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CHAPTER 1

OPTIMIZATION MODELS

Unit -A

UNIT STRUCTURE

1A.1 Introduction to Optimization Models

1A.2 Definitions of Optimization Models

1A.3 Features of Optimization Models

1A.4 Approaches to Optimization Models

1A.5 Exercise

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1A.1. INTRODUCTION TO OPTIMIZATION MODELS

Modeling is the essence of operation research. A model is an abstraction of idealized representation

of a real life problem. Modeling is a real life situation helps us to study the different behavior of

the problem corresponding to the description of the problem. A model can be a picture, a map, a

curve or an equation. The reliability of the decision drawn from the model may depend upon the

validity of the model or the basic assumptions on which the model is built.

Modeling a real life situation helps us to study the different behavior of the problem corresponding

to the description of the problem. Great efforts have been taken by expert to model business

situations, military operations, motions of planets and stars and so on. A model is an abstraction

or an idealized representation of a real life problem. The objective of a model is to provide means

for analyzing the behaviors of the system for further improvement. A map of multiple activity

chart, a project network, the representation of the behaviors of a queuing system, a model to

forecast the future and the present factors of time series, etc. are all examples a based on the past

etc. are all examples of models.

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An iconic model is a physical representation of a real project on a different scale. A toy automobile

is an iconic model of a real automobile. A globe is an iconic model of the earth. A photograph is

also an iconic model. In general, they enlarge or reduce the size of a real system. In other words,

they are the images of a real object.

Optimization is central to any problem involving decision making, whether in engineering or in

economics or in management. The task of decision making entails choosing between various

alternatives. This choice is governed by our desire to make the "best" decision. The measure of

goodness of the alternatives is described by an objective function or performance index.

Optimization theory and methods deal with selecting the best alternative in the sense of the given

objective function.

The area of optimization has received enormous attention in recent years, primarily because of the

rapid progress in computer technology, including the development and availability of user -friendly

software, high -speed and parallel processors, and artificial neural networks.

An iconic model is a physical representation of a real project on a different scale. A toy automobile

is an iconic model of a real automobile. A globe is an iconic model of the earth. A photograph is

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also an iconic model. I n general, they enlarge or reduce the size of a real system. In other words,

they are the images of a real object.

Advantages of Model:

An iconic model is concrete.

It is easy to construct the model.

It is easy to study the model than the system itself.

Disadvantages of Model:

This model is not suited for further manipulation.

It cannot be used to study the changes in the operation of the system.

It is not possible to make any modification of the model.

Adjustment with changing situations cannot be done in this model.

A model is a vehicle for arriving; a well -structured problem of reality. A commentator of a cricket

match describes the play as model to enable us to predict the future course of events of the play. It

is a descriptive model available for further analysis. It is not always possible to analyze a situation

only with the description of the situation. We have to formulate the problem into concrete

mathematical representation in the form of a curve, graph or equations. Models could be classified

as iconic model, analogue model and symbolic model.

1A.2 . DEFINITION OF OPTIMIZATION MODELS

Optimization is the act of obtaining the best result under given circumstances. In design,

construction, and maintenance of any engineering system, engineers and business managers have

to take many technological and managerial decisions at several stages. The ultimate goal of all

such decisions is either to minimize the effort required or to maximize the desired benefit. Since

the effort required or the benefit desired in any practical situation can be expressed as a function

of certain decision variables, optimization can be defined as the process of finding the conditions

that give the maximum or minimum value of a function . munotes.in

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One possible definition of optimization models is “Mathematical Models designed to help

institutions and individuals decide how to a llocate scarce resources , to activities and to make best

use of their circumstances. Hence optimization models are mathematical models designed to help

the managers make better decisions.

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1A.3 . FEATURES OF OPTIMIZATION MODELS

An optimization model has three main components:

An objective function. This is the function that needs to be optimized.

A collection of decision variables. The solution to the optimization problem is the set of values

of the decision variables for which the objective function reaches its optimal value.

A collection of constraints that restrict the values of the decision variables.

The Optimization M odel class provides a common API for defining and accessing variables and

constraints, as well as other properties of each model.

The existence of optimization methods can be traced to the days of Newton, Lagrange, and Cauchy.

The development of diffe rential calculus methods of optimization was possible because of the

contributions of Newton and Leibnitz to calculus. The foundations of calculus of variations, which

deals with the minimization of functional, were laid by Bernoulli, Euler, Lagrange, and Weirstrass.

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The method of optimization for constrained problems, which involves the addition of unknown

multipliers, became known by the name of its inventor, Lagrange. Cauchy made the first

application of the steepest descent method to solve unconstrained minimization problems. Despite

these early contributions, very little progress was made until the middle of the twentieth century,

when high -speed digital computers made implementation of the optimization procedures possible

and stimulated further researc h on new methods. Spectacular advances followed, producing a

massive literature on optimization techniques. This advancement also resulted in the emergence

of several well -defined new areas in optimization theory.

It is interesting to note that the major developments in the area of numerical methods of

unconstrained optimization have been made in the United Kingdom only in the 1960s. The

development of the simplex method by Dantzig in 1947 for linear programming problems and the

annunciation of the princip le of optimality in 1957 by Bellman for dynamic programming

problems paved the way for development of the methods of constrained optimization. Work by

Kuhn and Tucker in 1951 on the necessary and sufficiency conditions for the optimal solution of

programmi ng problems laid the foundations for a great deal of later research in nonlinear

programming. The contributions of Zoutendijk and Rosen to nonlinear programming during the

early 1960s have been significant. Although no single technique has been found to be universally

applicable for nonlinear programming problems, work of Carroll and Fiacco and McCormick

allowed many difficult problems to be solved by using the well -known techniques of

unconstrained optimization. Geometric programming was developed in the 1960s by Duffin,

Zener, and Peterson. Gomory did pioneering work in integer programming, one of the most

exciting and rapidly developing areas of optimization. The reason for this is that most real -world

applications fall under this category of problems. Dantzig and Charnes and Cooper developed

stochastic programming techniques and solved problems by assuming design parameters to be

independent and normally distributed.

There is no single method available for solving all optimization problems efficiently. Hence a

number of optimization methods have been developed for solving different types of optimization

problems. The optimum seeking methods are also known as mathematical programming munotes.in

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techniques and are generally studied as a part of operations research. Operations research is a

branch of mathematics concerned with the application of scientific methods and techniques to

decision making problems and with establishing the best or optimal solutions. The beginnings of

the subject of operations research can be traced to the early period of World War II. During the

war, the British military faced the problem of allocating very scarce and limited resources (such

as fighter airplanes, radars, and submarines) to several activities (deployment to numerous targets

and destinations). Because there were no systematic methods available to solve resource allocation

problems, the military called upon a team of mathematicians

to develop methods for solving the problem in a scientific manner. The methods developed by the

team were instrumental in the winning of the Air Battle by Britain. These methods, such as linear

programming, which were developed as a result of research on (military) operations, subsequently

became known as the methods of operations research.

1A.4 . APPROACHES TO OPTIMIZATION MODELS

Components of an Optimization Model

An optimization model has three main components:

An objective function. This is the function that needs to be optimized.

A collection of decision variables. The solution to the optimization problem is the set of values

of the decision variables for which the objective function reaches its optimal value.

A collection of constraints that restrict the values of the decision variables.

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Types of Optimization Models

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Optimization problems can be classified in terms of the nature of the objective function and the

nature of the constraints. Special forms of the objective function and the constraints give rise to

specialized algorithms that are more efficient. From this p oint of view, there are four types of

optimization problems, of increasing complexity.

An unconstrained optimization problem is an optimization problem where the objective function

can be of any kind (linear or nonlinear) and there are no constraints. Thes e types of problems are

handled by the classes discussed in the earlier sections.

A linear program is an optimization problem with an objective function that is linear in the

variables, and all constraints are also linear. Linear programs are implemented by the Linear

Program class.

A quadratic program is an optimization problem with an objective function that is quadratic in the

variables (i.e. it may contain squares and cross products of the decision variables), and all

constraints are linear. A quadratic program with no squares or cross products in the objective

function is a linear program. Quadratic programs are implemented by the Quadratic Program class.

A nonlinear program is an optimization problem with an objective function that is an arbitrary

nonlinear function of the decision variables, and the constraints can be linear or nonlinear.

Nonlinear programs are implemented by the Nonlinear Program class.

Decision Variables

Finding the optimal values of the decision variables is the goal of solving an optimization model.

In the optimization framework, variables are implemented by the Decision Variable class.

Each variable has a name , which may be generated automatically. The Lower Bound and Upper

Bound properties specify lower and upper bounds for the values the variable can take. After the

solution of the model has been computed, the value property returns the value of the variable in

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Specific models may have specialized vers ions of the decision variables. For example, both linear

and quadratic programs use variables of type Linear Program Vari able, which inherits

from Decision Variable and has an extra Cost property that represents the coefficient of the

variable in the linear portion of the objective function.

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Constraints

Constraints limit the possible values for the decision variables in an optimization model. There are

several types of constraints. The classes that implement them all inherit from the constraint class.

Each constraint also has a Name , which may again be generated automatically. The Lower

Bound and Upper Bound properties specify lower and upper bounds for the value of the constraint.

After the solution of the model has been computed, the Value property returns the value of the

constraint in the optimal solution.

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There are two types of constraints: linear and nonlinear . Linear constraints express that a linear

combination of the decision variables must lie within a certain range. Nonlinear constraints express

that the value of some arbitrary function of the decision variables must lie within a certain range.

In mathematics , computer science and operations research , mathematical

optimization (alternatively, mathematical programming or simply, optimization) is the selection

of a best element (with regard to some criterion) from some set of available alternatives.

In the simple st case, an optimization problem consists of maximizing or minimizing a real

function by systematically choosing input values fr om within an allowed set and computing

the value of the function. The generalization of optimization theory and techniques to other

formulations comprises a large ar ea of applied mathematics . More generally, optimization

includes finding "best available" values of some objective function given a defined domain (or

input), including a variety of different types of objective functions and different types of domains.

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1A.5 . EXERCISE

Question 1 . What do you understand by optimization models? Give examples.

Question 2 . What are the features of optimization models?

Question 3. Define optimization models. Give examples.

Question 4 . Explain the structure and types of optimization models which your own examples.

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CHAPTER 1

OPTIMIZATION MODELS

Unit -B

UNIT STRUCTURE

1B.1 Models and Modeling in Operations Research

1B.2 Definitions of Optimization Models

1B.3 Summing Up

1B.4 Exercise

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1B.1 . MODELS AND MODELING IN OPERATIONS RESEARCH

Models

Operations research (OR) is concerned with scientifically deciding how to best design and operate

man-machine systems, usually under conditions requiring the allocation of scarce resources.

No matter how OR is defined, the construction and use of models is at its core. Models are

representations of real systems. They can be iconic (made to look like the real system), abstract,

or somewhere in between.

Iconic models can be full -scale, scaled -down, or scaled -up in size. Sawmill headrig control

simulators are full-scale models. A model of the solar system is a scaled -down model, and a

teaching model of a wood cell or a water molecule is a scaled -up model. Models can be made of

the same material as the system they represent, such as the head rig control simulat or, or they can

be made of different materials, such as a plastic model of the solar system.

On the other end of the model spectrum are abstract mathematical models. OR professionals often

use mathematical models to make simplified representations of comp lex systems.

Regardless of the type of model used, modeling includes the following steps:

• Defining the problem and gathering data

• Constructing a model of the system

• Deriving a solution

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• Testing the model and solution (Is the model valid; that is, does it do what it is designed to

do?)

• Implementing the solution .

Modeling in Operations Research (OR)

If the actual system is simple enough to manipulate safely ; then OR techniques often are not

neces sary. For example, managers of the sawmill might wish to extend the hours for loading trucks

from 5:00 p.m. until midnight. Since the shipping shed is adjacent to the planer mill, they decide

to use the planer mill employees and possibly one or two overtime employees from shipping to

extend the loading hours for a trial period of 2 to 3 weeks. At the end of this time, they’ll have a

good idea of how the extended shipping hours affected the rest of the sawmill operations.

An OR professional also could model this problem, perhaps using a scheduling linear program or

a simulation model to examine the effect of extending the shipping hours. However, if the real

system can be manipulated without causing too much disruption, it would be preferable to do so

rather than to have an OR professional model it. Often, it can take weeks or months just to collect

enough data to realistically model a system. In this case, it probably would be less expensive to

manipulate the real system.

Meteorologists build models to study the weather because they can ’t experiment with the weat her

itself. One problem is repeatability. It ’s impossible to set up repeatable experiments with systems

over which the researcher has no control, such as the weather. Meteorologists can observe but not

test the system. With a model, however, they can test their theories. They can run the model under

different conditions and observe the weather to see whether the model is a good enough (although

simplified) representation of the real system. Using the same methods, an experiment using the

model is repeatable .

An important, but often overlooked, benefit to modeling is the insight one gets from the process.

Often, modelers learn a lot about a system during problem definition and data gathering. An

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process often reveals previously overlooked insights. Information gained from these new insights

can lead to problem resolution even before the model is completed.

An important step in model creation is validation. A model is valid if it does what it was intended

to do. The model creator may believe the model is valid, while other users may not. Thus,

validation is not as rigid a term as verification or proof and is somewhat subjective.

Operations research experts can stu dy actual systems, but usually use models instead .

What makes a good model?

A simple model is better than a complex one as long as it works as well. A model only needs to

perform its intended function to be valid. A model should be easy to understand. It’s important to

use the most relevant OR tool when constructing a model. A modeler should not try to shape the

problem to fit a particular OR method. For example, a linear programming (LP) expert may try to

use LP on a problem where there is no optimal solution. Instead, modelers should study the

problem and choose the most appropriate OR tool.

For complicated systems, users need to remember that models are only simplified representations.

If a user mistakenly considers a complicated model to be correct, he or she may disregard further

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study of the real system. Modelers and users of models never should rely only on a model’s output

and ignore the real system being modeled.

A good model should be easy to modify and update. New information from the real sys tem can be

incorporated easily into a well -planned model. A good model usually starts out simple and

becomes more complex as the modeler attempts to expand it enough to give meaningful answers.

1B.2 . ADVANTAGES AND APPLICATIONS OF OPTIMIZATION MODELS

• Advantages of Optimization Models

Most operations research studies involve the construction of a mathematical model. The model is a

collection of logical and mathematical relationships that represents aspects of the situation under

study. Models describe important relationships between variables; include an objective function

with which alternative solutions are evaluated, and constraints that restrict solutions to feasible

values.

Although the analyst would hope to study the broad implications of the problem using a systems

approach, a model cannot include every aspect of a situation. A model is always an abstraction that

is of necessity simpler than the real situation. Elements that are irrelevant or unimportant to the

problem are to be ignored; hopef ully leaving sufficient detail so that the solution obtained with the

model has value with regard to the original problem.

Models must be both tractable, capable of being solved, and valid, representative of the original

situation. These dual goals are often contradictory and are not always attainable. It is generally true

that the most powerful solution methods can be applied to the simplest, or most abstract, model.

• Applications of Optimization Models in OR

Operations Research uses any suitable tools or techniques available. The common frequently used

tools/techniques are mathematical procedures, cost analysis, electronic computation. However,

operations researchers given special importance to the development and the use of techniques like

linear progr amming, game theory, decision theory, queuing theory, inventory models and

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programming, integer programming, dynamic programming, sequencing theory, Markov process,

network scheduling (PERT/CPM), symbolic Model, information theory, and value theory. There

are many other Operations Research tools/techniques also exists. The brief explanations of some of

the above techniques/tools are as follows:

Linear Programming:

This is a constrained optimization technique, which optimize some criterion within some

constraints. In Linear programming the objective function (profit, loss or return on investment) and

constraints are linear. There are different methods available to solve linear programming.

Game Theory:

This is used for making decisions under conflicting situations where there are one or more

players/opponents. In this the motive of the players are dichotomized. The success of one player

tends to be at the cost of other players and hence they are in conflict.

Decision Theory:

Decision theory is concerned with making decisions under conditions of complete certainty about

the future outcomes and under conditions such that we can make some probability about what will

happe n in future.

Queuing Theory:

This is used in situations where the queue is formed (for example customers waiting for service,

aircrafts waiting for landing, jobs waiting for processing in the computer system, etc). The

objective here is minimizing the cost of waiting without increasing the cost of servicing.

Inventory Models:

Inventory model make a decision that minimize total inventory cost. This model successfully

reduces the total cost of purchasing, carrying, and out of stock inventory.

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Simulation:

Simulation is a procedure that studies a problem by creating a model of the process involved in

the problem and then through a series of organized trials and error solutions attempt to determine

the best solution. Sometimes this is a difficult/ time consuming procedure. Simulation is used when

actual experimentation is not feasible or solution of model is not possible.

Non-linear Programming:

This is used when the objective function and the constraints are not linear in nature. Linear

relations hips may be applied to approximate non -linear constraints but limited to some range,

because approximation becomes poorer as the range is extended. Thus, the non -linear

programming is used to determine the approximation in which a solution lies and then th e solution

is obtained using linear methods.

Dynamic Programming:

Dynamic programming is a method of analyzing multistage decision processes. In this each

elementary decision depends on those preceding decisions and as well as external factors.

Integer Programming:

If one or more variables of the problem, take integral values only then dynamic programming

method is used. For example, number or motor in an organization, number of passenger in an

aircraft, number of generators in a power generating plant, etc.

Markov Process :

Markov process permits to predict changes over time information about the behavior of a system

is known. This is used in decision making in situations where the various states are defined. The

probability from one state to another state is known and depends on the current state and is

independent of how we have arrived at that particular state.

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Network Scheduling:

This technique is used extensively to plan, schedule, and monitor large projects (for example

computer system install ation, R & D design, construction, maintenance, etc.). The aim of this

technique is minimize trouble spots (such as delays , interruption, production bottlenecks, etc.) by

identifying the critical factors. The different activities and their relationships of the entire project

are represented diagrammatically with the help of networks and arrows, which is used for

identifying critical activities and path.

There are two main types of technique in network scheduling, they are:

• Program Evaluation and Review Technique (PERT) – is used when activities time is not known

accurately/ only probabilistic estimate of time is available.

• Critical Path Method (CPM) – is used when activities time is know n accurately.

Information Theory:

This analytical process is trans ferred from the electrical communication field to O.R. field. The

objective of this theory is to evaluate the effectiveness of flow of information with a given system.

This is used mainly in communication networks but also has indirect influence in simulat ing the

examination of business organizational structure with a view of enhancing flow of information.

Today, almost all fields of business and government utilizing the benefits of Operations Research.

