Mba Operations Research

8/12/2019 Mba Operations Research

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Contents

Unit 9

Finite Queuing Models 153

Unit 10

Simulation 160

Unit 11

ProjectSchedulingand PERT-CPM 181

Unit 12

Game Theory 199

Edition: Fall 2008

BKID B0661


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INTRODUCTION

Operations Research is an important course of study for a management student. It is a branchof applied mathematics and has diverse applications. This subject essentially deals withproblems of choices and optimization of resource utilization.

This courseware comprises 12 units:

Unit 1: Introduction

Definition, Scope, Limitations of OR.

Unit 2: Linear Programming Problem

Formulation & Graphical Solution to L.P.P.

Unit 3: Simplex Method

Solution by Simplex method to L.P.P.

Unit 4: Duality in L.P.P

The economic interpretations of the final simplex table.

Unit 5: Transportation Problem

Distribution of Resources through Transportation Algorithm.

Unit 6: Assignment Problem

Allocation of Resources through Assignment Algorithm.

Unit 7: Integer Programming

Method of Solving L.P.P for integer values.

Unit 8: Infinite Queuing Models

The study of waiting line problems for infinite Queue.

Unit 9: Finite Queuing Models

The study of waiting line problems for finite Queue.


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Unit 10: Simulation

Non technical method of dealing with complex problems through Simulation.

Unit 11: Project Scheduling & PERT-CPM

Methods available for Planning, Scheduling & Monitoring of Projects.

Unit 12: Game Theory

Deals with competitive situations through Games Theory.


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Operations Research Unit 1

Sikkim Manipal University 1

Unit 1 Introduction To OR

Structure 1.1. Introduction

1.2 Historical Background

1.3. Scope of operation Research

1.4. Important Features of OR

1.5. Phases of OR

1.6. Types of OR Models

1.5.1 A broad classification of OR models

1.7. Methodology of OR

1.7.1 Definition of the problem

1.7.2 Construction of the model

1.7.3 Solution of the model

1.7.4 Validation the model

1.7.5 Implementation of the final result

1.8. Techniques or Tools of OR

1.9. The structure of Mathematical Model1.9.1 Decision variables and parameters

1.9.2 Objective functions

1.9.3 Constraints

1.9.4 Diet Problem

1.10. Limitations of OR

1.11. Summary

Terminal Questions

Answers of SAQs & TQs

1.1 Introduction

Optimization is the act of obtaining the best result under any given circumstance. In various

practical problems we may have to take many technical or managerial decisions at several


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stages. The ultimate goal of all such decisions is to either maximize the desired benefit or

minimize the effort required. We make decisions in our every day life without even noticing them.

Decision-making is one of the main activity of a manager or executive. In simple situations

decisions are taken simply by common sense, sound judgment and expertise without using anymathematics. But here the decisions we are concerned with are rather complex and heavily

loaded with responsibility. Examples of such decision are finding the appropriate product mix

when there are large numbers of products with different profit contributions and productional

requirement or planning public transportation network in a town having its own layout of factories,

apartments, blocks etc. Certainly in such situations also decision may be arrived at intuitively from

experience and common sense, yet they are more judicious if backed up by mathematical

reasoning. The search of a decision may also be done by trial and error but such a search may

be cumbersome and costly. Preparative calculations may avoid long and costly research. Doing

preparative calculations is the purpose of Operations research. Operations research does

mathematical scoring of consequences of a decision with the aim of optimizing the use of time,

efforts and resources and avoiding blunders.

Learning Objectives:

After studying this unit, you should be able to understand the following.

1. Know the significant features in O.R.

2. Understand the Methodology of O.R.

3. Define the structure of a mathematical model in O.R.

4. Know the significance of the Objective function

1.2Historical Background

During Second World War in United Kingdom a team of scientists from different disciplines

studied the strategic and tactical problems associated with air and land defence of the country.

Their objective was to determine the most effective utilization of limited military resources to win

the battle and the technique they developed was named as Operations research. After the war,

Operations research techniques rapidly developed in the fields of industrial, academic and

government organizations.

The application of Operations research methods helps in making decisions in such complicated

situations. Evidently the main objective of Operations research is to provide a scientific

basis to the decision-makers for solving the problems involving the interaction of various

components of organization, by employing a team of scientists from different disciplines,


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all working together for finding a solution which is the best in the interest of the

organization as a whole. The solution thus obtained is known as optimal decision.

Definition of Operations Research: Churchman, Ackoff and Aruoff have defined Operations

research as the application of scientific methods, techniques and tools to operation of a

system with optimum solutions to the problems. Here Optimum implies the one, which is

best of all possible alternatives. Another definition is that, Operations research is the use of

scientific methods to provide criteria for decisions regarding man, machine, systems

involving repetitive operations. This definition is more general and comprehensive and seems

to be more exhaustive than the previous definition.

Self Assessment Questions 1

1. The main objective of O.R is to provide a _______ ________ to the decision-makers.

2. O.R employs a team _________ from _________ __________

1.3Scope of Operations Research (OR)

In general, whenever there is any problem simple or complicated, the OR techniques may be

applied to find the best solution. In this section we shall try to find the scope of OR by seeing its

application in various fields of everyday life.

i) In Defence Operations: In modern warfare the defence operations are carried out by a

number of independent components namely Air Force, Army and Navy. The activities in each

of these components can be further divided in four sub-components viz.: administration,

intelligence, operations and training, and supply. The application of modern warfare

techniques in each of the components of military organizations requires expertise knowledge

in respective fields. Further more, each component works to drive maximum gains from its

operations and there is always a possibility that strategy beneficial to one component may

have an adverse effect on the other. Thus in defence operations there is a necessity to co-

ordinate the activities of various components which gives maximum benefit to the

organization as a whole, having maximum use of the individual components. The final

strategy is formulated by a team of scientists drawn from various disciplines who study the

strategies of different components and after appropriate analysis of the various courses of

actions, the best course of action, known as optimum strategy, is chosen.

ii) In Industry: The system of modern industries are so complex that the optimum point of

operation in its various components cannot be intuitively judged by an individual. The


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business environment is always changing and any decision useful at one time may not be so

good some time later. There is always a need to check the validity of decisions continually,

against the situations. The industrial revolution with increased division of labour and

introduction of management responsibilities has made each component an independent unithaving their own goals. For example: Production department minimize cost of production but

maximizes output. Marketing department maximizes output but minimizes cost of unit sales.

Finance department tries to optimize capital investment and personnel department appoints

good people at minimum cost. Thus each department plan their own objectives and all these

objectives of various department or components come to conflict with each other and may not

conform to the overall objectives of the organization. The application of OR techniques helps

in overcoming this difficulty by integrating the diversified activities of various components so

as to serve the interest of the organization as a whole efficiently.

OR methods in industry can be applied in the fields of production, inventory controls and

marketing, purchasing, transportation and competitive strategies etc.

iii) Planning: In modern times it has become necessary for every government to have careful

planning, for economic development of the country. OR techniques can be fruitfully applied to

maximize the per capita income, with minimum sacrifice and time. A government can thus use

OR for framing future economic and social policies.

iv) Agriculture: With increase in population there is a need to increase agriculture output. But

this cannot be done arbitrarily. There are a number of restrictions under which agricultural

production is to be studied. Therefore there is a need to determine a course of action, which

serves the best under the given restrictions. The problem can be solved by the application of

OR techniques.

v) In Hospitals: The OR methods can be used to solve waiting problems in out-patient

department of big hospitals. The administrative problems of hospital organization can also be

solved by OR techniques.

vi) In Transport: Different OR methods can be applied to regulate the arrival of trains and

processing times, minimize the passengers waiting time and reduce congestion, formulate

suitable transportation policy, reducing the costs and time of trans-shipment.vii) Research and Development: Control of R and D projects, product introduction planning etc.

and many more applications.


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d) Analysis of the available information and verification of the hypothesis using pre-established

measure of effectiveness.

e) Prediction of various results and consideration of alternative methods.

iii) Action Phase: It consists of making recommendations for the decision process by thosewho first posed the problem for consideration or by anyone in a position to make a decision,

influencing the operation in which the problem is occurred.


State True / False

1. O.R gives qualitative solution.

2. One of O.R phases is Action phase

1.6 Types of operation Research Models

A model is a representation of the reality. It is an idealized representation or abstraction of a real

life system. The objective of the model is to identify significant factors and their interrelationship.

A model is helpful in decision making as it provides a simplified description of complexities and

uncertainties of a problem in logical structure. Major advantage of models is it does not interfere

with real system.

