Or Module Haile
Or Module Haile
Or Module Haile
Chapter One
Unit objective:
Up on the completion of this unit, the learner would be able to:
Define operations research
Describe significance of OR
Explain models and their importance
Differentiate among different categories of models
Elucidate methodology in OR
Identify application areas of OR models
Describe techniques in OR
Decision-making in today’s social and business environment has become a complex task. The
uncertainty of the future and the nature of competition and social interaction greatly increase
the difficulty of managerial decision-making. Knowledge and technology are changing rapidly,
the new problems with little or no precedents these problems and provide leadership in the
advancing global age, decision-makers can not afford to make decisions by simply applying
their personal experiences, guesswork or intuition, because the consequences of the wrong
markets, producing the wrong products, providing inappropriate services, etc., will have major,
often disastrous consequences for organizations.
Operations Research as one of the quantitative aid to decision-making, offers the decision
maker a method of evaluating every possible alternative (act or course of action) by using
various techniques to know the potential outcomes. This is not to say, however, that
management decision-making is simply about the application of operations research techniques.
In general, while solving a real-life problem, the decision-maker must examine in both from
quantitative as well as qualitative perspective. Information about the problem from both these
perspectives needs to be brought together and assessed in the context of the problem. Based on
some mixes of the two sources of information, a decision should be taken by the decision-
maker.
The evaluation of each alternative is extremely difficult or time consuming for two reasons:
First, the amount and complexity of information that must be processed; second the number of
alternative solutions could be so large that a decision maker simply can not evaluate all of them
to select an appropriate one. For these reasons when there is lack of qualitative information,
decision makers increasingly turn to quantitative methods and use computers to arrive at their
optimal solution to problems involving large number of alternatives. The study of these
methods and how decision makers use them in the decision process is the essence of operations
research approach.
It is generally agreed that operations Research came is to existence as a discipline during World
War II when there was a critical need to manage scarce resources. The term “Operations
research” was coined as a result of research on military operations during this war. Since the
war involved strategic and tactical problems which were greatly complicated, to expect
adequate solutions from individual or specialists in a single discipline was unrealistic.
Therefore, group of individuals who collectively were considered specialists in mathematics,
Economics, statistics and probability theory, engineering, behavioral, and physical science were
formed as special unit within the armed forces to deal with strategic and tactical problems of
various military operations. The objective was the most effective utilization of most limited
military resources by the use of quantitative techniques.
After the war ended, scientists who had been active in the military OR groups made efforts to
apply the operations research approach to civilian problems, related to business, industry,
research and development, and even won Nobel prizes when they returned to their peacetime
disciplines.
There are three important factors behind the rapid development in the use of operations research
approach.
(i) The economic and industrial boom after World War II resulted in continuous
mechanization, automation, decentralization of operations and division of
management factors. This industrialization also resulted in complex managerial
problems, and therefore application of operations research to managerial decision-
making become popular.
(ii) Many operation researchers continued their research after war. Consequently, some
important advancement were made in various operations research techniques: linear
programming and its solution by a method known as simplex method, statistical
quality control, dynamic programming, queuing theory and inventory theory were
well developed during this time.
(iii) Analytic power was made available by high-speed computers. The use of computers
made it possible to apply many OR techniques for practical decision analysis.
1.3. Nature and Significance Operations Research
available to the decision-makers. In a theoretical sense, the optimum decision must be one that
is best for the organization as a whole. It is often called global optimum. A decision that is best
for one or more sections of the organization is usually called suboptimum decision. The OR
approach attempts to find global optimum by analyzing inter-relationships among the system
components involved in the problem. In other words, operations research attempts to resolve
the conflicts of interest among various sections of the organization and seeks the optimal
solution which may not be acceptable to one department but is in the interest of the organization
as a whole.
? Dear learner, discuss with your colleagues about how conflicting interests arise in the
organization and how OR tries to balance these interests?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
Because of the wide scope of application of operations research, giving a precise definition is
difficult. However, a few definitions of OR are given below.
Operations research is concerned with scientifically defining how to best design and
operate man-machine systems usually requiring the allocation of scarce resources.
Apart from being lengthy, the definition given by ORSUK, has been criticized, because it
emphasizes complex problems and large systems leaving the impression that it is a highly
technical approach suitable only to large organizations. The definition of ORSA contains an
important reference to the allocation of scarce resources. The key words used in the above
definitions are scientific approach, scarce resources, system and model. The UK definition
contains no reference to optimization, while the American definition has no reference to the
word, best.
ADU –Department of Business Administration Page 3 of 214
Operations Research
A few other definitions, which are commonly used and widely acceptable, are:
Operations Research, in the most general sense, can be characterized as the application of
scientific methods, techniques and tools, to problems involving operations of a system so as
to provide those in control of the operations with optimum solutions to the problems.
Operation research seeks the determination of the optimum course of action of a decision
problem under the restriction of limited resources. It is quite often associated almost
exclusively with the use of mathematical techniques to model and analyze decision
problems.
? Dear learners, would you discuss on the above definitions and define OR in your words?
___________________________________________________________________________
___________________________________________________________________________
________________________________________________________________________
? Dear learner, can you mention situations which can be and can not be analyzed by a single
individual?
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
Activity
1. Operations research is an aid for the executive in making his/her decisions by providing
the needed quantitative information, based on scientific method analysis. Discuss.
Both simple and complex systems can easily be studied by concentrating on some portion or
key features instead of concentrating on every detail of it. This approximation or abstraction,
maintaining only the essential elements of the system, which may be constructed in various
forms by establishing relationships among specified variables and parameters of the system, is
called a model. In general, models attempt to describe the essence of a situation or activity by
abstracting from reality so that the decision- maker can study the relationship among relevant
variables in isolation.
Models do not, and cannot, represent every aspect of reality because of the innumerable and
changing characteristics of the real life problems to be represented. Instead, they are limited
approximation of reality. For example, to study the flow of materials through a factory, a scaled
diagram on paper showing the factory floor, position of equipment, tools, and workers can be
constructed. It would not be necessary to give such details as the color of machines, the height
of the workers, or the temperature of the building. For a model to be effective, it must be
representative of those aspects of reality that are being investigated and have a major impact
on the decision situation.
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
__
A model is constructed to analyze and understand the given system for the purpose of
improving its performance. The reliability of the solution obtained from a model depends on the
validity of the model in representing the system under study. A model, allows the opportunity
to examine the behavioral changes of a system without disturbing the on-going operations.
Note: The key to model building lies in abstracting only the relevant variables that affect the
criteria of the measures of performance of the given system and expressing the relationship in a
suitable form. But oversimplification of problem can lead to a poor decision. Model enrichment
a) Physical Models
These models provide a physical appearance of the real object under study either reduced in
size or scaled up. Physical models are useful only in design problems because they are easy to
observe, build, and describe. Since these models can not manipulated and are not very useful
for prediction, problems such as portfolio analysis selection, media selection, production
scheduling, etc cannot be analyzed by physical models.
b) Symbolic models
These models use symbols (letters, numbers) and functions to represent variables and their
relationships to describe the properties of the system.
a) Descriptive models
Descriptive models simply describe some aspects of a situation, based on observation, survey,
questionnaire results or other available data of a situation and do not recommend anything.
Example: Organizational chart, plant layout diagram, etc.
b) Predictive Models
These models indicate “If this occurs, then that follow”. They relate dependent and independent
variables and permit trying out, “what if” questions. In other words, these models are used to
predict outcomes due to a given set of alternatives for the problem. These models do not have
an objective function as a part of the model to evaluate decision alternatives.
For example, S = a + bA +cI is a model that describes how the sales (S) of a product changes in
advertising expenditures (A) and disposal personal income (I). Here, a, b, and c are parameters
whose values must be estimated.
These models provide the “best” or “Optimal” solution to problems subject to certain
limitations on the use of resources. These models provide recommended courses of action. For
example, in mathematical programming, models are formulated for optimizing the given
objective function, subject to restrictions on resources in the context of the problem under
consideration and non negativity of variables. These models are also called prescriptive models,
because they prescribe what the decision maker ought to do.
a)Static Models
Static models represent a system at some specified time and do not account for changes over
time. For example, an inventory model can be developed and solved to determine an economic
order quantity for the next period assuming that the demand in planning period would remain
the same as that for today.
b) Dynamic models
In dynamic models, time is considered as one of the variables and allows the impact of changes
due to change in time. Thus, sequences of interrelated decisions over a period of time are made
to select the optimal course of action to optimize the given objective. Dynamic programming is
an example of a dynamic model.
a)Deterministic Models
If all the parameters, constants and functional relationships are assumed to be known with
certainty when the decision is made, the model is said to be deterministic. Thus, in such a case,
the outcome associated with a particular course of action is known. That is, for a specific set of
input values, there is a uniquely, determined output which represents the solution of the model
under consideration of certainty. The results of the models assume single value. Linear
programming models are examples of deterministic models.
Models in which at least one parameter or decision variable is a random variable are called
probabilistic (or stochastic). Since at least one decision variable is random, a dependant variable
which is the function of independent variable(s) will also be random. This means consequences
or pay off due to certain changes in the independent variable can not be produced with
certainty. However, it is possible to predict a pattern of values of both the variables by their
probability distribution. Insurance against risk of fire, accidents, sickness, etc are examples
where the pattern of events is studied in the form of a probability distribution.
a)Heuristic Model
These models employ some sets of rules which, though perhaps not optimal, do facilitate
solutions of problems when applied in a consistent manner.
b) Analytical Models
These models have a specific mathematical structure and thus can be solved by known
analytical or mathematical techniques. Any optimization model (which requires maximization
or minimization of an objective function) is an analytical model.
c)Simulation Models
These models have a mathematical structure but are not solved by applying mathematical
techniques to get a solution. Instead, a simulation model is essentially a computer-assisted
experimentation on a mathematical structure of a real-life problem in order to describe and
evaluate its behavior under certain assumptions over a period of time.
Simulation models are more flexible than mathematical ones and therefore, can be used to
represent a complex system which otherwise can not be represented mathematically. These
models do not provide general solution like those of mathematical Models.
? Dear learner, do you think that the above classification of models is mutually exclusive?
Support your response with evidence.
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
___
1.6.2. Advantage of Models
Models in general are used as an aid for analyzing complex problems. However, a model can
also serve other purposes as:
i) A model provides economy in representation of the realities of the system. That is,
models help decision makers to visualize a system so that he/she can understand the
system’s structure or operation in a better any. For example, it easier to represent a
factory lay out on paper than to construct it. It is cheaper to try out modifications of such
systems by rearrangement on paper.
ii) The problem can be viewed in its entirety, with all the components being considered
simultaneously.
iii) Models serve as aids to transmit ideas and visualization among people in the
organization. For example, process chart can help the management to communicate
about better work methods to workers.
iv) A model allows us to analyze and experiment in a complex situation to a degree that
would be impossible in the actual system and its environment. For example, the
experimental firing of satellite may be costly and require years of preparation.
v) Models simplify he investigation considerably and provide a powerful and flexible for
predicting the future state of the process or system.
For effective use of OR techniques, it is essential to follow some steps that are helpful for
decision-makers to make better solution. The flow diagram representing the methodology of
OR is shown as:
The first step in OR process is the identification of a problem that exists is a system
(organization). The system must be continuously and closely observed so that problems can be
identified as soon as they occur or anticipated. Problems are not always the results of crisis; but
instead frequently involve an anticipatory or planning situation. Once it has been determined
that a problem exists, the problem must be clearly and concisely defined. Because improperly
defining a problem can easily result in no solution or an inappropriate solution. Since the
existence of a problem implies that the objectives of the firm are not being met in some way,
the goals (objectives of the organization) must also be clearly defined.
? Dear learner, can you identify individual(s) who are responsible to identify problems and
problems they face while they identify these problems?
___________________________________________________________________________
_________
Model formulation involves an analysis of the system under study, determining objective of the
decision-maker, and alternative course of action, etc, so as to understand and describe, in
precise terms, the problem that an organization faces.
The major steps which have to be taken in to consideration for formulating the model are:
Problem Components. The first component of the problem to be defined is the decision
maker who is not satisfied with the existing state of affairs. The interaction with the
decision maker will help the OR specialist in knowing his/her objectives. That is, either
he/she has already obtained some solution of the problem and wants to retain it, or he
wants to improve it to a higher degree. If the decision maker has conflicting multiple
objectives, he/she may be advised to rank the objectives in the order of preference;
overlapping objectives may be eliminated.
Decision environment
It is desirable to know about the resources such as managers, employees equipments, etc which
are required to carry out the policies of the organization considering the social and ecological
environment in which the organization functions. Knowledge of such factors will help in
modifying the initial set of decision-maker’s objectives.
The problem arises only when there are several courses of action available for a solution. An
exhaustive list of course of action can be prepared in process of going through the above steps
of formulating the problem. Courses of action which are not feasible with respect to objectives
and resources may be ruled out.
Measure of effectiveness
After the problem is clearly defined and understood, the next step is to collect required data and
then formulate a mathematical model. Model construction consists of hypothesizing
relationships between variables subject to and not subject to control by decision-maker. Certain
basic components required in every decision problem model are:
o Controllable (decision) Variables - These are the issues or factors in the problem whose
values are to be determined (in the form of numerical values) by solving the model. The
possible values assigned to these variables are called decision alternatives (strategies or
courses of actions). Example, in LPP the number of units produced is a decision
variable.
o Uncontrollable variable. These are the factors whose numerical value depends up on the
external environment prevailing around the organization. The values of these variables
are not under the control of the decision-maker and are also termed as state of nature.
Objective function.
Constraints or Limitations
These are the restrictions on the values of the decision variables. These restrictions can arise
due to limited resources such as space, money, manpower, material, etc. The constraints may be
in the form of equations or inequalities.
Functional relationships
In a decision problem, the decision variables in the objective function and in the constraints are
connected by a specific functional relationship. A general decision problem model might take
the form:
A model is referred to as a linear model if all functional relationships among decision variables
X1, X2, Xn in f(x) and g(x) are of a linear form. But if one or more of the relationships are non –
linear, the model is said to be a non-linear model.
This involves obtaining the numerical values of decision variables. Obtaining these values
depends on the specific form or type, of mathematical models. Solving the model requires the
use of various mathematical tools and numerical procedures. In genera, there are two categories
of methods used for solving an OR model.
Optimization model. These models yield the best value for the decision variables both
for unconstrained and constrained problems. In constrained problems, these values
simultaneously satisfy all of the constraints and provide an optimal or acceptable value
for the objective function or measure of effectiveness. The solution so obtained is called
the optimal solution to the Problem.
Heuristic Model. These methods yield values of the variables that satisfy all the
constraints, but not necessarily provide optimal solution. However, these values provide
an acceptable value of the objective function.
Heuristic Methods are sometimes described as “rules of thumb” which work. These
methods are used when obtaining optimal solution is either very time consuming or the
model is complex.
Some times difficulties in problem solving arise due to lack of an appropriate methodology
for it and psychological perceptions on the part of the problem solver. The major difficulties
in problem solving:
(i) Failure to recognize the existence of a problem (iii) Failure to use all available
information
- Some people tend to personalize problems - The problem-solver fails to seek out
information
- Problems arise in context which people have
had no experience.
After solving the mathematical model, it is important to review the solution carefully to see
that values make sense and that the resulting decisions can be implemented. Some of the
reasons for validating the solution are:
(i) The mathematical model may not have enumerated all the limitations of the problem
under consideration.
(ii) Certain aspect of the problem may have been overlooked, omitted or simplified,
(iii) The data may have been incorrect estimated or recorded, perhaps when entered in to
the computer.
The decision-maker has not only to identify good decision alternatives but also to select
alternatives that are capable of being implemented. It is important to ensure that any solution
implemented is continually reviewed and updated in the light of a changing environment.
For a mathematical model to be useful, the degree to which it actually represents the system or
problem being modeled must be established. If during validation, the solution cannot be
implemented, one needs to (a) identify constraint that were omitted during the original problem
formulation or (b) find if some of the original constraints were incorrect and need to be
modified. In all such cases, one must return to the model formulation step and carefully make
the appropriate modifications to represent more accurately the given problem. A model must be
applicable for a reasonable time period and should be updated from time to time, taking in to
consideration the past, present, and future aspects of the problem.
The dynamic environment and changes within the environment can have significant
implications regarding the continuing validity of models and their solutions. Thus, a control
procedure has to be established for detecting significant changes in decision variables of the
problem so that suitable adjustments can be made in the solution without having to build a
model every time a significant change occurs.
A solution that works but is quite expensive compared to the potential savings from its
application should not be considered successful. Also a solution that is well within the budget
but which does not accomplish the objective is not successful either. The following are features
of good solution:
Technically appropriate. The solution should work technically; meet the constraints and
operate in the problem environment.
Reliable. The solution must be useful for a reasonable period of time under the
conditions for which it was designed.
Economically viable. The economic value should be more that what it costs to develop
and should be seen as wise investment in hiring OR talents.
There is no unique set of problems which can be solved by using OR models or techniques.
Some OR models or techniques include:
Allocation Models
Allocation models are used to allocate resources to activate in such a way that some measure of
effectiveness (objective function) is optimized. Mathematical programming is the broad term of
OR techniques used to solve allocation problems.
If the measure of effectiveness such as profit, cost, etc., is represented as a linear function of
several variables and if limitations of resources (constraints) can be expressed as a system of
When the solution values or decision variables for the problem are restricted to being integer
values, the problem is classified as an integer programming. The problem having multiple,
conflicting and incommensurable objective function (goals) subject to linear constraints is
called goal programming. If decision variables in the linear programming problem depend on
chance, such a problem is called is called a stochastic programming problem.
Inventory Model
Inventory Models deal with the problem of determination of how much to order at a point in
time and when to place an order. The main objective is to minimize the sum of three conflicting
inventory costs: the Cost of holding or carrying extra inventory, the cost of shortage or delay in
delivery of items when it is needed, a cost of ordering or set-up.
Linear Mathematical Probabilistic Inventory Net work Other linear and non-
Programming Techniques Techniques Techniques linear Techniques
Source: Taylor, 1990, Introduction to Management Science, 3rd edition, Brown Publisher
Note: This classification is loose as many of the techniques cross over between
classifications. Example, Net work, inventory models can be either deterministic or
probabilistic.
Summary
Management science is the application of a scientific approach to solving management
problems in order to help managers make better decisions.
Management science encompasses a logical, systematic approach to problem solving,
which closely parallels what is known as scientific method for attacking problems
and includes generally recognized ordered set of steps including: Observation,
definition of problems, model construction, model solution, solution testing,
implementation of solution results.
A management science model is an abstract representation of an existing problem
situation.
• Management science techniques roughly can be categorized as: Linear mathematical
programming, probabilistic techniques, inventory techniques, and network techniques,
other linear and nonlinear techniques.
Activity
1. Discuss what the management science approach to problem solving encompasses?
2. Explain what a model is and how it is used in management science?
3. The ultimate test of a manager who uses management science techniques is the ability
to transfer knowledge in this material in to the business world. What does it mean?
4. Suppose you are being interviewed by the manger of the commercial firm for a job in
a research department which deals with the application of quantitative techniques.
Explain the scope and purpose of quantitative technique and its usefulness to the firm.
Give some examples of the applications of quantitative techniques in industry.
5. Management science is an ongoing process. Why do you think is the reason?
6. Distinguish between model results that recommend a decision and model results that
are descriptive.
Chapter Two
2. LINEAR PROGRAMMING: Application and Model
Formulation
Unit Objectives:
Up on completion of this unit, the learner is expected to:
Recognize problems that can be solved using LP models.
Formulate an LP model in mathematical terms.
Solve Linear Programming Problems (LPP) using both graphic and simplex approach.
Explain special cases in both graphic and simplex techniques.
