M.c.a.(sem iii) operation research

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M.C.A.
SEMESTER - III
OPERATIONS RESEARCH
1. Nature of Operation Research
 History
 Nature of OR
 Impact of OR
 Application Areas
2. Overview of Modeling approach
 Formulating the problem
 Constructing a mathematical model
 Deriving a solution
 Testing a model and the solution
 Establishing control over the solution
 Implementation issues
3. Linear Programming
 Introduction
 Graphical solution
 Graphical sensitivity analysis
 The standard form of linear programming problems
 Basic feasible solutions
 Simplex algorithm
 Artificial variables
 Big M and two phase method
 Degeneracy
 Alternative optima
 Unbounded solutions
 Infeasible solutions
4. Dual Problem
 Relation between primal and dual problems
 Dual simplex method
5. Transportation problem
 Starting solutions. North-west corner Rule – lowest cost methods –
Vogels approximation method
 MODI Method

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6. Assignment problem
 Hungarian method
7. Travelling salesman problem
 Branch & Bound technique
8. Sequencing Problem
 2 machines n jobs
 3 machines n jobs
 n machines m job
9. Pert and CPM
 Arrow network
 Time estimates, earliest expected time, latest allowable occurrence
time, latest allowable occurrence time and stack
 Critical path
 Probability of meeting scheduled date of completion of project
 Calculation of CPM network
 Various floats for activities
 Project crashing
10. Integer programming
 Branch and bound algorithm
 Cutting plane algorithm
11. Deterministic Inventory Models
 Static EOQ models
 Dynamic EOQ models
12. Game theory
 Two person Zero sum games
 Solving simple games
13. Replacement theory
 Replacement of items that deteriorate
 Replacement of items that fail group replacement and individual
replacement.
Term work/Assignment : Each candidate will submit a journal in which at
least 10 assignments based on the above syllabus and the internal test
paper. Test graded for 10 marks and Practicals graded for 15 marks.

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Reference :
1. Gillet, B.E., “Introduction to Operation Research : a computer oriented
algorithmic approach” Tata McGraw Hill, NY
2. Hillier F., and Lieberman, G.J. “Introduction to Operation Research”,
Holden Day
3. Operations Research Applications and Algorithms Waynel L. Winston
Thomson
4. Optimization methods K.V. Mital & Mohan New Age
5. Operations Research : Principles and Practice 2nd
edition Ravindran
Wiley Production
6. Kambo, N.S., “Mathematical Programming Techniques”, McGraw
Hill
7. Kanti Swaroop, Gupta P.K. Man Mohan, “Operations Research”,
Sultan Chand and Sons
8. Taha, H.A. “Operations Research – An Introduction”, McMillan
Publishing Company, NY
9. Operation Research – S.D. Sharma
10. Operations Research by P.K.Gupta & Hira S. Chand
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1
NATURE OF OPERATION RESEARCH
(OR)
Unit Structure :
1.0 Objectives
1.1 Introduction
1.2 History of OR
1.3 Nature of OR
1.4 Application Areas
1.5 Unit End Exercise
1.0 OBJECTIVES
After going through this unit you will be able to:
 Understand the need of introducing OR
 Define OR
 Understand History of OR
 Describe Nature of OR
 Know the application areas of OR
1.1 INTRODUCTION
In any organization, managers have to take decisions while,
planning the activities. Generally decisions were taken based on past
experiences, the style of working of the manager and judgement.
However, decisions taken in this manner had the risk of making wrong
decisions which would affect the cost of the project. To overcome this
difficulty managers have started using scientific methods to evaluate the
various alternatives so that proper decisions can be taken.
With the increased complexities of life, the business has grown
tremendously in different directions and hence there is necessity to modify
the organisations relevant to business. Decision making is a requirement in
business and as any problem becomes complicated with large input data,
the analysis becomes difficult and hence an effective use of systematic
approach is needed. This has created the necessity of scientific methods
for decision making in business.

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These methods are called Quantitative methods, Operations
Research, Decision Science, System Analysis etc. Hence, Operations
Research can be understood as a scientific tool evolved as a supporting
help for the decision making process. This is extensively used in business,
also in government and defense organisations for solving complicated
problems.
Many definitions of OR have been suggested from time to time.
Some are :
1. “Operations Research” is the application of scientific methods,
techniques and tools to problems involving the operations of systems
so as to provide those in control of operations with optimum solution
to the problems - CHURCHMAN
2. OR is a scientific method of providing executive departments with a
quantitative basis for decisions under their control.
- P.M. MORSE & G.E. KIMBAIL
3. OR is the art of winning wars without actually fighting them. -
AUTHER CLARK
4. OR is the systematic application of quantitative methods, techniques
and tools to the analysis of problems involving the operation of
systems. - DAELLENBACK & GEORGE
Thus, we can conclude by saying OR arrives at optimal or near-
optimal solutions to complex decision making problems.
1.2 HISTORY OF OR
Many experts consider the start of OR in the III century B.C,
during the II Punie War, with analysis and solution that Archimedes
named for the defense of the city of Syracuse, besieged by the Romans.
In 1503, Leonardo Da Vinci a took part in the war against Prisa
because he knew the techniques to accomplish bombardments, to construct
ships, armored vehicles, cannons, catapults, and another warlike machines.
In 1885 Ferderick W. Taylor emphasised the application of
scientific analysis to methods of production.
In 1917, A. K. Erlang, a Danish mathematician, published his work
on the problem of congestion of telephone traffic.
In 1930, H.C. Levinson, an American astronomer, applied
scientific analysis to the problem of merchandising.
However, the development of OR started during the Second World
War. The name was also derived from its use for research on Military
Operations during the war. Since strategic and tactical decisions during the

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war are very complicated with time horizon for such decisions being
comparatively small, the necessity for group analysis and use of
mathematical, economic and statistical theories along with engineering,
behavioural and Physical Sciences was felt and utilized. American and
British groups worked on various research projects. Success and
usefulness of these projects led to the development of various techniques
for decision making and later the results prompted their uses in business
applications and civilian problems.
Hence, when the war ended, an effort was made to apply the OR
techniques to other areas of business and industry.
1.3 NATURE OF OR
The job of OR is to examine, formulate, analyse, treat, solve the
problems of management in a scientific way. OR study is always rooted in
mathematical analysis.
OR is also an interdisciplinary mathematical sciences. Employing
techniques from other mathematical sciences - such as mathematical
modeling, statistical analysis and mathematical optimization - OR arrives
at optimal or near-optimal solutions to complex decision making problem.
OR has overlap with other disciplines, notably industrial
engineering, management science, psychology and economics.
Some of the tools used in OR are statistics, optimization,
probability theory, queuing theory, game theory, decision analysis,
mathematical modeling and simulation. Because of the computational
nature of these fields, OR has also strong ties to computer science and
analytics.
The principal phases for implementing OR in practice include :
a) Definition of the problem :
It involves defining the scope of the problem under investigation
which is carried out by the entire OR team. The end result of the
investigation is to identify three principal elements of the problem -
i) Description of decision alternatives.
ii) Determination of the objective of the study and
iii) Specification of the limitations under which the modeled
system operates.
b) Model Construction :
Here the problem is translated into mathematical relationship.

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c) Model Solution :
Model is solved using various mathematical and statistical tools.
Using the input data.
d) Implementation of the solution of a model involves the translation
of the results into operating instructions.
1.4 APPLICATION AREAS
The techniques used in OR have very wide application in various
fields of business / industrial / government / social sector. Few areas of
applications are mentioned below :
Marketing and Sales :
1. Product Selection and Competitive strategies
2. Utilisation of salesmen, their time and territory control, frequency
of visits in sales force analysis.
3. Marketing advertising decisions for cost and time effectiveness.
4. Forecasting and decision trends.
5. Market Research decisions.
Production Management :
1. Product Mix and product proportioning.
2. Facility and production planning and scheduling.
3. Material handing facilities planning.
4. Design of Information systems.
5. Quality Control decisions.
Inventory Control :
Inventory control techniques of OR can help to develop better
inventory policies and bring down the investments in inventories.
Finance, Investments and Budgeting :
1. Profit planning
2. Cash flow analysis
3. Investment decisions and risk analysis.
4. Dividend Policies.
5. Portfolio Analysis.
Defence :
1. Optimum weaponry systems
2. Optimum level of force deployment
3. Transportation Cost
4. Assignment suitabilities

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Personnel Management :
1. Determination of optimum organisation level.
2. Job evaluation and assignment analysis.
3. Salary criteria
4. Recruitment policies
Research and Development :
1. Determination of areas of need of research and development.
2. Selection criteria for specific project.
3. Trade-off analysis for time-cost relationship and control of
development projects.
1.5 UNIT END EXERCISE
1. What is Operation Research?
2. Discuss the origin and development of OR
3. Discuss the applications of OR
4. Discuss how and why OR methods have been valuable in aiding
executive decisions.
5. What is the role of OR in decision making?
6. Explain the nature of OR.
7. What are the essential characteristics of OR?
8. “OR study is performed by a team of scientists” - Discuss.
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OVERVIEW OF MODELING APPROACH
Unit Structure :
2.0 Objectives
2.1 Introduction
2.2 Formulating the problem
2.3 Constructing a Mathematical Model
2.4 Deriving a solution
2.5 Testing model and the solution
2.6 Establishing control over the solution
2.7 Implementation Issues
2.0 OBJECTIVES
After going through this unit, you will be able to :
 Understand the importance of formulating the problem correctly.
 Define objective function and constraints.
 Construct a mathematical model of OR.
 Know what are the methods to derive a solution of the model of OR.
 Know importance of testing the model and solution.
2.1 INTRODUCTION
OR provides a scientific basis for the decision-makers of an
organisation for solving problems, by employing a systematic approach by
a team of scientists from different disciplines for finding a solution which
is in the best interest of the organisation as a whole. For this, it is very
essential that the problem at hand be clearly defined. It is impossible to get
right answer from a wrong problem. After formulating the problem, the
next step is to construct a model (mathematical) for the system under
study. After constructing the model, the utility of the solution can be
checked and also tools must be developed to indicate as to how and when
the model or its solution will have to be modified to take the changes into
account.

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2.2 FORMULATING THE PROBLEM
OR is a problem solving and a decision making science. Whenever
we have conflicts, uncertainty and complexity in any situation, OR can
help in the end to reduce costs and improve profits. So, it is very important
that the problem at hand be clearly defined. The problem may be set out
explicitly by the consumer, the sponsoring organisation or may be
formulated by the OR team.
In formulating a problem for OR study, analysis must be made of
the four major components.
a) The Environment :
OR team must study the environment involving men, machines,
materials, suppliers, consumers, the government and the public.
b) The decision maker :
Decision maker is the person who is actually responsible to take
the final decision. OR team must study the decision maker and his
relationship to the problem at hand.
c) Objectives :
Objectives should be defined very clearly by taking into account
the problem as a whole
d) Alternative courses of action and constraints :
The OR team has to determine which alternative course of action is
most effective under the constraints to achieve a certain set of objectives.
2.3 CONSTRUCTING A MATHEMATICAL MODEL
After formulating the problem, the next step is to construct a model
for the system under study.
A model is defined as an idealized representation of some real life
system which can analyze the behaviour of the system. In OR study, we
use mathematical model. A mathematical model consists of a set of
equations which describe the system or problem.
These equations represent i) the effectiveness function and ii) the
constraints. The effectiveness function usually called the objective
function is a mathematical expression of the objectives, i.e., mathematical
expression for the cost or profit of the operation. Constraints are
mathematical expressions of the limitations on the fulfillment of the
objectives.
The general form of a mathematical model is
E = f (xi, yi)

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Where E = objective function
xi = controllable variables
yi = uncontrollable variables
f = relationship between E and xi, yi
The values of controllable are to be determined from the solution
of the problem.
2.4 DERIVING A SOLUTION
A solution may be obtained from a model either by simulation or
mathematical analysis. Some cases may require the use of a combination
of simulation and mathematical analysis.
Mathematical Analysis :
OR team make use of various branches of mathematics such as
calculus, matrix algebra, finite differences, iteration etc.
Simulation methods :
Simulation is a quantitative procedure which describes a process
by developing a model of that process and then conducting a series of
organised trial and error experiment to predict the behaviour of the process
over time.
2.5 TESTING MODEL AND THE SOLUTION
Since a model is only a simplified representation of reality, the
results are to be tested against the real world experience in order to
establish the models credibility. The usefulness of a model is tested by
determining how well it predicts the effect of these changes. Such an
analysis is called sensitivity analysis. The utility or validity of the solution
can be checked by comparing the results obtained without applying the
solution with the results obtained when it is used.
2.6 ESTABLISHING CONTROL OVER THE SOLUTION
A solution which we felt optimum today, may not be so tomorrow
since the values or the variables (parameters) may change, new parameters
may emerge and the structural relationship between the variables may
undergo a change.
A solution derived from a model remains a solution only so long as
the uncontrollable variables retain their values and the relationship
between the variables does not change. The solution itself goes ‘out of
control’ if the values of one or more uncontrolled variables very or
relationship between variables undergo change. Therefore, controls must
be established to indicate the limits within which the model and its
solution can be considered as reliable. Also tools must be developed to

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indicate as to how and when the model or its solution will have to be
modified to take the changes into account.
2.7 IMPLEMENTATION ISSUES
This is the last and most important phase of OR. Once an
operationally feasible solution is obtained, it is put into practice. After the
solution has been implemented, the analyst observes the response of the
solution to the changes made.
Finally, no system is even completely static and it is always
necessary to monitor the environment within which a system operates to
ensure that changing conditions do not render a solution inappropriate. It
is important, to ensure that any solution implemented is continuously
reviewed and updated and modified in the light of a changing
environment. A changing economy, fluctuating demand, and model
improvements requested by decision-makers are only a few examples of
changes that might require the analysis to be modified.
Q.1 How is an OR model constructed?
Q.2 What is the role of decision-maker in formulating OR model?
Q.3 Describe the importance of the four components in formulating Or
model.
Q.4 Define objective function and Constraints.
Q.5 Which methods are used to find solution of OR model?
Q.6 What is sensitivity Analysis and how it is used in OR techniques?
Q.7 How do we establish control over the solution of an OR model?
Q.8 How do the solutions desired from OR model are put into practice?
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3
LINEAR PROGRAMMING MODEL –
GRAPHICAL METHOD
Unit structure :
3.0 Introduction
3.1 Objectives
3.2 Maximization Case (Problem)
3.3 Minimization case (problem)
3.4 Sensitivity Analysis of graphical solutions
3.5 Special cases in graphical solutions
3.6 Bibliography
3.0 OBJECTIVES
This unit should enable the learner to
 formulate a business problem into the language of mathematics
 find an optimal solution to a business problem by using graphical
method
 interpret the optimal solution which is in mathematical form for
decision making
 do sensitivity analysis of the optimal solution to determine its
utility in the light of changes in the problem environment
 understand and interpret special cases of optimal solution
3.1 INTRODUCTION
The World War II introduced mathematics and statistics to
business management which since then have been two best friends of
business managers. Linear programming came into business
management along with other Operations Research models from the
knowledge bank of World War II in the decade of 1950s. Linear
programming is a mathematical model used extensively in decision
making, a process of allocating available resources to maximize profits or
minimize costs in business operations.

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History of linear programming dates back to 1700s the time of
Jean Baptiste Joseph Fourier in France. Modern linear programming
model was developed by Leonid Kantorovich, a Russian mathematician,
in 1939. Linear programming model consists of a set of algebraic
inequalities and or equations of linear nature (first degree or first order) in
which the business problem is expressed. These inequalities are analyzed
in a series of steps to develop the solution. It was used during World War
II to plan operations to reduce costs to the army and increase losses to the
enemy. Postwar, many industries found its use in their daily planning.
Linear Programming model can deal with business problems with
maximization objective and minimization objective. The problems can
also differ from each other in terms of business constraints. These details
are discussed subsequently in this analysis. In business management
parameters like profit are maximized and parameters like cost are
minimized. For solving business problems, Linear Programming Model
follows a methodology. In this chapter focus will be on graphical
method of Linear Programming Model for solving business problems.
Graphical method being visual by nature enables understanding of various
related concepts quite easily. Inherent weakness of this method is its
inability to evaluate more than two variables while developing a decision.
But this method laid the foundation to the development of a more versatile
method called simplex method which will be discussed later. The
discussion will begin with a business problem of maximization nature. A
business case (Case1)will be taken through all steps of the methodology so
that the methodology and its application can be studied hand in hand.
Anatomy of Graphical method of Linear Programming Model
Structure or anatomy of the graphical method is discussed below in
a standardized form. The methodology of using graphical method is
broadly the same for maximization and minimization but may differ in
specifics. These specifics are clear when cases of maximization and
minimization are discussed separately. Case1 which is a maximization
problem isanalyzed and structured accordingly to facilitate development of
the solution.
3.2 MAXIMIZATION CASE (PROBLEM)
1. Study of the business problem: Now a profit maximization case is
being considered. A business problem calls for a decision by the
concerned manager.Such a decision should lead the business towards
prosperity or profitability. Careful study of the case reveals:
a. Nature of business objective in the case which is profit
maximization
b. Identification of decision variables.

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c. Expected profits or costs per unit of the product in monetary terms
d. Availability of resources in physical quantities and other
constraints like demand for the products in physical quantities in
the market.
The linear programming model is expected to give an optimal
solution for maximizing profit to meet the needs of business management.
A problem may have a large number of feasible solutions which fit within
the constraints but optimal solution is only one, or in some cases may have
some alternate ones. It can be said that an optimal solution is always
feasible, but every feasible solution is not optimal. These concepts are
revisited while discussing the specific case.
2. Formulation of the business problem: Formulation is presenting the
business problem in the language of mathematics. Formulation starts with
defining decision variables, identifying and writing objective
function,business constraints and non-negativity constraint.
a) Decision variables: these are unknown quantities which are to be
decided. They generally represent the amount of outputs produced by
the production system
b) Objective function: Business objective is always to maximize profit
which is a result of minimization of consumption of resources. But in
practice minimization of consumption of resources cannot take place
unchecked as there is always a minimum requirement of resources by
a unit of each product. Profit maximization has a limit posed by the
availability of resources for production. Hence business objective is
always optimality and not unlimited cost minimization or profit
maximization. The business objective can be either cost minimization
or profit maximization in a particular case under various constraints in
business.
The objective function is presented as a mathematical relationship in
the form of an equation, consisting of variables and coefficients. The
variables are unknown quantities of production volumes and
coefficients are per unit profit or cost contributions of individual
products. Development of objective function is dependent on some
important assumptionsnamely additivity, proportionality, divisibility
and non-negativity. Additivity states that the profits earned by
individual products, when produced in quantities suggested by the
optimal solution, are addable to determine total profit. It is an
assumption because the respective quantities suggested by the optimal
solution may not get sold in reality. It is a well-known fact that one
product steals the market of the other. Principle of proportionality
(also known as linearity) states that the total profit is proportional to
the total quantity of the product sold. This is not a reality as the price is

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affected by quantity discount. The assumption of divisibilitystates that
the decision variables may be expressed as divisions of units and not
necessarily as full integers. Assumption of non-negativity is that the
optimal solution gives non negative values for decision variables. In
graphical representation, the decision variables are in the first quadrant
of the graph. While formulating the objective function, these principles
are applied. Without these principles, Linear Programming Model
cannot be structured. In a business problem there may be infinite
number of objective functions as mathematical relationships as
explained earlier. But the business objective is to find a single
objective function which maximizes the profit.
c) Constraints: Constraints refer to the availability of resources, level of
demand for products in market, minimum per unit resource
requirements of individual products and non-negativity restriction on
variables. For the feasibility of the solution the variables are required
to be positive or zero.i) The resource constraints are written for
individual resources as relationships of linear inequality. On the left
hand side (LHS) of the relationship the required quantity of resource to
produce unknown quantities of products is expressed. The coefficients
of variables in the constraint relationships are called technology
constraints.The right hand side (RHS) should reveal the availability or
any other type of limit on the resource. The sign of inequality or
equality between LHS & RHS completes the expression of constraint.
There will be as many constraints as number of resources. There can
also be other constraints like demand constraints for individual
products. Decision or output of the model (which is the values of the
decision variables) should be within the framework of constraints. The
nature of the constraints in a maximization problem is generally “at
most” type. The sign between LHS and RHS is less than type.But
there are times when mixed constraints or constraints with different
signs occur.ii) Constraints on decision variables (non-negativity
constraints): The decision variables are restricted to non-negative
values in optimal solution. In graphical representation, the decision
variables are limited to first quadrant of the graph.
3. The graphical construction: The inequalities between LHS & RHS in
constraints are treated as equalities and constraints are written as equations
of first order and drawn on graph paper as straight lines. The
representation on paper brings visibility to the case. The graphical method
includes two basic steps.
a. Determination of what is known as Feasibility Region (FR) or
Solution Space which is a common feasibility area based on all
constraints.
b. Solution space is formed by a set of constraints which is called a
“convex set”. A set of constraints forming a common area form a
“convex set”, when a straight line joining any two points picked at
random in such an area lies completely within the area.

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c. Identification of optimality and optimal solution from the above
solution space.
a. Determination of Feasibility Region (FR) or solution space:
i. The first quadrant formed by the intersection of X axis and Y axis
is the area in which the construction is performed. Only the first
quadrant satisfies the non-negativity constraint.
ii. All constraints now considered as linear equations must be drawn
as straight lines.
iii. The area in which feasible solutions lie with respect to each
constraint drawn as straight line is marked as per the inequality
sign between LHS and RHS of the constraint.
iv. The area in which each solution is feasible with respect to all
constraints is Feasibility Region (FR) or solution space.
To draw constraint equations as straight lines:
Two points on the straight line are required to be fixed. These
points on the straight line are two feasible solutions to the equation.
Values of the decision variables of each feasible solution serve as
coordinates of the two points on the straight line. Let one of the decision
variables in the equation be equal to zero. Now the equation yields the
value for other decision variable. These two values are the coordinates of a
point lying on the straight line. Similarly, when other decision variable is
put equal to zero, the equation yields the value for the other variable.
These two coordinates fix another point on the straight line. Now the full
straight can be drawn on the graph paper by joining these two points.
Applying the same procedure as many constraints as exist in the
formulation can be drawn as straight lines. To identify feasibility region:
Once the straight lines are drawn on graph paper, the nature of the
constraint is made visible on the graph paper. In the constraint if the LHS
is more than RHS, then area of graph paper on the upper side of the
straight line (away from origin) is feasible. If LHS is less than RHS, the
area on lower side of the straight line (towards origin) is feasible. The
feasible areas for individual constraints are marked on the graph paper.
When all feasible areas are represented, the common area feasible for all
constraints in first quadrant is called feasibility region or solution space.
b. Optimal solution to Maximization problems:
Profit-slope method: As discussed above, FR is identified as the common
area satisfying all existing constraints. The objective function is assigned
an arbitrary positive value. A value easily divisible by coefficients of
decision variables (unit contributions) in the objective function is selected.
Based on the assigned value, two points lying on this straight line are
fixed. When these two points are joined by a dotted line, a straight line is
created which is parallel to itself which is called iso profit line.If the iso
profit line drawn as above is within the feasibility region, move this
straight line out of the FR away from the origin without changing its slope.

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For this purpose set squares may be used. The last point of contact with
the FR before exiting the FR is the optimal point. If the iso profit line
drawn as above is outside the FR, move this straight line towards the FR
without changing its slope. For this purpose set squares may be used. The
first point of contact with the FR is the optimal point.The coordinates of
this contact point give optimal solution.
Corner-points method (vertices method): once the feasibility region or
FR is identified, the optimal solution from a number of feasible solutions
within the FR is to be separated. The corner points or vertices (extreme
points) of the FR are denoted with convenient notations and coordinates of
every corner point (extreme point) are decided by solving simultaneous
equations. The corner points (extreme points) or vertices represent feasible
solutions at the high points of the FR as the objective function is
maximization type. The solution values of the feasible solutions at corner
points are put in the objective function written earlier. Whichever feasible
solution gives the best value (highest value) of the objective function is the
optimal solution.
Following is a business case taken up for step by step discussion
Case1.A firm produces products, A & B, each of which requires two
resources, namely raw materials and labour. Each unit of product A
requires 2 & 4 units and each unit of product B requires 3 & 3 units
respectively of raw materials and labour. Everyday 60 units of raw
materials and 96 units of labour are available. If the unit profit
contribution of product A is Rs.40/-, product B is Rs.35/- determine the
number of units of each of the products that should be made each day to
maximize the total profit contribution.
1. Study of the business problem: Case1 is analyzed as per above
guidelines and following facts are established:
a. This is a case of profit maximization. An optimal solution
maximizing profits is contemplated.
b. Company is considering producing products A & B.
c. Expected profit contribution by a unit of A is Rs 40/- and by a unit
of B is Rs 35/-
d. Availability of raw materials is 60 units per day and availability of
labour is only 96 units.
2. Formulation of the business problem:
a) The decision variables: Let an unknown quantity X1 units of A and
an unknown quantity X2 units of B be produced per day. X1&X2 are
the decision variables.