1.6.1. A broad classification of OR models

a) Physical Model b) Mathematical or symbolic model c) Models by nature of Environment

and d) Models by the extent of generality.

a. Physical Modes include all form of diagrams, graphs and charts. They are designed to deal

with specific problems. They bring out significant factors and inter-relationship in pictorial firm

so as to facilitate analysis. There are two types i) Iconic models and ii) Analog models.

Iconic model is an image of an object or system, represented on a small scale. These models

can simulate the actual performance of a product.

Analog models are small physical systems that has similar characteristics and work l

ike an objects it represents Eg: Toy

b. Mathematical Model or symbolic models employ a set of mathematical symbols to represent

the decision variable of the system. The variables are related by mathematical system Eg:

Allocation, sequencing, replacement models etc.


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c. By nature of Environment

We have i) Deterministic model in which every thing is defined and the results are certain.

Eg: EOQ model ii) Probabilistic Models in which the input and output variables follow a

probability distribution Eg: Games Theory.d. By the extent of Generality : The two models belonging to this class are i) General models

can be applied in general and does not pertain to one problem only. Eg: Linear programming

ii) Specific Model is applicable under specific condition only. Eg: Sales response curve or

equation as a function of advertising is applicable in the marketing function alone.


State True / False

1. Diagram belongs to physical model

2. Allocation problems are represented by iconic model.

1.7 Methodology of Operations Research

The basic dominant characteristic feature of operations research is that it employs mathematical

representations or model to analyze problems. This distinctive approach represents an adaptation

of the scientific methodology used by the physical sciences. The scientific method translates a

real given problem into a mathematical representation which is solved and retransformed into the

original context. The OR approach to problem solving consists of the following steps:

1. Definition of the problem.2. Construction of the model.

3. Solution of the model.

4. Validation of the model.

5. Implementation of the final result.

1.7.1 Definition of the problem

The first and the most important requirement is that the root problem should be identified and

understood. The problem should be identified properly, this indicates three major aspects: (1) a

description of the goal or the objective of the study, (2) an identification of the decision alternativeto the system, and (3) a recognition of the limitations, restrictions and requirements of the

system.


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1.8. Techniques or Tools of OR

1. Linear Programming:

It is used to find a solution for optimizing a given objective. Objectives may be to maximize

profit or minimize cost. Both objective function and constraints must be capable of beingexpressed as linear expression of decision variables. Its various use will be seen in Chapter-

2.

2. Inventory control Methods:

The production, purchasing and material managers are always confronted with question of

when to buy, how much to buy and how much to keep in stock. Inventory models aims at

optimizing inventory levels.

3. Goal Programming

Single objective function is taken in the linear programming and all other factors are

considered as constraints, but in actual practice there may be number of important objective

functions. Goal programming has several objective functions, each having a target value and

programme models are developed to minimize deviation from these targets.

4. The tools, namely, queuing model, sequence model, transportation and assignment model,

network analysis are discussed in detail in later chapters.


State True / False

1. O.R methodology consists of definition, solution and validation only.

2. The interaction between O.R team and Management reaches peak level in implementationphase.


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1.9 The Structure of Mathematical Model

Many industrial and business situations are concerned with planning activities. In each case of

planning, there are limited sources, such as men, machines, material and capital at the disposal

of the planner. One has to make decision regarding these resources in order to either maximizeproduction, or minimize the cost of production or maximize the profit etc. These problems are

referred to as the problems of constrained optimization. Linear programming is a technique for

determining an optimal schedule of interdependent activities, for the given resources.

Programming thus means planning and refers to the process of decision-making regarding

particular plan of action amongst several available alternatives.

Any business activity of production activity to be formulated as a mathematical model can best be

discussed through its constituents; they are:

-Decision Variables,

- Objective function,

- Constraints.

1.9.1 Decision variables and parameters

The decision variables are the unknowns to be determined from the solution of the model. The

parameters represent the controlled variables of the system.

1.9.2 Objective functions

This defines the measure of effectiveness of the system as a mathematical function of its decision

variables. The optimal solution to the model is obtained when the corresponding values of thedecision variable yield the best value of the objective function while satisfying all constraints.

Thus the objective function acts as an indicator for the achievement of the optimal solution.

While formulating a problem the desire of the decision-maker is expressed as a function of n

decision variables. This function is essentially a linear programming problem (i.e., each of its item

will have only one variable raise to power one). Some of the Objective functions in practice are:

- Maximization of contribution or profit

- Minimization of cost

- Maximization of production rate or minimization of production time - Minimization of labour turnover

- Minimization of overtime

- Maximization of resource utilization

- Minimization of risk to environment or factory etc.


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1.9.3 Constraints

To account for the physical limitations of the system, the model must include constraints, which

limit the decision variables to their feasible range or permissible values. These are expressed in

the form of constraining mathematical functions.

For example, in chemical industries, restrictions come from the government about throwing gases

in the environment. Restrictions from sales department about the marketability of some products

are also treated as constraints. A linear programming problem then has a set of constraints in

practice.

The mathematical models in OR may be viewed generally as determining the values of the

decision variables x J , J = 1, 2, 3, ------ n, which will optimize Z = f (x 1, x 2, ---- x n).

Subject to the constraints:

g i (x 1, x 2 ----- x n) ~ b i, i = 1, 2, ---- m

and x J 0 j = 1, 2, 3 ---- n where ~ is , or =.

The function f is called the objective function, where X j ~ b i, represent the i th constraint for i =

1, 2, 3 ---- m where b i is a known constant. The constraints x j 0 are called the non-negativity

condition, which restrict the variables to zero or positive values only.

1.9.4 Diet Problem

Formulate the mathematical model for the following:

Vitamin A and Vitamin B are found in food 1 and food 2. One unit of food 1 contains 5

units of vitamin A and 2 units of vitamin B . One unit of food 2 contains 6 units of vitamin

A and 3 units of vitamin B . The minimum daily requirement of a person is 60 units of vitamin

A and 80 units of Vitamin B . The cost per one unit of food 1 is Rs . 5/- and one unit of food

2 is Rs . 6/-. Assume that any excess units of vitamins are not harmful. Find the minimum cost of

the mixture (of food1 and food2) which meets the daily minimum requirements of vitamins.

Mathematical Model of the Diet Problem: Suppose x 1 = the number of units of food1 in the mixture,

x 2 = the number of units of food2 in the mixture.

Now we formulate the constraint related to vitamin-A. Since each unit of food 1 contains 5 units

of vitamin A, we have that x 1 units of food 1 contains 5 x 1 units of vitamin A. Since each unit

of food 2 contains 6 units of vitamin A, we have that x 2 units of food 2 contains 6 x 2 units of

vitamin A. Therefore the mixture contains 5 x 1 + 6 x 2 units of vitamin - A. Since the minimum


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requirement of vitamin A is 60 units, we have that

5 x 1 + 6 x 2 60.

Now we formulate the constraint related to vitamin B . Since each unit of food 1 contains 2

units of vitamin B we have that x 1 units of food 1 contains 2 x 1 units of vitamin - B . Since each

unit of food 2 contains 3 units of vitamin B , we have that x 2 units of food 2 contains 3 x 2 units

of vitamin B . Therefore the mixture contains 2 x 1 + 3 x 2 units of vitamin B . Since the

minimum requirement of vitamin B is 80 units, we have that

2 x 2 + 3 x 2 80.

Next we formulate the cost function. Given that the cost of one unit of food 1 is R's . 5/- and one

unit of food 2 is R's . 6/-. Therefore x 1 units of food1 costs Rs . 5 x 1, and x 2 units of food 2

costs Rs . 6 x 2. Therefore the cost of the mixture is given b y Cost = 5 x 1 + 6 x 2. If we write z for the

cost function, then we have z = 5 x 1 + 6 x 2. Since cost is to be minimized, we write min z = 5 x 1 +6 x 2.

Since the number of units ( x 1 or x 2) are always non-negative we have that

x 1 0, x 2 0. Therefore the mathematical model is

5 x 1 + 6 x 2 60

2 x 1 + 3 x 2 80

x 1 0, x 2 0, min z = 5 x 1 + 6 x 2.

1.10 limitations of OR The limitations are more related to the problems of model building, time and money factors.

i) Magnitude of computation: Modern problem involve large number of variables and hence to

find interrelationship, among makes it difficult.

ii) Non quantitative factors and Human emotional factor cannot be taken into account.

iii) There is a wide gap between the managers and the operation researches

iv) Time and Money factors when the basic data is subjected to frequent changes then

incorporation of them into OR models is a costly affair.

v) Implementation of decisions involves human relations and behaviour.


Fill in the blanks

i. OR imbibes _________ team approach

ii. Linear programming is tool of _______


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iii. The three phases of OR are ________

iv. To solve any problem through OR approach the first step is _______

v. _________ represents a real life system

vi. _________ represents the controlled variables of the system.