INTRODUCTION
In 1947, George Danzig developed the use of algebra for determining solutions to problems that
involved the optimal allocation of scarce resources. In spite of numerous potential applications
in business, response to this new technique was low due to substantial computational burden,
which is now removed with subsequent advances in computer technology and related software
during the last three decades.
The term linear implies that all the mathematical relations used in the problem are linear or
straight-line relations, while the term programming refers to the method of determining a
particular program or plan of action, i.e., the use of algorithms that is a well defined sequence
of steps that will lead to an optimal solution. Taken as a whole, the term linear programming
refers to a family of mathematical techniques for determining the optimum allocation of
resources and obtaining a particular objective when there are alternative uses of the limited or
constrained resources.
The technique of linear programming is applicable to problems in which the total effectiveness
can be expressed as linear function of individual allocations and the limitations on resources
give rise to linear equation or inequalities of the individual allocations.
The usefulness of this technique is enhanced by the availability of several user-friendly soft
wares such as STORM, TORA, QSB+, LINDO, etc. However, there is no general package for
building an LP model. Model building is an art of practice.
? Dear learner, can you make differences between solutions of LP models solved manually
and using computers utilizing soft wares?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
grouped as components and assumptions. The components relate to the structure of a model,
where as the assumptions reveal the conditions under which the model is valid.
2.1.1. COMPONENTS OF LP MODELS
There are four major components of LP models including: Objective function, decision
variables, constraints and parameters.
Decision variables
They represent unknown quantities to be solved for. The decision maker can control the
value of the objective, which is achieved through choices in the levels of decision variables.
For example, how much of each product should be produced in order to obtain the greatest
profit?
Constraints
However, the ability of a decision maker to select values of the decision variables in an LP
problem is subject to certain restrictions or limits coming from a variety of sources. The
restrictions may reflect availabilities of resources (e.g., raw materials, labor time, etc.), legal
or contractual requirements (e.g., product standards, work standards, etc.), technological
requirements (e.g., necessary compressive strength or tensile strength) or they may reflect
other limits based on forecasts, customer orders, company policies, and so on. In LP model,
the restrictions are referred to as constraints. Only solutions that satisfy all constraints in a
model are acceptable and are referred to as feasible solutions. The optimal solution will be
the one that provides the best value for the objective function.
Parameters
The objective function and the constraints consist of symbols that represent the decision
variables (e.g., X1, X2, etc.) and numerical values called parameters. The parameters are
fixed values that specify the impact that one unit of each decision variable will have on the
objective and on any constraint it pertains to as well as the numerical value of each
constraint.
The following simple example illustrates the components of LP models:
? Dear learner, can you give some of the examples of LP models which consist of
unit and system constraint and discuss the components of each constraint in the
model?
____________________________________________________________________
____________________________________________________________________
________________________________________________________
Linearity (proportionality)
The linearity requirement is that each decision variable has a linear impact on the objective
function and in each constraint in which it appears. In terms of a mathematical model, a
function or equation is linear when the variables included are all to the power 1 (not squared,
cubed, square root, etc.) and no products (e.g., x 1x2) appear. On the other hand, the amount of
each resource used (supplied) and its contribution to the profit (or cost) in the objective function
must be proportional to the value of each decision variable. For example, if production of one
unit requires 5 hours of a particular resource, then making 3 units of that product requires 15
hours ( 3x5) of that resource.
Divisibility (Continuity)
The divisibility requirement pertains to potential values of decision variables. It is assumed that
non-integer values are acceptable. However, if the problem concerns, for example, the optimal
number of houses to construct, 3.5 do not appear to be acceptable. Instead, that type of problem
would seem to require strictly integer solutions. In such cases, integer-programming methods
should be used. It should be noted, however, that some obvious integer type situations could be
handled under the assumption of divisibility. For instance, suppose 3.5 to be the optimal
number of television sets to produce per hour, which is unacceptable, but it would result in 7
sets per two hours, which would then be acceptable.
Certainty
This requirement involves two aspects of LP models. One aspect relates to the model
parameters, i.e., the numerical values. It is assumed that these values are known and constant.
In practice, production times and other parameters may not be truly constant. Therefore, the
model builder must make an assessment as to the degree to which the certainty requirement is
met. Large departures almost surely will have a significant effect on the model. The other
aspect is the assumption that all relevant constraints have been identified and represented in the
model.
Additivity
The value of the objective function and the total amount of each resource used (or supplied),
must be equal to the sum of the respective individual contributions (profit or cost) by decision
variables. For example, the total profit earned from the sale of two products A and B must be
equal to the sum of the profits earned separately from A and B. Similarly, the amount of a
resource consumed for producing A and B must be equal to the sum of resources used for A and
B respectively.
Non-negativity
It assumes that negative values of variables are unrealistic and, therefore, will not be considered
in any potential solutions. Only positive values and zero will be allowed and the non-negativity
assumption is inherent in LP models.
? Dear learner, can you give examples for each of the above assumptions? What will happen
if the assumptions are not met?
__________________________________________________________________________
__________________________________________________________________________
____________________________________________________________
? Dear learner, can you mention some other advantages and disadvantages of LP?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
In many cases, the decision variables are obvious; in others it might require brief discussion
with the appropriate manager. However, identifying the constraints and determining appropriate
values for the parameters can require considerable time and effort. Potential sources of
information include historical records, interviews with managers and staff, and data collection.
Validating the model will involve a critical review of the output, perhaps under a variety of
inputs, in order to decide if the results are reasonable.
Product Mix
Organizations often produce similar services that use the same resources. For example, labor,
material cost, etc. because of limited resources available during any time period a decision
must be made concerning how much of each product to produce or make available. Linear
programming answers what mix of output (or service) will maximize profit given the
availability of scarce resources
Diet problem
It usually involves the mixing of raw materials or other ingredients to obtain an end product that
has certain characteristics. For example, what mix of inputs will achieve the desired results in
the output for the least cost? Other applications that fall into this category include mixing feed
for livestock, mixing pet foods, mixing building materials (concrete, mortar, paint), and so on.
Blending problems
They are very similar to diet problems. Strictly speaking, however, blending problems have
additional requirement, i.e. to achieve a mix that have specific consistency. For example, how
many quarts of the different juices each with different sugar content proportion must be mixed
together to achieve one gallon that has a sugar content of 17 percent?
Portfolio selection
These problems generally involve allocating a fixed dollar amount among a variety of
investments, such as bonds, sockets, real states, etc. The goal usually is to maximize income or
total return. The problems take on an added dimension when certain other requirements are
specified (for example, no more than 40 percent of the portfolio can be invested in bonds).
Dear learner, the section below presents some examples that demonstrate how a linear
programming model is formulated. Although these examples are simplistic, they are realistic
and represent the type of problem to which linear programming can be applied. By carefully
studying each of these examples, the learner can become familiar with the process of
formulating linear programming models.
Product Mix
ABC private limited company is engaged in the production of power and traction transformers.
Both of these categories of transformers pass through three basic processes: core preparation,
core to coil assembly, and vapor phase drying. A power transformer yields a contribution of
Birr 50,000 and traction transformer contributes Birr 10,000. The time required in the
production of these two products in terms of hours for each of the processes is as follows.
Power transformer Traction Transformer
Core preparation 75 15
Core to Coil Assembly 160 30
Vapor Phase Drying 45 10
If the capacities available are 1000, 1500, and 750 machine hours in each processes
respectively, formulate the problem as LP?
Solution
Step1. Identify decision variables
Since the products to be produced are power and traction transformers using the available
resource to attain the objective set, we consider them as decision variables. This is because the
organization’s problem here is how many of each product to produce in order to attain the
objective, which requires the management decision.
LetX1 = the no of power transformers to be produced.
X2= the no of traction transformer to be produced
Step2. Determine Objective Function
From the problem above, we understand that the problem is maximization problem.
Hence,
Zmax = 50,000X1+ 10,000X2
This is because, each unit of X1 contributes Birr 50,000 and X2 contributes Birr
10,000 to objective function.
Solution
Decision variables
Four decision variables represent the monetary amount invested in each investment alternative.
Let X1, X2, X3 and X4 represent investment on municipal bonds, certificates of deposits,
Treasury bill, and income bonds respectively.
Objective function
The problem is maximization because from the word problem we know that the objective of the
investor is to maximization return from the investment in the four alternatives.
Therefore, the objective function is expressed as:
Zmax = 0.085 X1+ 0.1X2+ 0.065X3+0.13X4
Where,
Z = total return from all investment
0.085 X1=return from the investment in municipal bonds.
0. 100 X2=return from the investment in certificates in deposit
This problem contains all three types of constraints possible in LP problems: and =. As
this problem demonstrates there is no restriction on mixing these types of constraints.
X1 + X2 + X3 + X4 = 70,000,
X1+ X2+ X3+X4 0
Marketing Application
2. Supermarket store chain has hired an advertising firm to determine the types and amount of
advertising it should have for its stores. The three types of advertising available are radio and
television commercials, and news papers advertisements. The retail chain desires to know the
number of each type of advertisement it should purchase in order to maximize exposure. It is
estimated that each ad or commercial will reach the following potential audience and cost the
following amount.
Exposure
Type of Advertisement (people /ad or commercial) Cost
Television commercial 20,000 Birr 15,000
Radio commercial 12,000 6,000
News paper advertisement 9,000 4,000
Model constraints
Budget constraint
15,000x1 + 6,000x2 + 4,000x3 Birr 100,000
Capacity constraint
Television commercials and radio commercials are limited to 4 and 10 respectively and news
paper ads are limited to 7.
X1 4 television commercials
X2 10 radio commercials
X3 7 news paper ads
Policy constraint
The total number of commercials and ads can not exceed 15
X1 + X2 + X3 15
The complete linear programming model for this problem is summarized as:
Zmax = 20,000X1 + 12,000X2 + 9,000X3
Sub. to:
15,000X1 + 6,000X2 + 4,000X3 Birr 100,000
X1 4
X2 10
X3 7
X1 + X2 + X3 15
X1, X2 , X3 0
Chemical mixture
A chemical corporation produces a chemical mixture for the customer in 1000- pound batches.
The mixture contains three ingredients- Zinc, mercury and potassium. The mixture must
conform to formula specifications (i.e., a recipe) supplied by a customer. The company wants to
know the amount of each ingredient to put in the mixture that will meet all the requirements of
the mix and minimize total cost.
The customer has supplied the following formula specifications for each batch of mixture.
1. The mixture must contain at least 200 lb of mercury
2. The mixture must contain at least 300 lb of zinc
3. The mixture must contain at least 100 lb of potassium
The cost per pound of mixture is Birr4, of zinc Birr 8 and of potassium Birr 9.
Required: Formulate LPM for the problem
Solution
Decision variables
The model of this problem consists of three decision variables representing the amount of each
ingredient in the mixture.
X1 = number of lb of mercury in a batch
X2 = number of lb of zinc in a batch
X3 = number of lb of potassium in a batch
Objective function
Zmin = 4x1 + 8x2 + 9x3
Constraints
X1 200 lb… specification 1
X2 300 lb… specification 2
X3 100 lb… specification 3
Finally, the sum of all ingredients must equal 1000 pounds.
x1 + x2 + x3 = 1000 lb
Activity
The manager of a department store in Adama is attempting to decide on the types and amounts
of advertising the store should use. He has invited representatives from the local radio station,
television station, and newspaper to make presentations in which they describe their audiences.
a. The television station representative indicates that a TV commercial, which costs Birr
15,000, would reach 25,000 potential customers. The break down of the audience is as follows.
Male Female
b. The news paper representative claims to be able to provide an audience of 10,000 potential
customers at a cost of Birr 4000 per ad. The break down of the audience is as follows.
Male Female
c. The radio station representative says that the audience for one of the station’s commercials,
which costs Birr 6000, is 15,000 customers. The break down of the audience is as follows.
Male Female
Available space limits the number newspaper ads to 7. The store wants to know the optimal
number of each type of advertising to purchase to minimize total cost.
Required:
Formulate appropriate linear programming model.
2. An electronic company produces three types of parts for automatic machines. It purchases
casting of the parts from the local foundry and then finishes parts on drilling, shaping and
polishing machines.
The selling prices of parts A, B, and C respectively are Birr8, 10, 14. All parts made can be
sold. Casting for parts A, B and C respectively costs Birr 5, 6 and 10. The shop possesses only
one of each type of machine. Costs per hour to run each of the three machines are Birr 20 for
drilling, 30 for shaping, and 30 for polishing. The capacities (parts per hour) for each part on
each machine are shown in the following table:
Drilling 25 40 25
Shaping 25 20 20
Polishing 40 30 40
The management of the shop wants to know how many parts of each type it should produce per
hour in order to maximize profit for an hour’s run. Formulate this problem as an LP model so as
to maximize total profit to the company.
In this method, the two decision variables are considered as ordered pairs (X 1, X2), which
represent a point in a plane, i.e, X1 is represented on X-axis and X2 on Y-axis.
Important Definitions
Solution The set of values of decision variables xj (j = 1,2,…, n) which satisfy the constraints of
an LP problem is said to constitute solution to that LP problem.
Feasible solution The set of values of decision variables x j (j = 1,2,…, n) which satisfy all the
constraints and non- negativity conditions of an LP problems simultaneously is said to
constitute the Feasible solution to that LP problem.
Infeasible solution The set of values of decision variables x j (j = 1,2,…, n) which do not satisfy
all the constraints and non- negativity conditions of an LP problems simultaneously is said to
constitute the infeasible solution to that LP problem.
Basic solution For a set of m simultaneous equations in n variables ( n>m), a solution obtained
by setting ( n-m) variables equal to zero and solving for remaining m equations in m variables
is called a basic solution.
The (n-m) variables whose value did not appear in this solution are called non-basic variables
and the remaining m variables are called basic variables.
Basic feasible solution A feasible solution to LP problem which is also the basic solution is
called the basic feasible solution. That is, all basic variables assume non-negative values. Basic
feasible solutions are of two types:
Degenerate A basic feasible solution is called degenerate if value of at least one
basic variable is zero.
Non-degenerate A basic feasible solution is called non-degenerate if value of all m
basic variables are non- zero and positive.
Optimal Basic feasible solution A basic feasible solution which optimizes the objective
function value of the given LP problem is called an optimal basic feasible solution.
Unbounded solution A solution which can increase or decrease the value of OF of the LP
problem indefinitely is called an unbounded solution.
Example
In order to demonstrate the method, let us take a microcomputer problem in which a firm is
about to start production of two new microcomputers, X 1 and X2. Each requires limited
resources of assembly time, inspection time, and storage space. The manager wants to
determine how much of each computer to produce in order to maximize the profit generated by
selling them. Other relevant information is given below:
Type 1 Type 2
Profit per unit $60 $50
Assembly time per unit 4 hrs 10 hrs
Inspection time per unit 2 hrs 1 hrs
Storage space per unit 3 cubic feet 3 cubic feet
ADU –Department of Business Administration Page 33 of 214
Operations Research
? Dear learner, can you mention some of the advantages and disadvantages of graphic
method of solving LP models?
_________________________________________________________________________
_________________________________________________________________________
_______________________________________________________________
Once the constraints are plotted and feasible region is determined, we use either of the two
graphic methods of Graphic approach to find a solution for LP model consisting of only two
decision variables:
i) The extreme point enumeration method
ii) The objective (Iso-profit or cost) function line approach
The optimal solution to an LP problem will always occur at a corner point because as the
objective function line is moved in the direction that will improve its value (e.g., away from the
origin in our profit maximization problem), it will last touch one of these intersections of
constraints. Then we determine which two constraints intersect there (in our case inspection and
storage constraints) and solve the equations simultaneously to obtain the mix of the two
decision variables that gives the value of the objective function at the optimum. Simultaneously
solving inspection and storage equations, we find the quantity of type 1 microcomputer to be
produced (X1) = 9 and that of type 2 (X 2) = 4 giving the maximum profit of 60(9) + 50(4) =
Birr740
Note: The maximization problems the movement of the iso-lines is outward from the
origin; while for minimization the movement is inward to the origin.
2.3.1.2. The Extreme Point Approach
Corner or extreme point graphic method states that for problems that have optimal solutions, a
solution will occur at the corner point in the case of unique solution, while in the case of
multiple solutions, at least one will occur at a corner point as these multiple solutions will be
combinations of those points between two corner points. The necessary steps for this approach
is after graphing the problem, we determine the values of the decision variables at each corner
point of the feasible region either by inspection or using simultaneous equations. We then
substitute the values at each corner point into the objective function to obtain its value at each
corner point and select the one with the highest value of the objective function (for a maximum
problem) or lowest value (for a minimum problem) as the optimal solution.
i) Extreme Method of Solving Maximization Problems with < constraint
Maximization of objective function involves finding the point where the combination of
products results in maximum value of objective function. The constraints are connected
with < sign. The solution space lies below the slant line and is bounded by the line
segments. The origin and other points below the slant lines are in the solution space (i.e.,
feasible region).
Using this method for our example, simultaneously solving for corner points a, b, c, and d, we
find corresponding profit values of 500, 700, 740, and 660, respectively giving us the same
solution as the above one at C. Therefore, the optimal solution is x1= 9 units and x2 = 4 units
while the optimal value of objective function is 740.
Interpretation:
For a firm to maximize its profit (740), it should produce 9 units of the Model I microcomputer
and 4 units of model II.
? Dear learner, what are the values of decision variables at each corner point? (Hint:
Solve simultaneously those lines which intersected each other). Can you mention
some of the characteristics of the LP graphs consisting of sign?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
Solution
Coordinates
Corner point x y Zmin=0.1x+0.07y
A 0 9 0.63
C 15/4 1 0.445
? Dear learner, can you draw a graph for the above problem and determine values for each
corner point of the feasible region?
What difference can you make between LP problems consisting of > and <signs?
___________________________________________________________________________
? Dear learner, can you find slack and surplus amounts of the constraints in the above
problems?
_________________________________________________________________________
_________________________________________________________________________
_______________________________________
Following the above listed steps of graphical solution method, we find the following graph for
the above model:
The shaded area represents the set of all feasible solutions and as can be seen from the graph,
the solution is unbounded.
iii) Infeasibility
In some cases after plotting all the constraints on the graph, feasible area (common region) that
represents all the constraint of the problem cannot be obtained. In other words, infeasibility is a
condition that arises when no value of the variables satisfy all the constraints simultaneously.
Such a problem arises due to wrong model formulation with conflicting constraints.
For example,
Max Z = 3X1+2X2
Subject to: 2X1 + X2 < 2
3X1 + 4X2 > 12
X1, X2 > 0
Note: The constraint is said to be an active or binding or tight, if at optimality the left hand
side equals the right hand side. In other words, an equality constraint is always active. An
inequality sign may or may not be active.
For example
Max Z = 8X1+16X2
Subject to: X1 + X2 < 200 ……. C1
3X1 + 6X2 < 900 ……. C2
X2 < 125 ……. C3
X1, X2 > 0
In the problem above, using extreme point method and solving for values of corner points
simultaneously, the objective function assumes its maximum value of 2,400 at two corner
points B (50,125) and C (100,100). Therefore, the optimal solution is found on the line segment
connecting the two corner points. One benefit of having multiple optimal solutions is that for
other (perhaps qualitative) reasons, a manager may prefer one of them to the others, even
though each would achieve the same value of the objective function. In practical terms, one of
the two corner points is usually chosen because of ease in identifying its values.
Other corner points are computed by identifying the two corner points and keeping the value of
the objective function constant, where 50 X1 100 and 100 X1 125. By selecting one value
for one of the decision variables from the domain, we determine the value for another decision
variable keeping the objective function’s value. Let X1 = 80.