There are voluminous of applications of Operations Re search. Although it is not feasible to cover

all applications of O R in brief , the following are the abbreviated set of typical operations research

applications to show how widely these techniques are used today:

Accounting:

• Assigning audit teams effectively

• Credit policy analysis

• Cash flow planning

• Developing standard costs

• Establishing costs for byproducts

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Construction:

• Project scheduling, monitoring and control

• Determination of proper work force

• Deploymen t of work force

• Allocation of resources to projects

Facilities Planning:

• Factory location and size decision

• Estimation of number of facilities required

• Hospital planning

• MBA -H2040 Quantitative Techniques for Managers

• International logistic system design

• Transportation loading and unloading

• Warehouse location decision

Finance:

• Building cash management models

• Allocating capital among various alternatives

• Building financial planning models

• Investment analysis

• Portfolio analysis

• Dividend policy making

Manufacturing:

• Inventory control

• Marketing balance projection

• Production scheduling

• Production smoothing

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Marketing:

• Advertising budget allocation

• Product introduction timing

• Selection of Product mix

• Deciding most effective packaging alternative

Organizational Behavior / Human Resources:

• Personnel planning

• Recruitment of employees

• Skill balancing

• Training program scheduling

• Designing organizational structure more effectively

Purchasing:

• Optimal buying

• Optimal reordering

• Materials transfer

Resear ch and Development:

• R & D Projects control

• R & D Budget allocation

• Planning of Product introduction

Competitive Strategies

Project Management

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1B.3 . LET US SUM UP

In this UNIT – B of Chapter 1 , you have learnt Models and Modeling in Operations Research,

Advantages and Applications of Optimization Models .

Operations Research is relatively a new discipline, which originated in World War II, and became

very popular throughout the world. India is one of the few

first countries in the world who started using operations research. Operations Research is used

successfully not only in military/army operations but also in business, government and industry.

Now a day’s operations research is almost used in all the fields.

Proposing a definition to the operations research is a difficult one, because its boundary and content

are not fixed. The tools for operations search is provided from the subject’s viz. economics,

engineering, mathematics, statistics, psychology, etc., which helps to choose possible a lternative

courses of action. The operations research tool/techniques include linear programming, non -linear

programming, dynamic programming, integer programming, Markov process, queuing theory, etc.

Operations Research has a number of applications. Sim ilarly, it has a number of limitations, which

is basically related to the time, money, and the problem involves in the model building. Day -by-

day operations research gaining acceptance because it improves decision making effectiveness of

the managers. Almo st all the areas of business use the operations research for decision making.

1B.4 . EXERCISE

Question 1 . Discuss attributes of a good model and the process of modeling.

Question 2. Explain different modeling techniques of OR.

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CHAPTER 2

SEQUENCING MODELS

UNIT STRUCTURE

2.1 Introduction

2.2 Processing n - jobs through Two machine

2.3 Processing n - jobs through Three machines

2.4 Processing n - jobs through m - machines

2.5 Processing Two jobs through m - machines

2.6 Graphical Solution of Two jobs and m - machines Sequencing Problem

2.7 Exercise

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2.1. INTRODUCTION

The short -term schedules show an optimal order (sequence) and time in which jobs are

processed as well as show timetables for jobs, equipment, people, materials, facilities and all

other resources that are needed to support the production plan. The schedul es should use

resources efficiently to give low costs and high utilisations. Other purpose of scheduling is,

minimising customers wait time for a product, meeting promised delivery dates, keeping

stock levels low, giving preferred working pattern, minimisi ng waiting time of patients in a

hospital for different types of tests and so on.

The general scheduling or sequencing problem may be described as: Let there be n jobs to be

performed one at a time on each of m machines. The sequence (order) of the machine s in

which each job should be performed is given. The actual or expected time required by the

jobs on each of the machines is also given. The general sequencing problem, therefore, is to

find the sequence out of (n!)m possible sequences which minimise the total elapsed time

between the start of the job in the first machine and the completion of the last job on the last

machine.

In particular, if n = 3 and m= 3, then total number of possible sequences will be (3!)3 = 216.

Theoretically, it may be possible to find optimum sequence but it will require a big

computational time. Thus, one should adopt sequencing technique.

To find optimum sequence we first calculate the total elapsed time for each of the possible

sequen ces. As stated earlier, even if values of m and n are very small, it is difficult to get the

desired sequence with total minimum elapsed time. However, due to certain rules designed

by Johnson, the task of determining an optimum sequence has become quite e asy.

NOTATIONS, TERMINOLOGY AND ASSUMPTIONS

Notations

tij = Processing time (time required) for job i on machine j.

T = Total elapsed time for processing all the jobs. This includes idle time, if any.

Iij = Idle time on machine j from the end of job (j - 1) to the start of job i.

Terminology

• Number of Machines: The number of machines refer to the number of service

facilities through which a job must pass before it is assumed to be completed.

• Processing Ti me: It is the time required by a job on each machine.

• Processing Order: It refers to the order (sequence) in which machines are required

for completing the job.

• Idle Time on a Machine: It is the time for which a machine does not have a job to

process, i.e . idle time from the end of job (i -1) to the start of job i.

• Total Elapsed Time: It is the time interval between starting the first job and

completing the last job including the idle time (if any) in a particular order by the

given set of machines.

• No Passi ng Rule: It refers to the rule of maintaining the order in which jobs are to be

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machines M 1 and M 2 in the order M 1 M2, then each job should go to machine M 1 first

and then to M 2.

Assumptions

1. The processing time on different machines are exactly known and are independent of the

order of the jobs in which they are to be processed.

2. The time taken by the job in moving from one machine to another is negligible.

3. Once a job h as begun on a machine, it must be completed before another job can begin on

the same machine.

4. All jobs are known and are ready for processing before the period under consideration

begins.

5. Only one job can be processed on a given machine at a time.

6. Machines to be used are of different types.

7. The order of completion of jobs are independent of the sequence of jobs.

2.2. PROCESSING n JOBS THROUGH TWO MACHINES

Let there be n jobs, each of which is to be processed through two machines, M 1 and M 2 in the

order M 1 M2, i.e. each job has to be passed through the same sequence of operations. In other

words, a job is assigned on machine M 1 first and after it has been completely processed on

machine M 1, it is assigned to machine M 2. If the machine M 2 is not free at the moment for

processing the same job, then the job has to wait in waiting line for its turn on machine M 2,

i.e. passing is not allowed.

Since passing is not allowed, therefore, machine M 1 will remain busy in processing all the n

jobs one -by-one which machine M 2 may remain idle time of the second machine. This can be

achieved only by determining sequence of n jobs which are to be processed on two machines

M1 and M 2. The procedure suggested by Johnson for determining the optimal sequence can

be summarised as follows:

The Algorithm

Step 1 List the jobs along with their processing times on each machine in a table as shown

below:

Processing Time on Machine Job Number

1 2 3 n

M1 t11 t12 t13 t1n

M2 t21 t22 t23 t2n

Step 2 Examine the columns for processing times on machines M 1 and M 2, and find the

smallest processing time in each column, i.e . find out, min. (t 1j, t2j) for all j.

Step 3 (a) If the smallest processing time is on machine M 1, then schedule the job as early as

possible without moving jobs already schedules, i.e . place the job in the first available

position in the sequence. If the processing time is on machine M 2, then schedule the job as

late a s possible without moving any jobs already scheduled, i.e. place the job in the last

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(b) If there is a tie in selecting the minimum of all the processing times, then there may be

three situations:

a. Minimum among all pr ocessing times is same for the machine i.e. min (t 1j, t2j) = t 2k = t2r,

then proce ss the kth job first and the rth job last.

b. If the tie for the minimum occurs among processing times t 1j on machine M 1 only, then

select the job corresponding to the smallest job subscript first.

c. If the tie for the minimum occurs among processing times t 2j on machine M 2, then select

the job corresponding to the largest job corresponding to the largest job subscript las t.

Step 4 Remove the assigned jobs from the table. If the table is empty, stop and go to Step 5.

Otherwise, got to Step 2.

Step 5 Calculate idle time for machines M 1 and M 2:

a. Idle time for machine M 1 = (Total Elapsed Time) - (Time when the last job in a sequence

finishes on machine M 1)

b. Idle time for machine M 2 = Time at which the first job in a sequence finishes on machine

M1 + ∑𝑛

𝑗=2{(Time when the jth job in a sequence starts on machine M 2) - (Time when the (j

- 1)th job in a sequence finishes on machine M 2)}.

Step 6 The total elapsed time to process all jobs through two machines is given by

Total Elapsed time = Time when the nth job in a sequence finishes on Machine M 2.

= ∑ 𝑀2𝑗𝑛

𝑗=1 + ∑ 𝐼2𝑗𝑛

𝑗=1

where M 2j = Time required for processing jth job on machine M 2.

I2j = Time for which machine M 2 remains idle after processing (j - 1)th job and before starting

work in jth job.

EXAMPLE 1

Find the sequence that minimises the total elapsed time required to complete the following

tasks on two machines:

Task A B C D E F G H I

Machine I 2 5 4 9 6 8 7 5 4

Machine II 6 8 7 4 3 9 3 8 11

Solution:

The sm allest processing time between the two machines is 2 which corresponds to task A on

Machine I. Thus, task A is scheduled as early as possible to give the sequence as shown

below:

A

After the task A has been set for processing first, we are left with 8 tasks and their processing

times as given below:

Task B C D E F G H I

Machine I 5 4 9 6 8 7 5 4

Machine II 8 7 4 3 9 3 8 11

The minimum processing time in this reduced problem is 3 which corresponds to task E and

G both on machine II. Since the corresponding processing time of task E on machine I is less munotes.in

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than the corresponding processing time of task G on machine I, therefore, task E will be

scheduled in the last and task G shall be scheduled before it. Tasks E and G will not be

considered further. Thus, current partial sequence of scheduling tasks becomes:

A G E

A set of processing times now gets reduced to:

Task B C D F H I

Machine I 5 4 9 8 5 4

Machine II 8 7 4 9 8 11

The smallest processing time in this reduced problem is 4, which corresponds to task C and I

on machine I and to task D on machine II. Thus task C will be placed in the second sequence

cell and task I in the third sequence cell and task D in the sequence cell before task G. The

entries of the partial sequence are now:

A C I D G E

The set of processing time now gets reduced as follows:

Task B F H

Machine I 5 8 5

Machine II 8 9 8

the smallest processing time in this reduced problem is 5, which corresponds to tasks B and H

both on machine I. Since the corresponding processing times of B and h on machine II is

same, therefore, either of these two tasks can be placed in fourth and fifth s equence cells.

Thus, it indicates an alternative optimal sequence. the optimal sequences are, therefore, given

below:

A C I B H F D G E

A C I H B F D G E

The minimum elapsed time for machines I and II is calculated as shown in Table 1.

Task Sequence Machine I Machine II

Time In Time Out Time In Time Out

A

C

I

B

H

F

D

G

E 0

2

6

10

15

20

28

37

44 2

6

10

15

20

28

37

44

50 2

8

15

26

34

42

51

55

58 8

15

26

34

42

51

55

58

61

In table 1, the minimum elapsed time, i.e . time from start of task A to completion of last task

E is 61 hours. During this time the machine I remains idle for 61 - 50 = 11 hours. The idle

time for machine II is equal to the time at which the first task A in the sequence finishes on

machine I plus the last task E in the sequence starts on machine II minus the last but one ta sk

G finishes on machine II, i.e. 2 + 58 - 58 = 2 hours.

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2.3. PROCESSING n JOBS THROUGH THREE MACHINES

Johnson provides an extension of his procedure to the case in which there are three instead of

two machines. Each job is to be processed through three machines M 1, M 2 and M 3. The list of

jobs with their processing times is given below. An optimal solution to this problem can be

obtained if either or both of the following c onditions hold good.

Processing time

on Machine Job Number

1 2 3 4

M1

M2

M3 t11

t21

t31 t12

t22

t32 t13

t23

t33 t1n

t2n

t3n

1. The minimum processing time on machine M 1 is at least as great as the maximum

processing time on machine M 2, that is,

min t 1j ≥ max t1j, j = 1, 2, 3, ...n

2. The minimum processing time on machine M 3 is at least as great as the maximum

processing time on machine M 2, that is

min t3j ≥ max t2j, j = 1, 2, 3, ...n

If either or both the above conditions hold good, then the steps of the algorithm can be

summarised in the following steps:

THE ALGORITHM

Step 1: Examine processing times of given jobs on all three machines and if either one or

both the above conditions hold, then go to step 2, otherwise the algorithm fa ils.

Step 2: Introduce two fictitious machines, say G and H with corresponding processing times

given by

i. t Gj = t1j + t2j, j = 1, 2, 3, .....n.

that is, processing time on machine G is the sum of the processing times on machines M 1 and

M2, and

ii. t Hj = t2j + t3j, j = 1, 2, 3, .....n.

that is, processing time on machine H is the sum of the processing times on machines M 2 and

M3.

Step 3: Determine the optimal sequence of jobs for this n -job, two machine equivalent

sequencing problem with th e prescribed ordering GH in the same way as discussed earlier.

EXAMPLE

Find the sequence that minimises the total time required in performing the following jobs on

three machines in the order ABC. Processing times (in hours) are given in the following

table:

Job 1 2 3 4 5

Machine A 8 10 6 7 11

Machine B 5 6 2 3 4 munotes.in

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Machine C 4 9 8 6 5

Solution: Here, min (t Aj) = 6; Min (t Cj) = 4; max (t Bj) = 6. Since min (t Aj) ≥ (tBj) for all j is

satisfied, the given problem can be converted into a problem of 5 jobs and two machines. The

processing time on two fictitious machines G and H can be determined by the following

relationships:

tGj = tAj + tBj, j = 1, 2, 3, .....n.

and t Hj = tBj + tCj, j = 1, 2, 3, .....n.

The processing times for the new problem are g iven below:

Job 1 2 3 4 5

Machine G

Machine H 8 + 5 = 13

5 + 4 = 9 10 + 6 = 16

6 + 9 = 15 6 + 2 = 8

2 + 8 = 10 7 + 3 = 10

3 + 6 = 9 11 + 4 = 15

4 + 5 = 9

When the algorithm described for n jobs on two machines is applied to this problem, the

optimal sequence so obtained is given by

3 2 5 1 4

The total minimum elapsed time is given in Table 1:

Job

Sequence Machine A Machine B Machine C

Time In Time Out Time In Time Out Time In Time Out

3

2

5

1

4 0

6

16

27

35 6

16

27

35

42 6

16

27

35

42 8

22

31

40

45 8

22

31

40

45 16

31

36

44

51

Table 1 indicates that the minimum total elapsed time is 51 hours. The idle time for machines

A, B and C is 9 (=51 - 42) hours, 6 (= 51 - 45) hours and 9 (= 8 - 0) + (45 - 44) hours,

respectively.

2.4. PROCESSING n JOBS THROUGH m MACHINES

Let there be n jobs, each of which is to be processed through m machines, say M 1, M 2, .....M m

in the order M 1, M 2, .....M m. The optimal solution to this problem can be obtained if either or

both of the following conditions hold goo d.

(a) Min {t 1j} ≥ Max { t1j}; j = 2, 3, ..... m - 1

and or (b) Min {t mj} ≥ Max { tij}; j = 2, 3, ..... m - 1

that is, the minimum processing time on machines M 1 and M m is as great as the maximum

processing time on any of the remaining (m - 1) machines.

If either or both these conditions hold good, then the steps of the algorithm can be

summarised in the following steps:

Step 1: Find, Min {t 1j}, Min {t mj} and max {tij}and verify above conditions. If either or both

the conditions mentioned above hold, then go to step 2. Otherwise the algorithm fails. munotes.in

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Step 2: Convert m -machine problem into 2 -machine problem by introducing two fictitious

machines, say

(i) t Gj = t1j + t2j + t3j +......+t m - 1j j = 1, 2, 3, .....n.

i.e. processing time of n -jobs on machine G is the sum of the processing times on Machines

M1, M 2 ......M m - 1j

(ii) t Hj = t2j + t3j + t4j +......+ tmj j = 1, 2, 3, .....n.

i.e. processing time of n -jobs on machine H is the sum of the processing times on Machines

M1, M 2 ......M mj.

Step 3: The new processing times so obtained can now be used for solving n -job, two

machines equivalent sequencing problem with the prescribed ordering HG in the same way as

t2j + t3j +......+t m - 1j = k (constant)

for all j = 1, 2, 3, ..... m - 1, then the optimal sequence can be obtained for n -jobs and two

machines M 1 and M m in the order M 1 Mm as usual.

2. If t 1j = tmj and t Gj = tHj, for all j = 1, 2, 3, ....n, then total number of optim al sequences will

be n and total minimum elapsed time in these cases would also be the same.

3. The method described above for solving n -jobs and m -machines sequencing problem is not

a general method. It is applicable only to certain problems where the min imum cost (or time)

of processing the jobs through first and/or last machine is more than or equal to the cost (or

time) of processing the jobs through remaining machines.

EXAMPLES

1. Find an optimal sequence for the following sequencing problems of four jobs and five

machines when passing is not allowed of which processing time (in hours) is given below:

Job Machines

M1 M2 M3 M4 M5

A

B

C

D 7

6

5

8 5

6

4

3 2

4

5

3 3

5

6

2 9

10

8

6

Also find the total elapsed time.

Solution: Here,

Min (t M1, j) = 5 = t M1, C

Min (t M5, j) = 6 = t M5, D

and Max {t M2, j, tM3, j, tM4, j} = {6, 5, 6} respectively.

Since the condition of Min (t M5, j) ≥ Max { tM2, j, tM3, j, tM4, j} is satisfied, therefore the given

problem can be converted into a four jobs and two machines problem as G and H. The

processing times of four jobs denoted by t Gj and t Hj on G and H, respectively are as follows:

Job A B C D

Machine G

Machine H 17

19 21

25 20

23 16

14

where t Gj = ∑ 𝑡𝑖𝑗𝑚−1

𝑖=1 and t Hj = ∑ 𝑡𝑖𝑗𝑚

𝑖=2.

Now using the optimal sequence algorithm, the following optimal sequence can be obtained.

A C B D

The total elapsed time corresponding to the optimal sequence can be calculated as shown in

Table 1, using the individual processing times given in the original problem. munotes.in

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Table 1 shows that the minimum total elapsed time is 51 hours. The idle time for machines

M1, M2, M3, M4 and M 5 is 25, 33, 37 and 18 hrs respectively.

Table 1 Minimum Elapsed Ti me

Job Sequence Machine

M1 M2 M3 M4 M5

A

B

C

D 0 - 7

7 - 12

12 - 18

18 - 26 7 - 12

12 - 18

18 - 24

26 - 29 12 - 14

16- 21

24 - 28

29 - 32 14 - 17

21 - 27

28 - 33

33 - 35 16 - 26

27 - 35

35 - 45

45 - 51

2. Solve the following sequencing problem giving an optimal solution when passing is not

allowed.

