
In a cafeteria at a bus depot the customers will have to pass through three counters. At the first
counter the customers buy Coupons at the second they select and collect the snacks to be taken,
and at the, third they collect coffee or tea as required. The server at each counter takes on an
average two minutes although the distribution of the time of service is approximately exponential.
If the arrivals of customers to the cafeteria are approximately Poisson at an average rate of six
per hour, what is the average time spent by a customer waiting in the cafeteria ? What is the
average time of getting the service ? What is the most probable time spent in getting the service ?
Solution
This is a queuing process with service time following Erlang distribution.
No. of phases k = 3
23
1 =m
\m = 1/6 persons per minute
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l = 6 persons per hour
= 1/10 person per minute
(i) Average waiting time =)(k2
1k
lmm
l-
+
= 6 minutes
(ii) Average time spent in collecting coupons, snacks, etc
= m
1= 6 minutes
(iii) Most probable time spent in collecting coupons, snacks etc
= Modal value
= k
1k
m-
= 4 minutes
Self Assessment Questions 9
Write formula for the following characteristics.
1. E (m)
2. E (w)
8.11.Summary
The flow of customers form infinite population towards a service form secure on account of lack of
services facilities. The waiting line theory or Queuing Theory analysis the number of facilities
required and the cost of customers waiting time and suggest the optimum service level. It
contributes vital information required for balancing the cost of service and cost associated with
waiting time of the customer. There are different models which are used under different
conditions all these models were discussed in this unit.
Terminal Questions
1. If in a particular single-server system, the arrival rate, l = 5 per hour and service rate, m = 8
per hour. Assume the conditions for use of the single channel queuing model, find out:
a. The probability that the server is idle.
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b. The probability that there are at least two customers in the system.
c. Expected time that a customer is in the queue.
2. Customers arrive at the First Class Ticket counter of a Theater at a rate of 12 per hour. There
is one clerk serving the customers at a rate of 30 per hour. Assuming the conditions for use
of the single channel queuing model, evaluate:
a. The probability that an arriving customer has to wait for the service.
b. The expected no. of customers in the system.
c. The average waiting time of the customer in the system.
3. In a bank every 15 minutes one customer arrives for cashing the cheque. The clerk takes 10
minutes to service. Assuming the usual conditions find.
a. The ideal time of the clerk in 8 working hours.
b. Expected no. of customers in the queue.
c. The probability that a customer has to wait for 15 minutes or more to receive the service.
Answers To Self Assessment QuestionsSelf Assessment Questions 1
1. False 2. True 3. False
Self Assessment Questions 2
1. Utilization
2. Various alternatives
3.Simulation
Self Assessment Questions 3
1. Yes 2. Yes
Self Assessment Questions 4
1. Series, parallel
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2. Constant, varying
Self Assessment Questions 5
1. Jockeying
2. Balancing
Self Assessment Questions 6
1. Average length of Queue
2. F (m)
3. Simultaneous
Self Assessment Questions 71. m / m - l
2. l / m
Self Assessment Questions 8
1. Refer Section 8.9
Self Assessment Questions 9
1. Refer Section 8.10
Answer To Terminal Questions
1. a) 3 /8 b)125 / 512 c)12.5 minutes
2. a) 3 /5 b) 2 / 3 per hour c) 10 /3 minutes
3. a) 8 /3 hrs b) 4 /3 c) 0.4043