Then, Z = 8X1+16X2 2400 = 8(80) +16X2
16X2 = 1760
X2 = 110
? Dear learner, can you identify other combination of decision variables that will optimize
the objective function and can you check whether the two conditions for multiple solution
is satisfied?
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
2x+3y>42 y
2x+3y=42
If x=0,
2x+3y=42
2(0) +3y=42
2(0) +3y=42 24_(0,24)
y=42/3 IV
y=14
The coordinates (0,14)
If y=0, 18 _ I
2x+3y=42
2x+3(0)=42 A (2.5,14) (14,14)
x=42/2 B V
x=21 (0, 14) E (3, 12)
Constraint III 12 _
X+4y>36 Feasible
X+4y=36 region multiple solution line
If x=0, (0,9) (double lined)
x+4y=36
y=36/4 6_ C (14, 5.5)
y=9
The coordinates (0, 9) D
If y=0, (12,6) II III x
x+4(0) =36 (0, 0)
x =36 (6,0) (14,0) (21,0) (36,0)
The coordinates (36, 0) 6 12 18 24 36
1 2 3 4 5
Corner constraints Coordinates Objective functions
points intersected x y 1500x+2400y
A I&V 2.5 14 37,350
B IV&V 14 14 54,600
C III&IV 14 5.5 34,200
D II&III 12 6 32,400
E I&II 3 12 33,300
The minimum value of the objective function is 32,400. Therefore, the optimum solution is
found at the corner point D, where constraint II&III are intersected. Hence, optimal solution: x
=12, y = 6.
?
ADU Dear learner,
–Department howAdministration
of Business would the above optimal solution change if only objective
Page 42 of function
214 if
modified as Zmin1600x+2400y while all other constraints remain as they are?
_________________________________________________________________________
_________________________________________________________________________
______________________________________________________________________
Operations Research
i) Identify coordinates of each of the extreme (or i) determine the slope x1,x2 of the objective
corner points of feasible region by either drawing function and join intercepts to reveal the
perpendiculars on x and y axis or by solving the profit or cost line.
two intersecting equations.
ii) compute the profit(cost) at each extreme points ii) In case of maximization, maintain the
by substituting that point’s coordinates in to the same slope through a series of parallel
objective function. lines, and move the line up and to the right
until it touches the feasible region at only
one point.
But in case of minimization, move down
to the left until it touches only one point in
the region.
.
iii) Identify the optimal solution at that extreme iii) Compute the coordinates of the point
point with a highest profit in a maximization touched by iso-profit (or cost) line
on the problem or lowest cost in minimization
problem on the feasible region
. . iv) Compute the profit or cost
Summary
The term linear programming refers to a family of mathematical techniques for
determining the optimum allocation of resources and obtaining a particular objective
when there are alternative uses of the limited or constrained resources.
The linear programming models exhibit certain common characteristics: An objective
function
ADU –Department to Administration
of Business be maximized or minimized, a set of constraints, decision variables for
Page 43 of 214
measuring the level of activity, and linearity among all constraint relationships and the
objective function.
The graphic approach to the solution LP problems is not efficient means of solving
problems. For one thing, drawing accurate graphs is tedious. More over the graphic
approach is limited to models with only two decision variables.
Special cases that one face solving a problem graphically include: Mix of constraints,
unbounded solution, infeasibility, redundancy, multiple solution
Operations Research
2.5. Activity
1. Peniel House and Furniture manufacturer produces two products: Beds and Chairs. Each unit
of Bed requires 3 hrs in molding unit, 4hrs in painting unit, and 1 hr in finishing. On the other
hand, each unit of Chair requires 3 hrs in molding unit, 2 hrs in the paint shop and 2 hours in
finishing. Each week, there are 210 hrs available in molding, 200hrs in painting, and 120 hrs in
finishing unit. The demand for Beds can not exceed 40 units per week. Each unit of Bed
contributes Birr 20 to profit, while each unit of chair contributes Birr 30. Determine the number
of units of each product per week to maximize the profit? (Use Graphic Method to solve the
given LP problem).
2. A firm plans to purchase at least 200 quintals of scrap containing high metal quality metal X
and low quality metal Y. It decides that the scrap to be purchased must contain at least 100
quintals of X-metal and not more than 35 quintals of Y-metal. The firm can purchase the scrap
from two suppliers ( A and B) in unlimited quantities. The percentage of X and Y metals in
terms of weight in the scrap supplied by A and B is given below.
Metals Supplier A Supplier B
X 25% 75%
Y 10% 20%
The price of A’s scrap is Birr 200 per quintal and that of B is Birr 400 per quintal. The firm
wants to determine the quantities that it should buy from the two suppliers so that total cost is
minimized? (Use graphic approach)
3. Personal Mini Warehouses is planning to expand its successful Orlando business into Tampa.
In doing so, the company must determine how many storage rooms of each size to build. Its
objective and constrains follow:
The simplex method is an algebraic procedure that starts with a feasible solution that is not
optimal and systematically moves from one feasible solution to another until an optimal
solution is found. Incase of the graphical approach, optimal solution occurs at the extreme
points where the constraints intersect. Solutions where constraints intersect are called basic
solutions, and those satisfying all of the constraints together with non-negativity constraints are
called basic feasible solutions.
Constraints are generally expressed using inequalities either in ‘less than’ or ‘greater than’ or in
mixed form. Thus, constraints are not in standard form, meaning they should be converted into
equalities. To convert the inequality constraint into equality, we introduce slack or surplus
variables. In economic terminology, slack variables represent unused capacity and surplus
variables represent excess amount. The contribution (cost or profit) associated with the slack
and surplus variables is zero. An inequality of the ‘less than or equal to’ type is transformed
into equality by introducing a non-negative slack variable, as follows: -
Example
Non Standard form Standard form
X1+2X2 <= 6 X1+2X2+S1= 6
, where X1 and X2 are decision variables and S1 is a slack variable, added to the
smaller side of the inequality.
On the other hand, an inequality with ‘greater than or equal to’ type is changed into equality by
subtracting surplus variable as follows: -
Example
Non Standard form Standard form
X1+2X2 <=6 X1+2X2+S1=6
, where X1 & X2 are decision variables and S1 is a slack variable, added to the
smaller side of the inequality.
As stated above, since both the slack and surplus variables are insignificant with no
contribution in the objective function, they are represented with coefficient of zero in the
objective function.
To find a unique solution, the number of variables should not exceed the number of equations.
When the number of variables is more than the number of equations, the number of solutions is
unlimited. So as to get a unique solution, we have to set at least (n-m) variables to zero, where n
is the number of variables and m is the number of equalities. Those variables that are set to zero
are called non basic variables indicating they are not in the solution. The variables that are in
solution are called basic variable.
.
To demonstrate the simplex method, we will use the microcomputer problem with the following
objective function and constraints.
The Microcomputer Problem, which was discussed in graphic approach, can be standardized as:
Max. Z = 60X1+50X2+0S1+0S2+0S3
Subject to 4X1+10X2+S1 = 100
2X1+X2+S2 = 22
3X1+3X2+S3 = 39
, where X1 & X2 are decision variables and S1, S2 & S3 are slack variables.
Here, the number of variables (5) is greater than the number of equations (3). Therefore, the
decision variables are set to zero and we have
X1 = 0, X2 = 0, S1 = 100, S2 = 22, and S3 = 39.
This solution will serve as an initial feasible solution. An initial feasible solution is a first
solution used to generate other basic feasible solutions. The initial basic feasible solution is
obtained by setting all the decision variables to zero. As a result, the initial basic feasible
solution is entirely composed of the slack variables. X 1 & X2 are non basic variables since they
are not in solution. S1, S2 & S3 are basic variables since they are in solution.
A tableau is a system of displaying the basic feasible solution, the constraints of the standard
linear programming problem as well as the objective function in a tabular form. A tableau is
useful in summarizing the result of each iteration, i.e. the process of moving from one
solution/corner to another solution/corner in order to select the optimal solution.
Note that the slack variables have been assigned zero coefficients in the objective function. The
reason is that these variables typically contribute nothing to the value of the objective function.
Step 3. Design the initial feasible solution. An initial basic feasible solution is obtained by
setting
the decision variables to zero.
X1= X2 = ... = Xn = 0. Thus, we get S1 = b1, S2 = b2 ... Sm = bm.
Step 4. Set up the initial simplex tableau. For computational efficiency and simplicity, the
initial basic feasible solution, the constraints of the standard LPP as well as the
objective function can be displayed in a tabular form, called the simplex tableau as
shown below:
The interpretation of the data in the above tableau is given as under. Other simplex tableau will
have similar interpretations.
In the first row labeled "Cj", we write the coefficients of the variables in the objective
function. These values will remain the same in subsequent tableaus.
The second row shows the major column headings.
In the first column of the second row, under the label "Basic variables" (also called
Product mix column), the basic variables are listed.
In the second column of the second row, under the label "CB", the coefficients of the
current basic variables in the objective function are listed. Thus the coefficients of S 1,
S2... Sm, which are included in the initial feasible solution, are written in the CB
column.
The values listed under the non-basic variables (X 1, X2… Xn) in the initial simplex
tableau consists of the coefficients of the decision variables in the constraint set. They
can be interpreted as physical rates of substitution.
The values listed under the basic variables (S1, S2... Sm) in the initial simplex tableau
represents the coefficients of the slack variables in the constraints set.
In the next column (also called Quantity column), we write the solution values of the
basic variables.
To find an entry in the Zj row under a column, we multiply the entries of that column by
the corresponding entries of ‘CB’ column and add the results, i.e., Z j= ∑CBiXj. The Zj
row entries will all be equal to zero in the initial simplex tableau. The other Z j entries
represent the decrease in the value of objective function that would result if one of the
variables not included in the solution were brought into the solution. The Z j entry under
the “Quantity Column" gives the current value of the objective function.
The last row labeled "Cj-Zj", called the index row or net evaluation row, is used to
determine whether or not the current solution is optimal or not. The calculation of C j-Zj
row simply involves subtracting each Zj value from the corresponding Cj value for that
column, which is written at the top of that column. We observe that Cj -Zj, values are
meaningful for the non basic variables only. This is because for a basic variable, Zj= 1 x
Cj = Cj so that Cj-Zj=Cj-Cj= 0.
Note: The entries in the C - Z , row represent the net contribution to the objective function
j j
that results by introducing one unit of each of the respective column variables. A plus value
indicates that a greater contribution can be made by bringing the variable for that column into
the solution. A negative value indicates the amount by which contribution would decrease if
one unit of the variable for that column were brought into the solution. Index row elements are
also known as the shadow prices (or accounting prices)
Step 5. We test if the current solution is optimum or not. If all the elements or entries in the
Cj- Zj row (i.e., index row) are negative or zero, then the current solution is
optimum. If there exists some positive number, the current solution can be further
improved by removing one basic variable from the basis and replacing it by some
non-basic one. So start trying to improve the current solution in line with the
following steps.
Step 6. Further, iterate towards an optimum solution. To improve the current feasible solution,
we replace one current basic variable (called the departing variable) by a new non-
basic variable (called the entering variable).
We now determine the variable to enter into the solution mix, the entering variable. One
way of doing this is by identifying the column with the largest positive value in the C j -
Zj row of the simplex table. The column with the largest positive entry in the C j - Zj row
is called the key or pivot column. The non-basic variable at the top of the key column is
the entering variable that will replace a basic variable.
Next, we determine the departing variable to be replaced in the basis solution. This is
accomplished by dividing each number in the quantity column by the corresponding
number in the key column selected in identifying the entering variable. We compute the
ratio b1/a1j, b2/a2j... bm/amn. This is called replacement ratio.
The row with the minimum ratio is the key row or pivot row. The corresponding
variable in the key row (the departing variable) will leave the basis.
We identify the key or pivot element. This is the number that lies at the intersection of
the key column and key row of a given simplex tableau.
Step 7. Evaluate the new solution by constructing a second simplex tableau. After identifying
the entering and departing variable, all that remains is to find the new basic feasible
solution by constructing a new simplex tableau from the current one.
Now we evaluate or update the new solution in the following way:
New values for the key row are computed by simply dividing every element of the key
row by the key element to obtain a unit vector (1) in the key element.
The new values of the elements in the remaining rows for the new simplex table can be
obtained by performing elementary row operations on all rows so that all elements
except the key element (1) in the key column are zero, i.e. unit vector.
New entries in the CB column and XB column are entered in the new table of the
current solution.
Compute the values of the Cj - Zj row. If all the numbers in Cj - Zj row are either
negative or zero, an optimum solution has been obtained.
Step 8. If any of the numbers in C j - Zj row are positive, repeat the steps (6-7) again until an
optimum solution has been obtained.
Note: Rules for Ties. In choosing a key column and a key row, whenever there is a tie
between two numbers, the following rules may be followed:
The column farthest to the left may be selected if there is a tie between two numbers in
the index row.
The nearest ratio to the top may be selected whenever there is a tie between two
replacement ratios in the ratio column.
Example (Maximization Case)
Finding the initial Feasible Solution
As discussed above, the initial feasible solution is found by setting the decision variables to
zero.
Max Z = 60X1+50X2+0S1+0S2+0S3
Subject to 4X1+10X2+S1 = 100
2X1+ X2+S2 = 22
3X1+ 3X2+S3 = 39
X1, X2 > = 0
To be noted first is that the values of each basic variable (variables that are in solution) is
composed of a single 1 and the rest 0’s. This is called a unit vector. Basic variables will have a
unit vector. Moreover, ‘1’ will appear in the same row that the variable appears in. The unit
vector concept will help us in developing subsequent tableaus when we want to change the list
of variables that are in solution.
The Zj row in the quantity column indicates that the value of the objective function is 0.
The values in the Cj-Zj row indicate the net contribution of the variables if one unit of each
variable is added into that solution. For example, the 60 in column X 1 indicates that bringing
one unit of X1 would increase the value of the objective function by $60. The same is true for
the X2 column, i.e. bringing one unit of X2 would increase the value of the objective function by
$50. On the other hand, since the slack variables are at their maximum, their values in the C j-Zj
row are all 0. According to what has been said before, we have the following rule in order to
identify an optimal solution.
A simplex solution in a maximization problem is optimal if the C j-Zj row consists of entirely
zeros and negative numbers, i.e. there are no positive values in the row.
So for our case in the initial tableau, we have two positive values under the non-basic variables,
which indicate that further improvement of the solution is possible. As a result, we go for the
optimal solution by developing the second simplex tableau.
In answering the first question, the non-basic variable that should enter the solution should be
the one with the highest positive value in the C j-Zj row since bringing that non-basic variable
into the solution would make the highest contribution to the solution (objective function) and
result in the largest profit potential. As a result, the variable with the highest value in the C j-Zj
row of the initial simplex tableau is the X 1 column with a value 60. Therefore, X 1 should enter
the basis or the solution mix. The X1 column is now called the pivot column. The numbers in
this column (4 2 3) indicate the amount of the basic variables needed to get one unit of X 1. For
example, the number 4 indicates that 4 units of the first slack are needed to obtain one unit of
X1.
Since X1 has the highest profit potential, we need to make as much X 1 units as possible. The
amount of X1 that can be made depends on the values in the pivot column and the amount of
slack available shown under the quantity column. By dividing the values in the pivot column by
their respective values in the quantity column, we can identify the variable that is most limiting.
The values obtained by dividing will help us in determining the variable that should leave the
solution.
Cj 60 50 0 0 0 Quantity
Basic V. X1 X2 S1 S2 S3
S1 0 4 10 1 0 0 100 100/4 = 25
S2 0 2 1 0 1 0 22 22/2 = 11
S3 0 3 3 0 0 1 39 39/3 = 13
Zj 0 0 0 0 0 0
Cj-Zj 60 50 0 0 0
Pivot Column
Pivot Row Pivot Element
ADU –Department of Business Administration Page 52 of 214
Operations Research
In interpreting the ratios obtained by the division, I only the constraint was the first one, we
could make 25 units of X1. But there are also other constraints. Therefore, the one with the
smallest non-negative ratio is the most restrictive since it determines the amount of X 1. In this
particular case, there is only enough of the second constraint to make 11 units of X 1. In making
the 11 units of X1, the second resource (S2) will be down to zero indicating that S2 will leave out
the solution mix. The row of the leaving variable is called the pivot row. The intersection of the
pivot row and the pivot column is called the pivot element. As a rule,
The leaving variable is the one with the smallest non-negative ratio.
If a zero is obtained in the results of division, none of the corresponding variable is needed to
obtain one unit of the entering variable. If a negative value is obtained in the division, bringing
the entering variable into the solution would increase the amount of the basic variable. These
values will not limit the amount of the entering variable. Therefore, there is no need to divide
the quantity column by zero or negative.
Since we have determined the leaving and the entering variables and since the initial feasible
solution can be improved further, we need to develop the second tableau in order to find the
optimal solution. In developing the second tableau, we should compute for revised values of the
constraint equations, the Zj row and the Cj-Zj row and remember that the variables in solution
will have a unit vector, with a value of 1 in the intersection of the column and the row of the
basis, i.e. the pivot element.
To obtain a unit vector in the basis column, we perform elementary row operations resulting in
new row values by either multiplying/dividing all the elements in a row by a constant or
adding/subtracting the multiple of a row to or from another row.
In computing for the new row values from our initial simplex tableau, we first multiply the
second constraint by ½ obtaining the values as follows:
1X1+1/2X2+0S1+1/2S2+0S3 = 11, which results in making the pivot element 1.
Next, we multiply the above new row values by 4 and subtract it from the first constraint
obtaining the following results:
0X1+8X2+1S1+2S2+0S3 = 56.
Then, we multiply the new row values in the pivot row by 3 and subtract it from the third
constraint resulting as follows:
0X1+3/2X2+0S1-3/2S2+1S3 = 6.
Having these new row values, we develop the second simplex tableau as shown below.
Second Simplex Tableau
Cj
Basic V.
S1 0
X2 60
S3 0
Zj
Cj-Zj
To determine the leaving the variable, we divide the values in the pivot column by their
corresponding row values in the quantity column. The result obtained, as shown in the table
below indicates that S3 is the leaving variable with the smallest non-negative ratio. This means
that S3 is the most limiting resource for how much units of X2 can be made.
Basic V
S1
X2
S2
Pivot Row
Pivot Column Pivot Element
After identifying the entering and the leaving variables, we perform elementary row operations
as follows. To obtain a unit vector with 1 in the pivot element, we multiply the pivot row by 2/3
resulting as follows:
0X1+1X2+0S1+-1S2+2/3S3 = 4
Then, multiply the above new row values by 8 and subtract it from the first constraint (row)
resulting as follows:
0X1+0X2+1S1+6S2+-16/3S3 = 24.
For row 2, first multiply the new row values in the pivot row by ½ and subtract it from the
second row resulting as follows:
1X1+0X2+0S1+1S2+-1/3S3 = 9.
And the third tableau looks like the following.
+MAi in case of minimization. Use the same constant M for all artificial variables.
Form the simplex tableau for the modified problem.
Solve the modified problem using the simplex method.
Relate the solution of the modified problem to the original problem:
If the modified problem has no solution, then the original problem has no
solution.
If any artificial variables are non-zero in the solution to the modified problem,
then the original problem has no solution.
Note. The artificial variables are introduced for the limited purpose of obtaining an initial
solution and are required for the constraints of >= type or the constraints with '=' sign. It is not
relevant whether the objective function is of the minimization or the maximization type.
Obviously, since artificial variables do not represent any quantity relating to the decision
problem, they must be driven out of the system and must not remain in the final solution (and if
at all they do, it represents a situation of infeasibility). This can be ensured by assigning an
extremely high cost to them. Generally, a value M is assigned to each artificial variable, where
M represents a number higher than any finite number. For this reason, the method of solving the
problems where artificial variables are involved is termed as the 'Big M Method'. When the
problem is of the minimization nature, we assign in the objective function a coefficient of +M
to each of the artificial variables. On the other hand, for the problems with the objective
function of maximization type, each of the artificial variables introduced has a coefficient -M.