Machine Job

A B C D E

M1

M2

M3

M4 11

4

6

15 13

3

7

8 9

5

5

13 16

2

8

9 17

6

4

11

Solution: From the data of the problem it is observed that

Min (t M1, j) = 9 = t M1, C

Min (t M4, j) = 8 = t M4, B

and Max {t M2, j} = 6 = , tM2, E; Max { tM3, j} = 8 = , tM2, D.

Since both the conditions

Min (t M1, j) ≥ Max { tM2, j; tM3, j} ; j = 1, 2, ....., 5

are satisfied, therefore given problem can be converted into a 5 -jobs and 2 -machine problem

as G and H.

Further, it may be noted that, t M2, j + tM3, j = 10 (a fixed constant) for all j (j = 1, 2, ...., 5).

Thus the given problem is reduced to a problem of solving 5 -jobs through 2 -machines M 1 and

M4 in the order M 1 M4. This means machines M 2 and M 4 will have no effect on the optimality

of the sequences.

The processing times of 5 jobs on machine M 1 and M 4 is as follows:

Job A B C D E

Machine M 1

Machine M 4 11

15 13

8 9

13 16

9 17

11

Now using the algorithm described earlier, the optimal sequence so obtained as follows:

C A E D B

The total elapsed time corresponding to the optimal sequence is 83 hours as shown in table 1,

using the individual processing times given in the original problem:

Table 1 Minimum Total Elapsed Time

Job Sequence Machine

M1 M2 M3 M4

C

A

E

D

B 0 - 9

9 - 20

29 - 36

36 - 52

52 - 65 9 - 14

20 - 24

36 - 42

52 - 54

65 - 68 14 - 19

24 - 30

42 - 46

54 - 62

68 - 75 19 - 32

32 - 45

46 - 57

62 - 71

75 - 83 munotes.in

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2.5. PROCESSING TWO JOBS THROUGH m MACHINES

Let there be two jobs A and B each of which is to processed on m machines say M 1, M 2, .....,

Mm in two different orders. The technological ordering of each of the two jobs through m

machines is known in advance. Such ordering may not be same for both the jobs. The exact

or expected processing times on the given machines are known. Each machine can perform

only one job at a time. The objective is to determine an optimal sequence of processing the

jobs so as to minimise total elapsed time.

The optimal sequen ce in this case can be obtained by using graph. The procedure can be

illustrated by taking examples.

Example 1: Use the graphical method to minimise the time needed to process the following

jobs on the machines shown, i.e. each machine finds the job which should be done first. Also

calculate the total elapsed time to complete both jobs.

Machine

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

Job 1 {Sequence: A B C D E

Time (hrs) 3 4 2 6 2

Machine

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

Job 2 {Sequence: A B C D E

Time (hrs) 5 4 3 2 6

Solution

The solution procedure for solving the above problem can be summarised in the following

steps :

1. Draw the set of axes at r ight angle to each other where x -axis represents the processing

time of job 1 on different machines while job 2 remains idle and y -axis represents processing

time of job 2 while job 2 remain idle.

2. Mark the processing times for jobs 1 and 2 on x -axis and y-axis, respectively according to

the given order of machines as shown in the figure:

20 Idle time

for Job 1 E Idle time for job 1

14 D

12

A

9

C

5

B

0 3 7 9 15 17 Job 1

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2.6. GRAPHICAL SOLUTION OF TWO JOBS AND m - MACHINES

SEQUENCING PROBLEM

For example, machine A takes 3 hours for job 1 and 3 hours for job 2. Construct the rectangle

for machine A as shown in above Figure. Similarly, construct other rectangles for machines

B, C, D and E.

3. Construct various blocks starting from the origin by pairing the same machine until a point

marked 'finished' is obtained.

4. Draw a line starting from origin to the point marked 'finish' by moving horizontally,

vertically and diagonally along a line which makes an angle of 45 ° with the horizontal axis.

Moving horizontally along this line indicates that first job is under process while second job

is idle. Similarly, moving vertically along this line indicates that the second job is under

process while first job is idle. The diagonal movement along this line shows that both the jobs

are under process simultaneously.

Since simultaneous processing of both jobs on a machine is not possible, therefore, diagonal

movement is not allowed. In other words, diagonal movement through rectangle areas is not

allowed.

5. An optimal path is one that minimises the id le time for both the jobs. Thus, we must

choose the path on which diagonal movement is maximum as shown in the above figure.

6. Total elapsed time is obtained by adding the idle time for either job to the processing time

for that job. In this example, the idle time for the chosen path is found to be 5 hrs and 2 hrs

for jobs 1 and 2, respectively. The total elapsed time is calculated as follows:

Elapsed time, Job 1 = Processing time of Job 1 + Idle time for Job 1

= 17 + (2 + 3) = 22 hrs.

Elapsed time, Job 2 = Processing time of Job 2 + Idle time for Job 2

= 22 + (17 -15) = 24 hours.

2.7. EXERCISE

1. Explain the four elements that characterise a sequencing problem.

2. Explain the principal assumptions made while dealing with sequencing problems.

3. Give Johnson's procedure for determining an optimal sequence for processing n items on

two machines. Give justification of the rule used in the procedure.

4. What is no passing rule in a sequencing algorithm? Explain the principal assumptions

made while d ealing with sequencing problems.

5. Give three different examples of sequencing problems from your daily life.

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6. We have five jobs, each of which must be processed on the two machines A and B in the

order AB. Processing times in hours are given in the t able below:

Job 1 2 3 4 5

Machine A 5 1 9 3 10

Machine B 2 6 7 8 4

7. We have 5 jobs each of which must go through the machines A, B and C in the order ABC.

Processing times (in hours) is as follows:

Job 1 2 3 4 5

Machine A 5 7 6 9 5

Machine B 2 1 4 5 3

Machine C 3 7 5 6 7

8. What do you understand by the following terms in the context of sequence of jobs:

1. Job arrival pattern 2. Number of machines 3. The flow pattern in the shop

4. the criteria of evaluating the performance of a schedule.

9. Find an optimal sequence for the following sequencing problem of four jobs and five

machines (when passing is not allowed) of which processing time (in hrs) is as follows:

Job 1 2 3 4

Machine M 1 6 5 4 7

Machine M 2 4 5 3 2

Machine M 3 1 3 4 2

Machine M 4 2 4 5 1

Machine M 5 8 9 7 5

Also find the total elapsed time.

10. Two jobs are to be processed on four machines A, B, C and D. The technological order

for these jobs on machines is as follows:

Job 1: A B C D

Job 2: D B A C

Processing times are given in the following table:

Machines

A B C D

Job 1 4 6 7 3

Job 2 4 7 5 8

Find the optimal sequence of jobs on each of the machines.

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1

CHAPTER 3

REPLACEMENT MODELS

UNIT STRUCTURE

3.1 Introduction

3.2 Types of Failure

3.3 Replacement of items whose efficiency deteriorates with time

3.4 Replacement of items that fail completely

3.5 Exercise

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2

3.1. INTRODUCTION

The problem of replacement is felt when the job performing units such as men, machines,

equipments, parts etc. become less effective or useless due to either sudden or gradual

deterioration in their efficiency, failure or breakdown. By replacing them with new ones at

frequent intervals, maintenance and other overhead costs can be reduced. However, such

replacements would increase the need of capital cost for new ones.

For example,

1. A vehicle tends to wear out with time due to constant use. More money needs to be spent

on it on account of increased repair and operating cost. A stage comes when it becomes

uneconomical to maintain the vehicle and it is better to replace it with a new one. Here the

replacement decision may be taken to balance the increasing maintenance cost with the

decreasing money value of the vehicle as time passes.

2. In case of highway tube lights where time of failure is not predictable for individual tubes,

replacement is done only after individual failure. However, it may be economical to replace

such tubes on a schedule basis before their failure. Here the replacement decision may be

taken to balance between the wasted life of a tube before failure and cost incurred when a

tube fails completely during service.

Thus, the basic problem in such situations is to formulate a replacement policy to determine

an age (or period) at which the replacement of the given machinery/equipment is most

economical keeping in view all possible alternatives.

In this chapter, we shall discuss the replacement policies in the context of the following three

types of replacement situations:

1. Items such as machines, vehicles, tyres etc. whose efficiency deteriorates with age due to

constant use and which need increased operating and maintenance costs. In such cases

deterioration level is predictable and is represented by a. Increased maintenance/operational

cost, b. its waste or scrap value and damage to item and safety risk.

2. Items such as light bulbs and tubes, electric motors, radio, television parts etc. which do

not give any indication of deteriorations with time but fail all of a sudden and become

completely useless. Such cases require an anticipation of failures to specify the probability of

failure for any future time period. With this probability distribution and the cost information,

it is desired to formulate optimal, replacement policy to balance the wasted life of an item

replaced before failure against the costs incurred when the item fail in service.

3. The existing working staff in an organisation gradually reduces due to retirement, death,

retrenchment and other reasons.

3.2. TYPES OF FAILURE

The term 'failure' here will be discussed in the context of replacement decisions. There are

two types of failures: 1. Gradual failure and 2. Sudden failure.

Gradual Failure

Gradual Failure is progressive in nature. That is, as the life of an item increases its

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3

increased running (maintenance and operating) costs.

decrease in its productivity.

decrease in the resale or salvage value.

Mechanical items like pistons, rings, bearings etc., and automobile types fall under this

category.

Sudden Failure

This type of failure occurs in items after some period of giving desired service rather than

deterioration while in service. The period of desired service is not constant but follows some

frequency distribution which may be progressive, retrogressive or random in nature.

1. Progressive Failure: If the probability of failure of an item increases with the increase in its

life, then such failure is called progressive failure. For example, light bulbs and tubes fail

progressively.

2. Retrogressive Failure: If the probability of failure in the beginning of the life of a n item is

more but as time passes the chances of its failure becomes less, then such failure is said to be

retrogressive.

3. Random Failure: In this type of failure, the constant probability of failure is associat ed

with items that fail from random cases such as physical shocks, not related to age. For

example, vacuum tubes in air-born equipment have been found to fail at a rate independent of

the age of the tube.

3.3. REPLACEMENT OF ITEMS WHOSE EFFICIENCY

DETERIORATES WITH TIME

When operational efficiency of an item deteriorates with time (gradual failure), it is

economical to replace the same with a new one. For example, the maintenance cost of a

machine increases with time and a stage is reached when it may not be economical to allow

machine to continue in the system. Besides, there could be a number of alternative choices

and one may like to compare available alternatives on the basis of the running costs (average

maintenance and operating costs) involved. In this section, we shall discuss various

techniques for making such comparisons under different conditions. While making such

comparisons it is assumed that suitable expressions for running costs are available.

Model 1 Replacement Policy for items Whose Running Cost Increases with Time and

Value of Money Remains Constant During a Period

Theorem 1

The cost of maintenance of a machine is given as a function increasing with time and its

scrap value is constant.

a. If time is measured continuously, then the average annual cost will be minimised by

replacing the machine when the average cost to date becomes equal to the current

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4

b. If time is measured in discrete units, then the average annual cost will be minimised by

replacing the machine when the next period's maintenance cost becomes greater than the

current average cost.

PROOF: The aim here is to determine the optimal replacement age of an equipment whose

running cost increases with time and the value of money remains constant (i.e. value is not

counted) during that period.

Let C = capital or purchase cost of new equipment

S = Scrap (or salvage) value of the equipment at the end of t years

R(t) = running cost of equipment for the year t

n = replacement age of the equipment.

i. When time 't' is a continuous variable: If the equipment is used for t years, then the total

cost incurred over this period is given by

TC = Capital (or purchase) cost - Scrap value at the end of t years + Running cost for t years.

= C - S + ∫𝑅ሺ𝑡ሻ𝑑𝑡𝑛

Therefore, the average cost per unit time incurred over the period of n years is:

ATC n = ଵ

𝑛 C - S + ∫𝑅ሺ𝑡ሻ𝑑𝑡𝑛

------------------------ (1)

To obtain optimal value n for which ATC n is minimum, differentiate ATC n with respect to n,

and set the first derivative equal to zero. That is, for minimum of ATC n.

ௗ

ௗ𝑛{ATC n} = - ଵ

𝑛ଶ {C - S} + 𝑅 ሺ𝑛ሻ

𝑛 - ଵ

𝑛ଶ ∫𝑅ሺ𝑡ሻ𝑑𝑡𝑛

= 0

or R (n) = ଵ

𝑛 C - S + ∫𝑅ሺ𝑡ሻ𝑑𝑡𝑛

, n ≠0 ----------------- (2)

R (n) = ATC n

Hence, the following replacement policy can be derived with help of Eq. 2.

Policy

Replace the equipment when the average annual cost for n years becomes equal to the current

/annual running cost. That is,

R (n) = ଵ

𝑛 C - S + ∫𝑅ሺ𝑡ሻ𝑑𝑡𝑛

b. When time 't' is a discrete variable: The average cost incurred over the period n is given by

ATC n = ଵ

𝑛 C - S + ∑ 𝑅 ሺ𝑡ሻ𝑛

𝑡= ----------------------------(3)

If C - S and ∑ 𝑅 ሺ𝑡ሻ𝑛

𝑡= are assumed to be monotonically decreasing and increasing

respectively, then there will exist a value of n for which ATC n is minimum. Thus we shall

have inequalities

ATC n - 1 > ATC n < ATC n+1

or ATC n - 1 - ATC n > 0

and ATC n+1 - ATC n > 0

Eq. (3) for period n + 1, we get

ATC n = ଵ

𝑛+ଵ C - S + ∑ 𝑅 ሺ𝑡ሻ𝑛=

𝑡=ଵ = ଵ

𝑛+ଵ C - S + ∑ 𝑅 ሺ𝑡ሻ𝑛=

𝑡=ଵ + R (n + 1)

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= 𝑛

𝑛+ଵ C − ୗ + ∑ 𝑅 ሺ𝑡ሻ𝑛

𝑡=0

𝑛 + 𝑅 ሺ𝑛+ଵሻ

𝑛+ଵ = 𝑛

𝑛+ଵ . ATC n + 𝑅 ሺ𝑛+ଵሻ

𝑛+ଵ

Therefore,

ATC n+1 - ATC n = 𝑛

𝑛+ଵ ATC n + 𝐴 ሺ𝑛+ଵሻ

𝑛+ଵ - ATC n = 𝑅 ሺ𝑛+ଵሻ

𝑛+ଵ + ATC n [𝑛

𝑛+ଵ - 1] = 𝑅 ሺ𝑛+ଵሻ

𝑛+ଵ - ACn

𝑛+ଵ

Since ATC n+1 - ATC n > 0, we get

𝑅 ሺ𝑛+ଵሻ

𝑛+ଵ - ACn

𝑛+ଵ > 0

R (n + 1) - ATC n > 0 or R (n + 1) > ATC n ----------------------------- (4)

Similarly,

ATC n - 1 - ATC n > 0, implies that R (n + 1) < ATC n - 1. This provides the following

replacement policy.

Policy 1: If the next year, running cost R(n + 1) is more than average cost of nth year, ATC n,

then it is economical to replace at the end of n years. That is

R (n + 1) > ଵ

𝑛 C - S + ∑ 𝑅 ሺ𝑡ሻ𝑛

𝑡=

Policy 2: If the present year's running cost is less than the previous year's average cost, ATC n

- 1, then do not replace. That is

R (n) < ଵ

𝑛−ଵ C - S + ∑ 𝑅 ሺ𝑡ሻ𝑛−ଵ

𝑡=

EXAMPLE

A firm is considering replacement of a machine, whose cost price is Rs. 12, 200 and scrap

value is Rs. 200. The running (maintenance and operating) costs are found from experience to

be as follows:

Year 1 2 3 4 5 6 7 8

Running cost (Rs.) 200 500 800 1200 1800 2500 3200 4000

When should the machine be replaced?

Solution: We are given the running cost, R(n), the scrape value S - Rs. 200 and the cost of

the machine, C = Rs. 12, 200. In order to determine the optimal time n when the machine

should be replaced, first we calculate average cost per year during the life of the machine as

shown in table below:

TABLE

Year of

Service n Running

Cost (Rs.)

R(n) Cumulative

Running

Cost (Rs.)

Ʃ R(n) Depreciation

Cost (Rs.)

C - S Total Cost

(Rs.)

TC Average Cost

(Rs.)

ATC n

1 2 3 4 5 = 3 + 4 6 = 5/1

1

2

3

4

5 200

500

800

1200

1800 200

700

1500

2700

4500 12000

12000

12000

12000

12000 12, 200

12, 700

13, 500

14, 700

16, 500 12, 200

6350

4500

3675

3300 munotes.in

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6

7

8 2500

3200

4000 7000

10200

14200 12000

12000

12000 19, 000

22, 200

26, 200 3167

3171

3275

From the table, it may be noted that the average cost per year, ATC n is minimum in the sixth

year (Rs. 3, 167). Also the average cost in the seventh year (Rs. 3171) is more than the cost in

the sixth year. Hence, the machine should be replaced after every six years.

(2) The data collected in running a machine, the cost of which is Rs. 60, 000 are given below:

Year 1 2 3 4 5

Resale Value (Rs.) 42, 000 30, 000 20, 400 14, 400 9, 650

Cost of Spares (Rs.) 4, 000 4, 270 4, 880 5, 700 6, 800

Cost of Labour (Rs.) 14, 000 16, 000 18, 000 21, 000 25, 000

Determine the optimum period for replacement of the machine.

Solution: The costs of spares and labour together determine running (operational or

maintenance) cost. Thus, the running costs and the resale price of the machine in successive

years are as follows:

Year: 1 2 3 4 5

Resale Value (Rs.) 42, 000 30, 000 20, 400 14, 400 9, 650

Running Cost (Rs.) 18, 000 20, 270 22, 880 26, 700 31, 800

The calculations of average running cost per year during the life of the machine are shown in

table 1.

Year of

Service n Running

Cost (Rs.)

R(n) Cumulative

Running

Cost (Rs.)

Ʃ R(n) Resale

Value

(Rs.)

S Depreciation

Cost (Rs.)

C - S Total Cost

(Rs.)

TC Average

Cost

(Rs.)

ATC n

1 2 3 4 5 = 60, 000 6 = 3 + 5 7 = 6/1

1

2

3

4

5 18, 000

20, 270

22, 880

26, 700

31, 800 18, 000

38, 270

61, 150

87, 850

1, 19, 650 42, 000

30, 000

20, 400

14, 400

9, 650 18, 000

30, 000

39, 600

45, 600

50, 350 36, 000

68, 270

1, 00, 750

1, 33, 450

1, 70, 000 36, 000.00

34, 135.00

33, 583.00

33, 362.00

34, 000.00

The calculations in table 1 reveal that the average cost is lowest during the fourth year.

