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Table of Content
COURSE INTRODUCTION ................................................................................................................... 2
COURSE OBJECTIVES.......................................................................................................................... 3
UNIT ONE ................................................................................................................................................... 4
1. INTRODUCTION TO OPERATIONS RESEARCH .............................................................................. 4
1.1 LEARNING OBJECTIVES: .............................................................................................................. 4
1.2 GENERAL INTRODUCTION .......................................................................................................... 4
1.3 HISTORY OF OPERATIONS RESEARCH ..................................................................................... 5
1.4 MODEL............................................................................................................................................ 11
1.5 OPERATION RESEARCH IN PRACTICE .................................................................................... 13
1.6 OPERATIONS RESEARCH APPROACH ..................................................................................... 13
CHAPTER SUMMARY .................................................................................................................... 14
SELF ASSESSMENT QUESTIONS (SQA1) ................................................................................... 16
UNIT TWO................................................................................................................................................. 17
2. LEANER PROGRAMMING PROBLEM MODELS (LPPM) .............................................................. 17
2.1 LEARNING OBJECTIVES: ............................................................................................................ 17
2.2 GENERAL INTRODUCTION ........................................................................................................ 17
2.3 LINEAR PROGRAMMING PROBLEM MODELS ....................................................................... 18
2.4 FORMULATION OF LPPM ........................................................................................................... 22
2.5 APPROACHES TO SOLVE LPPM ................................................................................................ 26
2.6 MAXIMIZATION PROBLEM ........................................................................................................ 35
2.7 SPECIAL CASES IN SIMPLEX METHOD ................................................................................... 55
2.8 SENSATIVITY ANALYSIS (POST OPTIMALITY ANALYSIS) ................................................ 62
UNIT SUMMARY ............................................................................................................................ 79
SELF ASSESSMENT QUESTIONS (SAQ 2) .................................................................................. 81
UNIT THREE ............................................................................................................................................. 85
3. TRANSPORTATION AND ASSIGNMENT MODELS ....................................................................... 85
3.1 LEARNING OBJECTIVES ............................................................................................................. 85
3.2. INTRODUCTION ......................................................................................................................... 86
3.3 TRANSPORTATION PROBLEMS ................................................................................................ 86
3.3. SPECIAL CASES ......................................................................................................................... 113
3.4 ASSIGNMENT PROBLEMS ........................................................................................................ 121
UNIT SUMMARY .......................................................................................................................... 141
SELF ASSESMENT QUESTIONS (SQL-4) .................................................................................. 142
UNIT FOUR ............................................................................................................................................. 145
4. DECISION THEORY .......................................................................................................................... 145
4.1 LEARNING OBJECTIVES ........................................................................................................... 145
4.2 INTRODUCTION .......................................................................................................................... 145
4.2 DECISION THEORY .................................................................................................................... 146
4.3 DECICISIN CRITERIA ................................................................................................................. 150
UNIT SUMMARY .......................................................................................................................... 157
SELF ASSESMENT QUESTIONS (SQL-4) .................................................................................. 157
UNIT FIVE ............................................................................................................................................... 160
5. PROJECT MANAGEMENT................................................................................................................ 160
5.1 LEARNING OBJECTIVES ........................................................................................................... 160
5.2 INTRODUCTION .......................................................................................................................... 160
5.3 PLANNING AND SCHEDULING OF PROJECTS ..................................................................... 161
5.4 PROJECT DIAGRAMS (NET- WORK DIAGRAMS)................................................................. 163
5.5 COMPUTING ALGORITHMS ..................................................................................................... 170
UNIT SUMMARY .......................................................................................................................... 179
SELF ASSESMENT QUESTIONS (SQL-4) .................................................................................. 180
ANSWER KEY FOR SELF ASSESSMENT QUESTIONS ........................................................... 182

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COURSE INTRODUCTION

OPERATIONS RESEARCH/ MANAGEMENT SCIENCE, in essence is a branch of especially applied

mathematics, used to provide a scientific base for management to take timely and effective decisions
to their problems. It tries to avoid the dangers from taking decisions merely by guessing or by using
thumb rules.

Management is the multidimensional and dynamic concept. It is multidimensional, because

management problems and their solutions have consequences in several dimensions, such as human,
economic, social and political fields. Hence, any manager, while making decisions, considers all
aspects in addition to economic aspect, so that his/her solution should be useful in all aspects. The
general approach is to analyze the problem in economic terms and then implement the solution if it
is not aggressive or violent to other aspects like human, social and political constraints.

Management may be considered as the process of integrating the efforts of a purposeful group, or
organization, whose members have at least one common goal. You have studied various schools of
management in your Introduction to Management Module.

This is an introductory course on the use of quantitative models that can be used to describe and
solve certain kinds of problems that confront managers. It is written at a level that makes it suitable
for distance education students.

The main purposes of this module are to introduce readers to the models that comprise the field of
management science (operations research) and to help readers to develop ad improve their problem
solving skills. The presentation of the material is more students oriented. A rigorous math
background is not needed; an understanding of basic algebra is sufficient

This module, therefore, is prepared in a way that you as a distant learner shall be able to
understand the concept of Operations Research through its meaning, characteristics, and its
importance; the fundamentals of Operations Research and most importantly the models in the
subject, i.e. linear programming , transportation and assignment models , decision theory and
project management.

The module is organized so that you can guide yourself in understanding & internalizing the
conceptual and empirical elements of Operations Research. All the chapters and their sub topics ,
Illustrations, self-assessment questions, activities, tips, and exercises furthermore tutor marked
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assignments (TMAS) are designed in association to the contents of the subject matter but with an
objective that you will be able to associate the concepts of the course to the realities in your
environment. It is hence my hope that we will have an interesting and successful time together
covering the concepts of project and project management.

COURSE OBJECTIVES

The fundamental objective of the course module is:

 To familiarize the students with the meaning, characteristics & importance of Operations
Research.
 To provide the basic knowledge on how to formulate, analyze and interpret problems using
linear programming models.
 To provide the basic knowledge on how to formulate, analyze and interpret problems using
transportation and assignment models.
 To familiarize the approaches of decision making techniques under different decision making
environments.
 To acquaint the basics of project management concept and the techniques of project
planning and scheduling.

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UNIT ONE
1. INTRODUCTION TO OPERATIONS RESEARCH

1.1 LEARNING OBJECTIVES:

After completing this unit you should be able to:
 Explain the origin and historical development of Operations Research;
 Define operations research;
 Explain the advantages and disadvantages of quantitative and qualitative
decision making;
 Explain model and how they are used in Operations Research;
 Explain the approaches to problem solving.

1.2 GENERAL INTRODUCTION

The subject OPERATIONS RESEARCH is a branch of mathematics - especially applied mathematics,
used to provide a scientific base for management to take timely and effective decisions to their
problems. It tries to avoid the dangers from taking decisions merely by guessing or by using thumb
rules.

Management is the multidimensional and dynamic concept. It is multidimensional, because

management problems and their solutions have consequences in several dimensions, such as human,
economic, social and political fields. Hence any manager, while making decisions, considers all
aspects in addition to economic aspect, so that his/her solution should be useful in all aspects. The
general approach is to analyze the problem in economic terms and then implement the solution if it
does not aggressive or violent to other aspects like human, social and political constraints.

Management may be considered as the process of integrating the efforts of a purposeful group, or
organization, whose members have at least one common goal. You have studied various schools of
management in your Introduction to Management Module. Most important among them, which
uses scientific bases for decision making are:

(i) The Decision Theory or Decisional Management School and

(ii) The Mathematical or Quantitative Measurement School.

The above-mentioned schools of management thoughts advocate the use of mathematical methods
or quantitative methods for making decisions. Quantitative approach to management problems
requires that decision problems be defined, analyzed, and solved in a conscious, rational, logical
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and systematic and scientific manner - based on data, facts, information and logic, and not on mere
guess work or thumb rules. Here we use objectively measured decision criteria. Operations
Research is the body of knowledge, which uses mathematical techniques to solve management
problems and make timely optimal decisions. Operations Research is concerned with helping
managers and executives to make better decisions. Today’s manager is working in a highly
competitive and dynamic environment. In present environment, the manager has to deal with
systems with complex interrelationship of various factors among them as well as equally
complicated dependence of the criterion of effective performance of the system on these factors,
conventional methods of decision-making is found very much inadequate.

Though the common sense, experience, and commitment of the manager is essential in making
decision, we cannot deny the role-played by scientific methods in making optimal decisions.
Operations Research uses logical analysis and analytical techniques to study the behaviors of a
system in relation to its overall working as resulting from its functionally interconnected
constraints, whose parameters are recognized, quantified wherever possible relationships
identified to the extent possible and alterative decisions are derived.

Operations Research is discipline devoted to solving certain managerial types of problems using
quantitative models. The quantitative approach is widely employed in business. Ares of application
include forecasting, capital budgeting, capacity planning, scheduling, inventory management,
project management and production planning.

Some of the basics of Operations Research covered in this unit include answers to questions such
as: What is Operations Research? Who uses it? What are models? Including the process managers
use to solve problems using operations research & the origins and historical development of
Operations Research.

1.3 HISTORY OF OPERATIONS RESEARCH

Operations Research is a ‘war baby’. It is because, the first problem attempted to solve in a
systematic way was concerned with how to set the time fuse bomb to be dropped from an aircraft
onto a submarine. In fact the main origin of Operations Research was during the Second World War.
At the time of Second World War, the military management in England invited a team of scientists to
study the strategic and tactical problems related to air and land defense of the country. The problem
attained importance because at that time the resources available with England was very limited and
the objective was to win the war with available meager resources. The resources such as food,
medicines, ammunition, manpower etc., were required to manage war and for the use of the

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population of the country. It was necessary to decide upon the most effective utilization of the
available resources to achieve the objective.

It was also necessary to utilize the military resources cautiously. Hence, the Military Generals of
military, invited a team of experts that comprises scientists, doctors, mathematicians, business
people, professors, engineers etc., were presented with the problem of resource utilization that
they may discuss and come up with a feasible solution. These specialists had a brainstorming
session and arrived at a method of solving the problem, which they named as “Linear
Programming”. This method worked out well in solving the war problem. As the name indicates,
the word ‘Operations’ is used to refer to the problems of military and the word Research is use for
inventing new method. As this method of solving the problem was invented during the war, the
subject is given the name ‘OPERATIONS RESEARCH’ and abbreviated as ‘O.R.’ After the World War
there was a scarcity of industrial material and industrial productivity reached the lowest level.
There was industrial recession and to solve the industrial problem the method linear
programming was used to get optimal solution.

From then on words, lot of work is done in the field and today the subject of OR has numerous
methods to solve different types of problems. After analyzing the success of British military, the
United States military management started applying the techniques to various activities to solve
military, civil and industrial problems. They have given various names to this discipline. Some of
them are Operational Analysis, Operations Evaluation, Operations Research, System Analysis,
System Evaluation, Systems Research, Quantitative methods, Optimizations Techniques and
Management Science etc. But most widely used one is OPERATIONS RESEARCH. In industrial
world, most important problem for which these techniques used is how to optimize the profit or
how to reduce the costs. The introduction of Linear Programming and Simplex method of solution
developed by American Mathematician George B. Dontzig in 1947 given an opening to go for new
techniques and applications through the efforts and co-operation of interested individuals in
academic field and industrial field. Today the scenario is totally different. In one word we can say
that Operations Research play a vital role in every organization, especially in decision-making
process.

1.3.1 The Nature and Meaning of OR

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? Activity
How do you define the term operation research?
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 Comments:
Operations Research is a systematic analysis of a problem through scientific methods, carried
out by appropriate specialists, working together as a team, constituted at the instance of
management for the purpose of finding an optimum and the most appropriate solution ,to meet
the given objective under a given set of constraints.

Many definitions of OR have been offered by different scholars in different times, but the following
definitions provide a useful basis for an initial understanding of the nature of OR:

1. A scientific method of providing executive management with a quantitative base for

decisions regarding operations under their control (Mores-Kimball 1943).

2. Scientific approach to problem solving for executive management (Wagner 1969).

3. Optimal decision-making in, and modeling of, deterministic and probabilistic systems that
originate from real life. These applications, which occur in government, business,
engineering, economics, and the natural and social sciences, are largely characterized by
the need to allocate limited resources. In these situations, considerable insight can be
obtained from scientific analysis, such as that provided by OR (Hiller-Lieberman 1974).

4. A branch of applied mathematics where in the application is to the decision-making process

(Gross 1979).

5. Is a discipline devoted to solving certain managerial type problems using quantitative

models (Stevenson )

Working definitions
Operations Research is a systematic analysis of a problem through scientific methods,
carried out by appropriate specialists, working together as a team, constituted at the
instance of management for the purpose of finding an optimum and the most appropriate
solution, to meet the given objective under a given set of constraints.

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Key points

From the concept and definition given above, Operations Research is:
1. The application of scientific methods, techniques and tools to find an optimal
solution to a problem.
2. A management tool in the hands of a manager to take a decision
3. A scientific approach to decision making process
4. An “applied research” aims at finding a solution for an immediate problem facing a
society, industry or a business enterprise .This is not “fundamental research”
5. A decision-oriented research, using scientific methods, for providing management a
quantitative basis for taking decision regarding operations under its control
6. Applied decision theory uses scientific, mathematical and logical means to take
decisions.
The Five functions of Operations Research
A. objective: helps managers to make objective decision
B. scientific approach: helps decision makers to follow scientific approach to olve
managerial problems
C. inter disciplinary team work: it allows for team work
D. digital computers: over reliance on scientific calculators and other computing
machines
E. decision making : helps decision makers to solve managerial problems

1.3.2 Decision Making

Making appropriate decision is the most vital aspect in management .Every one of us takes
a number of decisions every day. Some are important; some are trivial. Some decisions
initiate a set of activities; some put an end to a certain activities. In business environment
right decisions at the right times ensure success. This shows the importance of decision
making.

 Problem is any variation between what was planned and what is actually
have/produced.

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 Problem solving can be defined as the process of identifying a difference between
some actual and some desired states of affairs and then taking action to resolve the
difference.
 Decision making defined as the process of selecting or choosing based on some
criteria, the best alternative among alternatives. It requires for all human being
because each of us make decision every day in our life. Thus, decision making is
universal. Decision making is a rational selection among alternatives.

The Decision Making Process:

Steps in the process of rational decision making:
1. Identify and define the problem
Problem is a necessary condition for a decision. i.e.: There would be no need for
decisions if problems did not exist.
2. Determine the set of alternative solutions.
3. Determine the criteria to evaluate alternatives: (Identifying those characteristics that are
important for making the decision)
4. Analyze the alternatives: (The advantages and disadvantages of each alternative)
5. Select the best alternative: (Select the best alternative that suits to solve our decision
problem. In selecting the best alternative, factors such as risk, timing and limiting
factors should be considered adequately)
6. Implement the solution: (Putting the decision into action)
7. Establishing a control and evaluation system :( Ongoing actions need to be monitored,
following the decision, evaluate the results and determine if a satisfactory solution has
been obtained.

1.3.3 The Decision-Making Environment

Decisions are made under the following three basic conditions:
 Decision under certainty
 Decision under risk
 Decision under uncertainty
i. Decision making under conditions of certainty
The decision maker has perfect knowledge about the outcome.

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In this situation, we are reasonably sure what will happen when we make a decision. The
information is available and is considered to be reliable and we know the cause & effect
relationships.
Example: If you decide to invest your money in saving account in the commercial Bank of
Ethiopia, You are certain that you will earn five percent (5%) interest.
ii. Decision making under condition of risk
Usually, decision makers cannot have a precise knowledge about the outcome of a decision.

Decision makers may only be able to attach a probability to the expected outcomes of each
alternative.

Under this situation, one may have factual information, but it may be incomplete.

Example: If we gamble by tossing a fair coin, the probability that a tail will turn up is 50%.
iii. Decision making under conditions of uncertainty
It is a case where neither there is complete data nor probabilities can be assigned to the
surrounding condition. It is the most difficult for a manager.

Some conditions that are uncontrollable by management include competition, government

regulations, technological advances, the overall economy, & the social & cultural tendencies
of society.

Example: A corporation that decides to expand its operation, launching a new product, or
developing of a new technology in a foreign country may know little about its culture, laws,
economic environment, or politics. The political situation may be so volatile that even experts
cannot predict a possible change in government.
1.3.4 Decision Making Approach
Decision making can be either quantitative or qualitative. In qualitative decision making
intuitions and subjective judgments are used. Past experience with similar problems is
often an important factor in choosing a qualitative approach, as in the complexity and
importance of a problem. Managers tend to use a qualitative approach to problem solving
when:
 The problem is fairly simple
 The problem is familiar
 The cost involved are not great
 Immediate decisions are not needed
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In quantitative decision making, managers use mathematical methods to solve the problem.
Conversely, managers generally prefer to use a quantitative approach when one or more of
the following conditions exist:
 The problem is complex
 The problem is not familiar
 The costs involved are substantial
 Enough time is available to analyze the problem

1.4 MODEL

? Activity
What is a model?
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 Comments:
Model is the abstractions of a reality

Model is an abstraction of a reality. It is a simplified and often an idealized representation

of real objects/situations. By its very nature a model is incomplete. A good model will
capture the important details of reality without including innumerable minor details that
would obscure rather than illuminate
Models can be classified in the following ways:
a. Physical (iconic) Model
Iconic models are the least abstract; they are physical models that look like the reality.
It is also called Static Model. It is given in two or three dimensions. It is a representation of
the real object as it is.

Example:
 Model of an airplane
 Photograph of a machine
 Layout drawing of a factory
 Glob
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b. Analogue Models:
They are also physical models but they are more abstract than iconic models. Instead of
replicating physical appearance as iconic models do, these models substitute some physical
analogy for important aspects of the reality. These models mostly showing inter and intra
relationships between two or more parameters. It may show the relationship between an
independent variable (input) with that of a dependent variable (output). It is two
dimensional
Example:
 histogram  Gantt charts
 frequency table  price-demand graph
 cause-effect diagram  world map with different colors
 flow charts  Organizational chart
c. Mathematical Models (Symbolic Models)
These are the most abstract models. They incorporate numbers and algebraic symbols to
represent important aspects of a problem often in equation form. Here a set of relations is
represented in the form of mathematical equations, using symbols to represent various
parameters.
Example:
Max Z=3000x1 +2500x2
Subject to:
2x1+x2 < 40
x1+3x2 < 45
x1 < 12
x1 , x2 > 0
Where:
The first line represents the objective function - A mathematical statement of the goal of an
organization, stated as intent to maximize or to minimize some important quantity such as
profits or costs.
Max Z=3000x1 +2500x2 is the objective function
 Lines three to six represent constraints - A restriction on the resources available to a
firm (stated in the form of an inequality or an equation.)

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2x1+x2 < 40
x1+3x2 < 45 Are constraints
x1 < 12
x1, x2 > 0 x1, x2 > 0 is non-negativity constraint
 x1 and x2 are decision variables

1.5 OPERATION RESEARCH IN PRACTICE

There are various techniques used in OR, some of these are listed here:

Probability theory Assignment problems

Linear Programming Queuing Theory
Transportation algorithm PERT/CPM Method etc

OR techniques are ’application specific’. Maximum benefit can be derived from selecting
most appropriate techniques for each specific area or problem. Appropriate selection of OR
technique is an equally important task. Each technique has its own advantages and
limitations .In all such cases, the ability of the Manager is tested in appropriate selection of
OR technique.

1.6 OPERATIONS RESEARCH APPROACH

? Activity
What are the specific steps that managers use to solve problems using operations
research?
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 Comments:
The approach in operations research consists: problem definition, model
construction analysis and implementation of the solution.

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The approach in OR is quite similar to the scientific approach commonly used in the physical
sciences. Both involve a logical sequence that includes careful definition of the problem use of
models and analysis leading to solution of the problem. Diagrammatically

Problem Definition

Model
Construction

Analysis

Implementation and
follow up
CHAPTER SUMMARY

Operations Research is a discipline devoted to the solution of management problems

using a scientific approach. The focus is on problem solving using quantitative models.
The approach is to define the problem, construct a model that reflects the important
aspects of the problem, analyze the model to identify an appropriate solution to the
problem, implement the solution and check to see that the problem has been solved.

The core of Operation Research is the scientific approach to problem solving. An

important part of this is the use of quantitative models. Models are abstractions of
reality: ideally a model will capture the important aspects of a situation with including
all of the minor details that would increase complexity and thereby reduce the chance
of finding a solution to the problem.

There are a variety of model types that are employed in decision making environment;
Operations Research models fall under the heading of symbolic models (that is
numbers and symbols are used to form mathematical models) using these models
tends to be a more objective approach than using qualitative models, although in

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symbolic models the more important qualitative aspects of a problem may be ignored.
Furthermore quantitative models enable users to take advantage of the tremendous
computational abilities and calculators.

Operation Research has applications in forecasting, inventory management, production

planning, capital budgeting, capacity planning and other important areas of
management decision making.

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SELF ASSESSMENT QUESTIONS (SQA1)

Give short answers for the following questions and compare your answers with the
answer keys provided at the end of this material.
1. Briefly describe the following model types and give an example of each: iconic,
analog and symbolic.
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2. Discuss the various steps used in solving Operations Research problems.

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3. Your manager has faced a difficulty making a choice between quantitative and
qualitative decision making approaches. As A student of Operations Research
please help him under what circumstances would he chose the quantitative
methods and qualitative approach.
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UNIT TWO
2. LEANER PROGRAMMING PROBLEM MODELS (LPPM)

2.1 LEARNING OBJECTIVES:

After completing this unit, you should be able to:

 Explain what linear programming is;

 Explain what is meant by the term constrained optimization;

 List and briefly explain the components and assumptions of linear

programming models;

 Recognize problems that can be solved using linear programming models

and formulate linear programming models;

 Explain the approaches to solving linear programming models;

 Solve problems using graphical method for solving linear programming

problems;

 Solve problems using simplex method for solving linear programming

problems;

 Use sensitivity analysis to evaluate a change in the RHS of a constraint and

changes in the value of an objective function coefficient.

2.2 GENERAL INTRODUCTION

Linear programming (LP) models enable users to find optimal solutions to certain
problems in which the solution must satisfy a given set of requirements or constraints.
It is a model, which is used for optimum allocation of scarce or limited resources to
competing products or activities under such assumptions as certainty, linearity, fixed
technology, and constant profit per unit.
Linear Programming is one of the most versatile, powerful and useful techniques for
making managerial decisions. Linear programming technique may be used for solving
broad range of problems arising in business, government, industry, hospitals, libraries,
etc. Whenever we want to allocate the available limited resources for various
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competing activities for achieving our desired objective, the technique that helps us is
LINEAR PROGRAMMING. As a decision making tool, it has demonstrated its value in
various fields such as production, finance, marketing, research and development and
personnel management.
Determination of optimal product mix (a combination of products, which gives
maximum profit), transportation schedules, assignment problem and many more are
the techniques managers use to make optimal decisions.

In this chapter, you will learn about linear programming models. Emphasis will be
given on familiarization of the model, problem recognition model and formulation of
the model, solving linear programming problem and post optimality analysis.

2.3 LINEAR PROGRAMMING PROBLEM MODELS

2.3.1 Meaning and Definition

? Activity
What is a linear programming problem model?
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 Comments:
Linear programming problem model is the mathematical form of a linear programming
problem.

Mangers encountered decision making situations in which the set of acceptable

solutions is somehow restricted. The restrictions may be imposed internally or
externally. An internal restriction might be the amount of raw materials, availability of
labour time, machine time, technical requirements and budgets that a department has
to produce its products. An external restriction might be labour regulations (e.g. safety

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equipment, training requirements overtime) that limit the options open to decision
maker. The restrictions are referred to as constraints.

The goal in linear programming is to find the best solution given the constraints
imposed by the problem; hence it is constrained optimization.

Linear Programming is one of the most versatile, powerful and useful techniques for
making managerial decisions. Linear programming technique may be used for solving
broad range of problems arising in business, government, industry, hospitals, libraries,
etc whenever we want to allocate the available limited resources for various competing
activities for achieving our desired objective. As a decision making tool, it has
demonstrated its value in various fields such as production, finance, marketing,
research and development and personnel management.

A model, which is used for optimum allocation of scarce or limited resources to

competing products or activities under such assumptions as certainty, linearity, fixed
technology, and constant profit per unit, is linear programming

It is the mathematical representation of LP problems that has been developed to help

management in decision making, involving the efficient allocation of scares resources to
achieve an optimization objective.
Diagrammatically,

Scares Resource
To be allocated to: Objectives Optimization

Subject to Maximization Minimization

Constraints

Non-negativity Resource
Constraints constraints
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2.3.2 Components of LP models

? Activity
What are the elements that form the structure of a linear programming problem model?
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 Comments:
The elements that form the structure of a linear programming problem model are known as
the components of a linear programming problem model. These are: objective function,
decision variables constraints and parameters.

A. Objective Function: Is the goal or objective of a management, stated as intent to

maximize or to minimize some important quantity such as profits or costs.
B. Decision Variables: represents the unknown values to be solved by the decision
makers. Decision variables are the variables whose values are unknown and are
searched.
C. Parameters: are fixed values that specify the impact of one unit on the objective function
and the constraint.
D. Constraints: are limitations or restrictions imposed by the problems. It includes:
 Recourse constraints: are restrictions that should be clearly identifiable and
measurable in quantitative terms, which arise from limitation of available resources.

Examples of limited resources:

 Plant capacity
 Raw materials availability
 Labor power
 Market demand, etc
 Individual constraints: are constraints which are subject to individual
decision variables.
 Non-negativity constraints: are constraints that require the decision variables not
to take on negative values.

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Generally speaking, a constraint also consists of four elements. These are:

I. A right hand side value (RHS): represents quantity that specifies the limit for that
constraint. It must be a constant, not a variable.
II. An algebraic sign: represents whether the limit is:
 An upper bound (< )that cannot be exceeded
 A lower bound (> )that is the lowest acceptable limit
 An equality (=) that must be met exactly.
III. The decision variable: to which the constraint applies.
IV. Parameters: The impact that one unit of each decision variable will have on the right
hand side quantity of the constraint.

2.3.3 Assumptions of LPP models

? Activity
What are the conditions under which the LPPM is valid?
__________________________________________________________________________
__________________________________________________________________________
_________________________________________________________________________

 Comments:
The assumptions of linear programming problem models are: linearity, certainty, divisibility
and non negativity.

The following are some important assumptions made in formulating a linear

programming model:
A. Linearity: The Objective Function and the constraints must be linear in nature in
order to deal with a Linear Programming Problems (LPP). Here the term linearity
implies proportionality and additively.
B. Certainty: It is assumed that the decision maker here is completely certain (i.e.,
deterministic conditions) regarding all aspects of the situation, i.e., availability of
resources, profit contribution of the Products, technology, courses of action and
their consequences etc.

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C. Divisibility: It is assumed that the decision variables are continuous. It means that
companies manufacture products in fractional units. For example, a company
manufactures 2.5 vehicles, 3.2 barrels of oil etc.
D. Non- Negativity: indicate all variables are restricted to non-negative values (i.e.,
their numerical value will be ≥ 0).i.e. negative values of variables are unrealistic or
meaningless.