In case of equality constraints, a slack variable is not acceptable since these constraints require
a precise amount. To overcome this problem, artificial variables are introduced which are added
to = and >= constraints. They have no physical representation for their purpose is to allow the
use of simplex process. As a result, they should not appear in the final optimal solution.
Assignment of coefficients for artificial variables depends on the type of the problem, whether
it is a maximization or a minimization problem. In case of maximization problem, a large
negative contribution or coefficient in the objective function would ensure the removal of
artificial variables from the optimal solution, commonly represented by -M. For minimization
problems, introducing a large positive coefficient would ensure the non-existent of artificial
variables in the optimal solution, commonly represented by M.
In addition to this, a slack is not allowed in >= constraints since they can only happen an excess
amount that is more than the minimum amount of a constraint. This excess amount is
represented by a surplus variable and is subtracted from the constraint.
Example – Maximization Problems with Mixed Constraints
Assume the following maximization problem with mixed constraints.
Max. Z = 6X1+8X2
Subject to X2 <= 4
X1+X2 = 9
6X1+2X2 >= 24
In solving for this problem, introducing a slack variable is not acceptable since they represent
unused capacity and there is no unused capacity in = and >= constraints. Therefore, for >=
constraints, we introduce a surplus variable and for both >= and = constraints we introduce
artificial variables resulting as follows.
Max Z = 6X1+8X2+0S1+0S3-MA2-MA3
Subject to X2+ S1 = 4
X1+X2+A2 = 9
6X1+2X2-S3+A3 = 24
X1, X2 > = 0
Wouldn’t it be possible to set A2 as A1 and A3 as A2 in the above standardized LPM form?
slack variables and/or surplus variables for the initial solution. Therefore the initial tableau for
the problem is represented as follows.
Since this is a maximization problem, the entering variable is the one with the maximum C j-Zj
value. Therefore, the variable with the maximum C j-Zj is X1, the decision variable. And the
leaving variable is with the minimum non-negative ratio which is A 3, the artificial variable.
Since this artificial variable is not needed we remove it from the next tableau.
Developing the Second Simplex Tableau
After identifying the entering and the leaving variable, the usual elementary row operations are
performed to obtain a unit vector in the entering variable column. The end result after
performing the row operations is as follows which shows the second tableau.
Second Simplex Tableau
Cj 6 8 0 0 -M Quantity
Basic V. X1 X2 S1 S3 A2
S1 0 0 1 1 0 0 4
A2 -M 0 2/3 0 1/6 1 5
X1 6 1 1/3 0 -1/6 0 4
Zj 6 2-2/3M 0 -1-M/6 -M 24-5M
Cj-Zj 0 6+2/3M 0 1+M/6 0
Again here, we identify the leaving the entering variables, i.e. the pivot row and the pivot
column respectively. The entering variable with the highest C j-Zj row value is X2 and the
leaving variable with the smallest ratio is S1.
Similarly, we identify the entering and the leaving variables which are S 3 and A2 respectively
representing the maximum Cj-Zj value and the minimum ratio.
This tableau represents the final tableau since we have only zeros and negative values in the C j-
Zj row which indicates that it is the optimal solution. So we have the following results for each
of the variables and the profit obtained.
X1 = 5, X2 = 4, S3 = 14, and Profit = 62
objective function point of view. If objective function Z is to be minimized, then a very large
positive price (called penalty) is assigned to each artificial variable. Similarly, if Z is to be
maximized, then a very large negative cost (also called penalty) is assigned to each of these
variables. Following are the characteristics of Big-M Method:
(ii) Next we determine the departing variable from the basic solution. If an artificial
variable goes out of solution, then we discard it totally and even this variable may
not form part of further iterations. Same procedure, as in maximization case, is
employed to determine the departing variable.
Step 5. We update the new solution now. We evaluate the entries for next simplex table in
exactly the same manner as was discussed earlier in the maximization case.
We introduce both surplus and artificial variables into both constraints as follows.
Min Z = 7X1+9X2+0S1+0S2+MA1+MA2
Subject to 3X1+6X2 –S1+A1 = 36
8X1+4X2-S2+A2 = 64
X1, X2 > = 0
So the subsequent tableaus for this problem are shown below. To remind in these tableaus is in
transforming from one tableau to another, we perform elementary row operations to obtain the
unit vector in the pivot column for the entering variable into the solution.
Initial Simplex Tableau
Cj 7 9 0 0 M M Quant
Basic X1 X2 S1 S2 A1 A2 ity
V.
A M 3 6 -1 0 1 0 36
1
A M 1 4 0 -1 0 1 64
2
Basic V. X1 X2 S1 S2 A1
A1 M 0 9/2 -1 3/8 1 12
X1 7 1 ½ 0 -1/8 0 8
Zj 7 7/2+9/2M -M 3/8M-7/8 M 56+12M
Cj-Zj 0 11/2-9/2M M 7/8-3/8M 0
The third tableau represents a final tableau since it is the optimal solution with entirely zeros
and non-negative values in the Cj-Zj row.
Therefore, the optimal solution is: X 1 = 20/3 and X2 = 8/3 and value of objective function is
212/3.
Summary
? Dear learner, why do you think that artificial variables did not appear in the optimal
solution? What is the guarantee not to get artificial variable in the optimal solution?
A situation with no feasible solution may exist if the problem was formulated incorrectly.
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
Whenever the optimality criteria is satisfied but still there exist an artificial variable in the basis
or solution mix, this is the indication of infeasibility.
Example:
Minimization case
Cj 5 8 0 0 M
BV X1 X2 S1 S2 A2 Q
5 X1 1 1 -2 3 0 200
8 X2 0 1 1 2 0 100
M A2 0 0 0 -1 1 20
Zj 5 8 -2 31-M M 1,800+200M
Cj - Zj 0 0 2 M-31 0
Even though all Cj - Zj are positive or 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.
Unbounded Solution
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
unboundedness 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 unbounded solution is to divide each quantity column number by the
corresponding pivot column number. 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.
Note: A negative ratio means that increasing that variable would increase resources. A zero
ratio means that increasing the variable would not use any resources.
Example:
Maximization case:
Cj 6 9 0 0
SV X1 X2 S1 S2 Q
9 X2 -1 1 2 0 30 RR
30/-1=-30
0 S2 -2 0 -1 1 10 Unacceptable RRs
Zj -9 9 18 0 270 10/-2=-5
Cj - Z j 15 0 -18 0
Pivot Column
The solution in the above case is not optimal because not all Cj - Zj 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 column
numbers to their corresponding numbers in the X1 or pivot column. Since both pivot column
numbers are negative, an unbounded solution is indicated.
Example
5 8 2 0 0 0
Cj
Q
SV X1 X2 X3 S1 S2 S3
8 X2 1/4 1 1 -2 0 0 10 RR
0 S2 4 0 1/3 -1 1 0 20 10/1/4=40
0 S3 2 0 2 2/5 0 1 10
20/4= 5 Tie for the smallest ratio
Indicates degeneracy.
Zj 2 8 8 16 0 0 80
10/2= 5
Cj - Z j 3 0 -6 -16 0 0
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. How
ever, the number of iterations required to arrive at the optimal solution can be minimized by
adopting the following rule:
Divide the coefficient of slack variables in the simplex table where degeneracy is
detected by the corresponding positive numbers of the key column in the row, starting
from left to right.
The row which contains smallest ratio comparing from left to right column wise
becomes the key row.
For the above example:
Column
Row s2 s3
s2 1 0
s3 0 1
ADU –Department of Business Administration Page 64 of 214
Operations Research
Divide the coefficients by the corresponding element of the key column, we obtain the
following:
Column
Row s2 s3
s2 ¼= ¼ 0/4=0
s3 0/2 =0 ½=1/2
Comparing the ratios from left to right column wise until they are not equal, the minimum ratio
occurs for the second row (0). Therefore, s3 is selected to leave the basis.
Note: 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.
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.
3. If there is a tie between slack or surplus variable, then selection can be made arbitrary.
Example:
If the equation is max Z:
Cj
SV X1 X2 S1 S3 Q
Zj
Cj - Z j 5 2 5 0
non basic variable (variable not in the basis) has a zero value in C-Z row, this is the indication
of existence of multiple solution. To obtain the other solution, we will make a non- basic
variable with a zero C-Z value to enter in to the basis and solve in the same way as the one we
did in the previous discussion.
Example:
Maximization problem
Cj 3 2 0 0
BV X1 X2 S1 S2 Q
2 X2 3/2 1 1 0 6
0 S2 1 0 1/2 1 3
Zj 3 2 2 0 12
Cj - Z j 0 0 -2 0
MaxZ=3X1+2X2
X1=0, X2=6, S2=3 and MaxZ=12 or: X1=3, X2=3/2 and MaxZ=12
The Cj - Zj value of the non-basic variable (X1) is 0. Thus, this shows the existence of
alternative optimal solution. Can you identify more alternative optimal solutions?
? Dear learner, can you raise examples of your own for each of the above special issues
and show how each relate to graphic approach?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
___
ADU –Department of Business Administration Page 66 of 214
Operations Research
Summary
The simplex method is an algebraic procedure that starts with a feasible solution that
is not optimal and systematically moves from one feasible solution to another until
an optimal solution is found
The variations described in general simplex approach include; maximization and
minimization problems, mixed constraint problems, problems with multiple optimal
solution, problems with no feasible solution, undounded problems, tied pivot
columns and rows.
Multiple optimal solution are identified by Cj-Zj(or Zj- Cj) =0 for a non basic
variable. To determine the alternate solution(s), enter the non basic variable(s) with
a Cj-Zj value equal to zero.
An infeasible problem is identified in the simplex procedure when an optimal
solution is achieved and one or more of the basic variables are artificial.
The unbounded problems are identified in the simplex procedure when it is not
possible to select a pivot row- that is, when the replacement ratio is negative or
undefined.
Degeneracy occurs when there is a tie for pivot row.
Activity
1. The crumb and Custard bakery makes cakes and pies. The main ingredients are flour and
sugar. The following linear programming model has been developed for determining the
number of cakes and pies (x1 and x2 to produce each day to maximize profit.
Max Z=x1 + 5x2
Subject to:
8x1+10x2< 25(flour, lb)
2x1+4x2< 16(sugar, lb)
x1< 5(demand for cakes)
x1,x2 >0
Solve this model using graphical and simplex method and compare the answers.
2. A transistor radio Co., manufactures models A, B, and C which have profit contribution of
Birr 16, Birr 30 and Birr 50 respectively. The weekly minimum production requirements are 20
for model A, 120 for model B, and 60 for model C. Each type of radio requires a certain amount
of time for the manufacturing of component parts, assembly and packaging. Specifically a
dozen unit of model A requires 3 hrs for manufacturing of component parts, 4b hrs for
assembly and 1 hr for packaging. The corresponding figure for a dozen unit of model B is 3.5, 5
and 1.5 and for a dozen unit of C are 5, 8, and 3 hrs. During the forth coming week, the
company has availability of 120 hrs of manufacturing, 160 hrs of assembly, and 45 hrs of
packaging time.
Required:
a) Formulate the scheduling problem as LPM.
b) Solve the problem using simplex method
3. A company has two plants, each of which produces and supplies two products: A and B. The
plants can each work up to 16 hrs a day. In plant 1, it takes 3 hrs to prepare and pack 1000
gallons of A and 1 hr to prepare and pack 1 quintal of B. In plant 2, it takes 2 hrs to prepare and
pack 1000 gallons of A and 1.5 hrs to prepare and pack a quintal of B. In plant 1, it costs Birr
15,000 to prepare and pack 1000 gallons of A and Birr 28,000 to prepare and pack a quintal of
B, whereas these costs are Birr 18,000 and Birr 26,000 respectively in plant 2. The company is
obliged to produce daily at least 10,000 gallons of A and 8 quintals of B.
a) Formulate the problem as LPM to find out as to how the company should organize its
production so that the required amounts of the two products be obtained at minimum cost.
b) Solve the problem using simplex method
Type of product Profit contribution per unit (Birr) Assembly time per product (hrs)
A 12 0.8
B 20 1.7
C 45 2.5
The company has a daily order commitment for 20 units of product A, and a total of 15 units of
product B and C.
Required:
a) Formulate the problem as LPM so as to maximize the total profit?
b) Solve the problem using the algebraic method
Chapter Three
3. Post optimality Analysis
Dear learner, in the previous chapters we discussed about how to solve LP problems using
either graphic or simplex approach which is a base for this section. This unit deals with the
other version of the simplex model, dual and the sensitivity of the optimal solution to changes
in different parameters.
Objectives:
Up on completing this chapter, the learners would be able to:
Explain what post optimality analysis is
Describe the dual and primal forms of LP problems
Conduct sensitivity analysis on different parameters
3.1. DUALITY
Section objective:
Up on completion of this sub section the learner will be able to:
Define duality
Describe the importance of duality
Explain how to convert dual in to primal and the vice versa
Formulate dual when the constraints are mixed
Elucidate how to read the solution for primal from dual’s solution and vice versa.
Analyze the impact of addition of a new product on the decision
The term ‘dual’ in a general sense implies two or double. Every linear programming problem
can have two forms. The original formulation of a problem is referred to as its Primal form.
The other form is referred to as its dual LP problem or in short dual form. In general, however,
it is immaterial which of the two problems called primal or dual, since the dual of the dual is
primal.
The dual involves setting up and solving an LP problem that is almost a ‘mirror image’ of an
LP problem that has been formulated. Both in its formulation and solution, the dual is the flip
flop version of the primal.
In the context of LP, duality implies that each LP problem can be analyzed in two different
ways, but having equivalent solution. For example, consider the problem of production
planning. By using the primal LP problem, the production manager attempts to optimize
resource allocation by determining quantities for each product to be produced that will
maximize profit. But through dual LP problem approach, he attempts to achieve production
plan that optimizes resource allocation so that each product is produced at that quantity such
that its marginal opportunity cost equals its marginal return. Thus, the main focus of dual is to
find for each resource its best marginal value or shadow price. This value reflects the scarcity of
resources, i.e., the maximum additional prices to be paid to obtain one additional unit of the
resources to maximize profit under the resource constraints. If resource is not completely used,
i.e., there is slack, then its marginal value is zero.
The shadow price is also defined as the rate of change in the optimal objective function value
with the respect to the unit change in the availability of a resource. Precisely for any constraint,
we have,
Analysis of the dual can also enable a manager to evaluate the potential impact of a new
product, and it can be used to determine the marginal values of resources (i.e. constraints).
Relative to a product, a manager would want to know what impact adding a new product would
have on the solution quantities and the profit; relative to resources, a manager can refer to a
dual solution to determine how much profit one unit of each resource equivalent to. Whereas
the primal gives solution results in terms of the amount of profit gained from producing
products, the dual provides information on the value of the constrained resources in achieving
that profit.
? Often the manager is less concerned about profit than about the use of resource
in dual. Do you agree? Dear learner, would you discuss with your class mates.
____________________________________________________________________
____________________________________________________________________
________________________________________________________________
The above pair of LP problems can be expressed in the general LP model form as
Primal Dual
Max Zx = Min Zy =
aij = aji
and xj 0; j = 1,2,…,n and yi 0; j = 1,2,…,m
The following rules which guide the formulation of the dual problem will give you the
summary of the general relationship between primal and dual LP problems.
1. If the primal’ objective is to minimize, the dual’s will be to maximize; and the vice
versa
2. The coefficient’s of the primal’s objective function become the RHS values for the
dual’s constraints.
3. The primal’s RHS values become the coefficients of the dual’s objective function.
4. The coefficients of the first “row” of the primal’s constraints become the coefficients of
the first “column” of the dual’s constraint …..
5. The ≤ constraints become ≥ and the vice versa.
Consider this Primal problem:
The following table shows how the primal problem is transformed into its dual.
Primal Dual
Objective Minimize 40x1 + 44x2 + 48x3 Maximize 20y1 + 30y2
function and Subject to Subject to
right-hand 1 20 1 40
side values 2 30 2 44
3 48
Primal Dual
Constraint 1 1x1 + 2x2 + 3x3 1 1y1 + 4y2
coefficients 2 4x1 + 4x2 + 4x3 2 2y1 + 4y2
3 3y1 + 4y2
We can see from the table that the original objective was to minimize, whereas the objective of
the dual is to maximize. In addition, the coefficients of the primal’s objective function become
the right-hand side values for the dual’s constraints, whereas the primal’s right-hand side values
become the coefficients of the dual’s objective function.
Note that the primal has three decision variables and two constraints; whereas the dual has two
decision variables and three constraints.
The constraint coefficients of the primal are constraint coefficients of the dual, except that the
coefficients of the first “row” of the primal become the coefficients of the first “column” of the
dual, and the coefficients of the second “row” of the primal become the coefficients of the
second “column” of the dual.
When the primal problem is a maximization problem with all < constraints, the dual is a
minimization problem with all > constraints.
3.1.2. Formulating the Dual when the Primal has Mixed Constraints
In order to transform a primal problem into its dual, it is easier if all constraints in a
maximization problem are of the < variety, and in a minimization problem, every constraint is
of the > variety.
To change the direction of a constraint, multiply both sides of the constraints by -1. For
example,
-1(2x1 + 3x2 > 18) is -2x1-3x2 < -18
If a constraint is an equality, it must be replaced with two constraints, one with a < sign and the
other with a > sign. For instance,
4x1 + 5x2 = 20
will be replaced by
4x1 + 5x2 < 20
4x1 + 5x2 > 20
Then one of these must be multiplied by -1, depending on whether the primal is maximization
or a minimization problem.
EXAMPLE:
Formulate the dual of this LP model.
Maximize z = 50x + 80x
Subject to:
C 3x + 5x ≤ 45
C 4x + 2x ≥16
C3 6x +6x = 30
x ,x ≥ 0
SOLUTION
Since the problem is a max problem, put all the constraints in to the ≤ form. Subsequently, C
and C3 will be first adjusted in to ≤ constraints.
- C will be multiplied by -1:
-1(4x + 2x ≥16) becomes -4x - 2x ≤ -16
- C3 is equality, and must be restated as two separate constraints. Thus, it becomes:
6x +6x ≤ 30 and 6x +6x ≥30. Then the second of these must be multiplied by
-1.
-1(6x +6x ≥30) becomes -6x -6x ≤ -30
After making the above adjustments, rewrite the LP model again.
Maximize z = 50x + 80x
Subject to:
C 3x + 5x ≤ 45
C -4x - 2x ≤ -16
C3 6x +6x ≤ 30
C4 -6x -6x ≥ -30
x ,x ≥0
? Dear learner, would you guess the reason why the RHS quantity consists of
negative values and do you think that it contradicts with rules of LP?
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
Cross -referencing the values in the primal and dual final simplex tableaus is shown as follows.
Example:
To show that the flip-flopping of values between the primal and the dual carries over to their
final simplex tableaus, let us look at the following tables. The first table contains the final
tableau for the dual and the second one contains the final tableau for the primal.
Let’s consider how we can obtain the answers to these questions from the dual solution. Notice
that the solution quantities of the dual are equal to the shadow prices of the primal (i.e. 40/3 and
10). Next, notice that the values of the solution quantities of the primal (i.e., 24, 9, and 4) can
be found in the bottom row of the dual. Now, in the primal solution, s 1 equals 24. In the dual,
the 24 appears in the y1 column. The implication is that a slack variable in the primal solution
becomes a real variable (i.e., a decision variable) in the dual. The reverse is also true: A real
variable in the primal solution becomes a slack variable in the dual. Therefore, in the primal
solution we have x1 = 9 and x2 = 4; in the dual, 9 appears under s 1 in the bottom row, and 4
appears under s2. Thus, we can read the solution to the primal problem from the bottom row of
the dual: In the first three columns of the dual, which equate to slack variables of the primal, we
can see that the fist slack equals 24 and the other two are zero. Under the dual’s slack columns,
we can read the primal values of the decision variables. In essence, then, the variables of the
primal problem become the constraints of the dual problem, and vice versa.