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b) The objective function for Case1 is written as below.
The model will finally give the values of X1 and X2 for maximum
profit at optimum level. Contribution per unit for A & B are Rs 40/-
and Rs 35/-. Then the objective is expressed in mathematical form as:
40X1 + 35X2 = Z (Z is the total profit per day) which is to be
maximized
c) Constraints for Case1 are written below:
The study of the case earlier has already revealed the required factors
for writing constraints. The constraint relationships are expressed
below:
i) Resource constraints:
2X1 + 3X2  60………..(1) Raw Materials Constraint
4X1 + 3X2  96………..(2) Labour Constraint
ii) Constraint on decision variables:
X1, X2 ≥ 0 Non negativity constraint
3. Graphical construction of constraints in problem in Case1:
Constraint relationships are expressed below as equations:
2X1 + 3X2 = 60………(1) Raw Materials Constraint
4X1 + 3X2 = 96……. (2) Labour Constraint
In constraint (1):
Put X1=0, then X2=20, now point A (0, 20) gives a feasible solution.
When X2=0, then X1=30, now point B (30, 0) gives another feasible
solution.
Points A& B are fixed on the 1st
quadrant of graph paper, and joined by a
straight line
See figure 3.1

20
In constraint (2):
Put X1=0, then X2=32, now point C (0, 32) gives a feasible solution.
When X2=0, then X1=24 now point D, (24, 0) gives another feasible
solution.
Points C & D are fixed on the 1st
straight line
See figure 3.2

21
Feasibility Region (Solution Space): In the above figure dotted area is
feasibility region marked by the high points O, A, E, D
Optimal solution for the problem in Case1 as per profit slope method is
shown in Fig 3.3 below:
Optimal solution for maximization problem in Case1 as per corner point
method is shown in table below. Optimal solution is X1 = 18, X2= 8,
maximum profit is Rs1000/-
HIGH
POINT ON
FR
X1 X2 40X1 + 35X2
= Z
MAX PROFIT
(OPTIMAL
SOLUTION)
O 0 0 0
A 0 20 700
E 18 8 1000 1000
D 24 0 960
Table 3.1

22
Maximization case (problem) with mixed constraints which form a
closed common area
Maximization problems with mixed constraints occur when the
nature of one or some of the constraints (not all) is more than type. Such
problems are solved by following the same methodology. Case1a below is
a maximization case (problem) with mixed constraints. Detailed
explanation of each step is not attempted while solving this case.
Important point to note is that a common closed area satisfying all
constraints is required.
Case1a.A firm produces products, A & B, each of which requires two
respectively of raw materials and labour. Every day at least 60 units of raw
materials and at most 96 units of labour must be used.If the unit profit
Objective function: 40X1 + 35X2 = Z (Z is the total profit per day) which
is to be maximized
Constraints for Case1a are written below:
The study of the case reveals the required factors for writing
constraints. The constraint relationships are expressed below:
2X1 + 3X2 ≥ 60………..(1) Raw Materials Constraint
X1 , X2 ≥ 0 Non negativity constraint
Above constraints are treated as equations to ploton graph paper.
Taking up constraints one by one,
In constraint (1):
solution.
straight line.
In constraint (2):
solution.

23
straight line
Graphical construction is done and optimal solution in the feasibility
region is shown below in figure F5 as per profit slope method. Optimal
solution is X1 = 0, X2= 32, maximum profit is Rs1120/-. Solution by
Corner point method can be worked out as per methodology followed
earlier.
Fig. 3.4
3.3 MINIMIZATION CASE (PROBLEM)
To understand analysis of minimization problem, Case2 is
considered below. As discussed earlier, all the steps leading to the
identification of FR are similar. FR is identified as the common area
satisfying all existing constraints. The nature of the constraints in a
minimization problem is generally “at least” type. The sign between LHS
and RHS is more than type. But there are times when mixed constraints
or constraints with different signs occur.

24
Optimal solution to Minimization problems:
Cost-slope method
A straight line parallel to the objective function is setup which is
called iso cost line.If the iso cost line drawn as above is within the
feasibility region, move this straight line out of the FR towards the origin
without changing its slope. For this purpose set squares may be used. The
last point of contact with the FR before exiting the FR is the optimal point.
If the iso cost line drawn as above is outside the FR, move this straight
line towards the FR without changing its slope. For this purpose set
squares may be used. The first point of contact with the FR is the optimal
point. The coordinates of this contact point give optimal solution.
Corner-points (vertices) method
The corner points or vertices (extreme points) of the FR are
denoted with convenient notations and coordinates of every corner point
(extreme point) are decided by solving simultaneous equations. Each
corner point is a feasible solution at low points of FR. Value of the
objective function for each solution is calculated and tabulated. The lowest
value of the objective function indicates optimal solution.
Case2.A firm produces products, A & B, each of which requires two
materials and at least 96 units of labour are to be used. If unit production
cost of product A is Rs.40/- and product B is Rs.35/- determine the
minimize the total cost of production.
The objectives function for Case2 is written as below.
Let an unknown quantity X1 units of A and an unknown quantity X2 units
of B be produced. The model will finally give the values of X1 and X2 for
minimum cost at optimum level. X1 and X2 are the decision variables.
Cost contribution per unit for A & B are Rs 40/- and Rs 35/-. Then the
objective is expressed in mathematical form as:
40X1 + 35X2 = G (G is the total cost per day) which is to be minimized
Graphical construction of constraints in problem in Case2:
Constraint relationships are expressed below:
2X1 + 3X2 ≥ 60………(1) Raw Materials Constraint
4X1 + 3X2 ≥ 96……. (2) Labour Constraint
The inequalities between LHS & RHS in constraints are treated as
equalities and constraints are written as equations of first order and
drawn on graph paper as straight lines.

25
In constraint (1):
solution.
straight line as shown in fig 3.5
In constraint (2):
solution.
straight line as shown in fig 3.6

26
Fig 3.6
Feasibility Region (Solution Space): In the above figure dotted area is
feasibility region marked by the low points C,E,B
Optimal solution for cost minimization problem in Case2 as per cost slope
method is shown in fig 3.7 below: Optimal solution is X1 = 18, X2= 8,
minimum cost is Rs1000/-

27
Fig. 3.7
Optimal solution for minimization problem in Case2 as per corner points
or vertices method is shown in fig Table 3.2 below:
HIGH
POINT ON
FR
X1 X2 40X1 + 35X2
= Z
OPTIMAL
SOLUTION
C 0 32 1120
E 18 8 1000 1000
B 30 0 1200
Table 3.2
Minimization case (problem) with mixed constraints
Minimization problems with mixed constraints occur when the
nature of one or some of the constraints (not all) is less than type. Such
problems are solved by following the same methodology. Case1b below is
a minimization case (problem) with mixed constraints. Detailed
explanation of each step is not attempted while solving this case.
Case2a.A firm produces products, A & B, each of which requires two

28
respectively of raw materials and labour. Everyday at least 60 units of raw
materials and at most 96 units of labour are to be used. If unit production
cost of product A is Rs.40/- and product B is Rs.35/- determine the
minimize the total cost of production.
The objectives function for Case2b is written as below.
Let an unknown quantity X1 units of A and an unknown quantity X2 units
of B be produced. The model will finally give the values of X1 and X2 for
minimum cost at optimum level. X1 and X2 are the decision variables.
Cost contribution per unit for A & B are Rs 40/- and Rs 35/-. Then the
40X1 + 35X2 = G (G is the total cost per day) which is to be minimized
Graphical construction of constraints
Constraint relationships are expressed below:
2X1 + 3X2 ≥ 60………(1) Raw Materials Constraint (at least
constraint)
4X1 + 3X2 ≤ 96……. (2) Labour Constraint (at most constraint)
The inequalities between LHS & RHS in constraints are treated as
equalities and constraints are written as equations of first order and drawn
on graph paper as straight lines.
In constraint (1):
solution.
straight line.
In constraint (2):
solution.
straight line.
Graphical construction is done and optimal solution in the feasibility
region is shown below as per cost slope method. Optimal solution is X1 =
0, X2= 20, minimum cost is Rs700/-. Solution by Corner point method can
be worked out as per methodology followed earlier.

29
Fig. 3.8
3.4 SENSITIVITY ANALYSIS OF GRAPHICAL
SOLUTIONS
Sensitivity analysis is testing the optimal solution to see within what
range of changes in the environment it remains useful. The changes in the
environment bring pressure on the basis of the optimal solution and
solution values to take up new values. If the basis changes the optimal
solution no longer is the same. If the basis remains the same and only the
solution values change, the optimal solution holds good. The sensitivity
analysis which is being discussed here considers following changes:
a. changes in the unit contributions(changes in coefficients of decision
variables in the objective function )
b. changes in limits on the resources
When the above changes take place in the market after the optimal
solution is established, sensitivity analysis (also known as post optimality
analysis is done) is performed.

30
Changes in the unit contributions (coefficients of decision variables in
the objective function, Cj)
Locate the optimal point which is an intersection of any two
constraints in the formulation. The straight line representing objective
function passes through this intersection. It is seen that when the values of
coefficients in objective function change the slope of the objective
function changes. The change in the slope is limited by the intersecting
constraints. The limit through which these values change determines
optimality.
The figure below shows graphical construction of the
maximization problem discussed earlier. Point E is the point at which
optimality occurs as shown by the objective function. The objective
function exits the FR at E. It can be seen that if the objective function
changes its slope at E through a specific the optimal point remains
unchanged. The angle through which the objective function can swivel is
limited by the constraints straight lines (1) and (2). The current slope of
objective function is determined by the coefficients. This slope is limited
by the slopes of constraints (1) and (2) The mathematical explanation is
given below:
General structure of objective function is C1X1+C2X2 = Z, where
C1 and C2 coefficients representing contributions of variables X1 and X2.
The constraints also are structured similarly with their technology
constraints and variables on LHS and resource limits on RHS
C1/ C2is the slope of objective function, limits of which are determined by
the slopes of constraints in the formulation. The Case1which is solved
earlier is being discussed again for sensitivity analysis
40X1 + 35X2 = Z (Z is the total profit per day) which is to be maximized
2X1 + 3X2  60………(1) Raw Materials Constraint
4X1 + 3X2  96……. (2) Labour Constraint
X1 and X2 ≥ 0

31
Fig. 3.9
Slope of the objective function for Case1, C1/ C2 = 40/35,
Slope of the constraint (1) = 2/3, Slope of the constraint (2) = 4/3
Lower (smaller) slope ≤ C1/ C2 ≤Upper (larger) slope
Hence, (2/3) ≤ (C1/C2)≤(4/3) (I)……………. C2 ≠ 0
Similarly, (3/2)≥(C2/C1)≥(3/4) (II) ……………..C1 ≠ 0
When C2=35, in (I) above, (2/3) ≤ (C1/35)≤(4/3) therefore, (2/3) X35 ≤
C1)≤(4/3) X35
Hence, (70/3) ≤ C1 ≤(140/3)
When C1=40, in (II) above, (3/2) ≥ (C2/40)≥(3/4) therefore, (3/2) X 40 ≥
(C2)≥ (3/4) X40
Hence, 60 ≥ C2 ≥30
When C1 or C2 assume zero value, the objective function becomes either
vertical or horizontal
From the above relationship (I), if C2 is 35, value of C1 can vary between
Rs (70/3)/unit and Rs (140/3)/unit

32
From the above relationship (II), if C1 is 40, value of C2 can vary between
Rs30/unit and Rs60/unit
Changes in the availability of resources
Fig 3.10
The optimum point E is the intersection of equations (1) & (2). As
the availability of resource R/M reduces, the intersection point E starts
moving along the line segment CD towards D. The figure fig. 3.10 shows
that the limit for raw material availability is between 48 units and 96 units.
When the R/M availability varies between these limits, the point of
intersection E,slides between C & D along the line segment CD. Within
these limits of resource availability the point of intersection of (1) & (2)
continues to give a solution with maximum profit. When the point of
intersection ‘E’ between the constraints (1) & (2), lies anywhere between
E - C or E - D the current basis is OK. While the product mix remains the
same, quantities change and value of the objective function changes.
Similarly, the other resource, labour is considered in fig. 3.11. When the
labour availability varies between 60 units and 120 units the point of
intersection E, moves along the line segment AB. This means for a change
in labour resource between 60 units to 120 units, the point of intersection
between (1) & (2) continues to provide a solution of maximum profit.

33
When the point of intersection ‘E’ between constraints (1) & (2), lies
anywhere between E & D current basis is OK. While the product mix
remains the same, quantities change and value of the objective function
changes.
Fig. 3.11
Unit worth or shadow price or marginal profitability of Resource1,
R/M:refer figure fig. 3.10, when availability of this resource is maximum
(R/M is 96 units) the product mix is (X1=0, X2=32) and the maximum
profit is Rs.1120/-. When R/M is 48 units, the lower limit, the product mix
is the (X1=24,X2=0) maximum profit is Rs.960/-. For resource R/M, for a
change of 48 units, profit change is Rs.160/-. Hence it can be said that the
unit worth of R/M is Rs 3.33.
Calculations are shown below:
(Change in profit)/(change in resource quantity) = (1120 – 960)/(96 – 48)
= (160/48) = 3.33

34
Unit worth of Resource2, Labour: refer figure fig. 3.11, when
availability of this resource is maximum (Labour is 120 units), the product
mix is (X1=30, X2=0) and the maximum profit is Rs.1200/-. When Labour
is 60 units, the lower limit, the product mix is (X1=0, X2=20) and the
maximum profit is Rs.700/-. For a change in resource units of 60 units, the
profit change is Rs.500/-. Hence it can be said that the unit worth of
Labour is Rs 8.33.
Calculations are shown below:
(Change in profit)/(change in resource quantity) = (1200 – 700)/(120 – 60)
= (500/60) = 8.33
3.5 SPECIAL CASES IN GRAPHICAL SOLUTIONS
Special cases in graphical solutions are some special type of
solutions. Some of them may be optimal and some of them may not be
optimal. Such solution communicates special information to the user.
1. Alternate optima are alternate optimal solutions. An optimal
solution which doesn’t have an alternate is a unique optimal solution.
Alternate optima can occur in both maximization and minimization cases
Corner point method: An FR, in which more than one decision points
(high or low) give same optimal solution, has alternate optima. Profit/cost
slope method: When the line drawn parallel to the objective function, runs
parallel to one of the binding constraints, alternate optima exists. Binding
constraint is one of the constraints forming feasibility region. The optimal
solution to the problem in Case1 is a unique solution.. Case3 with an
alternate solution is shown in fig. 3.12

35
Fig. 3.12
2. Unbounded solutions do not offer finite positive values for decision
variables. FR without any boundary on one side, or open at one end is
unbounded. If the objective function is maximization, the FR is
unbounded in maximization direction (away from origin). If the objective
function is minimization, the FR is unbounded in minimization direction
(towards origin). An unbounded solution is shown in fig. 3.13.It can be
seen that a maximization problem with all at-least type constraints is
shown as example in fig. 3.13. But the learner should know that a
minimization problem with at-most constraints also becomes unbounded.

36
Fig. 3.13
3. Infeasibility
A solution which cannot satisfy all constraints is an infeasible
solution. The FR cannot be identified as constraints do not form a
common area in the first quadrant of the graph. This leads to the condition
of infeasibility. Infeasible solution does not provide solution which can be
implemented. An infeasible solution is shown in fig. 3.14. The Case5-1
also depicts infeasibility where a common area satisfying also constraints
doesn’t exist. This is shown in Fig. 3.14.1

38
4. Redundancy
When one of the constraints remains outside the FR without
participating in its formation redundancy is said to be existing and the
constraint outside the FR is called redundant constraint. The redundant
constraint doesn’t affect the optimal solution. A solution with redundancy
is shown in fig. 3.15. In the case 7 explained below constraint (3) is
redundant because even if this is removed, feasibility region doesn’t
change.
Fig. 3.15
5. Degeneracy
When the FR is formed, one of the constraints makes a point
contact and thereafter remains outside the FR. In a situation of this kind
one of the decision variables assumes zero value. The optimal solution
provides positive value for only one decision variable. A degenerate is
shown in fig. 3.16.

40
3.6 BIBLIOGRAPHY
1. Jhamb L.C., Quantitative Techniques for Managerial Decisions Vol
I, Pune, Everest Publishing House, 2004
2. Vohra, N. D., Quantitative Techiques in Management, Third
Edition, New Delhi, Tata McGraw-Hill Education Private Limited,
2007
3. Taha Hamdy, Operations Research, Sixth Edition, New Delhi,
Prentice Hall of India, 2002
3.7 UNIT END EXERCISES
1. A production process for products A and B can produce A by using 2
units of chemicals and one unit of a compound and can produce B by
using 1 unit of chemicals and 2 units of compound. Only 800 units of
chemicals and 1000 units of the compound are available. The profits
available per unit of A and B are respectively Rs.30/- and Rs.20/-.
a. Find the optimal solution to the problem using graphical method.
b. Discuss the sensitivity of the above solution to changes in the
profit contributions of the products and availability of resources
2. In a carpentry shop it was found that 100 sq feet of ply wood scrap and
80 square feet of white pine scrap are in usable form for construction
of tables and book-cases. It takes 16 sq feet of plywood and 8 sq feet
of white pine to make a table. 12 sq feet of plywood and 16 sq feet of
white pine are required to construct a book-case. A profit of Rs.25 on
each table and Rs.20 on each book case can be realized. How can the
left over wood be most profitably used? Use graphical method to solve
the problem.
a. Find the optimal solution to the problem using graphical method.
b. Discuss the sensitivity of the above solution to changes in the profit
contributions of the products and availability of resources.
3. M/S BM Electric co. produces two types of electric dryers A and B
that are produced and sold on weekly basis. The weekly production
quantity cannot exceed 25 units for A and 35 for B due to limited
resources. The company employs 60 workers. Dryer A requires two
man-weeks of labor whereas B requires only one man-hr. Profit earned
by A is Rs.60/-and B is Rs.40/-. Formulate the above as a linear
programming problem and graphically solve for maximum profit.
4. A manufacturer produces two different models X & Y of the same
product. Model X makes a contribution of Rs.50/- per unit and model
Y, Rs.30/- per unit towards total profit. Raw materials R1 & R2 are
required for production. At least 18 kg of R1 and at least 12 kg of R2
must be used daily. Also at most 34 hours of labour are to be utilized.
A quantity of 2 kg of R1 is required for X and 1 kg of R1 is required for

41
Y. For each of X & Y, 1 kg of R2 is required. It takes 3 hrs to
manufacture X and 2 hours to manufacture Y. How many units of each
model should be produced to maximize the profit?
5. A company manufactures two types of belts A & B. The profits are
Rs. 0.40/- &Rs. 0.30/- per belt respectively. Time required for
manufacturing A is twice the time required for B. If the company were
to manufacture only B type of belts they could make 1000 units per
day. Leather available for production is only worth 800 belts per day
(A & B combined). Belt A requires a fancy buckle availability of
which is restricted to 400 units per day. Buckles required for B are
available only 700 per day. What should the daily production plan be
for belts A & B? Formulate the above problem as LPP and solve by
graphical method.
6. An advertising agency wishes to reach two types of audiences,
customers with monthly incomes greater than Rs 15,000/- (target
audience A) and customers with monthly incomes less than Rs
15,000/- (target audience B). The total advertising budget is
Rs.2, 00, 000 /-. One programme of TV advertising costs Rs. 50, 000/-.
One programme of radio advertising costs Rs.20, 000/-. For contract
reasons at least 3 programmes ought to be on TV and number of radio
programmes must be limited to 5. A survey indicates that a single TV
programme reaches 4, 50, 000 customers in target audience A and 50,
000 in target audience B. One radio programme reaches 20, 000
customers in target audience A and 80,000 in target audience B.
Determine the media mix to maximize the total reach.
7. A dealer wishes to purchase a number fans and sewing machines. he
has only Rs.5760/- to invest and has space utmost for 20 items. A fan
costs him Rs 360 and a sewing machine Rs 240. His expectation is that
he can sell a fan at a profit of Rs 22 and sewing machine at a profit of
Rs 18. Assuming that he can sell all the items that he can buy, how
should he invest his money in order to maximize his profit? Formulate
this problem as linear programming problem and then use graphical
method to solve it.
8. The inmates of an institution are daily fed two food items A & B. To
maintain their health the nutritional requirements for each individual
and the nutrient content of each food item are given. The problem is to
determine the combination of units of A & B per person at minimum
cost. When per unit cost of A & B is given as below

42
FOOD A FOOD B MINIMUM
DAILY
REQUIREMENT
CALCIUM 10 UNITS 4 UNITS 20 UNITS
PROTEIN 5 UNITS 5 UNITS 20 UNITS
CALORIES 2 UNITS 6 UNITS 12 UNITS
PRICE IN
RUPEES
0.6 1.00
9. Ashok Chemicals Company Manufactures two chemicals A & B
which are sold to the manufacturers of soaps & detergents. On the
basis of the next month’s demand the management has decided that the
total production for chemicals A & B should be 350 kilograms.
Moreover a major customer’s order of 125 kilograms for product A
also must be supplied. Product A requires 2 hrs of processing time per
kilogram and product B requires one hr of processing time per
kilogram. For the coming month 600 hrs of processing time is
available. The company wants to meet the above requirements at
minimum total production cost. The production costs are Rs2/- per
kilogram for product A and Rs3/- for B. Ashok chemicals company
wants to determine its optimal product mix and the total minimum cost
relevant to the above. Formulate the above as a linear programming
problem. Solve the problem with graphical method. Does the problem
have multiple optimal solutions?


43
4
SIMPLEX METHOD
Unit structure :
4.0 Objectives
4.1 Introduction
4.2 Maximization case (problem) with all “At-Most” Type of
Constraints
4.3 Minimization case (problem) - “Big M” Method
4.4 Minimization Case (Problem) - “Two Phases” Method
4.5 Sensitivity analysis of optimal solution
4.6 Study of special cases of simplex solution
4.7 Bibliography
4.0 OBJECTIVES
 Express the business problem in the language of mathematics in
standard form
 Use simplex method to find optimal solution to business problems
 Use “Big M” method and “two phase” method to solve
minimization problems
 Interpret the optimal solution which is in mathematical form for
decision making
 Do sensitivity analysis of the optimal solution to determine its
utility in the light of changes in the problem environment
 Understand and interpret special cases of optimal solution
4.1 INTRODUCTION
In the previous Section the learner is already introduced to linear
programming as a mathematical model for finding solutions to business
problems. Graphical method studied earlier in Chapter#3 is limited to
business problems with only two decision variables. Simplex algorithm

44
overcomes this problem by offering a more versatile method to solve
business problems. Prof George Danzig, Prof von Newman and Prof
Kantorovich are said to be the three founders of simplex algorithm. Their
researches in the area of mathematics during World War II largely
contributed to the development of simplex algorithm or method. This
algorithm is one of the most popular ones used in finding solutions to
business problems. Simplex algorithm can deal with maximization and
minimization problems under constraints of resources and demand. The
assumptions of additivity, proportionality, divisibility and non-
negativity discussed in section #3 hold good for developing an optimal
solution using simplex method.
The methodology follows the same route as graphical method for
structuring a business problem in mathematical form.
Anatomy of the Simplex Method
Structure or anatomy of the simplex method is discussed below in
a standardized form. The methodology of using simplex method is broadly
the same for maximization and minimization but may differ in specifics.
These specifics are clear when cases of maximization and minimization
are discussed separately. A specific business case which is a maximization
problem is analyzed and structured accordingly to facilitate development
of the solution.
4.2 MAXIMIZATION CASE (PROBLEM) WITH ALL
“AT-MOST” TYPE OF CONSTRAINTS
The linear programming model is expected to give an optimal
solution for maximizing profit to meet the needs of business management.
A problem may have a large number of feasible solutions which fit within
the constraints but optimal solution is only one, or in some cases may have
some alternate ones. It can be said that an optimal solution is always
feasible, but every feasible solution is not optimal. These concepts are
revisited while discussing a specific business case later. A business case
with similar constraints which are of less than or equal to type ( ) is
considered here. This type of constraints is also known as “at-most” type.
Problems with mixed constraints (“at-most”, “at-least” & “equal to”) are
discussed later. When the availability of a resource is limited to a quantity
and it is not available beyond that quantity, the resource constraint is “at-
most” type.
1.Study of the business case: it is same way as it was done in graphical
method. Following specific characteristics of the business case are
identified.

45
a. Nature of business objective in the business problem now being
considered is profit maximization.
b. Identification of decision variables. Decision variables are unknown
quantities to be decided for maximization of profits
c. Expected profits per unit of the product in monetary terms
d. Availability of resources in physical quantities and other constraints
like demand for the products in physical quantities in the market.
2.Formulation: as it was done in graphical method. The details are not
discussed once again since this is already covered in previous chapter.
Problem is expressed as a set of mathematical equations. The set of
equations formulated in terms of decision variables identified earlier
consist of
 objective function
 resource constraints
 non-negativity constraint
3. Standard form: This is a special requirement of simplex method. After
formulation the step is called standard form where the inequalities in the
constraints are converted to equalities. When the LHS is smaller than
RHS, as in an at-most constraint, a slack variable is added to the LHS to
establish equality between LHS and RHS. Value of the slack variable is
unknown so long as the values of decision variables are unknown. Initially
such cases where LHS is less than RHS are considered.
The constraint equations are rewritten in standard form, so also the
objective function. The non-negativity constraint will include all the
variables in the objective function.
It may be noted that in each equation all variables are required to
be represented. If a variable is absent it should have zero as coefficient.
4. Starting solution: Simplex method needs a feasible starting solution
which is improved in a series of steps called iterations to reach optimality.
A feasible starting solution is found by equating all non-slack variables to
zero in constraint equations. This is same as not starting production
activity in the work place. Then the slack variables assume the values of
balance resources in the receiving stores. These values of slack variables,
when all other variables are zero form the starting solution. The starting
solution is put in the first simplex table.