2.4 FORMULATION OF LPPM

Formulating linear programming problem models involves the following steps
1. Identify the decision variables and represent them in terms of X1, X 2…..
2. Determine the objective function:
 First decide whether the problem is maximization or minimization
problem.
 Second identify the coefficients of each decision variable.
 If the problem is a maximization problem, the profit per unit for each
variable must be determined.
 If the problem is a minimization problem the cost per unit must be
determined.
 Make sure the units of measurements of all coefficients in the objective
functions must be the same.

3. Identify the constraints:

 First express each constraint in words.
 Second identify the coefficients of the decision variables in the
constraints; and the RHS values of the constraints.
 Determine the limits for the constraints i.e. see whether the constraint is
of the form (<), (>) or (=).
 Write the equation.
4. Using the above information (step 1 to 3) build the model.

NOTE THAT:
The coefficients of the variables in the Objective Function are called the profit or cost coefficients.
They express the rate at which the value of the Objective Function increases or decreases by
including in the solution one unit of each of the decision variables.

The coefficients of the constraints’ variables are called the input- output coefficients that indicate the
rate at which the given resources are depleted or utilized.

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Example 1:
A firm that assembles computers and computer equipment is about to start production two
new microcomputers. Each type of microcomputers will require assembly time, inspection
time and storage space. The amount of each of these resources that can be devoted to the
production of these microcomputers is limited. The manager of the firm would like to
determine the quantity of each microcomputer to produce in order to maximize the profit
generated by sales of these microcomputers.
Additional information
In order to develop a suitable model of the problem, the manager has met with design and
manufacturing personnel. As a result of these meetings the manager has obtained the following
information.

TYPE ONE TYPE TWO

Profit per unit Birr 60 Birr 50

Assembly time per unit 4 hours 10 hours

Inspection time per unit 2 hours 1 hour

Storage space per unit 3 cubic feet 3 cubic feet

The manager also has acquired information on the availability of company resources. These
weekly resources are:

RESOURCE AMOUNTS AVAILABLE

Assembly time 100 hours

Inspection time 22 hours

Storage space 39 hours

The manager also met with the firms marketing manager and learned that demand for the
microcomputers was such that whatever combination of these two types of microcomputers is
produced, all of the outputs can be sold.
Required: Formulate the LPPM of the problem.
Solution
Step 1: identify the decision variable

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 the quantity/ amount/ units of each microcomputer( microcomputer type 1 and

microcomputer type 2) to be produced

Let x1 represents quantity of microcomputer type 1 to be produced

x2 represents quantity of microcomputer type 2 to be produced

Step 2: Identify the objective function

The problem is maximization problem, as indicated in the problem (5th line of the
problem)
To write the equation both the objective function and the constraints summarize the given
information in tabular form accordingly

RESOURCE TYPE ONE TYPE TWO AVAILABILITY

Assembly time per unit 4 hours 10 hours 10 hours

Inspection time per unit 2 hours 1 hour 1 hour

Storage space per unit 3 cubic feet 3 cubic feet 3 cubic feet

Profit per unit Birr 60 Birr 50

 Max Z = 60x1 + 50 x2

Step 3: identify each constraints and write the equation

 Assembly time 4x1 + 10x2< 100

 Inspection time 2x1 + x2< 22

 Storage space 3x1+ 3x2 < 39

 Non negativity constraint x1 & x2 > 0

Step 4: write the summarized form of the model

In summary, the mathematical model of the microcomputer
problem is:
Max Z+ 60x1 + 50x2
Subject to
4x1 + 10x2< 100
2x1 + x2< 22
3x1+ 3x2 < 39
x1 & x2 > 0

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Example 2:
A firm is engaged in breeding pigs. The pigs are feed on various products grown on the farm. In
view of the need to ensure certain nutrient constituents (call them vitamins, minerals and
proteins) it is necessary to buy two additional products say A and B. One unit of product A
contains 36 units of vitamins, 3 units of minerals and 20 units of proteins. One units of product
B contains 6 units of vitamins, 12 units of minerals and 10 units of proteins. The minimum
requirement of vitamins, minerals and proteins is 108units, 36units and 100 units respectively.
Product A costs birr 20 per unit and product B costs birr 40 per unit
Required: formulate the LPPM of the problem

? Activity:
Try to formulate by yourself, in the given space below before you check the answer.
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
Step 1: identify the decision variables
Product A and B to be purchased to prepare the food for pigs
Let x1 represent product A to be purchased
X2 represent products B to be purchased

Step 2: identify the objective function

The problem is a minimization problem, so the equation;
 Min Z = 20x1 + 40 x2

When we put the information in a tabular form

NUTRIENT
CONSTITUENTS
PRODUCT A PRODUCT B REQUIREMENT

Vitamins 36 units 6 units 108 units

Minerals 3 units 12 units 36 units

Proteins 20 units 10 units 100 units

Cost per unit Birr 20 Birr 40

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Step 3: identify each constraints and write the equation

 Vitamins 36x1 + 6x2> 108

 Minerals 3x1 +12 x2> 36

 Proteins 20x1+ 10x2 >100

 Non negativity constraint x1 & x2 > 0

Step 4: write the summarized form of the model

In summary, the mathematical model of the problem is:

Min Z=60x1 + 50x2
Subject to
36x1 + 6x2> 100
3x1 +12 x2> 22
20x1+ 10x2 >39
x1 & x2 > 0

2.5 APPROACHES TO SOLVE LPPM

There are two methods used to solve LP problems
1. Graphical methods

2. Simplex methods

1. Graphical methods: used to find solutions for LPP when the decision variables of the
problem are not greater than two.

Graphical LP is a two-dimensional model

To apply this method, we should follow the following steps.

A. Draw a graph including all the constraints

B. Identify the feasible region

C. Identify the corner points and its coordinates

D. Evaluate the objective function at each corner point and Obtain a point on
the feasible region that optimizes the objective function-Optimal solution

E. Interpret the results

Examples 1: Consider the microcomputer problem and solve it using graphical approach.

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Step one: Graph each constraints: To draw the graph first change the inequality to equality
i.e. replace the < and > sign into = sign.
Max Z=60x1 + 50x2
Subject to
36x1 + 6x2> 100 36x1 + 6x2= 100
3x1 +12 x2> 22 3x1 +12 x2= 22
20x1+ 10x2 >39 20x1+ 10x2 =39
x1 & x2 > 0

Then find the x and y intercepts (in our case x1 and x2 respectively) - points where each
constraint intersects the axis. To do so, set x1 = 0 to find values for x2 and set x2= 0 to find values
for x1.
For the first constraint:
4x1 + 10x2 = 100 4x1 + 10x2 = 100
4(0) + 10 x2 = 100 4x1 + 10(0) = 100
10x2= 100 4x1= 100
X2= 10 x1 = 25
The x1 and x2 intercepts are (0, 10) (25, 0). Similarly calculate for the second and third
constraints.
For the second constraint the intercepts are (0, 22) (11, 10)
For the third constraint the intercepts are (0, 13) (13, 0)
Graph the constraints using the intercepts calculated above. The graph is:

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As indicated in the graph, the corner points of the feasible region (the corner points of the
shaded region) are A, B, C, D and E. The next task is to find the coordinates of these corner
points, some are determined by observation and some are through simultaneous equation.
Then we test each corner points to find the points that results the optimal solution. These
activities are indicated in the following table.

Evaluating the corner points

Point Coordinates How determined Value of the objective function
X1 X2 Max Z=60x1 + 50x2
A 0 0 Inspection 60(0) + 50(0) = 0
B 11 0 Inspection 60(11) + 50(0) =660
C 9 4 Simultaneous equation 60(9) + 50(4) =740
D 5 8 Simultaneous equation 60(5) + 50(8) =700
E 0 10 Inspection 60(0) + 50(10)=500

Since the maximum value (because the objective function of the problem is maximization) is
740. The solution is
X1= 9, X2= 4 and The Maximum Profit is Birr 740.
When we interpret the result, the company should produce 9 units of microcomputer type
one and 4 units of microcomputer type two to get a maximum profit of Birr 740.
Example 2:
Max.Z  50 X 180 X 2
St :
X 12 X 2 32
X 12 X 2 82
X1, X 2  0

Note that:
Try to solve the problem by yourself, in separate piece of paper before you check the
answer.

Example 3: Consider two models of color TV sets; Model A and B, are produced by a company
to maximize profit. The profit realized is $300 from A, and $250 from set B. The limitations are:
A. availability of only 40hrs of labor each day in the production department
B. a daily availability of only 45 hrs on machine time
C. ability to sale 12 set of model A

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How many sets of each model will be produced each day so that the total profit will be as large
as possible?
Resources used per unit

Constraints Model A Model B Maximum Available hrs.

(X1) (X2)

Labor hr. 2 1 40

Machine hr. 1 3 45

Marketing hr. 1 0 12

Profit $300 $250

Note
Try to solve the problem by yourself, in separate piece of paper before you check the
answer.

Solution
1. Formulation of mathematical modeling of LPP
Max Z=300X1 +250X2
St:
2X1 +X2< 40
X1 +3X2< 45
LPP Model
X1 < 12
X1, X2 >0

2. Convert constraints inequalities into equalities

2X1 +X2 = 40
X1 +3X2= 45
X1 = 12
3. Draw the graph by intercepts
2X1 +X2 = 40 ==> (0, 40) and (20, 0)
X1 +3X2= 45==> (0, 15) and (45, 0)
X1 = 12==> (12, 0)
X1 X2 =0

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2X1 +X2 = 40
X2
X1=0
40 X1=12

X1 +X2 = 45
15

Feasible C(12, 11)

Region X2=0
X1
D
A 12 20 45

4. Identify the feasible area of the solution which satisfies all constrains.
5. Identify the corner points in the feasible region
A (0, 0), B (0, 15), C (12, 11) and D (12, 0)
5. Identify the optimal point

Corners Coordinates Max Z=300 X1 +250X2

A (0, 0) $0
B (0, 15) $3750
C (12, 11) $6350
D (12, 0) $3600

Interpretation:
12 units of product A and 11 units of product B should be produced so that the total profit will
be $6350
Example 4:
A manufacturer of light weight mountain tents makes two types of tents, REGULAR tent and
SUPER tent. Each REGULAR tent requires 1 labor-hour from the cutting department and 3
labor-hours from the assembly department. Each SUPER tent requires 2 labor-hours from the
cutting department and 4 labor-hours from the assembly department. The maximum labor

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hours available per week in the cutting department and the assembly department are 32 and
84 respectively. Moreover, the distributor, because of demand, will not take more than 12
SUPER tents per week. The manufacturer sales each REGULAR tents for $160 and costs $110
per tent to make. Whereas SUPER tent ales for $210 per tent and costs $130 per tent to make.

Required:
A. Formulate the mathematical model of the problem
B. Using the graphic method, determine how many of each tent the company should
manufacture each tent the company should manufacture each week so as to maximize its
profit?
C. What is this maximum profit assuming that all the tents manufactured in each week are sold
in that week?

Note
Try to solve the problem by yourself, before you check the answer.

Solution
_____________________________________________________________________
Labor hours per tent
Department REGULAR (X1) SUPER(X2) Maximum labor-hours
available per week

_____________________________________________________________________

Cutting department 1 2 32
Assembly department 3 4 84
Selling price per tent $160 $210
Cost per tent $110 $130
Profit per tent $50 $80

*The distributor will not take more than 12 SUPER tents per week. Thus, the manufacturer
should not produce more than 12 SUPER tents per week.
Decision variable: number of regular and super tent to be produced per week.
Let X1 =The No of REGULAR tents produced per week.
X2 =The No of SUPER tents produced per week.
X1 and X2 are called the decision variables
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Max.Z  50 X 180 X 2
St :
……….Cutting department constraint
X 12 X 2 32
LPP Model ……….Assembly department constraint
X 14 X 2 82
……….Demand constraint
X 2  12
X1, X 2  0 ……….Non-negativity constraints

Corners Coordinates Max Z=50 X1 +800X2

A (0, 0) $0
B (0, 12) $960
C (8, 12) $1360
D (20, 6) $1480
E (28, 0) $1400
Interprétation:
The manufacturer should produce and sale 20 REGULAR tents and 6 SUPERS tents to get a
maximum weekly profit of $1480.

Example 4
Suppose that a machine shop has two different types of machines; machine 1 and machine 2,
which can be used to make a single product .These machine vary in the amount of product
produced per hr., in the amount of labor used and in the cost of operation.
Assume that at least a certain amount of product must be produced and that we would like to
utilize at least the regular labor force. How much should we utilize each machine in order to
utilize total costs and still meets the requirement?
Solution
____________________________________________________________________________________
Resource used
Machine 1 (X1) Machine (X2) Minimum required hours

________________________________________________________________________________________
Product produced/hr 20 15 100
Labor/hr 2 3 15______
Operation Cost $25 $30____________________________________

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Min.Z  25 X 130 X 2
St :
20 X 115 X 2 100 LPP Model

2 X 13 X 2 15
X1, X 2  0
Constraint equation:
20X1 +15X2=100 ==> (0, 20/3) and (5, 0)
2X1+3X2=15 ==> (0, 5) and (7.5, 0)
X1 X2> 0

______________________________________________________________________________________________

Corners Coordinates MinZ=25 X1 + 30X2

A (0, 20/3) 200
B (2.5, 3.33) 162.5
C (7.5, 0) 187.5
_______________________________________________________________
X1 =2.5
X2=3.33 and
Min Z= 162.5

Example 5
A company owns two flour mills (A and B) which have different production capacities for HIGH,
MEDIUM and LOW grade flour. This company has entered contract supply flour to a firm every
week with 12, 8, and 24 quintals of HIGH, MEDIUM and LOW grade respectively. It costs the Co.

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$1000 and $800 per day to run mill A and mill B respectively. On a day, mill A produces 6, 2,
and 4 quintals of HIGH, MEDIUM and LOW grade flour respectively.
Mill B produces 2, 2 and 12 quintals of HIGH, MEDIUM and LOW grade flour respectively. How
many days per week should each mill be operated in order to meet the contract order most
economically standardize? Solve graphically.

Solution:
No of days per week of Minimum flour in
Mil A (X1) Mill B(X2) quintals
HIGH Capacity (in quintal) 6 2 12
MEDIUM Capacity (in quintal) 2 2 8
LOW Capacity (in quintal) 4 12 24
$1000 $800

Max.Z  100 X 1800 X 2

St :
6 X 12 X 2 12
2 X 12 X 2 8
4 X 1  12 X 2  24
X1, X 2  0
Solution
X1 =1
X2=3 and
Min Z= $3400

Note:
In maximization problems, our point of interest is the furthest point from the origin.
In minimization problems, our point of interest is the point nearest to the origin.

2. Simplex method
The graphical method to solving LPPs provides fundamental concepts for fully
understanding the LP process. However, the graphical method can handle problems
involving only two decision variables (say X1 and X2).

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In 19940’s George B. Dantzig developed an algebraic approach called the Simplex

Method which is an efficient approach to solve applied problems containing numerous
constraints; and involving many variables that cannot be solved by the graphical
method.
The simplex method is an ITERATIVE or “step by step” method or repetitive algebraic
approach that moves automatically from one basic feasible solution to another basic
feasible solution improving the situation each time until the optimal solution is reached
at.

2.6 MAXIMIZATION PROBLEM

2.6.1. Maximization having all the constraints in “<”form
Example: Consider the microcomputer problem which is solved with graphical method
so that you can compare and contrast the two approaches.
Max Z+ 60x1 + 50x2
Subject to
4x1 + 10x2< 100
2x1 + x2< 22
3x1+ 3x2 < 39
x1 & x2 > 0
To solve the problem using the simplex approach, follow the following steps.
Step 1
Formulate LPP Model: in our example it is already done.
Step 2
Standardize the problem
I.e. convert each constraint into a standard form by introducing a variable called Sack variable,
surplus variable and artificial variable
Slack Variables(S): A sack variable/s is /are added to the left hand side of a < constraint to
covert the constraint into its standard form. The value of the slack variable shows unused
resource.
Surplus variables(S): A surplus variable/s is/are added to the left hand side of a >constraints
to convert the constraint into its standard form. The value of the surplus variable shows the
excess resources used.

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Artificial variables (A): somewhat analogous to slack variables in that they are added to
equality and a > constraints in the same way that slack variables are added to a< constraints.
However artificial variables have no physical interpretation, they merely serve as a device to
enable us to use the simplex process.
Since the above problem constraints have all a < algebraic sign, we use slack variables for
standardizing the problem.

A slack variable is always added for a < constraint to convert the constraint to a standard form.

Slack variables represent unused resource or idle capacity. Thus, they don’t produce any
product and their contribution to profit is zero.

Slack variables are added to the objective function with zero coefficients.
Max Z+ 60x1 + 50x2+0s1+0s2+0s3
Subject to
4x1 + 10x2+s1=100
2x1 + x2+s2= 22
3x1+ 3x2 +s3=39
x1 , x2 ,s1,s2,&s3> 0

NOTE THAT;
To standardize, start from the constraint and finally move to the objective function.
When you add the slack variables it should be according to the existence of the constraints (how it is written
in the model). I.e. add S1 to the first constraint, s2 to the second constraints etc

Step 3
Obtain the initial simplex tableau
To make ready the data for analysis, the simplex method uses a table called the simplex
tableau or the simplex matrix.
In constructing the initial simplex tableau, the search for the optimal solution begins at the
origin. Indicating that nothing can be produced;
Thus, based on this assumption, no Microcomputer Type One and Microcomputer Type
Two is produced, which implies that x1 =0 and x2=0

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==>4x1+10x2 + s1 +0 s2+ 0 s3= 100 ==> 2x1+x2 +0s1 + s2+ 0s3= 22

2(0) +10(0) + s1 +0 s2+ 0 s3= 100 2(0) +0 + 0s1 + s2+ 0 s3= 22
s1= 100 – Unused assembly time. s2= 22 – Unused inspection time.

==> 3x1+ 3x2+0s1 +0s2+ s3= 39

3(0) +3(0) +0s1 +0 s2+ s3= 39
s3= 39 – Unused storage space.
Therefore, Max Z=60x1 +50x2 + 0 s1 +0 s2+ 0 s3
=60(0) +50(0) + 0(100) +0(22) + 0(39)
=0
NOTE THAT;
In general, whenever there are n variables and m constraints (excluding the non-negativity), where m is
less than n (m<n), n-m variables must be set equal to zero before the solution can be solved
algebraically.

a. Basic variables
a. are variables with non-zero solution values
b. are variables that are in the basic solution
c. have 0 values in the C-Z row
b. Non-basic variables
a. are variables with zero solution values
b. are variables that are out of the solution

From the above example n=5 variables (x1, x2, s1, s2, and s3) and m=3 constraints (assembly,
inspection and storage space constraints), excluding non-negativity.
Therefore, n-m=5-3=2 variables(x1 and x2) are set equal to zero in the 1st simplex tableau.
These are non-basic variables. 3 Variables (s1, s2, and s3) are basic variables (in the 1st simplex
tableau) because they have non-zero solution values.

Step 3
Construct the initial simplex tableau

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Slack variables
Initial simplex tableau

columns
variables column
Real or decision

Solution quantity
Profit per unit
Basic or DV
column

column

column
Profit per unit row

C 60 50 0 0 0
BV X1 X2 S1 S2 S3 Q
S1 0 4 10 1 0 0 100 R1
S2 0 2 1 0 1 0 22 R2 Constraint
equation rows
S3 0 3 3 0 0 1 39 R3
Z 0 0 0 0 0 0
Gross Profit row
C-Z 60 50 0 0 0
Net Profit row
/Indicator row/

Step 4:
Choose the “incoming” or “entering” variables
Note:
The entering variable is the variable that has the most (the largest) positive value in the C - Z
row (indicator row).
Note:
The entering variable is the variable that has the highest contribution to profit per unit.
X1 in our case is the entering variable. (Because the maximum number in the c-z row is 60)

The column associated with the entering variable is called key or pivot column X1 column in our case.
Step 5:
Choose the “leaving “or “outgoing” variable, and the “pivot element”
In this step, we determine the variable that will leave the solution for X1.
To identify the leaving variable, we should calculate the ratio first and then we should select
the minimum non negative ratio.

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Note:
The leaving variable is the variable that has the smallest replacement ratio.
S2 in our case is the leaving variable. (Because the smallest replacement ratio is 11

The row associated with the entering variable is called key or pivot row S2 row in our case.
The pivot element is the intersection point between the pivot row and pivot column;
No 2 in the above example
Replacement Ratio (RR) = Solution Quantity (Q)
Corresponding values in pivot column
In our case
100 =25
4
22 = 11 is the minimum
2
39 =13
3

It is interesting to note that the three ratios (25, 11, 13) corresponding to the intersections of
the constraints with the x1(look the graphical solution). Note that the smallest of the ratios
represents the extreme point of the feasible solution space; the other points lie beyond the
feasible solution space. Hence by selecting the smallest ratio, the simplex procedure stays
within the feasible solution space. It sometimes happens that some of the substation rates for
the variable we want to bring into solution are zero or negative. We don’t nee to divide the
quantity values by a negative or a zero substitution rates.

? Activity
Briefly describe the following points:
1. Leaving variable
2. Entering variable
3. Pivot element
_______________________________________________________________________________________
_______________________________________________________________________________________
_______________________________________________________________________________________

 
Comment:
The entering variable is the variable that has the highest contribution to profit per unit
 X1 in our case is the entering variable. (Because the maximum number in the c-z row is 60)
 The variable leaving the solution is called leaving variable or outgoing variable.
 The row associated with the leaving variable is called key or pivot row (s2 column in our case)
 The element that lies at the intersection of the pivot column and pivot row is called pivot element(No 2 in
our case)

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Step 6: Perform row operations: perform algebraic operations on the PIVOT COLUMN to
convert the pivot element into 1 and the remaining pivot column numbers into 0. Obtain the
new row values through the following two operations:
 Multiply (divide) all of the elements in a row by a constant
 Add or subtract the multiple of the row to or from another row
After identifying the entering, leaving and pivot element construct the second tableau by
replacing the S2 by X1 in the basic solution.
2nd tableau

C 60 50 0 0 0

SV X1 X2 S1 S2 S3 Q RR

S1 0 0 8 1 -2 0 56 56/8=7 R’1=R1+ (-4R’2)

X1 60 1 1/2 0 1/2 0 11 11/1/2=22 R’2=R2/2
S3 0 0 3/2 0 -3/2 1 6 6/3/2=4 R’3=R3+ (-3R’2)
Z 60 30 0 30 0 660
C-Z 0 20 0 -30 0

Note the following to see how it is calculated

Since the minimum replacement ratio (RR) is associated with the 2nd row we should start the
operation on the second row i.e. we should change the pivot element into 1. To do so we should
divide the whole elements of the 2nd row by the pivot element of the first tableau.
Look
2 1 0 1 0 22 when we divide it by 2 we get 1 ½ 0 ½ 0 11 so this will be
the new row values of the second tableau.
Then calculate the new values of the first row and the third row of the second tableau. You can
begin in ether row one or row three.
Let’s start with the first row
Bring the first row values of the initial tableau
4 10 1 0 0 100
The take the second row values of the 2nd tableau
1 ½ 0 ½ 0 11
Then in order to change 4 into 0, ask your self what algebraic operation should be
done. Very simple, multiply the whole elements of the second tableau’s value by -4.

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You get
-4 -20 -2 0 -44, then add this value to the first row value of initial tableau.
Look the summary
4 10 1 0 0 100 4 10 1 0 0 100 4 10 1 0 0 100
1 ½ 0 ½ 0 11 -4( 1 ½ 0 ½ 0 11) + -4 -2 0 -2 0 -44
0 8 1 -2 0 56 will be the new values
of the first row of the second tableau. To put it in a simple formula
R’1=R1 + (-4R’2)
Similarly calculate for row three of the second tableau
3 3 0 0 1 39 3 3 0 0 1 39 3 3 0 0 1 39
1 ½ 0 ½ 0 11 -3( 1 ½ 0 ½ 0 11) + -3 -3/2 0 -3/2 0 -33
0 3/2 0 -3/2 1 6 will be the new
values of the 3rd row value of the second tableau. To put it in a simple formula again it will be:
R’3=R3 + (-3R’2)

Note:
 Divide each element of the pivot row by the pivot element to find new values in the key or pivot row.
 Perform row operations to make all other entries for the pivot column equal to zero.

Then calculate the Z row values i.e. these are found by multiplying the values in each column by
the corresponding coefficients in the C column and adding them.

BV C X1 x2 s1 s2 s3 Q

S1 0 0(0 ) =0 0 (8 )=0 0 (1)=0 0(-2 ) =0 0(0)=0 0(56)=0

X1 60 60(1 )=60 60 (1/2) =30 60( 0 )=0 60(1/2)=30 60 (0)=0 60(11)=660

S3 0 0(0 )= 0 0(3/2) =0 0( 0 )=0 0(-3/2)=0 0( 1)=0 0(6)= 0

Z 60 30 0 30 0 660

Then finally calculate C – Z row. It is simple follow the formula.

Step 7: Repeat step 4-6 till optimum basic feasible solution is obtained. I.e. repeat these steps
till no positive value occurs in the C – Z row.

A simplex solution in a maximization problem is optimal when the C-Z row consists entirely
zeros and negative No (when there are no positive values in the C-Z row.

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3rd simplex tableau

BV C 60 50 0 0 0 Q

X1 X2 S1 S2 S3
S1 0 0 0 1 6 -16/3 24 R’’1=R’1+ (-8R’3)
X1 60 1 0 0 1 -1/3 9 R’’2= R’2+ (-1/2R’3)

X2 50 0 1 0 -1 2/3 4 R’’3=R’3/3/2

Z 60 50 0 10 40/3 740
C-Z 0 0 0 -10 - 40/3

Since the entire C – Z < 0 indicating that no additional potential for improvement
exists. I.e. optimal solution is reached at.
Therefore, the optimal solution is X1=9, X2=4, S1=24and Max Z=740

Interpretation of the Results:

The last step is interpreting the result: in order to achieve the maximum weekly profit of birr 740,
the company should produce 9 units of microcomputer type one and 4 units of microcomputer
type two. This will leave no slack in either inspection (s2=0) or storage space (s3 =0). How ever,
there will be 24 hours of assembly time that is unused.

Example 2
A Juice producing Company has available two kinds of food Juices: Orange Juice and
Grape Juice. The company produces two types of punches: Punch A and Punch B. One
bottle of punch A requires 20 liters of Orange Juice and 5 liters of Grape Juice. Where as
1 Bottle of punch B requires 10 liters of Orange Juice and 15 liters of Grape Juice.
From each bottle of Punch A; a profit of $4 is made and from each bottle of Punch B; a
profit of $3 is made. Suppose that the company has 230 liters of Orange Juice and 120
liters of Grape Juice available

Required:
a. Formulate this problem as a LPP
b. How many bottles of Punch A and Punch B the company should produce in order to
maximize profit? (Using the simplex method)

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c. What is this maximum profit?

? Activity:
Try to solve the problem by yourself, in separate piece of paper before you check the answer.

Solution:
Juice needed for one bottle of
Juice Punch A Punch Juice Available
___________________________________________________________________________________________________
Orange Juice (lt) 20 10 230
Grape Juice (lt) 5 15 120
Profit per tent $4 $3
Let X1= the No of bottles of punch A produced.
X2= the No of bottles of punch B produced.