The manager of the firm would reason in the following way. For each unit of Model I that the
firm sacrifices to produce computers for the department store, it will gain 4 hours of assembly
time, 2 hours of inspection time, and 3 cubic feet of storage space, which can be applied to the
store computers. However, it will also give up a unit profit of $60. Therefore, in order for the
firm to realistically consider the store’s offer, the amounts of scarce resources that will be given
up must produce a return to the firm that is at least equal to the foregone profit. Hence, the
value of 4 assembly hours + 2 inspection hours + 3 cubic feet of storage space > $60. By
similar reasoning, giving up one unit of Model II will require that the value received by giving
up 10 assembly hours + 1 inspection hour + 3 cubic feet of storage must equal or exceed the
Model II profit of $50 per unit. These, then, become the constraints of the dual problem. Thus:
Hence the constraints of the dual refer to the value of capacity (i.e., the scarce resources). The
formulation indicates that in order to switch from making units of Models I and II to making
computers for the department store, the value received from that switch must be at least equal to
the profit foregone on the microcomputer models. The variables y 1, y2 and y3 are the marginal
values of scarce resources (assembly time, inspection time, and storage space). Solving the
dual will tell us the imputed values of the resources given our optimal solution.
Naturally, the department store would want to minimize the use of the scarce resources, because
the computer firm almost certainly would base its charges on the amount of resources required.
Consequently, the objective function for the dual problem focuses on minimizing the use of the
scarce resources. Thus:
Looking at the optimal solution to the dual of the microcomputer problem, we can see the
marginal values of y2 and y3 in the Quantity column, but not the value of y 1. This is because the
optimal solution to the primal did not completely use up all of the assembly capacity.
Consequently, no amount of either x 1 or x2 would need to be given up to obtain one free hour of
assembly time. Thus, the marginal value of one hour of assembly time is $0.
Finally, the optimal dual solution always yields the same value of the objective function as the
primal optimal. In this case, it is 740. The interpretation is that the imputed value of the
resources that are required for the optimal solution equals the amount of profit that the optimal
solution would produce.
Conversely, if the dual constraint has not been satisfied, the new variable would have come into
solution. In this instance, the marginal value of scarce resources that would be required for the
new model exceeds the marginal contribution to profit that the new model would provide. That
is, Birr106.67 > Birr70.
Note: The value of this approach is that it is not necessary to rework the entire problem in order
to test the potential impact that adding a new decision variable would have on the optimal
solution.
Activity
Convert the following in to dual and solve it using simplex approach
1. Min Z= 4x1 +x2
Subject to:
3 x1+ x2 = 3
4x1+ 3x2 > 6
x1+ 2x2 < 3
x1,x2 > 0
Under this section, we will discuss about sensitivity analysis which involves an examination of
the potential impact of changes in any of the parameters (i.e. numerical values) of a problem
(e.g., a change in the amount of scarce resource available).
When LP models are formulated, the input data of the model (also called parameters) such as:
(i) profit (or cost) contribution (Cj) per unit of decision variable, ii) availability of resources(bi),
and (iii) consumption of resources per unit of decision variables (a ij), implicitly assumed
constant and known with certainty during a planning period. However, rarely does a manager
know all of these parameters exactly. In reality the model parameters are simply estimates (or
best “guesses”) that are subject to change. For this reason it is of interest to the manager to see
what effect a change in a parameter will have in the solution to the model. Changes may be
either to the reactions to anticipated uncertainties in the parameters or reactions to the
information gained from the dual. The analysis of the parameter changes and their effects on the
model solution is known as sensitivity analysis.
? Dear learner, discuss with your group members about why sensitivity analysis is
conducted?
__________________________________________________________________________
__________________________________________________________________________
_________________________________________________________________
Section Objective
Up on completion of this section, the learner will be able to:
Define sensitivity analysis
Describe the significance of sensitivity analysis
Conduct sensitivity analysis on Two parameters; namely
o Coefficient of OF- when the variable is in the solution mix and not
o RHS
Sensitivity analysis carries the LP analysis beyond the determination of the optimal solution
and begins with the final simplex tableau. Its purpose is to explore how changes in any of the
parameters of a problem, such as the coefficients of the constraints, coefficients of the objective
function, or the right hand side values; would affect the solution. For this, instead of resolving
the entire problem as anew problem with new parameters, we may consider the original optimal
solution as an initial solution for the purpose of knowing the ranges, both lower and upper,
within which a parameter may assume a value.
For instance, a manager may want to consider obtaining an additional amount of scarce
resource that might be available, in which case the manager would want to know the answers to
questions such as:
b. Will an increase in the right hand side of this constraint affect the objective function?
c. If so, how much of the resource can be used?
d. Given the answer to the preceding question, what will be the revised optimal solution?
Conversely, the decision maker might be facing a situation in which the amount of scarce
resource available is less than the original amount, in which case the issues would be:
a. Will the decreased level of the resource have an impact on the value of the objective
function?
b. If yes, how much of an impact will it have, and what will be the revised optimal
solution?
The process of studying the sensitivity of the optimal solution of an LP problem is also called
post optimality analysis because it is done after an optimal solution, assuming a given set of
parameters, has been obtained for the model.
The discussion here covers only two types of changes: changes in the right hand side levels of
the constraints, and changes in the coefficients of the objective function as change in the
coefficient of constraints is beyond the coverage of this material.
The most obvious way to ascertain the effect of a change in the parameter of the model is to
make the change in the original model, resolve the model, and compare the solution result with
the original solution. However, resolving a problem can be very time consuming and as it will
be demonstrated below, it is unnecessary. In most cases the effect of changes in the model can
be determined directly from the final simplex tableau.
to examine the shadow prices in the final simplex tableau. Shadow prices are the values in the Z
row in the slack columns.
The final tableau for the microcomputer is shown in the following table since sensitivity
analysis starts from final tableau.
Cj 60 50 0 0 0 Quantity
Basis 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
Negatives of
shadow prices
A shadow price is a marginal value; it indicates the impact that a one-unit change in the amount
of a constraint would have on the values of the objective function. As we can see from the table,
the shadow prices are $0 for s1 (Assembly time), $10 for s2 (Inspection time), and $40/3 for s 3
(Storage space). These tell us that an increase in the amount of assembly time would have no
effect on profit; if inspection time is increased by one hour, it will increase the profit by $10,
and if storage space is increased by 1 cubic foot, profit would increase by $40/3. The reverse
also holds. If we decrease them by such respective amounts, the decrease in profit will take the
same figure.
What shadow prices do not tell us is the extents to which the level of a scarce resource can be
changed and still have the same impact per unit. The ability to use additional amounts of a
resource will disappear at some point because of the fixed amounts of the other constraints.
Therefore, we need to determine the range over which we can change the right hand side
quantities and still have the same shadow price. This is called the range of feasibility, or the
right hand side range.
The key to computing the range of feasibility for the constraint lies in each slack column of the
final simplex tableau. To compute the range for each constraint, the entries in the associated
slack column must be divided into the values in the Quantity column. For example, for the
storage column values, the resulting ratios are:
24 9 4
-16/3 = -4.5 -⅓ = -27 ⅔ = +6
Here, the smallest positive ratio indicates how much the constraint level can be decreased
before it reaches the lower limit of its range of feasibility, and the smallest negative ratio (i.e.
the negative ratio closest to 0) indicates how much the storage constraint can be increased
before it reaches its upper limit of feasibility. The general rule applies when computing the
upper and lower limits on the range of feasibility for maximization problem is:
Cj 60 50 0 0 0 Quantity
Basis x1 x2 s1 s2 s3
s1 0 0 0 1 6 -16/3 24-16/3
x1 60 1 0 0 1 -1/3 9-1/3
x2 50 0 1 0 -1 2/3 4 + 2/3
Z 60 50 0 10 40/3 740
C-Z 0 0 0 -10 -40/3
Recall that the requirement of simplex method is that the quantity values can not be negative. If
any q1 value becomes negative, the solution no longer be feasible. Thus the inequalities
24-16 /3 > 0
9- /3> 0
4 + 2 /3 > 0
are solved for :
24-16 /3 > 0 9- /3> 0 4+2 /3 > 0
72-16 > 0 27- >0 12+2 >0
3 3 3
72-16 > 0 27- >0 12+2 >0
The value 66 can be eliminated since q3 must be less than 43.5; thus
33 < q3 < 43.5
As long as q3 remains in this range, the present basic solution variables remain positive and
feasible. However, the quantity value of those basic variables may change.
? Dear learner, can you determine the range within which q1&q2 vary with out causing
the basic solution mix change?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_____________________________________________________________________
Once we have computed these limits, we proceed to assess the impact that a contemplated
change would have on the optimal solution.
Example
The manager in the microcomputer problem is contemplating a change in the level of the
storage constraint – an increase of 3 cubic feet. Determine the revised optimal solution for the
change.
Solution
First, note that an increase of 3 cubic feet is within the range. Then, the effect of an increase of
three cubic feet is computed in the following way:
For an increase in storage space of 8 cubic feet, we take the upper limit of 4.5 cubic feet for the
computation. The resulting revised figures, then, would be s 1= 0, x1=7.5, x2=7, and Profit =800.
Note, however, that beyond the upper limit, s 3 would come into solution, replacing S 1,
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, and Profit = 800.
If a variable is not currently in the solution in a maximization problem, its objective function
coefficient would have to increase by an amount that exceeds the C-Z value for that variable in
the final tableau in order for it to end up as a basic variable in the optimal solution. The range
over which a non-basic variable’s objective function coefficient can change without causing
that variable to enter the solution mix is called its range of insignificance.
Range of Insignificance: the range over which a non-basic variable’s objective function
coefficient can change without causing that variable to enter the solution mix.
Maximize 3x + 2x + 5x
Subject to:
C1 x + 2x + 2x ≤ 18
C2 3x + 2x + 6x ≤ 12
C3 2x + 3x + 4x ≤ 12
x ,x x , ≥ 0
C 3 2 5 0 0 0
Basis x x x s s Quantity
s
s 0 0 0 0 1 1/5 -4/5 4/5
x 3 1 0 2 0 3/5 -2/5 12/5
x 2 0 1 0 0 -2/5 3/5 12/5
Z 3 2 6 0 1 0 12
C-Z 0 0 -1 0 -1 0
In the above tableau we see that x is a non- basic variable. To determine the range over which
the objective function coefficient of x would change without changing the optimal solution,
recall how a variable in a max problem enters the solution mix. Because x is not in solution, its
objective function coefficient would need to increase in order for it to come in to solution (i.e.,
to make its c-z value positive). The amount of increase must be greater than the absolute
value of its c-z value, which is │-1│. Therefore, its objective function coefficient must increase
by more than 1. Hence, the range of insignificance for x is 6 or less (i.e, its Z value). In general
the rule is: Range of insignificance for a non- basic variable is the Z value of the variable. The
allowable increase is the absolute value of its own C- Z value.
Range of Optimality: the range over which objective function coefficient of a variable that is
in solution can change with out changing the optimal values of the decision variables. Note,
however, that such a change would change the optimal value of the objective function.
For variables that are in solution, the determination of the range of optimality (the range over
which the objective function coefficient of a variable that is in the solution can change without
changing the optimal values of the decision variables) requires a different approach. Divide the
values in row C-Z by the corresponding row values of the variable in question and follow the
following rule for both maximization and minimization problems.
Allowable increase: The smallest positive ratio of C-Z value and the
variable’s substitution rate.
Allowable decrease: the smallest negative ratio of C-Z value and the
variable’s substitution rate.
ADU –Department of Business Administration Page 83 of 214
Operations Research
Note: If there is no positive ratio, it means that there is no upper limit on that variable’s
objective function coefficient.
Example
Determine the range of optimality for the decision variables in the microcomputer problem.
Solution
The final simplex tableau for that problem is repeated here for convenience.
Cj 60 50 0 0 0
Basis x1 x2 s1 s2 s3 Quantity
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
For x1 we find:
Column x1 x2 s1 s2 s3
C-Z value 0/1 = 0 0/0 = Undefined 0/0 = Undefined -10/1 = -10 -40/3 = +40
x1 row value -1/3
The smallest positive ration is +40. Therefore, the coefficient of X1 can be increased by Birr 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 (Birr60 +Birr40) = Birr100. Also, the
smallest negative ratio is -10; therefore, the x 1 coefficient can be decreased by as much as
Birr10 from its current value, making the lower end of the range equal to (Birr60 - Birr10 )=
Birr 50.
For x2 we find:
Column x1 x2 s1 s2 s3
C-Z value 0/0 = Undefined 0/1= 0 0/0 = Undefined -10/-1 = +10 -40/3 = -20
x2 row value ⅔
The smallest positive ration is +10. This tells us that the x 2 coefficient in the objective function
could be increased by Birr10 to (Birr50 + Birr10) = Birr 60. The smallest negative ratio is -20,
which tells us that the x2 coefficient could be decreased by Birr20 to (Birr50 - Birr20) = Birr30.
Hence, the range of optimality for the objective function coefficient of x2 is Birr30 to Birr 60.
This can also be done in the same way as change in the constraints by using the concept of
change starting from the final tableau. That is, the range of c j that will maintain the optimal
solution can be determined directly from the optimal simplex tableau.
Cj 60+ 50 0 0 0
Basis x1 x2 s1 s2 s3 Quantity
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
Consider a change for c1. This will change the c1 value from c1=60 to c1=60+ . This new
value is not only included on the top cj row, but also in the left-hand cj column. Since 60+ is
in left-hand column, it becomes a multiple of the column values when the new Zj row values
and the subsequent Cj – Zj row values.
The solution becomes optimal as long as the Cj –Zj row values remain negative. If Cj-Zj
becomes positive, the product mix will change, if it becomes zero, there will be an alternative
solution. Thus, for the solution to remain uniquely optimal,
-10- < 0
and
(-40+ )/3 <0
Both of these inequalities must be solved for .
-10- < 0 and (-40+ )/3 <0
- < 10 (-40+ )<0
> -10 < 40
Now recall that c1=60+ ; therefore, = c1-60. Now substituting c 1-60 for in the above
inequalities, we get:
> -10 < 40
C1-60 > -10 c1-60 < 40
C1> 50 C1<100
Therefore, the range of values for C1 over which the solution basis remain optimal (although) the
value of the objective function may change) is:
50<C1<100
? Dear learner, would you try to determine the range over which coefficient of X 2 in
the objective function change without changing the solution mix in the same way as
above?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
Summary
__ Every linear programming problem can have two forms. The original formulation
of a problem is referred to as its Primal form. The other form is referred to as its
dual LP problem or in short dual form.
The main focus of dual is to find for each resource its best marginal value or
shadow price. This value reflects the scarcity of resources, i.e., the maximum
additional prices to be paid to obtain one additional unit of the resources to
maximize profit under the resource constraints.
If the primal’ objective is to minimize, the dual’s will be to maximize; and the
vice versa; the coefficient’s of the primal’s objective function become the RHS
values for the dual’s constraints; the primal’s RHS values become the
coefficients of the dual’s objective function; the coefficients of the first “row”
of the primal’s constraints become the
coefficients of the first “column” of the dual’s constraint …..; the ≤
constraints become ≥ and the vice versa.
Sensitivity
ADU –Department of Businessanalysis is the study of sensitivity of the optimal solution
Administration Page 85 ofof
214an LP
problem due to discrete variations (changes) in its parameters. The degree of
sensitivity of the solution due to these changes can range from no change to all to
a substantial change in the optimal solution of the given LP problem.
The purpose of sensitivity analysis purpose is to explore how changes in any of
the parameters of a problem, such as the coefficients of the constraints,
coefficients of the objective function, or the right hand side values; would affect
the solution.
Operations Research
Activity
1. ABC Metal crafters company makes brass tray and buckets. The number of trays (x1 )and
buckets (x2) that can be produced daily is constrained by the availability of brass and labor, as
reflected in the following LP model.
Maximize Z = 6x1 + 10x2
Subject to:
x1 + 4x2 90 ( brass, lb)
2x1 + 2x2 960 (labor, hrs)
x1,x2 0
The final optimal simplex table for this model is as follows:
6 10 0 0
Basic variables cj x1 x2 s1 s2 quantity
x2 10 0 1 1/3 -1/6 20
x1 6 1 0 -1/3 2/3 10
2. A Manufacturing firm produces electric motors for washing machines and vacuum cleaners.
The firm has resource constraints for production time, steel and wire. The LP model
determining the number of washing machine motors (x1) and vacuum cleaner motors (x1) to
produce has been formulated as follows.
Maximize Z = 70x1 + 80x2
Subject to:
2x1 + x2 19 ( production, hrs)
ADU –Department of Business Administration Page 86 of 214
Operations Research
x1 + x2 14 (steel, lb)
x1 + 2x2 20 (wire, ft)
x1,x2 0
The final simplex table for this model is as follows.
70 80 0 0
Basic variables cj x1 x2 s1 s2 s3 quantity
x1 70 1 0 2/3 0 -1/3 6
s2 0 0 0 -1/3 1 -1/3 1
x2 0 0 1 -1/3 0 2/3 7
Zj 70 80 20 0 30 980
Cj –Zj 0 0 -20 0 -30
3. The evergreen products firm produces three types of pressed paneling from pine and spruce.
The three types of paneling are wide(x1), medium (x2) and small(x3). Each sheet must be cut and
pressed. The resource requirements are given in the following LP formulation.
4 10 8 0 0 0
Basic variables cj x1 x2 x3 s1 s2 s3 s4 quantity
s1 0 7/3 0 0 1 0 -4/3 -2/3 80
s2 0 -1/3 0 0 0 1 1/3 -4/3 70
x3 8 1/3 1 0 0 0 2/3 1/6 20
x2 10 1/3 1 0 0 0 -1/3 1/3 10
Zj 6 10 8 0 0 2 2 260
Cj –Zj 0 0 0 0 0 -2 -2
a. What is the marginal value of an additional pound of spruce? Over what range is this value
valid?
b. What is the marginal value of additional hours of cutting? Over what range is this value
valid?
c. Given a choice between securing more hours and or more pressing hours, which should the
management select?
d. If the amount of the spruce available to the firm were decreased from 160 to 100 pounds,
would this reduction affect the solution? Why? If it is what is the new solution?
e. What unit profit would have to be made from wide paneling before management consider
producing it?
f. Management is considering changing the profit of small paneling from Birr 8 to Birr 13.
would this change affect the solution?
4. A manufacturing firm produces four products. Each product requires material and machine
processing. The liner programming model formulated to determine the number of product 1
(x1)
Product 2 (x2) product 3 (x3), and product 4 (x4), to produce is as follows.
2 8 10 6 0 0
Basic variables cj x1 x2 x3 x4 s1 s2 quantity
x3 6 1 0 2 1 2/3 -1/3 80
x2 8 0 1 0 0 -1/3 2/3 40
a. What is the marginal value of an additional pound of material? Over what range is this value
valid?
b. What is the marginal value of additional hours of processing time? Over what range is this
value valid?
c. How much would the contribution to profit of x1 have to increase before x1 would be
produced?
d. Over what range the contribution of x2 change with out affecting the solution? Taking one of
the values within the range arbitrarily, calculate the value of the optimal solution?
e. Over what range machine processing hours change with out changing the solution mix?
taking one of the value within this range, calculate the revised optimal solution?