1.11. Summary

The OR approach needs to be equally developed in various agricultural problems on a regional or

international basis. With the explosion of population and consequent shortage of food, every

country faces the problem of optimum allocation of land in various crops in accordance with

climate conditions and available facilities. The problem of optimal distribution of water from a

resource like a reservoir for irrigation purposes is faced by each developing country, and a good

amount of scientific work can be done in this direction.

Terminal Questions

1. Define OR

2. What are the characteristic features of OR?

3. What is a model in OR? Discuss different models available in OR

4. Write short notes are different phases or OR

5. What are the limitations of OR

Answers to Self Assessment QuestionsSelf Assessment Questions 1

1. Scientific basis

2. Scientists, different, disciplines


1. Industry, Planning

2. To solve waiting problems


1. Imbibes

2. Decision-making


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Unit 2 Linear Programming

Structure

2.1 Introduction

2.2. Requirements

2.2.1 Basic assumptions of L.P.P

2.3. Linear Programming

2.3.1 Canonical forms

2.3.2 Examples of a linear programming problem

2.4. Graphical analysis2.4.1 Some basic definitions

2.5 Graphical Methods to solve L.P.P

2.5.1 Working Rule:

2.5.2 Examples 6 for mixed constraints LP problem

2.5.3 Examples 9 for Unbounded Solution

2.5.4 Examples 10 for Inconsistent:

2.5.5 Examples 11 for redundant Constraint:

2.6. Summary

Terminal Questions

Answers to SAQs to TQs

2.1 Introduction

One of the most important problems in management decision is to allocate limited and scarce

resource among competing agencies in the best possible manner. Resources may represent man,

money, machine, time, technology on space. The task of the management is to derive the best

possible output (or set of outputs) under given restraints on resources. The output may be

measured in the form of profits, costs, social welfare, effectiveness, etc. In many situations theoutput (or the set of outputs) can be expressed as a linear relationship among a number of

variables. The amount of available resources can also be expressed as a linear relationship among

some system variables. The management problem may be to optimize (maximize or minimize) the

out-put or the objective function subject to the set of constraints An optimization problem in which


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both the objective function and the constraints are represented by linear forms is a problem in linear

programming.

Learning Objectives: After studying this unit, you should be able to understand the following

1. Formulate the LPP and observe the feasible region.

2. Graphically analyze and solve a L.P.P.

2.2Requirements of L.P.P

i. Decisions variables and their relationship

ii. Well defined objective function

iii. Existence of alternative courses of actioniv. Non-negative conditions on decision variables.

2.2.1 Basic assumptions of L.P.P

1 . Linearity : Both objective function and constraints must be expressed as linear inequalities.

2. Deterministic: All coefficient of decision variables in the objective and constraints expressions

should be known and finite.

3. Additivity: The value of objective function for the given values of decision variables and the

total sum of resources used, must be equal to sum of the contributions earned from each

decision variable and the sum of resources used by decision variables respectively.

4. Divisibility: The solution of decision variables and resources can be any non-negative values

including fractions.


Fill in the blanks

1. Both objective function and constraints are expressed in _________ forms.

2. L.P.P requires existence of _________ __________ _________ ________.

3. Solution of decision variables can also be ___________

2.3 Linear Programming

The Linear Programming Problem (LPP) is a class of mathematical programming in which the

functions representing the objectives and the constraints are linear. Here, by optimization, we


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mean either to maximize or minimize the objective functions. The general linear programming

model is usually defined as follows:

Maximize or Minimize

Z = c 1 x 1 + c 2 x 2 + - - - - + c n x n

subject to the constraints,

a 11 x 1 + a 12 x 2 + ----- + a 1n x n ~ b 1

a 21 x 1 + a 22 x 2 + ----- + a 2n x n ~ b 2

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

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

a m1 x 1 + a m2 x 2 + ------- + a mn x n ~ b m

and x 1 0, x 2 0, -------------------- x n 0.

Where c j , b i and a ij (i = 1, 2, 3, .. m, j = 1, 2, 3 ------- n) are constants determined from the

technology of the problem and x j (j = 1, 2, 3 ---- n) are the decision variables. Here ~ is either

(less than), (greater than) or = (equal). Note that, in terms of the above formulation the

coefficient c j , a ij , b j are interpreted physically as follows. If bi is the available amount of resources

i, where a ij is the amount of resource i , that must be allocated to each unit of activity j , the worth

per unit of activity is equal to c j .

2.3.1 Canonical forms:

The general Linear Programming Problem (LPP) defined above can always be put in the following

form which is called as the canonical form:

Maximise Z = c 1 x 1+c 2 x 2 + ------ + c n x n

Subject to

a 11 x 1 + a 12 x 2 + ------ + a 1n x n b 1

a 21 x 1 + a 22 x 2 + ------ + a 2n x n b 2

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

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

a m1 x 1+a m2 x 2 + + a mn x n bm

x 1, x 2 , x 3, x n 0.

The characteristics of this form are:

1) all decision variables are non-negative.

2) all constraints are of type.


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3) the objective function is of the maximization type.

Any LPP can be put in the cannonical form by the use of five elementary transformations:

1. The minimization of a function is mathematically equivalent to the maximization of the

negative expression of this function. That is, Minimize Z = c 1 x 1 + c 2 x 2 + . + c n x n isequivalent to

Maximize Z = c 1 x 1 c 2 x 2 c n x n.

2. Any inequality in one direction ( or ) may be changed to an inequality in the opposite

direction ( or ) by multiplying both sides of the inequality by 1 .

For example 2x 1+3x 2 5 is equivalent to 2x 1 3x 2 5.

3. An equation can be replaced by two inequalities in opposite direction. For example, 2x 1+3x 2 =

5 can be written as 2x 1+3x 2 5 and 2x 1+3x 2 5 or 2x 1+3x 2 5 and 2x 1 3x 2 5.

4. An inequality constraint with its left hand side in the absolute form can be changed into two

regular inequalities. For example: | 2x 1+3x 2 | 5 is equivalent to 2x 1+3x 2 5 and 2x 1+3x 2 5

or 2x 1 3x 2 5.

5. The variable which is unconstrained in sign (i.e., 0, 0 or zero) is equivalent to the

difference between 2 non-negative variables. For example, if x is unconstrained in sign then x

= (x+ x ) where x+ 0 , x 0.

2.3.2 Examples Of A Linear Programming Problem:

Example 1: A firm engaged in producing 2 models, viz., Model A and Model B, performs only 3

operations painting, assembly and testing. The relevant data are as follows:

Unit Sale PriceHours required for each unit

Assembly Painting Testing

Model A Rs. 50.00Model B Rs. 80.00

1.01.5

0.20.2

0.00.1

Total number of hours available each week are as under assembly 600 , painting 100 , testing 30 .

The firm wishes to determine the weekly product-mix so as to maximize revenue.

Solution: Let us first write the notations as under:

Z : Total revenue

x 1 : Number of Units of Model A

x 2 : Number of Units of Model B

X 1, X 2 : Are known as decision variables

b 1 : Weekly hours available for assembly


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b 2 : Weekly hours available for painting

b 3 : Weekly hours available for testing.

Since the objective (goal) of the firm is to maximize its revenue, the model can be stated as

follows:

The objective function, Z = 50x 1 + 80x 2 is to be maximized subject to the constraints

1.0 x 1+1.5x 2 600 , (Assembly constraints)

0.2 x 1+0.2x 2 100, ( Painting constaints)

0.0 x 1+0.1x 2 30 , (Testing constraints)

and

x1 0, x 2 0, The Non-negativity conditions.

Example 2: A milk distributor supplier milk in bottles to houses in three areas A, B, C in a city. His

delivery charges per bottle is 30 paise in area A, 40 paise in area B and 50 paise in area C. Hehas to spend on an average, 1 minute to supply one bottle in area A, 2 minutes per bottle in area

B and 3 minutes per bottle in area C. He can spare only 2 hours 30 minutes for this milk

distribution but not more than one hour 30 minutes for area A and B together. The maximum

number of bottles he can deliver is 120. Find the number of bottles that he has to supply in each

area so as to earn the maximum. Construct a mathematical model.

Solution: The decision variables of the model can be defined as follows:

x1 : Number of bottles of milk which the distributor supplies in Area A.

x2 : Number of bottles of milk which the distributor supplies in Area B.

x3 : Number of bottles of milk which the distributor supplies in Area C.

The objective :

Maximize Z = 321 x10050x

10040x

10030 ++ in rupees.

constraints:

1. Maximum number of milk bottles is 120, that is x 1+x2+x3 120.

2. Since he requires one minute per bottle in area A, 2 minutes per bottle in area B and 3

minutes per bottle in area C and he cannot spend more than 150 minutes for the work,

1.x 1 + 2.x 2 + 3.x 3 150.