Hence, the machine should be replaced after every fourth year, otherwise the average cost per

year for running the machine would start increasing.

3.4. REPLACEMENT OF ITEMS THAT FAIL COMPLETELY

It is somehow difficult to predict that a particular equipment will fail at a particular time. This

uncertainty can be avoided by deriving the probability distribution of failures. Here, it is

assumed that the failures occur only at the end of the period, say t. Thus, the objective

becomes to find the value of t which minimises the total cost involved for the replacement.

Mortality Tables: These tables are used to derive the probability distribution of life span of

an equipment in question. Let munotes.in

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M (f) = Number of survivors at any time t

M (t - 1) = Number of survivors at any time t - 1

N = Initial number of equipments

Then the probability of failure during time period t is given by

P (t) = ெሺ𝑡−ଵ ሻ− ெሺ𝑡ሻ

ே

The probability that an equipment has survived to an age (t - 1). and will fail during the

interval (t - 1) to t can be defined as the conditional probability of failure. It is given by

Pc (t) = ெሺ𝑡−ଵ ሻ− ெሺ𝑡ሻ

ெሺ𝑡−ଵሻ

The probability of survival to an age t is given by

Ps(t) = ெሺ𝑡ሻ

ே

Mortality Theorem 1: A large population is subject to a given mortality law for a very long

period of time. All deaths are immediately replaced by births and there are no other entries or

exits. Show that the age distribution ultimately becomes stable and that the number of deaths

per unit time becomes constant and is equal to the size of the total population divided by the

mean age at death.

Proof: Without any loss of generality, it is assumed that death (or failure) occurs just before

the age of (k + 1) years, where k is an integer. That is, life span of an item lies betwee n t = 0

and t = k. Let us define f(t) = number of births (replacements) at time t, and

p(x) = probability of death (failure) just before the age x + 1, i.e. failure at time x.

and ∑ 𝑝ሺ𝑥ሻ𝑘

௫= = 1

If f(t - x) represents the number of births at time t - x, t = k, k +1, k + 2,....then the age of

newly borns attain the age x at time t illustrated in the figure below:

Age 0 x x + 1

Time t - x t t + 1

Hence the expected number of deaths of such newly borns before reaching the time t + 1 (i.e.

at time t) will be

Expected number of death = ∑ 𝑝ሺ𝑥ሻ𝑘

௫= f(t - x), t = k, k + 1, .....

Since all deaths (failures) at time t are replaced immediately by births (replacements) at time t

+ 1, expected number of births are:

f(t + 1) = ∑ 𝑝ሺ𝑥ሻ𝑘

௫= f(t - x), t = k, k + 1, ..... (1)

The solution to the difference Eq. (9) in t can be obtained by putting the value f(t) = Aα',

where A is some constant. Then Eq. (9) becomes

Aαt + 1 = A ∑ 𝑝ሺ𝑥ሻ𝑘

௫= αt - x (2)

Dividing both sides of Eq. (10) by Aαt - k, we get

αk + 1 = ∑ 𝑝ሺ𝑥ሻ𝑘

௫= αk - x = αk ∑ 𝑝ሺ𝑥ሻ𝑘

௫= α- x

= αk {p (0) + p (1) α-1 + p(2) α-2 + .....}

or αk + 1 - {p (0) αk + p (1) αk - 1 + p (2) αk - 2} = 0 (3) munotes.in

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Equation (3) is of degree (k + 1) and will, therefore, have exactly (k + 1) roots. Let us denote

the roots of Eq. (3) by α 0, α1, α2, ......., αk.

For α = 1, the L. H. S of equation (3), becomes

L.H.S = 1 - {p(0) + p (1) + p (2) + ....... + p (w)}

= 1 - ∑ 𝑝ሺ𝑥ሻ௪

௫= = 0 R. H.S

Hence, one root of Eq. (3) is α = 1. Let us denote this root by α 0. The general solution of Eq.

(3) will then be of the form

f (t) = A 0 αt

0 + A 1 αt

1 + ..... + A k αt

k

= A 0 + A 1 αt

1 + ..... + A k αt

k

where A 0, A1, A2, .... A k are constant whose values are to be calculated. (4)

Since one of the roots of Eq. (3), α0 = 1 is positive, according to the Descrae's sign rule all

other roots α 1, α2, ......., αk will be negative and their absolute value is less than unity, i.e. |α i| <

1, i = 1, 2, 3, ....., k. It follows that the value of these roots tends to be zero as t ∞. With

the result that Eq. (4) becomes f(t) = A 0. This indicates that the number of deaths (as well as

births) becomes constant at any time.

Now the problem is to determine the value of the constant A 0. For this we can proceed as

follows:

Let us define

g (x) = Probability of survival for more than x years.

or g (x) = 1 - Prob (survivor will die before attaining the age x)

= 1 - {p(0) + p(1) + ....p(x -1)}

Obviously, it can be assumed that g(0) = 1.

Since the number of births as well as deaths has become constant and equal to A 0, expected

number of survivors of age x is given by A 0 . g(x).

As deaths are immediately replaced by births, size N of the population remains constant.

That is,

N = A 0 ∑ ݃ሺ𝑥ሻ𝑘

௫= α1, α2, ......., αk

Or

A0 = ே

∑ 𝑔ሺ௫ሻ𝑘

𝑥=0 ............ (4)

The denominator in Eq. (4) represents the average age at death. This can also be proved as

follows:

From finite differences, we know that

∆(x) = (x + 1) - x = 1

∑ ݂ሺ𝑥ሻ

௫= ∆ℎ(x) = f (b + 1) h(b + 1) - f (a) h(a) - ∑ ሺ𝑥 + 1ሻ

௫= ∆݃(x)

= g (k + 1) (k + 1) - g (0) . 0 - ∑ ሺ𝑥 + 1ሻ𝑘

௫= ∆݃(x)

= g (k + 1) (k + 1) - ∑ ሺ𝑥 + 1ሻ𝑘

௫= ∆݃(x) ............ (5)

But g (k + 1) = 1 - {p(0) + p(1) + p(2) + ....... + p(k)} = 0

since no one can survive for more than (k + 1) years of age and

∆݃(x) = g(x + 1) - g(x)

= {1 - p(0) - p(1) - ...... - p(x)} - {1 - p(0) - p(1) .....p(x - 1)} = - p(x)

Substituting the value of g(k + 1) and ∆݃(x) in Eq. (5), we get

∑ ݃ሺ𝑥ሻ𝑘

௫= = ∑ ሺ𝑥 + 1ሻ𝑘

௫= p (x) = Mean age at death

Hence from Eq. (4), we get munotes.in

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A0 = ே

𝐴௩𝑟𝑔 𝑔 𝑡 ௗ𝑡ℎ .

STAFFING PROBLEM

The principles of replacement may be applied to formulate some useful recruitment and

promotion policies for the staff working in the organisation. For this, we assume that life

distribution for the services of staff in the organisation is already known.

Example 1:

An airline requires 200 assistant hostesses, 300 hostesses and 50 supervisors. Women are

recruited at the age of 21, and if still in service retire at 60. Given the following life table,

determine

(a) How many women should be recruited in each year?

(b) At what age should promotion take place?

Airline Hostesses' Life Record

Age

No. in

Servie 21

1000 22

600 23

480 24

384 25

307 26

261 27

228 28

206

Age

No. in

Servie 29

190 30

181 31

173 32

167 33

161 34

155 35

150 36

146

Age

No. in

Servie 37

141 38

136 39

131 40

125 41

119 42

113 43

106 44

99

Age

No. in

Servie 45

93 46

87 47

80 48

73 49

66 50

59 51

53 52

46

Age

No. in

Servie 53

39 54

33 55

27 56

22 57

18 58

14 59

11 ---

Solution: If 1000 women had been recruited each year for the past 39 years, then the total

number of them recruited at the age of 21 and those serving upto the age of 59 is 6, 480. Total

number of women recruited in the airline are: 200 + 300 + 50 = 550.

(a) Number of women to be recruited every year in order to maintain a strength of 55

hostesses 550 x (1000/6840) = 85 approx.

(b) If the assistant hostesses are promoted at the age of x, then up to age (x - 1), 200 assistant

hostesses will be required. Among 550 women, 200 are assistant hostesses. Therefore, out of

a strength of 1, 000 there will be: 200 x (1000/550) = 364 assistant hostesses. But, from the

life table given in the question, this number is available up to the age of 24 years. Thus, the

promotion of assistant hostesses is due in the 25th year.

Since out of 550 recruitments only 300 hostesses are needed, if 1,000 girls are recruited, then

only 1000 x (300/550) = 545 appro x. will be hostesses. Hence, total number of hostesses and

assistant hostesses in a recruitment will be: 545 + 364 = 909. This means, only 1000 - 909 =

91 supervisors are required. But from life table this number is available up to the age of 46

years. Thus promotion of hostesses to supervisors will be due in 47th year. munotes.in

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3.5. EXERCISE

1. In the theory of replacement models construct an equation for the cost of maintaining a

system as a function of the control variable t (the number of periods between group

replacements).

2. State some of the simple replacement policies and give the average cost functions for the

same explaining your notations.

3. The cost of maintenance of a machine is given as a function that the average annual cost

will be minimised by replacing the machine when the average cost to date becomes equal to

the current maintenance cost.

4. Discuss the importance of replacement models.

5. Explain how the theory of replacement is used in the following problems.

(i) Replacement of items whose maintenance cost varies with time.

(ii) Replacement of items that fail completely.

6. Find the cost per period of individual replacement of an installation of 300 lighting bulbs,

given the following

(a) cost of replacing individual bulb is Rs. 3

(b) Conditional probability of failure is given below:

Week Number : 0 1 2 3 4

Conditional Probability of Failure : 0 1/10 1/3 2/3 1

7. The following mortality rates have been observed for a special type of light bulbs.

Month : 1 2 3 4 5

% failing at the end of the month : 10 25 50 80 100

In an industrial unit there are 1,000 special type of bulbs in use, and it costs Rs.10 to replace

an individual bulb which has burnt out. If all bulbs were replaced simultaneously it would

cost Rs. per bulb. It is proposed to replace all bulbs at fixed intervals, whether or not they

have burnt out, and to continue replacing burnt out bulbs as they fail. At what intervals of

time should the manager replace all the bulbs?

8. An airline, whose staff are subject to the same survival rates as in the previous problem,

currently has a staff whose ages are distributed in the following table. It is estimated that for

the next two years ’ staff requirements will increase by 10% per year. If women are to be

recruited at the age of 21, how many should be recruited for the next year and at what a ge

will promotions take place? How many should be recruited for the following year and at what

age will promotions take place?

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Assistant

Age

Number 21

90 22

50 23

30 24

20 25

10

(Total

200)

Hostesses

Age

Number 26

40 27

35 28

35 29

30 30

28 31

26 32

20 33

18 34

16

Age

Number 35

12 36

10 37

8 38

--- 39

8 40

8 41

6

(Total

300)

Supervisors

Age

Number

Age

Number

42

5

51

---

43

4

52

4

44

5

53

3

45

3

54

5

46

3

55

---

47

3

56

3

48

6

57

2

49

2

58

---

50

---

59

2

(Total

50)

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CHAPT ER 4

INVENTORY MODELS

UNIT STRUCTURE

4.1 Introduction

4.2 Importance of Inventory Control

4.3 Inventory Control Decisions

4.4 Inventory Control with Deterministic Models

4.5 Selective Inventory Control techniques- ABC Analysis

4.6 VED Analysis

4.7 FSN Analysis

4.8 XYZ Analysis

4.9 Exercise

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4.1. INTRODUCTION

Inventory is one of the most expensive and important assets of many companies, representing

as much as 50% of total invested capital. Managers have long recognised that good inventory

control is crucial. On one hand, a firm can try to reduce costs by reducing on-hand inventory

levels. On the other hand, customers become dissatisfied when frequent inventory outages,

called stock outs occur. Thus, companies must make the balance between low and high

inventory levels. As you would expect, cost minimisation is the major factor in obtaining this

delicate balance.

Inventory is any stored resource that is used to satisfy a current or future need. Raw materials,

work- in-progress, and finished goods are examples of inventory. Inventory levels for finished

goods, such as clothes dryers, are a direct function of market demand. By using this demand

information, it is possible to determine how much raw materials (eg., sheet metal, paint, and

electric motors in the case of clothes dryers) and work- in-processes are needed to produce the

finished product.

Every organisation has some type of inventory planning and control system. A bank has

methods to control its inventory of cash. A hospital has methods to control blood supplies and

other important items. State and federal governments, schools, and virtually every

manufacturing and production organisation are concerned with inventory planning and

control. Studying how organisations control their inventory is equivalent to studying how

they achieve their objectives by supplying goods and services to their customers. Inventory is

the common thread that ties all the functions and departments of the organisation together.

Figure 4.1 illustrates the basic components of inventory planning and control system. The

planning phase involves primarily what inventory is to be stocked and how it is to be

acquired (whether it is to be manufactured or purchased). This information is then used in

forecasting the demand for the inventory and in controlling inventory levels. The feedback

loop in figure 4.1 provides a way of revising the plan and forecast based on experiences and

observation.

Through inventory planning, an organisation determines what goods and/or services are to be

produced. In cases of physical products, the organisation must also determine whether to

produce these goods or to purchase them from another manufacturer. When this has been

determined, the next step is to forecast the demand. Many mathematical techniques can be

used in forecasting demand for a particular product. The emphasis in this chapter is on

inventory control - that is, how to maintain adequate inventory levels within an organisation

to support a production or procurement plan that will satisfy the forecasted demand.

In this chapter, we discuss several different inventory control models that are commonly used

in practice. For each model, we provide examples of how they are analysed. Although we

show the equations needed to compute the relevant parameters for each model, we use

EXCEL worksheets to actually calculate these values.

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4.2. IMPORTANCE OF INVENTORY CONTROL

Inventory control serves several important functions and adds a great deal of flexibility to the

operation of a firm. Five main uses of inventory are as follows:

1. Decoupling function

2. Storing resources

3. Irregular supply and demand

4. Quantity discounts

5. Avoiding stockouts and shortages.

Figure 4.1 Inventory Planning and Control

1. Decoupling Function

One of the major functions of the inventory is to decouple manufacturing processes within

the organisation. If a company did not store inventory, there could be many delays and

inefficiencies. For example, when one manufacturing activity has to be completed before a

second activity can be started, it could stop the entire process. However, stored inventory

between processes could act as a buffer.

2. Storing Resources

Agricultural and seafood products often have definite seasons over which they can be

harvested or caught, but the demand for these products is somewhat constant during the year.

In these and similar cases, inventory could be used to store these resources.

In manufacturing process, raw materials can be stored by themselves, as work- in-process, or

as finished products. Thus, if your company makes lawn movers, you may obtain lawn

mower tyres from another manufacturer. If you have 400 finished lawn movers and 300 tyres

in inventory, you actually have 1, 900 tyres stored in inventory. 300 tyres are stored by

themselves, and 1, 600 (= 4 tyre per lawn mower X 400 lawn movers) tyres are stored on the

finished lawn mowers. In the same sense, labour can be stored in inventory. If you have 500

subassemblies and it takes 50 hours of labour to produce each assembly, you actually have

25, 000 labour hours stored in inventory in the subassemblies. In general, any resource,

physical or otherwise can be stored in inventory. Planning on what

inventory to

Stock and how to

acquire it.

Feedback Measurements to revise Plans and

Forecasts Controlling

Inventory Levels Forecasting

Parts/Product

Demand munotes.in

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3. Irregular supply and demand

When the supply and demand for the inventory item is irregular, storing certain amount as

inventory can be important. If the greatest demand for Diet-Delight beverage is during the

summer, the Diet-Delight company will have to make sure that there is enough supply to

meet this irregular demand. This might require that the company produce more of the soft

drink in the winter than is actually needed in order to meet the winter demand. The inventory

levels of Diet-Delight will gradually build up over the winter, but this inventory will be

needed in summer. The same is true for irregular supplies.

4. Quantity Discounts

Another use of inventory is to take advantage of quantity discounts. Many suppliers offer

discounts for larger orders. For example, an electric jigsaw might normally cost $10 per unit.

If you order 300 or more saws at one time, your supplier may lower the cost to $8.75.

Purchasing in larger quantities can substantially reduce the cost of products. There are,

however, some disadvantages of buying in larger quantities. You will have higher storage

costs and higher costs due to spoilage, damaged stock, theft, insurance, and so on.

Furthermore, if you invest in more inventory, you will have less cash to invest elsewhere.

5. Avoiding stock outs and shortages

Another important function in inventory is to avoid stockouts and shortages. If a company is

repeatedly out of stock, customers are likely to go elsewhere to satisfy their needs. Lost

goodwill can be an expensive price to pay for not having the right item at the right time.

4.3. INVENTORY CONTROL DECISIONS

Even though there are literally millions of different types of products manufactured in our

society, there are only two fundamental decisions that you have to make when controlling

inventory:

1. How much to order

2. When to order

Table 4.1 Inventory Cost Factors

Operating Cost Factors Carrying Cost Factors

Developing and sending purchase orders

Processing and inspecting incoming inventory

Bill Paying

Inventory Inquiries

Utilities, phone bills, and so on for the purchasing

department

Salaries and wages for purchasing department

employees

Supplies such as forms and paper for the purchasing

department Cost of Capital

Taxes

Insurance

Spoilage

Theft

Obsolescence

Salaries and wages for warehouse

employees

Utilities and building costs for the

warehouse

Supplies such as forms and papers for the

warehouse munotes.in

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The purpose of all inventory models is to determine how much to order and when to order.

As you know, inventory fulfils many important functions in an organisation. But as the

inventory levels go up to provide these functions, the cost of holding and storing inventories

also increases. Thus, we must reach a fine balance in establishing inventory levels. A major

objective in controlling inventory is to minimise total inventory costs. Some of the most

significant inventory costs are as follows:

1. Cost of the items

2. Cost of ordering

3. Cost of carrying or holding inventory

4. Cost of stock outs

5. Cost of safety stock, the additional inventory that may be held to help avoid stockouts.

The inventory models discussed in the first part of this chapter assume that demand and the

time it takes to receive an order are known and constant, and that no quantity discounts are

given. When this is the case, the most significant costs are the cost of placing an order and the

cost of holding inventory items over a period of time. Table 4.1 provides a list of important

factors that make up these costs.