 The LPP Model of the problem is:

Max Z=4X1 +3X2
St:
20X1 +10X2 < 230 Orange Constraint

5X1 +15X2 < 120 Grape Constraint

X1, X2 >0 Non-negativity constraint

 Standard form
Max Z=4x1 +3x2 + 0 s1 +0 s2+ 0 s3
St:
20 x1+3x2 + s1 +0 s2 = 230
Standard form
5x1+15x2 +0s1 + s2+ = 120
x 1 , x2 , s1 , s2, >0
Where, s1 =Unused orange juice
s2 =Unused grape juice
Initial simplex tableau

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B C 4 3 0 0

X1 X2 S1 S2 Q RR

S1 0 20 10 1 0 230 11.5

S2 0 5 15 0 1 120 24

Z 0 0 0 0 0
C-Z 4 3 0 0

2nd simplex tableau

B C 4 3 0 0

X1 X2 S1 S2 Q RR

X1 4 1 1/2 1/20 0 11.5 23

S2 0 0 25/2 -1/4 1 62.5 5

Z 4 2 1/5 0 46
C-Z 0 1 -1/5 0

Optimal simplex tableau

4 3 0 0
B C Q
X1 X2 S1 S2

X1 4 1 0 3/50 - 1/25 9

X2 3 0 1 -1/50 2/25 5

4 3 0.12 0.08
Z
51
C- Z
0 0 - 0.12 0.08

Solution
X1= 9 X2= 5 s1 =0 s2=0
Max Z=$51
3.6.2 Maximization with Mixed Constraints:
In the previous section, you have learned how to solve a maximization problem with
all <constraints. Here you will learn how to solve a maximization problem with
mixed constraints (< > =). For the most, part the technique is identical to that
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illustrated in the previous section. What is new is the introduction of the –M

coefficients in the objective functions and the use of artificial variables.
Example1:
Max Z=6x1 +8x2
Subject to:
x2 < 4
x1+ x2 = 9
6x1+ 2x2 >24
x1, x2 >0

Standard form
To standardize the model what we have learned in the previous example is not enough.
Here the model consists <, > & = sign constraints which demands their unique
standardization approach. For < constraint you know it already we add a slack
variable. For the remaining two consider the following two variables.
1. Surplus variables (S): refers to the excess utilization of resources
 A variable inserted in a greater than constraint to create equality. It
represents the amount of resource usage above the minimum required level.
 Surplus variable is subtracted from a > constraint in the process of
converting the constraint to standard form.
Neither the slack nor the surplus should be negative value in the initial tableau. They must
have a positive value. Look the following
1. 5x1+3x2+ S1 < 45
x1= 0 and x2= 0==> s1 = 45
==> s1=45 unused resource (all resources are idle)
2. 2x1+x2 >40
x1= 0 and x2= 0(No production)==> 5x1+2x2- s1 = 20
==> s1=-6(This is mathematically unaccepted)
To avoid these problem another variable is inserted which is known as an Artificial
variable (A)

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Artificial variable is a variable that has no meaning in a physical sense but acts as a tool to
create an initial feasible LP solution. It helps to make the model logical and meaningful.

It is added to constraints having = and > algebraic sign.

Has a coefficient of [-M] in the objective function for maximization type problem. It
represents an artificial very large negative value.
Consider the above example
2x1+x2 >40
x1= 0 and x2= 0(No production)==> 5x1+2x2- s1 = 20
s1=-6(This is mathematically unacceptable, because it violates the non negativity assumptions
of LPPM)
 5x1+2x2- s1 = 20
 5x1+2x2- s1 + A1= 20
 So when we set x1,x2 and s1 = 0
 The basic solution in the initial solution will be A1= 20 which is
mathematically correct.
Lets consider anther example having = sign
5x1 +9x2=30
To determine the values of the basic variables for the initial tableau, we set x1& x2=0
5x1 +9x2=30 ==> 5(0) +9(0) =30
==> 0=30 which is mathematically wrong. To avoid this problem
we add an artificial variable to standardize constraints with = sign.
5x1 +9x2=30 ==> 5x1 +9x2+ s1=30 ==> 5(0) +9(0 +s1) =30
==> S1=30 which is logical.
Note:
For the initial basis, use artificial variables for constraints that have > and = sign. For a <
constraint use a slack variable. Hence, surplus variables will not appear in an initial
solution.

Now based on the above logic when we standardize the above problem
Max Z=6x1 +8x2 + 0 s1 +0 s2+ 0 s3-M A2- M A3

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St:
x2 + s1 =4
x1+ x2 + A2 =9 Standard form
6x1+2x2 - s3 + A3 =24
All Variables >0

Initial simplex tableau

B C 8 0 0 -M -M

X1 X2 S1 S3 A2 A3 Q RR

S1 0 0 1 1 0 0 0 4 4/0=und.

A2 -M 1 1 0 0 1 0 9 9

A3 -M 6 2 0 -1 0 1 24 4
Z -7M -3M 0 +M -M -M -33M
C-Z 7M +6 3M+8 0 -M 0 0

Since 7m+ 6 is the largest, x1 will be the entering variable for the next tableau. Since an
artificial value (A3) leaves the solution it will removed from the next table, so A3 column will
not exist in the 2nd tableau

2nd tableau

B C 6 8 0 0 -M

X1 X2 S1 S3 A2 Q RR
S1 0 0 1 1 0 0 4 4
A2 -M 0 2/3 0 1/6 1 5 27/2

A3 6 1 1/3 0 -1/6 0 4 12
Z 6 2-3/3M 0 -1-1/6M -M 24-5M
C-Z 0 6+2/3M 0 1+1/6M 0

3rd tableau

BV C 6 8 0 0 -M

X1 X2 S1 S3 A2 Q RR

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X2 8 0 1 1 0 0 4 und
A2 -M 0 0 2/3 1/6 1 7/3 14

A3 6 1 0 1/3 -1/6 0 8/3 Neg.

Z 6 8 6+2/3m -1-1/6M M 48-7/3M
C-Z 0 0 -6-2/3m 1+1/6M 0

Final tableau

BV C 6 8 0 0 Q

X1 X2 S1 S3
X2 8 0 1 1 0 4
S3 0 0 0 -4 1 14

A3 6 1 0 -1 0 5
Z 6 8 2 0 62
C-Z 0 0 2 0

Since the C - Z < 0==>optimal solution

X1=3, X2 =2, X3=0, S1=0 and Max Z= 8

Example1:
Max Z=2x1 +x2+3x3
Subject to:
2x1+ x2 + x3 < 5
2x1+ 3x2 +4x3 = 12
x1, x2 , x3 > 0
Solution
Initial Simplex tableau

B C 2 1 3 0 -M Q

X1 X2 X3 S1 A1

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S1 0 1 1 2 1 0 5
A1 -M 2 3 4 0 1 12
Z -2M -3M -4M 0 -M -12 M RR
C- Z
2M+2 3M+1 4M+3 0 0 2.5

2nd simplex tableau

B C 2 1 3 0 -M Q

X1 X2 X3 S1 A1 RR

X3 3 1/2 1/2 1 1 0 5 10

A1 -M 2 3 4 0 1 12 4

Z 3/2 3/2-M 3 2M+3/2 -M -2M+15/2

C-Z 1/22 M-1/2 0 -2M-3/2 0

3rd simplex tableau

B C 2 1 3 0 Q
RR
X1 X2 X3 S1

X3 3 1/2 0 1 3/2 1.5 6

X2 1 0 1 0 -2 2
Not defined
Z 3/2 1 3 5/2 13/2
C-Z 1/2 0 0 -5/2

4th simplex tableau

B C 2 1 3 0 Q

X1 X2 X3 S1

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X1 2 1 0 2 3 3

X2 1 0 1 0 -2 2

Z 2 1 4 4 8
C-Z 0 0 -1 -4

The optimal solution is

X1 = 3, X2 = 2, X3= 0, S1 = 0 max profit = 8
2.6.3 Minimization Problems
For most part, manual solution of minimization problems using simplex are handled in the
same fashion as maximization problems with mixed constraints. The two exceptions are:
 The M coefficients in the objective function are given positive signs instead of
negative signs
 The selection of the variable to enter the solution is based on the largest negative
value (the negative number far from 0) in the C-Z row of a tableau.
 Solution is optimal when there is no negative value of C-Z.

Note:
TYPE OF CONSTRAINT TO PUT INTO STANDARD FORM
< ----------------------------------------Add a slack variable
= -----------------------------------------Add an artificial variable
> ---------------------- Subtract a surplus variable and add an artificial variable

Example:
1. Minimize Z=25x1 +30x2
Subject to:
20x1+15x2 > 100
2x1+ 3x2 > 15
x1 & x2 > 0
Solution
Step 1
Standardize the problem:

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Minimize Z=25x1 +30x2 +0s1+0s2 +MA1+MA2

Subject to:
20x1+15x2- s1+A1 = 100
2x1+ 3x2 –s2+A2 = 15
x1, x2 , s1, s2 ,A1 ,A2 > 0
Step 2
Initial simplex tableau:
The initial basic feasible solution is obtained by setting x1= x2= s1= s2=0
No production, x1= x2= s1=0==>20(0) +15(0) - 0+A1 = 100 ==> A1 = 100
x1= x2= s2=0==>0(0)+3(0) - 0+A2 =15==> A2 = 15
Initial simplex tableau

B C 25 30 0 0 M M Q

X1 X2 S1 S2 A1 A2
RR
A1 M 20 15 -1 0 1 0 100
100/20=5
A2 M 2 3 0 -1 0 1 15
Z 22M 18M -M M M M 115 M 15/2=7.5
C-Z
25 -22M 30- 18M M M 0 0

Note:
Once an artificial variable has left the basis, it has served its purpose and can therefore be
removed from the simplex tableau. An artificial variable is never considered for re-entry into
the basis.
2nd Simplex Tableau

B C 25 30 0 0 M Q

X1 X2 S1 S2 A2
X1 25 1 3/4 -1/20 0 0 5 R’1=R1/20

A2 M 0 3/2 1/10 -1 1 5 R’2=R2-2 R’

Z 25 75/4+3/2M -5/4+1/10M M M 125+5 M
C-Z 0 45/4-3/2M 5/4-1/10 M M 0

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3rd Simplex Tableau

B C 25 30 0 0 Q

X1 X2 S1 S2
X1 25 1 0 -1/10 1/2 5/2
X2 30 0 1 1/15 -2/3 10/3
Z 25 30 -1/2 -15/2 162.5
C-Z 0 0 1/2 15/2

C - Z > 0==>Optimal solution is reached

X1=5/2
X2=10/3 and Min Z=162.5

Example 2 (Minimization with mixed constraint)

Min Z=5x1 +3x2
Subject to:
2x1+4x2 < 12
2x1+ 2x2 = 10
5x1+ 2x2>10
X1 &X2
Solution
Min Z=5x1+3x2 +0s1+0s3 +MA2+MA3
Subject to
2x1+4x2+s1 = 12
2x1+ 2x2+A2 = 10
5x1+ 2x2-S3+ A3=10
X1, X2, S1, S3, A2 and A3 > 0
Min Z=5x1 +3x2, If no production
2x1+4x2+s1 = 12 ==>x1 =x2=0==>s1=0 (Solution Value in the initial simplex tableau)
2x1+2x2 +A1 =10 ==>x1 =x2=0==>A2 =15 (Solution Value in the initial simplex tableau)

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5x1+2x2 –s2 +A1=10 ==>x1=x2=s3=0==>A3=10(Solution Value in the initial simplex tableau)

x1, x2 , s1, s2 ,A1 ,A2 > 0
Initial Simplex tableau

B C 5 3 0 0 M M Q

X1 X2 S1 S2 A1 A2 RR

S1 0 2 4 1 0 0 0 12
6
A1 M 2 2 0 0 1 0 10
A2 M 5 2 0 -1 0 1 10 5
Z 7M 4M 0 M M M 20 M
C-Z 5 -7M 3- 4M 0 -M 0 0
2

2nd simplex tableau

B C 5 3 0 0 M Q

X1 X2 S1 S2 A1
0 S1 0 16/5 1 2/5 0 8
M A1 0 6/5 0 2/5 1 6
5 X1 1 2/5 0 -1/5 0 2
Z 5M 6/5M +2 0 2/5M - 1 M 10+6 M

C-Z 0 -6/5M +1 0 -2/5M+1 0

3rd simplex tableau

B C 5 3 0 0 M Q

RR
X1 X2 S1 S2 A1
X2 3 0 1 5/16 1/8 0 2.5
20
A1 M 0 0 -3/8 1/4 1 3
X1 5 0 0 -1/8 -1/4 0 1 12
Z 5 3 -3/8M +5/6 M/4-7/8 M 12.5+3 M
-
C- Z
0 0 3/8M -5/6 M/4+7/8 0

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4th Simplex tableau

B C 5 3 0 0 Q

X1 X2 S1 S2

X2 3 0 1 1/2 0 1

S2 0 0 0 -3/2 1 12

X1 5 0 0 -1/2 0 4
Z 5 3 -1 0 23
C-Z 0 0 1 0

X1=4, X 2=1, S1=0, S2=12 and Min Z=23

Exercise
Find the optimal solution using simplex method
1. Min Z=10x1 +5x2
Subject to:
2x1 + 5x2 > 150
3x1+ x2 > 120
x1, x2 >0
Ans: x1=450/13, x2 =210/13 and Min Z=$540
2. Min Z=4x1 +5x2
Subject to:
X1 + 2x2 > 80
3x1+ x2 > 75
x1, x2 >0
Ans: x1=14, x2 =33 and Min Z=$221

3. Min Z=7x1 +9x2

Subject to:
3x1 + 6x2 > 36
8x1+ 4x2 > 64
x1, x2 >0

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Ans: x1=20/3, x2 =8/3 and Min Z=212/3

Note:
To get an initial feasible solution
Types of constraint Presence of variables in the initial solution mix
1. < (Slack) Yes
2. > *(Surplus) No
*(Artificial) Yes
3. = (Artificial) Yes

? Activity
What are the special cases in simplex methods? Please try in the given space.
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
___________________________________

 Comment:
The special cases are: Tie, Infeasibility, Unbounded solution, Degeneracy, Multiple optimal
solutions

2.7 SPECIAL CASES IN SIMPLEX METHOD

I. Two incoming variables / or Tie for entering variables/

In order to break this tie, the selection for the key column (entering variable) can be made
arbitrary. However; the number of solution can be minimized by adopting the following
rules:
1. If there is a tie between two decision variables, then the selection can be made
arbitrary.
2. If there is a tie between a decision variable and a slack (or surplus) variable, then
select the decision variable to enter into basis first.

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3. If there is a tie between slack or surplus variable, then selection can be made
arbitrary.
Example:
If the equation is max Z:

B C Q

X1 X2 S1 S3

Z
C- Z 5 2 5 0

In such a case, X1 is the entering variable

II. Infeasibility
A situation with no feasible solution may exist if the problem was formulated improperly.
Infeasibility comes about when there is no solution that satisfies all of the problem’s
constraints.
In the simplex method, an infeasible solution is indicated by looking at the final tableau .In it,
all C - Z row entries will be the proper sign to imply optimality, but an artificial variable
(A) will still be in the solution mix.
Example:
Minimization case

B C 5 8 0 0 M Q

X1 X2 S1 S2 A2

X1 5 1 1 -2 3 0 200

X2 8 0 1 1 2 0 100

A2 M 0 0 0 -1 1 20

Z 5 8 -2 31-M M 1,800+200M
C-Z 0 0 2 M-31 0

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Even though all C - Z are positive and 0 (i.e. the criterion for an optimal solution in a
minimization case), no feasible solution is possible because an artificial variable (A2)
remains in the solution mix.

III. Unbounded Solutions

No finite solution may exist in problems that are not bounded .This means that a variable can
be infinitely large without violating a constraint.

In the simplex method, the condition of unbounded ness will be discovered prior to reaching
the final tableau. We will note the problem when trying to decide which variable to remove
from the solution mix.

The procedure in simplex solution is to divide each quantity column number by the
corresponding pivot column number to identify the leaving variable. The row with the
smallest positive ratio is replaced. But if the entire ratios turn out to be negative or
undefined, it indicates that the problem is unbounded.
Example:
Maximization case

B C 5 9 0 0 Q

X1 X2 S1 S2 RR

X2 5 -1 1 2 0 30 30/-1=-30
Unacceptable RRs
S2 0 -2 0 -1 1 10

Z -9 9 18 0 270 10/-2=-5

C-Z 15 0 -18 0

Pivot Column

The solution in the above case is not optimal because not all C- Z entries are 0 or negative, as
required in a maximization problem. The next variable to enter the solution should be X1.To
determine which variable will leave the solution, we examine the ratios of the quantity

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column numbers to their corresponding numbers in the X1 or pivot column. Since both pivot
column numbers are negative, an unbounded solution is indicated.

IV. Degeneracy /Tie for leaving basic variable (key row)/

If there is a tie for the smallest ratio, this is a signal that degeneracy exists. Degeneracy can
occur right in the first (initial tableau).This normally happens when the number of
constraints is less than the number of variables in the objective function.

B C 5 8 2 0 0 0 Q

X1 X2 X3 S1 S2 S3
RR
X2 8 1/4 1 1 -2 0 0 10
10/1/4=40
S2 0 4 0 1/3 -1 1 0 20
20/4=5 Tie for the smallest
S3 0 2 0 2 2/5 0 1 10 ratio indicates deg.
10/2=5
Z 2 8 8 16 0 0 80
C-Z 3 0 -6 -16 0 0

Problem can be overcome by trial and error method.

Degeneracy could lead to a situation known as cycling, in which the simplex algorithm
alternatives back and forth between the same non-optimal solutions, i.e., it puts a new
variable in, then takes it out in the next tableau, puts it back in ,and so on.
One simple way of dealing with the issue is to select either row (S2 or S3 in this case)
arbitrary. If we are unlucky and cycling does occur, we simply go back and select the other
row.
Remark
When there is a tie between a slack and artificial variable to leave the basis, the preference
shall be given to artificial variable to leave the basis and there is no need to apply the
procedure for resolving such cases.

V. Multiple Optimal Solutions

Multiple optimal solutions exist when non-basic variable contains zero on its C – Z row.
Example:

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Maximization problem

B C 3 2 0 0 Q

X1 X2 S1 S2

X2 2 3/2 1 1 0 6

S2 0 1 0 1/2 1 3

Z 3 2 2 0 12
C-Z
0 0 -2 0

Max Z=3X1+2X2
X1=0, X2=6, S2=3 and Max Z=12 or: X1=3, X2=3/2 and Max Z=12
The C – Z value of the Non-basic variable (X1) is 0.Thus, there is alternative optimal solution.

Exercise:
1. Solve the following LPP by the simplex algorithm
Min Z=6x1 +8x2
Subject to:
x1+ 2x2 > 80
3x1+ x2 > 75
x1, x2 >0
What are the values of the basic variables at each iteration?
Which are the non-basic variables at each iteration?
Ans: X1=14, X2=33, and Min Z=221

B C Q
X1 X2 X3 S1 S2 S3
X3 5 0 1 1 -2 0 0 5
X1 6 1 -3 0 0 0 1 12
S2 0 0 2 0 1 1 -1 10
Z 6 -13 5 5 0 21 97

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2. At the C-Z 0 16 0 -5 0 -21 3rd

iteration
of a particular LP maximization problem, the following tableau is established:

What special condition exists as you improves the profit and move to the next iteration?
Proceed to solve the problem for optimal solution
Ans: Degeneracy; X1=27, X2=5, X3=0, and Max Z=$177
3. Covert the following constraints and objective function into the standard form for use in
the simplex method
Min Z=4x1 +x2
Subject to:
3 x1+ x2 = 3
4x1+ 3x2 > 6
x1+ 2x2 < 3
x1, x2 > 0

Answer:
Min.Z=4x1 +x2 + 0 s1 +0 s2+ M A1+M A3
St:
3x1+ x2 + A1 = 3
4x1+ 3x2 -s1 +A2 =6
x1+ 2x2 + s2 =3
All Variables >0
4. Solve the following LPP using simplex method
Max Z=9x1 +7x2
Subject to:
2x1+ x2 < 40

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x1+ 3x2 < 30

x1, x2 >0
Ans: X1=18, X2=4, and MaxZ=$190
5. Solve the following LPP to show that it has alterative optimal solutions.
a. MaxZ=6x1 +3x2 Ans: i. X1=4, X2=0, and MaxZ=24
Subject to: ii.X1=5/2, X2=3, and MaxZ=24
2x1+ x2 < 8
3 x1+ 3x2 < 18
x2 < 3
x1, x2 >0
b MinZ=2x1 +8x2 Ans: i. X1=32/6, X2=10/6, and MinZ=24
Subject to: ii.X1=12, X2=0, and MinZ=24
5x1+ x2 > 10
2x1+ 2x2 > 14
x1+ 4x2 > 14
x1, x2 >0
6. Solve the following LPP to show that it has unbounded solution.
a. Max Z=-2x1 +3x2 b. Max Z=3x1 +6x2
Subject to: Subject to:
x1 <5 3x1+ 4x2 > 12
2 x1-3x2 < 6 -2x1+ x2 < 4
x1, x2 >0 x1, x2 >0

7. Solve the following LPP to show that it has no feasible solution.

a. MaxZ=-2x1 +3x2 Ans: X1=2, X2=0, A1=2 and MaxZ=4-2M
Subject to:
x1 - x2 > 4
x1+ x2 < 6
x1 < 2
x1, x2 > 0
b. MaxZ=3x1 +3x2 Ans: X1=0, X2=2, A2=2 and MaxZ=4-4M

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Subject to:
2x1+ x2 < 2
3x1 + 4x2 > 12
x1, x2 >

2.8 SENSATIVITY ANALYSIS (POST OPTIMALITY ANALYSIS)

? Activity
Why sensitivity analysis is done in LPPM.
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 Comment:
It is done to see the impact of changes of the parameters of a LPPM on the optimal
solution.

Sensitivity Analysis is the study of how changes in the coefficients of LPP affect the optimal
solution:
o Is analysis after the optimal solution is obtained
o The purpose is to see the effect of changes in RHS of a constraint, coefficients of OF,
changes in the coefficients of the constraints. Changes of the input-output coefficient,
add/delete of constraints and the addition of a new variable to the problem on the
value of the optimal solution.
It starts from the final tableau. It is concerned with the study of ‘Sensitivity’ of the optimal
solution of an LPP with changes in parameters of the variables .The degree of sensitivity of
the solution due to those variations can range from no change at all to a substantial change
in the optimal solution of the given LPP. Thus, in sensitivity analysis, we determine the range

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over which the LP model parameters can change with out affecting the current optimal
solution. The process of studying the sensitivity of the optimal solution of an LPP is called
post-optimal analysis/what-if analysis/.
In this module we will see the following two post optimality analysis
A. Changes of the RHS Quantity of a constraint (b)
B. Changes of the coefficients of the objective function (C)

2.8.1. Change in the Right Hand—Side Quantity (RHS)

The first step in determining how a change in RHS of a constraint would influence
the optimal solution is to examine the shadow prices in the final simplex tableau.
Shadow prices:
o Are a manager’s window for guessing the impact of one unit change has on
the value of the Objective function
o Is the maximum amount that should be paid for one additional unit of
resource
o Is a marginal value, i.e. it indicates the impact a one unit change in the amount
of a constraint would have on the objective function
For illustration purpose, let’s look at the microcomputer case.
Max Z= 60x1 + 50x2
Subject to:
Assembly time 4x1 + 10x2 < 100
Storage space 3x1+ 3x2 < 39
Inspection time 2x1 + x2 < 22
X1 & x2 > 0
Take the final tableau

60 50 0 0 0 Q
BV C X1 X2 S1 S2 S3

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S1 0 0 0 1 6 -16/3 24

X1 60 1 0 0 1 -1/3 9

X2 50 0 1 0 -1 2/3 4

Z 60 50 0 10 40/3 740
C-Z 0 0 0 - 10 -40/3

To deal with shadow prices you should focus on the slack variable column. Because each
slack variable’s column represents each constraint, i.e. S1 represents the first constraint; S2
represents the second constraints and so on.
The number in C - Z row (the magnitude) in its slack variable columns provide us shadow
prices. Therefore the shadow prices for the above example are:
S1=0
S2=10
S3= 40/3
What does this indicate? Please try it in a separate piece of paper!
You know that Shadow prices signify the change in the optimal value of the objective
function for 1 unit increases in the value of the RHS of the constraint that represent the
availability of scarce resources. Based on this logic, we can interpret it as follows:

1. If the amount of assembly time (S1) was increased by one hour, the effect would
be no effect on the profit (objective function)
2. If the amount of inspection time (S2) was increased by one hour, the effect would
to increase profit by 10 birr.
3. If the amount of storage space (S3) was increased by one cubic feet, the effect
would be to increase profit by 40/3 birr.

But what the shadow prices do not tell us is the extent to which the level of a scarce resource
can be changed and still have the same impact per unit. Therefore, we need to determine the

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range over which we can change the RHS quantities and still have the same shadow prices.
This is called the range of feasibility or right- hand side range.

Range of feasibility: is the range over which the RHS value of a constraint can be
changed with out changing the optimal solution mix of a problem (still have the same
shadow prices)

The values in the body of the final tableau in each slack column are used to compute the
range.
To compute the range of feasibility for the corresponding constraint the entries in the
associated slack column must be divided into the values in the quantity column.
Example:
Determine the range of feasibility for each of the constraints in the microcomputer problem.

S1 S2 S3 Q S1 ( assembly time) S2(inspection time) S3( storage space )

1 6 -16/3 24 24/1= 24 24/6= 4 24/-16/3= -4.5
0 1 -1/3 9 9/0= und 9/1= 9 9/-1/3= -27
0 -1 2/3 4 4/0= und 4/-1= -4 4/2/3=6

0 10 40/3 740 100 hours 22 hours 39 cubic feet

0 -10 -40/3
Upper limit None( no upper limit) 22 + 4=26 39 + 4.5= 43.5
Range
Lower Limit
100 - 24= 76 22 - 4= 18 39 - 6= 33

To calculate the limits, use the following general rules. (Take only the magnitude,
don’t consider the sign)

For maximization problem (maximization with all < constraints)

To calculate the upper range, take the smallest negative ratio (a number closest to zero.
To calculate the lower range, take the smallest positive ratio (a number closest to zero.

For minimization problem with all > constraints

(The rules will be reversed)
To calculate the upper range, take the smallest positive ratio (a number closest to zero.
To calculate the lower range, take the smallest negative ratio (a number closest to zero.
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Based on this:
The upper rage for assembly time (S1) will be 100hours the original availability plus the
smallest negative ratio. But, since there is no negative ratio .There is no upper limit. But for
the lower limit 100 minus smallest positive ratio. Therefore, it will be 100 - 24= 76. The
others are also calculated similarly.