3. Further, since he cannot spend more than 90 minutes for areas A and B. 1.x 1+2.x 2 90.

4. Non-negativity x 1 0, x 2 0.

The problem can now be stated in the standard L.P. form is


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Maximize Z = 0.3x 1 + 0.4x 2 + 0.5x 3

Subject to

x1 + x2 + x3 120

x1 + 2x 2 + 3x 3 150

x1 + 2x 2 90

and

x1 0, x 2 0.

Example 3: An oil company has two units A and B which produce three different grades of oil

super fine, medium and low grade oil. The company has to supply 12, 8, 24 barrels of super fine,

medium and low grade oils respectively per week. It costs the company Rs. 1,000 and Rs. 800

per day to run the units A and B respectively. On a day Unit A produces 6, 2 and 4 barrels and

the unit B produces 2, 2 and 12 barrels of super fine, medium and low grade oil per day. Themanager has to decide on how many days per week should each unit be operated in order to

meet the requirement at minimum cost. Formulate the LPP model.

Solution: The given data can be presented in summary as follows:

Product Capacity Requirements

Super fineMediumLow gradeCost

Unit A Unit B 12

824

624

Rs. 1,000

22

12Rs. 800

Let x 1 and x 2 be the number of days the units A and B be operated per week respectively. Then

the objective of the manager is to,

Minimize the cost function

Z = 1000 x 1 + 800 x 2

Subject to the constraints 6x 1+2x 2 12 (Super fine)

2x 1+2x 2 8 (medium)

4x 1+12x 2 24 (low grade)

and x 1 0, x 2 0.


State True / False

a. One of the characteristics of canonical form in the objective function must be of maximisation.

b. 2x 3y 10 can be written as


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Combining we can sketch the area as follows:

The 3 constraints including non-negativity are satisfied simultaneously in the shaded regionOCEB. This region is called feasible region .

2.4.1 Some Basic Definitions

E


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Definition: Any non-negative value of (x 1, x2) (i.e.: x 1 0, x 2 0) is a feasible solution of the LPP

if it satisfies all the constraints. The collection of all feasible solutions is known as the feasible

region.

Definition : A set X is convex if for any points x 1, x 2 in X , the line segment joining these points isalso in X .

(That is, x 1, x 2 X , 0 l 1 l x 2 + (1- l ) x 1 X ). By convention, a set containing only a single

point is also a convex set.

l x 2 + (1- l ) x 1 (where 0 l 1) is called a convex combination of x 1 and x 2.

A point x of a convex set X is said to be an extreme point if there do not exist x 1, x 2 X ( x 1 x 2)

such that x = l x 2 + (1- l ) x 1 for some l with 0 < l < 1.

Definition: A linear inequality in two variables is known as a half plane. The corresponding equality or

the line is known as the boundary of the half- plane.

Definition: A convex polygon is a convex set formed by the inter-section of finite number of

closed half-planes.

Convex regions

Non-convex regions

Note: The objective function is maximized or minimized at one of the extreme points which is the

Optimum solution. Extreme points are referred to as vertices or corner points of the convexregions.

Definition: A redundant constraint is a constraint which does not affect the feasible region.

Definition: A basic solution of a system of m equations and n variables (m < n) is a

solution where at least n-m variables are zero.

E E

E E

E

E E

E

E

E E


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Definition: A basic feasible solution of a system of m equations and n variables (m < n) is a

solution where m variables are non-negative ( 0) and n-m variables are zero.

Definition : Any feasible solution that optimizes the objective function is called an optimal

feasible solution . Example : Find all basic solutions for the system x 1 + 2 x 2 + x 3 = 4, 2 x 1 + x 2 + 5 x 3 = 5.

Solution : Here A =

512

121 , X =

3

2

1

x x x

and b =

5

4 .

i) If x 1 = 0, then the basis matrix is B =

51

12 . In this case 2 x 2 + x 3 = 4, x 2 + 5 x 3 = 5.

If we solve this, then x 2 =

3

5 and x 3 =

3

2 . Therefore x 2 =

3

5 , x 3 =

3

2 is a basic feasible

solution.

ii) If x 2 = 0, then the basis matrix is B =

52

11 . In this case, x 1 + x 3 = 4, 2 x 1 + 5 x 3 = 5.If we

solve this, then x 1 = 5 and x 3 = -1. Therefore x 1 = 5, x 3 = -1 is a basic solution. (Note that

this solution is not feasible, because x 3 = -1 < 0).

iii) If x 3 = 0, then the basis matrix is B =

1221

. In this case, x 1 + 2 x 2 = 4.

2 x 1 + x 2 = 5. If we solve this, then x 1 = 2, and x 2 = 1. Therefore x 1 = 2, x 2 = 1 is a basic

feasible solution.

Therefore (i) ( x 2, x 3) = (5/3, 2/3), (ii) ( x 1, x 3) = (5, -1), and

(iii) ( x 1, x 2) = (2, 1) are only the collection of all basic solutions.


a. The collection of all feasible solutions is known as the _________ region.

b. A linear inequality in two variables is known as a _________.


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2.5 Graphical Methods To Solve The Linear Programming Problems

A LPP with 2 decision variables x 1 and x 2 can be solved easily by graphical method. We consider

the x 1 x2 plane where we plot the solution space, which is the space enclosed by theconstraints. Usually the solution space is a convex set which is bounded by a polygon; since a

linear function attains extreme (maximum or minimum) values only on boundary of the region, it is

sufficient to consider the vertices of the polygon and find the value of the objective function in these

vertices. By comparing the vertices of the objective function at these vertices, we obtain the optimal

solution of the problem.

The method of solving a LPP on the basis of the above analysis is known as the graphical

method. The working rule for the method is as follows:

2.5.1 Working Rule :

Step I: Write down the equations by replacing the inequality symbols by the equality symbol in the

given constraints.

Step II : Plot the straight lines represented by the equations obtained in step I.

Step III: Identify the convex polygon region relevant to the problem. We must decide on which

side of the line, the half-plane is located.

Step IV: Determine the vertices of the polygon and find the values of the given objective function

Z at each of these vertices. Identify the greatest and least of these values. These are

respectively the maximum and minimum value of Z.

Step V: Identify the values of (x 1, x2) which correspond to the desired extreme value of Z. This is

an optimal solution of the problem.

Example 4: We can solve the L.P.P. discussed in Example I.

Maximize Z = 50x 1 + 80x 2Subject to the constraints

1.0x 1 + 1.5x 2 600

0.2 x 1 + 0.2x 2 100

0.0x 1 + 0.1x 2 30and x 1 0, x 2 0

Let the horizontal axis represent x 1 and the vertical axis x 2. Plot the constraint lines and mark the

feasibility region as has been shown in the figure.


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Feasible region of the two dimensional LPP

Any point on the thick line or inside the shaded portion will satisfy all the restrictions of the

problem. Then ABCDE is the feasibility region carried out by the constraints operating on the

objective function. This depicts the limits within which the values of the decision variables are

permissible. The inter-section points C and D can be solved by the linear equations

x2 = 30; x 1 + 1.5 x 2 = 600, and 0.2x 1 + 0.2x 2 = 100 and x 1 + 1.5x 2 = 600 i.e. C (150, 300)

and D (300, 180).

After doing this, the next step is to maximise revenues subject to the above shaded area. We

work out the revenues at different corner points as tabulated below:

Atpoint

Feasible solution of theproduct-mix

Correspondingrevenue Total

revenuex 1 x 2 From x 1 From x 2

AB

CD

E

00

150300

500

0300

300180

0

00

750015000

25000

02400

2400014,400

0

024000

3150029400

25,000

From the above table we find that revenue is maximum at Rs. 31,500 when 150 units of x 1 and

300 units of x 2 are produced.

Example 5: For conducting a practical examination, the chemistry department of a college

requires 10, 12 and 7 units of three chemicals X, Y, Z respectively. The chemicals are available in

two types of boxes: Box A, Box B. Box A contains 3, 2 and 1 units of X, Y, Z respectively and

costs Rs. 300. Box B contains 1, 2 and 2 units of X, Y, Z respectively and costs Rs. 200. Find


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how many boxes of each type should be bought by the department so that the total cost is

minimum.

Solution: First, we summarize the given data in the following table:

Units Units in Box A Units in Box B Units required

X

YZ

3

21

1

22

10

127

Cost Rs. 300 Rs. 200

Let x 1 be the number of boxes of A-type to be bought and x 2 be the number of boxes of B-type.

Then the total cost is,

Z = 300x 1 + 200x 2.