46
5. Structuring of Simplex table
Simplex table has a standardized structure which is given below.
1. Number of columns = number of variables in the standard form.
2. Number of rows = number of constraints in the formulation.
3. Each column has a variable as header. The variables as headers should
appear in the same sequence as they appear in the objective function in the
standard form.
4. Above each variable its contribution is written.
5. Each row represents a constraint in the formulation.
6. In the first row, coefficients of the variables in first constraint, in the
same order as they appear in the objective function are entered. This is
repeated for all rows representing constraints.
7. From the above table where all coefficient values are filled in, a special
formation is detected which is known as "identity matrix". This matrix is
formed by some variables located as headers of the columns of the
simplex tables. In this matrix, if a variable has 1 as coefficient in row, in
other rows its coefficient is zero. Similarly, another variable has 1 as
coefficient in another row and zero in others. The variables forming the
identity matrix are identified as "basic variables". Basic variables form
the basis in the simplex table. These variables are headers of the rows.
The variable which has 1 as coefficient in first row is the header of first
row and so on. In the column adjacent to the one where basic variables are
written, the respective contributions of basic variables are written. The
contributions and the respective basic variables form the basis in the
simplex table. The basic variables form the output of the simplex model.
Their values known as solution values are listed in a column called
solution values column which is next to the column headed by the last
variable in the objective function. The first simplex table consists of the
starting solution found earlier. It can be seen that basic variables are the
ones which give starting solution. Write the values of the basic variables
in the starting solution in solution values column.
Every solution offered by simplex must be tested for optimality.
6. Calculation of Z row values: the Z values are written for every column
in the simplex table in the Z row which is just below the row of the last
basic variable. These values are the contributions of individual variables
in a particular solution. This is calculated as sum of the products of the
coefficient of individual variable in a particular row and the contribution
of basic variable of that particular row. This may be understood by
looking at the analysis of Case1 below.
7. Optimality test: The index row is written below the Z row and (Cj-Zj)
values are entered in the index row. When the index row (Net Evaluation

47
Row or NER) values, which are calculated as (Cj-Zj), are non-positive the
solution is optimal.
If solution is optimal, the process stops and solution offered by the
simplex table which can be read in bi column is put to use. If the solution
is not optimal, the simplex method improves the solution as described
below.
8. Simplex Iterations: Each step of improvement is called an iteration of
simplex method. Each of the iterations is represented in a simplex table.
The iterations continue until optimality is reached.
a. Identify most positive NER value and mark the column in which
this value is present as key column. In the event of a tie, the NER
value towards LHS (towards the basis) is selected. For every basic
variable write (bi/ai) ratio in a new column next to (bi) column.
The (ai) value is the coefficient in the key column for every basic
variable and the (bi) value is the solution value of for every basic
variable. These ratios are called replacement ratios or exit ratios.
The least non-negative ratio is selected and its row is called key
row. The intersection of key column and key row gives the key
element.
b. Set up a new table as described earlier. Make the change in the
basis of the table. Remove the variable with least non negative
minimum ratio and in its place bring in the variable indicated by
key column. Other variables in the basis remain unchanged.
c. The new values of the row in which basic variable is changed are
written first. This is done by dividing each value in the cells in the
row by the key element.
d. the values in the cells of other rows are written by using the
formula given below:
 New value for a cell in a particular row rj = value in the same
cell in the previous table – value in the intersection of key
column and the particular row rj in previous table X value in
the corresponding cell in the row where basic variable has been
changed in the new table.
 A sample calculation can be seen in the discussion of Case1
below.

48
Case1: A firm produces products A, B & C each of which passes through
three departments Fabrication, Finishing & Packaging. Each unit of
product A requires 3, 4 & 2, a unit of product B requires 5, 4 & 4 while
each unit of C requires 2, 4 & 5 hours respectively in the three
departments. Everyday 60 hours are available in Fabrication, 72 hours are
available in Finishing and 100 hours in Packaging department. If the unit
contribution of product A is Rs.5/-, product B is Rs.10/-, product C is
Rs.8/-, determine the number of units of each of the products, that should
be made each day to maximize the total contribution.
1.Study of the business case: Given Case1 is analyzed as per the
e. This is a case of profit maximization. An optimal solution
f. Company is considering producing products A, B & C.
g. Expected profit contribution by a unit of A is Rs.5/-, by a unit of B
is Rs.10/- and by a unit of C is Rs. 8/-
h. Resource requirements of each product in units, decision variables
and resources availability in appropriate units are tabulated below:
Products Fabrication
hrs
Finishing
hrs
Packaging
hrs
Decision
variables
(unknown
production
quantities
in units)
Expected
profit
contribution
in Rs/unit
A 3 4 5 X1 5
B 5 4 4 X2 10
C 2 4 5 X3 8
Availability
of resources
per day in
units
60 72 100
2.Formulation: Let unknown quantity X1 units of A, unknown quantity
X2 units of B and an unknown quantity X3 units of C be produced. The
model will finally give the values of decision variables X1, X2 & X3 for
maximum profit at optimum level. Contribution per unit for A, B & C are
Rs 5/-, Rs 10/- & Rs 8/-. Then the objective is expressed in mathematical
form as:

49
Decision variables: Unknown production quantities X1, X2 and X3 are
decision variables.
The objective function: The objective function for Case1 is written as
below.
5X1 + 10X2 + 8X3= Z (Z is the total profit per day) which is to be
maximized
Constraints: The study of the case earlier has already revealed the
required factors for writing constraints. The constraint relationships are
expressed below:
3X1 + 4X2 +5X3  60……….. (1) Fabrication Constraint
5X1 + 4X2 +4X3 72…..……. (2)Finishing Constraint
2X1 + 4X2 +5X3 100……….. (3)Packaging Constraint
X1 , X2, X3 ≥ 0………………..Non negativity constraint
3. Standard form: The mathematical formulation suitable for simplex
method is written in terms of decision variables, slack variables and their
coefficients. While X1, X2, X3 are decision variables S1, S2 & S3 are slack
variables introduced to establish equality between LHS & RHS
Objective function: 5X1 + 10X2 + 8X3+0S1 +0S2+0S3= Z (Z is the total
profit per day) which is to be maximized.
Constraints:
3X1 + 4X2 +5X3+S1 +0S2+0S3 = 60……….. (1) Fabrication Constraint
5X1 + 4X2 +4X3+ 0S1 +S2+0S3 = 72…..……(2)Finishing Constraint
2X1 + 4X2 +5X3+ 0S1 +0S2+S3 = 100………(3)Packaging Constraint
X1, X2, X3, S1 , S2, S3 ≥ 0…………………Non negativity constraint
4. Starting solution: From the above standard form, a feasible starting
solution is written as follows. Putting X1 = 0, X2= 0, X3= 0, Hence, S1= 60,
S2=72, S3=100
From the information collected, following simplex table ST-1 is
constructed:

50
Contributions >>> 5 10 8 0 0 0
Contribution
of basic
variables
X
basic
variables
X1 X2 X3 S1 S2 S3 bi
Solutio
n values
0 S1 3 4 5 1 0 0 60
0 S2 5 4 4 0 1 0 72
0 S3 2 4 5 0 0 1 100
Z 0 0 0 0 0 0
 = C - Z 5 10 8 0 0 0
 The identity matrix can be seen as marked by the shaded area.
 The variables forming the identity matrix are put in the basis.
6. Calculation of Z row values:
Calculations for the Z value for column of X1:
a) Coefficient of X1 in the first row of S1=3
b) Contribution of basic variable in first row,S1=0
c) Product of the above two values in first row is 0
d) Similarly the products for second and third rows are also 0
e) Hence Z value for first column is 0
f) Z values for other variables are similarly calculated
7. Optimality test:
 The table shows that the index row (Net Evaluation Row or NER)
values are not non-positive.
 Hence the conclusion is that the solution is not optimal but needs
further improvement.
8. Simplex Iterations:
The table below shows key column, key row, and key element,
information required to perform iteration and move towards optimality.

51
ST-1
C 5 10 8 0 0 0
C
contribution
of basic
variables
X
basic
variables
Solution
values
bi/ai
Replacement
ratio or
minimum
ratio θ
0 S1 3 4 5 1 0 0 60 15
0 S2 5 4 4 0 1 0 72 18
0 S3 2 4 5 0 0 1 100 25
Z 0 0 0 0 0 0
 = c - z 5 10 8 0 0 0
Using the key information from the solution in ST1, the solution is further
improved and written in ST2. The calculations for a new cell value are
shown below
Sample calculation for cell S2-X1:
New value for a cell (S2-X1) in a particular row rj (row of S2) = value in
the same cell of previous table (5) – value in the intersection of key
column and the particular row rj in previous table (4) X value in the
corresponding cell in the row where basic variable has been changed in the
new table (3/4).
New value for (S2-X1) = [5 – 4 X (3/4)] = 2
ST-2
C 5 10 8 0 0 0
Contribution
of basic
variables
X
basic
variables
Solution
values
10 X2 3/4 1 5/4 1/4 0 0 15
0 S2 2 0 -1 -1 1 0 12
0 S3 -1 0 0 -1 0 1 40
Z 15/2 10 25/2 5/2 0 0
 = C - Z -5/2 0 -9/2 -5/2 0 0
It is found that all NER values are non-positive hence the conclusion is
that the solution now is optimal.

52
The values of the basic variables which form the optimal solution are
given in solution values column.
Learner may put the values of decision variables in the objective function
to calculate the maximum profit as per the optimal solution.
Maximization case (problem) with mixed constraints
A constraint with LHS smaller than RHS (at-most constraint) is
put in standard form by using slack variables. But when a constraint with
LHS larger than RHS (at-least constraint) appears in the formulation a
new variable called surplus variable is used. A surplus variable is
removed from LHS to make it equal to the RHS. In the standard form of
this type, there is no natural positive slack variable. But the simplex
algorithm needs a slack variable in the standard form of every constraint in
the formulation. To satisfy this requirement, a new variable called
artificial variable is added to the LHS. This artificial variable, introduced
to put the problem in standard form, is removed from the simplex
iterations before the optimality is reached.
To enable removal of artificial variable, it is given a very high
value “M” as coefficient. In the objective function of a maximization
problem, this coefficient is “negative M” (or –M). The simplex
methodology in the process of maximizing the objective function, removes
the artificial variable associated with –M from simplex system.
A business problem is taken through simplex methodology discussed
earlier.
Case2. A firm produces products, A & B, each of which requires two
materials and at most 96 units of labour must be used .If the unit profit
i. This is a case of profit maximization. An optimal solution
j. Company is considering producing products A & B.

53
k. Expected profit contribution by a unit of A is Rs.40/-, by a unit of
B is Rs.35/-
l. Resource requirements of each product in units, decision variables
and resources availability in appropriate units are tabulated below:
Products Raw
materials
Labour Decision variables
(unknown
production
quantities in units)
Expected profit
contribution in
Rs/unit
A 2 4 X1 40
B 3 3 X2 35
Availability
of resources
per day in
units
60 96
2 .Formulation: Let unknown quantity X 1 unit of A, unknown quantity
X2 units of B be produced. The model will finally give the values of
decision variables X1 & X2 for maximum profit at optimum level.
Contribution per unit for A & B are Rs40/-,& Rs35/-
Decision variables: Unknown production quantities X1 & X2 are decision
variables.
The objective function: The objective function for Case1 is written as
below.
Constraints: The study of the case earlier has already revealed the
required factors for writing constraints. The constraint relationships are
expressed below:
2X1 + 3X2 ≥ 60……….(1) Raw Materials Constraint (at-least)
4X1 + 3X2  96……….(2) Labour Constraint (at-most)
method is written in terms of decision variables, slack variables, surplus
variables, artificial variables and their coefficients. While X1& X2 are
decision variables, S1 is surplus variable, S2 is slack variable introduced to
establish equality between LHS & RHS, and A1 is an artificial variable

54
required as per simplex method. A1 serves as an artificial slack variable. M
is a finite high value introduced with a negative sign as a coefficient to A1
in objective function to drive out artificial variable A1.
Objective function: 40X1 + 35X2+0S1+0S2 - MA1 = Z (Z is the total
profit) which is to be maximized
Constraints:
2X1 + 3X2 –S1 +0S2+A1= 60……….(1) Raw Materials Constraint
4X1 + 3X2+0S1+S2 + 0A1= 96……….(2) Labour Constraint
X1 , X2,S1,S2,A1 ≥ 0 Non negativity constraint
4. Starting solution: Put all non-slack variables equal to zero, X1=0 ,
X2=0,S1=0 now, A1=60, S2=96
constructed:
Simplex table ST-1
Contributions >>> 40 35 0 0 -
M
C
contributio
n of basic
variables
x
basic
variable
s
X1 X2 S1 S2 A1 bi
Solution
values
θ
-M A1 2 3 -1 0 1 60 20
0 S2 4 3 0 1 0 96 32
Z -2M -3M M 0 -
M
 = c - z 40+2
M
35+3M -M 0 0
The above table doesn’t give optimal solution as NER values are
not non positive. As per the methodology next table is written with first
iteration.

55
Simplex table ST2
Contributions >>> 40 35 0 0 M
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 A1 bi
Solution
values
θ
35 X2 2/3 1 -1/3 0 20
0 S2 2 0 1 1 36
Z (70/3) 35 -70/3 0
 = c - z -50/3 0 70/3 0
In the above table ST2 column of A1 is not written as artificial variable A1
has been removed from the iterations. The NER indicates that the solution
is not optimal hence key information is collected from ST2 as shown
below:
Simplex table ST2
Contributions >>> 40 35 0 0
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
θ
35 X2 2/3 1 -1/3 0 20 -60
0 S2 2 0 1 1 36 36
Z (70/3) 35 -70/3 0
 = c - z -50/3 0 70/3 0
Note that θ the replacement ratio (-60) is ignored as this value is not non-
negative. All negative values and ∞ in the column of θ are ignored. Next
iteration is given in ST3

56
Simplex table ST3
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
35 X2 4/3 1 0 1/3 32
0 S1 2 0 1 1 36
Z 140/3 35 0 35/3
 = c - z -20/3 0 0 -35/3
The above solution is optimal as all NER values are non-positive.
given in solution values column. Learner may put the values of decision
variables in the objective function to calculate the maximum profit as per
the optimal solution.
Maximization case (problem) with mixed constraints (a case with an
equality constraint)
This type of problem is resolved broadly in the same manner as the
problem with same type of constraints. The methodology differs in writing
the standard form. When there is an equality constraint, LHS is already
equal to RHS. Hence neither a slack variable nor a surplus variable is
required. When the standard form is written, every constraint needs a slack
variable, real or artificial. Hence the equality constraint is written in the
standard form with an artificial variable. This is explained with an
example below:
Case2-1
Q. Solve the following problem using simplex method (NDV 93 –
equality constraint)
Objective function: Maximize Z =2X1 + 4X2,
Subject to the constraints
2 X1 + X2  18
3X1 + 2X2 ≥ 30
X1 + 2X2 = 26
X1, X2 ≥ 0

57
The above problem is already formulated and X1 & X2 are
decision variables. Hence steps 1&2 are already done. Now standard
form is to be written
3. Standard form:
2 X1 + X2 + S1 +0S2 +0A1+0A2 = 18
3 X1 + 2X2 +0S1 – S2 + A1+0A2 = 30
X1 + 2X2 +0S1+0S2 +0A1+A2 = 26
X1,,X2, S1,S2, A1, A2 ≥ 0
S1 is a slack variable; S2 is a surplus variable and artificial variables A1 &
A2 serve as artificial slack variables.
4. Starting solution:
Put all non-slack variables equal to zero.X1 = 0,,X2 = 0, S2 =0, hence, S1 =
18, A1 = 30, A2 = 26
constructed:
Simplex table ST-1
Contributions >>> 2 4 0 0 -M -M
C
contribution
of basic
variables
X
basic
variables
X1 X2 S1 S2 A1 A2 bi
Solution
values
θ
0 S1 2 1 1 0 0 0 18 18
-M A1 3 2 0 -1 1 0 30 15
-M A2 1 2 0 0 0 1 26 13
Z -4M -4M 0 M -M -M
 = C -
Z
2+4M 4+4M 0 -
M
0 0
not non positive. As per the methodology first iteration is performed and
next table ST2 is written.

58
Simplex table ST2
Contributions >>> 2 4 0 0 -M -M
C
contribution
of basic
variables
x
basic
variables
X1 X
2
S
1
S2 A1 A2 bi
Solution
values
θ
S1 3/2 0 1 0 0 -1/2 5 10/3
A1 2 0 0 -1 1 -1 4 2
X2 1/2 1 0 0 0 1/2 13 26
Z 2-
2M
4 0 M -M M+2
 = C -
Z
2M 0 0 -M 0 -2M-2
not non positive. As per the methodology second iteration is performed
and next table ST3 is written.
Simplex table ST3
Contributions >>> 2 4 0 0 -
M
-M
C
contribution
of basic
variables
x
basic
variables
Solution
values
0 S1 0 0 1 3/4 2
2 X1 1 0 0 -
1/2
2
4 X2 0 1 0 ¼ 2
Z 2 4 0 0
 = C - Z 0 0 0 0
In the above ST3 columns of A1 & A2 are ignored as these artificial
variables are driven out of the system.
The above solution is optimal as all NER values are non-positive.
given in solution values column. The unknown values of the decision
variables are substituted in the objective function to calculate the
maximum profit as per the optimal solution.

59
Objective function: Maximize Z =2X1 + 4X2 ,
Z =2 (2) + 4 (2) = 12,
Maximum value of Z as per the above optimal solution is 12 in
appropriate units.
4.3 MINIMIZATION CASE (PROBLEM) - “BIG M”
METHOD
The methodology for solving minimization problems follows the
same flow as in maximization case but certain steps are different.
Minimization problems deal with parameters like cost. The objective
function is minimization type with constraints of at-least or at-most type.
Minimization problems also come with mixed constraints. The
formulation is done in the same manner as in maximization. Hence these
cases are not discussed separately. For minimization type of problems,
study of the business case and mathematical formulation are done in the
same way as in the case of a maximization problem. Construction of
objective function in standard form and application of optimality test are
different as compared to maximization problems. This chapter covers two
methods for solving minimization problem. The methods are “Big M”
method and “two phases” method. The Big M method is taken up first.
The concept of big M as a finite large quantity is already understood and
applied while solving maximization problem with mixed constraints.
“Two phases” method will be taken up later.
1. Study of the business case: done similarly as in the case of
maximization
2. Formulation: done similarly as in the case of maximization. But
nature of the objective function is now minimization of cost.
3. Standard form:
a) Objective function: objective being minimization type in order to
drive out the artificial variable from the system, the artificial
variable should be made very high. For this purpose artificial
variable is associated with a positive high value M.
b) Constraints: done similarly as in the case of maximization.
4. Starting solution: done similarly as in the case of maximization.
5. Structuring of Simplex table: done similarly as in the case of
maximization.
6. Calculation of Z row values: done similarly as in the case of
maximization.

60
7. Optimality test: when all the values in NER, index row or ∆ row
are non-negative (when calculated as  = C – Z) the solution for a
minimization problem is optimal. If solution is optimal, the process
stops and solution offered by the simplex table which can be read
in bi column is put to use. If the solution is not optimal, the
simplex method improves the solution as described below.
8. Simplex Iterations: iterations are carried out as in the case of
maximization to improve the solution until optimality is reached.
Minimization case (problem) with all “at-least” type of constraints
A business problem is taken through simplex methodology
discussed earlier.
Case3 is considered below. All the steps leading to the
construction of first simplex table are discussed earlier. The nature of the
constraints is “at least” type. A definite quantity of resources should be
used while running the process. Hence surplus variables and artificial
variables enter the standard form of formulation.
Case3: Food F1 contains 20 units of vitamin A and 40 units of vitamin B
per gramme; Food X2 contains 30 units each of vitamin A and B per
gramme. The minimum daily requirements for an individual are 900 units
of A and 1200 units of B. How many grammes of each food must be
consumed to satisfy daily vitamin requirements at minimum cost? If X1
costs 60 paise per gram and X2 costs 80 paise per gram, find the optimal
solution to minimize the costs using simplex method.
 This is a case of cost minimization. An optimal solution minimizing
costs is contemplated.
 Company is considering producing food products F1 & F2. Quantities
of food products are decision variables.
 Expected cost contribution by a unit of F1 is Rs.0.60/-, by a unit of F2
is Rs.0.80/-
 Availability of each vitamin in each food product in units, decision
variables and minimum requirement of each vitamin in appropriate
units are tabulated below:

61
Vitamin
A in units
Vitamin
B in units
Decision
variables
(unknown
production
quantities
in units)
Expected
cost
contribution
in Rs/unit
Food product F1 20 40 X1 Rs.0.6/-
Food product F2 30 30 X2 Rs.0.8/-
Minimum daily
requirement of
Vitamins in units
900 1200
2 .Formulation: Let unknown quantity X1 units of F1, unknown quantity
X2 units of F2 be produced. The model will finally give the values of
decision variables X1 & X2 for minimum cost at optimum level. Cost
contribution per unit for A & B are Rs 0.60/- & Rs 0.80/-. The cost
contributions are written in terms of paise for simplicity of calculations.
variables.
Objective function: 60X1 + 80X2 = G (G is the total cost per day which
is to be minimized)
Constraints:
20X1+30X2 ≥ 900 ……….(1) Vitamin A Constraint (at-least)
40X1+30X2 ≥ 1200..…….(2) Vitamin B Constraint (at-least)
X1 , X2 ≥ 0………………Non-negativity constraint
method is written in terms of decision variables, surplus variables,
artificial variables and their coefficients. While X1& X2 are decision
variables, S1 & S2 are surplus variables introduced to established equality
between LHS & RHS, and A1 & A2 are artificial variable required as per
simplex method. A1 & A2 serve as artificial slack variables. M is a finite
high value introduced with a positive sign as a coefficient to A1 & A2 in
objective function to drive out artificial variable A1.
Objective function: 60X1 + 80X 2+ 0S1+0 S2+ MA1+MA2 = G (G is the
total cost per day which is to be minimized)
Constraints:
20X1 + 30X2 - S1+ 0S2+A1+0A2 = 900 ……….(1) Vitamin A Constraint
(at-least)

62
40X1 + 30X2+ 0S1 - S2+ 0A1 + A2= 1200..…….(2) Vitamin B Constraint
(at-least)
X1, X2, S1, S2, A1, A2 ≥ 0………………Non negativity constraint
4. Starting feasible solution: Put all non-slack variables equal to zero,
then
X1=0, X2=0, S1=0, S2=0; A1=900, A2=1200.
5. Structuring of Simplex table: setup the simplex table as per the guide
lines already laid down. Put the starting feasible solution in the first
simplex table ST1.
Contributions >>> 60 80 0 0 M M
C
contribution
of basic
variables
x
basic
variables
Solution
values
θ
M A1 20 30 -1 0 1 0 900 45
M A2 40 30 0 -1 0 1 1200 30
Z 60M 60M -M -M M M
 = C -
Z
60-
60M
80-
60M
M M 0 0
The optimality test for a minimization problem is applied to the
solution in ST1. As all NER values are not non-negative the starting
solution is not optimal. Hence the first iteration is carried out to improve
the solution in ST1. The key column, key row and key element are
identified from the table ST I. This being minimization problem, the NER
value selected is most negative unlike in the maximization case. The rule
for selection of replacement ratio continues to be the same as in
maximization case.
It can be seen that column of X1 is the key column, row of A2 is the
key row and the value 40 (cell A2-X1) is the key element. Using the
information from ST1 following table ST2 is written.

63
ST2
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 A1 bi
Solution
values
θ
M A1 0 15 -1 1/2 1 300 20
60 X1 1 3/4 0 -1/40 0 30 40
Z 60 15M + 45 -
M
(M/2)-
(3/2)
M
 = C - Z 0 35-15M M (3/2)-
(M/2)
0
The optimality test for a minimization problem is applied to the
solution in ST2. As all NER values are not non-negative the starting
solution is not optimal. Hence the second iteration is carried out to
improve the solution in ST2. The key information from the table ST 2 is
collected. This being minimization problem, the NER value selected is
most negative unlike in the maximization case. The rule for selection of
replacement ratio continues to be the same as in maximization case.
It can be seen that column of X2 is the key column, row of A1 is the key
row and the value 15 (cell A1-X2) is the key element. Using the
information from ST2 following table ST3 is written.
ST3
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
θ
80 X2 0 1 -
1/15
1/30 20
60 X1 1 0 1/20 -1/10 15
Z 60 80 -7/3 -1/3
 = C - Z 0 0 7/3 1/3
The solution in ST3 is optimal as the optimality rule for
minimization problem is that the NER values should be non-negative.
When the values of the decision variables of the optimal solution are put
in the objective function the total minimum cost is obtained

64
60X1 + 80X2 = G
60 X (15) + 80 (20) = 2500
Rs 25.00/- is the minimum cost as cost contributions were written in terms
of paise.
4.4 MINIMIZATION CASE (PROBLEM) - “TWO
PHASES” METHOD
The two phase method, under simplex methodology, is the other method
to be studied for minimization problem. Study of the business case and
formulation are done as per earlier methodology.
Standard form: the constraints are written but while writing the objective
function artificial variables are not associated with M. the coefficients are
only 1 each.
As the name suggests there are two phases to this method. The objective
function is divided into two parts.
Phase I: the objective function is maximized to remove artificial variables
by associating artificial variables with -1. All other variables will have
zero as coefficients. The simplex table with optimal solution
(maximization) for Phase I doesn’t contain artificial variables. Phase II:
Putting original values as coefficients for other variables in objective
function, objective function is minimized. The optimal solution
(minimization) to Phase II gives the optimal solution to the original
problem.
The business case 3, a minimization problem considered for “big M
method” is considered again as case 4 for Two Phases method.
Case4: Food F1 contains 20 units of vitamin A and 40 units of vitamin B
per gramme; Food X2 contains 30 units each of vitamin A and B per
gramme. The minimum daily requirements for an individual are 900 units
of A and 1200 units of B. How many grammes of each food must be
consumed to satisfy daily vitamin requirements at minimum cost? If X1
costs 60 paise per gram and X2 costs 80 paise per gram, find the optimal
solution to minimize the costs using simplex method.