Now let’s see the application (relaxation and tightening of constraints)

Example 1
The manager in the microcomputer problem is contemplating one of two possible changes
in the level of the storage space constraints. One change would be an increase of 3 cubic
feet in its level and the other would be an increase of 8 cubic feet in its level.
Required: Determine the revised optimal solution for each possible change.
Solution
For simplicity fist let’s see the effects of a one unit change in storage space. To do so take
the S3 column value, because storage space is associated with S3

Basis S3 Q

S1 -16/3 24
X1 -1/3 9
X2 2/3 4

Z 40/3 740

The impact is
 Cause S1 (assembly time) to decrease by 16/3 units ( i.e. S1 will become 18.6)
 Cause x1(micro computer type one) to because by 1/3 units
 Cause x2(micro computer type two) to increase by 2/3 units
 Cause profit to increase by 40/3
Now let’s come to our problem. First let’s see the impacts of a three unit change.

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Look if a one unit has the above impact; it is very simple to calculate the impact of a three
unit change .But what you have to note is, first check the change is whether or not within the
rage
We have already calculated that the range for storage space is between 33 to 43.5. So the
original value is 39 and the increase is 3 therefore the changed value is 39+3= 42 which is
within the range. So the impact of the shadow prices is valid. Therefore, the effects of an
increase of three cubic feet can be computed in this way.

Basic S3 Quantity Current Change Revised solution

solution
S1 -16/3 24 24 24+3(-16/3) 8
X1 -1/3 9 9 9+3(-1/3) 8
X2 2/3 4 4 4+3(2/3) 6
Z 40/3 740 740 740+3(40/3) 780

Case two an increase of 8 cubic feet in its level:

For an increase in storage space of 8 cubic feet, since the upper limit of the feasible range for
storage is 43.5 (39+4.5) the amount of increase above 4.5 cubic feet will be excess or slack.
Moreover, the new optimal solution will occur at the point where the constraint level equals
its upper limit.
Therefore, the additional amount needed to achieve the upper limit is used to compute the
revised solution.

Basic S3 Quantity Current Change Revised solution

solution

S1 -16/3 24 24 24+4.5 (-16/3) 0

X1 -1/3 9 9 9+4.5 (-1/3) 7.5
X2 2/3 4 4 4+4.5(2/3) 7

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Z 40/3 740 740 740+4.5 (40/3) 800

Note that beyond the upper limit S3 would come into solution replacing S1 which would no
longer be slack. The amount of slack would be 8-4.5= 3.5 cubic feet. Consequently the revised
solution would be
S3= 3.5
X1= 7.5
X2= 7
Example 2
Suppose the manager in the microcomputer problem is contemplating a decrease in storage
space due to an emergency situation. There are two possibilities being considered
 A decrease of 6 cubic feet in its level
 A decrease of 9 cubic feet in its level
A decrease of 6 cubic feet: since the change is within the range we can calculate it straight
forward.

Basic S3 Quantity Current Change Revised solution

solution
S1 -16/3 24 24 24 - 6(-16/3) 56
X1 -1/3 9 9 9 - 6(-1/3) 11
X2 2/3 4 4 4 - 6(2/3) 0
Z 40/3 740 740 740-6 (40/3) 660

A decrease of 9 cubic feet: since the change is beyond the lower limit the substitution rates
and the shadow prices do not hold. Consequently for a decrease in the level of a constraint
beyond the lower limit a new simplex solution must be generated.

Example 2
Max Z=3x1+4x2
Subject to:
3 x1+5x2 < 15
2x1 + x2 < 8
x2 < 2

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x1, x2 > 0
The optimal tableau is given as:

B C 3 4 0 0 0 Q
X1 X2 S1 S2 S3

X1 3 1 0 -0.143 0.714 0 3.57

S3 0 0 0 -0.286 0.428 1 1.143

X2 4 0 1 0.286 -0.428 0 0.857

Z 3 4 0.714 0.428 0
C- Z 0 0 -0.714 -0.428 0

Required:
1. Determine the shadow price for each constraint
2. Determine the RHS ranges over which the shadow prices are valid

 Analysis of the 1st constraint (S1)

Q S1 Q/ S1

3.57 -0.143 -24.96

1.143 -0.286 -3.99

0.857 0.286 3.00

Lower Limit= b1 - the smallest positive number in the Q/ S1column

=15-3=12
Upper Limit= b1 +the smallest negative number in the Q/ S1 column
=15 + (3.99) =18.99
Therefore, 12< b1< 18.99 (The range of resource 1 over which the shadow price $0.714 per
unit is valid).

 Analysis of the 2nd constraint (S2)

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Q S2 Q/ S2

3.57 0.714 5

1.143 0.428 2.67

0.857 -0.428 -2

Lower Limit=b2-the smallest positive number in the Q/ S2 column

=8-(2.67) =5.33
Upper Limit=b2+ the smallest negative number in the Q/ S2 column
=8 +2 =10
Therefore, 5.33< b1< 10 (The range of resource 2 over which the shadow price $0.428 per unit
is valid).
 Analysis of the 3rd constraint (S3)

Q S3 Q/ S3

3.57 0 -

1.143 1 1.143

0.857 0 -

Lower Limit=b3-
the smallest positive number in the Q/ S3 column
=2-(1.143)= 0.857
Upper Limit=b3 + the smallest negative number in the Q/ S3 column
=2 +∞ =∞
Therefore, 0.857< b3< ∞ (The range of resource 3 over which the shadow price $0 per unit is
valid).
Exercise:
1. Max Z=50x1+40x2
Subject to:
3 x1+5x2 < 150 (Assembly time)
x2 < 20 (Portable display)
8x1 + 5x2 < 300 (Warehouse space)

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x1, x2 > 0
The final simplex tableau is:

B C 50 40 0 0 0 Q
X1 X2 S1 S2 S3

X2 40 0 1 8/25 0 -3/25 12

S2 0 0 0 -8/25 1 3/25 8

X1 50 1 0 -5/25 0 5/25 30

Z 50 40 14/5 0 26/5 1980

C-Z 0 0 -14/5 0 -26/5

Determine the shadow prices for the three constraints for the High Tech Co.
Answer:
The Z values for the three slack variables are 14/5, 0 and 26/5, respectively.
Thus, the shadow price for the assembly time constraint is 14/5(i.e.1 additional assembly time
over the existing 150 is $2.8)
The shadow price for the portable display constraint is 0.
The shadow price for the Warehouse space constraint is26.5
Therefore, we see that obtaining more warehouse space will have the biggest positive impact
on High Tech’s profit.

Note:
With a greater than or equal to constraint, the value of the shadow price will be less than or
equal to zero because a 1 unit increase in the value of the RHS cannot be helpful; it makes it
more difficult to satisfy the constraint. As a result, for a maximization problem the optimal
value of the objective function can be expected to decrease when the RHS side of a greater
than or equal to constraint is increased.
2. Solve the following LPP using simplex method. A firm that manufactures both lawn
mowers and snow blowers:
X1 =the number of lawn mowers
X2 =the number of snow blowers
Max Z=$30x1+$80x2

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Subject to:
2 x1+4x2 < 1000 Labor hours available
6x1 + 2x2 < 1,200lb of steel available
x2 < 20 snow blower engine available
x1 , x2 > 0
a. What is the best product mix? What is the optimal profit?
Answer:
x1=100, x2=200 and profit =$19,000
b. What are the shadow prices? When the optimal solution has been reached, which
resource has the highest marginal value?
Answer:
The shadow price for 1 additional labor=$15
The shadow price for 1 additional pound of steel=0
The shadow price for 1 additional snow blowers engine made available =$20
Thus, snow blower engine have the highest marginal value at the optimal
solution.
c. Over what range in each of the RHS values are these shadows valid?
Answer:
The shadow price for labor hours is valid from 800 hours to 1,066.66 hours
The shadow price for pounds of steel is valid from 1,000pounds up to an infinite number
of pounds
The shadow price for snow blower engines ranges from 180 engines up to 250 engines

2.8.2 Changes of the coefficients of the objective function (c)

To calculate changes in the coefficients of objective function we should consider only
Decision variables. Decision variables can be:
i. Basic (variable in the solution)
ii. Non-basic (variable out-of the solution)
Note:
Instead of resolving the entire problem as a new problem with new parameters, we
may take the original optimal solution table as an initial solution table for the

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purpose of knowing ranges both lower and upper within which a parameter may
assume value.
There are two cases to consider changes for a variable that is not currently in the
solution mix and changes for a variable that is currently in the solution mix.
a. Range for the coefficients of basic decision variables
The range of optimality: is the range over which a basic decision variable coefficient
in the objective function can change without changing the optimal solution mix.
However, this change will change only the optimal value of the objective function.
(The profit figure)

Example:
Given the final tableau of the microcomputers problem, determine the range of
optimality.
Solution
The final simplex tableau for the problem is repeated here for convenience

B C 60 50 0 0 0 Q

X1 X2 S1 S2 S3

S1 0 0 0 1 6 16/3 24

X1 60 1 0 0 1 -1/3 9

X2 50 0 1 0 -1 2/3 4

Z 60 50 0 10 40/3 740
C- Z 0 0 0 -10 - 40/3

Analysis of X1
Steps:
a. Take the C – Z row of the optimal solution.
b. Take the X1 row values.
c. C – Z row
X1 row

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Column X1 X2 S1 S2 S3
C – Z row 0 =0 0 =und 0 =und -10 =-10 -43/3 =+40
X1 row 1 0 0 1 -1/3

The general rule for calculating the upper and lower range is
Allowable increase: the smallest positive ratio of C-Z value and the variable’ substitution rate
Allowable decrease: the smallest negative ration of C- Value and the variable substitutions rate

The positive ratio is +40. Therefore, the coefficients of X1 can be increased by 40 without
changing the optimal solution. The upper end of its range of optimality is this amount added
to its current (original) value. Thus its upper end is 60+ 40 = 100. Also, the smallest negative
ratio is -10; therefore the x1 coefficient can be decreased by as much as 10 from its current
value, making the lower end of the range equal to 60-10= 50

Analysis of X2
Steps:
a. Take the C – Z row of the optimal solution
b. Take the X2 row values.
c. C – Z row
X1 row
Column X1 X2 S1 S2 S3
C – Z row 0 =und 0 =0 0 =und -10 =10 -43/3 =-20
X2 row 0 1 0 -1 2/3
The smallest positive ratio is + 10. This tells us that the X2 coefficient in the objective
function could be increased by 10 to 50+ 10 = 60. The smallest negative ratio is -20, which
tells us that x2 coefficient could be decreased by 20 to 50-20 =30. Hence the range optimality
for the objective function coefficient of x2 is 30 to 60.
Example:
Max Z=5x1 +4.5x2 +x3
Subject to:
15 x1+15.8x2 < 150
5x1+6.4x2+15x3 < 77
2.8x2+11.8x3 < 36

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x1, x2 , x3 > 0

The optimal tableau for this solution is:

B C 5 4.5 1 0 0 0 Q

X1 X2 X3 S1 S2 S3

X1 5 1 1.053 0 0.067 0 0 10

X3 1 0 0.67 1 -0.022 0.067 0 1.8

S3 0 0 1.924 0 0.258 -0.773 1 15.12

Z 5 5.342 1 0.311 0.067 0 51.8

C-Z 0 -0.842 0 -0.311 -0.067 0

Determine the range of optimality for the coefficients of the basic-decision variables.
Solution:
Analysis of basic decision variable
The analysis will be conducted on products on X1 and X3 which are in the basic solution.
Divide each C - Z row entry for variables not in the solution (for instance, by X2, S1 and S2
values) the remaining values result undefined so we can skip it
I. Analysis of X1
Steps:
a. Take the C – Z row of the optimal solution of the non-basic variables
b. Take the X1 row of the non-basic variables
c. C – Z row
X1 row
X2 S1 S2

C – Z row -0.842 -0.311 -0.067

X1 row 1.053 0.067 0

C - Z row -0.8 -4.64 -∞

X1 row

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 Upper Limit
The smallest positive number in the C- Z row tells how much the profit of
X1
X1 can be increased before the solution is changed.
Upper Limit= C (for X1) +the smallest positive value of C - Z row
=5+∞=∞ X1 row

Note: C (for X1) =5(Look in the OF of the LP problem)

 Lower Limit
The smallest negative ratio (the number closest negative amount closest to 0)

Lower Limit= C (for X1) +The largest negative value of C - Z row

=5+ (-
0.8) X2 S1 S2 = 4.2

X1 row

Therefore, the range of optimality for the coefficient of X1 is 4.2< C (for X1) < ∞
(The coefficient of X1 in the objective function can change between 4.2 and ∞ without changing the
optimal solution mix X1=10, X3=1.8 and S3=15.12)

II. Analysis of X3

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C- Z row -0.842 -0.311 -0.067 Upper

Limit= C X3 row 0.67 -0.022 0.067 (for X3)
+The C - Z row -1.26 14.13 -1 smallest
positive X1 row value of C
- Z row
=1- (1) =0 X1 row

Note: C (for X3) =5(Look in the OF of the LP problem

 Lower Limit= C (for X3 )+The largest negative value of C- Z row

=1+ (14.13) = 15.13 X1 row

Therefore, the range of optimality for the coefficient of X3 is 0 < C (for X3) < 15.13
(The coefficient of X3 in the objective function can change between 0 and 15.13 without changing the
optimal solution mix X1=10, X3=1.8 and S3=15.12).However, this change will change only the optimal
value of the objective function (i.e. Max Z will change)

Exercise:
Max Z=50x1 +120x2
Subject to:
2 x1+4x2 < 80
3x1+x2< 60
x1, x2 > 0
Required: Determine the range of optimality for the coefficient of the basic variables
for the given problem.
Optimal Solution

B C 80 60 0 0 Q

X1 X2 S1 S2

S2 0 5/2 0 -1/4 1 20
X2 60 3/2 1 1/2 0 40

Z 90 60 30 0 $2,400

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C-Z -10 0 -30 0

Ans: The range of optimality for X2’s profit coefficient is: $66.6 < C (for X2) < ∞
b. The range for the non-basic variables
If there is a variable C, not participating in the optimal basis, then, in order for the variable
to be included in the optimal solution, its coefficient in the objective function will have to
change from the existing C to a new level C.

C (new)> Z

The range of insignificance is the range over which Cj rates for non-basic variables
can vary without causing a change in the optimal solution mix (variable) is called the
range of insignificance
Example:
Max Z=5x1 +4.5x2 +x3
Subject to:
15 x1+15.8x2 < 150
5x1+6.4x2+15x3 < 77
2.8x2+11.8x3 < 36
x1, x2 , x3 > 0

The optimal tableau for this solution is:

5 4.5 1 0 0 0
B C
X1 X2 X3 S1 S2 S3 Q
X1 5 1 1.053 0 0.067 0 0 10

X3 4.5 0 0.67 1 -0.022 0.067 0 1.8

S3 0 0 1.924 0 0.258 -0.773 1 15.12

Z 5 5.342 1 0.311 0.067 0 51.8

C–Z 0 -0.842 0 -0.311 -0.067 0

Calculate the range of insignificance for the coefficient of non-basic variable(X2)

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Solution
Because X2 is not in the solution, its objective function coefficient would need to increase in
order for it to come into solution. The amount of increase must be greater than its C-Z value
which is -0.842. Therefore, its objective function coefficient must increase by more than
0.842. Because its current value is 4.5 as indicated above, the current solution will remain
optimal as long as its objective function coefficient does not exceed 4.5 +0.842= 5.342. Since
an increase would be needed to bring it into solution, any lesser value of its objective
function coefficients would keep it out of solution, hence the range of insignificance for X 2 is
5.342or less
Mathematically;
Range of insignificance = the OF coefficient of + (C - Z) value of
For X2 X2 X2
= C(for X2)=4.5 and C-Z ( for X2) =-0.842
= 4.5 + (0.842)
= 5.342
If the profit contribution of X2 is greater than 5.342, then X2 will be included in the
solution.
Thus, ∞< C (new for X2)< 5.342 is the range of insignificance for X2.
C (profit contribution) for X2 can vary with in the given range without causing a change in
the optimal solution mix.

UNIT SUMMARY
Linear Programming Models are used to find optimal solutions to constrained optimization
problems. In order for linear programming models to be used the problems must involve a
single objective, a linear objective function and linear constraints and have known and
constant numerical values. We can solve linear programming problems using either
graphical approach or simplex approach. For solutions.
The graphical method for solution is used when the problem deals with 2 decision variables.
The inequalities are assumed to be equations (in equalities should be changed to equations).
As the problem deals with 2 variables, it is easy to write straight lines as the relationship
between the variables and constraints are linear. In case the problem deals with three
variables, then instead of lines we have to draw planes and it will become very difficult to
visualize the feasible area.

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If at all there is a feasible solution, feasible area of polygon exists then .The boundaries of the
regions are lines or planes. There will be corners or extreme points on the boundary and
there will be edges joining the various corners. The closed figure is known as polygon.

In contrast to the graphical approach, the simplex approach is not limited to problems with
to decisions variables. It is a general purpose method to solve problems with any number of
decision variables.

The simplex procedure involves developing a series of tableaus, each of which describes the
solution at a corner point of the feasible solution space begging with the origin after
standardizing the problem.

The standard form of LP problem should have the following characteristics:

I. All the constraints should be expressed as equations by slack or surplus and/or
artificial variables
II. The right hand side of each constraint should be made non-negative; if it is not, this
should be done by multiplying both sides of the resulting constraint by -1.
Example: 2X1+3X2-4X3+X3<-50, we multiply both sides by negative
III. Three types of additional variables, namely
a. Slack Variable(S)
b. Surplus variable (-S), and
c. Artificial variables (A) are added in the given LP problem to convert it into
standard form for two reasons:
i. to convert an inequality to have a standard form of an LP model, and
ii. to get an initial feasible solution represented by the columns of an identity
matrix
The summery of the extra variables needed to add in the given LP problem to convert it into
standard form is given below:

Types of Extra variables to be added Coefficient of extra variables Presence of variables in the
constraint in the objective function initial solution mix
Max Z Min Z
< Add only slack variable 0 0 Yes

> Subtract surplus variable and 0 0 No

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Add artificial variable -M +M Yes

= Add artificial variable -M +M Yes

Moreover you should note that when you solve problems using the simplex
methods

1. Test of optimality
i. If all C – Z < 0, then the basic feasible solution is optimal (Maximization case)
ii. If all C – Z > 0, then the basic feasible solution is optimal (Minimization case)
2. Variable to enter the basis
i. A variable that has the most positive value in the C – Z row
(Maximization case)
ii. A variable that has the highest negative value in the C - Z row (Minimization case)
3. Variable to leave the basis
The row with the non-negative and minimum replacement ratio
(For both maximization and minimization cases i.e.: RR > 0

Sensitivity analysis (post optimality analysis) is an analysis concerned with providing the
decision maker with greater insight about the sensitivity of the optimal solution to changes
in various parameters of the problem. Such changes might involve the specified levels of
constraints or coefficients of the objective function. Interest in changes may arise due to
improved information relating to a problem or because of the desire to know the potential
impact of changes that are contemplated.

SELF ASSESSMENT QUESTIONS (SAQ 2)

1. A furniture company produces a variety of products. One department specializes in wood
tables, chairs and bookcases. These are made using three resources: labour, wood and
machine time. The department has 60 hours of available each day, 16 hours of machine time
and 400 board feet of wood. A consultant has developed a linear programming model for the
department.
X1 = quantity of tables
X2= quantity of chairs

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X3= quantity of book cases

Maximize Z= 40x1+ 30x2 45x3[profit]
Subject to
Labor 2x1+x2+2.5x3 < 60 hours
Machine .8x1+.6x2+1.0x3 < 16 hours
Wood 30x1+20x2+30x3 < 400feet
Tables x1 >10 units
X1+x2+x3 > 0
Required: answer the following questions based on the above given information
a. What is the main purpose of the model?
b. What are the decision variables?
c. What are the system constraints? (name them)
d. What is the meaning of the number 2.5 in the labour constraint?
e. Explain what is meant by x1> 10.
f. What does the term 40x1in the objective function represents?
2. Explain the similarities and differences between graphical methods of solving linear
programming problem with simplex method.

3. Briefly define or explain each of these terms.

Shadow price Incoming variable

Range of feasibility Pivot row
Range of insignificance Pivot column
Range of optimality Key element
Slack variable Maximization problem
Surplus variable Minimization problem
Artificial variable Mixed constraint problem
Outgoing variable
4. A farmer has 1000 acres of land on which he can grow corn, wheat and soybean. Each acre of
corn costs birr 100 for preparation, requires 7 man-days work and yields a profit of birr 30. An
acre of wheat costs birr 120 to prepare, requires 10 man-days of work and yields a profit of birr
20. An acre of Soybeans costs birr 70 to prepare, requires 8 man-days of work and yields a profit
of birr 20. If the farmer has birr 100,000 for preparation and can count on 8000 man-days of
work, how many acres should be allocated to each crop to maximize profit.
Required: A. formulates the LPPM of the problem.
B Fined the optimal solution
5. Food X contains 20 units of vitamin A and 40 units of Vitamin B per gram. Food Y contains 30
units each of Vitamin A and B. The daily minimum human requirement of vitamin A and B are

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900 units and 1200 units respectively. How many grams of each type of food should be
consumed so as to minimize the cost, if food X costs birr 0.60 per gram and food Y costs birr 0.80
per gram.
Required: A. formulate the LPPM of the model
B. Fined the optimal solution using graphical solution
6. An air force is experimenting with three types of bombs P, Q, and R in which three kinds of
Explosive viz, A, B and C will be used. Taking the various factors into consideration, it has been
decided to use the maximum 600 kg of explosive A, at least 480 kg of explosive B and Exactly
540 of explosive C. Bomb P requires 3,2,2 kg of A,B,C respectively. Bomb Q requires 1, 4, 3 kg of
A, B, and C respectively. Bomb R requires 4, 2, 3 kg of A, B, C respectively. Now bomb P will give
the equivalent of 2 ton explosive, bomb Q will give 3 ton explosive and bomb R will give 4 ton
explosive. Under what production schedule can the Air Force make the biggest bang?
Required: formulate the LPPM of the model
7. A furniture workshop makes desks, chairs, and cabinet and bookcases. The work is
carried out in three major debarments—designing, Fabrication, and Finishing. Time required
per unit of product in hours is

Department Desk Chair Cabinet Book Case Time available per week
Designing 4 2 3 3 800
Fabrication 10 6 8 7 1200
Finishing 10 8 8 8 800

Cost of operation per unit of the products is (birr)

Department Desk Chair Cabinet Book case

Designing 15 8 12 12
Fabrication 30 18 24 21
Finishing 35 28 25 21
Selling 175 95 145 130
price/unit

In order to maintain to maximize profits what should be the production program

Required: formulate the LPPM of the model
8. solve the following problems using graphical methods

a. Max Z=15X1-10X2 b. Max Z=2X1+X2

St: St:

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4X1+6X2 < 360 X1+2X2 < 10

3X1+0X2< 180 X1 +X2 < 6
0X1+5X2< 280 X1 - X2 < 2
X1, X2 > 0 X1 -2X2 < 1
X1, X2 >0
c. Max Z=10X1+15X2 d. .Min Z=4X1+2X2
St: St:
2X1+X2 < 26 3X1+X2 > 27
2X1+4X2< 56 - X1 -X2 < -27
-X1+X2< 5 X1 + 2 X2 > 30
X1, X2 > 0 X1, X2 >0
9. Solve the following problems using simplex methods
a. . Max Z=3x1 +5x2 b. Max Z=20x1 +10x2
Subject to: Subject to:
x2 < 6 5x1+4x2 < 250
3x1+2x2 < 18 2x1+5x2 < 150
x1, x2 > 0 x1, x2 > 0
C. Min Z=7x1 + 9x2
Sub
3x1+6x2 >36
8x1+4x2 >64
X1& x2 > 0

D. Max Z=x1 + 1.5x2 +5x3+2x4

Sub
3x1 + 2x2 +x3+4x4 <6
2x1 + x2 +5x3+x4 <4
2x1 + 6x2 -4x3+8x4 =0
x1 + 3x2 -2x3+4x4 =0
X1& x2 > 0
10. Given the following LPPM and its final simplex tableau answer the questions that follow.
Objective function: _____________________
Subject to
2X1 +X2+2X3 > 72
X1+X2-X3 > 18
X1, X2 & X3 > 0

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The final (optimal) simplex tableau is given as follows assuming the X1 represents food one, X2
represents food two and X3 represents food three. More over the first constraint represents
vitamin requirement and the second constraint represent protein requirement.

B C 20 10 14 0 0 RHS
X1 X2 X3 S1 S2

X3 14 1/3 0 1 -1/3 1 /3 18
X2 10 4/3 1 0 -1/3 -2/3 36

Z
C-Z

Required:
a. write the objective function_____________________________
b. what is the shadow prices of the second constraints_________________
c. what is the range of feasibility for the first constraint___________________
d. what is the range of feasibility for the second constraint_________________
e. what is the range of optimality / insignificance for food one______________
f. what is the range of optimality/ insignificance for food three_____________
g. What impact on the solution will have a decrease of 9 units in vitamin constraints?
Variable Revised value
X3 ______________
X2 ______________
Z ______________

UNIT THREE
3. TRANSPORTATION AND ASSIGNMENT MODELS

3.1 LEARNING OBJECTIVES

 Describe the nature of transportation and assignment problems;
 Use the transportation method to solve problems;
 Deals with special cases of transportation problems;
 Use the assignment problems to solve problems;
 Deal with special cases of assignment problems.

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3.2. INTRODUCTION

One important application of linear programming has been in the area of the physical
distribution (transportation) of resources, from one place to another, to meet a specific set of
requirement.

This chapter describes two special –purpose algorithms: the transportation model and the
assignment model. Model formulation and manual solution are covered for each of these classes
of problems.

3.3 TRANSPORTATION PROBLEMS

? Activity
What is the purpose of transportation problem model?
_____________________________________________________________________________________________________________________
_____________________________________________________________________________________________________________________
__________________________________________________________________________________________________________________

 Comment:
Transportation problem deals with the distribution of goods from several points of supplies
(sources) to a number of points of demands (destinations) with a minimum cost of
transportation

Transportation problem deals with the distribution of goods from several points of supplies
(sources) to a number of points of demands (destinations). It is usually applied to distribution
type problems in which supplies of goods that are held at various locations are to be distributed
to other receiving locations.
Objective of the model: is to identify a distribution plan that would minimize the cost of
transporting the goods from the supply areas to the demand areas taking into account supply
capacities and demand requirements as well as transportation costs.
Application areas: we can use transportation model for various types of problems. For
example:
 Shipments from factories to warehouses
 Shipments between departments within a company
 To compare location alternative

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 Production scheduling etc

Necessary conditions for the model:
A transportation problem typically involves a set of sending locations which are referred to as
origins(supply areas ) and a set of receiving locations which are referred to as destinations (
demand areas). In order to develop a model of a transportation problem it is necessary to have
the following information:
1. Supply quantity (capacity) of each origin
2. Demand quantity of each destination
3. Unit transportation cost for each origin- destination route
Assumptions:
1. All goods be homogeneous
2. Transportation costs are a direct linear function of the quantity shipped over any route
3. The total quantity available for shipment is equal to the total quantity demanded

The characteristics of transportation problem are as follows:

1. A limited supply of one commodity is available at certain sources or origins.
2. There is a demand for the commodity at several destinations
3. The quantities of supply at each source and the demand at each destination are constant.
4. The shipping or transportation costs per unit from each source to each destination are
assumed to be constant.
5. No shipments are allowed between sources or between destinations. All supply and demand
quantities are given in whole number or integers.
6. The problem is to determine how many units shipped from each source to each destination so
that all demands are satisfied at the minimum total shipping costs.