Obiviously x 1 0, x 2 0.From the details tabulated in the table, we find that x 1 and x 2 are subject to the following

constraints:

3x 1 + x2 10

2x 1 + 2x 2 12

x1 + 2x 2 7

Now, we consider the lines L 1: 3x1 + x2 = 10, L 2: 2x1 + 2x 2 = 12 L 3: x1 + 2x 2 = 7. These

lines are shown in fig.

We note that for the co-ordinates (x 1, x2) of a point satisfy the inequalities. The convex region

bounded by these lines and the co-ordinate axes is an unbounded region, this is shaded in fig.


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We check that a point (x 1, x2) that lies inside or on the boundary lines of this region satisfies the

conditions x 1 0, x 2 0 and the constraints.

We find that the vertices for the region of interest here are P, Q, R, S. Where P is the point at

which L meets the x 2 axis, Q is the point of inter-section of L 1 and L 2, R is the point of inter-

section of L 2 and L 3 and S is the point at which L 3 meets the x 1 axis. We find that P(0, 10), Q(2,

4), R(5, 1) and S(7, 0).:

At P (0, 10), Z = 300 0 + 200 10 = 2000

At Q (2, 4), Z = 300 2 + 200 4 = 1400

At R (5, 1), Z = 300 5 + 200 1 = 1700

At S (7, 0), Z = 300 7 + 200 0 = 2100

Evidently, Z is minimum at the vertices Q (2, 4) for which x 1 = 2, x 2 = 4. Thus the cost is minimum

if 2 boxes of type A and 4 boxes of type B are bought. The minimum cost is Rs. 1400.

2.5.2 Examples 6 on mixed constraints LP problem : By using graphical method, find themaximum and minimum values of the function Z = x 3y where x and y are non-negative and are

subject to the following conditions:

3x + 4y 19,

2x y 9

2x + y 15

x y 3

Solution: First, we write the constraints (conditions) to be satisfied by x, y in the following

standard (less than or equal) form:

3x 4y 19

2x y 9

2x + y 15

x + y 3

Now, consider the equations:

3x 4y = 19, 2x y = 9, 2x + y = 15, x + y = 3 which represents straight lines in the xy plane. Let us denote them by L 1, L2, L3 and L 4 respectively. These are shown in fig.:


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From the figure, we note that the lines L 1, L2, L3 and L 4 form a quadrilateral ABCD that lies in the

first quadrant of the xy plane. We readily see that the region bounded by this quadrilateral is

convex. As such, the points (x, y) that lie within or on the boundary lines of this quadrilateral

satisfy the inequalities x 0, y 0 and the constraints. The co-ordinates of the vertices A, B, C, D

of the quadrilateral are obtained by solving equations taken two of them at a time, we find that A

(1, 4), B (5, 1), C (6, 3), D (4, 7)

we get the solution

Zat A(1, 4) = 1 3 4 = 11

Zat B(5, 1) = 5 3 1 = 2

Zat C(6, 3) = 6 3 3 = 3

Zat D(4, 7) = 4 3 7 = 17

Evidently, Z is maximum at the vertex B and minimum at the vertex D. The maximum value of Z isZat B(5, 1) = 2, which corresponds to x = 5, y = 1, and the minimum values of Z is 17 at D(4, 7)

which corresponds to x = 4, y = 7.

Examples 7: Use the graphical method to solve the following LP problem:

Maximize Z = 7x 1+3x 2


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The co-ordinates of the extreme points of the feasible region are

41,

25 A

23,

25B and

23,0C . The value of the objective function at each of these extreme points is as follows:

Extreme point Co-ordinates (x 1, x2)Objective function

valueZ= 7x 1 + 3x 2

A

B

C

41,

25

23,

25

23,0

7

22

9/2

The maximum value of the objective function Z= 22 occurs at the extreme points

2

3,

2

5B .

Hence the optimal solution to the given LP problem is23x,

25x 21 == and Max. Z = 22.

In linear programming problems may have:

i) a unique optimal solution or

ii) many number of optimal solutions or


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iii) an unbounded solution or

iv) no solutions.

Example 8: Maximize Z = 100x 1 + 40x 2

Subject to10x 1 + 4x 2 2000

3x 1 + 2x 2 900

6x 1 + 12x 2 3000

and x 1, x2 0

Solutions: The given constraints can be rewritten as

1250x

500x

1450

x

300

x

1500x

200x

21

21

21

+

+

+

The values of (x 1 x2) at the points are 0(0, 0), A(200, 0) B(125, 187.5) and C(0, 250). The feasible

region is OABC. The values of Z at the points are

Z at O(00) = 0

Z at A(200, 0) = 20000

Z at B(125, 187.5) = 20000

Z at C(0, 250) = 10,000


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Thus the maximum value of Z occurs at 2 vertices at A and B. Any point on the line joining A and

B will also give the same maximum value of Z. Therefore, there are infinite number of feasible

solutions which yield the same maximum value of Z.

Suppose a linear programming problem has an unbounded feasible solution space.

If the set of all values of the objective function at different feasible solutions is not bounded above

(respectively, bounded below), and if the problem is a maximization (respectively, minimization)

problem, then we say that the given problem has an unbounded solution .

In the following, we present an example with unbounded solution.

2.5.3 Example 9 for Unbounded Solution:


Subject to

x1 x 2 2

x1 + x2 4

and x 1, x2 0

The intersection point A of the straight lines x 1 x 2 = 2 and x 1+x2 = 4 is A(3, 1). Here the

solution space is unbounded. The vertices of the feasible region are A(3, 1) and B (0, 4). Value of

objective at these vertices are

Z atA(31) = 2 3+3 1 = 9


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Z at B(0, 4) = 2 0+4 3 = 12.

But there are points in the convex region for which Z will have much higher values. For example E

(10, 9) lies in the shaded region and the value of Z there at 47. In fact, the maximum values of Z

occurs at infinity. Thus the problem has an unbounded solutions.

2.5.4 Example 10 for Inconsistent:


Subject to

x1 x 2 1

x 1 + x2 0

and x 1, x 2 0.

There being no point (x 1, x2) common to both the shaded regions, the LPP cannot be solved.

Hence the solution does not exist, since the constraints are inconsistent.

2.5.5 Example 11 for redundant Constraint:

A company making cold drinks has 2 bottling plants located at towns T 1 and T 2. Each plant

produces three drinks A, B and C and their production capacity per day is shown below:

Cold drinks Plant atT1 T2

ABC

600010003000

200025003000


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Thus the minimum value of Z is Rs. 80,000 and it occurs at B. Hence the optimal solution to the

problem is x 1 = 12 days, x 2 = 4 days.

Example 12: Final the maximum and minimum value of Z = 2x + 3y

Subject to x + y 30x y 0

y 3

0 x 20

0 y 12.

Solution: Any point (x, y) satisfies the conditions x 0, y 0 lies in the first quadrant only.

The desired point (x, y) lies with in the feasible convex region ABCDE.

Its vertices are A (3, 3) B (10, 3) C (20, 10), D (18, 12) and B (12, 12). The values of Z at the

five vertices are

Zat A (3, 3) = 2 3 + 3 3 =15

Z at B (20, 3) = 49Z at C (20, 10) = 70

Z at D (18, 12) = 72

Z zt E (12,12) = 60

Since the maximum value of Z is 72 which occurs at the vertix D (18, 12). Therefore the solution


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of the LPP is x = 18, y = 12 and the minimum value of z is 15 at x = 3, y = 3.


State True / False

1. The feasible region is a convex set.

2. The optimum value occurs anywhere in feasible region.

2.6 Summary

In LPP we first identify the decision variables which are some economic or physical quantities

whose values are of interest to the management. The problems must have a well-defined

objective function expressed in terms of the decision variable. The objective function may have tobe maximized when it expresses the profit or contribution. In case the objective function indicates

a cost, it has to be minimized. The decision variables interact with each other through some

constraints. These constraints occur due to limited resources, stipulation on quality, technical,

legal or variety of other reasons. The objective function and the constraints are linear functions of

the decision variables. A LPP with two decision variables can be solved graphically. Any non-

negative solution which satisfies all the constraints is known as a feasible solution of the problem.

The collection of all feasible solutions is known as a feasible region. The feasible region of a LPP

is a convex set. The value of the decision variables which maximise or minimize the objectives

function is located on the extreme point of the convex set formed by the feasible solutions.

Sometimes the problem may be infeasible indicating that no feasible solution of the problem

exists.

Terminal Questions .

