operations research 8

7/27/2019 Operations Research 8

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Operations Research Unit 8

Sikkim Manipal University 127

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.


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.


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


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.


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.


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.


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.


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


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


1. Utilization

2. Various alternatives

3.Simulation


1. Yes 2. Yes


1. Series, parallel


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2. Constant, varying


1. Jockeying

2. Balancing


1. Average length of Queue

2. F (m)

3. Simultaneous

Self Assessment Questions 71. m / m - l

2. l / m


1. Refer Section 8.9


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

operations research 8

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