Example:
Let’s consider that a firm has three factories /sources of supply/ & four warehouses /point of
demand/. The firm's production capacity at the three factories, the demand for the four
distribution centers located at various regions & the cost of shipping each unit from the factories
to the warehouses through each route is given as follows:

Factories Weekly capacity

F1 5000
F2 6000
F3 2500

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Warehouse Demand
W1 6000
W2 4000
W3 2000
W4 1500

Cost per unit to

From W1 W2 W3 W4

F1 3 2 7 6
F2 7 5 2 3
F3 2 5 4 5
Solution
The solution algorithm to a transportation problem is summarized into the following steps:
Step 1:
Formulate the transportation table: set up a transportation table with M- rows representing
the number of supply areas and N- columns representing the number of demand areas.
The formulation of the problem is similar to the linear programming. Here the objective function
is to minimize the total transportation cost and the constraints are the supply and demand
available at each source and destination respectively. The origins are listed down the left side of
the table, and the respective supply quantities are listed down the right side of the table.
Step 2:
Obtain an initial basic feasible solution: For this there are 3 methods to find the initial
feasible solution.

 North-West Corner Method (NWCM)

 Least Cost Method (LCM)
 Vogel’s Approximation Method (VAM)

The initial solution obtained by any of the above three methods must satisfy the following
condition:
The solution must be feasible: i.e. it must satisfy all the supply and demand constraints
The number of positive allocations must equal to m+n-1, where m=the number of rows
(or origins or supply centers) and n= the number of columns (or destination centers or demand
centers)
Example:

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M=3 origins and N=4 destinations ==>M+N-1=3+4 -1=6 (i.e. the transportation model should
have 6 occupied cells). If the number of occupied cells < m+n-1==> degenerate solution you
will see it in special issue of a transportation problem.

Step 3:
Test the initial solution for optimality: If the current solution is optimal, then stop.
Otherwise, determine the new improved solution.
For this there two method to find the optimal solution
1. Stepping stone methods
2. Modified distribution method( MODI)

Step 4:
Repeat step 3 until an optimal solution is reached
Now based on the above logic lets solve the problem.
3.3.1 TRNSPORTATION TABLE
Summarize all the given information in the tabular form as follows.
Destinations (dd) =j

Origin W1 W2 W3 W4 Factory
(Supply) Capacity =i

F1 3 2 7 6 5000

F2 7 5 2 3 6000

F3 2 5 4 5 2500

Requirements of the 6000 4000 2000 1500 13500

Warehouses
( Units of demand)

Note that we can solve the problem using LPPM, because transportation problem is a special
type of LPPM. So we can change the above problem into LPPM as follows:

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Let xij =The amount of commodity to be transported form source i (i =1,2,3 ) to destination j( j=
1,2,3,4).
Then the objective function of the problem (minimization of the total transportation cost) can be
formulated as:
Min Z = 3x11 +2x12 + 7x13 +6 x14 + 7x21 +5x22 +2x23 + 3x24 + 2x31+5x32 +4x33+5x34
Subject to the constraints
a. Supply constraints:
x11 +x12 +x13 +x 14 =5000 F1 supply constraint
x21 + x22 + x23 +x24 =6000 F2 supply constraint
x31 +x32 +x33+x34 = 2500 F3 supply constraint
b. Demand constraints:
x11 + x21 + x31 = 6000 W1 demand constraint
x12 + x22 + x32 = 4000 W2 demand constraint
x13 + x23 +x33 = 2000 W3 demand constraint
x14 +x24 + x34 = 1500 W4 demand constraint
xij > 0 for all i& j
In the above LPP, there are m x n = 3x4 =12 decision variables & m + n = 3+4 =7 constraints.
Thus, if this problem is solved by the simplex method, then it may take considerable
computational time.

To conceptualize the problem easily let’s represent of the problem using a Net work flow
diagram

Origin Destination centers

(Sources of Supply)
(Point of demand centers
F1 5000 3
2 W1 6000
6 7

7
F2 6000 5
W2 4000
3 2
2
5
F3 2500 4
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W3 2000
5

W4 1500

This LPP has 12 shipping routes. The objective is to identify the minimum cost route (Least cost
route). From these routes which rout results the minimum total transportation cost. Using Try
and error is very tiresome and inefficient. But by using the transportation problem models you
can identify the routes easily.

3.3.2. METHODS OF FINDING INITIAL FEASIBLE SOLUTION

There are several methods available to obtain an initial feasible solution. Here we shall discuss
only three different methods to obtain the initial feasible solution:

A. North- West Corner Method (NWCM)

This method does not take into account the cost of transportation on any route of
transportation.

The NWCM gets its name because the starting point for the allocation process is the Upper Left-
hand (Northwest) corner of the transportation table. Therefore, allocate to the Northwest
corner as many units as possible.

Northwest corner rule:

The following set of principles guides the allocation:
1. Begin with the upper left hand cell (Left, upper most in the table), & allocate as many units
as possible to that cell. This will be the smaller amount of either the row supply or the
column demand. Adjust the row & column quantities to reflect the allocation.

2. Subtract from the row supply & from the column demand the amount allocated
3. If the column demand is zero, move to the cell next to the right, if the row supply is zero,
move down to the cell in the next row.

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If both are zero, move first to the next cell on the right then down one cell.
4. Once a cell is identified as per step (3), it becomes a northwest cell. Allocate to it an amount as
per step (1)
5. Repeat, the above steps (1) - (4) until all the remaining supply and demand is gone.

Example:
1) Consider the following transportation problem:

T Store 1 Store 2 Store 3 Store 4 Supply

F

Plant 1 19 30 50 10 7

Plant 2 70 30 40 60 9

Plant 3 40 8 70 20 18

Demand 5 8 7 14 34

Required:
a. Develop an initial feasible solution using the NWCM
b. Compute the total cost for this solution.

? Activity
Using the above rules try to find the transportation schedule using NWCM in a piece of paper and
check your answer, before you read the solution.

Solution
a. Table: Initial feasible solution

To Store 1 Store 2 Store 3 Store 4 Supply

From
Plant 1 19 30 50 10 7
5 2
Plant 2 70 30 40 60 9
6 3
Plant 3 40 8 70 20 18
4 14

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Demand 5 8 7 14 34

Check that the solution is feasible or not:

==>m + n-1; m=3 and n=4  3+4-1= 6 cells occupied (Feasible solution)
The total transportation cost of the initial feasible solution derived by the NWCM is:

Route Unit Per unit Total

From To Shipped cost ( $) = Cost ( $)
Plant 1 Store 1 5 19 95
plant 1 Store 2 2 30 60
Plant 2 Store 3 6 30 180
Plant 2 Store 4 3 40 120
Plant 3 Store 4 4 70 280
Plant 3 Store 4 14 20 280

Total Cost= $ 1015

Note: NWCM does not consider the cost factor for allocation.
Example 2
1. Determine an initial basic feasible solution to the following transportation problem using
NWCM and Compute the total cost for this solution
Destination

A B C Supply
S1 2 7 14 5
S2 3 3 1 8
S3 5 4 7 7
S4 1 6 2 14
Demand 7 9 18 34

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Answer: X11=5, X21=2, X22=6, X32=3, X33=4, X43=4, and Total cost =$10

Assignment
Consider that Harley's Sand & Gravel Pit has contracted to provide topsoil for three residential
housing developments. Topsoil can be supplied form three different “farms" as follows:

_______________________________________________________________
Weekly Capacity
Farm (Cubic Yards)
A 100
B 200
C 200
_________________________________________________________________
Demand for the topsoil generated by the construction projects is:
_______________________________________________________________________________
Weekly Demand
Project (Cubic Yards)
1 50
2 150
3 300
_______________________________________________________________
The manager of the sand & gravel pit has estimated the cost per cubic yard to ship over each of
the possible routes:

Costs per cubic yard to

From Project # 1 Project #2 Project #3

Farm A $4 $2 $8
Farm B 5 1 9
Farm C 7 6 3

Required
Develop the initial feasible solution using NWCM & compute the total cost for this solution.
B. The Least- Cost Method (LCM)

LCM is the method that uses a minimum cost in the allocation.

It begins a solution by sequentially assigning to the cells with the minimum cost as many units
as possible. The first allocation be made to the cell with the lowest cost.

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The Least- Cost Method yields not only an initial feasible solution but also one that is close to
optimal in small problems.

? Activity
What is the difference between NWCM and LCM?
_______________________________________________________________________________
_______________________________________________________________________________
_______________________________________________________________________________


Example
Comment:
NWCM: we don’t use cost for allocation it uses direction (North West as base for allocation)
LCM: uses minimum transportation as base for allocation

1.Suppose that a firm has three factories / sources of supply /& four warehouses/point of
demand/ .The firm's production capacity at the three factories, the demand for the four
destination centers located at various regions & the cost of shipping each unit from the
factories to the warehouses through each route is given as follows:
Destinations

W1 W2 W3 W4 Factory
Capacity
F1 3 2 7 6 5000
F2 7 5 2 3 6000
F3 2 5 4 5 2500
Demand 6000 4000 2000 1500 13500

Required:
a. Develop an initial feasible solution using NWCM & Compute the total cost
b. Develop an initial feasible solution using least-cost method & compute the total cost.
Solution:

a. Initial feasible solution using NWCM

W1 W2 W3 W4 Factory
Capacity

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F1 3 2 7 6 5000
5000
Factory 7 5 2 3
F2 6000
1000 4000 1000
F3 2 5 4 5 2500
1000 1500
Demand 6000 4000 2000 1500 13500

m= 3, n =4 ==> 3+4 -1 =6 occupied cells (Feasible)

From To Units Shipped Unit Cost Total Cost

F1 W1 5000 3 $ 15000
F2 W1 1000 7 7000
F2 W2 4000 5 20000
F2 W3 1000 2 2000
F3 W3 1000 4 4000
F3 W4 1500 5 7500

Total transportation cost =$55,500

b.
Initial feasible solution using LCM

W1 W2 W3 W4 Factory
Capacity
F1 3 2 7 6 5000
1000 4000

F2 7 5 2 3 6000
2500 2000 1500
F3 2 5 4 5 2500
2500
Demand 6000 4000 2000 1500 13500

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Routes Units Unit Total

From To Shipped X Cost =Cost
F1 W1 1000 3 $ 3000
F1 W2 4000 2 8000
F2 W1 2500 7 17500
F2 W3 2000 2 4000
F2 W4 1500 3
45000
F3 W1 2500 2
5000

Total transportation cost =$42,000

m= 3, n=4 ==> 3+4-1 =6 occupied calls (Feasible)

Least- Cost method is better than the NWCM because it considers cost factories.
2. Develop the initial feasible solution for the following TP using the least-cost method (LCM)

De D E F G Supply
Source

A 1 5 3 4 100

B 4 2 2 5 60

C 3 1 2 4 120
DD 70 50 100 60 280

Solution
The 1st allocation be made to the cell with the least-cost. Cells AD & CD both have the lowest cost
f $1. Cell AD is selected 1st because more units can be allocated to it (70) than to cell CE (50).

The initial solution by the least -cost method

Destination D E F G Supply
Source
A 1 5 3 4 100
70 30

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B 4 2 2 5 60
30 30
C 3 1 2 4 120
50 70
demand 70 50 100 60 280

m=3, n=4 ==> 3+4-1 = 6 occupied cells (feasible)

Routes Units Unit Total

From To Shipped X Cost =Cost
A D 70 1 $ 70
B F 30 2 60
B G 30 5 150
C E 50 1 50
C F 70 2 140
Total transportation cost =$470

3. Develop an initial feasible solution using LCM

R S T Supply

A 1 2 3 100

B 4 1 5 110

Demand 80 120 60

Solution

R S T Supply
A 1 2 3 100
80 10 10
B 4 1 5 110
110
Dummy 0 0 0 50
50
Demand 80 120 60

Total transportation cost=80x1+10x2+10x3+110x1+50x0=$240

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? Activity
Solve problems that are solved by NWCM using LCM & vise-versa. What differences
you observe between NWCM and LCM. Which method is preferable? Why?
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 Comment:
The total cost of transportation is the minimum in LCM as compared to NWCM. LCM is
preferable because it minimizes the steps to reach the optimal solution.

C. Vogel's Approximation Method (VAM) or Penalty Method

VAM is preferred to the other two methods described above. In this method each allocation is
made on the basis of the opportunity (or penalty or extra) cost that would have incurred if
allocation in certain cells with minimum unit transportation cost were missed.

In this method allocation are made so that the penalty cost is minimized. The advantage of this
method is that it gives an initial solution which is nearer to an optimal solution or is the optimal
solution itself.

VAM determines the penalty for not using the minimum cost routes, where the objective is to
avoid large penalties so that the penalty from not using the routes is minimized.

The steps in VAM are as follows:

1. Calculate penalties for each row (column) by taking the smallest & the next smallest unit
transportation cost in the same row (column). This difference indicates the penalty or extra
cost which has to be paid if one fails to allocate to the cell with the minimum unit
transportation cost
2. Select the row or column with the largest penalty & allocate as much unit as possible in the
cell having the least cost in the selected row or column satisfying the conditions. If there is a
tie in the values of penalties, then t can be broken by selecting the cell where maximum
allocation can be made.

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3. Adjust the supply & demand & cross out the satisfied row or column. If a row or column is
satisfied simultaneously, only one of them is crossed out & the remaining row (column) is
assigned a zero supply (demand) .Any row or column with zero supply or demand should
not be used in computing future penalties.
4. Repeat step 1 to 3 until the entire available supply at various sources & demand at various
destinations are satisfied.
Example:
1. Determine an initial basic feasible solution to the following transportation problem using VAM.
Warehouse

A B C D Supply
F1 2 2 0 4 25 Row difference or Row penalty
Factory
5 20 or opportunity cost

5 9 8 3 2 0 - - -
F2 25
15 5 5
F3 6 4 3 2 10 2 2 2 2 5
10
Demand 20 15 20 5 60 1 2 2 - -

Column difference 3 2 3 1
or Column penalty
or opportunity cost 3 2 - 1

1 5 - 1

5 9 - -
m= 3, n=4 ==> 3+4-1 =6 Occupied cells (feasible)
5 - - -
The transportation cost associated with this solution is:
Total cost= 5x2 + 20x0+15x5x9 =+95x3+10x4= $185

2. A dairy firm has three plants located in different regions. The daily milk production at each
plant is as follows:
Plant 1: 6 million liters.
Plant 2: 1 million liters, &
Plant 3: 10 million liters
Each day the firm must fulfill the needs of its four distribution centers. Minimum
requirement at each center is as follows.
Distribution center 1: 7 million liters
" " 2: 5 " "
" " 3: 3 " "
" " 4: 2 " "

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Cost of shipping one million liters form each plant to each distribution center is given in the
following table in hundreds of dollar.

Distribution Center

D1 D2 D3 D4

P1 2 3 11 7
Plant
P2 1 0 6 1
P3 5 8 15 9

Find the initial basic feasible solution by:

a. North-west corners method
b. LCM
c. VAM if the object is to minimize the total transportation cost
Answer:
a. Total cost = $11, 600
b. Total cost= $11,200

D1 D2 D3 D4 Supply Row Penalty

P1 2 3 11 7 6 1 1 5 -
1 5
P2 1 0 6 1 1 1 - - -
1
P3 5 8 15 9 10 3 3 4 4
6 3 1
Demand 7 5 3 2 17
1 3 5 6
3 5 4 2
3 - 4 2
5 - 15 9

M+N -1 +3+4-1 =6 ==>the solution is non-degenerate.

The total transportation cost=$(1x2+5x3+1x1+6x5+3x15+1x9) x100 = $ 10,200

? Activity
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What is the base for allocation in VAM?
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3.3.3 OPTIMALITY TESTS

Once an initial solution is available, the next step is to check it for optimality. An optimal solution
is one in which there is no opportunity cost. That is, there is no other set of transportation
routes (allocations) that will reduce the total opportunity cost. Thus we have to evaluate each
unoccupied cell (represents unused route) in the transportation table in terms of opportunity
cost.

The purpose of the optimality test is to see if the proposed solution just generated can be
improved or not. The solution to be checked for optimality must be non-degenerate i.e. the no of
occupied cells must be >m+n-1.

The Procedure for testing optimality is analogous to that of the simplex method. A distinction is
made between basic variables, those associated with occupied cells & non-basic variables,
those associated with the empty cells.

For each empty cell, the effect of changing it to an occupied cell is examined. If any of these
changes are favorable, the solution is not optimal & a new solution must be designed. A
favorable change means an increase in the value of the objective function in maximization
problems or a decrease in minimization problems.

Optimum solution to a TP can be obtained by following two methods. These methods are much
simpler compared to simplex method of an LPP.

A. Stepping stone Method

B. Modified Distribution method (MODI Method)
A. Stepping-stone method
The Stepping-stone method is an iterative technique for moving from an initial feasible
solution to an optimal solution in transportation problems.

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For the stepping- stone method to be applied to a transportation problem, one rule about the no
of shipping routes being used must be observed. The rule is:

“The No of occupied routes (or squares) must always be equal or greater than to one less
than the sum of the no of rows plus the no of columns." i.e. Occupied shipping routes
(squares) > No of rows + No of columns - Non degenerate solution.

Rules for drawing each closed loop:

1. Select an unused square (cell) to be evaluates.
2.Beginning at this cell, trace a closed loop going clockwise draw an arrow to an occupied cell in
the same row ( or column).
3. Move vertically or horizontally (but never diagonally) to another occupied cell “stepping –
over” unoccupied or occupied cells (if necessary) without changing them. Follow the same
procedure to other occupied cells until returning to the original empty cell.

4. Begin with a plus (+) sign at the unused cell, place alternative (-) signs and plus signs on each
corner square of the closed path just traced. I.e. At each turn of the loop (the loop may cross
over itself at times), plus and minus signs are alternately placed in the cells, starting with a +
sign in an empty cell.

5. There must be exactly one cell with a + sign and exactly one cell with a - sign in any row or
column in which the loop turns.

6. An even no of at least four cells must participate in a loop and the occupied cells can be
visited once and only once.

7. Repeat steps 1 to 4 until an improvement index has been calculated for all unused squares (cells).
If all indices computed are greater than or equal to zero, an optimal solution has been reached.
If not, it is possible to improve the current solution and decrease total shipping costs.

Note:
In a non-degenerate problem, there is only one possible way of drawing the loop for each empty
cell.

How to find the value of a cell evaluator

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The value of a cell evaluator is the sum of the per unit shipping costs in the gaining cells less the
sum of the per unit shipping costs in the losing cells of the closed loop. This evaluation process
must be extended to all unoccupied cells.

If one or more of the cell evaluators is negative, the existing solution is not optimal. i.e.: For
minimization (cost) problems, all the cell evaluators
Project Project Project ss must be positive for optimality.
Fa

A B C
r

 Analysis of test:
Check all the empty cells and select for improvement the one with the largest improvement
potential.
 If the solution is not optimal, the next step in the transportation method is to find a
better solution. The operations in this step are:
a. Identify the “incoming" cell (the empty cell to be occupied) - In a minimization case, the
incoming cell is located by identifying the most negative cell evaluator
b. Design an improved solution, by shifting units form cell to cell

Note:
 A cell evaluator of 0 indicates the existence of another solution just as good as the
current solution. Thus, in the final solution, if cell evaluators of 0 exist, this indicates the
existence of multiple optimal solutions.
 If two or more cells have the same value, then either may be selected.
 If two or more of the "losing" cells contain the same no of units, both will become empty
simultaneously and a “degenerate" solution will result.

 For the minimization case; when one or more cell evaluators are negatives, the cell with
the largest negative should be brought into solution because that route has the largest
potential for improvement per unit.
 The loop starts and ends at the selected unoccupied cell. Every corner element of the
loop must be an occupied cell.
Example:
1. Use NWCM to find initial feasible solution and test the solution for optimality.

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F1 4 2 8 100
F2 5 1 9 200
F3 7 6 3 200
dd 50 150 300 500

Project Project Project

A B C ss

F1 4 2 8 100
50 50

F2 5 1 9 200
100 100

F3 7 6 3 200
200
dd 50 150 300 500

Project Project Project ss

A B C
F1 4 2 8 100
50 50 - +
m=3, n=3==> 3+3-1=5(Non-degenerate
F2 5 1 9 200
100 + 100- solution)
F3 7 6 3 200 Total transportation cost =
200
$[50x4+50x2+100x1+100x9+200x3] = $1900
Dd 50 150 300 500
Evaluation path for the unoccupied cells
Table: Test of optimality

Unoccupied cells Cell evaluators

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(F2 ,A) +5-4+2-1=+2

(F1 ,C) +8-9+1-2=-2
(F3 ,A) +7-4+2-1+9-3=+10
(F3 ,B) +6-1+9-3=+11

The negative value for cell (F1, C) indicates an improved solution is possible. For each unit we
can shift into that cell, the total cost will decrease by $2. The next question is how many units
can be reallocated into that cell while retaining the balance of supply and demand for that table?

The Stepping- stone path for cell (F1, C) is:

The + Signs in the path indicate units to be added, the - signs indicate units to be subtracted. The
limit on subtraction is the smallest quantity in a negative position along the cell path. There
are two quantities in negative positions, 50 and 100. Because 50 is the smaller quantity, that
amount will be shifted in the following manner: Subtract 50 units from each cell on the path
with a - sign and add 50 units to the quantity of each cell with a + sign in it.
With each iteration (new solution), it is necessary to evaluate the empty cells to see if further
improvements is possible.

The distribution plan after reallocation of 50 units is:

A B C ss
F1 4 2 8 100
50 50
F2 5 1 9 200
150 50

F3 7 6 3 200
200

dd 50 150 300 500

Table: Test of optimality

Unoccupied cells Cell evaluators

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(F1 ,B) +2 -8+9-1 =+2

(F2 ,A) +5 -4+8-9 =0
(F3 ,A) +7-4+8-3 =+8
(F3 ,A) +6-1+9-3= +11

Because none of these no is negative, this is an optimal solution. Therefore, the total cost for the
distribution plan is:
The total transportation cost = $ (50x4 +50x8 150x1+50x9 +200x3) = $1,800
Exercise:
Consider the following transportation problem

Store 1 Store 2 Store3 Supply

A 12 20 15 50
B 9 11 4 15
warehouses

C 20 14 8 55
DD 25 50 45 120

a. Develop an initial feasible solution using the NWCM. And compute the total cost for this
solution.
b. Evaluate the solution using the stepping-stone method. Is the solution optimal? Explain.
c. What is the total cost for the optimal solution?

B. Modified Distribution Method (MODI Method)

It is another algorithm for finding the optimal solution to a transportation problem.

The MODI method allows us to compute improvement indices quickly for each unused cell with
out drawing all of the closed paths. Because of this, it can often provide considerable time
savings over the stepping-stone method for solving transportation problems.
[

MODI provides a new means of finding the unused route with the largest negative improvement
index. Once the largest index is identified, we are required to trace only one closed path. Just as
with the stepping-stone approach, this path helps to determine the maximum No of units that
can be shipped via the best unused route.

Steps in the MODI Method

The steps to evaluate unoccupied calls are as follows:

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1. For an initial basic feasible solution, calculate Ui and Vj ;for rows and columns and
set

Cij = Ui + Vj for all occupied calls (i , j)

i.e.: Cell cost= Raw Index +Column Index

2. For unoccupied cells, calculate opportunity cost by using the relation:

Kij = Cij– Ui - Vj ; for all j and j.

I.e. cell cost –row index- column index

3. Examine the sign of each Kij :

i. If Kij > 0, then current basic feasible solution is optimal.
ii. If Kij = 0, then the current basic feasible solution will remain be unaffected but
an alternative solution exists.
iii. If one or more Kij < 0, then an improved solution be obtained entering
unoccupied cell (i, j), in the basis. An unoccupied cell having the largest
negative value of Kij is chosen for entering into the solution mix (new
transportation schedule)
4. Solve the problem as you did using the stepping-stone method.
i.e. construct a closed path (or loop) for the unoccupied cell with largest negative
opportunity cost. Start the close path with the selected unoccupied cell and mark a
plus sign (+) and in this cell, trace a path along the rows (or columns) to an occupied
cell, mark the corner with minus sign (-) and continue down the column (or row) to
an occupied cell and mark the corner with plus sign (+) and minus sign (-)
alternatively. Close the path back to the selected unoccupied call.
Locate the smallest quantity allocated to a cell marked with a minus sign. Allocate
this value to the selected unoccupied cell and add it to other occupied cells marked
with plus signs and subtract it from the occupied cells marked with minus signs.

5. Obtain a new improved solution by allocating units to the unoccupied call and
calculate the new transportation cost.

6. Test the revised solution for optimality.

Note:
Any initial feasible solution will do: NWCM, VAM Solution, or any arbitrary assignment.

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The stepping- stone method is efficient for small sized transportation problems. For larger
problems, however, the MODI method is recommended.

? Activity
What differences observe between stepping stone and MODI METHOD?
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
Example:
Comment:
Stepping stone uses closed path to evaluate empty cells
MODI uses row index and column index values to evaluate the empty cells

1.Obtain an optimal solution to the transportation problem by MODI method given below:

Project Project Project Supply

No 1 No 2 No 3

Farm 1 4 2 8 100
Farm 2 5 1 9 200
Farm 3 7 6 3 200
DD 50 150 300 500

Solution
Note:
Both the MODI and the stepping - stone method will yields the same values.
Remark:
Conventionally, we begin by assigning a value of zero as the index for row 1 (U1=0). Once row
index has been established, it will enable us to compute column index numbers for all occupied
cells in that row. Similarly, once a column index number has been determined, index numbers
for all rows corresponding to occupied cells in that column can be determined.

Consider the initial feasible solution of the given example by NWCM as shown below:
Initial solution, NWCM

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Project Project Project Supply Ui

No 1 No 2 No 3

Farm 1 4 2 8 100 U1=0

50 50

Farm 2 5 1 9 200 U2=1

100 100
Farm 3 7 6 3 200 U3=-7
200

Demand 150 300 500

50

Vj V1=4 V2=2 V3=10

To determine Cij, use the occupied cells.