1. Use graphical method and solve the L.P.P.

Maximize Z= 5x 1 + 3x 2

subject to: 3x 1 + 5x 2 15

5x 1 + 2x 2 10

x1, x2 0

2. Mathematically formulate the problem. A firm manufactures two products; the net profit on

product 1 is Rs. 3 per unit and the net profit on product 2 is Rs. 5 per unit. The manufacturing


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Unit 3 Simplex Method

Structure

3.1. Introduction

3.2. Standard form of L.P.P

3.2.1 The standard form of the LPP

3.2.2 Fundamental Theorem of L.P.P.

3.3. Solution of L.P.P Simplex method

3.3.1 Initial basic feasible solution of a LPP

3.3.2 To Solve problem by Simplex Method

3.4. The Simplex Algorithm

3.4.1 Steps

3.5. Penalty cost method or Big M-method

3.6. Two phase method

3.7. Maximisation Examples

3.8. Summary

Terminal Questions

Answers to SAQs & TQs

3.1 Introduction

The simplex method provides an efficient technique which can be applied for solving LPP of any

magnitude involving two or more decision variables. In this method the objective function used to

control the development and evaluation of each feasible solution to the problem.

The simplex algorithm is an iterative procedure for finding the optimal solution to a linear

programming problem. In the earlier methods, if a feasible solution to the problem exists, it is

located at a corner point of the feasible region determined by the constraints of the system. The

simplex method, according to its iterative search, selects this optimal solution from among the set

of feasible solutions to the problem. The efficiency of this algorithm is, because it considers only

those feasible solutions which are provided by the corner points, and that too not all of them. We

consider a minimum number of feasible solutions to obtain an optimal solution.


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Learning Objectives:

After studying this unit, you should be able to understand the following

1. To write the standard form of LPP from the given hypothesis

2. Apply the simplex algorithm to the system of equations3. Understand the big M-technique

4. Know the importance of the two phase method.

5. Formulate the dual from the primal (and vice versa).

3.2 The Standard Form Of LPP

The characteristics of the standard form are:

1. All constraints are equations except for the non-negativity condition which remain inequalities

( , 0) only.2. The righthand side element of each constraint equation is non-negative.

3. All variables are non-negative.

4. The objective function is of the maximization or minimization type.

The inequality constraints can be changed to equations by adding or substracting the lefthand

side of each such constraints by a non-negative variable. The non-negative variable that has to

be added to a constraint inequality of the form to change it to an equation is called a slack

variable . The non-negative variable that has to be substracted from a constraint inequality of the

form to change it to an equation is called a surplus variable . The right hand side of a constraintequation can be made positive by multiplying both sides of the resulting equation by (-1) wherever

necessary. The remaining characteristics are achieved by using the elementary transformations

introduced with the canonical form.

3.2.1 The Standard Form Of The LPP

Any standard form of the L.P.P. is given by

Maximize or Minimize i x iCzn

1i ==

Subject to: .m...........2,1i )0b(bS xa iii jij

n

1 j ==

=

& x j 0, j = 1, 2, --- n.

S i 0, i = 1, 2, --- m.


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3.2.2 Fundamental Theorem Of L.P.P.

Given a set of m simultaneous linear equations in n unknowns/variables, n m, AX = b, with r(A)

= m. If there is a feasible solution X 0 , then there exists a basic feasible solution.


State True / False

1. We add surplus variable for of constraint

2. The right hand side element of each constraint is non-negative.

3.3 Solution Of The Linear Programming Program Simplex Method

Consider a LPP given in the standard form,

To optimize z = c 1 x1 + c 2 x2 + ---+ c n xn

Subject to

a 11 x1 + a 12 x2 + -- + a n x n S 1 = b1

a 21 x1 + a 22 x2 + ----+ a 2n xn S 2 = b 2

.

.

a m1 x1 + a m2 x2 + -- + a mn xn S m = b m

x1, x2, --- xn, S 1, S 2 ---, S m 0.

To each of the constraint equations add a new variable called an artificial variable on the lefthand side of every equation which does not contain a slack variable. Then every constraint

equation contains either a slack variable or an artificial variable.

The introduction of slack and surplus variables do not alter either the constraints or the objective

function. So such variables can be incorporated in the objective function with zero coefficients.

However, the artificial variables do change the constraints, since these are added only to one side

i.e., to the left hand side of the equations. The new constraint equations so obtained is equivalent

to the original equations if and only if all artificial variables have value zero. To guarantee such

assignments in the optimal solutions, artificial variables are incorporated into the objective function

with very large positive coefficient M in the minimization program and very large negative coefficient M in the maximization program. These coefficients represent the penalty incurred in making an

unit assignment to the artificial variable.

Thus the standard form of LPP can be given as follows :

Optimize Z = C T X


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Subject to AX = B,

and X 0

Where X is a column vector with decision, slack, surplus and artificial variables, C is the vector

corresponding to the costs, A is the coefficient matrix of the constraint equations and B is thecolumn vector of the righthand side of the constraint equations.

Example 1: Consider the LPP

Minimize Z = 4x 1 + x2Subject to 3x 1 + x2 = 3

4x 1 + 3x 2 6

x1 + 2x 2 3,

x1, x2 0

Rewriting in the standard form,

Minimize Z = 4x 1 + x2 + 0.S 1 + 0.S 2 + M (A1 + A2)Subject to 3x 1+ x2 + A1 = 3

4x 1 + 3x 2 S 1 + A2 = 6x1 + 2x 2 + S 2 = 3

x1, x2, S 1, S 2, A1, A2 0

When S 2 is slack variable, S 1 is a surplus variable and A 1 & A2 an artificial variables.

Representing this program in matrixes, we have

Minimize Z = (4 1 0 0 M M)

2

1

2

1

2

1

A ASS

xx

Subject to

-

001021100134010013

2

1

2

1

2

1

A ASSxx

=

363

and

2

1

2

1

2

1

A ASSxx

0


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3.3.1 Initial basic feasible solution of a LPP

Consider a system of m equations in n unknowns x 1 , x2 - - xn,

a 11 x1 + a 12 x2 + - - + a 1n xn = b 1

a 21 x1 + a 22 x2 + - - + a 2n xn = b 2

a m1 x1 + a m2 x2 + - - + amn xn = b n

Where m n.

To solve this system of equations, we first assign any of n m variables with value zero. These

variables which have assigned value zero initially are called the non-basic variables, the

remaining variables are called basic variables. Then the system can be solved to obtain the

values of the basic variables. If one or more values of the basic variables are also zero valued,

then solution of the system is said to degenerate . If all basic variable, have non-zero values,

then the solution is called a non-degenerate solution.

A basic solution is said to be feasible, if it satisfies all constraints.

Example 2: Consider the system of equations

2x 1 + x2 x 3 = 2

3x 1 + 2x 2 + x3 = 3

where x 1, x2, x3 0.

Since there are 3 variables and two equations, assign 3 2 = 1 variable, the value zero initially.

Case (i) : Let x 3 = 0 i.e., x 3 be a non-basic variable, then equation becomes

2x 1 + x2 = 2

3x 1 + 2x 2 = 3

Solving, we get

x1 = 1, x 2 = 0.

\ The solution degenerates, but is feasible.

Case (ii) : Let x2 be a non-basic variable i.e., x 2 = 0, then solution is

x1 = 1 and x 3 = 0

Here also the solution degenerates but feasible. Case (iii) : Let x1 be non-basic i.e., x 1= 0

Solution is x 2 =31 , x3 =

31 .

The solution non-degenerates, but is not feasible.


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Consider a LPP given in the standard form

Optimize Z = C TX

Subject to AX = B,

X

0.The initial solution of such a problem denoted by X 0, is obtained by treating all decision and surplus

variables as non-basic variables i.e., they have assigned value zero, all slack and, artificial variables as

basic variables and have assigned values which are on R.H.S. of the corresponding constraint

equations.

3.3.2 To Solve problem by Simplex Method

1. Introduce stack variables (S is) for type of constraint.

2. Introduce surplus variables (S is) and Artificial Variables (A i) for type of constraint.

3. Introduce only Artificial variable for = type of constraint.

4. Cost (C j) of slack and surplus variables will be zero and that of Artificial variable will be M

Find Z j - C j for each variable.

5. Slack and Artificial variables will form Basic variable for the first simplex table. Surplus

variable will never become Basic Variable for the first simplex table.

6. Z j = sum of [cost of variable x its coefficients in the constraints Profit or cost coefficient of

the variable].

7. Select the most negative value of Z j - C j. That column is called key column. The variable

corresponding to the column will become Basic variable for the next table.

8. Divide the quantities by the corresponding values of the key column to get ratios select the

minimum ratio. This becomes the key row. The Basic variable corresponding to this row will

be replaced by the variable found in step 6.

9. The element that lies both on key column and key row is called Pivotal element.

10. Ratios with negative and a value are not considered for determining key row.

11. Once an artificial variable is removed as basic variable, its column will be deleted from next

iteration.

12. For maximisation problems decision variables coefficient will be same as in the objective

function. For minimization problems decision variables coefficients will have opposite signs

as compared to objective function.