65
The case
1. Study of the business case and formulation are done and expressed
below:
2. Formulation
variables.
is to be minimized)
Constraints:
20X1+30X2 ≥ 900 ……….(1) Vitamin A Constraint (at-least)
40X1+30X2 ≥ 1200...........(2) Vitamin B Constraint (at-least)
X1 , X2 ≥ 0………………Non-negativity constraint
artificial variables and their coefficients. While X1& X2 are decision
variables, S1 & S2 are surplus variables introduced to established equality
between LHS & RHS, and A1 & A2 are artificial variable required as per
simplex method. A1 & A2 serve as artificial slack variables.
Objective function: 60X1 + 80X 2+ 0S1+0 S2+ A1+A2 = G (G is the total
cost per day which is to be minimized)
Constraints:
20X1 + 30X2 - S1+ 0S2+A1+0A2 = 900 ……….(1) Vitamin A Constraint
(at-least)
40X1 + 30X2+ 0S1 - S2+ 0A1+A2 = 1200..…….(2) Vitamin B Constraint
(at-least)
X1, X2, S1, S2, A1, A2 ≥ 0………………Non negativity constraint
4. Starting feasible solution: Put all non-slack variables equal to zero,
then
X1=0, X2=0, S1=0, S2=0; A1=900, A2=1200.
5. Structuring of Simplex table: setup the simplex table as per the guide
simplex table ST1.

66
Phase I:
Objective function: Maximize G* = 0X1 + 0X 2+ 0S1+0 S2 - A1 - A2 to
drive out A1 & A2
The simplex table ST1 is constructed as per the guidelines
ST1
Contributions >>> 0 0 0 0 -1 -1
C
contribution
of basic
variables
x
basic
variables
Solution
values
θ
-1 A1 20 30 -1 0 1 0 900 45
-1 A2 40 30 0 -1 0 1 1200 30
Z -60 -60 1 1 -1 -1
 = C -
Z
60 60 -1 -1 0 0
The above table doesn’t give optimal solution as all NER values
are not non positive. As per the methodology first iteration is performed
and next table ST2 is written.
ST2.
Contributions >>> 0 0 0 0 -1 -1
C
contribution
of basic
variables
x
basic
variables
Solution
values
θ
-1 A1 0 15 -1 1/2 1 -1/2 300 20
0 X1 1 3/4 0 -
1/40
0 1/40 30 40
Z 0 -15 1 -1/2 -1 ½
 = C - Z 0 15 -1 1/2 0 -3/2
The above table doesn’t give optimal solution as all NER values
are not non positive. As per the methodology first iteration is performed
and next table ST3 is written. As A1 & A2 are already driven out the
columns of A1 & A2 are dropped.

67
ST3.
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
0 X2 0 1 -1/15 1/30 20
0 X1 1 0 1/20 -
1/20
15
Z 0 0 0 0
 = C - Z 0 0 0 0
It can be seen that the solution in ST3 is optimal as all NER values are
non-positive.
Phase II:
The solution in ST3 is now minimized in Phase II. Objective function is
now changed by rewriting the coefficients of variables. The artificial
variables are already driven out.
Objective function: Minimize G = 60X1 + 80X 2+ 0S1+0 S2
ST4.
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
80 X2 0 1 -1/15 1/30 20
60 X1 1 0 1/20 -
1/20
15
Z 60 80 -7/3 -1/3
 = C - Z 0 0 7/3 1/3
It can be seen that the solution in ST4 is optimal as all NER values are
non-negative.
Learner may put the solution values of basic variables in the
objective function and find the optimal minimum cost.
is to be minimized). Substituting the values of the decision variables,

68
60 X 15 + 80 X 20 =900+1600 = 2500. As the cost contributions are
written in paise, the total minimum cost is Rs 25/-
4.5 SENSITIVITY ANALYSIS OF OPTIMAL SOLUTION
Post optimality analysis is a comprehensive study of an optimal simplex
solution to interpret its features in order to solve a business problem. The
post optimality analysis consists of interpretation of the optimal solution
and sensitivity analysis.
Interpretation of the optimal solution:
Basis of the optimal solution: the basic variables (unknown quantities)
occupy the basis and solution values column gives the respective values of
the variables. Any variable not in the basis (non-basic variable) is zero.
NER values: NER values of decision variables are opportunity costs. A
negative opportunity cost for a decision variable in a maximization
problem (when ∆ ═ C-Z) indicates the amount by which the total
maximum profit (Z) would come down if a unit of this particular product
is produced.
NER values of resources represented by slack variables/surplus variables
are called Shadow prices or unit worth of resources. In a maximization
problem (when ∆ ═ C-Z), negative NER values for slack variables
indicate that the resource is scarce or just enough.
Special cases of optimal solution: this is discussed separately after
sensitivity analysis
Sensitivity analysis:
Sensitivity analysis is testing the optimal solution to see within
what range of changes in the environment it remains use full. The changes
in the environment bring pressure on the basis of the optimal solution and
solution values to take up new values. If the basis changes the optimal
solution no longer is the same. If the basis remains the same and only the
solution values change, the optimal solution holds good. The sensitivity
analysis which is being discussed here considers following changes:
a. changes in the unit contributions or Cj values (changes in coefficients of
decision variables in the objective function)
b. changes in limits on the resources (bi values in the first solution)
c. plan to introduce a new product

69
When the above changes take place in the market after the optimal
solution is established, sensitivity analysis (also known as post optimality
analysis) is performed.
In the current context, post optimality analysis which includes
interpretation of the optimal solution is also discussed.
Specific business cases are taken up to discuss interpretation of the
optimal solution, sensitivity analysis and special cases of the simplex
solution.
4.6 STUDY OF SPECIAL CASES OF SIMPLEX
SOLUTION
Following cases are said to be special cases of simplex solutions.
Each case has its own features which are important for its applicability to
a business problem. These cases are discussed in previous chapter # 3
1. Alternate optima: Alternate optima to an optimal solution are other
optimal solutions which also give the same optimal value of the objective
function. An optimal solution which doesn’t have alternate optimum is
called a unique optimal solution.
Test for uniqueness of an optimal solution: if one or more of non-basic
variables have zero value in NER then one or more alternate solutions
exist. To find the alternate solution, the column of the non-basic variable
with zero value in NER is treated as key column and θ values are
calculated. Key element is identified and the iteration is carried out to
improve the solution. The optimal solution found after necessary iterations
is an alternate one.
2. Unbounded solutions: when least non-negative replacement ratio (θ) is
not available for finding “out-going variable” (departing variable) the
solution is unbounded. Unbounded solution cannot reach optimality.
Hence, it is not useful as a solution to a business problem. Wrong
formulation of the LPP is the cause of unbounded solution.
3. Infeasibility: a solution which doesn’t satisfy all constraints (resource
constraints and non-negativity constraint) is said to be infeasible. This
solution does not have any applicability. An optimal solution is infeasible
when at least one basic variable assumes negative-value in the solution
values column. Wrong formulation of the LPP is the cause of infeasibility.

70
4. Redundancy: a resource which is in abundance and is not a constraint
while finding the optimal solution is redundant. Presence of a slack
variable in the basis indicates redundancy of that particular resource. If
this slack present in the basis has a positive value in the solution column,
the redundant resource is abundant.
5. Degeneracy: basic variables in a degenerate solution are one less than
number of constraints. An optimal solution is degenerate if one of the
basic variables assumes zero value in solution values column. While
performing simplex iterations, if tie is experienced between replacement
ratios the optimal solution will be degenerate.
Two following business cases are taken up one after the other for step
by step discussion of topics mentioned above
Case5:
The simplex tableau for a maximization problem of linear programming is
given below.
PRODUCT MIX
Cj Xi
X1 X2 S1 S2 QUANTITY
(bi)
5 X2 1 1 1 0 10
0 S2 1 0 -1 1 3
Cj 4 5 0 0
Zj 5 5 5 0
∆ ═ Cj-
Zj
-1 0 -5 0
X1 & X2 are decision variables representing production quantities of
products P & Q. S1 is slack variable representing resource (machine) A (in
hours/week) and S2 is slack variable representing resource (machine) B (in
hours/week). Answer the following questions giving reasons in brief:
i. Is the solution optimal?
Answer: as all NER values are non-positive (when ∆ ═ C-Z), the
solution to a maximization problem is optimal.
ii. Is there an alternate optimal solution?
Answer: X1 & S1 are non-basic variables and they have negative
values in the NER. Hence the optimal solution is unique. There are
no alternate optima.
iii. Is the solution degenerate?
Answer: As no basic variable has zero solution value the solution
is not degenerate.

71
iv. Is the solution feasible?
Answer: As the solution is optimal and no basic variable has
negative solution value the optimal solution is feasible.
v. Is the solution unbounded?
Answer: An unbounded solution doesn’t yield an optimal solution.
The iteration stops as departing variable cannot be found. As the
above solution is optimal it is not unbounded.
vi. If S1 is slack in machine A (in hours/week) and S2 is slack in
machine B (in hours/week), which of these machines being used to
the full capacity when producing according to this solution?
Answer: as the NER value for S1 is negative, the resource machine
A is scarce or fully utilized (being used to the full capacity)
vii. How many units of two products X1 & X2 are being produced and
what is the total profit?
Answer: Decision variable X1 is non-basic hence its solution value
is zero. Product P is not produced. Decision variable X2 is basic
variable and its solution value is 10 as read from the solution
values column. Production quantity for product Q is 10 units.
Value of maximum profit as per the optimal solution is obtained by
substituting solution values of X1 & X2. Max profit Z = 4X1 + 5X2
substituting the values of X1 & X2,
Z = 4 X 0 + 5 X 10 = 50 assuming units to be rupees
viii. Machine A has to be shut down for repairs for 2 hours next week.
What will be the effect on profit?
Answer: Unit worth of resource machine A is Rs 5/- as per the
NER. Hence, if A is not available for two hours the total profit will
come down by Rs 10/-
ix. How much would you prepare to pay for another hour (per week)
of capacity each on machine A & machine B?
Answer: the unit worth or shadow price is read from NER.
Shadow price for a unit of A is Rs 5/-. Hence the maximum rent
that can be paid is Rs 5/- for a unit of A and zero for be as it is
available in abundance.
x. A new product is proposed to be introduced which would require
processing time of 45 minutes on machine A and 30 minutes on
machine B. It would bring a profit of Rs.3/- per unit. Do you think
it would be advisable to introduce this product?
Answer: resource requirements of new product are worked out
below

72
UNITS
REQUIRED IN
HOURS
UNIT
PRICE
RESOURCE
REQUIREMENT
OF NEW
PRODUCT IN
Rs
MACHINE A 3/4 5 (3/4) x 5 = 15/4
MACHINE B 1/2 0 0
TOTAL
RESOURCE
CONSUMPTION
EXPEPECTED
DUE TO NEW
PRODUCT
Rs. (15/4) = Rs
3.55/-
Total resource worth required by new product is Rs 3.55/- and profit
expected from new product is Rs 3/-. Hence the proposed idea to
introduce a new product is not acceptable.
Case6:
An engineering company BMS Ltd. produces 3 products A, B & C using 3
machines M1, M2 & M3. The resource constraints on M1, M2 & M3 are 96,
40 & 60 hours respectively. The profits earned by the products A, B & C
are Rs.2/-, Rs.5/- & Rs.8/- per unit respectively. A simplex optimal
solution to maximize the profit is given below where X1, X2 & X3 are
quantities of products A, B & C produced by the company and S1, S2 & S3
represent the slack in the resources M1, M2 & M3. Study the solution given
below and answer the following questions.
c X
VARIABLES
IN THE
BASIS
X1 X2 X3 S1 S2 S3 b
Solution
values
5 X2 1/3 1 0 1/6 -1/3 0 8/3
8 X3 5/6 0 1 -1/12 2/3 0 56/3
0 S3 7/3 0 0 -1/13 -1/3 1 44/3
 = C - Z -19/3 0 0 -1/6 -11/3 0
(i) Indicate the shadow price of each resource. Which of the resources are
abundant and which are scarce?

73
Answer:
a. Shadow price of resource M1 is (Rs1/6) per hour
b. Shadow price of resource M2 is (Rs11/3) per hour
c. Shadow price of resource M3 is 0 because S3 is present in the
basis.
d. Resource M1 is scarce
e. Resource M2 is scarce
f. Resource M3 is abundant. Unutilized machine hours are
indicated by S3 value (44/3)hours
(ii) What profit margin for product A do you expect the marketing
department to secure if it is to be produced, and justify your advice.
Answer: If profit margin of A is raised to Rs.2 + Rs. 19/3, product A can
be produced.
At the moment A is not being produced as its opportunity cost is -19/3.
Opportunity cost of A will be zero if profit margin of A becomes Rs.[2 +
19/3] = Rs 8.33. Then an alternate solution emerges where product A is
produced. If Opportunity cost of A becomes positive due to rise in profit,
the current solution becomes non-optimal.
It can be seen that if the profit for A remains within the range of Rs 0 to
Rs 8.33 the optimal solution remains unchanged.
(iii) Within what range the profit of product B can change for the above
solution to remain optimal?
Answer: The sensitivity of the solution is tested by calculating ratios as
shown below:
 = C - Z -19/3 0 0 -1/6 -11/3 0
X2 1/3 1 0 1/6 -1/3 0
 / Row X2 -19 0 0 -1 11 0
Select least positive (first positive value when counted from zero) and
least negative (first negative value when counted from zero)  / Row X2
ratios and add them to profit margin of X2.
5 + [-1] = 4
5 + [11] = 16
Range of profit margin of X2 which maintains the optimality of solution is
Rs4/- to Rs16/-

74
(iv) How would an increase of 10hours in the resource M2 affect the
optimality?
Answer: A table is set up below and ratios of solution values and
respective coefficients from the column of the concerned resource are
calculated. The least positive and least negative values of the ratios are
identified.
The least positive value is subtracted from the current availability
of resource and least negative value is added to get the range of the
resource availability to maintain optimality.
Select least positive (first positive value when counted from zero) and
least negative (first negative value when counted from zero) ratios.
Amount of M2 resource is 40 hours. The range of M2 which would
not upset the optimality is [40 – 56/2] to [40 + 8] = 12hrs to 48hrs. Hence,
as an increase of 10 hrs makes the value of M2 to be 50hrs, the change
would upset the optimality because revised RHS will be out of range.
Range of availability of resource M2 which maintains the optimality of
solution is 12hrs to 48hrs
Learner may similarly calculate the range of resource availability to
maintain optimality for M1 & M3
(v) If the company BMS Ltd. wishes to raise production which of the three
resources should be given priority for enhancement?
Answer: as the shadow price of M2 is highest procuring this resource from
suppliers market on rental is most expensive. Hence M2 should be given
priority. Capacity enhancement should be up to the level M2 is not
redundant.
b
Solution
values
Coefficients of
s2 column
(Solution
Values)/ (Coefficients of
s2 column)
8/3 -1/3 -8
56/3 2/3 56/2
44/3 -1/3 -44

75
4.7 BIBLIOGRAPHY
1. Jhamb L.C., Quantitative Techniques for Managerial Decisions Vol I,
India, Everest Publishing House, 2004
2. Vohra, N. D., Quantitative Techniques in Management, Third Edition,
New-Delhi, Tata McGraw-Hill Education Private Limited, 2007
3. Taha Hamdy, Operations Research, Sixth Edition, New Delhi, Prentice
Hall of India, 2002
1. Solve the examples in the previous chapter using simplex method.
2. A firm uses three machinesM1, M2 and M3 in the manufacture of three
products A, B and C. Each unit of A requires 3, 2 and 1hrs on M1, M2
and M3 respectively. Each unit of B requires 4, 1 and 3 hours on M1,
M2 and M3 respectively while each unit of C requires 2hrs each on M1,
M2 and M3 respectively. The machine hours available on the three
machines are 90, 54 & 93 hours respectively. The profit contributions
of the three products are rupees 30, 40 & 35 per unit respectively.
a) Formulate the above case as linear programming problem
b) Obtain optimal solution to the problem by using simplex method.
Which of the three products shall not be produced by the firm?
Why?
c) Calculate the percentage of capacity utilization by the firm in the
optimal solution
d) What are the shadow prices of the machine hours?
e) Is the optimal solution degenerate?
f) Is the optimal solution unique?
g) If a customer demands the product not allowed by the optimal
solution what should be the approach of the Operations
Manager?
3. Study the solution in the table and answer the questions given below
CJ  40 80 0 0 0
C V X1 X2 S1 S2 S3 bi
0 S1 2 0 1 0 -3 18
0 S2 1 0 0 1 0 15
80 X2 0 1 0 0 1 10
Z 0 80 0 0 80
Δ
(C-Z)
40 0 0 0 -80

76
a. Is the above solution optimal? If it is not, find the optimal solution.
b. What is the maximum profit as per optimal solution?
c. Is the optimal solution found by you unique? If it is not, find the
alternate solution.
d. Is the optimal solution degenerate?
e. Is the optimal solution infeasible?
f. Which of the variables are non-basic?
4. An engineering company M/S QM Works Pvt. Ltd. produces two
products X1 & X2 using two resources A & B. The resources are
machine hours expressed in hours per week. Profits earned by X1 & X2
are Rs3/- & Rs4/- per unit respectively. The company wants to
maximize their profit and uses simplex method of linear programming
to plan weekly production of X1 & X2. The simplex tableau for the
maximization problem is given below
Answer the following questions giving reasons in brief:
BASIS
cj xj
x1 x2 s1 s2 SOLUTION
VALUE
B
4 x2 1 1 1 0 6
0 s2 1 0 -1 1 2
 = c - z -1 0 -4 0
a. If S1 is slack in resource A and S2 is slack in resource B which of
the resources is scarce and which is abundant?
b. How many units of two products X1 & X2 are being produced as
per this solution and what is the total profit?
c. If a customer wants a unit of X1 what price the company M/S QM
Works Pvt. Ltd should quote to ensure no reduction in profit?
d. If resource A is not available for 2hrs next week, what will be the
drop in profits of M/S QM Works Pvt. Ltd.?
e. If M/S QM Pvt. Ltd. decide to procure resources from outside what
price per unit of A & B they will pay?
4. Reddy Mikks produces both exterior and interior paints from two raw
materials M1 & M2. The following table provides the basic data of the
problem

77
Tons of raw materials per ton of
Exterior paint Interior paint Maximum daily
availability
Raw material
M1
6 4 24
Raw material
M2
1 2 6
Profit per Ton
in thousand
Rupees
5 4
A market survey restricts the maximum daily demand of interior paint to 2
tons. Additionally the demand for interior paint can not exceed that of
exterior paint by more than 1 ton.
Reddy Mikks wants to determine the optimum (best) product mix of
interior and exterior paints that maximizes the total daily profit.
5. Hardware Fabricators a manufacturer of boilers, produces and sells 3
models of boilers. While market demand poses no restrictions,
capacity to produce is currently constrained by the limited supply of
special grade steel to 1500 kg per week and machine processing time
to 1200 hrs to week. Model I requires 6 kg of steel and 3 hrs of
machine processing. Similarly these figure for Model II are 3 kg and 4
hrs and for Model III these figures are 5 kg and 5hrs each. The profit
contributions per boiler for these models are Rs. 60/-, Rs.40/- &
Rs.80/- respectively. In order to determine the optimal product mix to
maximize weekly profit contribution, a linear programming model is
formulated from which by using the simplex method, the following
table was obtained.
a) Fill in all the numerical values in starred positions.
b) Is the current solution optimal? If not find the optimal solution.
c) Analyze the sensitivity of optimal solution to the following changes,
giving the new solution
60 40 80 0 0
CB XB b X1 X2 X3 S1 S2
* X1 100 * -1/3 * 1/3 -1/3
* X3 180 0 1 1 -1/5 2/5
∆ = C-Z * * * * *

78
i) Due to machine breakdown, the machine hours available gets reduced to
1050 hrs.
ii) An additional quantity of 150 kg of steel can be obtained.
iii) The second model does not feature in the current optimal solution.
What should be the minimum increase per unit contribution on this
model for this to feature in the optimal solution?
iv) What will be optimal solution if the constraints of steel and machine
hours are changed to 2000 kg and 1500 hrs respectively?
6. Standard manufacturers produce three products P, Q & R which
generate profits of Rs.20/-, Rs 12/- & Rs 8/- per unit. Three operations
are needed for each product on three machines M1, M2 & M3. The
maximum working hors available for each type of these three
machines are 1200, 900 & 400 respectively. On eof the simplex
method solutions is given in the following table.
20 12 8 0 0 0c x b
X1 X2 X3 S1 S2 S3
0 S1 160 0 0 4/5 1 -4/5 4/5
12 X2 120 0 1 3/5 0 2/5 -3/5
20 X1 140 1 0 1/5 0 -1/5 4/5
Z 20 12 56/5 0 4/5 44/5
∆= C-Z 0 0 -16/5 0 -4/5 -44/5
(a) Which machine is not fully utilized? If the balance working hrs of this
machine are shifted to M2, what will be the effect on the solution?
(i) Retaining the optimality, find the range of working hrs of the
third machine.
(ii) Within what range of profit of each product the solution will
remain optimal?
(iii) Keeping the shadow prices intact, find the range of working
hrs of M2.
(iv) Without altering the optimality, is it possible to reduce the
availability of the working hrs of the M2 to 200 hrs?
(v) If it is decided to increase capacities of all the three machines
by 25% of their respective present capacities what will be the
new product mix?


79
5
DUALITY AND DUAL SIMPLEX METHOD
Unit structure
5.0 Objectives
5.1 Introduction
5.2 Construction of dual from primal
5.3 To study the optimal solutions of primal and dual
5.4 To Find Optimal Solution to the Dual
5.5 Dual simplex method
5.6 Bibliography
5.0 OBJECTIVES
 Understand the concept of duality and its interpretation from the
primal
 Interpret the worth of resources from the optimal solution of primal
 Understand the concept of shadow prices
 Use the concept of duality to solve a minimization problem by
using maximization method
 Use the concept of duality to reduce computational complexity of
simplex algorithm when number of constraints is much more than
number of decision variables in a Primal LPP.
 Understand the methodology of the dual simplex method
 Use the dual simplex method to remove infeasibility from an
infeasible optimal solution
5.1 INTRODUCTION
 Duality is a characteristic of a linear programming problem. Duality
suggests that each LPP formulation has its mirror image called dual.
The original formulation is called primal while its image is called
dual. There is strong connection between these two formulations.
Some of these connections are apparent but some of them need closer
investigation to be realized. This later characteristic of LPP leads to
the interpretation of the optimal solution of the primal. The
interpretation of the optimal solution of the primal LPP based on the

80
duality is the economic interpretation of the optimal solution.
An LPP is discussed here to understand the concept of duality.
 From the earlier discussions it is observed that simplex methodology
produces some special cases of solutions. One such special case is
infeasibility which is generally the result of wrong formulation. Dual
simplex method which is a variation of classical simplex method is
used to remove the infeasibility from the infeasible simplex solution.
5.2 CONSTRUCTION OF DUAL FROM PRIMAL
The formulation of a business problem is called primal of the given
LPP. Its dual is constructed from the parameters of the primal itself.
Before proceeding to carryout construction, a business problem is taken up
for discussion. The discussion reveals the basis for development of the
dual.
Case1. A firm produces products, A & B, each of which requires two
respectively of raw materials and labour. Everyday 60 units of raw
materials and 96 units of labour are available. If the unit profit
1. Study of the business problem: Case1 is analyzed as per above
This is a case of profit maximization. An optimal solution maximizing
profits is contemplated.
Company is considering producing products A & B.
Expected profit contribution by a unit of A is Rs 40/- and by a unit of B is
Rs 35/-
Availability of raw materials is 60 units per day and availability of labour
is only 96 units.
2. Formulation of the business problem:
The decision variables: Let an unknown quantity X1 units of A and an
unknown quantity X2 units of B be produced per day. X1 & X2 are the
decision variables.
The objectives function for Case1 is written as below.
The model will finally give the values of X1 and X2 for maximum profit at
optimum level. Contribution per unit for A & B are Rs 40/- and Rs 35/-.
Then the objective is expressed in mathematical form as:

81
Constraints for Case1 are written below:
The study of the case earlier has already revealed the required factors for
writing constraints. The constraint relationships are expressed below:
i) Resource constraints:
2X1 + 3X2  60……….. (1) Raw Materials Constraint
4X1 + 3X2  96……….. (2) Labour Constraint
ii) Constraint on decision variables:
Let the above formulation be called the primal of the LPP.
Products A&B are produced in quantities X1 & X2. Each unit of A
generates profit of Rs40/- and B generates a profit of Rs35/-. While
running this process, 60 units of Raw Materials and 96 units of labour are
consumed.
Now, the manager has an option of renting out all the resources to
the market. If this option is exercised, the production process cannot be
run and respective profits are not generated. Hence it is either produce or
rent out.
Let the unknown per unit rent values be Y1 & Y2 for resources
Raw materials and Labour in the market. The manager can intelligently
exercise his option if he knows the actual rent values. If the total minimum
rent which can be earned by renting out all available resources is equal to
the total maximum profit which can be earned by running the production
process the Manager can decide to rent the resources out. Hence to
exercise this option the manager has to find out values of Y1 & Y2 which
minimize the total rent. Now, the original LPP is restructured to obtain the
values of Y1 & Y2.
The objectives function can be written as
60Y1 + 96Y2 = G (G is the total rent) which is to be minimized
If product A requires 2 units of raw materials and 4 units of labour and
earns Rs40/- as profit then the rent earned by 2 units of raw materials and
4 units of labour must be at least Rs40/-.
Similarly if each unit of product B requires 3 units each of raw materials
and labour then the rent earned by 3 units of raw materials and 3 units of
labour must be at least Rs35/-.
Based on the above logic,
Constraint relations may be written as below.
2Y1 + 4Y2 ≥ 40……… (1) Product A unit profit Constraint
3Y1 + 3Y2 ≥ 35…….. (2) Product B unit profit Constraint
Y1 , Y2 ≥ 0 Non negativity constraint

82
The formulation structured from the primal LPP is called its dual.
Solution to the above dual problem can be found by processing it
through all the steps of the simplex algorithm
5.3 TO STUDY THE OPTIMAL SOLUTIONS OF
PRIMAL AND DUAL
To find optimal solution to the primal
Formulation of the business problem is done already for the primal. The
Constraints are written below:
Resource constraints:
2X1 + 3X2  60………..(1) Raw Materials Constraint
Constraint on decision variables:
Standard form:
2X1 + 3X2 +S1 +0S2 = 60……….(1) Raw Materials Constraint
4X1 + 3X2+0S1+S2 = 96……….(2) Labour Constraint
X1 , X2, S1, S2 ≥ 0 Non negativity constraint
Starting solution: Put all non-slack variables equal to zero, X1= 0 , X2= 0,
S1= 60, S2= 96
Structuring of Simplex table
constructed:
ST1
C
contribution of
basic variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
θ
0 S1 2 3 0 1 60 3
0
0 S2 4 3 1 0 96 2
4
Z 0 0 0 0
 = c - z 40 35 0 0

83
The solution in ST1 is not optimal as all values are not non-positive
in NER. Required information for carrying out the iteration is collected
from ST1 and ST2 is constructed
ST2
The solution in ST2 is not optimal as all values are not non-positive in
NER. Required information for carrying out the iteration is collected from
ST2 and ST3 is constructed
ST3
The solution in ST3 is optimal as all values are non- positive in NER.
As per the above optimal solution,
Max daily profit is Z = 40X1 + 35X2 which is the objective function, in
terms of unknown decision variables. Substituting the values of the
decision variables in the objective function,
Z = 40 X 18 + 35 X 8 = 1000
5.4 TO FIND OPTIMAL SOLUTION TO THE DUAL
Standard form: The mathematical formulation suitable for simplex
artificial variables and their coefficients.
Objective function: 60Y1 + 96Y 2+ 0S1+0 S2+ MA1+MA2 = G (G is the
total rent per day which is to be minimized)
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
θ
0 S1 0 3/2 1 -1/2 12 8
40 X1 1 3/4 1/4 0 24 32
Z 40 30 0 10
 = c - z 0 5 0 -10
contributions >>> 40 35 0 0
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solution
values
35 X2 0 1 2/3 -1/3 8
40 X1 1 0 -1/2 ½ 18
Z 40 35 10/3 25/3
 = c - z 0 0 -10/3 -25/3

84
Constraints:
2Y1 + 4Y2 - S1+ 0S2+A1+0A2 = 40 ………. (1) Product A unit profit
Constraint (at-least)
3Y1 + 3Y2+ 0S1 - S2+ 0A1 + A2=35……..... (2)Product B unit profit
Constraint (at-least)
Y1, Y2, S1, S2, A1, A2 ≥ 0………………Non negativity constraint
Starting feasible solution: Put all non-slack variables equal to zero, then
Y1=0, Y2=0, S1=0, S2=0; A1=40, A2=35.
Structuring of Simplex table: setup the simplex table as per the guide
simplex table ST1.
ST1
Contributions >>> 60 96 0 0 M M
C
contribution
of basic
variables
x
basic
variables
Y1 Y2 S1 S2 A1 A2 bi
Solution
values
θ
M A1 2 4 -1 0 1 0 40 10
M A2 3 3 0 -1 0 1 35 35/3
Z 5M 7M -
M
-M M M
 = C - Z 60-5M 96-7M M M 0 0
From the above table it can be seen that the solution is not
optimal as all NER values are not non-negative. Required information is
collected from the above ST1 table and the simplex iteration is carried out
and simplex table ST2 is written.
ST2
C
contributi
on of
basic
variables
x
basic
variables
Y1 Y2 S1 S2 A2 bi
Solution
values
θ
96 Y2 1/2 1 -1/4 0 0 10 20
M A2 3/2 0 3/4 -1 1 5 10/3
Z 48+(3/2)M 96 (3/4)M -24 -M M
 = C - Z 12-( 3/2M) 0 24-(3/4)M M 0
From the above table it can be seen that the solution is not
optimal as all NER values are not non-negative. Required information is
collected from the above ST2 table and the simplex iteration is carried out
and simplex table ST3 is written.