Chapter Four
4. Transportation and Assignment Models
Dear learners, in the previous chapters we have discussed about linear programming,
which is more general. In this chapter we will discuss about special types of linear
programming models called Transportation and Assignment Models
Unit objectives:
Up on completion of this unit, you will be able to:
Differentiate between linear programming and transportation and assignment models
Explain advantages of transportation and assignment models
Identify different techniques to find solution for both transportation and assignment
models.
Discuss special cases in transportation and assignment models
4.1. Transportation Models
Sub section objectives
Up on completion of this section, the learner will be able to:
Explain transportation models
Identify application areas of transportation models
Formulate LP model for transportation problems
Describe methods of finding initial feasible solution
Check for optimality of transportation model using stepping stone and MODI
techniques.
Elucidate special cases in transportation models
Introduction
Dear learner, in the previous chapter we have been thoroughly discussed about LP models and
their applications. In this chapter we will try to see special types of LP models which are
Transportation and Assignment Models. Though it is possible to solve the two problems using
simplex method, the process would result in rather large simplex tableaus and numerous
simplex iterations. Because of the unique characteristics of each problem, however, alternative
solutions methods requiring considerably less mathematical computation than the simplex
method have been developed.
A transportation problem typically involves a set of sending locations, which are referred to as
origins, and a set of receiving locations, which are referred to as destinations. In order to
develop a model of a transportation problem, it is necessary to have the following information:
a. Supply quantity (capacity) of each origin.
b. Demand quantity of each destination.
c. Unit transportation cost for each origin-destination route.
Assumptions
The transportation algorithm requires the assumption that:
All goods be homogeneous, so that any origin is capable of supplying any destination,
and
Transportation costs are a direct linear function of the quantity shipped over any route.
Though it will be modified later in our discussion, we shall add one additional
requirement that the total quantity available is equal to the total demand.
Example
Let’s consider an example. Harley’s Sand and Gravel Pit has contracted to provide topsoil for
three residential housing developments. Topsoil can be supplied from three different “farms”
as follows:
Farm Weekly capacity (cubic yards)
A 100
B 200
C 200
Demand for the topsoil generated by the construction projects is:
Project Weekly demand(cubic yards)
1 50
2 150
3 300
The manager of the sand and gravel pit has estimated the cost per cubic yard to ship over each
of the possible routes:
Cost per cubic yard to
From Project #1 Project #2 Project #3
Farm A Birr4 Birr 2 Birr 8
Farm B 5 1 9
Farm C 7 6 3
This constitutes the information needed to solve the problem.
The next step is to arrange the information into a transportation table. This is shown in the
following table.
5 1 9
Farm B 200
7 6 3
Farm C 200
drawback is that it does not take transportation costs into account. Consequently, such a
solution may require much additional effort to obtain the optimal solution.
The northwest corner method gets its name because the starting point for the allocation process
is the upper left-hand (Northwest) corner of the transportation table. For the Harley problem,
this would be the cell that represents the route from Farm A to Project #1. The following set of
principles guides the allocation:
i. Begin with the upper left-hand cell, and allocate as many units as possible to that
cell. This will be the smaller of the row supply and the column demand. Adjust
the row and column quantities to reflect the allocation.
ii. Remain in a row or column until its supply or demand is completely exhausted
or satisfied, allocating the maximum number of units to each cell in turn, until all
supply has been allocated (and all demand has been satisfied because we assume
total supply and demand are equal).
Initial Feasible Solution for Harley using Northwest-corner method
To:
Project Project Project Supply
From: #1 #2 #3
4 2 8
Farm A 50 50 100
(first) (second)
5 1 9
Farm B 100 100 200
(third) (fourth)
7 6 3
Farm C 200 200
(last)
The total cost is found by multiplying the quantities in “completed” (i.e. nonempty) cells by the
cell’s unit cost and, then, summing those amounts. Thus:
As noted earlier, the main drawback of the northwest-corner method is that it does not consider
cell (route) costs in making the allocation. Consequently, if this allocation is optimal, that can
be attributed to chance rather than the method used.
i. Identify the cell that has the lowest unit cost. If there is a tie, select one arbitrarily.
Allocate a quantity to this cell that is equal to the lower of the available supply for
the row and the demand for the column.
ii. Cross out the cells in the row or column that has been exhausted (or both, if both
have been exhausted), and adjust the remaining row or column total accordingly.
iii. Identify the cell with the lowest cost from the remaining cells. Allocate a quantity to
this cell that is equal to the lower of the available supply of the row and the demand
for the column.
iv. Repeat steps (ii) and (iii) until all supply and demand have been exhausted.
The initial feasible solution for the Harley problem completed using the above steps is shown
below.
Initial Feasible Solution for the Harley problem using the Intuitive approach
To:
Project Project Project Supply
From: #1 #2 #3
4 2 8
Farm A 50 50 100
5 1 9
Farm B 150 50 200
7 6 3
Farm C 200 200
We can easily verify that this is a feasible solution by checking to see that the row and column
totals of the assigned cell quantities equal the supply and demand totals for the rows and
columns. Now let us compute the total cost of this solution and compare it to that of the
northwest corner solution.
Total cost = 50(4) + 50(8) + 150(1) + 50(9) + 200(3) = $1800
Compared to the plan generated using the Northwest-corner method, this one has a total cost
that is $100 less. This is due to the fact that the previous one did not involve the use of cost
information in allocating units.
4.1.2.3. Vogel’s Approximation Method (VAM)
The third method for determining an initial solution, Vogel’s Approximation Method (also
called VAM), is based on the concept of penalty cost or regret. If a decision maker incorrectly
chooses from several alternative courses of action, a penalty may be suffered (and the decision
maker may regret the decision that was made). In transportation problem, the courses of action
are the alternative routes and a wrong decision is allocating to a cell that does not contain the
lowest cost.
This method is preferred over the other two methods because the initial feasible solution
obtained with VAM is either optimal or very close to the optimal. With VAM the basis of
allocation is unit cost penalty i.e. that column or row which has the highest unit cost penalty
(difference between the lowest and the next highest cost) is selected first for allocation and the
subsequent allocations in cells are also done keeping in view the highest unit cost penalty.
Steps in VAM
1. Construct the cost, requirement, and availability matrix i.e. cost matrix with column and
row information.
2. Compute a penalty for each row and column in the transportation table. The penalty is
merely the difference between the smallest cost and the next smallest cost element in
that particular row or column.
3. Identify the row and column with the largest penalty. In this identified row (column),
choose the cell which has the smallest cost and allocate the maximum possible quantity
to this cell. Delete the row (column) in which capacity (demand) is exhausted. When
there is a tie for penalty, select one arbitrarily. After allocation, cross that row or column
and disregard it from further consideration.
4. Repeat steps 1 to 3 for the reduced table until the entire capabilities are used to fill the
requirement at different warehouses.
5. From step 4 we will get initial feasible solution. Now for initial feasible solution find the
total cost.
The solution for our problem is as follows:
To: Penalty
From: Project Project Project Supply Hc NHC Hc- NHC
#1 #2 #3
4 2 8
Farm A 100
8 4 4
5 1 9
Farm B 15 200
9 5 4 selected
1
7 6 3
Farm C 200
7 6 1
NHC 5 2 8
Penalt 2 4 1
HC: Highest cost
NHC: Next Highest cost
Table2 Penalty
To: Hc NHC Hc- NHC
From: Project Project Project #3 Supply
#1 #2
4 2 8
8 Farm A 100
4 4
9 5 4
5 1 9
Farm B 15 200 50
0
ADU –Department of Business Administration Page 95 of 214
7 6 3
Farm C 200 200
2
7 3 4 selected
Table 3
Third To:
From: Project Project Project #3 Supply Penalty
allocation 3#1 #2 4 selected
4 2 8
Farm A 50 100 50
4
5 1 9
Farm B 15 200 50
0
7 6 3
Farm C 200 200
Table 4
Since there is no penalty for the remaining cells, we allocate for these cells according to their
cost.
To:
From: Project Project Project #3 Supply
#1 #2
4 2 4 8
Farm A 50 50 100
50 Fourth allocation
5 1 50 9
Farm B 15
5 200
0 50
ADU –Department of Business 7Administration
6 3 Page 96 of 214
Farm C 200 200
Fifth allocation
Therefore the final allocation will be:
To:
From: Project Project Project #3 Supply
#1 #2
4 2 4 8
Farm A 50 50 100
5 1 50 95
Farm B 15 200
0
7 6 3
Farm C 200 200
Total cost:
1x150 + 3x200+4x50 + 8x50 + 9x50
= Birr 1800
? Dear learners, could you make comparison among costs of the three
approaches? Why do you think cost of North West corner method has the
highest cost?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
The stepping-stone method involves a good deal of more effort than the MODI method.
However, it provides an intuitive understanding of the evaluation process. Moreover, when a
solution is not optimal, the distribution plan must be revised by reallocating units into and out
of various cells, and only the stepping-stone method can be used for the reallocation.
Reconsider the initial feasible solution we found using the northwest-corner method. Only the
unoccupied cells need to be evaluated because the question at this point is not how many units
to allocate to a particular route but only if converting a cell from zero units to nonzero (a
positive integer) would decrease or increase total costs. The unoccupied cells are A-3, B-1, C-1,
and C-2. They must be evaluated one at a time, but in no particular order.
Note that it is not necessary to actually alter the quantities in the various cells to reflect
the one-unit change, the + and – signs suffice.
The general implication of the plus and minus signs is that cells with + signs mean one unit
would be added, cells with a – sign indicate one unit would be subtracted. The net impact of
such a one-unit shift can be determined by adding the cell costs with signs attached and noting
the resulting value. Thus, for cell B-1, we have a net change of +2 (i.e., 5+2-4-1) which means
that for each unit shifted into cell B-1, the total cost would increase by $2.
Computed in similar way, the evaluations of cells C-1, A-3, and C-2 result in +10, -2, and +11
respectively. The negative value for cell A-3 indicates an improved solution is possible: For
each unit we can shift into that cell, the total cost will decrease by $2. The following table
shows how empty cell C-1 can be evaluated using the Stepping stone method.
There is one index number for each column and one for each row. These can be conveniently
displayed along the left and upper edges of a matrix. The index numbers are determined in such
a way that for any occupied cell, the sum of the row index and the column index equal the cell’s
unit transportation cost:
The index numbers are determined sequentially in a manner dictated by the position of
occupied cells. The process always begins by assigning a value of zero as the index number of
row 1.
The method will be illustrated by developing index numbers for the initial feasible solution for
the Harley problem generated by the northwest-corner method. We begin assigning a value of
zero for row 1. Once a 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. The complete set of row and column index numbers is shown in the
following table.
Cell evaluations for Northwest Corner solution for the Harley Problem using the MODI method
k1 k2 k3
+4 +2 +10
To:
Project Project Project Supply
From: #1 #2 #3
4 2 8
r1 0 Farm A 50 50 100
-
5 1 2 9
r2 -1 Farm B 100 100 200
7 6 3
+
2
+10 +11
Operations Research
The cell evaluations (improvement potentials) for each of the unoccupied cells are determined
using the relationship:
Cell evaluation = Cell cost – Row index – Column index
eij = cij – ri – kj
For example, the cell evaluations for A-3 is 8 – 0 – 10 = -2. Similarly, the evaluation for B-1 is
+2, for C-1, +10, and for C-2, it is +11. Note that they agree with the values we computed
earlier using the stepping-stone method.
When cell evaluations are positive or zero, an optimal solution has been found. If one or more
is negative, the cell with the largest negative should be brought into solution because that route
has the largest potential for improvement per unit. In this case, we found that cell A-3 had an
evaluation of –2, which represented an improvement potential of $2 per unit. Hence, an
improved solution is possible.
After reallocating the units using the stepping stone method, the empty cells will be A-2, B-1,
C-1 and C-2. Suppose we use the MODI method for evaluation. After assigning all the row and
column indices, we find the cell evaluations to be as follows:
Cell A-2: 2 – 0 – 0 = +2
Cell B-1: 5–1–4= 0
Cell C-1: 7-(-5)- 4 = +8
Cell C-2: 6-(-5)- 0 = +11
Because none of these numbers is negative, this is an optimal solution. You may recall that this
was the same solution obtained using the intuitive method for the initial feasible solution. At
that point, it was determined that the total cost for the distribution plan was $1800.
5 1 9
+1 Farm B 150 50 200
0
7 6 3
-5 Farm C +11
200 200
+
8
? Dear learner, which of the two approaches for improvement is lengthy and
difficulty to use?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
In the case of the transportation problem, the existence of an alternate solution is signaled by an
empty cell’s evaluation equal to zero. In fact, you may have noted that cell B-1 had an
evaluation equal to zero in the final solution of the Harley problem. We can find out what that
alternate solution is by reallocating the maximum number of units possible around the stepping-
stone path for that cell. After the cell evaluation of, and reallocating units to cell B-1, we find
the following table to be an alternate optimal solution.
To:
Project Project Project Supply
From: #1 #2 #3
4 2 8
Farm A 100 100
5 1 9
Farm B 50 150 200
7 6 3
Farm C 200 200
Degeneracy
A solution is degenerate if the number of occupied cells is less than the number of rows plus the
number of columns minus one. i.e., there are too few occupied cells to enable all the empty
cells to be evaluated. In the case of the stepping-stone method, this means that there will be at
least one empty cell for which an evaluation path cannot be constructed. For the MODI method,
it means that it will be impossible to determine all of the row and column index numbers.
Obviously, some modification has to be made to determine if such a degenerate solution is
optimal. The modification is to treat some of the empty cells as occupied cells. This is
accomplished by placing a delta () in one of the empty cells. The delta represents an extremely
small quantity (e.g., 0.001 unit); it is so small that supply and demand for the row and column
involved will be unaffected even without modifying other quantities in the row or column, and
so small that total cost will not change.
The purpose of the delta is to enable evaluation of the remaining empty cells. The choice of
location for the delta can be somewhat tricky: some empty cells may be unsuitable if they do
not enable evaluations of the remaining empty cells. Moreover, the 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 delta will be the “smallest quantity in 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 delta.
The technique can be demonstrated for the degenerate alternate solution of the Harley problem.
Suppose that after some experimentation, cell A-1 has been selected for the location of delta.
The resulting index numbers generated using MODI and the improvement potential for empty
cells based on delta in cell A-1 are shown in the following table. This confirms that the solution
is optimal
+4 0 +8
To:
Project Project Project Supply
From: #1 #2 #3
4 2 8
0 Farm A Δ + 100 100
2
5 1 9
+1 Farm B 50 150 0 200
7 6 3
-5 Farm C 200 200
+ +11
8
Unacceptable Routes
In some cases, an origin-destination combination may be unacceptable. This may be due to
weather factors, equipment breakdowns, labor problems, or skill requirements that either
prohibit, or make undesirable, certain combinations (routes).
Suppose that in the Harley problem route A-3 was suddenly unavailable because of recent
flooding. In order to prevent that route from appearing in the final solution (as it originally did),
the manager could assign a unit cost to that cell that was large enough to make that route
uneconomical and, hence, prohibit its occurrence. One rule of thumb would be to assign a cost
that is 10 times the largest cost in the table (or a very big +M). Then, this revised problem could
be solved using either of the methods we have discussed earlier. Note that the prohibited route
may appear in a non-optimal solution, but it will be eliminated by the time the optimal solution
is reached.
Dear learner, would you find the optimal solution if route A-3 is not available? How do you
compare the cost with the original cost?
is added if supply is less than demand and a dummy column is added if demand is less than
supply. The dummy is assigned unit costs of zero for each cell, and it is given a supply (if a
row) or a demand (if a column) equal to the difference between supply and demand. Quantities
in dummy routes in the optimal solution are not shipped. Rather, they serve to indicate which
supplier will hold the excess supply, and how much, or which destination will not receive its
total demand, and how much it will be short.
Let’s consider an example. Suppose that Farm C in the Harley problem has experienced an
equipment breakdown, and it will be able to supply only 120 cubic yards of topsoil for a period
of time. Therefore, total supply will be 80 units less than total demand. This will require adding
a dummy origin with a supply of 80 units. The final solution is shown in the following table.
We interpret the solution indicating that Project #3 will be short by 80 units per week until the
equipment is repaired. Note, though, that this analysis has considered only transportation costs,
and that other factors, such are shortage costs or schedules of the projects, may dictate some
other course of action.
If the intuitive approach is used to obtain the initial feasible solution when a dummy is
involved, make assignments to the dummy last. Hence, begin by assigning units to the cell with
the lowest nonzero cost, then the next lowest nonzero cost, and so on. For the Harley problem
this would mean that units would be assigned first to cell B-2 because its cost of $1 is the
lowest nonzero cell cost.
To:
Project Project Project Supply
From: #1 #2 #3
4 2 8
Farm A 50 50 100
5 1 9
Farm B 150 50 200
7 6 3
Farm C 120 120
0 0 0
Dummy 80 80
? Dear learner, would you complete the above case by evaluating the
assignments? What would assignments in the dummy row or column in the
optimal solution indicate?
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
______________________________
Maximization
Some transportation type problems concern profits or revenues rather than costs. In such cases,
the objective is to maximize rather than to minimize. Such problems can be handled by adding
one additional step at the start: Identify the cell with the largest profit and subtract all the other
cell profits from that value. Then replace the cell profits with the resulting values. These values
reflect the opportunity costs that would be incurred by using routes with unit profits that are less
than the largest unit profit. Replace the original unit profits with these opportunity cost
solution. This will be identical to maximizing the total profit. For example, suppose in the
Harley problem, the cell values had been unit profits instead of unit costs. Cell B-3 had the
largest dollar value: $9. Hence, each cell’s dollar amount would be subtracted from 9. For cell
A-1, the resulting opportunity cost would have been 9-4 = 5 and so on. Cell B-3 would have an
opportunity cost of 0 making it the most desirable route.
The remainder of the steps for developing an initial feasible solution, evaluation of empty cells,
and reallocation are identical to those used for cost minimization. When the optimal distribution
plan has been identified, use the original cell values (i.e., profits) to compute the total profit for
that plan.
? Dear learner, would you complete the above case by evaluating the assignments?
Which one would you consider the original price or opportunity cost of each cell to
calculate the total profit in the optimal solution?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
ACTIVITY
1. A transportation problem involves the following costs, supply, and demand.
To
From 1 2 3 4 Supply
1 $5 750 300 450 12
2 00 800
65 400 600 17
0
40
3 700 500 550 11
0
Demand 10 10 10 10
a. Find the initial solution using the northwest corner method, the minimum cell cost
method, and Vogel's Approximation Method. Compute total cost for each.
a. Using the VAM initial solution, find the optimal solution using the modified
distribution method (MODI).
2. Oranges are grown, picked, and then stored in warehouses in Tampa, Miami, and
Fresno. These warehouses supply oranges to markets in New York, Philadelphia,
Chicago, and Boston. The following table shows the shipping costs per truckload ($100s),
supply, and demand.
To
From New York Philadelphia Chicago Boston Supply
Tampa 9 14 12 17 200
Miami 11 10 6 10 200
Fresno 12 8 15 7 200
Demand 130 170 100 150
Because of an agreement between distributors, shipments are prohibited from Miami to
Chicago.
a. Set up the transportation tableau for this problem and determine the initial solution
using the minimum cell cost method.
b. Solve using MODI.
c. Are there multiple optimum solutions? Explain. If so, identify them.
d. Formulate this problem as a linear programming model.
3. Steel mills in three cities produce the following amounts of steel:
Location Weekly Demand (tones)
Bethlehem 130
Birmingham 210
Gary 320
620
These mills supply steel to four cities where manufacturing plants have the following
demand.
To
From 1 2 3 4
A $14 9 16 18
B 11 8 7 16
C 16 12 10 22
a. Set up the transportation tableau for this problem and determine the initial
solution using VAM.
b. Solve using the stepping-stone method.
c. Calculate the optimal allocation and total cost
Section objectives:
Up on completion of this section, the learner will be able to:
ADU –Department of Business Administration Page 107 of 214
Operations Research
? Dear learner, can you make any difference between assignment and transportation
problems?