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13. Values of artificial variables will always is M for both maximisation and minimization

problems.

14. The process is continued till all Z j - C j 0.


State True / False

1. A basic solution is said to be a feasible solution if it satisfies all constraints.

2. If one or more values of basic variable are zero then solution is said to be degenerate.

3.4 The Simplex Algorithm

To test for optimality of the current basic feasible solution of the LPP, we use the following

algorithm called simplex algorithm. Let us also assume that there are no artificial variable existingin the program.

3.4.1 Steps

1) Locate the most negative number in the last (bottom) row of the simplex table, excluding that

of last column and call the column in which this number appears as the work column.

2) Form ratios by dividing each positive number in the work column, excluding that of the last

row into the element in the same row and last column. Designate that element in the work

column that yields the smallest ratio as the pivot element. If more than one element yields

the same smallest ratio choose arbitrarily one of them. If no element in the work column is non

negative the program has no solution.

3) Use elementary row operations to convert the pivot element to unity (1) and then reduce all

other elements in the work column to zero.

4) Replace the x -variable in the pivot row and first column by x-variable in the first row pivot

column. The variable which is to be replaced is called the outgoing variable and the variable

that replaces is called the incoming variable. This new first column is the current set of basic

variables.

5) Repeat steps 1 through 4 until there are no negative numbers in the last row excluding the

last column.

6) The optimal solution is obtained by assigning to each variable in the first column that value in

the corresponding row and last column. All other variables are considered as non-basic and

have assigned value zero. The associated optimal value of the objective function is the

number in the last row and last column for a maximization program but the negative of this

number for a minimization problem.


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The initial table is given by

X1 x2 x3 x4 S 1 S 2 S 3 Ratio

2 4 1 1 0 0 0

S 1 0 1 3* 0 1 1 0 0 434

S 2 0 2 1 0 0 0 1 0 313

S 3 0 0 1 4 1 0 0 1 313

Z j-c j 2 4 1 1 0 0 0 0

work column * pivot element

S 1 is the outgoing variable, x 2 is the incoming variable to the basic set.

The first iteration gives the following table :

x1 x2 x3 x4 S 1 S 2 S 3 Ratio

2 4 1 1 0 0 0

x2 431 1 0

31

31 0 0

34

S 2 035 0 0

31

31 1 0

35

S 3 0 31 0 4*

32

31 0 1

35

125

Z j-C j 32 0 1

31

34 0 0

316

Work column


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x3 enters the new basic set replacing S 3, the second iteration gives the following table :

x1 x2 x3 x4 S 1 S 2 S 3 Ratio

2 4 1 1 0 0 0

x2 431 1 0

31

31 0 0

34 4

S 2 035 * 0 0

31

31 1 0

35 1

x3 1 121 0 1

61

121 0

41

125

Z j-C j

43 0 0

61

45 0

41

423 1/2

Work column

x1 enters the new basic set replacing S 2, the third iteration gives the following table:

x1 x2 x3 x4 S 1 S 2 S 3

2 4 1 1 0 0 0

x2 4 0 1 052

52

51 0 1

x1 2 1 0 0

51

51

53 0 1

x3 1 0 0 1203

101

201

41

21

Z j-C j 0 0 0207

1011

209

41

213

Since all elements of the last row are non-negative, the optimal solution is Z =2

13 which is

achieved for x 2 = 1, x 1 = 1, x 3 = 21 and x 4 = 0.

5) A manufacturing firm has discontinued production of a certain unprofitable product line.This created considerable excess production capacity. Management is considering to

devote this excess capacity to one or more of three products: call them product 1, 2 and 3.

The available capacity on the machines which might limit output are given below :


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Machine Type Available Time(in machine hours per week)

Milling Machine 250Lathe 150

Grinder 50

The number of machine-hours required for each unit of the respective product is given below :

Productivity (in Machinehours/Unit)

Machine Type Product 1 Product 2 Product 3

Milling Machine 8 2 3

Lathe 4 3 0

Grinder 2 1

The unit profit would be Rs. 20, Rs. 6 and Rs. 8 for products 1, 2 and 3. Find how much ofeach product the firm should produce in order to maximize profit ?

Let x 1, x2, x3 units of products 1, 2 and 3 are produced in a week.

Then total profit from these units is

Z = 20 x 1 + 6 x2 + 8 x3

To produce these units the management requires

8x 1 + 2x 2 + 3x 3 machine hours of Milling Machine

4x 1 + 3x 2 + 0 x3 machine hours of Lathe

and 2x 1 + x3 machine hours of Grinder

Since time available for these three machines are 250, 150 and 50 hours respectively, we

have

8x 1 +2x 2 + 3x 3 250

4x 1 + 3x 2 150

2x 1 + x3 50.

Obviously x 1, x2, x3 0

Thus the problem is to

Maximize Z = 20x 1 + 6x 2 + 8x 3

Subject to 8x 1 + 2x 2 + 3x 3 250

4x 1 + 3x 2 150

2x 1 + x3 50,

x1, x2, x3 0.


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Maximize Z = 20x 1 + 6x 2 + 8x 3 + 0S 1 + 0S 2 + 0S 3

Subject to 8x 1 + 2x 2 + 3x 3 + S 1 = 250

4x 1 + 3x 2 + S 2 = 1502x 1 + x3 + S 3 = 50,

x1, x2, x3, S 1, S 2, S 3 0.

The initial basic solution is

X0 =

3

2

1

S

S

S=

50150250

The initial simplex table is given by

x1 x2 x3 S 1 S 2 S 3 Ratio

20 6 8 0 0 0

S 1 0 8 2 3 1 0 0 25031

8250 =

S 2 0 4 3 0 0 1 0 1504

150 = 37.5

S 3 0 2* 0 1 0 0 1 502

50 = 25

ZJ C j 20 6 8 0 0 0 0

Work column * pivot element

x1 enters the basic set of variables replacing the variable s 3. The first iteration gives the

following table:

x1 x2 x3 s 1 s 2 s 3 Ratio20 6 8 0 0 0

s 1 0 0 2 1 1 0 4 502

50 = 25


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s 2 0 0 3* 2 0 1 2 503

50

x1 20 1 021 0 0

21 25

Z j C j 0 6 2 0 0 10 500

Work column * pivot element

x2 enters the basic set of variables replacing the variable s 2. The second iteration gives the

following table:

x1 x2 x3 S 1 S 2 S 3 Ratio

20 6 8 0 0 0

S 1 0 0 031 1 2/3

38

350 50

X2 6 0 1 32 0

31

32

350

X1 20 1 021 * 0 0

21 25 50

Zj

Cj

0 0 2 0 2 6 600

work column * pivot element.

x3 enters the basic set of variables replacing the variable x 1. The third iteration yields the following

table:

x1 x2 x3 S 1 S 2 S 3

20 6 8 0 0 0

S 1 0 32 0 0 1

32 3 0

X2 634 1 0 0

31 0 50

X3 8 2 0 1 0 0 1 50

Z j c j 4 0 0 0 2 8 700

Since all z j c j 0 in the last row, the optimum solution is 700.


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i.e., the maximum profit is Rs. 700/- which is achieved by producing 50 units of product 2 and

50 units of product 3.

3.4.2. Self Assessment Questions 3

State Yes / No

1. The key column is determined by Z j - C j row.

2. Pivotal element lies on the crossing of key column and key row

3. The ve and infinite ratios are considered for determining key row.

3.5 Penalty Cost Method Or Big-M Method

Consider a L.P.P. when atleast one of the constraints is of the type or = . While expressing in

the standard form, add a non negative variable to each of such constraints. These variables are

called artificial variables. Their addition causes violation of the corresponding constraints, since

they are added to only one side of an equation, the new system is equivalent to the old system of

constraints if and only if the artificial variables are zero. To guarantee such assignments in the

optimal solution, artificial variables are incorporated into the objective function with large positive

coefficients in a minimization program or very large negative coefficients in a maximization

program. These coefficients are denoted by M.

Whenever artificial variables are part of the initial solution X 0, the last row of simplex table will

contain the penalty cost M. The following modifications are made in the simplex method tominimize the error of incorporating the penalty cost in the objective function. This method is called

Big M-method or Penalty cost method.

1) The last row of the simplex table is decomposed into two rows, the first of which involves

those terms not containing M, while the second involves those containing M.

2) The Step 1 of the simplex method is applied to the last row created in the above modification

and followed by steps 2, 3 and 4 until this row contains no negative elements. Then step 1 of

simplex algorithm is applied to those elements next to the last row that are positioned over

zero in the last row.

3) Whenever an artificial variable ceases to be basic, it is removed from the first column of the

table as a result of step 4, it is also deleted from the top row of the table as is the entire

column under it.

4) The last row is removed from the table whenever it contains all zeroes.