For instance, C11=4, C12=2, C22=1, C23=9, and C33=3

Cij= Ui + Vj
==>C11= U1 +V1==>4=0+ V1==> V1=4 , U1=0 by convention
==>C12= U1 +V2==>2=0 +V2==> V1=2
==>C22= U2 +V2==>1= U2+ 0==> U2=-1
==>C23= U2 +V3==>9= -1+V3==> V3=10
==>C33= U3 +V3==>3= U3+10 ==> U3= -7

Note:
Cij≠ Ui + Vj (For unoccupied cells)
For instance, from the above information, C32 ≠ U3 + V2==>6≠-7+2

Table: Test of optimality

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Unoccupied cells Cell evaluators

Kij = Cij– Ui - Vj
(1,3) C13 – U1 -V3=8-0-10= -2
(2,1) C21 – U2 -1=5+1-4=+2
(3,1) C31– U3 -V1=7+7-4=10
(3,2) C32– U3 -V2=6+7-2=+11

In this case, we found hat cell (1, 3) had an evaluation of -2, which represented an improvement
potential of and $ 2 per unit. Hence, an improved solution is possible.
The stepping-stone path for call (1, 3) is:

Project Project Project Supply

No 1 No 2 No 3

Farm 1 4 2 8 100
50 50
Farm 2 5 1 9 200
100 100
Farm 3 7 6 3 200
200

Deman 150 300 500

d 50

The distribution plan after reallocation of 50 units is:

Projec Project Project Supply Ui

t No 2 No 3
Farm 1 4
No 1 2 8 100 U1=0
50 50

Farm 2 5 1 9 200 U2=1

+-100 100+-

Farm 3 7 6 3 200 U3=-5

200
Deman 150 300 500
50
d

Vj V1=4 V2=0 V3=8

Cij= Ui + Vj
==>C11= U1 +V1==>4=0+ V1==> V1=4 , U1=0 by convention
==>C13= U1 +V3==>8=0 +V3==> V3=8

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==>C23= U2 +V3==>1= U2+ 0==> U2=1

==>C22= U2 +V2==>1= 1+V2==> V2= 0
==>C33= U3 +V3==>3= U3+8 ==> U3= -5
Table: Test of optimality

Unoccupied cells Cell evaluators

Kij = Cij– Ui - Vj
(1,2) C12– U1 -V2=2-0-0= 2
(2,1) C21 – U2 -V1=5-1-4=0
(3,1) C31– U3 -V1=7+5-4=8
(3,2) C32– U3 -V2=6+5-0=+11

Because none of the cell evaluators is negative, this is an optimal solution.

Thus, the total cost for the distribution plan =$1800
Assignment
1. Obtain an optimal solution to the transportation problem by MODI method given below.

D1 D2 D3 D3 Supply

Farm 1 19 30 50 10 7

Farm 2 70 30 40 60 9

Farm 3 40 8 70 20 18

DD 8 7 34
5 14

2. A company has four ware houses a, b, c and d. It is required to deliver a product from these
warehouses to three customers A, B and C. The warehouses have the following amounts in
stock:
Ware house: a b c d
No of units: 15 16 12 13

And the customer’s requirements are:

Customer: A B C
No of units: 18 20 18

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The table below shows the costs of transporting one unit from warehouse to customer.

A b c d

A 8 9 6 3

B 6 11 5 10

C 3 8 7 9

Find the optimal transportation routes.

3. A manufacturer has distribution centers at X, Y and Z. These centers have availability 40, 20,
and 40 units of his product. His retail outlets at A, B, C, D and E require 25,10,20,30 and 15
units respectively. The transportation cost (in $) per unit between each centre outlet is given
below:

Retail outlets
Distribution
center

A B C D E

X 55 30 40 50 40

Y 35 30 100 45 60

Z 40 60 95 35 30

Determine the optimal distribution to minimize the cost of transportation.

3.3. SPECIAL CASES

I. Degeneracy
A condition that occurs when the No of occupied cells in any solutions less than the No of rows
plus the No of columns minus 1 in a transportation table

I.e. No of occupied calls < m+ n -1 .....................Degeneracy

If the No of occupied cells = m+n-1, then the solution is non- degenerate.

The degeneracy in the transportation problems may occur at two stages:

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1. when obtaining an initial solution

2. During improvement (or at any stage while moving to wards optimal
solution.
To resolve degeneracy, we processed by allocating a very small quantity close to zero to one or
more unoccupied cell 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.

 = Almost zero

In a minimization transportation problem, it should be allocated to a cell that has the smallest

transportation cost. Insert  when it is able to create a closed loop for each occupied cell.

The purpose of epsilon/delta is to enable evaluation of the remaining empty cells. The choice of
location for the epsilon/delta can be somewhat tricky: some empty cells may be unsuitable if
they do not enable evaluations of remaining empty cells. Not all choices would be acceptable.
Actually, the No of epsilon/deltas needed will equal the difference between the No of completed
cells and m+n-1.Howerver; you will only be exposed to the most common case in which one
more completed cell is needed.

The epsilon/delta cannot be placed in a cell which later turns out to be in a negative position of
a cell path involved in reallocation because epsilon/delta will be the “smallest quantity a
negative position “ and shifting that minute quantity around the cell path will leave the solution
virtually unchanged. Consequently, a certain amount of trial and error may be necessary before
a satisfactory location can be identified for epsilon/delta.

Example
1. Solve the following transportation problem.

1 2 Supply
1 3 3 50

2 4 6 30
Deman 50 30 80
d

Solution:
Using NWCM and MODI, the initial solution is:

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1 2 Supply Ui

1 3 3 50 U1=0
50 
2 4 6 30 U2=3
30
Demand 50 30 80
Vj V1=3 V2=3

Cij= Ui + Vj
==>C11= U1 +V1==>3=0+ V1==> V1=3, U1=0 by convention
==>C12= U1 +V2==>3=0 +V2==> V2=3
==>C22= U2 +V2==>6= U2+3==> U2= 3
==>C33= U3 +V3==>3= U3+8 ==> U3= -5
Note: m=2 and n=2==>2+2-1=3==>Occupied cells=2< 3 (Degeneracy)
Table: Test of optimality

Unoccupied cell Cell evaluator

Kij = Cij– Ui - Vj
(2,1) C21 – U2 -V1=4-3-3 =-2

The optimal solution is:

1 2 Supply Ui

1 3 3 50 U1=0
50 30

2 4 6 30 U2=1
30
Demand 50 30 80

Vj V1=3 V2=3

Cij= Ui + Vj
==>C11=U1+V1==>3=0+V1==>V1=3, U1=0 by convention
==>C21= U2+V1==>4= U2+3==> U2=3
==>C12= U1 +V2==>3= 0+ V2==> V2= 3

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Table: Test of optimality

Unoccupied cells Cell evaluators

Kij = Cij– (Ui + Vj)
(2,2) C22– U2 -V2=6-1-3= +2

The total cost= $(20x3+30x3+30x40=$270

II. Alternative Optimal solutions
The esist6ane of alternative optimal solution can be determined by an inspection of the
opportunity costs, Kij for the unoccupied cells. If an unoccupied cell in an optimal solution has
opportunity cost of zero, then an alternative optimal solution can be formed with another set of
allocations without increasing the total transportation cost.

Exercise
XYZ Tobacco Company purchases tobacco and stores in warehouses located in the following four
cities.

warehouse location Capacity (Tones)

City A 90
City B 50
City C 80
city D 60

The warehouses supply tobacco to cigarette companies in three cities that have the following
demand:

Cigarette Company Demand (Tones)

L 120
P 100
Q 110

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The following railroad shipping costs per tone (in hundred dollars) have been determined:

Warehouse L p Q

A 7 10 5

B 12 9 4

C 7 3 11

D 9 5 7

Because of rail road construction, shipments are temporarily from warehouse at city A to L
Cigarette Company.
A. Find the optimum distribution for XYZ Tobacco Company.
B. Are there multiple optimum solutions? If there are alternative optimum solutions,
identify them.

III. Prohibited Transportation Routes.

The situation may arise such as road hazards (snow, floods, etc), traffic regulation, equipment
break down, labour problem etc, when it is not possible to transport goods from certain sources
to certain destinations. To solve such kinds of problem, assign a cost 10 times the largest cost in
the table and then it will be uneconomical to transport to that cell. Then do as usual. (Initial
feasible solution by NWCM/LCM/VAM and optimal solution by stepping stone /MODI)
Example

A B C D Supply
S1 1 5 3 3 34
S2 3 3 1 2 15
S3 0 2 2 4 12
S4 2 7 2 4 19
demand 21 25 17 17

Note that due to uncontrollable reason it is impossible to transport from S3 to C. So to solve the
problem fist adjust it as follows

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A B C D Supply
S1 1 5 3 3 34
S2 3 3 1 2 15
S3 0 2 70 4 12
S4 2 7 2 4 19
demand 21 25 17 17

The cell cost of S3C i.e. 2 will be replaced by 70 i.e. 7 * 10, because 7 is the largest cost in the
table.
Then do as usual.

IV. Unequal Demand and Supply

To solve a problem with an equal supply and demand total, first,

convert the unbalanced TP into a balanced TP by using dummy destination/dummy source.
* If total Supply > Total demand, then create a fictitious or artificial destination called dummy
destination i.e.: total Supply > Total demand===> Add dummy column
* But if excess demand (Supply < demand) exists add a dummy source (add a dummy row)
Note: the cost of “shipments” to the dummy is usually set at zero ==> No real cost

Example
Consider the following TP
Destination
R S T SS
A 1 2 3 100
B 4 1 5 110
210
Origin

DD 80 120 60
260

a. Obtain the basic feasible solution using VAM

b. Obtain the optimal solution
c. What is the optimal shipping cost?
Solution:

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a. Initial feasible solution

To
R S T ss
Opportunity cost
1 2 3 100
A 10 10 1 1 1 1
80
4 1 5 110
B 3 3 - -
110
Dummy 0 0 0 50 0 - - -
From

dd 80 120 60 260
Opportunity 1 1 3
3 1 2
cost 1 2 3
1 2 -

b. Test of optimality.
Table: Test of optimality

Unoccupied cells Cell evaluators

(B,R) +4-1+2-1= +4
(B,T) +5-1+2-3= +3

(D,R) +0-1+3-0= +2

(D,S) +0-2+3-0= +1

Since none of the cell evaluators is negative, the above feasible solution is optimal.
Thus, accordingly the distribution is as follows
A Supplies 80 units to warehouse R
B Supplies 10 units to warehouse S
C Supplies 10 units to warehouse T
B Supplies 110 units to warehouse S

c. The total optimal shipping cost is = $240

V. Maximization Problem
If you are faced with transportation type problem concerned with profit (revenue). First change
the problem to a minimization type. The steps are:
 Select the largest unit profit from the total cells
 Subtract each cell value from the largest
 Do the same procedure as usual
Example

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T A B C SS
F

1 6 8 10 150

2 7 11 11 175

3 4 5 12 275

DD 200 100 300 600

To change the above maximization problem to a minimization type first identify the largest unit
profit from the cell (12) then subtract all the cells values from this largest selected value as
indicated below( 12-6=6, for cell1A, 12- 8=4 etc ). These yields opportunity cost figures.
Then do as usual. For example if you use LCM for initial solution and MODI for optimality test
the solution will be:

T A B C SS
F

1 6 4 2 150
U1=0
150

2 5 1 1 175 U2=-1
50 100 25

3 8 7 0 275
275 U3=-2

DD 200 100 300 600

v1 = 6 v2 = 2 v3 = 2

Unoccupied cells Cell evaluators

(1,B) 4-0-2=2
(1,C) 2-0-2=0
(3,A) 8+2-6=4

(3,B) 7+2-2=7

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All the improvement index are o and +ve, we reached the optimal solution. The optimal solution
is:

Routes Units Unit Total

From To Shipped X Cost =Cost
1 A 150 6 $ 900
2 A 50 7 350
2 B 100 11 1100
2 C 25 11 275
3 C 275 12 3300

Total profit gained =5925

3.4 ASSIGNMENT PROBLEMS

The Assignment Problem(AP) refers to the class of LPPs that involves determining the most
efficient assignment of people to projects, salespeople to territories, contracts to bidders
,jobs to machines, and so on. The objective is to assign a number of resources to an equal
number of activities so as to minimize total costs or total time or maximize total profit of
allocation.
The problem of assignment arises because available resources such as men, machines, etc have
varying degrees of efficiency for performing different activities such as job. Therefore, cost,
profit or time of performing the different activities is different.

Assumptions:
The AP is a special case of TP under the condition that the number of origins is equal to the
number of destinations. Viz. m=n .Hence assignment is made on the basis of 1:1relationship.

Following are the assumptions:

 Number of jobs is equal to the number of machines or persons
 Each man or machine is loaded with one and only one job.
 Each man or machine is independently capable of handling any of the jobs being presented.
 Loading criteria must be clearly specified such as “minimizing operating time” or
“maximizing profit” ,or “minimizing production cost” or “minimizing throughout
(production cycle) time ” etc.

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Remark:
 The AP is considered as a special TP in which the supply at each source and the demand at each
destination are always one unit.
 Since the supply and demand are always equal to one unit in each row and column, there is no
need to write them in the assignment table.

Example:
Service costs of different team assignment ($ in thousands

Table: The assignment table

Z Z1 Z2 Z3
Zone Z1 Z2 Z3
Se SS
Seri
S1 20 15 31 ====>
S1 20 15 31 1
S2 17 16 33
S2 17 16 33 1
S3 18 19 27
S3 18 19 27 1
DD 1 1 1

The above problem can be presented as a LPP as follows:

MinZ = 20x11 +15x12 + 31x13 +17x21 +16x22 +33x23 +18x31+19x32 +27x33
Subject to the constraints
a. Supply constraints:
x11 +x12 +x13 =1 S1 constraint
x21 + x22 + x23 =1 S2 constraint
x31 +x32 +x33 = 1 S3 supply constraint

b. Demand constraints
x11 + x21 + x31 = 1 Z1 constraint
x12 + x22 + x32 = 1 Z2 constraint
x13 + x23 +x33 = 1 Z3 constraint
xij either 0 or 1 for all i , j
Since all xij can be either 0 or 1, there will be one assignment in each supply constraint and one
assignment in each demand constraint.

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As in the transportation problem, assignment problems can be balanced or not. In a balanced

case, the number of objects to be assigned equals the number of objects to which they are
assigned. Unbalanced problem can be balanced by adding a dummy (dummies) with zero cost
coefficients.

? Activity
What differences observe between transportation problem and assignment problem?
_________________________________________________________________________
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 Comment:
TP- transporting products from supply area to consumption area with a minimum
transportation cost
AP -assigning in a one to one relationships jobs to men, machine to jobs etc

3.4.1 Methods of solving assignment problems

An assignment problem can be solved by the following methods:
1. Enumeration method
2. Simplex method
3. Transportation method
4. Hungarian method

1. Hungarian Method/Flood’s Technique/

The Hungarian Method (developed by Hungarian mathematician D.Konig) of assignment
provides us with efficient method of finding the optimal solution without having to make a direct
comparison of every solution. It works on the principle of reducing the given cost matrix (the
principle of matrix reduction) to a matrix of opportunity costs, which means that by
subtracting and adding appropriate numbers in the cost table or matrix ,we can reduce the
problem to a matrix of opportunity costs.

Opportunity costs show the relative penalties associated with assigning resource to an activity
as opposed to making the best or least-cost assignment. If we can reduce the cost matrix to the
extent of having at least one zero in each row and column, then it will be possible to make
optimal assignments.

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Steps in solving assignment problems:

Step.1. Develop the cost table from the given problem

If the number of rows does not equal the number of columns and vice versa, then a dummy row
or dummy column must be added. The assignment costs for dummy cells are always zero.

Step 2. Find the opportunity cost table

The transformation of the cost matrix that you have learned in TP is termed as a total-
opportunity cost matrix in assignment problem.
It involves two operations:
a. Perform row operation
i.e. Locate the smallest element in each row of the given cost table and then
subtract that from each element of that row.
b. Perform column operation
i.e. In the reduced matrix obtained from 2(a) ,locate the smallest element in each
column and then subtract that from each element of that column. Notice that
each row and each column now have at least one zero value.
Step 3.Test for an optimal assignment
Test the table resulting from step 2 to see whether an optimal assignment can be made. The
procedure is:
a. Draw the minimum number of Horizontal and /or Vertical lines necessary to cover all
zeros costs. Draw the lines by trial and error but always try to cover two or more
zeros with one line.
b. Count the number of lines
If the number of lines equals either the number of rows or columns in the table, an
optimal assignment can be made. If the number of lines is less than the number of
rows or columns, an improvement is possible (we proceed to step 4).

Step 4.Improve the present opportunity cost table (matrix)

This is done by the following operations:
a. Find the smallest entry in the uncovered cells (cells with no lines through them) and subtract it
from all entries in the uncovered cells.
b. Add the same smallest entry to those cells in which the lines intersect (cells with two lines them)

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c. Cells with one line through them are transferred (i.e. unchanged to the improved table).
In those problems where the first improvement does not yield an optimal solution, we keep on
improving the solution by repeating step 4 until an optimal solution is achieved.

Step 5: Make an optimal assignment

An optimal assignment should be made to cells with a zero entry, maintaining the one-to-one
requirement

If more than one optimal solution exists, a trial-and –error approach can be used to find all
possible combination assignments in the zero cells.

Note that multiple optimal solutions are possible.

? Activity
Write the specific steps to solve an assignment problem
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

Example:
1. A computer center has three programmers. The center wants three application programs to
be developed. The head of the computer center, after studying carefully the programs to be
developed, estimate the computer time in minutes required by the experts for the
application programs as follows:

Programs
(Estimated time in minute)
A B C
1 120 100 80
Programmers
2 80 90 110
3 110 140 120

Assign the programmers to the programs in such a way that the total computer time is the
minimum.

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Solution:
Steps 1 and 2:
a. Perform row reduction

The minimum time element in row 1, 2, and 3 is 80, 80 and 110 respectively. Subtract those
elements from all elements in their respective row. The reduced time matrix is: Table: After row
reduction

A B C
Row minimum

-80 1 40 20 0
-80 2 0 10 30
-110 3 0 30 10

b. Column reduction
Since column B has no one ‘0’, perform also column reduction. The minimum time element in
columns A, B and C is 0, 10 and 0 respectively. Subtract these elements from all elements in their
respective column to get the reduced time matrix.

Table: After column reduction

A B C

1 40 10 0
2 0 0 30
3 0 20 10

Step 3: Test for an optimal assignment

a. Draw the minimum number of horizontal and /or vertical lines necessary to cover all zero
times (costs).

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Table: Test of optimal assignment

A B C

1 40 10 0
2 0 0 30
3 0 20 10

b. Count the number of lines

If the number of lines is equal to the number of rows/columns, the optimal solution is obtained.
Thus proceed directly to step 5.

Step 5.Make an optimal assignment

An optimal assignment should be made to cells with a zero entry, maintaining the one-to-one
requirement.

Table: optimal assignment

A B C

1 40 10
0
2 30
0 0
3 20 10
0

Note:
In optimal assignment, start with row/column having one zero and cancel the alternative
zeros(x)

The pattern of assignment among programmers and programs with their respective time (in
minute) is given below:

Programmer Program Time (in minutes)

1 C 80

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2 B 90
3 A 110
Total time=280 minutes

Cross Check Formula

Optimal solution = summation of row minimum + summation of columin minimum +
revised reduction
i.e., considering the above problem:
280=270 +10+0
280=280

2. A department has five employees with five jobs to be performed .The time (in hours) each
man will take to perform each job is given in the effectiveness matrix.

Employees
I II III IV V

A 10 5 13 15 16

Jobs B 3 9 18 13 6
C 10 7 2 2 2
D 7 11 9 7 12
IV
E 7 9 10 4 12

How should the jobs be allocated, one per employees, so as to minimize the total man-hours?
Solution Table:
V After row reduction

-5 A 5
I 0 II 8 III 10
IV 11
V

-3 B 0 6 15 10 3

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-2 C 8 5 0 0 0
-7 D 0 5 2 0 5

-4 E 3 5 6 0 8

Since the number of lines less than the number of rows/columns, an improvement is possible.
Step 4: Improve the present opportunity cost table
This is done by the following operations;
a. Select the smallest entry (element) among all uncovered elements by the lines and subtract it
from all entries in the uncovered cells.
b. Add the same smallest entry to those cells in which lines intersect (cells with two lines them).
c. Cells with one line through them are unchanged to the improved table

Table: After improvement

A
I 7 0 8 12 11
B 0 4 13 10 1
C 10 5 0 2 0
D 0 2 0 0 2

E 3 3 4 0 6
II

III
Since the number of lines equals to the number of rows/columns, the solution is optimal.
Table: Optimal assignments
IV

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I II III IV V

A 7 8 12 11
0
B 4 13 10 1
II 0
C
10 5 2 0
III 0

D 2 2
IV 0 0 0
E 3 3 4 6
V 0

The pattern of assignments among jobs and employees with respective time (in hours) is given
below:

Job Employees Time (in minutes)

A II 5
B I 3
C V 2
D III 9

E IV 4

Total time=23 Hours

3. A manager has prepared the following table, which shows the costs for various combinations
of job-machine assignments:

Job Machine (Cost in ’000s))

1 20 15 31
2 17 16 33
A B C
3 18 19 27

a. What is the optimal (minimum-cost) assignment for this problem?

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b. What is the total cost for the optimum assignment?

Solution:
Table: After row reduction Table: After column reduction

A B C A B C

-15 1 5 0 16
1 5 0 7
-16 2 1 0 17
2 1 0 8
-18 3 0 1 9
3 0 1 0
Table: After
improvement

A B C

1 4 0 6
2 0 0 7
3 0 2 0

Since the number of lines and the number of rows/ columns are equal the solution is optimal

Table: Optimal Assignment

A B C

1 4 6
0

2
0 0 7

3 2 0
0
Job Machine Cost(in $)

1 B 15000

2 A 17000

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3 C 27000

Total optimal assignment=$59000

3.4.2 SPECIAL CASES IN THE ASSIGNMENT PROBLEMS

Certain situations can arise in which the model deviates slightly from that previously described.
Among those situations are the following:

I. Multiple Optimal Solutions

While making an assignment in the reduced assignment matrix, it is possible to have two or
more ways to strike off a number of zeros. Such situation indicates multiple optimal solutions
with the same optimal value of objective function. In such cases the more suitable solution
may be considered by the decision-maker.

In multiple optimal solutions, no unique 0 will exist at some point, resulting in more than one
choice for assignment and hence, more than one optimal solution. It should be noted that all
optimal solutions will yield the same value of the objective function.
Example:1
1. Given this final assignment table, identify two optimal solutions.

Job Machine (Estimated time in minute)

1 2 3

A 4 0 0
B 0 3 2
C 1 0 0

Solution
The first assignment must be B-1, because B-1 is the only 0 that appears in a single row or
column. Having made that assignment, there are two choices for the remaining two rows, and
two choices for the remaining two columns. This results in two possible solutions, as shown:

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Job Machine (Estimated time in minute)

Job Machine (Estimated time in minute)
1 2 3
1 2 3

A 4 A 4
0 0 0
B 3 2 B 3 2
0
0 0
C 1 C 1
0 0 0 0

Exampl
e: 2
The foreman of a machine shop wants to determine a minimum cost matching for operators
and machines. The foreman has determined hourly cost for of four operators for the four
machines, as shown in the following cost table.

Machine(Estimated cost in $)
A B C D

1 70 80 75 64
2 55 52 58 54
3 58 56 64 68
Operator

4 62 60 67 70

Required:
a. Determine the minimum-cost assignment for this problem
b. What is the total cost for the optimal assignment?
c. Is there an alternative optimal assignment? What is it? Calculate the total cost for the alternate
optimal assignment.
Solution:

A B C D
D

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Table: After row reduction 1 4 16 5 0

2 1 0 0 2

A B C D 3 0 0 2 12
D 4 0 0 1 10
1 6 16 11 0
2 3 0 6 2 Table: After column reduction
3 2 0 8 12
4 2 0 7 10

b.

Operator Machine Cost(in $)

4 A 62
3 B 56
2 C 58
1 D 64
Total cost =$240

c. Yes!
Alternative optimal assignment

Operator Machine Cost(in $)

1 D 64
2 C 58
3 A 58
4 B 60
Total cost=$240

II. Maximization case in assignment problems

There may arise situations when the assignment problem calls for maximization of profit,
revenue, etc as the objective function. Such problem may be solved by converting the given
maximization problem into a minimization problem by the following procedure:

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i. Find the largest profit coefficient in the entire.

ii. Subtract each entry in the original table from the largest profit coefficient.
The transformed assignment problem so obtained can be solved by using the Hungarian
method.

Example
1.A company has four territories open, and four salesmen available for an assignment. The
territories are not equally rich in their sales potential. Based on the past performance, the
following table shows the annual sales (in $) that can be generated by each salesman in each
territory. Find the optimal assignment and the maximum expected total sales.

Territory

I II III IV

A 42 35 28 21
Salesmen

B 30 25 20 15

C 30 25 20 15

D 24 20 16 12

Solution:

Convert maximization problem into minimization problem by subtracting all elements from the
highest element (i.e. 42)
Thus, the equivalent cost table is:

I II III IV I II III IV

A 0 7 14 21 A 0 3 6 9

B 12 17 22 27 B 0 1 2 3

C 12 17 22 27 C 0 1 2 3

D 18 22 26 30 D 0 0 0 0

Thus, after improvement of the table, the optimal assignment is:

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I II III IV

A 0 2 4 7

B 0 0 0 1

C 0 0 0 1

D 2 1 0 0

The pattern of two alternative optimal assignments among territories and salesmen with
respective sale is given below:

Assignment set II
Assignment set I

Salesman Territory Sales($) Salesman Territory Sales($)

A I 42 A I 42
B III 20 B II 25
C II 25 C III 20
D IV 12 D IV 12
Total= $ 99 Total= $ 99

Exercise
Five salesmen are to be assigned to five territories. Based on the past performance, the following
table shows the annual sales man in each territory. Find the optional assignment.

T1 T2 T3 T4 T5

S1 26 14 10 12 9

S2 31 27 30 14 16

S3 15 18 16 25 30

S4 17 12 21 30 25

S5 20 19 25 16 10

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III. Unbalanced Assignment Problem (unequal number of

The Hungarian method of assignment requires that the number of columns and rows in the
assignment matrix be equal. However, when the given cost matrix is not a square matrix, the
assignment problem is called an unbalanced problem. In such cases a dummy row(s) or
column(s) are added in the matrix (with zeros as the cost elements) to make it a square matrix.
After making the given cost matrix a square matrix, the Hungarian method may be used to solve
the problem.

Example
MEGA printing press, a publisher headquartered in Addis Ababa, wants to assign three recently
hired college graduates, Marta, Bakcha and Hirut to regional sales districts in Mekelle, Bahir
Dare, and DireDawa. But the firm also has an opening in Gambela and would send one of the
three there if it were more economical than a move to Mekelle, Bahir Dar and Dire Dawa. It will
cost Br. 1,000 to relocate Marta to Gambela, Br. 800 to relocate Baklcha there, and Br. 1,500 to
move Hirut. What is the optimal assignment of personnel to offices?

Office Mekelle Bahir Dare Dire Dawa

Hire

Marta Br.800 Br 1,100 Br 1,200

Bekcha Br. 500 Br 1,600 Br 1,300
Hirut Br. 500 Br 1,000 Br 2,300

Solution
To balance the problem, we add a dummy row (person) with a zero relocation cost to each city.

City C1 C2 C3 C4(Gambela)

Person

P1 800 1,100 1,200 1,000

P2 500 1,600 1,300 800

P3 500 1,000 2,300 1,500

Dummy 0 0 0 0

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Table: after row reduction Table: Optimal Assignment

C1 C2 C3 C4
C1 C2 C3 C4

P1 0 300 400 200

P1 100 0 100 0
P2 0 1,100 800 300
P2 0 700 400 0
P3 0 500 1800 1000
P3 0 100 1400 700
Dummy 0 0 0 0
Dummy 400 0 0 100

Thus, an optimal assignment can be made at zero cells (squares).