___________________________________________________________________________
___________________________________________________________________________
__________________________________________________________________
There are many situations where the assignment of people or machines and so on, may be
called for. Assignment of workers to machines, clerks to various counters, salesmen to different
sales areas, service crews to different districts, are typical examples of these. The assignment is
a problem because people posses varying abilities fro performing different jobs and, therefore,
the costs of performing the jobs by different people are different. Obviously, if all persons
could do a job in the same time or at the same cost then it would not matter who of them is
assigned the job. Thus, in assignment problem, the question is how should the assignment be
made in order that the total cost involved is minimized (or the total value is maximized when
pay-offs are given in terms of, say, profits).
Example
A computer centre has got four Expert Programmers. Centre needs four application
programmers to be developed. The head of dept., estimate the computer required by the
respective experts to develop the application programs as follows. Make appropriate
selection of experts.
Progra
m A B C D
Expert
1 120 100 80 90
2 80 90 110 70
3 110 140 120 100
4 90 90 80 90
the condition that the number of origins is equal to the number of destinations,
i.e. m=n
Hence assignment is made on the basis of 1:1
Number of jobs equal to 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 production cycle time e.t.c
2.
A.P
The Assignment Problem (A.P.) is a special case of Transportation Problem
(T.P.) in which the number of sources and destinations are the same, and the
objective is to assign the given job (task) to most appropriate machine
(person) so as to optimize the objective like minimizing cost.
Cost Vector (Cij)
Cost vector (Cij) is the cost of assigning ith job (Task) to j th machine
(person),
Effective Matrix
A cost Matrix in A.P. is called an “Effectiveness Matrix” when there is at
least one zero in each row and column. Following is an example of
Effectiveness Matrix.
1 2 3 4
1 0 1 3 4
2 5 0 6 0
3 7 8 0 9
4 0 4 3 8
3. Mathematical Modeling of an A.P.
Let there be ‘n’ jobs in a manufacturing plant. Let there be ‘n’ machines to process the
product. A job i (i = 1, 2,…, n) when processed in a machine j (j = 1,2,…, n), it is
assumed to incur a cost of Cij.
The assignment is made in such a way that one job is associated with one machine (see
assumption). Hence we have the following:
i = 1, 2, 3… n
j = 1, 2, 3…, n
This special structure of A.P. makes solution much easy compared to the conventional T.P.
Remark:
1. It may be noted that assignment problem is a variation of transportation problem with two
characteristics (i) the cost matrix is square matrix, and (ii) the optimum solution fro the problem
would always be such that there would be only one assignment in a given row or column of the
cost matrix.
2. In assignment problem if a constant is added or subtracted from every element of any row or
column in the given cost matrix then an assignment that minimizes the total cost in one matrix
also minimizes the total cost in the other matrix.
Step 6. Given the optimal solution, make the job assignments as indicated by the ‘zero’
elements. This done as follows:
a) Locate a row which only ‘zero’ element. Assign the job corresponding to this element to
its corresponding person. Cross out the zero’s, if any, in the column corresponding to
the element, whish is indicative of the fact that the particular job and person are no more
available.
b) Repeat (a) for each of such rows which contain only one zero. Similarly, perform the
same operation in respect of each column containing only one ‘zero’ element, crossing
out the zero(s), if any, in the row in which the element lies.
c) If there is no row or column with only a single ’zero’ element left, then select a
row/column arbitrarily and choose one of the jobs (or persons) and make the
assignment. Now cross the remaining zeros in the column and row in respect of which
the assignment is made.
d) Repeat steps (a) through (c) until all assignments are made.
e) Determine the total cost with reference to the original cost table.
Example
Solve the assignment problem given in Illustrative Example 1 for optimal solution using HAM.
The information is reproduced in the following table
Step 2: For each column of this table, the minimum value is subtracted from all the other
values. Obviously, the columns that contain a zero would remain unaffected by this operation.
Here only the fourth column values would change. The table below shows this.
3 1 4 0 3
4 0 4 19 1
Step 3: Draw the minimum number of lines covering all zeros. As a general rule, we should
first cover those rows/columns which contain larger number of zeros. The above table is
reproduced in the next table and the lines are drawn.
Assignment of Jobs
Job
Worker A B C D
1 5 0 11 14
2 15 0 X 21 0
3 1 4 0 3
4 0 4 19 1
Assignments are made in the following order. Rows 1, 3, and 4 contain only one zero each. So
assign 1-B, 3-C and 4-A. Since worker 1 has been assigned job B, only worker 2 and job E are
left for assignment. The final pattern of assignments is 1-B, 2-D, 3-C, and 4-A, involving a total
time of 40+55+48+41=184 minutes. This is the optimal solution to the problem-the same as
obtained by enumeration and transportation methods.
Example
Using the following cost matrix, determine (a) optimal assignment, and (b) the cost of
assignments.
Reduced Cost Table 1
Job
Machinist 1 2 3 4 5
A 10 3 3 2 8
B 9 7 8 2 7
C 7 5 6 2 4
D 3 5 8 2 4
E 9 10 9 6 10
B 7 5 6 0 5
C 5 3 4 0 2
D 1 3 6 0 2
E 3 4 3 0 4
Iteration 2: Obtain column reductions and draw the minimum number of lines to cover all
zeros.
Reduced Cost Table2
Job
Machinist 1 2 3 4 5
A 7 0 0 0 4
B 6 4 5 0 3
C 4 2 3 0 0
D 0 2 5 0 0
E 2 3 2 0 2
Since the number of lines covering all zeros is less than the number of columns/rows, we
modify the Table 6.13. The least of the uncovered cell values is 2. This value would be
subtracted from each of the uncovered values and added to each value lying at the intersection
of lines (corresponding to cells A-4, D-4, A-5 and D-5). Accordingly, the new table would
appear as shown as follows.
Iteration 3
Job
Machinist 1 2 3 4 5
A 7 0 0 X 2 6
B 4 2 3 0 3
C 2 0 X 1 0 X 0
D 0 2 5 2 2
E 0 X 1 0 0 X 2
The optimal assignments can be made as the least number of lines covering all zeros in Table
6.14 equals 5. Considering rows and columns, the assignments can be made in the following
order:
i. Select the second row. Assign machinist B to job 4. Cross out zeros at cells C-4 and E-4.
ii. Consider row 4, Assign machinist D to job 1. Cancel the zero at cell E-1.
iii. Since there is a single zero in the row, put machinist E to job 3 and cross out the zero at
A-3.
iv. There being only a single left in each of the first and third rows, we assign job 2 to
machinist A and job 5 to C.
The total cost associated with the optimal machinist-job assignment pattern A-2, B-4, C-5, D-1
and E-3 is 3+2+4+3+9 = 21
Example:
A company has 4 machines to do 3 jobs. Each job can be assigned to one and only one
machine. The cost of each job on each machine is given below. Determine the job
assignments which will minimize the total cost.
Machine
W X Y Z
A 18 24 28 32
Job
B 8 13 17 18
C 10 15 19 22
It happens sometimes that a worker cannot perform a certain job or is not to be assigned a
particular job. To cope with this situation, the cost of performing that job by such person is
taken to be extremely large (which is written as M). Then the solution to the assignment
problem proceeds in the manner discussed earlier. The effect of assigning prohibitive cost to
such person-job combinations is that they do not figure in the final solution.
Example:
You are given the information about the cost of performing different jobs by different
persons. The job-person marking X indicates that the individual involved cannot perform
the particular job. Using this information, state (i) the optimal assignment of jobs, and (ii)
the cost of such assignment,
Job
J1 J2 J3 J4 J5
person
P1 27 18 X 20 21
P2 31 24 21 12 17
P3 20 17 20 X 16
P4 22 28 20 16 27
Solution: - Balancing the problem not assigning a high cost to the pairings P1-J3 and P3-
J4, we have the cost given in the table below.
Job
J1 J2 J3 J4 J5
P1 27 18 M 20 21
person
P2 31 24 21 12 17
P3 20 17 20 M 16
P4 22 28 20 16 27
P5 0 0 0 0 0
dummy
Now we can derive the reduced cost table as shown in table shown below. Note that the
cells with prohibited assignments continue to be shown with the cost element M, since M is
defined to be extremely large so that subtraction or addition of value does not practically
affect it. To test optimality, lines are drawn to cover all zeros. Since the number of lines
covering all zeros is less than n, we select the lowest uncovered cell, which equals 4. With
this value, we can obtain the revised reduced cost table, shown in the final table.
Reduced Matrix
Job
J1 J2 J3 J4 J5
P1 9 0 M 2 3
person
P2 19 12 9 0 5
P3 4 1 44 M 0
P4 6 12 4 0 11
P5 0 0 0 0 0
dummy
Reduced Matrix
Job
pe
J1 J2 J3 J4 J5
ADU –Department of Business Administration Page 116 of 214
Operations Research
P1 9 0 M 2 3
P2 19 12 9 0 5
P3 4 1 4 M 0
rson
P4 6 12 4 0 11
P5 0 0 0 0 0
dummy
Assignment of Job
Job
J1 J2 J3 J4 J5
P1 9 0 M 2 3
person
P2 15 8 9 0 1
P3 4 1 4 M 0
P4 2 8 0 0 X 7
P5 0 0X 0X 0 0 X
dummy
The number of lines covering zeros is equal to 5(=n), hence the optimal assignment can be
made. The assignment is: P1-J2, P2-J4, P3-J5, P4-J3, while job J1 would remain
unassigned. This assignment pattern would cost 18+12+16+20=66 in the aggregate.
Unique Vs Multiple Optimal Solutions
In the process of making assignments, it was stated earlier that we select a row/column with
only a single zero to make an assignment. However, a situation may wherein the various
rows and columns, where assignment are yet to be done, have all multiple zeros. In such
cases, we get multiple optimal solutions to the given problem. In any of the problems
discussed so far, we have not experienced such a situation. Hence, each one of them has had
a unique optimal solution. When a problem has a unique optimal solution, it means that no
other solution to the problem exists which yields the same objective function value (cost,
time, profit e. t. c) as the one obtained from the optimal solution derived. On the other hand
in a problem with multiple optimal solutions, there exists more than one solution which all
is optimal and equally attractive. Consider the following example.
Example:
Solve the following assignment problem and obtain the minimum cost at which all the jobs
can be performed.
Solution: This problem is unbalanced since number of jobs is 5 while the number of
workers is 4. We first balance it by introducing a dummy worker E, as shown in table
below.
ADU –Department of Business Administration Page 117 of 214
Operations Research
Step 1: Obtain reduced cost values by subtracting the minimum value in each row from
every cell in the row. This is given in Table below.
Reduced Cost 1
Job (cost in ’00 Br.)
Worker 1 2 3 4 5
A 7 0 14 2 3
B 22 13 9 0 5
C 4 1 4 16 0
D 4 12 4 0 11
E 0 0 0 0 0
Since there is at least one zero in each row and column, we test it for optimality.
Accordingly, lines are drawn. All zeros are covered by 4 lines, which is less than 5 (the
order of the given matrix). Hence, we proceed to improve the solution. The least uncovered
value is 4. Subtracting from every uncovered value and adding it to every value lying at the
intersection of lines, we get the revised values as shown below.
The solution given in the reduced cist 2 table is optimal since the number of lines covering
zeros matches with the order of the matrix. We can, therefore, proceed to make
assignments. To begin with, since each of the columns has multiple zeros, we cannot start
making assignments considering columns and have, therefore, to look through rows. The
first row has a single zero. Thus, we make assignment A-2 and cross out zero at E-2.
Further, the second and the third rows have one zero each. We make assignments B-4 and
C-5, and cross out zeros at D-4 and E-5. Now, both the rows left two zeros each and so have
both the columns. This indicates existence of multiple optimal solutions. To obtain the
solutions, we select zeros arbitrarily and proceed as discussed below.
i. Select the zero at D-1, make assignment and cross out zeros at D-3 and E-1 (as
both, worker D and job 1, are not available any more). Next, assign worker E to job 3,
corresponding to the only zero left. Evidently, selecting the zero at E-3 initially would
have the effect of making same assignments.
ii. Select the zero at D-3, make assignment and cross at D-1 and E-3. Next, make
assignment at the only zero left at E-1. Obviously, selecting the zero at E-1 making
assignment in the first place would lead to the same assignments.
To conclude, the problem has two optimal solutions as given below.
Example:
A company plans to assign 5 salesmen to 5 districts in which it operates. Estimates of sales
revenue in thousands of birr for each salesman in different districts are given in the
following table. In your opinion, what should be the placement of the salesmen if the
objective is to maximize the expected sales revenue?
District
Salesman D1 D2 D3 D4 D5
S1 40 46 48 36 48
S2 48 32 36 29 44
S3 49 35 41 38 45
S4 30 46 49 44 44
S5 37 41 48 43 47
Solution: Since it is a maximization problem, we would first subtract each of the entries in
the table from the largest one, which equals 49 here. The resultant data are given in Table
below.
Since the number of lines covering all zeros is fewer than n, we select uncovered cell value,
which equals 4. With this, we can modify the table as given in the Reduced Cost Table 3.
Steps 4, 5, 6: Find improved solution. Test for optimality and make assignments.
More than one optimal assignment is possible in this case because of the existence of
multiple zeros in different rows and columns. The possible assignments are:
S1-D2, S2-D5, S3-D1, S4-D3, S5-D4; or
S1-D2, S2-D1, S3-D5, S4-D3, S5-D5; or
S1-D2, S2-D1, S3-D1, S4-D4, S5-D3; or
S1-D2, S2-D1, S3-D5, S4-D4, S5-D
Each of these assignment patterns would lead to expected aggregated sales equal to 231
thousand birr.
STAR
T
Is it a No Add dummy
balanced row(s)/column(s)
problem?
Yes
STO
P
Fig. Schematic Presentation of Hungarian Assignment Method.
Summary
ADU –Department of Business Administration Page 122 of 214
Operations Research
Activity
1. An electronics firm produces electronic components, which it supplies to various
electrical manufacturers. Quality control records indicate that different employees produce
different numbers of defective items. The average number of defects produced by each
employee for each of six components is given in the following table.
Component
Employee A B C D E F
2
1 30 24 16 26 30
2
2
1
2 22 28 14 30 20
3
2
3 18 16 25 14 12
2
3
4 14 22 18 23 21
0
2
5 25 18 14 16 16
8
6 32 14 10 14 18 2
Determine the optimal assignment that will minimize the total average
number of defects per month.
2. The Bunker manufacturing firm has five employees and six machines, and wants to
assign the employees to the machines so as to minimize cost. A cost table showing the
cost incurred by each employee on each machine is given below.
Machine
Employe A B C 0 E F
1 $12 7 20 14 8 10
2 10 14 13 20 9 11
3 5 3 6 9 7 10
4 9 11 7 16 9 10
5 10 6 14 8 10 12
Course
Instructor A B C D
1 80 75 90 85
2 95 90 90 97
3 85 95 88 91
4 93 91 80 84
5 91 92 93 88
The department head wants to know the optimal assignment of instructors to courses that
will maximize the overall average evaluation. The instructor who is not assigned to teach
a course will be assigned to grade exams. Solve this problem using the assignment
method.
UNIT Five
5. DECISION THEORY/ANALYSIS
Unit objective:
After completing this unit, the learner should be able to:
? Dear learner, from your previous courses, can you define decision making? What
steps are involved in it? Can you mention conditions in which decision at different
levels are made?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
Dear learner, in the previous units dealing with LP, models were formulated and solved in order
to aid the manager in making decision. The solutions to the models were represented by values
for the decision variables. However, these LP models are formulated under the assumption that
certainty existed. In actual practice, however, many decision making situations occur under
conditions of uncertainty. For example, the demand for a product may be not 100 units next
week, but 50 or 200 units, depending on the market (which is uncertain).
4. Degree of certainty: - the approach often used by a decision maker depends on the
degree of certainty that exists. There can be different degrees of certainty. One extreme
ADU –Department of Business Administration Page 125 of 214
Operations Research
is complete certainty and the other is complete uncertainty. The later exists when the
likelihood of the various states of nature are unknown. Between these two extremes is
risk (probabilities are unknown for the states of nature). Knowledge of the likelihood of
each of the states of nature can play an important role in selecting a course of active.
5. Decision criteria: - the decision maker’s attitudes toward the decision as well as the
degree of certainty that surrounds a decision. Example; maximize the expected payoffs.
A payoff table is a device a decision maker can use to summarize and organize information
relevant to a particular decision. It includes a list of alternatives, the possible future states of
nature, and the payoffs associated with each of the alternative/state of nature combinations.
If probabilities for the states of nature are available, these can also be listed. The general
format of the table is illustrated below:
States of nature
S1 S2 S3
A1 V11 V12 V13
Alternatives A2 V21 V22 V23
A3 V31 V32 V33
Under complete uncertainty, the decision maker either is unable to estimate the probabilities
for the occurrence of the different state of nature, or else he or she lacks confidence in
available estimates of probabilities, and for that reason, probabilities are not included in the
analysis.
A decision making situation includes several components- the decision themselves and the
actual event that may occur future, known as state of nature. At the time the decision is
made, the decision maker is uncertain which state of nature will occur in the future, and has
no control over them.
Decisions made under these circumstances are at the opposite end of the spectrum from the
certainty case just mentioned. Once the decision has been organized in to a payoff table,
several criteria are available making the actual decision. There are several approaches
(criteria) to decision making under complete uncertainty. Some of these discussed in this
section include: maximax, maximin,minimax regret, Hurwicz, and equal likelihood.
5.4.1. MAXIMAX
With the maiximax criterion, the decision maker selects the decision that will result in the
maximum of the maximum payoffs.( In fact this is how this criterion derives its name-
maximum of maximum). Tha maximax is very optimistic. The decision maker assumes that
the most favorable state of nature for each decision alternative will occur. For example, the
investor would optimistically assume that good economic conditions will prevail in the
future. The best payoff for each alternative is identified, and the alternative with the
maximum of these is the designated decision.
For the previous problem:
S1 S2 S3 Row Maximum
A1 4 16 12 16*maximum
A2 5 6 10 10
A3 -1 4 15 15
Note: If the pay off table consists of costs instead of profits, the opposite selection
would be indicated: The minimum of minimum costs. For the subsequent decision criteria
we encounter, the same logic in the case of costs can be used.
? Dear learner, how would the decision in the above example change if the
values in the table stand for costs instead of profit?
________________________________________________
________________________________________________
________________________________________________
___
5.4.2. Maximin Criteria
This approach is the opposite of the previous one, i.e. it is pessimistic. This strategy is a
conservative one; it consists of identifying the worst (minimum) payoff for each alternative,
and, then, selecting the alternative that has the best (maximum) of the worst payoffs. In
effect, the decision maker is setting a floor on the potential payoff by selecting maximum of
the minimum; the actual payoff can not be less than this amount. It involves selecting best
of the worst.
ADU –Department of Business Administration Page 127 of 214
Operations Research
S1 S2 S3 Row minimum
A1 4 16 12 4
A2 5 6 10 5*maximum
A3 -1 4 15 -1
Note: If it were cost, the conservative approach would be to select the maximum cost
for each decision and select the minimum of these costs.
? Dear learner, how would the decision in the above example change if the
values in the table stand for costs instead of profit?
________________________________________________
________________________________________________
______________________________________________
An approach that does take all payoffs in to consideration is Minimax regret. 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 column (i.e.,
given a state of nature). Hence, opportunity loss amounts are found by identifying the best
payoff in a column and, then, subtracting each of the other values in the column from that
payoff. Therefore, this decision avoids the greatest regret by selecting the decision
alternative that minimizes the maximum regret.