ECONOMIC ORDER QUANTITY: DETERMINING HOW MUCH TO ORDER

The Economic Order Quantity (EOQ) model is one of the oldest and most commonly known

inventory control techniques. Research on its use dates back to a 1915 publication by Ford

W. Harris. This model is still used by a large number of organisations today. This technique

is relatively easy to use, but it makes a number of assumptions. Some of the more important

assumptions are as follows:

ASSUMPTIONS OF EOQ

1. Demand is known and constant.

2. The lead time - that is, the time between the placement of the order and the receipt of the

order - is known and constant.

3. The receipt of inventory is instantaneous. In other words, the inventory from an order

arrives in one batch, at one point in time.

4. Quantity discounts are not possible.

5. The only variable costs are the cost of placing an order, ordering cost, and the cost of

holding or storing inventory over time, carrying or holding cost.

6. If orders are placed at the right time, stock outs and shortages can be completely avoided.

With these assumptions, inventory usage has a sawtooth shape, as in Figure 4.2. Here, Q

represents the amount that is ordered. If this amount is 500 units, all 500 units are arrived at

one time when an order is received. Thus, the inventory level jumps from 0 to 500 units. In

general, the inventory level increases from 0 to Q units when an order arrives.

Because demand is constant over time, inventory drops at a uniform rate over time. Another

order is placed such that when the inventory level reaches 0, the new order is received and the

inventory level again jumps to Q units, represented by the vertical lines. This process

continues indefinitely over time.

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ORDERING AND INVENTORY COSTS

The objective of most inventory models is to minimise the total cost. With the assumptions

just given, the significant costs are the ordering costs and inventory carrying cost. All other

costs, such as the cost of the inventory itself, are constant. Thus, if we minimise the sum of

the ordering and carrying costs, we also minimise the total cost.

To help visualise this, Figure 4.3 graphs total cost as a function of the order quantity, Q. As

the value of Q increases, the total number of orders placed per year decreases. Hence, the

total ordering cost decreases. However, as the value of Q increases, the carrying cost

increases because the firm has to maintain larger average inventories.

The optimal order size, Q*, is the quantity that minimises the total cost. Note in figure 12.3

that Q* occurs at the point where the ordering cost curve and the carrying cost curve intersect.

This is not by chance. With this particular type of cost function, the optimal quantity always

occurs at a point where the ordering cost is equal to the carrying cost. OPTIMISING INVENTORY AT PROCTER & GAMBLE

Procter & Gamble (P&G) is a world leader in consumer products with annual sales of

over $76 billion. Managing inventory in such a large and complex organisation requires

making effective use of the right people, organisational structure, and tools. P&G's

logistics planning personnel coordinate material flow, capacity, inventory, and logistics

for the firm's extensive supply chain network, which comprises P&G-owned

manufacturing facilities, 300 contract manufacturers, and 6, 900 unique product-

category market combinations. Each supply chain requires effective management based

on the latest available information, communication and planning tools to handl e

complex challenges and trade-offs on issues such as production batch sizes, order

policies, replenishment timing, new product introductions and assortment management.

Though the effective use of inventory optimisation models, P&G has reduced its

total inventory investment significantly. Spreadsheet based inventory models that

locally optimise different portions of the supply chain drive nearly 60 % of P&G's

business. For more complex supply chain networks (which drive about 30% of

P&G's business), advanced multi-stage models yield additional average inventory

reductions of 7%. P&G estimates that the use of these tools was instrumented in

driving $1.5 billion in cash savings in 2009, while maintaining or increasing

service levels.

Source: Based on I. Farasyn et. al. "Inventory Optimisation at P&G: Achieving real benefits

through User Adoption of Inventory Tools," Interfaces 41, 1(Januar y-February 2011): 66- 78 munotes.in

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Figure 12.2 Inventory Usage Over time

Inventory Level Order Quantity = Q = Maximum

Q Inventory Level

0

Time

Figure 4.3 Total Cost as a Function of Order Quantity

Optimal Order Quantity (Q*) Order Quantity in Units

Now that we have a better understanding of inventory costs, let us see how we can determine

the value of Q* that minimises the total cost. In determining the annual carrying cost, it is

convenient to use the average inventory. Referring to figure 4.2, we see that the on-hand munotes.in

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inventory ranges from a high of Q units to a low of zero units, with a uniform rate of decrease

between these levels. Thus, the average inventory can be calculated as the average of the

minimum and maximum inventory levels. That is,

Average Inventory = (0 +Q)/2 = Q/2 ----------------------------- (1)

We multiply this average inventory by a factor called the annual inventory carrying cost per

unit to determine the annual inventory cost.

FINDING THE ECONOMIC ORDER QUANTITY

We pointed out that the optimal order quantity, Q*, is the point that minimises the total cost,

where total cost is the sum of ordering cost and carrying cost. We also indicated graphically

that the optimal order quantity was at the point where the ordering cost was equal to the

carrying cost. Let us now define the following parameters:

Q* = Optimal Order Quantity (i.e. EOQ)

D = Annual Demand, in units, for the inventory item

Co = Ordering cost per order

Ch = Carrying or holding cost per unit per year

P = Purchasing cost per unit of the inventory item

The unit carrying cost, C h, is usually expressed in one of two ways, as follows:

1. As a fixed cost, for example, C h is $0.50 per unit per year.

2. As a percentage (typically denoted by l) of the item's unit purchase cost or price . For

example, C h is 20% of the item's unit cost. In general

Ch = I X P----------------------- (2)

For a given order quantity, Q, the ordering, holding and total costs can be computed using the

following formulae:

Total Ordering Cost = (D/Q) x C o----------------------- (3)

Total Carrying Cost = (Q/2) x C h ----------------------- (4)

Total Cost = Total Ordering Cost + Total Carrying Cost + Purchase Cost

= (D/Q) x C o + (Q/2) x C h + P x D --------------------- (5)

Observe that the total purchase cost (i.e. P x D) does not depend on the value of Q. This is so

because regardless of how many orders we place each year, or how many units we order ea ch

time, we will still incur the same annual total purchase cost.

The presence of Q in the denominator for the first term makes equation (5) a non-linear

equation with respect to Q. Nevertheless, because the total ordering cost is equal to the total

carrying cost at the optimal value of Q, we can set the terms in equation (3) and (4) equal to

each other and calculate EOQ as

Q* = √ʹ𝐷 Co/ Ch --------------------------------------- (6)

4.4. INVENTORY CONTROL WITH DETERMINISTIC MODELS

Examples 1

1. The production department for a company requires 3,600 kg of raw material for manufac turing a

particular item per year. It has been estimated that the cost of placing an order is Rs. 36 and the cos t of munotes.in

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carrying inventory is 25% of the investment in the inventories. The price is Rs. 10/kg. The purchase

manager wishes to determine an ordering policy for raw material.

Solution

From the data of the problem we know that

D = 3600 kg per year; C o = Rs. 36 per order; C h = 25% of the investment in inventories

= 10 X 0.25 = Rs. 2.50 per kg per year.

a. The optimal lot size is given by

Q* = √ʹ𝐷 Co/ Ch = √ʹ 𝑋 ͵00 𝑋 ͵ / 250 = 321.99 kg per order.

b. Optimal order cycle time

t* = 𝑄∗

= 321.99/3600 = 0.894 year

c. The minimum yearly variable inventory cost

TVC = √ʹ𝐷 CoCh = √ʹ𝑋 ͵00 𝑋 ͵ 𝑋 ʹ.ͷ = Rs. 804.98 per year

d. The minimum yearly total inventory cost

TC* = TVC* + DC = Rs. 804-98 + (3600 kg) (Rs. 10/kg) = Rs. 36, 804.98 per year.

2. A company operating 50 weeks in a year is concerned about its stocks of copper cable.

This costs Rs. 240/- a meter and there is a demand for 8, 000 meters a week. Each

replenishment costs Rs. 1,050 for administration and Rs. 1, 650 for delivery, while holding

costs are estimated at 25% of value held a year. Assuming no shortages are allowed, what is

the optimal inventory policy for the company?

How would this analysis differ if the company wanted to maximise profit rather than

minimise cost? What is the gross profit if the company sell cable for Rs. 360 a meter?

Solution

From the data of the problem, we have

Demand rate (D) = 8, 000 x 50 = 4, 00, 000 metres a year

Purchase Cost (C) = Rs. 240 a unit; Ordering cost (C o) = 1, 050 + 1, 650 = Rs. 2, 700

Holding Cost = 0.25 x 240 = Rs. 60 a meter a year

a. Optimal order quantity Q* = √ʹ𝐷 Co/ Ch = √ଶ 𝑋 ସ, 𝑋 ଶ,

= 6, 000 meters

b. Total variable inventory cost, TVC = Q*. Ch = 6, 000 x 60 = Rs. 3, 60, 000 a year

c. Total inventory cost, TC = D.C + TVC = 4, 00, 000 x 240 + 3, 60, 000 = Rs. 9, 63, 60, 000

With a turnover in excess of 96 million rupees a year inventory costs are only Rs. 3, 60, 000

or 0.36%. This figure is usually low but any well run organisation should try to make all the

waiving it can, however, small.

If the company wanted to maximise profit rather than minimise cost, the analysis used would

remain exactly the same. This can be demonstrated by defining selling price (SP) per unit so

that gross profit per unit becomes,

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= D x SP -

DC +

𝑄 Co + 𝑄

ଶ Ch

When this equation is solved to maximise profit with respect to Q as discussed earlier, we get

the same result by applying usual method.

If company sells the cable for Rs. 360 a meter, its revenue is Rs. 360 x 4, 00, 000 = Rs. 14, 40, 00,

000 a year. Direct cost of Rs. 9, 63, 60, 000 is subtracted from this to get a gross pr ofit of Rs. 4, 76,

40, 000 a year.

3. A chemical company is considering the optimal batch size for reorder of concentrated sulphuric

acid. The management accountant has supplied the following information:

(1) The purchase price of H 2SO 4 is Rs. 150 per gallon.

(2) The clerical and data processing costs are Rs. 500 per order.

All the transport is done by rail. A charge of Rs. 2, 000 is made each time the special line to the

factory is opened. A charge of Rs. 20 gallons is also made. The company uses 40, 000 gall ons per

year. Maintenance costs of stock are Rs. 400 per gallon per year.

Each gallon requires 0.5 sq ft of storage space. If warehouse space is not used, it can be rented out to

another company at Rs. 200 per sq ft per annum. Available warehouse space is 1,000 sq ft , the

overhead costs being Rs. 5, 000 p.a. Assume that all free warehouse space can be rented out.

(a) Calculate the economic reorder size

(b) Calculate the minimum total annual cost of holding and reordering stock.

Solution

Based on the data of the problem, the relevant cost components which will vary over the time period

due to change in lot size quantity (Q) and which will remain fixed is summarised as follows:

Relevant Cost for EOQ

Calculation Irrelevant Cost for EOQ

Calculation

Ordering Cost

Carrying cost Clerical and data

processing, Rs. 500;

Rail transport, Rs. 2000

Maintenance cost, Rs.

200;

Rented cost, Rs. 200/2 =

Rs. 100 Rail transport, Rs. 20 per

gallon because a fixed

money of Rs. 40, 000 x 20

= Rs. 8, 00, 000 will incur

irrespective of size of Q.

Overhead cost, Rs. 5, 000

Thus the relevant costs needed for calculating EOQ are: C o = Rs. 2, 500; C h = Rs. 300.

Q* (EOQ) = √ଶCo

Ch = √ଶ 𝑥 ସ, 𝑥 ଶ,ହ

ଷ = Rs. 817 (approx.) gallons.

Hence, the minimum total annual costs are:

Ordering:

Q∗ Co = ସ,

଼ଵ x 2, 500 = Rs. 1, 22, 399.02

Carrying: Q∗

ଶ Ch = ଼ଵ

ଶ x 300 = Rs. 1, 22, 250.00

Total = Rs. 2, 44, 949.02

Rail transport: 40, 000 x 2 = Rs. 8, 00, 000

Storage Overhead cost: = Rs. 5, 000

Purchase Cost: 40, 000 x 150 = Rs. 60, 00, 000

Total = Rs. 70, 49, 949.02 munotes.in

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4. A contractor has to supply 10, 000 bearings per day to an automobile manufacturer. He fin ds that

when he starts production run, he can produce 25, 000 bearings per day. The cost of holding a bearing

in stock for a year is Rs. 2 and the set-up cost of a production run is Rs. 180. How f requently should

production run be made?

Solution

From the data of the problem in usual notation, we have

Co = Rs. 1, 800 per production run; C h = Rs. 2 per year

p = Rs. 25, 000 bearings per day d = 10, 000 bearings per day

D = 10, 000 x 300 = 30, 00, 000 units/year (assuming 300 working days in the year).

a. Economic batch quantity for each production run is given by

Q* = √ଶ

ℎሺ

−𝑑ሻ = √ଶ 𝑥 ଷ,, 𝑥 ଵ,଼

ଶሺଶହ,

ଶହ,− ଵ,ሻ = 1, 04, 446 bearings

b. Frequency of production cycles

t* = 𝑄∗

𝑑 = 1, 04, 446/10, 000 = 10.44 days.

5. A commodity is to be supplied at a constant rate of 25 units per day. A penalty cos t is being

charged at the rate of Rs. 10 per unit per day late for missing the scheduled del ivery date. The cost of

carrying the commodity in inventory is Rs. 16 per unit per month. The production process is such that

each month (30 days) a batch of items is started and are available for delivery an y time after the end

of the month. Find the optimal level of inventory at the beginning of each month.

Solution:

From the data in usual notations, we have

D = 25 units / day; C h = 16/30 = 0.53 per unit per day; C o = Rs. 10 per unit per day;

t = 30 days

Thus optimal inventory level is given by

M* = [ଵ

.ହଷ+ଵ] (25) (30) = 712 units.

4.5. SELECTIVE INVENTORY CONTROL TECHNIQUES

In practice when a firm maintained large number of items in its inventory, obvi ously all items cannot,

and need not be controlled (i.e. keeping record of time interval between success ive reviews of

demand; order frequencies; expected demand rate; order quantities etc.) with equal att ention. All

items are not of equal importance to the firm in such terms as sales, profits, avai lability etc. One way

of exercising proper degree of control overall and various types of items held in stocks is to classify

them into groups (or classes) on the basis of the degree of control or intensity of management effort

that they require.

By selectively applying inventory control policies to these different groups, i nventory objectives can

be achieved with lower inventory levels than with a single policy applied to all items. These

techniques are also known as selective multi-item inventory control techniques.

In this section, we shall consider certain group classifications such as: ABC, VED, FSN, HML, XYZ

etc.

Classification Basis of Classification Purpose

ABC

{Always, Better, Control} Value of consumption To control raw material,

components and work- in-

progress inventories in the

normal course of business.

HML

{High, Medium, Low} Unit price of the material Mainly to control purchases munotes.in

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XYZ Value of items in storage To review the inventories and

their uses at scheduled intervals.

VED

{Vital, Essential, Desirable} Criticality of the component To determine the stocking

levels of spare parts.

FSN

{Fast, Slow, Non-moving} Consumption pattern of the

component To control obsolescence

ABC ANALYSIS

The ABC analysis consists of separating the inventory items into three groups: A, B and C according

to their annual cost volume consumption (unit cost x annual consumption). Although the break points

between these groups vary according to individual business conditions, a common break down might

be as follows:

Category or Group Percentage of the item % of the Total Annual Value

of the inventories (Rs.)

A 10 - 20 70 - 85

B 20 - 30 10 - 25

C 60 - 70 5 - 15

This type of classification is also known as the principle of law of Vital Few and Trivial Many. The

ABC analysis facilitates analysis of yearly consumption value of items in t he store to identify the vital

few items which are generally referred to as A category items. Generally, these it ems accounting for

about 70% of the total money value of consumption. Items accounting for about 25% of the total

money value of consumption are called B category items and the remaining ones accounting for about

5% consumption value as C category items.

Carrying on the ABC analysis of the store items helps identifying the few items that are vital from

financial point of view and require careful watch, scrutiny and follow-up. The application of ABC

analysis extends overall of the aspects of materials management like purchasing, i nventory control,

value analysis etc.

After the items are so classified, the inventory control policies are made on the b asis of the

classification 'A' category items require special managerial attention, theref ore, fixed-interval

inventory control system might be used for these items. 'C' category items ca n be managed in a little

casual manner. For these items, a fixed-order quantity system might be used. The order quan tities can

be relatively large without incurring excessive costs. A large reserve stock can also be maintained. 'B'

items are not so costly as to require special managerial attention, but these are not so cheap as to

ignore overstocking, therefore (s, S) inventory control system might be used for these items.

The procedure of ABC analysis is summarised in the following steps:

Step 1

Obtain data on the annual usage (or consumption) in units and unit cost of each inventory uni t.

Multiple the annual usage in units and the value of each item to get annual value for each of these

items.

Annual Value = Unit Cost x Annual Consumption

Step 2

Arrange these inventory items in a decreasing order of their value computed in step 1.

Step 3

Express the annual value of each item as percentage of the total value of a ll items. Also compute the

cumulative percentage of annual consumption rupees spent.

Step 4 munotes.in

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Obtain the percentage value for each of the items. That is, if there are 50 item s involved in

classification, then each item would represent 100/50 = 2 percent of the total i tems. Also cumulate

these percentage values.

Step 5

Draw a graph between cumulative percentage of items (on x-axis) and cumulative annual per centage

of usage value (on y-axis), and mark cut-off points where the graph changes slope as shown in figure.

Example 1

A company produces a mix of high technology products for use in hospitals. The annual sales

data are as follows:

Product Type Number of

Units Unit Price

(Rs.) Product Type Number of

Units Unit Price

(Rs.)

1 1, 000 2.50 10 600 1.62

2 250 0.55 11 25 33.00

3 150 6.50 12 4 15.50

4 300 1.00 13 1, 000 5.00

5 100 1.50 14 2, 850 2.50

6 700 1.43 15 10 0.83

7 500 7.00 16 355 0.98

8 15 4.98 17 50 1.37

9 1, 000 0.75 18 393 1.85

For inventory control reasons, the company wants to classify these items into three groups A,

B and C on the basis of annual sales value of each item. You please help the company.

Solution

The annual sales volume (in Rs.) for each product and the item ranking on the basis of this

volume is shown in Table.