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5) If non zero artificial variables are present in the final basic set, then the program has no

solution. In contrast, zero valued artificial variables in the final solution may exist when one or

more of the original constraint equations are redundant.

Examples: 6) Use Penalty Cost method to

Maximize z = 2x 1 + 3x 2

Subject to x 1 + 2x 2 2

6x 1 + 4x 2 24,

x1, x2 0.Rewriting in the standard form, we have

Maximize z = 2x 1 + 3x 2 + 0S 1 + 0S 2 M A 1

Subject to x 1 + 2x 2 + S 1 = 2

6x 1 + 4x 2 S 2 + A1 = 24,

x1, x2, S 1, S 2, A1 0.The initial simplex table is

x1 x2 S 1 S 2 A 1

2 3 0 0 M Ratio

S 1 0 1* 2 1 0 0 212 = 2

A1 M 6 4 0 1 1 24

6

24= 4

2 3 0 0 0 0

6M 4M 0 M M 24M

Work column


x1 x2 S 1 S 2 A1

2 3 0 0 M

x1 2 1 2 1 0 0 2

A i M 0 8 3 1 1 12

0 1 2 0 0 4


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0 8M 6M 1M 0 12M

Since all elements of the last two rows are non negative, the procedure is complete. But

existence of non zero artificial variable in the basic solution indicates that the problem has no

solution.

3.5.1 Self Assessment Questions 4

State Yes / No

1. The value of artificial value is M

2. Artificial variables enters as Basic Variable.

3.6 Two Phase Method The drawback of the penalty cost method is the possible computational error that could result

from assigning a very large value to the constant M. To overcome this difficulty, a new method is

considered, where the use of M is eliminated by solving the problem in two phases. They are

Phase I: Formulate the new problem by eliminating the original objective function by the sum of

the artificial variables for a minimization problem and the negative of the sum of the artificial

variables for a maximization problem. The resulting objective function is optimized by the simplex

method with the constraints of the original problem. If the problem has a feasible solution, the

optimal value of the new objective function is zero (which indicates that all artificial variables are

zero). Then we proceed to phase II. Otherwise, if the optimal value of the new objective functionis non zero, the problem has no solution and the method terminates.

Phase II : Use the optimum solution of the phase I as the starting solution of the original problem.

Then the objective function is taken without the artificial variables and is solved by simplex

method.

Examples:

7) Use the two phase method to

Maximise z = 3x 1 x 2

Subject to 2x 1 + x2 2x1 + 3x 2 2

x2 4,

x1, x2 0



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Maximize z = 3x 1 x 2 + 0S 1 MA1 + 0.S 2 + 0.S 3

Subject to 2x 1 + x2 S 1 + A1 = 2

x1 + 3x 2 + S 2 = 2

x2 + S 3 = 4,x1, x2, S 1, S 2, S 3, A1 0.

Phase I :

Consider the new objective,

Maximize Z* = A 1

Subject to 2x 1 + x2 S 1 + A1 = 2

x1 + 3x 2 + S 2 = 2

x2 + S 3 = 4,

x1, x2, S 1, S 2, S 3, A1

0.Solving by Simplex method, the initial simplex table is given by

x1 x2 S 1 A1 S 2 S 30 0 0 1 0 0 Ratio

A1 1 2* 1 1 1 0 0 222 = 1

S 2 0 1 3 0 0 1 0 212 = 2

S 3 0 0 1 0 0 0 1 4

2 1 1 0 0 0 2 Work column * pivot element

x1 enters the basic set replacing A 1.

The first iteration gives the following table:

x1 x2 x1 A1 S 2 S 3

0 0 0 1 0 0

X1 0 1

2

1

2

1

2

1 0 0 1

S 2 0 025

21

21 1 0 1

S 3 0 0 1 0 0 0 1 4

0 0 0 1 0 0 0


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Since all elements of the last row are non negative, the current solution is optimal.

The maximum value of the objective function

Z = 6 which is attained for x 1 = 2, x 2 = 0.

8) Maximize z = 3x 1 + 2x 2,

subject to 2x 1 + x2 2,

3x 1 + 4x 2 S 2 + A1 12,

x1, x2 0.


Maximize z = 3x 1 + 2x 2 + 0S 1 + 0.S 2 MA1

Subject to 2x 1 + x2 + S 1 = 2

3x 1 + 4x 2 S 2 + A1 = 2

x1, x2, S 1, S 2, A1 0.Solving by two phase method.

Phase I :

Consider the new objective function

Maximize z* = A 1

Subject to 2x 1 + x2 + S 1 =2

3x 1 + 4x 2 S 2 + A1 = 12,

x1, x2, S 1, S 2, A1 0.

The initial Simplex table is given byx1 x2 S 1 S 2 A1 Ratio

0 0 0 0 1

S 1 0 2 1* 1 0 0 212 = 2

A1 1 3 4 0 1 1 124

12 = 3

3 4 0 1 0 12

Work column


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x1 x2 S 1 S 2 A 1

0 0 0 0 1

X2 0 2 1 1 0 0 2 A1 1 5 0 4 1 1 4

5 0 4 1 0 4

Since all elements of the last row an non negative, the procedure is complete.

But the existence of non zero artificial variable in the basic set indicates that the problem has no

solution.

3.7 Minimization Examples

Example 10

Minimize = Z = 3x 1 + 8x 2

Subject to

x1 + x2 = 200

x1 80

x2 60

x1 , x2 0

Solution

In a standard form

Minimize Z = 3x 1 + 8x 2 + MA1 + OS 1 + MA2 + OS 2

Subject to

x1 + x2 +A1 = 200

x1 S 1 + A2 = 80

x2 + S 2 = 60

S 1, S 2, A1, A2 0

Simplex Table 1

C j

- 3 - 8 0 0 - M - M

C.B B.V x 1 x 2 S 1 S 2 A1 A3 Qty Ratio

- M A1 1 1 0 0 1 0 200 200

- M A2 1

P.E

0 - 1 0 0 0 80 80 K R


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0 S 2 0 1 0 1 0 1 60 a

Z j - C j -2M +3 K C -M + 8 M 0 0 0

Simplex Table 2 C j - 3 - 8 0 0 - M

C.B B.V x 1 x2 S 1 S 2 A1 Qty Ratio Transformation

- M A1 0 1 1 0 1 120 120 R 11 = R1-R 21

- 3 x1 1 0 -1 0 0 80 a R21 = R 2

0 S 2 0 1

P.E

0 1 0 60 60 K

RR31 = R 3

Z j - C j 0-M + 8

K C

-M +

3

0 0

Simplex Table 3

C j - 3 - 8 0 0

C.B B.V x 1 x 2 S 1 S 2 A1 Qty Ratio Transformation

- M A1 0 0 1

P.E

- 1 1 60 60R 11 = R1

- 3 x1 1 0 - 1 0 0 80 -ve R 21 = R2

- 8 x2 0 1 0 1 0 60 a R 31 = R 3 R 11

Z j - C j 0 0- M + 3

KCM - 8 0 0

Simplex Table 4

C j - 3 - 8 0 0

C.B B.V x 1 x 2 S 1 S 2 Qty Ratio Transformation

0 S 1 0 0 1 - 1 60 - ve R 11 = R 1

- 3 x1 1 0 0 - 1 140 -ve R 21 = R 2 + R11

- 8 x2 0 1 0 1P.E

60 80 KR

R 31 = R11

Z j - C j 0 0 0 - 5

Simplex Table 5


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C j - 3 - 8 0 0

C.B B.V x 1 x 2 S 1 S 2 Qty Ratio Transformation

0 S 1 0 1 1 0 120 R 11 = R 1 + R31

- 3 x1 1 1 0 0 200 R 21

= R 2 + R31

0 S 2 0 1 0 1 60 R 31 = R 3

Z j - C j 0 4 0 0

Since all Z j - C j 0, the optimum solution is x 1 = 200 x 2 = 0

Min Z = 60

3.8. Summary

In this unit we solved the L.P.P by simplex method. The constraints for which slack, surplus and

artificial variables to be introduced and the method of solving L.P.P is explained with examples.

Terminal Questions

1. Maximize z = 3x 1 x 2

Subject to2x 1 + x2 2

x1 + 3x 2 3

x2 4,

x1, x2 0.

2. Minimize Z = 6x 1 + 7x 2

Subject to the constraints

x1 + 3x 2 12

3x 1 + x2 12

x1 + x2 8

x1 + x2 0

Answers To Self Assessment Questions


1. False 2. True


1. True 2. True


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1. Yes 2. Yes 3. No

Self Assessment Questions 4 1. Yes 2. Yes

Answer For Terminal Questions

1. Z = 9 x 1 = 3 x2 = 0

2. Z

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