Person City
Dummy(No person) Dire Dawa

Hirut Mekelle

Bekcha Gambela

Marta Bahir Dare

Cost =Br. (0+500+800+1,100)=Br.2,400

IV. Restrictions on Assignments

In certain instances, it may happen that a particular match or pairing may be either undesirable
or otherwise unacceptable. For example, an employee may not have the skills necessary to
perform a particular job or a machine may not be equipped to handle a particular operation. In
such cases, the cost of performing that particular activity by a particular resource is considered
to be very large (written as M or  ) so as to prohibit the entry of this pair of employee-job into
the final solution. When such a restriction is present, a letter (M) is often placed in the table in
the position that would represent a paring. Analysis is performed as usual except the M is
ignored throughout the analysis. That is, M is not used in any reductions, nor is any value added
to it or subtracted from it during the course of the analysis.
Example
1. In the modification of a plant layout of a factory four new machines M1, M2, M3 and M4 are to
be installed in a machine shop. There are five vacant places A, B, C, D and E available. Because
of limited space, machine M2 can not be placed at C and M3 cannot be placed at A. the cost of
placing of machine at place i (in $) is shown below.

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Location

A B C D E

M1 9 11 15 10 11

Machine M2 12 9 - 10 9

M3 - 11 14 11 7

M4 14 8 12 7 8

Find the optimal assignment schedule.

Solution:

As the cost matrix is not balanced, add one dummy row (machine) with a zero cost element
in that row. Also assign a high cost, denoted by M, to the pair (M2, C) and (M3, A).

Apply the Hungarian method to solve the problem

A B C D E

M1 9 11 15 10 11

M2 12 9 M 10 9

M3 M 11 14 11 7

M4 14 8 12 7 8

M5 0 0 0 0 0

The total minimum cost ($) and optimal assignments made are as follows:

Machine Location Costs ($)

M1 A 9
M2 B 9
M3 E 7
M4 D 7
M5 (Dummy) C 0
Total = $32

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Exercise:
1. A car rental company has one car at each of five depots a, b, c, d and e. A customer in each of
the five towns A, B, C, D and E requires a car. The distance in (in kilometers) between the
depots and towns where the customers are, is given in the following distance matrix:

Depots

a b c d E

A 160 130 175 190 200

Towns

B 135 120 130 160 175

C 140 110 155 170 185

D 50 50 90 80 110

E 55 35 70 80 105

How should the cars be assigned to the customers so as to minimize the distance traveled?
Answer:
Minimum distance = 570 km

2. An airline company has drawn up a new flight schedule involving five flights. To assist in
allocating five pilots to the flights, it has asked them to state their preference scores by giving
each flight a number out of 10 .The higher the number , the greater is the preference. Certain
of these flights are unsuited to some pilots owing to domestic reasons. These have been
marked with a X
Flight number

a b c d E
A 8 2 X 5 4

B 10 9 2 8 4
Pilot

C 5 4 9 6 X

D 3 6 2 8 7
E 5 6 10 4 3

What should be the allocation of the pilots to flights in order to meet as many performances as
possible?

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(Hint: The problem is to maximize the total preference score)

Pilot Flight No Preference Score

A 1 8
B 2 9
C 4 6
D 5 7
E 3 10
Total 40

UNIT SUMMARY

The unit describes two types of problems that lend themselves to solution using linear
programming techniques: Transportation Problem and Assignment Problem. The
transportation type problems often involve the distribution of goods, whereas
assignment type problems involve the matching or pairing of two sets of items. Usually
such problems have different costs for different distribution alternatives or different
pairings and the objective is to identify the plan which minimizes total cost.

For each types of problem, a procedure for formulating the problem in a way that lends
itself to solution using simple algorithm is described. Solving such problems manually
provides additional insights into models and their solutions.

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SELF ASSESMENT QUESTIONS (SQL-4)

1. Briefly describe the following points

A. Assignment problem E. Northwest corner method
B. Column reduction F. Row reduction
C. Degeneracy G. Stepping stone method
D. MODI H. Transportation problem
I. Dummy
2. Determine an initial feasible solution to the following transportation problem
Using NWCM, LCM and VAM and compare and contrast each methods.

A B C D Supply
S1 1 5 3 3 34
Source S2 3 3 1 2 15
S3 0 2 2 4 12
S4 2 7 2 4 19
demand 21 25 17 17

3. The apex company is the distributor for Television receivers. It owns three ware
houses with stocking capacity as follows:

Ware house Location Sets in Stock

 A 100
 B 25
 C 75

It has the following order for set deliveries:

Market Location Orders

 X 80
 Y 30
 Z 90

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Delivery costs for warehouses to each customer are largely a function of mileage or
distance. The per unit costs have been determined to be

X Y Z
A 5 10 2
B 3 7 5
C 5 8 4

Required: determine the optimal distribution schedule using LCM & MODI
methods
4. Solve the following transportation problem assuming that the objective of the
firm is to maximize profit.

I II III IV SS
A 40 25 22 33 100
B 44 35 30 30 30
C 38 38 28 30 70
DD 40 20 60 30

5. In a modification of plant layout of a factory four new machines M1, M2, M3 & M4 are to
be installed in a machine shop. There are five vacant classes A, B, C, D and E available.
Because of limited space, Machine 2 can’t be placed at C and Machine 3 can’t be
placed at A. the cost (in birr) is shown below. Find the optimum assignment schedule
and total cost.

A B C D E
M1 9 11 15 10 11
M2 12 9 - 10 9
M3 - 11 14 11 7
M4 14 8 12 7 8

6. Solve the following Assignment problem. The data given in the table refer to production
in units.

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Machine
Operators A B C D
1 10 5 7 8
2 11 4 9 10
3 8 4 9 7
4 7 5 6 4
5 8 9 7 5
7. XYZ Tobacco Company purchases tobacco and stores in warehouses located in the following
four cities.

warehouse location Capacity (Tones)

City A 90
City B 50
City C 80
city D 60

The warehouses supply tobacco to cigarette companies in three cities that have the following
demand:

Cigarette Company Demand (Tones)

L 120
P 100
Q 110

The following railroad shipping costs per tone (in hundred dollars) have been determined:

Warehouse L p Q
location
A 7 10 5
B 12 9 4
C 7 3 11
D 9 5 7

Required: find the optimal transportation schedule .Give all possible answers

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UNIT FOUR
4. DECISION THEORY

4.1 LEARNING OBJECTIVES

 Describe what decision making is;
 List the steps in decision making;
 Construct a pay of table;
 Make decisions using maxi-max, maxi-min, mini-max regret, the principle of insufficient
reason and expected value criteria
 Determine the expected value of perfect information

4.2 INTRODUCTION

Decision theory represents a generalized approach to decision making which often serves as the
bases for a wide range of managerial decision making. The decision model includes a list of courses of
action that are available & the possible consequences of each course of action. The decisions are
classified according to the degree of certainty as deterministic models, where the manager assumes
complete certainty and each strategy results in a unique payoff, and Probabilistic models, where
each strategy leads to more than one payoffs and the manager attaches a probability measure to these
payoffs. The scale of assumed certainty can range from complete certainty to complete uncertainty
hence one can think of decision making under certainty (DMUC) and decision making under
uncertainty (DMUU) on the two extreme points on a scale. The region that falls between these
extreme points corresponds to the concept of probabilistic models, and referred as decision-making
under risk (DMUR). Hence we can say that most of the decision making problems fall in the
category of decision making under risk and the assumed degree of certainty is only one aspect of a
decision problem.

In this chapter you will learn decision theories concerned with decision under certainty, decision
under risk and decision under uncertainty.

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4.2 DECISION THEORY

Decision theory deals with decision making under conditions of risk and uncertainty. For our
purpose, we shall consider all types of decision models including deterministic models to be under
the domain of decision theory. In management literature, we have several quantitative decision
models that help managers identify optimal or best courses of action.
Before we go to decision theory, let us just discuss the issues, such as
( i) What is a decision?
(ii) Why must decisions be made?
(iii) What is involved in the process of decision-making?
(iv) What are some of the ways of classifying decisions? This will help us to have clear concept of
decision models.

WHAT IS A DECISION

? Activity
What is decision making?
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________

 Comment:
Decision making is the process of selecting the best choices from the available
alternatives.

A decision is the conclusion of a process designed to weigh the relative utilities or merits of a set of
available alternatives so that the most preferred course of action can be selected for implementation.
Decision-making involves all that is necessary to identify the most preferred choice to satisfy the
desired goal or objective. Hence decision-making process must involve a set of goals or objectives, a
system of priorities, methods of enumerating the alternative courses of feasible and viable courses
and a system of identifying the most favorable alternative. One must remember that the decisions are
sequential in nature. It means to say that once we select an alternative, immediately another question
arises.
For example if you take a decision to purchase a particular material, the next question is how much.
The next question is at what price. The next question is from whom… Like that there is no end.

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STEPS IN DECISION THEORY APPROACH

? Activity
What are the steps to make to make decisions? Please try your answer in the
given space.
________________________________________________________________
________________________________________________________________
________________________________________

 Comment
The common steps in decision making are list of possible alternatives, list of
future events; construct pay of tables, assessment of degree of certainty and
making a choice

Though the steps to make the decision differ from problem to problem, the general steps in decision
theory include the following

1. List the viable alternatives (strategies) that can be considered in the decision.
2. List all future events that can occur. These future events (not in the control of decision maker) are
called as states of nature.
3. Construct a payoff table for each possible combination of alternative course of action and state of
nature.
4. An assessment of the degree of certainty of possible future events
5. Choose the criterion that result in the largest payoff.
List of alternatives:
The list of alternatives must be a set of mutually exclusive and collectively exhaustive decisions
that are available to the decision maker.
For example suppose that a real state developer plan to develop a building. After careful
analysis, the developer lists the following acceptable alternatives.
Residential
Hospital
Hotel
State of nature:
State of nature refers to a set of possible future conditions or events beyond the control of the
decision maker that will be the primary determinants of the eventual consequence of the
decision. Suppose in the case of the real state developer, the main factor that that will influence

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the profitability is the state of the economic development that will be achieved in the future.
Suppose that the developer views the possibilities as:

Low economic growth

Medium economic growth
High economic growth

Payoffs:
In order for a decision maker to be able to rationally approach a decision problem, it is
necessary to have some idea of the payoffs that would be associated with each decision
alternative and the various states of nature. The payoffs might be profits, revenue, costs, or other
measure of value. They may be weekly, monthly, or annual amounts.

Pay off table:

The pay off table is a device a decision maker can use to summarize and organize information
relevant to particular decision. It includes a list of alternatives, the possible future state of nature
and payoffs associated with each of the alternative / state of nature combinations.

Now let’s assume the real state developer, set up the following payoff

Low economic Medium economic High economic

Growth Growth Growth

4 16 12
Residential
5 6 10
Hospital
-1 4 15
Hotel

The pay of table for the real state developer’s decision is shown in the above table. The three
alternatives under considerations are listed down the left side of the table and the three possible
states of nature are listed across the top of the table. The pay offs that are associated with each
alternative /state of nature combinations are shown in the body of the table. Suppose that the
values represent profits in million birr. Hence if the residential is chosen and if the economic

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growth is low the developer realize a profit of birr 4,000,000. Similarly if hotel is selected and if
the economic growth is low, the developer will lose birr 1,000,000, similarly there are nine
payoffs. But the question is which alternative is the best for the decision maker? To answer the
question, it is a must to determine the degree of certainty i.e. is it certainty, risk or uncertainty.

DECISION MAKING ENVIRONMENT

Base on the availability of information during the decision making, there are three decision
making environment. These are:
A. Decision making under certainty
B. Decision making under risk
C. Decision making under uncertainty

A. DECISION MAKING UNDER CERTAINTY (DMUC)

Decision making under certainty assumes that all relevant information required to make
decision is certain and well known. It uses a deterministic model, with complete knowledge,
stability and no ambiguity. To make decision, the manager will have to be quite aware of the
strategies available and their payoffs and each strategy will have unique payoff resulting in
certainty. The decision-making may be of single objective or of multiple objectives.

If the developer knows that the economic growth is low, he/she select hospital as the best
alternatives because it results a profit of birr 5,000,000 and the like.

B. DECISION MAKING UNDER UNCERTANITY (DMUU)

Decision making under uncertainty is formulated exactly in the same way as decision making
under risk, the only difference is that no probability to each strategy is attached. In decision
making under uncertainty, remember that no probabilities are attached to set of the states of
nature. Sometimes we may have only positive elements in the given matrix, indicating that the
company under any circumstances will have profit only. Sometimes, we may have negative
elements, indicating potential loss. While solving the problem of decision making under
uncertainty, we have two approaches, the first one is pessimistic approach and the second one
is optimistic approach.

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C. DECISION MAKING UNDER RISK (DMUR)

Decision-making under risk (DMUR) describes a situation in which each strategy results in
more than one outcome or payoffs and the manager attaches a probability measure to these
payoffs. This model covers the case when the manager projects two or more outcomes
for each strategy and he or she knows, or is willing to assume, the relevant probability
distribution of the outcomes. The following assumptions are to be made: (1) Availability of
more than one strategy, (2) The existence of more than one states of nature, (3) The relevant
outcomes and (4) The probability distribution of outcomes associated with each strategy. The
optimal strategy in decision making under risk is identified by the strategy with highest
expected utility (or highest expected value).

4.3 DECICISIN CRITERIA

The process of selecting one alternative from a list of alternatives is governed by a decision
criterion, which embodies the decision maker’s attitudes towards the decision.
4.3.1 Decision Criterion under Certainty
The simplest of all circumstances occurs when decision making takes place in an environment of
complete certainty. For example in the case of real state problem, an unexpected early
announcement concerning the economic development indicate low economic growth could
reduce the problem to a situation of certainty since the developer focus on the first column of
the pay of table.
4.3.2 Decision Criterion under Uncertainty
Decisions made under these circumstances are at the opposite end of the spectrum from the
certainty case. We shall consider four approaches to decision making under complete
uncertainty. They are:
1 . Maxi-max: (optimistic view) under this method the best payoff for each alternative is
identified and the alternative with the maximum of these is the designated decision.( identifying
the best from the best alternatives) it is the optimistic view because the decision maker assume
the best things will occur. ( the maxi- max will be converted to mini-min principle under cost
scenario)
For the real state problem, the maxi-max solution is shown below.

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Low economic Medium economic High economic

Growth Growth Growth Row max

4 16 12 16 max
Residential
5 6 10 10
Hospital
-1 4 15
Hotel 15

Therefore, the developer will select Residential as the best alternative, since it provides the
highest pay off i.e. 16,000,000.

2 .Maxi-min: (pessimistic view) it is the conservative strategy; it consists of identifying the

worst (minimum) payoff for each alternative and then selecting the alternative that has the best
(maximum) of the worst pay off. Many people view the maxi-min criterion as pessimistic
because they believe the decision maker must assume the worst will occur. ( the maxi- max will
be converted to mini-min principle under cost scenario)
For the real state problem, the maxi-max solution is shown below.

Low economic Medium economic High economic

Growth Growth Growth Row mini

4 16 12 4
Residential
5 6 10 5 max
Hospital
-1 4 15 -1
Hotel

Therefore, the developer will select Hospital as the best alternative, since it provides the highest
pay off from the worst alternatives i.e. 5,000,000.
3. Mini-max regret: both the maxi-max and maxi-min strategies can be criticized because
they focus only on a single extreme payoff and exclude the other payoffs. This approach does
take all payoffs in to account. In order to use this approach it is necessary to develop an
opportunity loss table. The opportunity loss reflects the difference between each payoff and
the best possible payoff in a given state of nature. For the real state problem, the maxi-max
solution is shown below

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Low economic Medium economic High economic

Growth Growth Growth

4 16 12
Residential
Hospital 5 6 10

Hotel -1 4 15
Column
Max 5 16 15

The opportunity loss table is calculated as follows

5-4 16-16 15-12

5-5 16-6 15-10

5-(-1) 16-4 15-15

The result of the opportunity loss calculation

Maximum loss (regret)

1 0 3 3 Mini

0 10 5 10
6 12 0 12

Therefore, the developer will select Residential as the best alternative; since it provides the
minimum loss pay off from the worst alternatives (highest loss) i.e. 3,000,000.
4. The principle of insufficient reason: it offers a method that incorporates more of
the information. It treats the states of nature as if each were equally likely and it focuses
on the average payoff for each row selecting the one that has the highest row average .we
calculate the payoffs by assuming all the states of natures have equal chance of existence
i.e. 1/3 since the number of state of natures are three in our case.

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Alternatives State of nature as if all are equally likely

LEG MEG LEG
RES (4*1/3)+(16*1/3)+(12*1/3)=32/3=10.6
HOT (5*1/3)+(6*1/3)+(10*1/3)=21/3=7
HOS (-1*1/3)+(4*1/3)+(15*1/3)=18/3=6

Therefore, the developer will select Residential as the best alternative; since it provides the
maximum pay off from the available alternatives i.e. 10.6 million.

? Activity:
What are the different approaches for making decisions under uncertainty?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
___________________________________________________________________

 Comment:
Maxi-max, max-min, mini-max regret and the principle of insufficient reason

4.3.3 Decision Criterion under Risk

The essential difference between decision making under complete uncertainty and decision
making under Risk (partial uncertainty) is the presence of probabilities for the occurrence of the
various states of nature in partial uncertainty. The probabilities may be subjective estimates
from managers or from experts in a particular field or they may reflect historical frequencies. If
they are reasonably correct they provide a decision maker with additional information that can
dramatically improve the decision making process.

Note that
The sum of the probabilities for all states of nature must be 1.

The Expected monetary value (EMV): This approach provides the decision maker with a
value which represents an average pay off for each alternative the best alternative is , then, the
one that has the highest expected pay offs . Thus the expected monetary value is:

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EMVi = PjVij
Where:
EMVi = the expected monetary value for the i th alternative
Pj = the probability of the j th state of nature
Vij =the estimated payoff for alternative i under state of nature j
For example considering the real stare developer problem, assuming the manager estimates the
state of nature as LEG 20%, MEG 50% & HEG 30%. We ca compute the expected monetary value
for the real state developer’s alternatives as follows
EMVR= .2(4) +.5(16) + 3(12) =12.40
EMVHT= .2(5) +.5(6) + 3(10) =7.00
EMVHS= .2(-1) +.5(4) + 3(15) =6.30
Because the residential alternatives has the largest expected monetary value, it would be
selected using this criterion
The Expected opportunity loss (EOL): This approach provides an alternative method for
incorporating probabilities into the decision making. The approach is nearly identical to the EMV
approach except that a table of opportunity loses is used rather than a table of loss. Hence the
opportunity loses for each alternative are weighted by the probabilities of their respective states
of nature to compute a long run average opportunity loss and the alternative with the smallest
expected loss is selected as the best choice. For real state problem, the expected opportunity loss
can be calculated as follows (use previously calculated opportunity loss table)
EOLR =.2(1) +.5(0) + 3(3) =1.10
EOLHT = .2(0) +.5(10) + 3(5) =6.5
EOLHT = .2(6) +.5(112) + 3(0) =7.2
Because the residential alternatives have the minimum expected opportunity loss, it would be
selected .note that the EOL approach resulted in the same alternative as the EMV approach. This
is more than coincidence; the two methods will always result in the same choice because they
are equivalent ways of combining the values maximizing the pay offs is equivalent to minimizing
the opportunity loss.
Expected value of perfect information: it can some times be useful for a decision maker to
determine the potential benefit of knowing for certain which state of nature is going to prevail.
For instance, a decision maker might have the option of delaying a decision until it is evident
which state of nature is going to materialize. The obvious benefit of waiting would be to move

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the decision maker into the realm of certainty, thereby allowing the decision maker to obtain the
maximum possible payoff.

The expected value of perfect information (EVPI) is a measure of the difference between the
certain payoff that could be realized under a condition of certainty and the expected under
conditions involving risk.

Consider the payoff that the real state developer could expect under certainty. If the developer
knew that the economic growth is low, the hospital alterative would be choose with a payoff of
5, 000,000. If the developer knew a medium economic growth would exist, the residential
alternative would be chosen for a payoff of 16,000,000 and if the developer knew that a high
economic growth would happen, hotel would be chosen for a payoff of 15,000,000.

Hence if it were possible to remove the uncertainty surrounding the states of nature, the
decision maker could capitalize on that knowledge. Obviously before investing time or money in
eliminating the probabilities, it will be impossible for the decision maker to say which states of
nature will turn out to be the one that will occur. However, what can be said is that the
probability that perfect information will indicate that LEG will happen t is 20%, that the
probability that perfect information will indicate a MEG will happen is 50% and the probability
of perfect information indicating a HEG will happen is 30%. Thus, these probabilities which are
the original state of nature probabilities can be used to weight the payoffs, one of which will
occur under certainty. This is called the expected pay off under certainty (EPC), and is computed
in the following way for the real state problem.
EPC =.2(5) +.5(16) + 3(15) =13.50

The difference between this figure and the expected payoff under risk (i.e. the EMV) is the
expected value of perfect information. Thus,
EVPI= EPC- EMV
For the real state problem, with EPC 13.5- and EMV = 12.4 we find EVPI= 13.5-12.4= 1.1
The EVPI represents an upper bound on the amount of money the real state developer would be
justified in spending to obtain perfect information. Thus the real state developer would be
justified in spending up to 1100,000 to find out for certain which h state of nature will prevail

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Note that
The EVPI is exactly equal to the previously computed EOL. In fact these two quantities will
always be equal. The EOL indicates the expected opportunity loss due to imperfect
information which is another way of saying the expected payoff that could be achieved by
having perfect information. Hence there are two equivalent ways to determine the expected
value of perfect information:
o Subtract the EMV from the EPC or
o Compute the EOL
For cost scenario
EVPI=EMV-EPC

Example Two:

An investor has three alternatives which results different payoffs under the three possible
market conditions. The conditional payoffs (in birr) for each alternative – state of nature
combination are given below.

Investment Estimated conditional returns per share under different market

alternatives conditions
Market conditions (state of nature )
LESS ATTRACTIVE MEDIUM VERY ATTRACTIVE
BANK -2,000 4,000 8,000
BREWERY 6,000 12,000 18,000
CEMENT 7,000 10,0000 8,000

Required:
1. Determine which alternative will be selected if the decision maker uses
a. Maxi-max criterion
b. Maxi- min criterion
c. The principle of insufficient reason
d. Mini-max regret criterion

2. Assume that the probabilities for less attractive and very attractive market conditions
are 30% and 40% respectively. then

a. chose the best alternative using expected monetary value ( EMV) criterion
b. chose the best alternative using the expected opportunity loss ( EOL) criterion
c. calculate the expected value of perfect information (EVPI)

Solution Table
Question 1. Uncertainty Question 2. Risk
Criterion Decision Value Criterion Decision Value
A. Maxi- max
B. Maxi- min
C. Insufficient Reason
D. Mini-max

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UNIT SUMMARY

Decision theory is a general approach to decision making. It is very useful for a decision maker
who must choose from a list of alternatives knowing that one of a number of possible future
states of nature will occur and that this will have an impact on the payoff realized by a particular
alternative.
Decision models can be categorized according to the degree of uncertainity that exists relative
to the occurrence of the states of nature. This can range from complete knowledge about which
state will occur to partial knowledge (probabilities) to no knowledge (no probabilities or
complete uncertainty).

When complete uncertainty exists the approach a decision maker takes in choosing among
alternatives depends on how optimistic or pessimistic the decision maker is and it also depends
on other circumstances related to the eventual outcome or payoff.

Under complete certainty decisions are relatively straight forward. Under partial uncertainty
expected values often are used to evaluate alternatives. An extension of the use of expected
values enables decision makers to assess the value of improved or perfect information about
which state of nature will occur.

SELF ASSESMENT QUESTIONS (SQL-4)

1. Briefly describe the following points.

A. Decision criterion H. Mini-max regret

B. Expected monetary value I. Opportunity loss
C. Expected opportunity lose J. Payoff table
D. Expected payoff under K. Principle of insufficient
certainty reason
E. Expected value of perfect L. Risk
information M. State of nature
F. Maxi-max N. Uncertainty
G. Maxi-min

2. Given this profit payoff table:

States of nature
#1 #2 #3
a 12 18 15
Alternatives b 17 10 14
c 22 16 10
d 14 14 14

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i) Determine which alternative would be chosen using each of these decision criteria:
a) Maxi-max
b) Maxi-min
c) Mini-max regret
d) Principle of insufficient reason
ii) Suppose the payoffs in the above problem had been costs rather than profits. Determine
which alternative would be chosen, using these decision criteria:
a) Maxi-max
b) Maxi-min
c) Mini-max
d) Principles of insufficient reason
iii) For the payoff table given in problem 4, suppose the manager has assigned probabilities of
20% to the occurrence of state #1, 50% to the occurrence of state #2, and 30% to the occurrence
of state #3.
a) Which alternative would be chosen using maximum expected value as the criterion,
treating the payoff as profits?
b) Calculate the expected value of perfect information.
3. Consider the following payoff table (profits in 00,000).
State of nature

S1 S2 S3 S4
Alternatives A1 7 14 13 10
A2
3 6 12 15
A3
-8 -1 11 22

A. Which alternative would be chosen for each of the following decision criteria
I. Maxi-max
II. Mini-max
III. Mini-max regret
IV. Principle of insufficient reason
B. Suppose the probabilities were available for the state of nature in the previous
problem were 20%, 10%, and 40% for S1, S2 & S3 respectively. Which
alternative will be selected using expected pay of and expected opportunity
loss methods.
C. What is the value of perfect information?
4. An assistant buyer of a sports wear department in Debrebirhan store must decide on
cloths orders for the summer season. The profits for various states of nature are
estimated as follows.

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Alternatives

Style A Style B style B

Style A 80 20 60
Market acceptance Style B 30 90 70
Style B
45 55 96

A. Determine which alternative would be selected for each of these decision

criteria.
I. Maxi-min
II. Maxi-max
III. Mini-max regret
IV. principle of insufficient reason
B. Suppose that the assistant buyer is able to estimate the probability for market
acceptance of style A, and style C are 30% and 50% respectively
I. which alternative would maximize the expected profit
II. Which alternative would be selected using expected opportunity loss
approach?
C. Determine the expected value of perfect information

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UNIT FIVE

5. PROJECT MANAGEMENT

5.1 LEARNING OBJECTIVES

 Describe the nature and characteristics of projects;

 Describe the techniques and tools of project management;
 Give a general description of CPM technique;
 Construct net work diagrams;
 List the kinds of information that a CPM analysis can provide;

5.2 INTRODUCTION
Managers typically oversee a variety of operations. Some of these involve routine, repetitive
activities, where as others tend to vary with the task. Projects fall under the latter heading; they
are unique one time operations designed to accomplish a specific set of objectives during a
limited time frame.
A project is a combination of various activities. For example, Construction of a house can be
considered as a project. Similarly, conducting a public meeting may also be considered as a
project. In the above examples, construction of a house includes various activities such as
searching for a suitable site, arranging the finance, purchase of materials, digging the foundation,
construction of super structure etc.
Conducting a meeting includes printing of invitation cards, distribution of cards, arrangement of
platform, chairs for audience etc. In planning and scheduling the activities of large sized projects,
the two network techniques — PERT and CPM — are used conveniently to estimate and evaluate
the project completion time and control the resources to see that the project is completed within
the stipulated time and at minimum possible cost. But for a simple and small projects Gantt
charts will be employed

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? Activity:

What is project? What makes different projects different from ordinary business operations?
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
_____
_________________________________________________________________________
5.3 PLANNING AND SCHEDULING OF PROJECTS

5.3.1 Planning and scheduling using Gantt chart

Initially, projects were represented by milestone chart and bar chart. But they had little use in
controlling the project activities. Bar chart simply represents each activity by bars of length
equal to the time taken on a common time scale .This chart does not show interrelationship
between activities. It is very difficult to show the progress of work in these charts. An
improvement in bar charts is milestone chart. In milestone chart, key events of activities are
identified and each activity is connected to its preceding and succeeding activities to show the
logical relationship between activities. Here each key event is represented by a node (a circle)
and arrows instead of bars represent activities.