85
ST3
C
contribution
of basic
variables
x
basic
variables
Y1 Y2 S1 S2 bi
Solution
values
θ
96 Y2 0 1 -1/2 1/3 25/3
60 Y1 1 0 1/2 -2/3 10/3
Z 60 96 -18 -8
 = C - Z 0 0 18 8
The solution in ST3 is optimal as the optimality rule for
minimization problem is that the NER values should be non-negative.
When the values of the decision variables of the optimal solution are put
in the objective function the total minimum cost is obtained
60Y1 + 96Y2 = G
60 X (10/3) + 96 X (25/3) = 1000
Rs 1000/- is the minimum rent. The minimum rent values which enable a
decision in favor of renting out the resources is given by decision
variables Y1 =10/3, Y2 = 25/3 in rupees per unit.
The maximum profit of the primal is same as the minimum rent of the
dual
Study of primal and dual models of same LPP
To understand the inherent relationship between primal and dual, the
formulations and final solutions of primal and dual are exhibited side by
side below
40X1 + 35X2 60Y1 + 96Y2
2
4
3
3
60
96
2Y1 + 4Y2
3Y1 + 3Y2
40
35
X1
+
X
X2
≤
X
≥
≥

86
Scrutiny of the two models (formulations and optimal solutions)
highlights the following facts:
1. If the primal is a maximization problem, the dual is a minimization
problem.
2. The number of variables in the dual is equal to the number of
constraints in the primal and number of constraints in the dual is equal
to number of variables in the primal.
3. Right hand side constants of the primal constraints become the
coefficients of objective function in dual
4. Coefficients in the objective function in the primal become RHS
constants of the dual constraints.
5. If the primal has all constraints of ≤ type as in a maximization
problem, all constraints in the dual become ≥ type. Sign of inequality
between primal and dual are reversed.
a)
a)
b)
contributions >>> 40 35 0 0
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 bi
Solutio
n
values
35 X2 0 1 2/3 -1/3 8
40 X1 1 0 -1/2 ½ 18
Z 40 35 10/3 25/3
 = c - z 0 0 -10/3 -25/3
C
contribution
of basic
variables
x
basic
variables
Y1 Y2 S1 S2 bi
Solution
values
96 Y2 0 1 -1/2 1/3 25/3
60 Y1 1 0 1/2 -2/3 10/3
Z 60 96 -18 -8
 = C - Z 0 0 18 8

87
The column coefficients of the primal constraints become row coefficients
in the dual
a) Shadow prices of the primal are the solution values of decision
variables in the dual.
b) Shadow prices of the dual are the values of decision variables in the
primal
6. The above relationship leads to the economic interpretation of the
dual.
7. It also shows that the values of the decision variables can be found by
solving its dual. If there is a minimization LPP it can be solved by
solving its dual which is maximization LPP.
8. If a primal has four constraints and two decision variables, the dual
when it is written will have only two constraints and four decision
variables. Solving the dual instead of primal reduces computational
difficulty while solving the LPP. Unit end exercise 3 illustrates this
point.
Some important rules for construction of dual of a primal LPP
1. If the objective function of the primal LPP is maximization, all
inequalities must be ≤. If there are mixed constraints, convert ≥ signs
into ≤ signs by multiplying both LHS and RHS by (-1). This is
explained below
Primal LPP with mixed constraints
Objective function: Maximize Z = 40X1 + 35X2
Constraints:
2X1 + 3X2 ≥ 60……….(1) Raw Materials Constraint (at-least)
X1 X2 ≥ 0 Non negativity constraint
As per the rule above make all signs ≤
-2X1 - 3X2 ≤ - 60……….(1) Raw Materials Constraint (at-least)
X1 X2 ≥ 0 Non negativity constraint
2. If there any of the constraints has an equal to sign, the equal to
constraint is replaced by two constraints, one with an ≤ sign and the other
with ≥ sign. An example is shown below:
Primal LPP with mixed constraints
Constraints:
2 X1 + X2  18
3X1 + 2X2 ≥ 30
X1 + 2X2 = 26

88
As per the rule 2 , let the constraints be rewritten and the formulation will
be
Constraints:
2 X1 + X2  18
3X1 + 2X2 ≥ 30
X1 + 2X2 ≤ 26
X1 + 2X2 ≥ 26
As per the rule 1 above let all signs be ≤. Hence formulation is rewritten
2 X1 + X2  18
-3X1 - 2X2 ≤ - 30
X1 + 2X2 ≤ 26
- X1 - 2X2 ≤ - 26
X1 X2 ≥ 0 Non negativity constraints
now, the dual can be constructed keeping in mind salient features of
dual-primal relationship.
2. If the primal is a minimization LPP, the dual is maximization. All rules
hold good with obvious adjustments
a) All signs of the constraints of the minimization primal should
be ≥.
b) The mixed constraints must be handled as explained in the case
of maximization primal
5.5 DUAL SIMPLEX METHOD
The simplex method stops in its tracks when all NER values
are non-positive for a maximization problem. As this solution is optimal
further iterations are not required. But if any of the solution values is
negative, the solution is unusable as it is infeasible.
Dual simplex method starts when normal simplex method is unable to
proceed further as solution is optimal though infeasible. An optimal
solution with infeasibility is the starting point for dual simplex algorithm.
Utility of Dual Simplex Algorithm:
a. To remove infeasibility from an optimal solution to a maximization
LPP
b. To solve a minimization LPP which starts as optimal and infeasible
when converted to maximization
The Key Row: to find the key row “dual feasibility condition” is applied
which says that the departing variable is the one having the most negative
value, with ties broken arbitrarily. If all basic variables are nonnegative
the algorithm ends.

89
The largest negative value in the solution column is chosen for
determining departing variable. The key row is the one which contains the
largest negative solution value.
The Key Column: to find the key column, “dual optimality condition”
is applied which says that the entering variable is determined from among
the non-basic variables as the one corresponding to the minimum ratio
between NER value and the coefficient with negative value. This can be
summed up as the non-basic variable which has the following minimum
ratio: j j
rj
c z

, where 0rj  .
To determine the key column, ratios between NER value and the
coefficient in the key row are written for every NER value. Only
negative coefficients in key row for non-basic variables are selected for
calculation of above mentioned ratios.
The Key Element: as per the methodology the intersection of key column
and key row.
Iteration: as per the methodology with the changes introduced to deal
with optimal solution with negative solution values until the infeasibility is
removed
An LPP formulation is taken as a sample case to understand the dual
simplex methodology.
To remove infeasibility from an optimal solution to a maximization
LPP
Objective function: Maximize Z = -3X1 - 2X2,
Constraints:
-3X1 + X2 ≥ 3
4X1 + 3X2 ≥ 6
X1 + 2X2 ≤ 3
Step1: formulation
Change all signs to ≤ so that there will be only slack variables in
standard form.
Objective function: Maximize Z = -3X1 - 2X2,
Constraints:
-3X1 - X2 ≤ -3
-4X1 - 3X2 ≤ - 6
X1 + 2X2 ≤ 3

90
Step2: Standard form:
Objective function: Maximize Z = -3X1 - 2X2 + 0S1+ 0S2 + 0S3
Constraints:
--3X1 - X2 + S1+ 0S2 + 0S3 = -3
-4X1 - 3X2 + 0S1 + S2 + 0S3 = - 6
X1 + 2X2 +0S1 + S2 + S3 = 3
X1, X2, S1, S2, S3 ≥ 0 Non negativity constraint
Step3: Initial feasible solution: X1=0, X2=0, S1= -3, S2= -6, S3=3.
Step4: The above initial solution is put in simplex table ST1
ST1
Contributions >>> -3 -2 0 0 0
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 S3 bi
Solution
values
θ
0 S1 -3 -1 1 0 0 -3
0 S2 -4 -3 0 1 0 -6
0 S3 1 2 0 0 1 3
Z 0 0 0 0 0
 = c - z -3 -2 0 0 0
∆/RowS2 3/4 2/3
a) It may be seen that the initial solution itself is optimal but infeasible.
Hence Dual Simplex Method is used to remove the infeasibility.
b) In ST1 row S2 is considered as key row as the solution value for S2 is
most negative
c) Ratio (∆/Row S2) is calculated for negative coefficients in row S2 for
non-basic variables. As column of X2 has the minimum ratio, X2 is the
incoming variable. From the above findings, key element is identified
and ST2 is setup.
ST2
C
contribution
of basic
variables
x
basic
variables
X1 X2 S1 S2 S3 bi
Solution
values
θ
0 S1 -5/3 0 1 -
1/3
0 -1
-2 X2 4/3 1-3 0 -
1/3
0 2
0 S3 -1/3 0 0 1/3 1 1
Z -1/3 0 0 -
2/3
0
 = c - z -8/3 -2 0 0 0
∆/RowS1 8/5 0

91
The above solution is optimal but still infeasible. Hence iteration is
carried out to write ST3
ST3
C
contribution
of basic
variables
x
basic
variable
s
X1 X2 S1 S2 S3 bi
Solutio
n
values
θ
-3 X1 1 0 -3/5 1/5 0 3/5
-2 X2 0 1 4/5 -3/5 0 6/5
0 S3 0 0 -1/5 2/5 1 6/5
Z -3 -2 1/5 3/5 0
 = c - z 0 0 -1/5 -3/5
Solution in ST3 is optimal but no longer infeasible. The Dual Simplex
algorithm ends here. Hence maximum value of Z is -21/5
To solve a minimization LPP which starts as optimal and infeasible
when converted to maximization
Objective function
Minimize G = 3X1 + X2
Subject to:
X1 + X2 ≥ 1
2X1 + 3X2 ≥ 2
x1, x2 ≥ 0
Convert the minimization problem into maximization problem
a. Minimization objective function is converted to maximization by
multiplying both LHS and RHS by -1 and writing G as Z
b. All constraints are made ≤ relationships by multiplying by -1 if the
sign is ≥
Objective function:
Maximize Z = -3X1 - X2
Subject to:
-X1 - X2 ≤ -1
-2X1 -3X2 ≤ -2
x1, x2 ≥ 0
Standard form:
Objective function: Maximize Z = -3X1 - X2 + 0S1+ 0S2
Constraints:
-X1 - X2 + S1+ 0S2 = -1
-2X1 - 3X2 + 0S1 + S2 = -2
X1, X2, S1, S2 ≥ 0 Non negativity constraint
Initial feasible solution: X1=0, X2=0, S1= -1, S2= -2,

92
The above initial solution is put in simplex table ST1
ST1
Contributions >>> -3 -1 0 0
C
contribution
of basic
variables
X
basic
variables
X1 X2 S1 S2 bi
Solution
values
0 S1 -1 -1 1 0 -1
0 S2 -2 -3 0 1 -2
Z 0 0 0 0
 = c - z -3 -1 0 0
∆/RowS1 3/2 1/3
a) It may be seen that the initial solution itself is optimal but infeasible.
Hence Dual Simplex Method is used to remove the infeasibility.
b) In ST1 row S2 is considered as key row as the solution value for S2 is
most negative
c) Ratio (∆/Row S2) is calculated for negative coefficients in row S2 for
non-basic variables. As column of X2 has the minimum ratio, X2 is the
incoming variable. From the above findings, key element is identified
and ST2 is setup.
ST2
C
contribution
of basic
variables
X
basic
variables
X1 X2 S1 S2 bi
Solution
values
0 S1 -1/3 0 1 -1/3 -1/3
-1 X2 2/3 1 0 -1/3 2/3
Z -2/3 -1 0 1/3
 = c - z -7/3 0 0 -1/3
∆/RowS1 7 1
The above solution is optimal but still infeasible. Hence the iteration is
carried out to write ST3
ST3
C
contribution
of basic
variables
X
basic
variables
X1 X2 S1 S2 bi
Solution
values
0 S2 1 0 -3 1 1
-2-1 X2 1 1 -1 0 1
Z -1 -1 1 0
 = c - z -2 0 -1 0

93
Solution in ST3 is optimal but no longer infeasible. The Dual Simplex
algorithm ends here. The decision variables are X1 = 0, X2=1.
Min value of G = 1
5.6 BIBLIOGRAPHY
1. Jhamb L.C., Quantitative Techniques for Managerial Decisions Vol I,
Pune, Everest Publishing House, 2004
2. Vohra, N. D., Quantitative Techniques in Management, Third Edition,
New-Delhi, Tata McGraw-Hill Education Private Limited, 2007
3. Taha Hamdy, Operations Research, Sixth Edition, New Delhi, Prentice
Hall of India, 2002
1. TOYCO assembles three types of toys namely trains, trucks and cars
using three operations. The daily limits on the available times for the
three operations are 430, 460 & 420 minutes respectively. and the
profits per toy train, truck & car are Rs.3/- Rs2/- & Rs5/- respectively.
The assembly times per train at the three operations are 1, 3 & 1
minutes respectively. The corresponding times per truck and per car
are (2, 0, 4) and (1, 2, 0) minutes (a zero time indicates the operation is
not used). TOYCO wants to maximize the profit
a) Find the optimal solution
b) Write the dual of the above problem
2. A company manufactures and sells three models of large sized
pressure cookers for canteen use. While market demands pose no
constraints, supplies of aluminum limited to 750 kgs. per week and
availability of machine time limited to 600 hours per week restrict the
product mix. Resource usage of three models and their profitability are
given below:
M1 M2 M3
Aluminum/unit 6 3 5
Machine time/unit 3 4 5
Cost contribution Rs/unit 60 20 80
Formulate the problem as an LPP and solve for optimal solution by using
the concept of dual.
3. Write the dual of the following LPP and find the optimal solution
Objective function: Minimize Z =5X1 - 6X2 + 6X3
Constraints:
3X1 + 4X2 + 6X3 ≥ 9
X1 + 3X2 + 2X3 ≥ 5
7X1 - 2X2 - X3 ≤ 10

94
X1 - 2X2 + 4X3 ≤ 4
2X1 + 5X2 - 3X3 = 3
X1, X2, X3 ≥ 0 Non negativity constraint
Write the dual of the following LPP and find the optimal solution
Objective function: Maximize Z = X1 - X2 + 3X3
Constraints:
X1 + X2 + X3 ≤ 10
2X1 - X3 ≤ 2
2X1 - 2X2 + 3X3 ≤ 6
X1, X2, X3 ≥ 0 ( Non negativity constraint)
4. Solve following problems using dual simplex method
a) Maximize Z = - 3X1 - 2X2
Constraints:
X1 + X2 ≥ 1
X1 + 2X2 ≥ 10
X1 + X2 ≥ 7
X2 ≤ 3
X1, X2, ≥ 0 ( Non negativity constrain)
b) Minimize Z = 3X1 + X2
Constraints:
X1 + X2 ≥ 1
2X1 + 3X2 ≥ 2
X1, X2, ≥ 0 ( Non negativity constrain)
c) Minimize Z = 10X1 +6X2 + 2X3
Constraints:
- X1 + X2 + X3 ≥ 1
3X1 + X2 - X3 ≥ 2
X1, X2, X3 ≥ 0 ( Non negativity constraint)


95
6
TRANSPORTATION PROBLEM
Unit Structure :
6.0 Objectives
6.1 Introduction
6.2 Mathematical Model of Transportation Problem
6.2.1 General Mathematical Model of Transportation Problem
6.3 The Transportation Algorithm
6.4 Check your progress
6.5 Steps of MODI Method (Transportation Algorithm)
6.6 Degeneracy and its Resolution
6.7 Additional Problems
6.8 Let us sum up
6.9 Future References
6.0 OBJECTIVES
After going through this chapter you will be able to :
 Understand what is a Transportation Problem.
 Formulate a Transportation problem
 Know the different methods of finding Initial Basic Feasible
solution to a Transportation problem
 Know the Modified Distribution (MODI) method of obtaining the
optimum solution to the Transportation Problem.
 Distinguish among the different types of a Transportation Problem.
 Solve the Transportation Problem.
6.1 INTRODUCTION
One important application of linear programming is in the area of
physical distribution (transportation) of goods and services from several
supply centres to several demand centres. It is easy to express a

96
transportation problem mathematically in terms of an LP model, which
can be solved by the simplex method. But because it involves a large
number of variables and constraints, it takes a long time to solve it.
However, transportation algorithms, namely the Stepping Stone Method
and the MODI (modified distribution) Method, have been developed for
his purpose.
The structure of transportation problem involves a large number of
shipping routes from several supply origins to several demand
destinations. The objective is to determine the number of units of an item
(commodity or product) which should be shipped from an origin to a
destination in order to satisfy the required quantity of goods or services at
each destination centre, within the limited quantity of goods or services
available at each supply centre, at the minimum transportation cost and /
or time.
The transportation algorithm discussed in this chapter is applied to
minimize the total cost of transporting a homogeneous commodity
(product) from supply centres to demand centres. However, it can also be
applied to the maximization of some total value of utility, for example,
financial resources are distributed in such a way that the profitable return
is maximized.
There are various types of transportation models and the simplest
of them was first presented by F L Hitchcock (1941). It was further
developed by T. C. Koopmans (1949) and G B Dantzig (1951). Several
extensions of transportation model and methods have been subsequently
developed.
6.2 MATHEMATICAL MODEL OF TRANSPORTATION
PROBLEM
Let us consider Example 6.1 to illustrate the mathematical model
formulation of transportation problem of transporting single commodity
from three sources of supply to four demand destinations. The sources of
supply are production facilities, warehouses, or supply points,
characterized by available capacities. The destinations are consumption
facilities, warehouses or demand points, characterized by required levels
of demand.
Example 6.1
A company has three production facilities S1, S2 and S3 with
production capacity of 7, 9 and 18 units (in 100s) per week of a product,
respectively. These units are to be shipped to four warehouses D1, D2, D3
and D4 with requirement of 5, 6, 7 and 14 units (in 100s) per week,
respectively. The transportation costs (in rupees) per unit between
factories to warehouses are given in the table below.

97
D1 D2 D3 D4 Capacity
S1 19 30 50 10 7
S2 70 30 40 60 9
S3 40 8 70 20 18
Demand 5 8 7 14 34
Formulate this transportation problem as an LP model to minimize the
total transportation cost.
Model formulation :
Let xij = number of units of the product to be transported from
factory i(i = 1, 2, 3) to warehouse j (j = 1, 2, 3, 4)
The transportation problem is stated as an LP mode as follows :
Minimize (total transportation cost) Z = 11 12 1319 30 50  x x x
14 21 22 23 24 31 32 33 3470 30 40 60 40 8 70 20        x x x x x x x x x
(i) Capacity constraints
11 12 13 14
21 22 23 24
31 32 33 34
7
9
18
x x x x
x x x x
x x x x
   
   
   
(ii) Requirement constraints
11 21 31
12 22 32
13 23 33
14 24 34
5
8
7
14
0 .ij
x x x
x x x
x x x
x x x
x for i and j
  
  
  
  
    
and
in the above LP model, there are 3 4 12m n    decision
variables, xij and m + n = 7 constraints, where m are the number of rows
and n are the number of columns in a general transportation table.
6.2.1 General Mathematical Model of Transportation Problem :
Let there be m sources of supply, S1, S2, ….Sm having ai (i = 1, 2,
…, m) units of supply (or capacity) respectively, to be transported among
n destinatiorn, D1, D2, …, Dn with bj(j = 1, 2, …, n) units of demand (or
requirement) respectively. Let cij be the cost of shipping one unit of the
commodity from source i to destination j for each route. If xij represents
number of units shipped per route from source i to destination j, the

98
problem is to determine the transportation schedule so as to minimize the
total transportation cost satisfying supply and demand conditions.
Mathematically, the problem in general may be stated as follows :
Minimize (total cost)
1 1
m n
ij ij
i j
Z c x 
 
 (1)
1
, 1, 2,....,
n
ij i
j
x a i m

      (Supply constraints) (2)
1
, 1, 2,....,

     
m
ij j
i
x b j n (Demand Constraints) (3)
And 0ijx  for all i and j. (4)
For easy presentation and solution, a transportation problem data is
generally presented as shown in Table 6.1.
Existence of feasible solution : A necessary and sufficient condition for
the existence of a feasible solution to the transportation problem is
Total supply = Total demand
1 1
m n
i j
i j
a b 
 
 (also called rim conditions)
That is, the total capacity (or supply) must equal total requirement
(or demand).
Table 6.1 General Transportation Table
To
from
D1 D2 … Dn Supply
ai
S1 11c 12c … 1nc 1a
S2 21c 22c … 2nc 2a
     
Sm 1mc 2mc … mnc ma
Demand
bj
b1 b2 … bn
1 1
m n
i j
i i
a b 
 

x1
1
x1
2
x1
n
x2
1
x2
2
x2
n
xm1 xm2 xmn

99
In this problem, there are (m +n) constraints one for each of supply
and destination and m × n variables. Since all (m + n) constraints are
equations. Since the transportation model is always balanced (total supply
= total demand), one of these equations is extra (redundant). The extra
constraint equation can be derived from the other constraint equations
without affecting the feasible solution. it follows that any feasible solution
for a transportation problem must have exactly (m + n - 1) non-negative
basic variables (or allocations) xij satisfying rim conditions.
Remarks :
1. When the total supply equals total demand, the problem is called a
balanced transportation problem, otherwise an unbalanced
transportation problem. The unbalanced transportation problem can be
made balanced by adding a dummy supply centre (row) or a dummy
demand centre (column) as the need arises.
2. When the number of positive allocations (values of decision variables)
at any stage of the feasible solution is less than the required number
(rows + columns - 1), i.e. number of independent constraint equations,
the solution is said to be degenerate, otherwise non-degenerate.
3. Cells in the transportation table having positive allocation will be
called occupied cells, otherwise empty or non-occupied cells.
6.3 THE TRANSPORTATION ALGORITHM
The algorithm for solving to a transportation problem may be
summarized into the following steps :
Step 1 Formulate the Problem and Arrange the data in the Matrix
Form :
The formulation of the transportation problem is similer to the LP
problem formulation. Here the objective function is the total transportation
cost and the constraints are the supply and demand available at each
source and destination, respectively.
Step 2 Obtain an Initial Basic Feasible Solution :
In this chapter, following three different methods are discussed to
obtain an initial solution :
 North-West Corner Method.
 Least Cost Method, and
 Vogel’s Approximation (or Penalty) Method.
The initial solution obtained by any of the three methods must
satisfy the following conditions :