Product ranking as per sales volume

Product Type Number of

Units Unit Price

(Rs.) Annual Sales

Volume

(Rs.) Ranking

1 1000 2.50 2, 500.00 4

2 250 0.55 137.50 14

3 150 6.50 975.00 6

4 300 1.00 300.00 12

5 100 1.50 150.00 13

6 700 1.43 1001.00 5 munotes.in

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7 500 7.00 3500.00 3

8 15 4.98 74.70 16

9 1000 0.75 750.00 9

10 600 1.62 972.00 7

11 25 33.00 825.00 8

12 4 15.50 77.50 15

13 1000 5.00 5000.00 2

14 2850 2.50 7125.00 1

15 10 0.83 8.30 18

16 355 0.98 347.90 11

17 40 1.37 54.80 17

18 393 1.85 727.05 10

The cumulative percentage of products and cumulative percentage of sales for each product is

given in table for the purpose of ABC classification:

ABC Classification

Rank Product Product

Cumulative

% of

products Annual

Sales

Volume

(Rs.) Cumulative

Annual

Sales

Volume

(Rs.) Cumulative %

Product Class

of Sales

1 14 5.56 7, 125.00 7,125.00 29.05 A product

11.11%

2 13 11.11 5, 000.00 12,125.00 49.43 products

and 49.43 Rs.

3 7 16.67 3, 500.00 15,625.00 63.70 B

products:

38.89%

4 1 22.22 2, 500.00 18,125.00 73.90 Products

and 42.91 Rs.

5 6 27.78 1, 001.00 19,125.00 77.97

6 3 33.33 975.00 20,101.00 81.95

7 10 38.89 825.00 21,073.00 85.92

8 11 44.44 750.00 21,898.00 89.28

9 9 50.00 727.05 22,648.00 92.34

10 18 55.56 347.90 23,375.00 95.30 C Products

44.44%

11 16 61.11 300.00 23,722.00 96.72 products

and 7.66 Rs.

12 4 66.67 150.00 24,022.00 97.94

13 5 72.22 137.50 24,172.95 98.56

14 2 77.78 27.50 24,310.45 99.12

15 12 83.33 24.70 24,387.95 99.43

16 8 88.89 54.80 24,462.65 99.74

17 17 94.44 8.30 24,517.45 99.96

18 15 100.00 24,525.75 100.00

The percentage of products and percentage of annual sales volume can also be plotted on the

graph.

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4.6. VED Analysis

This analysis helps in separating the inventory items into three groups according to their

criticality, usually called V, E, and D items in that order, VED classification calls for

classification of items as Vital, Essential and Desirable.

V items are considered vital for smooth running of the system and without these items the

whole system becomes inoperative. Thus adequate stock is required all the time.

E items are considered essential to the efficient running of the system and non-availability of

these items reduces the efficiency of the system.

D items neither stop the system nor reduce its efficiency, but availability of such items will

lead to increase in efficiency and reduction of failure.

This classification is largely useful in controlling inventory of spare parts. It can also be used

in case of such raw materials whose availability is rare.

ABC analysis and VED analysis can also be combined to control the stocking of spare parts

based on the desired customer service level as shown in the table.

ABC

Classification VED Classification

V E D

A Constant control,

Regular follow-up

Low stocks and

ordering more

frequently Average stock; No

risk of stock outs. No stock

B Average stocks

No risk of stock outs Average stock; some

risk can be taken Very low stocks;

Some risk can be

taken.

C High Stocks

Restricted orders; No

risk Average Stock; some

risk can be taken. Low stocks; some

risk can be taken.

HML Analysis

Based on the unit price of items, the HML classification separates inventory items, as High

price, Medium price and Low price. This analysis is helpful to control purchase of various

items for inventory.

4.7. FSN Analysis

The consumption pattern or inventory items forms the basis for FSN analysis. Items are

classified as Fast-moving, Slow-moving and Non-moving. Sometimes items are also

classified as FSND: Fast-moving; Slow-moving; Normal-moving and Dead (or Non-moving). munotes.in

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This classification is based on the movement (or consumption pattern) and therefore helps to

controlling obsolescence of various items by determining the distribution and handling

patterns. Cut-off points of the three classes are usually in terms of number of issues in the

previous few years.

4.8. XYZ Analysis

This classification is based on the closing value of items in storage. Items whose inventory

values are high and moderate are classified as X-items and Y-items respectively, while ite ms

with low inventory value are termed as Z-items.

This analysis is usually undertaken once a year during the annual stock taking exercise. This

helps in identifying the items which are being stocked extensively.

This classification can also be combined with ABC classification of items to control

inventory of items as shown in the table below:

ABC

Classification XYZ Classification

X Y Z

A Attempt to reduce

stocks Attempt to convert

Z-items Items are within

control

B Stock and

consumption is

reviewed more often Items are within

control Review stock at least

twice a year

C Dispose off the

surplus items Check and maintain

the control Review stock

annually

XYZ-FSN classification exercise helps in the timely prevention of obsolescence.

XYZ

Classification FSN Classification

F S N

X Light inventory control Reduce stock to very

low level Quick disposal of

items at optimum

price

Y Normal inventory control Low level of stocks Should be disposed

as early as possible.

Z Can reduce clerical work

by increasing stocks. Low level of stocks Can afford to dispose

at lower prices.

4.9. EXERCISE

Questions

1. Why is inventory an important consideration for managers?

2. State the purpose of inventory control.

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3. Why would not a company all the time store large quantities of inventory to avoid

shortages and stockouts?

4. Describe the major decisions that must be made in inventory control.

5. State some of the assumptions made in EOQ.

6. Discuss in detail the major inventory costs that are used in EOQ.

7. Classify the following 14 items in ABC categories.

Code Number Monthly

Consumption (Rs.) Code Number Monthly

Consumption (Rs.)

D-179-0 451 D-196 214

D-195-0 1,052 D-198-0 188

D-186-0 205 D-199 172

D-191 893 D-200 170

D-192 843 D-204 5,056

D-193 727 D-205 159

D-195 412 D-212 3,424

How the policies with regard to safety stocks, order quantity, material control and inventory

system will be different for the items classified as A, B and C?

8. The following information is provided for an item:

Annual demand: 12, 000 units; Ordering Cost: Rs. 60/order

Carrying Cost: 10%; Unit cost of item: Rs. 10 and Lead Time: 10

days.

There are 300 working days in a year. Determine EOQ and number of orders per year. In the

past two years the demand rate has gone as high as 70 units/day. For a reordering system

based on the inventory level, what should be the buffer stock? What should be the reorder

level at this buffer stock? What would be the carrying costs for a year?

9. The demand for an item in a company is 18,000 units per year, and the company can

produce the item at a rate of 3,000 per month. The cost of one set-up is Rs. 500 and the

holding cost of one unit per month is 15 paise. The shortage cost of one unit is Rs. 240 per

year. Determine the optimum manufacturing quantity and the number of shortages. Also

determine the manufacturing time and the time between set-ups.

10. A product is sold at the rate of 50 pieces per day and is manufactured at a rate of 250

pieces per day. The set-up cost of the machines is Rs.1, 000 and the storage cost is found to

be Re 0.0015 per piece per day. With labour charges of Rs. 3.20 per piece, material cost at

Rs.2.10 per piece and overhead cost of Rs.4.10 per piece, find the minimum cost batch size if

the interest charges are 8% (assume 300 working days in a year). Compute the optimal

number of cycles required in a year for the manufacture of this product.

xxx-----xxx-----xxx

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CHAPTER 5

SIMULATION TECHNIQUES

UNIT STRUCTURE

5.1 Introduction

5.2 Types of Simulation

5.3 Steps of Simulation Process

5.4 Advantages and Disadvantages of Simulation

5.5 Stochastic Simulation and Random Numbers

5.6 Monte Carlo Distribution

5.7 Simulation of Inventory Problems

5.8 Simulation of Queuing Problems

5.9 Applications of Simulation

5.10 Exercise

LEARNING OBJECTIVES

After studying this chapter, you should be able to

Make distinction between analytical and simulation models.

Appreciate what simulation is and what we can do with it.

Know several definitions of simulation and appreciate the importance of simulation

modelling.

Understand advantages and disadvantages of simulation.

Apply Monte Carlo simulation techniques for solving various types of problems.

Develop random number intervals and use them to generate outcomes.

Know several types of computer languages that are helpful in the simulation process.

Know certain causes of simulation analysis failure and how to avoid them.

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5.1. INTRODUCTION

Mathematical models cannot be manipulated to destroy its acceptability as a reasonable

representation of a system under study. However, a model once constructed may be used to

predict consequences of taking alternative actions. In particular, we could 'experiment' on the

model by 'trying' alternative actions or parameters and compare their consequences. This

'experimentation' allows to answer 'what if' questions relating the effects of assumptions on

the model response. Comparing the consequences of substituting various parameters into the

model is referred to as simulating the model.

As the complexity of a model increases, simulation seeks to replicate the uncertainty in the

model and assess model's response to events made to occur (i.e. simulated) with a frequency

characterised by pre-specified probabilities.

Simulation is used in almost all fields, restricted only by our imagination and our ability to

translate such imagination into a set of computer directives. The following are some simple

examples which will make us appreciate what simulation is and what we can do with it.

Location of ambulances

Often, it is not known exactly where the demand for ambulance services will arise and,

therefore, simulation is required to test alternative scheduling rules of the ambulances, their

locations (in a geographical area), the response time to an emergency call and, of course, the

overall service quality and the costs incurred if ambulances were to be configured in a certain

way (in types, quantity, location, scheduling and staffing).

Design of computer systems

In general, it is difficult to specify future needs. Simulation is useful to highlight the basic

characteristics of a computer system configuration (its speed, size, computing qualities,

memory and so on.) in terms of the equipment used, the management of the computer system,

the costs of such configurations and the resulting computational service which it provides to

users.

Shop floor Management

The management of shop floor in factories is extremely complex because it involves many

persons, machines, products parts, robots, material handling systems, Televisions and

computer aided vision and inspection systems. Even if we were able to manage some specific

aspects of the shop floor, integrating it into a simple mathematical model is very difficult,

provided we can construct a model (however complex it may be) of the shop floor.

Simulation is useful t analyse how a model could respond to certain policies (such as

scheduling jobs to be performed on the basis of their due-dates, with earliest dates jobs

performed first, by comparing alternative configuration and routing control policies, and so

on). Simulation can also be used to test alternative work flows, the introduction of a new

machinery, and in some cases the use of completely new manufacturing and management

technologies (such as Just- In-Time manufacturing, flexible manufacturing, etc.) In such

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i.e. it provides, an experience and a learning tool which will give us greater confidence in the

productivity and the management of the future system, at a relatively small cost.

Design of Optimal Replenishment Policy

In inventory control, the problem of optimal replenishment policy arises due to the

probabilistic (stochastic) nature of demand and lead time. Thus, instead of manually trying

out an appropriate replenishment alternative for each level of demand and lead time for a

specific period and then selecting the best one, we process data (called experiment) on the

computer and obtain the results in a very short time at a very small cost.

Design of a Queuing System

In queuing theory, the problem of balancing the cost of waiting against the cost of idle time

of service facilities in the system arises due to the probabilistic nature of the inter-arrival

times of customers and the time taken to complete the service to the customer. Thus, instead

of trying out in actual manually with data to design a queuing system, we process the data on

computers and obtain the expected value of various characteristics of the queuing system

such as idle time of servers, average waiting time, queue length etc.

Other problems such as aircraft management routing, scheduling of bank tellers and location

of bank branches, the deployment of fire stations, routing and dispatching when roads are not

secured (where materials sent might not, potentially, reach their destination), the location and

the utilisation of recreational facilities (such as parks, public swimming pools etc.) and many

other problems have been studied or could be studied through simulation.

SIMULATION DEFINED

Simulation is one of the oldest operations research technique known to man - that is, the

representation of the real world by numbers and other symbols that can be readily

manipulated. To gain a better grasp of the real world, games such as chess to simulate the

battles, backgammon to simulate racing and other games to simulate hunting and diplomacy

were already invented in antiquity.

Today, a modern game like monopoly simulates the competitive arena of real estate. Many

have played baseball with a deck of cards which has hits, strikeouts and walks with a

cardboard, diamond and plastic chips as runners. The distribution of hits, runs and outs etc.,

in a deck of cards serves as a realistic reflection of the overall average with which eac h will

occur in real life. Other games people play, generate experiences, understanding and gain

knowledge which is basically what we will be trying to do through simulation.

What is new in modern simulation? It is the availability of computers which make it possible

for us to deal with an extraordinary large quantity of details which can be incorporated into a

model and the ability to manipulate the model over many 'experiments' (i.e. replicating all the

possibilities that may be imbedded in the external world, and events would seem to recur).

The modern use of the word simulation can be traced to the mathematicians Von Neumann

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first to 'break' the Casino at Monte Carlo (which they did not succeed!) and subsequently,

applying it to the solution of nuclear shielding problems that were either too expensive for

physical experimentation, or too complicated for treatment by known mathematical

techniques.

An agreed definition for the world simulation has not been reached so far, however, a few

definitions are stated as:

a. A simulation of a system or an organism operation of a model or simulator which is a

representation of the system or organism. The model amenable to manipulation which would

be impossible, too expensive or unpractical to perform on the entity it portrays. The

operation of the model can be studied and for it, properties concerning the behaviour of the

actual system can be inferred.

This definition is broad enough to be applied equally to military war games, business games,

economic models etc. In this view simulation involves logical and mathematical constructs

that can be manipulated on a digital computer using iterations or successive trials.

Simulation is the process of designing a model of a real system and conducting

experiments with this model for the purpose of understanding the behaviour (with the

limits imposed by a criterion or set of criteria) for the operation of the system.

Simulation is a numerical technique for conducting experiments on a digital

computer, which involves certain types of mathematical and logical relationships

necessary to describe the behaviour and structure of a complex real-world system over

extended periods of time.

- Naylor et al.

Few other definitions of simulations are as under:

'X simulated Y' is true if and only if

i. X and Y are formal systems.

ii. Y is taken to be the real system.

iii. X is taken to be an approximation to the real system, and

iv. The rules of validity in X are non-error-free, otherwise X will become the real

system.

Simulation is the use of a system model that has the designed characteristics of reality

in order to produce the essence of actual operation.

For Operations Research, simulation is a problem solving technique which uses a computer-

aided experimental approach to study problems that cannot be analysed using direct and

formal analytical methods. As a result, simulation can be thought of as a last resort technique.

It is not a technique which should be applied in all cases. However, table 1 highlights what

simulation is and what it is not.

Simulation - what it is/not

It is It is not

- A technique which uses computers.

- An approach for reproducing the

processes by which events of chance

and change are created in a computer. - An analytical technique which provides

exact solution.

- A programming language but it could be

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- A procedure for testing and

experimenting on models to answer

what if....then so and so....types of

questions. programming of simulation.

5.2. TYPES OF SIMULATION

There are several types of simulation. Few of them are listed below:

1. Deterministic/probabilistic simulation

The deterministic simulation is used when a process is very complex or consists of multiple

stages with complicated (but known) procedural interactions between them. The measures of

performance of such a system would be extremely detailed and time consuming. Formulating

such a process as a simulation with fixed procedures (algorithms) allows the determination of

the outcome and measure of performance in a straight forward manner.

In probabilistic simulation, one or more of the independent variables (e.g., the arrival rate of

customers at a service-window is probabilistic, that is, it follows a certain probability

distribution.

2. Time dependent and time independent simulation

In time independent simulation it is not important to know exactly when the event is likely to

occur. For example, in inventory control situation, we may know that the demand is three

units per day, but it is not necessary to know when during the day the item was demanded.

On the other hand, in time dependent simulation it is important to the precise time when the

event is likely to occur. For example, in queuing situation the precise time of arrival should

be known (to know if the customer will have to wait).

3. Visual interactive simulation

Visual interactive simulation uses computer graphics displays to present the consequences of

change in the value of input variation in the model. The decisions are implemented

interactively while the simulation is running. These simulations can show dynamic systems

that evolve over time in terms of animation. The decision maker watches the progress of the

simulation in an animated form on a graphics terminal and can alter the simulation as it

progresses.

4. Business Games

Business game simulation model involves several participants who need to play a role in a

game that simulates a realistic competitive situation. Individuals or teams compete to achieve

their goal, such as profit maximisation, in competition or cooperation with the other

individuals or teams. The few advantages of business games are:

i. Participants learns much faster and the knowledge and experience gained are more

memorable than passive instruction.

ii. Complexities, interfunctional dependencies, unexpected events, and other such factors can

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iii. The time compression - allowing many years of experience in only minutes or hours 0 lets

that participants try out actions that they would not be willing to risk in an actual situation

and see the result in the future.

iv. Provide insight into the behaviour of organisation. They dynamics of team decision-

making style highlight the roles assumed by individuals on the teams, the effect of

personality types and managerial styles, the emergence of team conflict and cooperation, and

so on.

5. Corporate and financial simulations

The corporate financial simulation is used in corporate planning, especially the financial

aspects. The models integrate production, finance, marketing, and possibly other function

into one model either deterministic or probabilistic when risk analysis is desired.

5.3. STEPS OF SIMULATION PROCESS

The process of simulating a system consists of following steps:

1. Identify the problem

The simulation process is used to solve any problem only when the assumptions required for

analytical models are not satisfied or there is no appropriate model developed for a system

under study. For example, a queuing situation may be of interest but the arrival and/or service

pattern do not meet the assumptions required to use queuing theory.

2. Identify the decision variables and decide the performance criterion (objective)

In the context of an inventory control situation, the demand (consumption rate), lead time and

safety stock are identified as decision variables. These variables shall be responsible to

measure the performance of the system in terms of total inventory cost under the decision

rule - when to order.

3. Construct a simulation model

For developing a simulation model, an intimate understanding of the relationships among the

elements of the system being studied is required. For this purpose, the influence diagram

(drawn in a variety of different ways) is useful because simulation models for each of these

diagrams may be formulated until one seems better or more appropriate than the other. Even

after one has been chosen, it may be modified again and again before a final version is

acceptable.

4. Testing and validating the model

Any simulation model must represent the system under study. This requires comparison of

the model with the actual system - a validation process. A validated model should behave

similar to the system under study. However, this validation condition may not be sufficient on

model's predictive abilities.

The validation process requires

1. Determine whether the model is internally correct in a logical and programming sense

called internal validity and

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The first step involves checking the equations and procedures in the model for accuracy, both

in terms of mistakes (or errors) and in terms of properly representing the system under study.

The verification of internal validity can be simplified if the model is developed in modules

and each module is tested as it is developed. Focusing on a particular module rather than

trying to evaluate the logic of the entire model all at once facilitates identifying the source of

errors and correcting them.

After verifying internal validity, the model is tested by substituting historical values into the

model and seeing if it replicates what happens in reality. If the model passes this test, extreme

values of the input variables are entered and the model is checked for the expected output.

A visual display of simulation results gives a better feel for what is happening in the model.

The decision maker can make changes in the assumptions or input data and see the effect on

the outputs. This improves the validity testing process.