The Gantt chart is a popular tool for planning and scheduling simple projects. It enables a
manager to initially schedule project activities and then to monitor progress over time by
comparing planned progress to actual progress.

In order to prepare the chart, the vice president who was in charge of the project had to first
identify the major activities that would be required . Next the time estimates for each activity
were made and the sequence of activities was to occur, their planned duration and when they
were to occur. Then as the project progressed, the manager was able to see which activities were
ahead of schedule and which activities were delaying the project. This enabled the manager to
direct attention where it was needed most to speed up the project in order to finish on schedule.

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Advantages:
It is simple
Disadvantages:
Fail to reveal certain relationships among activities that can be crucial to effective project
management

Example:
 If one of the early activities in a project suffers a delay, it would be important for the
manager to be able to easily determine which latter activities would have to be delayed
because they could not start until that activity was completed.
 Some activities may safely be delayed with out affecting the project schedule
Consequently Gantt charts are most useful for simple projects say where activities are
simultaneous or where a string of sequential activities is involved.
On more complex projects Gantt chart can be useful for initial project planning which then gives
way to the use of networks.

A Gantt chart for a bank’s plan to establish a new direct marketing department is illustrated as
follows.

May 2011 Jun 2011 Jul 2011

ID Task Name Start Finish Duration
5/1 5/8 5/15 5/22 5/29 6/5 6/12 6/19 6/26 7/3 7/10 7/17

1 Locate new facilities 4/28/2011 6/8/2011 4/28/2011

6w Locate new facilities
Interview prospective staffs
2 4/28/2011 5/11/2011 2w

3 Hire and train staffs 5/11/2011 5/24/2011 2w

4 Select and order furniture 4/28/2011 7/6/2011 10w

5 Receive furniture and set up 7/6/2011 7/12/2011 1w

6 Start up 7/12/2011 7/18/2011 1w

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5.3.2 Planning and scheduling using CPM/ PERT

Program Evaluation and Review Technique (PERT) and Critical Path Method (CPM) are two of
the most widely used techniques for planning and scheduling the large projects. By using PERT/
CPM, managers are able to obtain:
1. A graphical display of project activities
2. An estimate of how long the project will take
3. An indication of which activities are the most critical to timely completion of the project
4. An indication of how long any activity can be delayed with out lengthening the project

5.4 PROJECT DIAGRAMS (NET- WORK DIAGRAMS)

One of the main features of PERT and related techniques is their use of a network or precedence
diagram to depict major project activities and their sequential relationships.
Recall the bank ‘s case that we used for the Gantt chart example. The network diagram for that
same problem is

Eve
A F nt
Eve
nt B Eve
nt
C
D Eve
nt E

Eve
nt

The diagram is composed of a number of arrows and nodes. Where the arrows represent the
project activities and the nodes represent the events.
To construct the net work diagram, consider the following points.
1. The project can be sub-divided into a set of predictable independent activities each of which
has a clear beginning and ending.
2. Each activity can be sequenced as to its predecessors or successors.
3. The network is not cyclical i.e. each activity is executed once and only once during the life
of the project.
4. Activity times may be estimated either as a single point estimate (CPM) or as a 3- point
estimate (PERT)
 CPM- Critical Path Method
 PERT- Program Evaluation and Review Technique

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The duration of the activities is independent of each other

5.4.1 EVENT
Events of the network represent project milestone, such as the start or the completion of an
activity (task) or activities, and occur at a particular instant of time at which some specific part
of the project has been or is to be achieved.

Event: An event is a point in time that marks the beginning or ending of activity.

Events are commonly represented by circles (nodes) in the network diagram.

The events can be further classified into the following two categories (types):

1. Merge event: An event which represents the joint completion of more than one activity
is known as merge event.

Eve
nt

Fig: merge event

2. Burst event: An event which represents the initiation (beginning) of more than one
activity is known as burst event.

Eve
nt

Fig: Burst event

Events in the network diagram are identified by numbers. Each event should be identified by a
number higher than that allotted to its immediately preceding event to indicate progress of
work. The numbering of events in the network diagram must start from left (start of the
project) to right (completion of the project) and top to the bottom. Care should be taken that
there is no duplication in the numbering.

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ACTIVITY

? Activity:

What is activity in project? What is the difference with events?

__________________________________________________________________________________
__________________________________________________________________________________
_________________________________________________________________________________

 Comment

Activity indicates time consuming tasks. It is represented by arrows in the network diagrams (for AOA
case.)
But events are milestones which indicate the starting and finishing time of the activities of the project.
It is represented by nodes in the network diagrams (for AOA case.)

Activity is a time-consuming job or task that is a key subpart of the total project.
Activities of the net work represent project operations or tasks to be conducted. As such each
activity except dummy consumes time and resources and incurs costs. An arrow is commonly
used to represent an activity with its head indicating the direction of progress in the project.

Activities are identified by the numbers of their starting (tail or initial) event and ending (head or
terminals) event. An arrow extended between two events; the tail event represents the start of
the activity and the head event, represents the completion of the activity.

Activity
i j
Starting Event Completion Event
Fig: Activity -Node relationship in network diagram.

Activities can be further classified into the following three categories:

I. Predecessor activity: predecessor activity is an activity which must be completed before one
or more other activities start.

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II. Successor activity: Successor activity is an activity which started immediately after one or
more of other activities are completed.

III. Dummying activity: Dummying activity is an activity which does not consume either any
resource or time.

A dummy activity in the network is added only to represent the given precedence relationships
among activities of the project and is needed when
a) Two or more parallel activities in a project have same head and tail events, or
b) Two or more activities have some (but not all) of their immediate predecessor
activities in common.

A dummy activity is depicted by dotted line in the network diagram.

A B C
Activity Activity Activity
1 2 4

D Dummy Activity
B is the predecessor of C and D Activity
3
B is the successor of A;
C and D are the successor for B

A. Rules of Network Construction

Following are some of the rules which have to be followed while constructing a network:

 In network diagram, arrows represent activities and circles the events. The length of an
arrow is of on significance.

 Each activity should be represented by one arrow and must start and end in a circle called
event. The tail of an activity represents the start and head the completion of work.

 The event numbered 1 denotes start of the project and is called initial event. All activities
emerging (or taking off) from event 1 should not be preceded by any other activity or
activities. Event carrying the highest number denotes the completion events. A network
should have only one initial event and only one terminal event.

 The general rule of numbering the event is that the number at an activity’s head should
always be larger than that at its tail. That is, events should be numbered such that for each
activity (i, j), i< j.

 An activity must be uniquely identified by its starting and completion event which implies
that

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a. an event number should not get repeated or duplicated

b. two activities should not be identified by the same completion event
c. activities must be represented either by their symbols or by the corresponding
ordered pair of starting-completion events.
 The logical sequence ( or interrelationship ) between activities must follow the following
rules :
o An event cannot occur until all the incoming activities into it have been completed.

o An activity cannot start unless all the preceding activities, on which it depends, have been
completed.

o Though a dummy activity does not consume ether any resource or time, even then it has to
follow the rules 6(a) and 6(b).

B. Errors and Dummies in Network

I. Looping and Dangling: Looping (cycling) and dangling are considered as faults in a
network. Therefore, these must be avoided.
o Looping

A case of endless loop in a network which is also known as looping is shown in figure below,
where activities A, B, and C from a cycle.

C 3
1 Loopin
B
A
2

Fig: Looping in network diagram

In this case it is difficult to number three events associated with activity A, B, and C so as to
satisfy rule 6 of constructing the network.

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o Dangling
A case of disconnect activity before the completion of all activities which is also known as
dangling.

1 A 2 B 4

C Dangling

Fig: Dangling in network diagram

In this case, activity C does not give any result as per the rules of the network. The dangling may
be avoided by adopting rule 5 of constructing the network.

II. Dummy or Redundant Activity.

The following is the one of the cases in which the use of dummy activity may help in drawing the
network correctly as per the various rules
When two or more parallel activities in a project have the same head and tail events:
i.e: two events are connected with more than one arrow.

In Fig. a), activities B and C have a common predecessor activity A. At the same time, they have
activity D as a common successor.
To get the correct network a dummy activity for the ending event of B to show that D may not
start before both B and C are completed is shown in figure b) or c).
B 2

A B Dummy
A D 1
B 3
D
1 2 A
1
C 3
D
C Dummy
2
C
C
Fig: a) Parallel Activities Fig: b) Dummy Activity

Note that
The diagram is composed of a number of arrows and nods
 The arrows represent the project activities
 The nodes represent the starting and finishing time the activities , which are called
events
 The activities have associated time estimates
 Where as the events do not have time estimates

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Example: 1
An assembly is to be made from two parts X and Y. Both parts must be turned on a lathe and Y
need not be polished. The sequence of activities together with their predecessors is given below:

Activity Description Predecessor Duration in

Activity months
A Open work order _ 5

B Get material for X A 10

C Get material for Y A 3

D Turn X on Lathe B 7

E Turn Y on Lathe B, C 3

F Polish Y E 12

G Assemble X and Y D, F 15

H Pack G 2

Required: Draw a network diagram for the project.

Solution
The network diagram for the project is show as follows:

B 3 D
1 A 2
Dummy
6
G 7
H 8
C
F
4 E 5

Example: 2 Listed in the table are the activities and sequencing necessary for a maintenance job
on the heat exchangers in a refinery.

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Activity Description Predecessor Activity

A Dismantle pipe connections -

B Dismantle header, closure, and floating A

front
C Remove tube bundle B

D Clean bolts B

E Clean header and floating head front B

F Clean tube bundle C

G Clean shell C

H Replace tube bundle F, G

I Prepare shell pressure test D, E, H

J Prepare tube pressure test and I

reassemble

Required: Draw a network diagram for the project

Solution
The network diagram for the project is show as follows:

5
F D2
C 4
A 2 B 3
G
6 H
1 D 8 I 9
J 1
E 0

D1
7

5.5 COMPUTING ALGORITHMS

Many real –life project networks are much larger than the simple network illustrated in the
preceding example; they often contain hundreds or even thousands of activities. Because the
necessary computations can become exceedingly complex and time consuming large networks

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are generally analyzed by computer programs rather than manually. The intuitive approach just
demonstrated does not lend itself to computerization because in many instances path sequences
are not readily apparent. Instead an algorithm is used to develop four pieces of information
about net work activities.

The objective of critical path analysis is to estimate the total project duration & to assign starting
& finishing times to all activities involved in the project. This helps in checking actual progress
against the scheduled duration of the project.

The duration of individual activities may be uniquely determined (in case of CPM) or may
involve the three time estimates (in case of PERT) out of which the expected duration of an
activity is computing. Having done this, the following factors should be known to prepare
project scheduling
i. Total completion time of the project
ii. Earliest & latest start time of each activity
iii. Float for each activity: i.e. the amount of time by which the completion of an activity can
be delayed without delaying the total project completion time.
iv. Critical activities & critical path.

Consider the following notation for the purpose of calculating various times of event & activities
A. Earliest start time for activity (ES): It is the time at which the activity can start
without affecting the total project time
B. Latest start time for activity ( LS): It is the latest possible time by which an activity
must start without affecting the total project time
C. Earliest finish time for activity (EF): It is the earliest possible time at which an
activity can finis without affecting the total project time.
D. Latest finish time for activity (LF): It is the latest time by which an activity must
get completed without delaying the project completion.
E. Duration of activity: Expected time estimate

In a network diagram, there should only be one initial event & one final event, while other
events are numbered consecutively with integers 1,2,……..,n, Such that i < j , fir any two event i &
j connected by an activity which starts at i & finish at j .

For calculating the above mentioned times, there are two methods; namely;
1) Forward pass method &
2) Backward pass method

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I. Forward Pass Method (For Earliest Event Time)

In this methods calculation begin from the initial event 1, proceed through the network visiting
event in an increasing order of event no & end to the final event, say N. At each event we
calculate earliest occurrence event time (E) & earliest start & finish time for each activity that
begins at that event. When calculations end at the final event N, its earliest occurrence time
gives the earliest possible completion time of the entire project.
The method may be summarized as follows:
1) Set the earliest occurrence time of initial event 1 to zero, i.e.
E1 = 0, i =1
2) Calculate earliest start time for each activity that begins at event i where i(=1). This is
equal to earliest occurrence time of event, i (tail event
3) Calculate the earliest finish time pf each activity that begins at event i. This equal to the
earliest start time of the activity plus the duration of the activity.
4) Proceed to the next event ,
5) Calculate the earliest occurrence time for the event j. This is the maximum of the earliest
finish times of all activities ending into that event.

II. Backward Pass Method (For Latest Allowable Event Time)

In this method calculations begin from final event N, proceed through the network visiting
events in the decreasing order of event no & end at the initial event 1. At each event, we
calculate the latest occurrence event time (L) for the corresponding event, latest finish & start
time for each activity that is terminating at the event, such that the earliest finish time for the
project remains the same. The method may be summarized as follows.
1. Set the latest occurrence of last event N equal to its earliest occurrence time ( known
from forward pass method),
2. Calculate latest finish time of each activity which ends at event j. This is equal to latest
occurrence time of final event N.
3. Calculate the latest start times of all activities ending at j. It is obtained by subtracting the
duration of the activity from the latest finish time of the activity.
4. Proceed backward to the event in the sequence that decreases j by 1.
5. Calculate the latest occurrence time of event i (i<j). This is the minimum of the latest
start time of all activities starting from that event.

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Float (or slack) of an Activity and Event

The float of an activity is the amount of time by which it is possible to delay its completion time
with out affecting the total project completion time.

The value of the slack for each activity can be calculated

Slack = LS- ES
or
Slack = LF-EF

Critical Path

Certain activities in a network diagram of a project are called critical activities because delay in
their execution will cause further delay in the project completion time. Thus, all activities having
zero total slack value are identified as critical activities. The path that connects all these critical
activities is known as critical path.

The critical path is the continuous chain of critical activities in a network diagram. It is the
longest path starting from first to the last event & is shown by the thick line or double lines in
the network diagram.

The length of the critical path is the sum of the individual times of the critical activities living on
it & defines the minimum time required to complete the project.
Example
Consider the assembly example and
a. Calculate the ES, EF, LS, LF & slack of the activities of the project
b. Identify the critical path of the project
c. Determine the minimum time to complete the project

Solution
For simplicity bring the problem and the network diagrams, to calculate the above requirements
it a must first to draw the network diagram.

Activity Description Predecessor Activity Duration in months

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A Open work order _ 5

B Get material for X A 10

C Get material for Y A 3

D Turn X on Lathe B 7
E Turn Y on Lathe B, C 3
F Polish Y E 12
G Assemble X and Y D, F 15

H Pack G 2

B 3 D
10
1 A 2 7
Dummy
6 G 7 H 8
5 C
F 12 15 2
3
4 E 5
3

Earliest time Last time

Activity Duration Start Finish Start Finish Slack
(ES) (EF) (LS) (LF)
A 5 0 5 0 5 0

B 10 5 15 5 15 0

C 3 5 8 12 15 7
D 7 15 22 23 30 8
E 3 15 18 15 18 0
F 12 18 30 18 30 0
G 15 30 45 30 45 0
H 2 45 47 45 47 0

The critical path indicates the path that connects the critical activities i.e., activities with 0 slack
values. These are A, B, E, F, G, and H. so the path is [A B E F G H]
The minimum time required to complete the project is the sum of the time required to complete
the critical activities
Therefore we add [5+10+3+12+15+2= 47] so the project requires 47 weeks.

Example: consider the following problem and determine the following

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a. Calculate the ES, EF, LS, LF & slack of the activities of the project
b. Identify the critical path of the project
c. Determine the minimum time to complete the project

Activity Immediate Duration

Predecessor in days
A - 8
B - 4
C A 6
D A 11
E B 9
F E 3
G E,D,F 1

SOLUTION
A. to calculate ES, EF, LS, LF & slack first we should draw the network of the project

C
4 F
A 2
D
1
G
B
E 5 6

Earliest time Last time

Activity Duration Start Finish Start Finish Slack

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(ES) (EF) (LS) (LF)

A 8 0 8 0 8 0

B 4 0 4 6 10 6
C 6 8 14 10 16 2
D 11 8 19 8 19 0
E 9 4 13 10 19 6
F 3 14 17 16 19 2
G 1 19 20 19 20 0

PERT

? Activity:

What is PERT? What is its difference with CPM

__________________________________________________________________________________
__________________________________________________________________________________
_________________________________________________________________________________

 Comment

PERT refers program evaluation and review technique. Its major difference is it uses probabilistic
time estimates rather than deterministic time estimates as CPM do.

The preceding discussion (CPM) assumed that activity times were known and not subject to
variation. Although the assumption is appropriate in some situations there are many others in
which it is not. Consequently those situations require a probabilistic approach. The main
determinant of the way PERT & CPM networks are analyzed and interpreted is whether activity

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time estimates are probabilistic or deterministic. If time estimates can be made with high degree
of confidence that actual time will not differ significantly, we say the estimates are deterministic.
On the other hand, if estimated times are subject to variation, we say the estimates are
probabilistic. Probabilistic time estimates must include an indication of the extent of probable
variation. This section describes analysis of networks with probabilistic time estimates.
The probabilistic approach involves three time estimates for each activity instead of one.
I. Optimistic time: the length of time required under optimum conditions (it is represented by a
letter ‘a’)
II. Pessimistic time: the amount of time that will be required under the worst condition (it is
represented by a letter ‘b’)
III. Most likely time: the most probable amount of time that will be required. (It is represented
by a letter m)
Of special interest in network analysis are the average of expected time for each activity, te and
the variance time 2. The expected time is computed as weighted average of the three time
estimates.
Te= a +4m+b
6
The standard deviation of each activities time is estimated as one- sixth of the difference
between the pessimistic and optimistic time estimates. (Analogously, essentially all of the area
under a normal distribution lies within +3 standard deviations of the mean, which is a range of
six standard deviations.) The variance is found by squaring the standard deviations. Thus:
2 = [ (b-a) ] 2 or (b - a) 2
6 36
The size of the variance reflects the degree of uncertainty associated with an activities time. The
larger the variance, the greater the uncertainty

It is also desirable to compute the standard deviation of the expected time for each path. This
can be accomplished by summing the variance of the activities on a path and then taking the
square root of that number.
That is:
 Path = (variance of activities on path )
Example

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A small project is composed of 7 activities whose time estimates are listed below.

Activity Predecessor TIME IN WEEKS

a m b

A - 1 1 7

B - 1 4 7
C - 2 2 8
D C 1 1 1
E B 2 5 14
F A 2 5 8
G E,F 3 6 15
H D,G 1 4 1

Required
1. Draw the network
2. calculate the expected variance
3. find the expected project completed time
4. calculate the probability that the project will be completed at least 3 weeks than
expected
5. if the project due date is 18 weeks , what is the probability of not meeting the
due date

A 3 D
6
B G
2 E H
C 1 7 8

4 F

The time required to complete the project 20 weeks

ADGH= 13 weeks

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BEGH=20 weeks (the maximum which indicates the critical path )

CFH= 11weeks

Activity Predecessor Te  2

A - 2 1 1

B - 4 1 1
C - 3 1 1
D A 1 0 0
E B 6 2 4
F C 5 1 1
G D,E 7 2 4
H F,G 3 0 0

Critical activities Variance

B 1
F 4
G 4
H 0
2 9

 Path = 9=3
The probability of completing the project at least 3 weeks earlier i.e. 17 weeks
TL= 17 WEEKS TE = 20 WEEKS

TL-TE= 3 WEEKS
Z= TL-TE/ 2= -3/3=-1
From the probability table, the probability of completing the project is 15.9%
If TL = 18 weeks, the probability of completing is 18-20=2/3. From the table the probability
is 27%. The probability of not meeting the due date is 100-27=73%.

UNIT SUMMARY

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Projects are composed of a unique set of activities established to realize a given set of objectives
during a limited life span. The non routine nature of project activities places a set of demands on
the project manager which are different in many respects than those required for the manager of
more traditional operations activities both in planning and coordinating the work.

PERT and CPM are two commonly used technique for developing projects. Although each
technique was developed independently and more expressly different purposes time and
practice have erased most of the differences so that now little distinction can be made between
the two. Both depict the sequential relationships that exist among activities and reveal to
managers which activities must be completed on time in order to achieve timely completion of
the project. Managers can use that information to direct their attention toward the most critical
activities.
CPM is used when a deterministic time is fairly established for the duration of the activities.
While PERT is used for probabilistic time estimates is used i.e. when activity time is subject to
some uncertainty.

Two slightly different conventions can be used when constructing a network diagrams. One
designates the arrows as activities; the other designates the nodes as activities. We have seen
only one approach, activities on arrow.

SELF ASSESMENT QUESTIONS (SQL-4)

1. Give brief explanation about the following terms
A. Critical path B. Critical activity

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C. Early start H. Predecessor

D. Early finish I. Activity
E. Latest start J. Node
F. Latest finish K. Arrow
G. Slack L. Node
2. The following are the activities which are to be performed for developing anew computer
system

Activity Predecessor Duration in days

A - 6
B A 8
C A 8
D B 20
E B 18
F C 10
G E 8
H D,F 2
I G,H 7
Required
A. drew the network referring to the data
B. Number the events as discussed earlier.
C. Calculate the ES, EF, LS, LF and slack for each activities
D. Identify the critical path
3. A company manufacturing plant and equipment for chemical processing is in the process
of quoting tender called by public sector undertaking. Help the manager to find the project
completion time to participate in the tender.

S. No Activities Duration in days

1 A 3 -

2 B 4 -

3 C 5 A

4 D 6 A

5 E 7 C

6 F 8 D

7 G 9 B

83 H 3 E,F,G

Required

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A. drew the network referring to the data

B. Number the events as discussed earlier.
C. Calculate the ES, EF, LS, LF and slack for each activities
D. Identify the critical path

4. Considering the following project activities answer the following questions

A. Drew the network referring to the data
B. Number the events as discussed earlier.
C. Calculate the ES, EF, LS, LF and slack for each activities
D. Identify the critical path

S. No Activities Duration in years Predecessors

1 A 3 -

2 B 4 -

3 C 5 B

4 D 6 A, C

5 E 7 A

6 F 8 D, E

5. Construct a network Diagram and find the critical path, early start time, late start time,
early finish time, and late finish time for the following data.

Job Description Successor Times in

minutes
S Start A 0
A Processing letter B,C 4
B Dictating letter D 3
C Order execution E 7
D Supervision E 6
E Dispatching F 2
F Finish None 0

ANSWER KEY FOR SELF ASSESSMENT QUESTIONS

UNIT ONE [SAQ 1]
1. Iconic models are physical models
Analog models are physical models but are more abstract than iconic model

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Symbolic models use symbols to represent a reality

2. Definition of the problem, use of models and analysis leading to solution of the problem and
implementation and follow up.

UNIT TWO [SAQ 2]

1.
A. Maximization
B. X1, X2 &X3
C. Labour, machine and wood
D. Unit of labour hours required to produce one unit of x3( book case )
E. A minimum of 10 units of tables should be produced
F. The profit that is secured by producing and selling one unit of X1 (table )
4 Let x1 represent corn
X2 represent wheat
X3 represent soybeans
Max Z= 30x1+20x2+20x3
Sub to
X1+x2+x3 < 1000
100x1+120x2+70X3< 100,000
7x1+10x2+8x3< 8000
X1, x2&x3 > 0

5. Let x1 represent food X

X2 represent food Y
Min Z= 0.6x1+0.8x2
Sub to
20X1+30x2 > 900
40x1+30x2 >1200
X1, x2&x3 > 0
6. Let x1 represent bomb P
X2 represent bomb Q
X3 represent bomb R
Max Z= 30x1+20x2+20x3
Sub to
3 X1+x2+4x3 < 600
2x1+4x2+2X3 >480
2x1+3x2+3x3 =540
X1, x2&x3 >

7. Let x1 represent desk

X2 represent chair
X3 represent cabinet
X4 represent book case
Max Z= 95x1+41x2+84x3 +76x4
Sub to
4X1+2x2+3x3+ 3x4< 8ooo

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10x1+6x2+8X3 +7x4< 1200

10x1+8x2+8x3+ 8x4< 800
X1, x2&x3 > 0
8. a) x1=60, x2= 20 Z= 1100
b) x1=4, x2= 2, Z= 10
c) x1=4, x2= 2, Z= 230
d) x1=3, x2= 18,Z= 48
9. a) x1=2, x2= 6, s1=0, s2=0 Z= 36
b) X1=50, x2= 0, s1=0, s2=50 Z= 1000
C) X1=20/3, x2= 8/3, s1=0, s2=0 Z= 212/3
d) X1=0, x2= 0, X3= 8/11, X4=4/11,s1=42/11, s2=0 Z= 48/11
10. a. Min Z= 20x1+10x2+14x3
b. -2
c. [18- ]
d. [0-72]
e. [18- ]
f. [8-20]
g. x1=21
x2= 39
Z= 622
UNIT THREE [SAQ 3]
3 ) TC= 740
4 ) TP=4230
5 ) TC=32
6 ) TP=38
7 ) TC=1320
UNIT FOUR [SAQ 4]
2. I. a) 22- alternative C
b) 14-alternative D
C) 5-alternative C
D) 16- alternative C
II.a) 10- alternative B&C
b) 14-alternative D
c) 2-alternative C&D
d)13.6-alternative B
III.a) 16.6 alternative C
b) 18.6
3. A. I. 20- alternative B
II. 96- alternative B
III. 40 - alternative C
IV. 68.6- alternative B
B. I. 72- alternative B
II. 18- alternative B
C. 90

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UNIT FIVE [SAQ 5]

2. Project completion time = 47 DAYS and the critical path is = A – B – E – G - I.

3. Project completion time = 20 weeks and the critical path = A – D – F – H.
4. Project completion time = 15 years and the critical path = A – D – F

REFERENCES

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1. Williams J. Stevenson, Introduction to Management Science, Richard D Irwin INC, 1989

2. David R. Anderson and Sweeney, Introduction to Management Sciences: Quantitative
Approach to Decision Making, west publishing company, 1988
3. Dr. E. Turban and DR.J. Meredith, Fundamentals of Management Sciences, Richard D.
Irwin, INC,1991
4. K.Chawala, V.Gupta and K. Sharma, Operations Research: Quantitative Analysis for
Management , kalyani publisher
5. P. Rama Murthy , Operations Research, New Age International, 2007

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