100
i) The solution must be feasible, i.e. it must satisfy all the supply and
demand constraints (also called rim conditions).
ii) The number of positive allocations must be equal to m + n - 1, when m
is the number of rows and n is the number of columns.
Any solution that satisfies the above conditions is called non-
degenerate basic feasible solution, otherwise, degenerate solution.
Step 3 Test the Initial Solution for Optimality :
The modified Distribution (MODI) method is discussed to test the
optimality of the solution obtained in Step 2. If the current solution is
optimal then stop. Otherwise, determine a new improved solution.
Step 4 Updating the Solution :
Repeat Step 3 until an optimal solution is reached.
Example 6.2
Explain with an example the North West Corner rule, the Least
Cost method and the Vogel’s Approximation Method of obtaining an
initial feasible solution for a transportation problem.
Answer
Consider the example of three manufacturing plans with three
markets. The table given below indicates the capacities of each plant,
requirements of each market and unit shipping costs from each plant to
each market.
Plants Markets Plant capacities
1W 2W 3W
1P 3 4 2 30
2P 2 1 5 25
3P 4 3 3 20
Market
requirements
20 20 35
The problem is to make assignment in such a way so as to
minimise the total shipping cost.
1. North-West Corner Rule
The initial feasible solution of the above problem by the North-
West corner rule involves the following steps.
1. According to this rule, the first allocation is made in the cell in first
column of the first row. So we start with the cell on the intersection of

101
1P and 1W with shipping cost 3. The column total (i.e. market
requirement) corresponding to this cell is 20 while the row total (Plant
capacity) is 30. So we allocate 20 units to this cell. Not the market
requirement of 1W has been satisfied, we move horizontally in the row
1P to the cell 1P 2W .
2. The market requirements of column 2W is 20 while a total of 10 units
are left in row 1P . Thus, 10 units are assigned to this cell. With this,
the supply of the row 1P is exhausted.
3. Now, move vertically to the cell 2P 2W . For this cell, the market
requirement remains 10 and the plant capacity being 25, assign 10
units to this cell and exhaust the requirement of market 2W . Then,
move to the cell 2P 3W and allocate remaining 15 units of plant 2P to
this cell, exhausting the capacity of plant 2P leaving requirement of 20
units for market 3W .
4. Now capacity of plant 3P is 20 units which will satisfy the
requirement of market 3W so we allocate 20 units to cell 3P 3W
exhausting all plant capacities and market requirements.
These assignments are shown in the table below :
1W 2W 3W
1P
3 4 2
30 / 10 / 0
2P
2 1 5
25 / 15 / 0
3P
4 3 3
20 / 0
Market
requirements
20 / 0 20 / 10 / 0 35 / 20 / 0
2. The Least-Cost Method
This method involves following steps for finding initial feasible solution :
i) The least cost method starts by making the first allocation to that
cell whose shipping cost per unit is lowest.
20 10
10 15
20

102
ii) This lowest cost cell is loaded or filled as much as possible in view
of the plant capacity of its row and the market requirements of its column.
In our example, the cell 2P 2W has the least shipping cost, so we allocate
20 units to this cell exhausting the requirements of market 2W .
iii) We move to the next lowest cost cell and make an allocation in
view of the remaining capacity and requirement of its row and column. In
case there is a tie for the lowest cost cell during any allocation we make an
allocation in the cell which will satisfy either the maximum market
requirement or exhaust the plant capacity. In this example, there is a tie
between cells 1P 3W and the cell 2P 1W . So we choose cell 1P 3W because
it will exhaust maximum capacity i.e. 30 units.
iv) The above procedure is repeated till all rim requirements are
satisfied.
The initial feasible solution of the given problem using the least-
cost method is given in the following table.
1W 2W 3W
1P
3 4 2
30/0
2P
2 1 5
25/5/0
3P
4 3 3
20/15/0
Market
requirements
20/15/0 20/0 35/5/0
3. Vogel’s Approximation Method
Various steps involved for finding a initial feasible solution are
given below:
i) For each row of the transportation matrix, calculate difference
between the smallest and the next smallest element. Similarly for
each column of the matrix, compute the difference of the smallest and
next smallest element of the column.
ii) Identify the row or column with largest difference. If a tie occurs, use
a row or column which will either exhaust the maximum plant
capacity or satisfy the maximum market requirements. In the given
example, the largest difference is 2, thus the column designated as
2W is selected.
5
15
20
30
5

103
iii) Allocate the maximum feasible quantity in the box corresponding to
the smallest cost in that row or column. Eliminate that row or column
where an allocation is made. The lowest cost in column 2W is 1,
hence the cell 2P 2W is selected for allocation, 20 units are assigned
to this cell and column 2W is deleted as the requirement of market
2W is satisfied.
iv) Re-compute row and column differences for the reduced
transportation table. The above procedure is repeated until all the rim
requirements are satisfied. The assignments made by VAM are given
in the following table.
Plants Markets Plant Diff.
capacity
1W 2W 3W
1P
3 4 2
30/0 1/1/1
2P
2 1 5
25/5/0 1/3
3P
4 3 3
20/15/0 0/1/1
Market
requirements
Diff.
20/15/0
1/1/1
20/0
2
35/5/0
1/1/1
The minimum total transportation cost associated with this solution is
Total cost = (3 × 8 + 4 × 4 + 6 × 1 + 4 × 4 + 6 × 2 + 6 × 3) × 10 = Rs. 920
6.4 CHECK YOUR PROGRESS
Determine an initial basic feasible solution to the following
transportation problems using (a) north-west corner method and (b)
Vogel’s method.
1.
Destination
E E F G Supply
A 11 13 17 14 250
Source B 16 18 14 10 300
C 21 24 13 10 400
Demand 200 225 275 250
5
15
20
30
5

104
2. Consider the transportation problem having the following cost and
requirements table :
Destination
1 2 3 4 Supply
1 5 8 3 6 30
Source 2 4 5 7 4 50
3 6 2 4 5 40
Demand 30 20 40 30
6.5 STEPS OF MODI METHOD (TRANSPORTATION
ALGORITHM)
The steps to evaluate unoccupied cells are as follows :
Step 1 For an initial basic feasible solution with m + n -1 occupied cells,
calculate ui and vj for rows and columns. The initial solution can be
obtained by any of the three methods discussed earlier.
To start with, any one of uj’s or vj’s is assigned the value zero. It is
better to assign zero for a particular ui or vj where there are maximum
number of allocation in a row or column respectively, as it will reduce
arithmetic work considerably. Then complete the calculation of ui’s and
vj’s for other rows and columns by using the relation
cij = ui + vj for all occupied cells (i, j).
Step 2 For unoccupied cells, calculate opportunity cost (the difference that
indicates the per unit cost reduction that can be achieved by an allocation
in the unoccupied cell) by using the relationship
dij = cij - (ui + vj) for all i and j.
Step 3 Examine sign of each dij
i) If dij > 0, then current basic feasible solution is optimal.
ii) If dij = 0, then current basic feasible solution will remain unaffected
but an alternative solution exists.
iii)If one or more dij < 0, then an improved solution can be obtained by
entering unoccupied cell (i, j) in the basis. An occupied cell having the
largest negative value of dij is chosen for entering into the solution mix
(new transportation schedule).
Step 4 Construct a closed-path (or loop) for the unoccupied cell with
largest negative opportunity cost. Start the closed path with the selected
unoccupied cell and mark a plus sign (+) in this cell, trace a path along the
rows (or column) to an occupied cell, mark the corner with minus sign (-)
and continue down the column (or row) to an occupied cell and mark the

105
corner with plus (+) sign and minus sign (-) alternatively. Close the path
back to the selected unoccupied cell.
Step 5 Select the smallest quantity amongst the cell marked with minus
sign on the corners of closed loop. Allocate this value to the selected
unoccupied cell and add it to other occupied cells marked with plus sign
and subtract it from the occupied cells marked with minus sign.
Step 6 Obtain a new improved solution by allocating units to the
unoccupied cell according to Step 5 and calculate the new total
transportation cost.
Step 7 Test the revised solution further for optimality. The procedure
terminates when all 0ijd  , for unoccupied cells.
Remarks :
1. The closed-loop (path) starts and ends at the selected unoccupied cell.
It consists of successive horizontal and vertical (connected) lines
whose end points must be occupied cells, except for an end point
associated with entering unoccupied cell. This means that every corner
element of the loop must be an occupied cell.
It is immaterial whether the loop is traced in a clockwise or anti-
clockwise direction and starting up, down, right or left (but never
diagonally). However, for a given solution only one loop can be
constructed for each unoccupied cell.
2. There can be only one plus (+) sign and only one minus (-) sign in any
row or column.
3. The closed path indicates changes involved in reallocating the
shipments.
6.6 DEGENERACY AND ITS RESOLUTION
A basic feasible solution for the general transportation problem
must consist of exactly m + n -1 (number of rows + number of columns -
1) positive allocations in independent positions in the transportation table.
A solution will be called degenerate when the number of occupied cells is
less than the required number, m + n - 1. In such cases, the current
solution cannot be improved because it is not possible to draw a closed
path for every occupied cell. Also, the values of dual variables ui and vj
which are used to test the optimality cannot be computed. Thus, we need
to remove the degeneracy to improve the given solution. The degeneracy
in the transportation problems may occur at two stages :
a) When obtaining an initial basic feasible solution we may have less
than m + n - 1 allocations.

106
b) At any stage while moving towards optimal solution. This happens
when two or more occupied cells with the same minimum allocation
become unoccupied simultaneously.
Case 1 Degeneracy at the initial solution :
To resolve degeneracy at the initial solution, we proceed by
allocating a very small quantity close to zero to one or more (if needed)
unoccupied cells so as to get m + n -1 number of occupied cells. This
amount is denoted by a Greek letter  (epsilon) or  (delta). This quantity
would not affect the total cost as well as supply and demand values. In a
minimization transportation problem it is better to allocate  to
unoccupied cells that have lowest transportation costs, whereas in
maximization problems it should be allocated to a cell that has a high
payoff value. In some cases,  must be added in one of those unoccupied
cells which makes possible the determination of ui and vj uniquely.
The quantity  is considered to be so small that if it is transferred
to an occupied cell it does not change the quantity of allocation. That is,
0;
0 ;
ij ij ijx x x
k
 
 
  
It is also then obvious that  does not affect the total transportation
cost of the allocation. Hence, the quantity  is used to evaluate
unoccupied cells and to reduce the number of improvement cycles
necessary to reach an optimal solution. Once purpose is over,  can be
removed from the transportation table.
The minimum total transportation cost associated with this solution is
Total cost = (3 × 8 + 4 × 4 + 6 × 1 + 4 × 4 + 6 × 2 + 6 × 3) × 10 =
Rs. 920
Cost 2 Degeneracy at subsequent iterations :
To resolve degeneracy which occurs during optimality test, the
quantity may be allocated to one or more cells which have become
unoccupied recently to have m + n - 1 number of occupied cells in the new
solution.
6.7 ADDITIONAL PROBLEM
Example 6.3
A manufacturer wants to ship 22 loads on his product as shown
below. The matrix gives the kilometers from sources of supply to the
destinations.

107
Destination
D1 D2 D3 D4 D5 Supply
S1 5 8 6 6 3 8
Source S2 4 7 7 6 5 5
S3 8 4 6 6 4 9
Demand 4 4 5 4 8 22
25
Shipping cost is Rs. 0 per load per km. What shipping schedule should be
used to minimize total transportation cost?
Solution :
Since the total destination requirement of 25 units exceeds the total
resource capacity of 22 by 3 units, the problem is unbalanced. The excess
requirement is handled by adding a dummy plant, Sexcess with a capacity
equal to 3 units. We use zero unit transportation cost to the dummy plant.
The modified transportation table is shown in Table 6.3.1.
Initial solution 6.3.1
S1 5 8 6 6 3 8
S2 4 7 7 6 5 5
S3 8 4 6 6 4 9
Sexcess 0 0 0 0 0 3
Demand 4 4 5 4 8 25
1v  2v  3v  4v  5v 
The initial solution is obtained by using Vogel’s approximation
method as shown in Table 6.3.1. Since the solution includes 7 occupied
cells, therefore, the initial solution is degenerate. In order to remove
degeneracy we assign  to unoccupied cell  2 5S ,D which has minimum
cost among unoccupied cells as shown in Table 6.3.2.
4
4
5
3
1
3
5
1u 0
2u 
3u 
4u 

108
Table 6.3.2
D1 D2 D3 D4 D5 Supply ui
S1 5
+3
8
+5
6
(-)
6
+2
3
(+)
8 u1 = 0
S2 4 7
+2
7
-1
6
(+)
5
(-)
5 u2 = 2
S3 8
+5
4 6
-1
6
+1
4 9 u3 = 1
Sexcess 0
+2
0
+1
0
(+) -2
0
(-)
0
-7
3 u4 = - 4
Demand 4 4 5 4 8 25
vj v1 = 2 v2 = 3 v3 = 6 v4 = 4 v5 = 3
Determine ui and vj for occupied cells as shown in Table 6.3.2.
Since opportunity cost in the cell (Sexcess, D3) is largest negative, it must
enter the basis and the cell (S2, D5) must leave the basis. The new solution
is shown in Table 6.3.3.
Table 6.3.3
D1 D2 D3 D4 D5 Supply ui
S1 5
+3
8
+5
6
(-)
6
+2
3
(+)
8 u1 = 0
S2 4 7
+4
7
+1
6 5
+2
5 u2 = 0
S3 8
+3
4 6
-1
6
(+)
-1
4
(-)
9 u3 = 1
Sexcess 0
+2
0
+3
0
(+)
0
(-)
0
+3
3 u4 = - 6
Demand 4 4 5 4 8 25
vj v1 = 4 v2 = 3 v3 = 6 v4 = 6 v5 = 3
4
4
5
3
1
5
3

4
4
5
3
1
5
3


109
Repeat the procedure of testing optimality of the solutions given in
Table 6.33. The optimal solution is shown in Table 6.3. 4
Table 6.3. 4
S1 5 8 6 6 3 8
S2 4 7 7 6 5 5
S3 8 4 6 6 4 9
Sexcess 0 0 0 6 0 3
Demand 4 4 5 4 8 25
The minimum total transportation cost associated with this solution
is
Total cost = (3 × 8 + 4 × 4 + 6 × 1 + 4 × 4 + 6 × 2 + 6 × 3) × 10 =
Rs. 920
Example 6.4
Solve the following transportation problem for minimum cost:
Destinations Origins Requirements
A B C D
1 7 4 3 4 15
2 3 2 7 5 25
3 4 4 3 7 20
4 9 7 5 3 40
Availabilities 12 8 35 25
Solution
Since this is an unbalanced transportation problem, introduce a
dummy activity E with availability of 20 units. Apply vogel’s method to
find initial feasible solution.
4
4 2 3
3
1
8

110
A B C D E Requirement 1 2 3 4 5 6
1
7 4 3 4 0
15
0
3 1 1 1 1 1
2
3 2 7 5 0
25
17
12
2 1 2 2 2 2
3
4 4 3 7 0
20
0
3 1 1 1 4 -
4
9 7 5 3 0
40
20
0
3 2 2 - - -
Availability
12
0
8
0
35
15
0
25
5
0
20
0
100
Differene 1 2 0 1 0
1 2 0 1 0
1 - 0 1 -
1 - 0 1 -
- - 0 1 -
- 4 1 -
The initial feasible solution is given below :
A B C D E Requirements
1
7 4 3 4 0
15
2
3 2 7 5 0
25
3
4 4 3 7 0
20
4
9 7 5 3 0
40
Availability 12 8 35 25 20
12
15
8 5
20
20 20
15
12 8 5
20
20 20

111
This solution is tested for optimality. Since there are 7 allocations we
place a small entity (e) in the cell (E, 1). Assuming   1 i ju 0,u and v are
computed below :
A B C D E iu
1
7 4 3 4 0
0
2
3 2 7 5 0
2
3
4 4 3 7 0
0
4
9 7 5 3 0
0
jv 1 0 3 3 0
6 4 1
2 -2
3 4 4 0
8 7 2
ij matrix
Since one of the ij is -ve including the most-negative marginal
cost cell (E,2) in the loop and test the solution for optimality.
7 4 4 0
3 2 7 - 5
0
+
4 4 3 7 0
  min 5 , 20
5
    

9 7 5
+
3
-
0
20
12
15
8 5
20 20
e
12 8
20 20
5
15 e
20

112
Assume 0    i i ju u and v are calculated below : iu
7 4 3 4 0
0
3 2 7 5 0
0
4 4 3 7 0
0
9 7 5 3 0
0
jv 3 2 3 3 0
4 2 1
4 2
1 2 4 0
6 5 2
ij matrix
Since all ij 0, this is optimal solution and the allocation is given by
FROM C A B E C D E
TO 1 2 2 2 3 4 4
QTY. 15 12 8 5 20 25 15
COST 45 36 16 0 60 75 0 = 232 Total Cost
Example 6.5
Consider the following transportation cost table. The costs are
given in Rupees, the supply and demand are in units. Determine an
optimal solution.
Destination
Source 1 2 3 4 5 Supply
I 40 36 26 38 30 160
II 38 28 34 34 198 280
III 36 38 24 28 30 240
Demand 160 160 200 120 240
12
15
8
e
5
20
25 15

113
Solution
Since demand is greater than supply by 200 units, we introduce a
dummy source with a supply of 200 units. We, then, find initial solution
by vogel’s method.
Destination
Source 1 2 3 4 5 Difference
I 40 36 26 38 30 160/0 4 4 4 4 4
II 38 28 34 34 198 280/120/0 6 6 6 0 164
III 36 38 24 28 30 240/120/0 4 4 4 4 6
Dummy 0 0 0 0 0 200/40/0 0 0
Demand 160/0 160/0 200/
80/0
120/
0
240/
200/
40
Difference 36 28 24 28 30
28 24 28 30
8 2 6 0
2 6 0
Thus the initial solution is given by the table given above. To test
whether the initial solution is optimal, we apply optimality test. If the
solution is not optimal, then we will try to improve the solution to make it
optimal. To test the optimality of the initial solution, we find whether the
number of allocations is equal to m + n – 1 or not. Also, these allocations
should be independent.
Let us introduce the variables iu and jv . To determine the values
of these variables, let 3 0 u  , the values of other variables are shown in
the following table.
16
0
12
0
16
0
12
0
16
0
4080
40

114
1 2 3 4 5 i
I
40 36 26 38 30
0
II
38 28 34 34 198
10
III
36 38 24 28 30
0
Dummy
0 0 0 0 0
-30
jv 30 18 24 28 30
Let us now find      ij ij i jC (u v )   for the non basic cells.
1 2 3 4 5 i
I
10 18 2 10
0
II
-2 -
- 4
+ 158
10
III
6 20 + -
0
Dummy
12 6 2
-30
jv 30 18 24 28 30
Since there are negative ij s this solution is not optimal. Including the
most negative ij in the loop, the next improved solution is given below.
We test this solution for optimality. Since the total number of basic cells is
less than 8 (= m + n – 1), we put e which is a very small quantity in the
least cost independent cell and compute values of i and jv
80
16
0
40
160
120
16
0
120
40
80
16
0
40
160
120
16
0
120
40
120

115
Destinations
Source 1 2 3 4 5 i
I
40 36 26 38 30 30
II
38 28 34 34 198 34
III
36 38 24 28 30 30
Dummy
0 0 0 0 0 0
jv 0 -6 -6 0 0
The values of      ij ij i jC (u v )   are given in the table below :
1 2 3 4 5 i
I
10 12 2 8 30
II
4 6 164 34
III
6 14 -2 + - 30
Dummy
6 6 - + 0
jv 0 -6 -6 0 0
Since there is one negative ij this solution is not optimal.
Including this negative value in the loop, the next improved solution is
given below which is tested for optimality.
40
200
16
0
160
16
0
120
40
e
16
0
40
160
200
16
0
120
40
e

116
Destinations
Source 1 2 3 4 5 i
I
40 36 26 38 30 0
II
38 28 34 34 198 6
III
36 38 24 28 30 0
Dummy
0 0 0 0 0 - 30
jv 30 22 24 28 30
Now we find ij for non basic cells
Destination
sources 1 2 3 4 5 i
I
10 14 2 10 0
II
2 4 162 6
III
6 16 0
Dummy
8 6 2 -30
jv 30 22 24 28 30
200
16
0
40
160
e
16
0
120
40
200
16
0
40
160
16
0
120
40e

117
Since all ij s are positive, the above table will give optimal
solution. The optimal solution is given below.
From Source To Destination No. of Units Cost per unit Rs.
I 5 160 30
II 2 160 28
II 4 120 34
III 3 200 24
III 5 40 30
and the associated total cost is given by
160 × Rs. 30 + 160 × Rs. 28 + 120 × Rs. 34 + 200 × Rs. 24 + 40 × Rs. 30
= Rs. 19,360.
Example 6.6
Solve the following transportation Problem :
GODOWNS
1 2 3 4 5 6 Stock
Available
F 1 7 5 7 7 5 3 60
A
C 2 9 11 6 11 - 5 20
T
O 3 11 10 6 2 2 8 90
R
Y 4 9 10 9 6 9 12 50
DEMAND 60 20 40 20 40 40
Note : it is not possible to transport any quantity from factory, 2 to
Godown 5. State whether the solution derived by you is unique.

118
Solution
The initial solution is found by VAM below :
Factory Godowns Availability Diff.
1 2 3 4 5 6
1
7 5 7 7 5 3
60/40/0 2/4/0
2
9 11 6 11  5
20/10/0 1/3
3
11 10 6 2 2 8
90/70/30/0 0/4/2/5
4
9 10 9 6 9 12
50/0 3/0
Demand 60
50
0
20
0
40
10
0
20
0
40
0
40
0
Diff. 2 5 0/1 4 3 2
The initial solution is tested for optimality. Since there are only 8
allocations and we require 9 (m + n – 1 = 9) allocations, we put a small
quantity e in the least cost independent cell (2,6) and apply the optimality
test. Let 3 0u   and then we calculate remaining iu and jv
1 2 3 4 5 6 i
1
7 5 7 7 5 3 -2
2
9 11 6 11  5 0
3
11 10 6 2 2 8 0
4
9 10 9 6 9 12 0
jv 9 7 6 2 2 5
40
20 40
1010
2030
50
e
50
30 20 40
10 10
20 40

119
Now we calculate  ij ij i jC u v        for non basic cells which
given in the table below :
0 3 7 5
4 9 
2 3 3
3 3 4 7 7
ij
Since all ij are positive, the initial solution found by VAM is an
optimal solution. The final locations are given below :
Factory to Godown Unit Cost Value
1 2 20 5 100
1 6 40 3 120
2 1 10 9 90
2 3 10 6 60
3 3 30 6 180
3 4 20 2 40
3 5 40 2 80
4 1 50 9 450
Total cost Rs. 1,120
The above solution is not unique because the opportunity cost of
cell (1,2) is zero. Hence alternate solution exists. Students may find that
the alternate solution is as given below :
Factory to Godown Unit Cost Value
1 1 10 7 70
1 2 20 5 100
1 6 30 3 90
2 3 10 6 60
2 6 10 5 50
3 3 30 6 180
3 5 40 2 80
3 4 20 2 40
4 1 50 9 450
Total cost Rs. 1,120
Example 6.7
STRONGHOLD Construction company is interested in taking
loans from banks for some of its projects P, Q, R, S, T. The rates of
interest and the lending capacity differ from bank to bank. All these
projects are to be completed. The relevant details are provided in the
following table. Assuming the role of a consultant, advise this company as
to how it should tae the loans so to the total interest payable will be the

120
least. Are there alternate optimum solutions? If so, indicate one such
solution.
Interest rate in percentage for
PROJECT
Max. Credits
(in thousands)
Bank P Q R S T
Pvt. Bank 20 18 18 17 17 Any amount
Nationalized Bank 16 16 16 15 16 400
Co-operative Bank 15 15 15 13 14 250
Amount required
(in thousands)
200 150 200 125 75
Solution
The total amount required by five projects is Rs. 750 thousands.
Since private bank can give credit for any amount, we allocate [Rs. 750 –
(Rs. 400 + Rs. 250) = Rs. 100] thousand to private banks. The balanced
problem is given below. The initial solution is found using VAM.
P Q R S T Max. Cred. Diff.
Pvt. Bank
20 18 18 17 17 100/0 0/1/0/0
Nationalised
Bank
16 16 16 15 16 400/200/50/0 1/0/0/0
Co-operative
Bank
15 15 15 13 14 250/125/50/0 1/1/0
amount
required
200
0
150
100
0
200
150
0
125
0
75
0
difference 1 1 1 2 2
1 1 1 - 2
1 1 1 - -
4 2 2 - -
Let us test initial solutions for optimality.
There are m + n – 1 = 7 independent allocations.
Let us now introduce 1 2 3i ju ,v ,i , ,      and 1 2 5j , ,... .   
Let 2 0u   and calculate remaining  1 3iu i ,   and jv ’s  1 2 5j , ,...   
Calculate  ij ij i jC u v        for non allocated cells.
10
0
20
0
50 15
0
5
0
125 7
5

121
P Q R S T
iu
Pvt. Bank 2
20 18
0
18
1
17
0
17
2
Nationalised Bank
16 16 16
1
15
1
16
0
Co-operative Bank 0
15
0
15 15 13 14
-1
jv 16 16 16 14 15
Since none of the ij is negative, the initial solution obtained as
above is optimal one.
Total interest (as per above allocation)
= 100 × .18 + 200 × .16 + 50 × .16 + 150 × .16 + 50 × .15 + 125 × .13 +
75 × .14
= Rs 116.25 thousands
= Rs. 1,16,250
Further, since some of the ij ’s are zero, therefore the above solution is
not unique. To find out an alternative solution, let us include the cell
(CB,Q) as the basic cell so the new solution is given below :
P Q R S T
Pvt. Bank
20 18 18 17 17
Nationalised Bank
16 16 16 15 16
Co-operative Bank
15 15 15 13 14
10
0
200 50 15
0
50 12
5
75
10
0
200 15
0
50 12
5
75

122
Since there are only 6 allocations, let us introduce e-a small
quantity in the least independent cell (NB,S). We now introduce
1 2 3iu ,i , ,   and 1 2 5iv , j , ,... .   Let 3 u u Calculate  ij ij i jC u v       
for non basic cells.
P Q R S T
iu
Pvt. Bank 3
20 18
1
18
1
17
0
17
3
Nationalised
Bank
16
-1
+ 
16 16
- 
15
0
16
2
Co-operative
Bank
1
15
- 
15
1
15
+ 
13 14
0
jv 14 15 14 13 14
Since ij for cell (NB,Q) is negative, this solution is not optimal.
We include the cell (NB,Q) in the basic cell
0 = [min (50 –  , e –  ) =  ] = e
So the new solution given below is tested for optimality.
P Q R S T
iu
Pvt. Bank 2
20 18
0
18
1
17
0
17
2
Nationalised Bank
16 16 16
1
15
1
16
0
Co-operative Bank 0
15 15
0
15 13 14
-1
jv 16 16 16 14 15
10
0
200 e 20
0
50 12
5
75
10
0
200 e20
0
50 12
5
75

123
Since all ij are non-negative, the above solution is optimal one
and the total cost is
= 100 × 018 + 200 × .16 + 200 × .16 + 50 × .15 + 125 × .13 + 75 × .14 =
Rs. 116.25 thousands = Rs. 1,16,250
Note : The alternative solution can be found by taking any cell with zero
ij as the basic cell.
Example 6.8
Solve the following transportation problem to maximize profit and
give criteria for optimality :
Profit (Rs.) / Unit
destination
Origin Supply
1 2 3 4
A 40 25 22 33 100
B 44 35 30 30 30
C 38 38 28 30 70
Demands 40 20 60 30
Solution
As the given matrix is a profit matrix and the objective is to
maximize profit, we first of all convert the profit matrix into a loss matrix
for solving it by transportation method. This is done by multiplying the
given matrix by minus sign to get the following matrix.
Destination Profit (Rs.) / Unit
destination
Origin Supply
1 2 3 4
A -40 -25 -22 -33 100
B -44 -35 -30 -30 30
C -38 -38 -28 -30 70
Demands 40 20 60 30
Further, the above loss matrix is not a balanced one as (supply >
demand), we will therefore introduce a dummy destination and find the
initial feasible solution using VAM.