5. Designing of the experiment

Experimental design refers to controlling the conditions of the study, such as the variables to

include. This is in contrast to situations where observations are taken but the conditions of the

study are not controlled.

It requires to determine factors considered fixed and variable in the model, levels of the

factors to use, what the resulting dependent measures are going to be, how many times the

model will be replicated and length of time of each replication and so on. For example, in a

queuing simulation we may consider arrival and service rates constant but vary the number of

servers and the evaluate the customer waiting times (dependent variable).

6. Run the simulation Model

Run the model on the computer to get the results in the form of operating characteristics.

7. Evaluate the Results

Examine the results of problem as well as their reliability and correctness. If the simulation

process is complete, then select the best course of action (or alternative) otherwise make

desired changes in model decision variables, parameters or design and return to step 3.

5.4. ADVANTAGES AND DISADVANTAGES OF SIMULATION

Advantages

The increased acceptance of simulation at various managerial levels is due to a number of

following factors:

1. This approach is suitable to analyse large and complex real-life problems which cannot be

solved by usual quantitative methods.

2. It is useful for sensitivity analysis of complex systems. In other words, it allows the

decision-maker to study the interactive system variables and the effect of changes in these

variables on the system performance in order to determine the desired one.

3. Simulation experiments are done with the model, not on the system itself. It also allows to

include additional information during analysis that most quantitative models do not permit. In

other words, simulation can be used to 'experiment' on a model of a real situation without

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4. Simulation can be used as a pre-service test to try out new policies and decision rules for

operating a system before running the risk of experimentation in the real system.

5. The only 'remaining tool' when all other techniques become intractable or fail.

Disadvantages

The primary reasons of not often using simulation are as under:

1. Sometimes simulation models are expensive and take a long time to develop. For example,

a corporate planning model may take a long time to develop and prove expensive also.

2. It is the trial and error approach that produces different solutions in repeated runs. This

means it does not generate optimal solutions to problems.

3. Each application of simulation is ad hoc to a great extent.

4. The simulation model does not produce answers by itself. The user has to provide all the

constraints for the solutions which he wants to examine.

5.5. STOCHASTIC SIMULATION AND RANDOM NUMBERS

In simulation, probability distributions are used to quantify outcomes in numerical terms by

assigning a probability to each of possible outcomes. For example, if you flip a coin, the set

of possible outcomes is {H, T}. A random variable assigns a number to the possible

occurrence of each outcome. In simulation random variables are controlled numerically and

are used to simulate elements of uncertainty which are defined in a model. This is done by

generating (using the computer) outcomes with the same frequency as those encountered in

the process being simulated. In this manner many experiments (also called simulation runs)

can be performed, leading to a collection of outcomes that have a frequency (probability)

distribution similar to that of the model under study.

To use simulation, it is necessary to learn generating the sample random events that make up

the model. This helps to use the computer to reproduce the process through which chance is

generated in actual situation. Thus a problem which involves many inter relationships among

random variables can be evaluated as a function of given parameters. Thus process generation

(simulating chance processes) and modelling are the two fundamental techniques of that are

needed in simulation.

The first elementary and important type of process is the random process, which requires the

selection of samples (or events) from a given distribution so that repetition of this selection

process will yield a frequency distribution of sample values that matches the original

distribution. When these samples are generated through some mechanical or electronic device

- called pseudo random numbers. Alternately it is possible to use table of random numbers

where the selection of number in any consistent manner will yield numbers that behave as if

they were drawn from a uniform distribution.

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Random numbers can also be generated using random number generator (which are inbuilt

feature of spread sheets and many computer languages) tables, a roulette wheel etc.

Random numbers between 00 and 99 are used to obtain values of random variables that have

a known discrete probability distribution in which the random variable of interest can assume

one of a finite number of different values. In some applications, however, the random

variables are continuous, that is, they can assume any real value according to a continuous

probability distribution. For example, in queuing theory applications, the amount of time a

server spends with a customer might follow an exponential distribution.

5.6. MONTE CARLO DISTRIBUTION

The Monte Carlo simulation technique involved conducting repetitive experiments on the

model of the system under study with some known probability distribution to draw random

samples (observations) using random numbers. If a system cannot be described by a standard

probability distribution such as normal, Poisson, exponential, gamma etc. an empirical

probability distribution can be constructed. The Monte Carlo simulation technique consists of

following steps:

1. Setting up a probability distribution for variables to be analysed.

2. Building a cumulative probability distribution for each random variable.

3. Generate random numbers and then assign an appropriate set of random numbers to

represent value or range (interval) of values for each random variable.

4. Conduct the simulation experiment using random sampling.

5. Repeat Step 4 until the required number of simulation runs has been generated.

6. Design and implement a course of action and maintain control.

Random Number Generation

Monte Carlo Simulation requires the generation of a sequence of random numbers. This

sequence of random numbers helps in choosing random observations (samples) from the

probability distribution.

Arithmetic Computation

The nth random number r n consisting of k-digits generated by using multiplicative

congruential method is given by

rn≡ p.r n-1 (modulo m)

where p and m are positive integers, p < m, r n-1 is a k-digit number and modulo m means that

rn is the remainder when p.r n-1 is divided by m. This means r n and p.r n-1 differ by an integer

multiple of m. To start the process of generating random numbers, the first random number

(also called seed) r 0 is specified by the user. Then using above recurrence relation a sequence

of k-digit random number with period h < m at which point the number r 0 occurs again can

be generated.

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For illustration, let p = 35, m = 100 and arbitrarily start with r 0 = 57. Since m - 1 = 99 and is

the 2-digit number, therefore, it will generate 2-digit random numbers:

r1 = p. r 0 (modulo m) = 35 x 57 (modulo 100) = 1, 995/100 = 95, remainder

r2 = p. r 1 (modulo m) = 35 x 95 (modulo 100) = 3, 325/100 = 25, remainder

r3 = p. r 2 (modulo m) = 35 x 25 (modulo 100) = 875/100 = 75, remainder

The choice of r 0 and p for any given value of m require great care, and the method used is

also not a random process because sequence of numbers generated is determined by the input

data for the method.

Thus, the numbers generated through this process are pseudo random numbers because there

are reproducible and hence, not random.

The above defined recurrence relation can also be used to generate random numbers as

decimal fraction between 0 and 1 with a desired number of digits. For this, the recurrence

relations u n = r n/m is used to generate uniformly distributed decimal fraction between 0 and 1.

Computer generator

The random numbers that are generated by using computer software are uniformly distributed

decimal fractions between 0 and 1. The software works on the concept of cumulative

distribution function for the random variables for which we are seeking to generate random

numbers.

For example, for the negative exponential function with density function f(x) = 𝜆e-𝜆𝑥, 0<

𝑥 < ∞ , the cumulative distribution function is given by

F(x) = ∫𝜆e − 𝜆𝑥𝑥

0 dx = 1 - e - 𝜆𝑥

e - 𝜆𝑥 = 1 - F(x) or

Taking logarithm on both sides, we have

- 𝜆𝑥 = log [1- F(x) ]

x = - (1/ 𝜆) log [1- F(x) ] or

If r = F(x) is a uniformly distributed random decimal fraction between 0 and 1, then the

exponential variable associated with r is given by

xn = - (1/ 𝜆) log (1- r ) = - (1/ 𝜆) log r.

This is an exponential process generator since 1-r is a random number and can be replaced by

r.

5.7. SIMULATION OF INVENTORY PROBLEMS

Examples

1. Using random numbers to simulate a sample, find the probability that a packet of 6

products does not contain any defective product, when the production line produces 10 per

cent defective products. Compare your answer with the expected probability.

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Solution

Given that 10 percent of the total production is defective and 90 per cent is non-defective. If

we have 100 random numbers (0 to 99), then 90 or 90 percent of them represent non-

defective products and remaining 10 or 10 percent of them represent defective products.

Thus, the random numbers 00 to 89 are assigned to variables representing non-defective

products and 90 to 100 are assigned to variables representing defective products.

If we choose a set of 2-digit random numbers in the range 00 to 99 to represent a packet of 6

products as shown below, then we would expect that 90 percent of the time they would fall in

the range 00 to 89.

Sample Number Random Number

A

B

C

D

E

F

G

H

I

J 86 02 22 57 51 68

39 77 32 77 09 79

28 06 24 25 93 22

97 66 63 99 61 80

69 30 16 09 05 53

33 63 99 19 87 26

87 14 77 43 96 43

99 53 93 61 28 52

93 86 52 77 65 15

18 46 23 34 25 85

Here it may be noted that out of ten simulated samples 6 contain one or more defectives and 4

contain no defectives. Thus, the expected percentage of non-defective products is 40 percent.

However, theoretically the probability that a packet of 6 products containing no defective

product is (0.9)6 = 0.53144 = 53.14%.

2. A bakery keeps stock of a popular brand of cake. Previous experience shows the daily

demand pattern for the item with associated probabilities, as given below:

Daily demand (number) : 0 10 20 30 40 50

Probability : 0.01 0.20 0.15 0.50 0.12 0.02

Use the following sequence of random numbers to simulate the demand for next 10 days.

Random Numbers: 25, 19, 65, 76, 12, 05, 73, 89, 19, 49.

Also estimate the daily average demand for the cakes on the basis of simulated data.

Solution:

Using the daily demand distribution, we obtain a probability distribution as shown in table 1.

Daily Demand Distribution

Daily Demand Probability Cumulative

Probability Random Number

Interval

0

10

20

30

40

50 0.01

0.20

0.15

0.50

0.12

0.02 0.01

0.21

0.36

0.86

0.98

1.00 00

01 - 20

21 - 35

36 - 85

86 - 97

98 - 99

Conduct the simulation experiment for demand by taking a sample of 10 random numbers

from a table of random numbers, which represent the sequence of 10 samples. Each random

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The simulation calculations for a period of 10 days are given in the following table.

Simulation Experiments

Days Random Number Demand

1

2

3

4

5

6

7

8

9

10 40

19

87

83

73

84

29

09

02

20

Total

Expected Demand 30

10

40

30

30

30

20

10

10

10

--------

220

220/10 = 22 units per day

5.8. SIMULATION OF QUEUING PROBLEMS

Examples

1. S dentist schedules all his patients for 30-minute appointments. Some of the patients take

more or less than 30 minutes depending on the type of dental work to be done. The following

summary shows the various categories of work, their probabilities and time actually needed

to complete the work:

Category of Service Time Required

(Minutes) Probability of Category

Filling

Crown

Cleaning

Extraction

Check-up 45

60

15

45

15 0.40

0.15

0.15

0.10

0.20

Simulate the dentist's clinic for 4 hours and determine the average waiting time for the

patients as well as the idleness of the doctor. Assume that all the patients show up at the

clinic at exactly their scheduled arrival time starting at 8.00 a.m. Use the following random

numbers for handling the above problem:

40 82 11 34 25 66 17 79

Solution:

The cumulative probability distribution and random number interval for service time are

shown in table 1.

Category of

Service Service Time

Required

(minutes) Probability Cumulative

Probability Random

Number

Interval

Filling

Crown

Cleaning 45

60

15 0.40

0.15

0.15 0.40

0.55

0.70 00 - 39

40 - 54

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Extraction

Check-up 45

15 0.10

0.20 0.80

1.00 70 - 79

80 - 99

The various parameters of a queuing system such as arrival pattern of customers, service

time, waiting time in the context of the given problem are shown in the table:

Arrival Pattern and Nature of Service

Patient

Number Scheduled

Arrival Random

Number Category of

Service Service Time

(Minutes)

1

2

3

4

5

6

7

8 8.00

8.30

9.00

9.30

10.00

10.30

11.00

11.30 40

82

11

34

25

66

17

79 Crown

Check-up

Filling

Filling

Filling

Cleaning

Filling

Extraction 60

15

45

45

45

15

45

45

Computation of Arrivals, Departures and Waiting of Patients

Time Event

(Patient Number) Patient Number

(Time to Exit) Waiting

(Patient Number)

8.00

8.30

9.00

9.15

9.30

10.00

10.30

10.45

11.00

11.30

11.45

12.00 1 arrive

2 arrive

1 departs; 3 arrive

2 depart

4 arrive

3 departs; 5 arrive

6 arrive

4 depart

7 arrive

5 depart; 8 arrive

6 depart

End 1 (60)

1(30)

2(15)

3(45)

3(30)

4(45)

4(15)

5(45)

5(30)

6(15)

7(45)

7(30) ---

2

3

---

4

5

5, 6

6

6, 7

7, 8

8

8

The dentist was not idle during the entire simulated period. The waiting times for the patients

were as follows:

Computation of Average Waiting Time

Patient Arrival Time Service Starts at Waiting time

(minutes)

1

2

3

4

5

6

7

8 8.00

8.30

9.00

9.30

10.00

10.30

11.00

11.30 8.00

9.00

9.15

10.00

10.45

11.30

11.45

12.30 0

30

15

30

45

60

45

60

_____

280

The average waiting time = 280/8 = 35 minutes.

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5.9. APPLICATIONS OF SIMULATION

There is a wide range of applications of computer-based simulation models because it is an

approach rather than an application of specific techniques.

The major use of computer-based Monte Carlo simulation model has been in the solution of

complex queuing problems.

A number of job shop simulation programmes have been developed involving deterministic

times for the individual operations of a given order. Due to different processing times for

similar operations and different order operations sequences, it is difficult to predict the

waiting time for a particular job at any given work centre. For better scheduling, orders must

be scheduled with a provision of waiting at the various work centres they will pass through.

Simulation can help in estimating accurately such waiting times.

A good deal of work has been done in the development of inventory simulation models such

as determination of optimal reorder level and lot size under conditions of probabilistic

demand and lead time, optimal review period and ordering policy for continuous review

inventory models.

A number of network simulation models have also been developed. For example, with a

randomly selected activity times the critical path can be evaluated. Repeating this process

many times, the probability distribution of project completion time can be obtained as well as

the probability that each given activity is on the critical path.

A variety of other problems which could be solved by simulation include:

1. Military studies of logistics, support planning and weapon systems effectiveness.

2. Studies of individual and group behaviour.

3. Financial studies involving risky investments.

4. Testing of decision rules for hospital admission and operating policies.

5.10. EXERCISE

Questions

1. Distinguish between solutions derived from simulation models and solution derived

from analytical models?

2. What are random numbers? Why are random numbers useful in simulation models and

solutions derived from analytical models?

3. What is Monte Carlo Simulation? Describe the idea of experimentation (Random Sampling)

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4. Describe the kind of problems for which Monte Carlo will be an appropriate method of

solution.

5. Discuss the applications of simulation.

6. A book store wishes to carry a particular book in stock. Demand is not certain and there is a

lead time of 2 days for stock replenishment. The probabilities of demand are given below:

Demand (units/day) : 0 1 2 3 4

Probability : 0.05 0.10 0.30 0.45 0.10

Each time an order is placed, the store incurs an ordering cost of Rs.10 per order. The store

also incurs a carrying cost of Rs.0.5 per book per day. The inventory carrying cost is

calculated on the basis of stock at the end of each day. The manager of the book store wishes

to compare two options for his inventory decision.

A: Order 5 books when present inventory plus any outstanding order falls below 8 books.

B: Order 8 books when present inventory plus any outstanding order falls below 8 books.

Currently (beginning of 1st day) the store has a stock of 8 books plus 6 books ordered two

days ago and are expected to arrive next day. Carryout simulation run for 10 days to

recommend an appropriate option. You may use random numbers in the sequences. Using

first number for day one.

89, 34, 78, 63, 61, 81, 19, 16, 13, 73.

7. A company trading in motor vehicle spare parts wishes to determine the levels of stock it

should carry for the items in its range. Demand is not certain and there is a lead time for stock

replenishment. For an item A, the following information is obtained:

Demand (units/day) : 3 4 5 6 7

Probability : 0.10 0.20 0.30 0.30 0.10

Carrying cost (per unit/day) : Rs. 2

Ordering Cost (per order) : Rs. 50

Lead Time for replenishment : 3 days

Stock on hand at the beginning of the simulation exercise was 20 units.

Carry out a simulation run over a period of 10 days with the objective of evaluating the

inventory rule: Order 15 units when present inventory plus any outstanding order falls below

15 units.

You may use random numbers in the sequence of: 0, 9, 1, 1, 5, 1, 8, 6, 3, 5, 7, 1, 2, 9 using

the first number for day one. Your calculation should include the total cost of operating this

inventory rule for 10 days.

8. A firm has a single channel service station with the following arrival and service time

probability distributions:

Inter arrival Time

(minutes) Probability Service Time

(minutes) Probability

10

15

20 0.10

0.25

0.30 5

10

15 0.08

0.14

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25

30 0.25

0.10 20

25

30 0.24

0.22

0.14

The customer's arrival at the service station is a random phenomenon and the time between

the arrival varies from 10 minutes to 30 minutes. The service time varies from 5 minutes to

30 minutes. The queuing process begins at 10.00 a.m. and proceeds for nearly 8 hours. An

arrival goes to the service facility immediately, if it is free. Otherwise it will wait in a queu e.

The queue discipline is first-come-first-served.

If the attendant's wages are Rs.10 per hour and the customer's waiting time costs Rs. 15 per

hour, then would it be an economical proposition to engage a second attendant? Answer

using Monte Carlo simulation technique.

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CHAPTER 6

QUEUING THEORY

UNIT STRUCTURE

6.1 Introduction

6.2 The structure of a Queuing System

6.3 Elements of Queuing System

6.4 Applications and Limitations

6.5 Operating Characteristics of Queuing System

6.6 Basic Characteristics of Queuing System

6.7 Service Channels

6.8 Multi- Server Queuing Models

6.9 Exercise

LEARNING OBJECTIVES

After studying this chapter, you should be able to

identify and examine situations that generate queuing problems.

describe the trade-off between cost of service and cost of waiting time.

understand various components (or parts) of a queuing system and description of each

of them.

analyse a variety of performance measures (operating characteristics) of a queuing

systems

derive relationships among variety of performance measure using probabilistic

distributions.

understand distinction among several queuing models and derive performance

measures for each of them.

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6.1. INTRODUCTION

A common situation occurring in everyday life is that of queuing or waiting in a line. Queues

are usually seen at bus stops, ticket booths, doctors' clinics, bank counters, traffic lights and

so on. Queues are also found in workshops where the machines wait to be repaired; at a tool

crib where the mechanics wait to receive tools, in a warehouse where items wait to be used,

incoming calls wait to mature in the telephone exchange, trucks wait to be unloaded,

airplanes wait either to take off or land and so on.

In general, a queue is formed at a queuing system when either customers (human beings or

physical entities) requiring service wait due to number of customers exceeds the number of

service facilities, or service facilities do not work efficiently and take more time than

prescribed to serve a customer.