124
1 2 3 4 Dummy Supply Diff.
A
- 40 -25 -22 -33 0 100/70/50/0 7/7/7/1
B
- 44 -35 -30 -30 0 30/0 9
C
-38 -38 -28 -30 0 70/50/40/0 0/0/8/2
Demands 40/10/0 20/0 60/20/0 30/0 50/0
Diff. 4/2/2 3/13 2/6/6/6 3/3 0/0/0/0
The initial feasible solution after introducing the variables iu and
jv is tested for optimality as below :
1 2 3 4 Dummy
iu
A
- 8

- 40
+7
-25
-
-22
-33 0 0
B
- 44 -35
4
-30
15
-30
12
0 -12
C
-
-38 -38
+
-28
9
-30
6
0 -6
Demands 40/10/0 20/0 60/20/0 30/0
Since some of the ij ij i iC (u v )         for non-allocated cells
shown in the above table are not positive, the solution obtained above is
not an optimal solution. In order to obtain an optimal solution, place a
small assignment  in the cell with most negative value of ij . In the
above table,  is placed in the most negative cell (A,1) and a loop is
formed including this cell. Now, the maximum value which  can take is
10 units. After putting this value of  in the cell (A,1), other allocation
cells in the loop will also get affected as shown in the following table.
30
20
10
30
4020
50
3020 50
30
10 4020

125
1 2 3 4 Dummy
iu
A
+
- 40
7
-25
-
-22 -33 0 0
B
-
- 44
1
-35
- 4
+
-30
7
-30
4
0 -4
C
8
-38
-38
+
-28
9
-30
6
0 -6
jv -40 -32 -22 -33 0
Calculate again ij ij i iC (u v )         after determining the value of
iu and jv for non-allocated cells. Cell (B,3) has most negative value in
the above table. We place a small allocation  and determine its value as
described above which comes out to be 10 units. Putting the value of  ,
we get the following table.
1 2 3 4 Dummy
iu
A
- 40
11
-25
4
-22 -33 0 0
B
- 44 -35 -30
7
-30
4
0 -4
C
4
-38 -38 -28
5
-30
2
0 -2
jv -40 -36 -26 -33 0
In the above table, we find that the values of ij ij i iC (u v )        
are positive for all non-allocated cells. Hence we can say that allocation
shown by the table are optimal. The final allocation after converting the
loss matrix into profit matrix are given below :
10
30 50
20
20
5020
10 30 50
30
10
5020

126
1 2 3 4 Dummy
A
40 25 22 33 0
B
44 35 30 30 0
C
38 38 28 30 0
and the profit is given by
Profit = 20 × Rs. 40 + 30 × Rs. 33 + 20 × Rs. 44 + 10 × Rs. 30 + 20 × Rs.
38 + 50 × Rs. 28
= Rs. 800 + Rs. 990 + Rs. 880 + Rs. 300 + Rs. 760 + Rs. 1,400
= Rs. 5,130.
Note : The above problem can also be converted into a minimization
problem by subtracting each element of the given matrix from 44 (the
highest element).
Example 6.9
A manufacturer of jeans is interested in developing an adverting
campaign that will reach four different age groups. Advertising campaigns
can be conducted through TV, Radio and Magazines. The following table
gives the estimated cost in paise per exposure for each age group
according to the medium employed. In addition, maximum exposure
levels possible in each of the media, namely TV, Radio and Magazines are
40, 30 and 20 millions respectively. Also the minimum desired exposures
within each age group, namely 13-18, 19-25, 26-35, 36 and older, are
30,25,15 and 10 millions. The objective is to minimize the cost of
attaining the minimum exposure level in each age group.
Media Age Groups
13-18 19-25 26-35 36 & older
TV 12 7 10 10
Radio 10 9 12 10
Magazines 14 12 9 12
i) Formulate the above as a transportation problem, and find the
optimal solution.
ii) Solve this problem if the policy is to provide at least 4 million
exposures through TV in the 13-18 age group and at least 8 million
exposures through TV in the age group 19-25
30 5020
1020
5020

127
Solution
i) As a first step, let us formulate the given problem as a
transportation problem :
Media Age Groups
13-18 19-25 26-35 36 & older Exposures
available
(in million)
TV 12 7 10 10 40
Radio 10 9 12 10 30
Magazines 14 12 9 12 20
Minimum number of
exposures required (in
millions)
30 25 15 10 80/90
It is apparent from the above that this transportation problem is an
unbalanced on it is balanced by introducing a dummy category before
applying Vogel’s Approximation method.
Media
13-18 19-25 26-35 36 & older Dummy
category
Max. Exp.
(in million)
Penalty
TV
12 7 10 10 0
40/15/0 7/3/0/0/0
Radio
10 9 12 10 0
30/0 9/1/0/0
Magazines
14 12 9 12 0
20/10/0 9/3/3
Min. no. of
Exposures
required
(million)
30/0 25/0 15/5/0 10/0 10/0 90
Penalty 2 2 1 0 0
2 2 1 0 -
2 - 1 0 -
2 - 2 0 -
- - 10 10 -
The total cost for
this allocation
works out to Rs.
71.5 lakhs
2
5
5
30
1
0
10
10

128
The solution given by VAM is degenerate since there are only six
assignments. Let us put an E in the least cost independent cell to check for
optimality. Let 1 0u  and we calculate remaining iu and jv s .
Media
13-18 19-25 26-35 36 &
older
Dummy
category
iu
TV
12 7 10
1
10 0 0
Radio 0
10 9 12 10 0 -1
Magazines
14 12 9 12 0 -1
jv 11 7 10 10 1
Let us not calculate       ij ij i jC u v   for empty cells.
ij S
1  -1
3 3 1
4 6  3 
Since one of the ij s is negative, the solution given above is not
optimal. Let us include the cell with negative ij as a basic cell and try to
improve the solution. The reallocated solution is given below which is
tested for optimality.
25 5
30 E
10
1
0
10

129
13-18 19-25 26-35 36 &
older
Dummy
category
iu
TV
12 7 10
1
10 0 0
Radio 0
10 9 12 10 0 0
Magazines
14 12 9 12 0 0
jv 10 7 9 10 0
Let us calculate ij for empty cells.
2 1
2 3 0
4 5 2
Since all the entries in the last table are non-negative, the second
solution is optimal.
Through TV, 25 million people must be reached in the age group
19-25 and 10 million people in the age group 36 & older.
Through Radio, 30 million people must be reached in the age
group 13-18.
Through Magazines, 15 million people must be reached in the age
group 26-35.
And the total minimum cost of attaining the minimum exposure level is
Rs. 71 lakhs.
Note : since one of ij in the second solution is zero. This solution is not
unique, alternate solution also exists.
25 5
30 E
10
1
5
5

130
ii) The required solution is given by :
12 7 10
1
10 0 40
10 9 12 10 0 30
14 12 9 12 0 20
30 25 15 10 10
The total cost for this allotment is 25 × 7 = 175
10 × 10 = 100
15 × 9 = 135
30 × 10 = 300
= 710
i.e. Rs. 71 Lakhs.
Example 6.10
A company wishes to determine an investment strategy for each of
the next four years. Five investment types have been selected, investment
capital has been allocated for each of the coming four years, and
maximum investment levels have been established for each investment
type. An assumption is that amounts invested in any year will remain
invested until the end of the planning horizon of four years. The following
table summarises the data for this problem. The values in the body of the
table represent net return on investment of one rupee upto the end of the
planning horizon. For example, a rupee invested in investment type B at
the beginning of year 1 will grow to Rs. 1.90 by the end of the fourth
years, yielding a net return of Re. 0.90.
Investment type Rupees
available
(in 000’s)
A B C D E
Investment made
at the beginning
of year
NET RETURN DATA
1 0.80 0.90 0.60 0.75 1.00 500
2 0.55 0.65 0.40 0.60 0.50 600
3 0.30 0.25 0.30 0.50 0.20 750
4 0.15 0.12 0.25 0.35 0.10 800
Maximum Rupees
Investment (in
000’s)
750 600 500 800 1,000
510
1
5
5
25
30

131
The objective in this problem is to determine the amount to be
invested at the beginning of each year in an investment type so as to
maximize the net rupee return for the four-year period.
Solve the above transportation problem and get an optimal
solution. also calculate the net return on investment for the planning
horizon of four-year period.
Solution
We note that this transportation problem is an unbalanced one and
it is a MAXIMISATION problem. As a first step, we will balance this
transportation problem.
Step 1:
Investment type
Years A B C D E Available Rupees
(in 000’s)
Net Return Data (Re.)
1 0.80 0.90 0.60 0.75 1.00 500
2 0.55 0.65 0.40 0.60 0.50 600
3 0.30 0.25 0.30 0.50 0.20 750
4 0.15 0.12 0.25 0.35 0.10 800
Dummy 0 0 0 0 0 1,000
Maximum rupees
investment in (‘000)
750 600 500 800 1,000 3,650
Step 2:
We shall now convert the above transportation problem (a profit
matrix) into a loss matrix by subtracting all the elements from the highest
value in the table viz. Re. 1.00
Investment type
Years A B C D E Available Rupees
(in 000’s)
Net Loss Data (Re.)
1 0.20 0.10 0.40 0.25 0 500
2 0.45 0.35 0.60 0.40 0.50 600
3 0.70 0.75 0.70 0.50 0.80 750
4 0.85 0.88 0.75 0.65 0.90 800
Dummy 1.00 1.00 1.00 1.00 1.00 1,000
Maximum rupees
investment in (‘000)
750 600 500 800 1,000 3,650

132
For convenience, let us express the net loss data in the body of the
above table in paise. Thereafter, we shall apply VAM to get an initial
solution.
Years A B C D E Rs.
available
(000’s)
Difference
1
20 10 40 25 0
500/0 10 - - - -
2
45 35 60 40 50
600/0 5 5 - - -
3
70 75 70 50 80
750/0 20 20 20 - -
4
85 88 75 65 90
800/750/
250/0
10 10 10 10 10
Dummy
100 100 100 100 100
100/500/0 0 0 0 0 0
Maximum
rupee
investment
in (000)
750/
500/0
600/0 500/0 800/
50/0
1000/
500/0
differences 25 25 20 15 50
25 40 10 10 30
15 - 5 15 10
15 - 25 35 10
15 - 25 - 10
500
600
750
250 500 50
500500

133
The initial solution got above by VAM is given below :
Years A B C D E
1
20 10 40 25 0
2
45 35 60 40 50
3
70 75 70 50 80
4
85 88 75 65 90
Dummy 100 100 100 100 100
We will now test the above solution for optimality.
The total number of allocations should be 1 1 9n n        but there
are only 8 allocations in the above solution which are one less than 9,
hence the initial solution found above is degenerate. We introduce a small
quantity e in an independent least cost cell which is (1, B) in this case t
make the total number of allocations equal to 9. we introduce 'iu s and
'jv s such that   , 1, 5            ij ij i jC u v for i j    we assume that
4 0u   and various ,i ju s v s and 'ij s are calculated as below.
Years A B C D E
1 20
-
10
50 45
+
1 85u  
2 20
35
45 35 25 2 60u  
3 0 -5 10
50
10 3 15u  
4
85
-7
75 65 5
4u  
Dummy 100
-10
+
10 20
-
100
5u  
1 85u   2u   3 5u   4u   5 8u  
500e
600
750
250 500 50
500500
500e
600
750
250 500 50
500 500

134
Since some of the 'ij s are negative, the above initial solution is
not optimal. Introducing in the cell (Dummy, B) with most negative ij
an assignment  . The value of  and the new solution as obtained from
above is shown below. The values of ,i ju s v s are also calculated. The
solution satisfies the conditions of optimality. The condition
  0ij ij i jC u v            for non allocated cells is also fulfilled.
Investment Type
Years A B C D E
1 20 10 50 45
0
1 85u  
2 10
35
35 25 10 2 50u  
3 0 575 10
50
100 3 15u  
4
85
3
75 65
5 4u  
Dummy 100 100
10 20
100
5u  
1 85u   2u   3 5u   4u   5 8u  
Since all ij are positive, hence, the second solution obtained
above is optimal. The allocation is given below :
In the year Investment type Amount (in 000’s)
1 E 500
2 B 600
3 D 750
4 A 250
4 C 500
4 D 50
The net return on investment for the planning horizon of four years
period is given by :
500 × 1.0 + 600 × 0.65 + 750 × 0.50 + 250 × 0.15 + 500 × 0.25 +
50 × 0.35 = Rs. 1,445 thousands.
500
e
600
750
250 500 50
500500

135
Example 6.11
A leading firm has three auditors. Each auditor can work upto 160
hours during the next month, during which time three projects must be
completed. Project 1 will take 130 hours, project 2 will take 140 hours, the
project 3 will take 160 hours. The amount per hour that can be billed for
assigning each auditor to each project is given in Table 1 :
Project
Auditor 1
Rs.
2
Rs.
3
Rs.
1 1,200 1,500 1,900
2 1,400 1,300 1,200
3 1,600 1,400 1,500
Formulate this as a transportation problem and find the optimal
solution. Also find out the maximum total billings during the next month.
Solution
The given information can be tabulated in following transportation
Example:
Project
Auditor 1
Rs.
2
Rs.
3
Rs.
Time
available (hours)
1 1,200 1,500 1,900 160
2 1,400 1,300 1,200 160
3 1,600 1,400 1,500 160
Time required (hrs) 130 140 160
The given problem is an unbalanced transportation. Introducing a
dummy project to balance it, we get.
Auditor 1
Rs.
2
Rs.
3
Rs.
Dummy
(Rs.)
Time
available
(hours)
1 1,200 1,500 1,900 0 160
2 1,400 1,300 1,200 0 160
3 1,600 1,400 1,500 0 160
Time required (hrs) 130 140 160 50 480
The objective here is to maximize total billing amount of the
auditors. For achieving this objective, let us convert this maximisation
problem into a minimization problem by subtracting all the elements of
the above payoff matrix from the highest payoff i.e. Rs. 1,900.

136
Auditor 1
Rs.
2
Rs.
3
Rs.
Dummy
(Rs.)
Time
available
(hours)
1 700 400 0 1,900 160
2 500 600 700 1,900 160
3 300 500 400 1,900 160
Time required (hrs) 130 140 160 50 480
Now, let us apply vogel’s Approximation Method to the above
matrix for finding the initial feasible solution.
Project (Figure of payoff’s in Rs. 00’s)
Auditor 1 2 3 Dummy Time
available
Difference
1
7 4 0 19 160/0 4/-/-/-
2
5 6 7 19 160/50/0 1/1/13/13
3 3 5 4 19 160/30/0 1/2/14/-
Time
Required
Difference
130/0
2/2/-/-
140/110/0
1/1/1/
160/0
4/-/-
50/0
0/0/0
The initial solution is given below. It can be seen that it is a
degenerate solution since the number of allocations is 5. In order to apply
optimality test, the total number of allocations should be 6 (= m + n – 1).
To make the initial solution a non-degenerate, we introduce a very small
quantity in the least cost independent cell which is cell of Auditor 3,
Project 3.
Auditor 1 2 3 Dummy Time
available
1
7 4 0 19 160
2
5 6 7 19 160
3 3 5 4 19
160
Time
Required
130 140 160 50
110
3
0
13
0
5
0
160
160
110
3
0
13
0
5
0
e

137
Introduce 'iu s and 'jv s such that  ij ij i jC u v         
 , 1 3; 1,2,3,for i to j dummy          . To determine the values of 'iu s and
'jv s , we assume 0iu   , values of other variables i.e. 'iu s , ij s are
calculated as follows :
6.8 LET US SUM UP
Thus The Transportation Problem is a particular case of the Linear
Programming Problem. There are 3 methods of obtaining the initial basic
feasible solution to the transportation problem.
1) The North West Corner Rule (NWCR)
2) The Least Cost Method / The Matrix Minima Method (MMM)
3) Vogel’s Approximation Method (VAM)
The initial basic feasible solution obtained by VAM is very close
to the optimum solution. Further the MODI method is applied to obtain
the optimum solution of the Transportation Problem.
The variations in the Transportation Problem are :
1) The Maximisation Case : Before obtaining the initial basic feasible
solution to the Transportation problem, it is necessary that the objective
function is to minimize. The maximisation Problem can be converted to
the Minimisation Problem by any one of the following method.
i) Multiply all the elements of maximisation variable by - 1.
ii) Subtract all the elements of maximisation variable from the highest
element.
2) The Unbalanced Problem : The Transportation Problem is a
balanced problem when the row total = column total. When it is an
unbalanced Transportation Problem it is necessary to introduce a dummy
row or a dummy column according to the requirement of the problem.
3) Restrictive Allocation.
4) Multiple Optimum Solutions.
6.9 FUTURE REFERENCES
1. Budnick Frank : Mcleavey Dennis
2. Mojena Richard : Principles of Operations Research for Management
: Richard D. Irwin INC.
3. Kapoor V.K : Operations Research Problems and Solutions.
Sultanchand and Sons.
4. Sharma, J.K : Operations Research Theory and Applications
Macmillan India Limited.

138
5. Sharma S. D : Operations Research Kedar Nath Ram Nath Publishers.
6. Taha Hamdy A : Operations Research an Introduction Prentice Hall
of India.
7. Verma A. P : Operations Research S. K. Kataria & Sons.
8. Vohra N.D : Quantitative Techniques in Management. Tata McGraw-
Hill Publishing Company Limited, New Delhi.
1. A steel company has three open hearth furnaces and five rolling
mills. Transportation costs (rupees per quintal) for shipping steel from
furnaces to rolling mills are shown in the following table :
M1 M2 M3 M4 M5 Supply
F1 4 2 3 2 6 8
F2 5 4 5 2 1 12
F3 6 5 4 7 7 14
Demand 4 4 6 8 8
What is the optimal shipping schedule?
2. Consider the following unbalanced transportation problem.
To
I II III Supply
A 5 1 7 10
From B 6 4 6 80
C 3 2 5 15
Demand 75 20 50
Since there is not enough supply; some of the demands at these
destinations may be satisfied. Suppose there are penalty costs for every
unsatisfied demand unit which are given by 5, 3 and 2 for destinations I, II
and III, respectively. Find the optimal solution.
3. A company produces a small component for an industrial product
and distributes it to five wholesalers at a fixed delivered price of Rs. 2.50
per unit. Sales forecasts indicate that monthly deliveries will be 3,000,
3,000, 10,000, 5,000 and 4,000 units to wholesalers I, II, III, IV and V
respectively. The monthly production capacities are 5,000, 10,000 and
12,500 at plants W, X and Y, respectively. Respective direct costs of
production of each unit are Re 1.00, Re 0.90, and Re 0.80 at plants W, X
and Z. Transportation costs of shipping a unit from a plant to a wholesaler
are as follows.

139
Wholesaler
I II III IV V
W 0.05 0.07 0.10 0.25 0.15
Plant X 0.08 0.06 0.09 0.12 0.14
Y 0.10 0.09 0.08 0.10 0.15
Find how many components each plant supplies to each wholesaler
in order to maximize its profit.
4. Obtain an optimum basic feasible solution to the following
degenerate transportation problem.
To
A B C Supply
X 7 3 4 2
From Y 2 1 3 3
Z 3 4 6 5
Demand 4 1 5
5. A manufacturer wants to ship 8 loads of his product shown in the
table. The matrix gives the mileage from origin to destination. Shipping
costs are Rs. 10 per load per mile. What shipping schedule should be
used?
D1 D2 D3 Supply
O1 50 30 220 1
O2 90 45 170 3
O3 250 200 50 4
Demand 4 2 2
6. A cement company has three factories which manufacture cement
which is then transported to four distribution centres. The quantity of
monthly production of each factory, the demand of each distribution
centre and the associated transportation cost per quintal are given below :
Distribution centresFactory
W X Y Z
Monthly production (in
quantals)
A 10 8 5 4 7,000
B 7 9 15 8 8,000
C 6 10 14 8 10,000
Monthly
demand
(in
quintals)
6,000 6,000 8,000 5,000 25,000

140
i) Suggest the optimum transportation schedule.
ii) Is there any other transportation schedule which is equally
attractive? If so, write that.
iii) If the company wants that at least 5,000 quintals of cement are
transported from factory C to distribution centre Y, will the transportation
schedule be any different? If so, what will be the new optimum schedule
and the effect on cost?
7. A company has three plants of cement which has to be transported
to four distribution centres. With identical costs of production at the three
plants, the only variable costs involved are transportation costs. The
monthly demand at the four distribution centres and the distance from the
plants to the distribution centres (in kms) are given below :
Distribution centres
Factory W X Y Z
Monthly production
(tonnes)
A 500 1,000 150 800 10,000
B 200 700 500 100 12,000
C 600 400 100 900 8,000
Monthly
demand
(tonnes)
9,000 9,000 10,000 4,000
The transport charges are Rs. 10 per tonne per kilometer. Suggest
optimum transportation schedule and indicate the total minimum
transportation cost. If, for certain reasons, route from plant C to
distribution centre X is closed down, will the transportation scheme
change? If so, suggest the new schedule and effect on total cost.
8. A company has three factories A, B and C and four distribution
centres W, X, Y and Z. With identical costs of production at the three
factories, the only variable costs involved are transportation costs. The
monthly production at the three factories is 5,000 tonnes, 6,000 tonnes and
2,500 tonne respectively. The monthly demand at the four distribution
centres is 6,000 tonnes, 4,000 tonnes, 2,000 tonnes and 15,000 tonnes,
respectively. The transportation cost per tonne from different factories to
different are given below :
Distribution centres
Factory W X Y Z
A 3 2 7 6
B 7 5 2 3
C 2 5 4 5
i) Suggest the optimum transportation schedule. What will be the
minimum transportation cost?
ii) If the transportation cost from factory C to centre Y increases by
Rs. 2 per tone, will the optimum transportation schedule change? Why?

141
iii) If the company wants to transport at least 1,000 tonnes of the
product from factory A to centre Z, will the solution in (i) above change?
If so, what will be the new schedule and the effect on total transportation
cost?
9. A company has three plants and four warehouses. The supply and
demand in units and the corresponding transportation costs are given. The
table below has been taken from the solution procedure of the
transportation problem :
Distribution Centres
Plant I II III IV
Supply
A
5 10 4 5
10
B
6 8 7 2
25
C
4 2 5 7
20
Demand 25 10 15 5 55
Answer the following questions, giving brief reasons :
1. Is this solution feasible?
2. Is this solution degenerate?
3. Is this solution optimum?
4. Does this problem have more than one optimum solution? If so, show
all of them?
5. If the cost for the route B-III is reduced from Rs. 7 to Rs. 6 per unit,
what will be the optimum solution?
10. A company has three plants in which it produces a standard
product. It has four agencies in different parts of the country where this
product is sold. The production cost varies from factory to factory and the
selling price from market to market. The shipping cost per unit of the
product from each plant to each of the agencies is known and is stable.
The relevant data are given in the following table :
a) Plant Weekly production capacity
(Units)
Unit production cost
(in Rs.)
1 400 18
2 300 24
3 800 20
10
20 5
55 10

142
b) Shipping cost (in Rs.) per unit :
To Agency
1 2 3 4
1 2 5 7 3
From Plant 2 8 4 6 2
3 3 4 4 5
c) Agency Demand (Units) Selling price (Rs.)
1 300 32
2 400 35
3 300 31
4 500 36
Determine the optimum plan so as to maximize the profits.
11. ABC Enterprises is having three plants manufacturing dry-cells,
located at different locations. Production cost differs from plant to plant.
There are five sales offices of the company located in different regions of
the country. The sales prices can differ from region to region. The
shipping cost from plant to each sales office and other data are given by
following table :
PRODUCTION DATA TABLE
Production cost per unit Max. capacity in number of
units
Plant number
20 150 1
22 200 2
18 125 3
Shipping costs :
Sales
office 1
Sales
office 2
Sales
office 3
Sales
office 4
Sales
office 5
Plant 1 1 1 4 9 4
Plant 2 9 7 8 3 3
Plant 3 4 5 3 2 7
Demand & sales price
Demand 0 100 75 45 125
Sales Price 0 32 31 34 29
Find the production and distribution schedule most profitable to the
company.