Queuing theory can be applied to a variety of operational situations where it is not possible to

predict accurately the arrival rate (or time) of customers and service rate (or time) of service

facility or facilities. In particular, it can be used to determine the level of service (eit her the

service rate or the number of service facilities) that balances the following two conflicting

costs:

i. Cost of offering the service

ii. Cost incurred due to delay in offering service

The first cost is associated with the service facilities and their operations and the second

represents the cost of customer's waiting time.

Obviously, an increase in the existing service facilities would reduce the customer's waiting

time. Conversely, decreasing the level of service should result in long queue(s). This means

an increase (decrease) in the level of service increases (decreases) the cost of operating

service facilities but decreases (increases) the cost of waiting. Figure 1 illustrates both t ypes

of costs as a function of level of service. The optimum service level is one that minimises the

sum of the two costs.

Since cost of waiting is difficult to estimate, it is usually measured in terms of loss of sales or

goodwill when the customer is a human being who has no sympathy with the service. But, if

the customer is a machine waiting for repair, then cost of waiting is measured in terms of the

cost of lost production.

Figure 1 Queuing Costs Vs Level of Service

Total expected Cost

Cost of Operating the

Service facility per unit

time

Cost of waiting

customers per unit time

Optimum Service

Level

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Many practical situations in which study of queuing theory can provide solution to waiting

line problems are listed in the following table.

Situations Customers Service Facilities

Petrol Pumps

Hospital

Airport

Post Office

Job Interviews Automobiles

Patients

Aircraft

Letters

Applicants Pumps/Passionel

Doctors/Nurses/Rooms

Runways

Sorting System

Interviewers

6.2. THE STRUCTURE OF A QUEUING SYSTEM

A queuing system is composed of the following components (or parts) each of which is

described below:

1. Calling population (or input source)

2. Queuing process

3. Queue discipline

4. Service Process (or mechanism)

Arrival Process Queue Discipline Departure

(serviced

customers)

Balk Renege Jockey

Customers are defined as those in need of services. Customers requiring services are

generated at different times by a calling population, commonly known as input source.

The manner in which customers arrive at the service facility individually or in batches, at

scheduled or unscheduled time is called the arrival process. The customer's entry into the

queuing system depends upon the queue conditions.

The service is provided by one or more service facilities. This may be person (bank teller,

barber etc.). or a space (airport runway, hospital bed, parking lot etc.). Customers from a

queue are selected for services according to certain rules known as queue discipline. A

service facility may be without server (self-service) or include one or more servers operating

either in a series (as a team or in parallel) (multiple service channels). The rate (constant or

random) at which service is rendered is known as the service process. After the service is

rendered, the customer leaves the system.

If server is idle at the time of customer's arrival, then customer is served immediately,

otherwise the customer is asked to join a queue or waiting line, which may have single,

multiple or ever priority lines. The order of service may be random, by appointment, first-

cum-first served or some other order. The queue can also be considered to be either finite or

infinite. Input Source

(Population) Queue or

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6.3. ELEMENTS OF QUEUING SYSTEMS

Population of Customers can be considered either limited (closed systems) or unlimited

(open systems). Unlimited population represents a theoretical model of systems with a large

number of possible customers (a bank on a busy street, a motorway petrol station). Example

of a limited population may be a number of processes to be run (served) by a computer or a

certain number of machines to be repaired by a service man. It is necessary to take the term

"customer" very generally. Customers may be people, machines of various nature, computer

processes, telephone calls, etc.

Arrival defines the way customers enter the system. Mostly the arrivals are random with

random intervals between two adjacent arrivals. Typically the arrival is described by a

random distribution of intervals also called Arrival Pattern.

Queue represents a certain number of customers waiting for service (of course the queue may

be empty). Typically the customer being served is considered not to be in the queue.

Sometimes the customers form a queue literally (people waiting in a line for a bank teller).

Sometimes the queue is an abstraction (planes waiting for a runway to land). There are two

important properties of a queue: Maximum Size and Queuing Discipline.

Maximum Queue Size (also called System capacity) is the maximum nu mber of customers

that may wait in the queue (plus the one(s) being served). Queue is always limited, but some

theoretical models assume an unlimited queue length. If the queue length is limited, some

customers are forced to renounce without being served.

Queuing Discipline represents the way the queue is organised (rules of inserting and

removing customers to/from the queue). There are these ways:

1) FIFO (First In First Out) also called FCFS (First Come First Serve) - orderly queue.

2) LIFO (Last In First Out) also called LCFS (Last Come First Serve) - stack.

3) SIRO (Serve In Random Order).

4) Priority Queue, that may be viewed as a number of queues for various priorities.

5) Many other more complex queuing methods that typically change the customer ’s position

in the queue according to the time spent already in the queue, expected service duration,

and/or priority. These methods are typical for computer multi-access systems.

Most quantitative parameters (like average queue length, average time spent in the system) do

not depend on the queuing discipline. That ’s why most models either do not take the queuing

discipline into account at all or assume the normal FIFO queue. In fact the only parameter

that depends on the queuing discipline is the variance (or standard deviation) of the waiting

time. There is this important rule (that may be used for example to verify results of a

simulation experiment):

The two extreme values of the waiting time variance are for the FIFO queue (minimum) and

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Theoretical models (without priorities) assume only one queue. This is not considered as a

limiting factor because practical systems with more queues (bank with several tellers with

separate queues) may be viewed as a system with one queue, because the customers alway s

select the shortest queue. Of course, it is assumed that the customers leave after being served.

Systems with more queues (and more servers) where the customers may be served more

times are called Queuing Networks.

Service represents some activity that takes time and that the customers are waiting for. Again

take it very generally. It may be a real service carried on persons or machines, but it ma y be a

CPU time slice, connection created for a telephone call, being shot down for an enemy p lane,

etc. Typically a service takes random time. Theoretical models are based on random

distribution of service duration also called Service Pattern. Another important parameter is

the number of servers. Systems with one server only are called Single Chan nel Systems,

systems with more servers are called Multi Channel Systems.

Output represents the way customers leave the system. Output is mostly ignored by

theoretical models, but sometimes the customers leaving the server enter the queue again

("round robin" time-sharing systems).

Queuing Theory is a collection of mathematical models of various queuing systems that take

as inputs parameters of the above elements and that provide quantitative parameters

describing the system performance.

Because of random nature of the processes involved the queuing theory is rather demanding

and all models are based on very strong assumptions (not always satisfied in practice). Many

systems (especially queuing networks) are not soluble at all, so the only technique that may

be applied is simulation.

Nevertheless queuing systems are practically very important because of the typical trade-off

between the various costs of providing service and the costs associated with waiting for the

service (or leaving the system without being served). High quality fast service is expensive,

but costs caused by customers waiting in the queue are minimum. On the other hand long

queues may cost a lot because customers (machines e.g.) do not work while waiting in the

queue or customers leave because of long queues. So a typical problem is to find an optimum

system configuration (e.g. the optimum number of servers). The solution may be found by

applying queuing theory or by simulation.

6.4. APPLICATIONS AND LIMITATIONS

Applications

The queueing theorie is used to analyze computer, telecommunication systems, traffic

systems (traffic flue), logistic and manufacturing systems.

Some examples :

-Where customers are involved such as restaurants, supermarket, airport, …

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-applicable for the problem of machine breakdown and repair

-applicable for the sheduling of jobs in production control

-provide solution of inventory control problems

Limitations

-Most of the queueing models are still complex and are not easy to understand

-Many times form of theoratical distribution applicable to given queueing situations is not

known

-The study of queueing problems become more difficult if the queueing discipline is not in

"first in, first out".

6.5. OPERATING CHARACTERISTICS OF QUEUING SYSTEM

The operating characteristics of a queuing system refer to the numerical values of the

probability distributions of various decision variables like arrival rate, number of facilities,

service time, time length, priority system etc.

1. Queue Length

Probability distribution of queue length can be obtained with the help of the given probability

distribution of the arrival and service process. A large queue indicates poor service facility o r

a need for more space. On the other hand, small queue indicates excess of service facilities.

2. Waiting time in queue

This is the time spent by a customer in the queue before the commencement of his service.

Long waiting times may increase the customer's dissatisfaction and potential loss of future

revenue.

3. Waiting time in system

Waiting time in a system is the total time spent by a customer in the queue plus the service

time. Long waiting times may indicate a change in priority rules is needed.

6.6. BASIC CHARACTERISTICS OF A QUEUING SYSTEM

Even though there are many types of queuing systems, all such systems are fully defined and

classified based on the following characteristics.

Input Process or Arrival Pattern of Customers

This is concerned with the pattern in which customers arrive and join the queuing system.

The arrival of customers towards service station and the behaviour of customers in the queue

are considered in the input process.

1. Arrival time distribution

The arrival of the customers to the service station is governed by certain probability loss.

Generally, it is assumed that the arrival rate follows Poisson distribution with mean arrival

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count the number of arrivals per unit time. The number of customers may be from finite or

infinite sources. Customers arrive at the service facility in batches of fixed size or of variable

size or one by one. Sometimes, more than one arrival (in bulk or batches) can enter in th e

system simultaneously.

2. Behaviour of the customer

The reaction of a customer upon entering the system is called the behaviour of the customer.

Customers generally behave in the following four ways:

a. Balking

When the customer comes to the service station he may find a lengthy queue already waiting

to receive the service, so he may not join the queue. This is defined as balking.

b. Reneging

A customer who is waiting in the queue may leave the queue due to his impatience. This is

defined as reneging.

c. Jockeying

A customer who is waiting in a queue may leave that queue and join with another queue,

which is smaller in length. It is defined as jockeying.

d. Collusion

Two or more customers may go to the service station but only one customer will receive the

service. This is collusion.

An arrival pattern, that does not change time and in which a steady state condition occurs, is

called a stationary arrival time, whereas in an arrival pattern that is time dependent, the input

is called a non-stationary arrival time.

6.7. SERVICE CHANNELS

Service facilities or service channels may be in series, parallel or mixed. Series arran gement

includes a sequence of a number of service facilities, arranged such that a customer must go

through one facility after another in a particular sequence. Each service facility works

independently of others. Basically there are two structures of waiting line situations:

1. Single-channel service system

2. Multi-channel service system

1. Single-channel service system

There may be only one service station a server to provide service. This is defined as a single-

channel service system. A single server model involves a single server and a single line of

customers.

To further specify the model, we make the following assumptions:

1. The customer population is infinite and all customers are patient.

2. Customers arrive accord ing to a Poisson distribution, with a mean arrival rate of λ.

3. The service distribution is exponential, with a mean service rate of µ.

4. Customers are served on a first-come-first basis.

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2. Multi-channel service system

Sometimes two or more service stations or servers may provide the same type of service,

which is defined as a multi-channel service system. With a multiple server model, customers

form a single line and choose one of s servers when one is available. The service system has

only one phase. In addition to the assumptions for the single-server model or multi-server

model, we assume that there are s identical servers, and the service distribution for each

server is exponential, with a mean service time of 1/ µ..

FOUR BASIC STRUCTURES OF WAITING LINE SITUATIONS

The time gap between the commencement of service and the completion of service time for a

customer is called the service time for a customer.

a. Service time may be i. constant or ii. scattered.

b. Service time distribution may be i. Constant time, ii. Poisson, iii. Exponential and

iv. Erlang.

Anyway, the most frequently used distribution to describe service time is the negative

exponential distribution.

Capacity of the service station

Sometimes the service station may accommodate any number of customers (infinitely large).

It is then called an infinite system. But sometimes, the service station may accommodate only

a specific number of customers and the system is called a finite system.

Examples

1. A television repairman finds that the time spent on his jobs has an exponential

distribution with a mean of 30 minutes. If he repairs sets in the order in which they

came in, and if the arrival of sets follows a Poisson distribution approximately with an

average rate of 10 per 8-hour day, what is the repairman's expected idle time eac h day?

How many jobs are ahead of the average set just brought in?

Solution:

From the data of the problem, we have

λ = 10/8 = 5/4 sets per hour; and µ = (1/30) 60 = 2 sets per hour

a. Expected idle time of repairman each day

Number of hours for which the repairman remains busy in an 8-hour day (traffic intensity) is

given by

(8) (λ/µ) = (8) (5/8) = 5 hours

Hence, the idle time for a repairman in an 8-hour day will be (8 - 5) = 3 hours.

b. Expected (or average) number of TV sets in the system

Ls = λ

µ− 𝜆 = ହ/ସ

ଶ−ሺఱ

రሻ = 5/3 = 2(approx.) TV sets.

2. Arrivals at telephone booth are considered to be Poisson with an average time of 10

minutes between one arrival and the next. The length of phone call is assumed to b e

distributed exponentially, with mean 3 minutes.

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b. The telephone department will install a second booth when convinced that an arrival

would expect waiting for at least 3 minutes for a phone call. By how much sh ould the

flow of arrivals increase in order to justify a second booth?

c. What is the average length of the queue that forms from time to time?

d. What is the probability that it will take him more than 10 minutes altogether to wait

for the phone and complete his call?

Solution :

From the data of the problem, we have

λ = 1/10 = 0.10 person per minute and µ = 1/3 = 0.33 person per minute

a. Probability that a person has to wait at the booth.

P(n> Ͳሻ = 1 - P 0 = λ/µ = 0.10/0.33 = 0.3

b. The installation of second booth will be justified only if the arrival rate is more than the

waiting time.

Let λ' be the increased arrival rate. Then expected waiting time in the queue will be

Wq = λ′

µሺµ − λ′ሻ ;

3 = λ′

.ଷଷሺ.ଷଷ − λ′ሻ ; or λ′ = 0.16

where W q = 3 (given) and λ = λ' (say) for second booth.

Hence, the increase in the arrival rate is 0.16 - 0.10 = 0.06 arrivals per minute.

c. Average length of non-empty queue

L = µ

µ− 𝜆 = 0.33/0.23 = 2 customers (approx.)

d. Probability of waiting for 10 minutes or more is given by

P(t≥ͳͲ) = ∫𝜆

µሺµ − 𝜆ሻ∞

ଵ e-(µ - λ) tdt

= ∫ሺͲ.Ͳ͵ሻሺͲ.ʹ͵ሻሻ∞

ଵ e-0.23 tdt = 0.069 [e-0.23 t/- 0.23] 10∞ = 0.03

This shows that 3 percent of the arrivals on an average will have to wait for 10 minutes or

more before they can use the phone.

6.8. MULTI-SERVER QUEUING MODELS

Examples

1. A tax consulting firm has 4 service counters in its office to receive people who have

problems and complaints about their income, wealth and sales taxes. Arrivals average

80 persons in an 8-hour service day. Each tax advisor spends an irregular amount of

time servicing the arrivals which have been found to have an exponential distribution.

The average service time is 20 minutes. Calculate the average number of customers in

the system, average number of customers waiting to be serviced, average time a

customer spends in the system, and average waiting time for a customer. Calculate how

many hours each week does a tax adviser spend performing his job. What is the

probability that a customer has to wait before he gets service? What is the expected

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Solution:

Given that λ = 10/hour; µ = 3/hour, 4 and ρ = λ/s µ = 5/6

a. Probability of no customer in the system,

P0 = ∑ଵ

𝑛!𝑥−ଵ

𝑛= (λ/µ)n + ଵ

𝑠!(λ/µ)s (sµ/sµ- λ) -1

= [∑ଵ

𝑛!ଷ

𝑛= (10/3)n + ଵ

ସ!(10/3)4 (12/12-10) ]-1

= [1 + 10/3 + 1/2 (10/3)2 + 1/6(10/3)4 12/12 - 10]-1

= 0.021

b. Average number of customers in the system

Ls = L q + λ/ µ = [ଵ

ሺ𝑠−ଵ ሻ! (λ/µ)s 𝜆µ

ሺ𝑠µ− 𝜆 ሻଶ] P0 + λ/ µ

= [ଵ

ଷ! (10/3)4 ଷ

ሺଵଶ−ଵሻଶ] x 0.021 + 10/3 = 6.57

c. Average number of customers waiting in the queue (queue length),

Lq = L s - 𝜆

µ = 6.57 - (10/3) = 3.24 customers

d. Average time a customer spends in the system

Ws = W q + 1/µ = L q/ λ + 1/µ = 3.24/10 + 1/3 = 0.657 hour or 39.42 minutes

e. Average time a customer waits for service in the queue

Wq = L q/λ = 3.24/10 = 0.324 hour or 19.44 minutes.

f. time spent by a tax counsellor, i.e. utilisation factor,

ρ = λ/ sµ = 5/6 = 0.833 hour or 50 minutes

The expected time spent in servicing customers during an 8-hour day is 8 x 0.833 = 6.66

hours. Thus, on an average, a tax advisor is busy for 6.66 x (40/8) = 33.30 hours based on a

40 hours ’ week.

g. Probability that a customer has to wait

Pw (n ≥ 𝑠ሻ = ଵ

𝑠!(λ/µ)s (sµ/sµ- λ) . ρ 0

= ଵ

ସ!(10/3)4 (4 x 3/4 x 3-10) (0.021) = 0.622

h. The expected number of idle advisers at any specified time can be obtained by adding the

probability of 3 idle, 2 idle and 1 idle adviser. That is

Expected number of idle advisers = 4 P 0 + 3P 1 + 2P 2 + P 3

= 4 (0.021) + 3(0.070) + 2(0.118) +0.131 - 0.661

This means less than one 0.661 adviser is idle on an average at any instance of time.

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6.9. EXERCISE

Questions

1. Derive the difference equations for the queuing model {(M/M/1) : ( ∞/FCFS)}. How would

you proceed to solve the model?

2. In a single server, Poisson arrival and exponential service time queuing system show that

probability P n of n customers in steady-state satisfies the following equations:

λP0 = µP 1 ; n = 0

(λ + µ) P1 = µP 2 ; n = 1

(λ + µ) Pn = µP n + 1 + λP n-1 ; n ≥ ʹ

3. At what average rate must a clerk at a super market work in order to ensure a probability of

0.90 so that the customer will not wait longer than 12 minutes? It is assumed that there is

only one counter at which customers arrive in a Poisson fashion at an average rate of 15 pe r

hour. The length of service by the clerk has an exponential distribution.

4. Consider a self-service store with one cashier. Assume Poisson arrivals and exponential

service times. Suppose that nine customers arrive on the average every 5 minutes and the

cashier can serve 10 in 5 minutes. Find

a. Average number of customers queuing for service.

b. Probability of having more than 10 customers in the system and

c. Probability that a customer has to queue for more than 2 minutes.

If the service can be speed up to 12 in 5 minutes by using a different cast register, what will

be the effect on the quantities (a0, (b) and (c).

5. For the single server, finite (or limited) queuing system, find the

a. Average number of customers in the system and

b. Average queue length.

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