143
12. A company has seven manufacturing units situated in different
parts of the country. Due to recession it is proposing to close four of these
units and to concentrate production on the three remaining units.
Production in the remaining units will actually be increased from present
levels and will require an increase in the personnel employed in them.
Personnel at the units to be closed have signified a willingness to move to
any of three remaining units and the company is willing to provide them
with removal costs. The technology of production is different to some
degree at each unit and retraining expenses will be incurred on transfer.
Not all existing personnel can be absorbed by transfer and a number of
redundancies will arise. Cost of redundancy is given is a general figure at
each unit to be closed.
Number employed A B C D
Retraining cost : 200 400 300 200
Rs. (thousand) per person
Transfer to: A B C D
Unit E 0.5 0.4 0.6 0.3
Unit F 0.6 0.4 0.6 0.3
Unit G 0.5 0.3 0.7 0.3
Removal costs : Rs. (thousand) per person
Transfer to: A B C D
Unit E 2.5 3.6 3.4 3.7
Unit F 2.4 4.6 3.4 1.7
Unit G 2.5 2.7 3.3 2.7
Redundancy payments : Rs. (thousand) per person
A B C D
6.0 5.0 6.0 7.0
Additional personnel required at
units remaining open :
E F G
350 450 200
Use the transportation method to obtain an optimum solution to the
problem of the cheapest (least cost) means to transfer personnel from the
units to be closed to those will be expanded.


144
7
ASSIGNMENT PROBLEM
Unit Structure :
7.0 Objectives
7.1 Introduction
7.2 Mathematical Model of Assignment Problem
7.3 Solution methods of assignment problem
7.3.1 Enumeration method
7.3.2 Simplex method
7.3.3 Transportation method
7.3.4 Hungarian method
7.4 Check your progress
7.5 Additional Problems
7.6 Let us sum up
7.7 Future References
7.0 OBJECTIVES
After going through this chapter you will be able to :
 Understand what is an assignment problem.
 Distinguish between the transportation problem and an assignment
problem.
 Know the algorithm to solve an assignment problem.
 Formulate an assignment problem.
 Identify different types of assignment problems.
 Solve an assignment problem.
7.1 INTRODUCTION
An assignment problem is a particular case of transportation
problem where the objective is to assign a number of resources to an equal
number of activities so as to minimize total cost or maximize total profit
of allocation.

145
The problem of assignment arises because available resources such
as men, machines, etc., have varying degrees of efficiency for performing
different activities. Therefore, cost, profit or time of performing the
different activities is different. Thus, the problem is : How should the
assignments be made so as to optimize the given objective.
Some of the problems where the assignment technique may be
useful are : Assignment of workers to machines, salesmen to different
sales areas, clerks to various checkout counters, classes to rooms, vehicles
to routes, contracts to bidders, etc.
7.2 MATHEMATICAL MODEL OF AN ASSIGNMENT
PROBLEM
Given n resources (or facilities) and n activities (or jobs), and
effectiveness (in terms of cost, profit, time, etc.) of each resource (facility)
for each activity (job), the problem lies in assigning each resource to one
and only one activity (job) so that the given measure of effectiveness is
optimized. The data matrix for this problem is shown in Table 7.1.
From Table 7.1, it may be noted that the data matrix is the same as
the transportation cost matrix except that supply (or availability) of each
of the resources and the demand at each of the destinations is taken to be
one. It is due to this fact that assignments are made on a one-to-one basis.
Table 7.1 Data Matrix
Resources (workers) Activities (jobs)
1J 2J … nJ
supply
1W 11c 12c … 1nc 1
2W 21c 22c … 2nc 1
. . . . .
. . . . .
. . . . .
nW n1c n2c … nnc 1
Demand 1 1 … 1 n
Let ijx denote the assignment of facility i to job j such that



ij
1 if facility i is assigned to job j
x
0 otherwise

146
Then, the mathematical model of the assignment problem can be stated as:
Minimize
n n
ij ij
i 1 j 1
Z c x
 
 subject to the constraints.
n
ij
j 1
x 1

 for all i (resource availability)
n
ij
i 1
x 1

 , for all j (activity requirement) and 0ijx or 1,
for all i and j where ijc represents the cost of assignment of resource i to
activity j.
From the above discussion, it is clear that the assignment problem
is a variation of the transportation problem with two characteristics : (i)
the cost matrix is a square matrix, and (ii) the optimal solution for the
problem would always be such that there would be only one assignment in
a given row or column of the cost matrix.
Remark : In an assignment problem if a constant is added to or subtracted
from every element of any row or column in the given cost matrix, an
assignment that minimizes the total cost in one matrix also minimizes the
total cost in the other matrix.
7.3 SOLUTION METHODS OF ASSIGNMENT
PROBLEM
An assignment problem can be solved by the following four
methods :
 Enumeration method
 Simplex method
 Transportation method
7.3.1 Enumeration Method :
In this method, a list of all possible assignments among the given
resources (men, machines, etc.) and activities (jobs, sales areas, etc.) is
prepared. Then an assignment involving the minimum cost (or maximum
profit), time or distance is selected. If two or more assignments have the
same minimum cost (or maximum profit), time or distance, the problem
has multiple optimal solutions.
In general, if an assignment problem involves n workers / jobs,
then there are in total n! possible assignments. For example, for an n = 5
workers/ jobs problem, we have to evaluate a total of 5! or 120

147
assignments. However, when n is large, the method is unsuitable for
manual calculations. Hence, this method is suitable only for small n.
7.3.2 Simplex Method :
Since each assignment problem can be formulated as a 0 or 1
integer linear programming problem, such a problem can also be solved by
the simplex method. As can be seen in the general mathematical
formulation of the assignment problem, there are n × n decision variables
and n + n or 2n equalities. In particular, for a problem involving 5 workers
/ jobs, there will be 25 decision variables and 10 equalities. It is, again,
difficult to solve manually.
7.3.3 Transportation Method :
Since an assignment problem is a special case of the transportation
problem, it can also be solved by transportation methods discussed in
Chapter 6. However, every basic feasible solution of a general assignment
problem having a square payoff matrix of order n should have n + n - 1 =
2n - 1 assignments. But due to the special structure of this problem, any
solution cannot have more than n assignments. Thus, the assignment
problem is inherently degenerate. In order to remove degeneracy, (n-1)
number of dummy allocations (deltas or epsilons) will be required in order
to proceed with the algorithm solving a transportation problem. Thus, the
problem of degeneracy at each solution makes the transportation method
computationally inefficient for solving an assignment problem.
7.3.4 Hungarian Assignment Method (HAM) :
It may be observed that none of the three working methods
discussed earlier to solve an assignment is efficient. A method, designed
specially to handle the assignment problems in an efficient way, called the
Hungarian Assignment Method, is available, which is based on the
concept of opportunity cost. It is shown in Table 7.1 and is discussed here.
For a typical balanced assignment problem involving a certain number of
persons and an equal number of jobs, and with an objective function of the
minimization type, the method is applied as listed in the following steps :
Step 1 Locate the smallest cost element in each row of the cost table. Now
subtract this smallest element from each element in that row. As a result,
there shall be at least one zero in each row of this new table, called the
Reduced Cost Table.
Step 2 In the Reduced Cost Table obtained, consider each column and
locate the smallest element in it. Subtract the smallest value from every
other entry in the column. As a consequence of this action, there would be
at least one zero in each of the rows and columns of the second reduced
cost table.

148
Step 3 Draw the minimum number of horizontal and vertical lines (not the
diagonal ones) that are required to cover all the ‘zero’ elements. If the
number of lines drawn is equal to n (the number of rows/columns) the
solution is optimal and proceed to step 6. If the number of lines drawn is
smaller than n, go to step 4.
Step 4 Select the smallest uncovered (by the lines) cost element. Subtract
this element from all uncovered elements including itself and add this
element to each value located at the intersection of any two lines. The cost
elements through which only one line passes remain unaltered.
Step 5 Repeat steps 3 and 4 until an optimal solution is obtained.
Step 6 Given the optimal solution, make the job assignments as indicated
by the ‘zero’ elements. This is done as follows :
a) Locate a row which contains only one ‘zero’ element. Assign the
job corresponding to this element to its corresponding person. Cross out
the zero’s, if any, in the column corresponding to the element, which is
indicative of the fact that the particular job and person are no more
available.
b) Repeat (a) for each of such rows which contain only one zero.
Similarly, perform the same operation in respect of each column
containing only one ‘zero’ element, crossing out the zero(s), if any, in the
row in which the element lies.
c) If there is row or column with only a single ‘zero’ element left,
then select a row/column arbitrarily and choose one of the jobs (or
persons) and make the assignment. Now cross the remaining zeros in the
column and row in respect of which the assignment is made.
d) Repeat steps (a) through (c) until all assignment are made.
e) Determine the total cost with reference to the original cost table.
Example 7.1
Solve the assignment optimal solution using HAM.
Table 7.1 Time Taken (in minutes) by 4 workers
Job
Worker
A B C D
1 45 40 51 67
2 57 42 63 55
3 49 52 48 64
4 41 45 60 55

149
The solution to this problem is given here in a step-wise manner.
Step 1 The minimum value of each row is subtracted from all elements in
the row. It is shown in the reduced cost table, also called Opportunity Cost
Table, given in Table 7.2.
Table 7.2 Reduced Cost Table 1
Job
Worker
A B C D
1 5 0 11 27
2 15 0 21 13
3 1 4 0 16
4 0 4 19 14
Step 2 For each column of this table, the minimum value is subtracted
from all the other values.
Obviously, the columns that contain a zero would remain unaffected by
this operation. Here only the fourth column values would change. Table
7.3 shows this.
Job
Worker
A B C D
1 5 0 11 14
2 15 0 21 0
3 1 4 0 3
4 0 4 19 1
Step 3 Draw the minimum number of lines covering all zeros. As a
general rule, we should first cover those rows/columns which contain
larger number of zeros. Table 7.3 is reproduced in Table 7.4 and the lines
are drawn.
Job
Worker
A B C D
1 5 0 11 14
2 15 0 21 0
3 1 4 0 3
4 0 4 19 1
Step 4 Since the number of lines drawn is equal to 4 (=n), the optimal
solution is obtained. The assignments are made after scanning the rows
and columns for unit zeros. Assignments made are shown with squares, as
shown in Table 7.5

150
Table 7.5 Assignment of Jobs
Job
Worker
A B C D
1 5 0 11 14
2 15 0 21 0
3 1 4 0 3
4 0 4 19 1
Assignments are made in the following order. Rows 1, 3 and 4
contain only one zero each. So assign 1-B, 3-C and 4-A. Since worker 1
has been assigned job B, we cross the zero in the second column of the
second row. After making these assignments, only worker 2 and job D are
left for assignment. The final pattern of assignments is 1-B, 2-D, 3-C and
4-A, involving a total time of 40 + 55 + 48 + 41 = 184 minutes. This is the
optimal solution to the problem - the same as obtained by enumeration and
transportation methods.
Example 7.2
Using the following cost matrix, determine (a) optimal job
assignment, and (b) the cost of assignments.
JobMachinist
1 2 3 4 5
A 10 3 3 2 8
B 9 7 8 2 7
C 7 5 6 2 4
D 3 5 8 2 4
E 9 10 9 6 10
Iteration 1 Obtain row reductions.
JobMachinist
1 2 3 4 5
A 8 1 1 0 6
B 7 5 6 0 5
C 5 3 4 0 2
D 1 3 6 0 2
E 3 4 3 0 4
Iteration 2 Obtain column reductions and draw the minimum number
of lines to cover all zeros.

151
JobMachinist
1 2 3 4 5
A 7 0 0 0 4
B 6 4 5 0 3
C 4 2 3 0 0
D 0 2 5 0 0
E 2 3 2 0 2
Since the number of lines covering all zeros is less than the number
of columns/ rows, we modify the Table 7.7. The least of the uncovered
cell values is 2. This value would be subtracted from each of the
uncovered values and added to each value lying at the intersection of lines
(corresponding to cells A-4, D-4, A-5 and D-5). Accordingly, the new
table would appear as shown in Table 7.8.
Iteration 3
JobMachinist
1 2 3 4 5
A 7 0 0 2 6
B 4 2 3 0 3
C 2 0 1 0 0
D 0 2 5 2 2
E 0 1 0 0 2
The optimal assignments can be made as the least number of lines
covering all zeros in Table 7.8 equals 5. Considering rows and columns,
the assignments can be made in the following order :
i) Select the second row, Assign machinist B to job 4. Cross out zeros
at cells C-4 and E-4.
ii) Consider row 4. Assign machinist D to job 1. Cancel the zero at cell
E-1.
iii) Since there is a single zero in the fifth row, put machinist E to job 3
and cross out the zero at A-3.
iv) There being only a single zero left in each of the first and third rows,
we assign job 2 to machinist A and job 5 to C.
The total cost associated with the optimal machinist job
assignment pattern A-2, B-4, C-5, D-1 and E-3 is 3 + 2 + 4 + 3 9 = 21.

152
7.4 CHECK YOUR PROGRESS
1. What is an assignment problem? Give two applications.
2. Give the mathematical formulation of an assignment problem. How
does it differ from a transportation problem?
3. Explain the conceptual justification that an assignment problem can be
viewed as a linear programming problem.
4. Explain the difference between a transportation problem and an
assignment problem.
5. State and discuss the methods for solving an assignment problem. How
is the Hungarian method better than other methods for solving an
assignment problem?
6. Solve the following assignment problem by (a) enumeration method,
and (b) Hungarian assignment method.
Time (in minutes)
Worker Job 1 Job 2 Job 3
A 4 2 7
B 8 5 3
C 4 5 6
7. ABC Company is engaged in manufacturing 5 brands of packed
snacks. It has five manufacturing setups, each capable of
manufacturing any of its brands one at a time. The costs to make a
brand on these setups vary according to the following table.
1S 2S 3S 4S 5S
1B 4 6 7 5 11
2B 7 3 6 9 5
3B 8 5 4 6 9
4B 9 12 7 11 10
5B 7 5 9 8 11
Assuming five setups are 1S , 2S , 3S , 4S and 5S and five brands are
1B , 2B , 3B , 4B and 5B . Find the optimum assignment of products on
these setups resulting in the minimum cost.
8. Five employees of a company are to be assigned to five jobs which can
be done by any of them. Because of different number of years with the
firm, the workers get different wages per hour. These are : Rs. 15 per
hour for A, B and C each, and Rs. 13 per hour for D and E each. The
amount of time taken (in hours) by each employee to do a given job is
given in the following table. Determine the assignment pattern that (a)

153
minimizes the total time taken, and (b) minimizes the total cost of
getting five units of work done.
EmployeeJob
A B C D E
1 7 9 3 3 2
2 6 1 6 6 5
3 3 4 9 10 7
4 1 5 2 2 4
5 6 6 9 4 2
9. A departmental head has four subordinates and four tasks to be
performed. The subordinates differ in efficiency and the tasks differ in
their intrinsic difficulty. His estimates of the times that each man
would take to perform each task is given below in the matrix :
Tasks
I II III IV
A 8 26 17 11
B 13 28 4 26
Subordinates C 38 19 18 15
D 19 26 24 10
How should the tasks be allocated to subordinates so as to minimize the
total man-hours?
10. An automobile dealer wishes to put four repairmen to four different
jobs. The repairmen have somewhat different kinds of skills and they
exhibit different levels of efficiency from one job to another. The
dealer has estimated the number of man-hours that would be required
for each job-man combination. This is given in matrix form in the
following table :
jobs
A B C D
1 5 3 2 8
2 7 9 2 6Men
3 6 4 5 7
4 5 7 7 8
Find the optimal assignment that will result in minimum man-hours
needed.
7.5 ADDITIONAL PROBLEMS
Example 7.3
Solve the following unbalanced assignment problem of minimizing
total time for doing all the jobs :

154
Job
Operator 1 2 3 4 5
1 6 2 5 2 6
2 2 5 8 7 7
3 7 8 6 9 8
4 6 2 3 4 5
5 9 3 8 9 7
6 4 7 4 6 8
Solution
Since the number of operators is unequal to the number of jobs, a
dummy job is created. The time consumed by any operator for the dummy
job is 0.
Job
Operator 1 2 3 4 5 6
1 6 2 5 2 6 0
2 2 5 8 7 7 0
3 7 8 6 9 8 0
4 6 2 3 4 5 0
5 9 3 8 9 7 0
6 4 7 4 6 8 0
Subtracting the smallest value in each column from all the values
of that column drawing minimum number of lines to cover all zeros we
have :
1 2 3 4 5 6
1 4 0 2 0 1 0
2 0 3 5 5 2 0
3 5 6 3 7 3 0
4 4 0 0 2 0 0
5 7 1 5 7 2 0
6 2 5 1 4 3 0
We find that no more than four lines are needed to cover all zeros,
which is not equal to number of assignments. We note that all rows
contain at least one zero, and 1 is the smallest value not covered by a line.
Subtracting 1 from every uncovered value and adding 1 to every value at
the intersection of two lines, we find
1 2 3 4 5 6
1 4 0 2 0 1 1
2 0 3 5 5 2 1
3 4 5 2 6 2 0
4 4 0 0 2 0 1
5 6 0 4 6 1 0
6 1 4 0 3 2 0

155
Drawing minimum number of lines to cover all zeros, we find
1 2 3 4 5 6
1 4 0 2 0 1 1
2 0 3 5 5 2 1
3 4 5 2 6 2 0
4 4 0 0 2 0 1
5 6 0 4 6 1 0
6 1 4 0 3 2 0
Now the minimal number of lines is equal the number of
assignments that can be made. Optimal assignment can be made by
sciecting one zero in each row so that no two selected zeros are in the
same column.
1 2 3 4 5 6
1 4 0 2 0 1 1
2 0 3 5 5 2 1
3 4 5 2 6 2 0
4 4 0 0 2 0 1
5 6 0 4 6 1 0
6 1 4 0 3 2 0
Thus the optimal assignment is
Operator 1 to job 4, Operator 2 to job 1, Operator 3 to dummy 6
Operator 4 to job 5, Operator 5 to job 2, Operator 6 to job 3
Time = 2 + 2 + 0 + 5 + 3 + 4 = 16 units.
Example 7.4
A solicitors’ firm employs typists on hourly piece-rate basis for
their daily work. There are five typists for service and their charges and
speeds are different. According to an earlier understanding only one job is
given to one typist and the typist is paid for full hours even if he works for
a fraction of an hour. Find the least cost allocation for the following.
Typist Rate per hour
(Rs.)
No. of pages
Typed / hour
Job No. of pages
A 5 12 P 199
B 6 14 Q 175
C 3 8 R 145
D 4 10 S 298
E 4 11 T 178

156
Solution
The following matrix gives the cost incurred if the ith typist (i =
A,B,C,D,E) executes the jth
job (j = P,Q,R,S,T)
Job
Typist P Q R S T
A 85 75 65 125 75
B 90 78 66 132 78
C 75 66 57 114 69
D 80 72 60 120 72
E 76 64 56 112 68
Subtracting the minimum element of each row from all its
elements in turn matrix reduces to
Job
Typist P Q R S T
A 20 10 0 60 10
B 24 2 0 66 12
C 18 9 0 57 12
D 20 12 0 60 12
E 20 8 0 56 12
Now subtract the minimum element of each column from all its
elements in turn, of matrix reduces to :
Job
Typist P Q R S T
A 2 2 0 4 0
B 6 4 0 10 2
C 0 1 0 1 2
D 2 4 0 4 2
E 2 0 0 0 2
Since there are only 4 lines (<5) to cover all zeros, optimal
assignment cannot be made. The minimum uncovered element is 2.
We subtract the value 2 from all uncovered elements, add this
value to all junction values and leave the other elements undisturbed. The
revised matrix looks as

157
Job
Typist P Q R S T
A 2 2 2 4 0
B 4 2 0 8 0
C 0 1 2 1 2
D 0 2 0 2 0
E 2 0 2 0 2
Since the minimum no. of lines required to cover all the zeros is
only 4(<5), optimal assignment cannot be made at this stage also.
The minimum uncovered element is 1. Repeating the usual process
again, we get the following matrix :
Job
Typist P Q R S T
A 2 1 2 3 0
B 4 1 0 7 0
C 0 0 2 0 2
D 0 1 0 1 0
E 3 0 3 0 0
Since the minimum number of lines to cover all zero is equal to 5,
this matrix will given optimal solution. The optimal assignment is made in
the matrix below.
Job
Typist P Q R S T
A 2 1 2 3 0
B 4 1 0 7 0
C 0 0 2 0 2
D 0 1 0 1 0
E 3 0 3 0 3
Cost Rs.
Thus typist A is given Job T : 75
Thus typist B is given Job R : 66
Thus typist C is given Job Q : 66
Thus typist D is given Job P : 80
Thus typist E is given Job S : 112
Rs. 399
Note : In this case the above solution is not unique. Alternate solution also
exists.

158
Example 7.5
To stimulate interest and provide an atmosphere for intellectual
discussion, a finance faculty in a management school decides to hold
special seminars on four contemporary topics leasing, portfolio
management, private mutual funds, swaps and options. Such seminars
should be held once per week in the afternoons. However, scheduling
these seminars (one for each topic, and not more than one seminar per
afternoon) has to be done carefully so that the number of students unable
to attend is kept to a minimum. A careful study indicates that the number
of students who cannot attend a particular seminar on a specific day is as
follows :
Leasing Portfolio
Management
Private
Mutual funds
Swaps and
Options
Monday 50 40 60 20
Tuesday 40 30 40 30
Wednesday 60 20 30 20
Thursday 30 30 20 30
Friday 10 20 10 30
Find an optimal schedule of the seminars. Also find out the total
number of students who will be missing at least one seminar.
Solution
This is an unbalanced minimization assignment problem.
We first of all balance it by adding a dummy topic.
Leasing Portfolio
Management
Private
Mutual
funds
Swaps
and
Options
Dummy
Monday 50 40 60 20 0
Tuesday 40 30 40 30 0
Wednesday 60 20 30 20 0
Thursday 30 30 20 30 0
Friday 10 20 10 30 0
Subtracting the minimum element of each column from all
elements of that column, we get the following matrix :
Leasing Portfolio
Management
Private
Mutual
funds
Swaps
and
Options
Dummy
Monday 40 20 50 0 0
Tuesday 30 10 30 10 0
Wednesday 50 0 20 0 0
Thursday 20 10 10 10 0
Friday 0 0 0 10 0

159
The minimum number of lines to cover all zeros is 4 which is less
than the order of the square matrix (i.e.5), the above matrix will not give
the optimal solution, Subtract the minimum uncovered element (=10) from
all uncovered elements and add it to the elements lying on the intersection
of two lines, we get the following matrix.
Leasing Portfolio
Management
Private
Mutual
funds
Swaps
and
Options
Dummy
Monday 30 20 40 0 0
Tuesday 20 10 20 10 0
Thursday 10 10 0 10 0
Friday 0 10 0 20 10
Since the minimum number of lines to cover all zeros is 5 which is
equal to the order to the matrix, the above matrix will give the optimal
solution which is given below :
Leasing Portfolio
Management
Private
Mutual
funds
Swaps
and
Options
Dummy
Monday 30 20 40 0 0
Tuesday 20 10 20 10 0
Thursday 10 10 0 10 0
Friday 0 10 0 20 10
And the optimal schedule is
No. of students missing
Monday : Swaps and options 20
Tuesday : No seminar 0
Wednesday : Portfolio Management 20
Thursday : Pvt. mutual funds 20
Friday : leasing 10
70
Thus, the total number of students who will be missing at least one
seminar = 70
Example 7.6
Four operators 1 2 3, ,O O O  and 4O are available to a manager who
has to get four jobs 1 2 3, ,J J J  and 4J done by assigning one job to each
operator. Given the times needed by different operators for different jobs
in the matrix below :

160
1J 2J 3J 4J
1O 12 10 10 8
2O 14 12 15 11
3O 6 10 16 4
4O 8 10 9 7
i) How should the manager assign the jobs so that the total time
needed for all four jobs is minimum?
ii) If job 2J is not to be assigned to operator 2O what should be the
assignment how much additional total time will be required?
Solution
i) This is an assignment problem whose objective is to assign one job
to one operator so that total time needed for all four jobs is
minimum. To determine appropriate assign of jobs and operators,
let us apply the assignment algorithm. Subtract the minimum
element of each row from all elements of that row to get the
following matrix :
Job
Operators
1J 2J 3J 4J
1O 4 2 2 0
2O 3 1 4 0
3O 2 6 12 0
4O 1 3 2 0
Now subtract the minimum element of each column from all
elements of that column
Job
Operators
1J 2J 3J 4J
1O 3 1 0 0
2O 2 0 2 0
3O 1 5 10 0
4O 0 2 0 0
The minimum number of lines drawn to cover all zeros is equal to
4. Since the number of lines drawn viz., 4 is equal to the number of jobs or
the number of operators, so we proceed for making the optimal
assignment.

161
The optimal assignment is made as below :
Job
Operators
1J 2J 3J

M.c.a.(sem iii) operation research

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