EXAMPLE:
S1 S2 S3
A1 4 16 12
5 6 10
-1
ADU –Department of Business Administration 4 15 Page 128 of 214
Operations Research
A2
A3
S1 S2 S3
A1 5-4=1 16-16=0 15-12=3
A2 5-5=0 16-6=10 15-10=5
A3 5-(-1)=6 16-4=12 15-15=0
S1 S2 S3 Max. Loss
A1 5-4=1 16-16=0 15-12=3 3*minimum
A2 5-5=0 16-6=10 15-10=5 10
A3 5-(-1)=6 16-4=12 15-15=0 12
EXAMPLE
S1 S2 S3 S4 S5 Row Average
A1 28 28 28 28 4 23.2*maximum
5 5 5 5 28 9.6
5
ADU –Department of Business Administration 5 5 5 28 9.6 Page 129 of 214
Operations Research
A2
A3
The basis for the criterion of insufficient reason is that under complete uncertainty, the decision
maker should not focus on either high or low payoffs, but should treat all payoffs (actually, all
states of nature), as if they were equally likely. Averaging row payoffs accomplishes this.
? Dear learner, how would the decision in the above example change if the values in the
table stand for costs instead of profit?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
_
The Hurwitz criterion requires that for each alternative, the maximum payoff is multiplied by
and the minimum payoff be multiplied by 1-.
Example: If = 0.4 for the above example,
A1 = (0.4x16) + (0.6x4)
= 8.8
A2 = (0.4x10) + (0.6x5)
=7
A3 = (0.4x15) – (0.6x1)
= 5.4
Decision: A1 is selected
A limitation of Hurwicz criterion is the fact that must be determined by the decision maker.
Regardless of how the decision maker determines, it is still a completely a subjective measure
of the decision maker’s degree of optimism. Therefore, Hurwicz criterion is a completely
subjective decision making criterion.
? Dear learner, can you mention conditions under which Hurwicz criterion criteria can be
considered as maximin or minimax criterion
_________________________________________________________________________
ADU –Department of Business Administration Page 130 of 214
_________________________________________________________________________
_________________________________________________________________________
__
Operations Research
Dear learner, the decision making criteria just presented were based on the assumption that no
information regarding the likelihood of the states of the nature was available. Thus, no
probabilities of occurrence were assigned to the states of nature, except in the case of the equal
likely hood criterion.
It is often possible for the decision maker to know enough about the future state of nature to
assign probabilities to their occurrences. The term risk is often used in conjunction with partial
uncertainty, presence of probabilities for the occurrence of various states of nature. 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 the decision
maker with additional information that can dramatically improve the decision making process.
Given that probabilities can be assigned, several decision criteria are available to aid the
decision maker. Some of these are discussed below.
EXAMPLE:
Probability
0.20 0.50 0.30
S1 S2 S3 Expected payoff
A1 4 16 12 12.40*maximum
A2 5 6 10 7
A3 -1 4 15 6.30
Decision: A1 will be chosen.
Note that it does not necessarily follow that the decision maker will receive a payoff
equal to the expected monetary value of a chosen alternative. Similarly, the expected
payoffs for either of the other alternatives do not equal any payoffs in those rows. What,
then, is the interpretation of the expected payoff? Simply a long-run average amount; the
approximate average amount one could reasonably anticipate for a large number of identical
situations.
Note: The EOL approach resulted in the same alternative as the EMV approach
(Maximizing the payoffs is equivalent to minimizing the opportunity losses).
Note: The expected value approach is particularly useful for decision making when a
number of similar decisions must be made; it is a long-run approach. For one-shot decisions,
especially major ones, other methods (perhaps, maximax or maximin) may be preferable. In
addition, non monetary factors, although not included in a payoff table, may be of considerable
importance. Unfortunately, there is no convenient way to include them in an expected value
analysis.
??? Dear learner, can you make differences between decision making situations under
uncertainty and risk?
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
ADU –Department of Business Administration Page 132 of 214
___
Operations Research
Decision tree, like probably tree is composed of squares, circles, and lines:
The squares indicate decision points
Circles represent chance events( circles and squares are called nodes)
o The lines (branches) emanating from squares represent alternatives.
o The lines from circles represent states of nature
The tree is read from right to left.
It should be noted that although decision trees represent an alternative approach to payoff
tables, they are not commonly used for problems that involve a single decision. Rather, their
greatest benefit lies in portraying sequential decisions (i.e., a series of chronological decisions).
In the case of a single decision, constructing a decision tree can be cumbersome and time
consuming.
Example
Pay off table for Real Estate investment
State of Nature
Good economic Good economic
Decision conditions conditions
(Purchase) 0.6 0.4
Apartment
decision tree for the above2example will
Thebuilding be:
Poor economic $30,000
conditions (0.4)
The circles ( ) and squares ( ) in the above figure are referred to as nodes. The
squares are decision nodes, and the branches emanating from a decision node reflect the
alternative decisions possible at that point. For example, in the above figure, node 1
signifies a decision to purchase an apartment building. Office building, or ware house.
The circles are probability nodes, and the branches emanating from them indicate the
state of nature that can occur: good economic conditions or poor economic conditions.
The decision tree represents the sequence of events in a decision situation. First, one of
the three decision choices is selected at node 1. Depending on the branch selected, the
decision maker arrives at probability node 2, 3, or 4, where one of the states of nature will
prevail, resulting in one of six possible payoffs.
Determining the best decision using a decision tree involves computing the expected
value at each probability node. This is accomplished by starting with the final outcomes
(payoffs) and working backward through the decision tree toward node 1. First, the
expected value of the payoffs is computed at each probability node.
EV(node 2) = .60($ 50,000) + .40($ 30,000) = $42,000
EV(node 3) = .60($100,000) + .40($-40,000) = $44,000
EV(node 4) = .60($ 30,000) + .40($ 10,000) = $22,000
These values are now shown as the expected payoffs from each of the three branches
emanating from node 1 in figure below. Each of these three expected values at nodes 2, 3,
and 4 is the outcome of a possible decision that can occur at node 1. Moving toward node
1, we select the branch that comes from the probability node with the highest expected
payoff. In figure below, the branch corresponding to the highest payoff, $44,000 is from
node 1 to node 3. This branch represents the decision to purchase the office building. The
decision to purchase the office building, with an expected payoff of $44,000, is the same
result we achieved earlier using the expected value criterion. In fact, when only one
decision is to be made (i.e., there is not a series of decisions), the decision tree will always
yield the same decision and expected payoff as the expected value criterion. As a result, in
these decision situations the decision tree is not very useful. However, when a sequence or
series of decisions is required, the decision tree can be very useful.
Good economic
$42,000
Conditions (0.6)
$50,000
Apartment 2
building
Poor economic
Conditions(0.4) $30,000
$44,000
Good economic
Office building Conditions (0.6)
3 $100,000
1
Purchase
Poor economic
Conditions -$40,000
Good economic
Warehouse $22,000 Conditions (0.6)
$30,000
4
Poor economic
Conditions(0.4) $10,000
In order to demonstrate the use of a decision tree for a sequence of decisions, we will alter
our real estate investment example to encompass a ten-year period during which several
decisions must be made. In this new example, the first decision facing the investor is whether
to purchase an apartment building or land. If the investor purchases the apartment building,
two states of nature are possible. Either the population of the town will grow (with a
probability of 0.60), or the population will not grow (with a probability of 0.40). Either
state of nature will result in a payoff. On the other hand, if the investor chooses to purchase
land, three years in the future another decision will have to be made regarding the
development of the land. The decision tree for this example, shown in figure below,
contains all the pertinent data, including decisions, states of nature, probabilities, and
payoffs.
At decision node 1 in figure below, the decision choices are to purchase an apartment
building and to purchase land. Notice that the cost of each venture ($800,000 and
$200,000, respectively) is shown in parentheses. If the apartment building is purchased,
two states of nature are possible at probability node 2. The town may exhibit population
growth, with a probability of .60, or there may be no population growth or a decline, with a
probability of .40. If the population grows, the investor will achieve a payoff of $2,000,000
over a ten-year period. (Note that this whole decision situation encompasses a ten-year
time span.) However, if no population growth occurs, a payoff of only $225,000 will result.
If the decision is to purchase land, two states of nature are possible at probability node 3.
These two states of nature and their probabilities are identical to those at node 2; however,
the payoffs are different. If population growth occurs for a three-year period, no payoff will occur,
but the investor will make another decision at node 4 regarding development of the land. At that
point either apartment will be built at a cost of $800,000 or the land will be sold with a payoff
of $450,000. Notice that the decision situation at node 4 can occur only if population growth
occurs first. If no population growth occurs at node 3, there is no payoff and another decision
situation becomes necessary at node 5: the land can be developed commercially at a cost of
$600,000 or the land can be sold for $210,000. (Notice that the sale of the land results in less
profit if there is no population growth than if there is population growth.)
If the decision at decision node 4 is to build apartments, two states of nature are possible. The
population may grow, with a conditional probability of .80, or there may be no population
growth, with a conditional probability of .20. The probability of population growth is higher
(and the probability of no growth is lower) than before because there has already been
population growth for the first three years, as shown by the branch from node 3 to node 4. The
payoffs for these two states of nature at the end of the ten-year period are $3,000,000 and
$700,000, respectively, as shown in figure below.
If the investor decides to develop the land commercially at node 0. 5, then two states of nature
can occur. Population growth can occur, with a probability of .30 and an eventual payoff of
$2,300,000, or no population growth can occur, with a probability of .70 and a payoff of
$1,000,000. The probability of population growth is low (i.e., .30) because there has already
been no population growth, as shown by the branch from node 3 to node 5.
$ 2,000,000
.60
$ 225,000
2 $ 3,000,000
.40
.80
6
6
.20
,00
1 0,0 $ 700,000
4 00
$ 450,000
Sell land
.60 .30
3 .40 ,00
$ 2,000,000
7 0,0
00
.70
,00
0,0
5 00 $ 1,000,000
Sell land
$ 210,000
This decision situation encompasses several sequential decisions that can be analyzed using the
decision tree approach outlined in our earlier (simpler) example. As before, we start at the end
of the decision tree and work backward toward a decision at node 1.
First we must compute the expected values at nodes 6 and 7.
EV (node 6) = .80($3,000,000) + .20($ 700,000) = $2,540,000
EV (node 7) = .30($2,300,000) + .70($1,000,000) = $1,390,000
Both of these expected values (as well as all other nodal values) are shown in boxes in Figure
4.4.
At decision nodes 4 and 5, we must make a decision. As with a normal payoff table, we make
the decision that results in the greatest expected value. At node 4 we have a choice between two
values: $1,740,000, the value derived by subtracting the cost of building an apartment building
($800,000) from the expected payoff of $2,540,000, or $450,000, the expected value of selling
the land computed with a probability of 1.0. The decision is to build the apartment building,
and the value at node 4 is $1,740,000.
This same process is repeated at node 5. The decisions at node 5 result in payoffs of $790,000
(i.e., $1,390,000 - 600,000 = $790,000) and $210,000. Since the value $790,000 is higher, the
$1,290,000
.60
Purchase $ 225,000 $ 7540,000
2 $ 3,000,000
Apartment .40 .80
Building 6
6
.20 .20
$1,740,000 ,00
1 0,0
$1,160,000 00
$ 700,000
4
$ 450,000
Sell land
$1,390,000
.60 .30
.40 ,00 $ 2,300,000
3
7 0,0
00
$1,360,000 $790,000 .70
,00
0,0
5 00 $ 1,000,000
Sell land
$ 210,000
In this section we will present a process for using additional information m the decision-
making process by applying Bayesian analysis, a probabilistic technique. This process will
be demonstrated using the real estate investment example employed throughout this
chapter. To
briefly review this example, a real estate investor is considering three alternative in-
vestments, which will occur under one of two possible economic conditions (states of
nature) shown in table above.
Expected values at nodes 6 and 7
2 540 IPopulation
. .000growth
G .80
.20
Population growth
Purchase apartment building (-$800,000)
Decision (Purchase)
S 50,000 100,000 30,000
Recall that using the expected value criterion, we found the best decision to be the purchase of
the office building, with an expected value of $44,000. We also computed the expected value of
perfect information to be $28,000. Therefore, the investor would be willing to pay up to
$28,000 for information about the states of nature, depending on how close to perfect the
information was.
Now suppose that the investor has decided to hire a professional economic analyst who will
provide additional information about future economic conditions. The analyst is constantly
researching the economy, and the results of this research are what the investor will be
purchasing.
The economic analyst will provide the investor with a report predicting one of two outcomes.
The report will be either positive, indicating that good economic condition are most likely to
prevail in the future, or negative, indicating that poor economic conditions will probably occur.
Based on the analyst's past record in forecasting future economic conditions, the investor has
determined conditional probabilities of the different report outcomes given the occurrence of
each state of nature in the future. We will use the following notations to express these
conditional probabilities:
g = good economic conditions
p = poor economic conditions
P = positive economic report
N = negative economic report
The conditional probability of each report outcome given the occurrence of each state of nature
is shown below.
P(P/g) = .80
P(N/g) = .20
P(P/p) = .10
P(N/p) = .90
For example, if future economic conditions are in fact good (g), the probability that a positive
report (P) will have been given by the analyst, P(P/g), is .80. The other three conditional
probabilities can be interpreted similarly. Notice that these probabilities indicate that the analyst
is a relatively accurate forecaster of future economic conditions.
The investor now has quite a bit of probabilistic information available - not only the conditional
probabilities of the report, but also the prior probabilities that each state of nature will occur.
These prior probabilities that good or poor economic conditions will occur in the future are
P(g) = .60
P(p) = .40
Given the conditional probabilities, the prior probabilities can be revised to form posterior
probabilities by means of Bayes's rule. If we know the conditional probability that a positive
report was presented given that good economic conditions prevail, P(P / g), the posterior
probability of good economic conditions given a positive report, P(g/P), can be determined
using Bayes's rule, as follows.
P(g /P) = P(P / g)P(g)
P(P/g)P(g) + P(P/p)P(p)
(.80) (.60)
(.80) (.60) + (.10) (.40)
= .923
The prior probability that good economic conditions will occur in the future is .60.
However, by obtaining the additional information of a positive report from the analyst, the
investor can revise the prior probability of good conditions to a .923 probability that good
economic conditions will occur. The remaining posterior (revised) probabilities are
P(g/N) = .250
P(p /P) = .077
P(p/N) = .750
The original decision tree analysis of the real estate investment example is shown in Figures 4.1
and 4.2. Using these decision trees, we determined that the appropriate decision was the
purchase of an office building, with an expected value of $44OOO. However, if the investor hires
an economic analyst, the decision regarding which piece of real estate to invest in will not be
made until after the analyst presents the report. This creates an additional stage in the decision-
making process, which is shown in the decision tree in Figure 4.5.
The decision tree shown in Figure 4.5 differs in two respects from the decision trees in Figures
4.1 and 4.2. The first difference is that there are two new branches at the beginning of the
decision tree that represent the two report outcomes. Notice, however, that given either report
outcome, the decision alternatives, the possible states of nature, and the payoffs are the same as
those in the first two figures.
The second difference is that the probabilities of each state of nature are no longer the prior
probabilities given in Figure 4.1; instead they are the revised posterior probabilities computed
in the previous section using Bayes's rule. If the economic analyst issues a positive report, then
the upper branch in figure below (from node 1 to node 2) will be taken.
If an apartment building is purchased (the branch from node 2 to node 4), the probability of
good economic conditions is .923, whereas the probability of poor conditions is .077. These are
the revised posterior probabilities of the economic conditions given a positive report. However,
before we can perform an expected value analysis using this decision tree, one more piece of
probabilistic information must be determined-the initial branch probabilities of a positive and a
negative economic report.
The probability of a positive report, P(P), and of a negative report, P(N), can be determined
according to the following logic. Recall from Chapter 10 that the probability that two
dependent events, A and B, will both occur is
4 $50,000
Pig p1=.923
Apartment
2 building
P/P/p) = 077 $30,000
Office building 5 PigP)= 923
$100,00
Positive
report P(P/P)=077 -$ 40,000
Warehouse 6 P(g/p)= 923
$ 30,000
1
8
3
9
Operations Research
$ 10,000
P(p/N)= 750
If event A is a positive report and event B is good economic conditions, then according to the
above formula,
P(Pg) = pcp / g)P(g)
We can also determine the probability of a positive report and poor economic conditions the
same way.
P(Pp) = P(P/p)P(p)
Next consider the two probabilities P(Pg) and P(Pp). These are, respectively, the probability of
a positive report and good economic conditions and the probability of a positive report and
poor economic conditions. These two sets of occurrences are mutually exclusive, since both good
and poor economic conditions cannot occur simultaneously in the immediate future.
Conditions will be either good or poor, but not both. To determine the probability of a positive
report, we add the mutually exclusive probabilities of a positive report with good economic
conditions and a positive report with poor economic conditions, as follows.
P(P) = P(Pg) + P(Pp)
Now, if we substitute into this formula the relationships for P(Pg) and P(Pp) determined
earlier, we have
P(P) = P(P/g)P(g) + P(P/p)P(p)
You might notice that the right-hand side of this equation is the denominator of the
Bayesian formula we used to compute P(g/P) in the previous section. Using the conditional and
prior probabilities that have already been established, we can determine that the probability of
a positive report is
P(P) = P(P/g)P(g) + P(P/p)P(p) = (.80) (.60) + (.10) (.40)
= .52
Similarly, the probability of a negative report is
P(N) = P(N/g)P(g) + P(N/p)P(p)
= (.20) (.60) + (.90) (.40)
= .48
Now we have all the information needed to perform a decision tree analysis. The decision tree
analysis for our example is shown in Figure 11.6. To see how the decision tree analysis is
conducted, consider the result at node 4 first. The value $48,460 is the expected value of the
purchase of an apartment building given both states of nature. This expected value is computed
as follows.
$50,000
$30,000
$100,000
-$40,000
$30,000
$10,000
$50,000
$30,000
$100,000
-$40,000
ADU –Department of Business Administration Page 143 of 214
$30,000
$10,000
$15,000
Operations Research
P (p/N) =.750
This amount, $63,190, is the expected value of the investor's decision strategy given that a
report forecasting future economic condition is generated by the economic analyst.
No matter how large the decision analysis, the steps of this tabular approach can be
followed the same way as in this relatively small problem. This approach is more
systematic than the direct application of Bayes's rule, making it easier to compute the
posterior probabilities for larger problems.
EVSI
Efficiency = EVPI
= $ 19,190
28,000
=.68
Thus, the analyst's economic report is viewed by the investor to be 68% as efficient as
perfect information. In general, a high efficiency rating indicates that the information is
very good, or close to being perfect information, and a low rating indicates that the addi-
tional information is not very good. For our example, the efficiency of .68 is relatively
high; thus it is doubtful that the investor would seek additional information from an
alternative source. (However, this is usually dependent on how much money the decision
maker has available to purchase information.) If the efficiency had been lower, however,
the investor might seek additional information elsewhere.
Summary
Decision theory problems are characterized by, list of alternatives, states of nature,
payoffs, degree of certainty, decision criteria.
There are several approaches (criteria) to decision making under complete uncertainty.
Some of these discussed in this section include: maximax, maximin,minimax regret,
Hurwitz, and equal likelihood.
Decision making under risk (with probabilities) involves several decision criteria
including Expected Monetary Value (EMV), Expected Opportunity Loss (EOL),
Expected Value of Perfect Information (EVPI).
Decision trees represent an alternative approach to payoff tables; which are used for
problems that involve a series of chronological decisions by portraying sequential
decisions graphically.
The circles ( ) and squares ( ) in the above figure are referred to as nodes. The
squares are decision nodes, and the branches emanating from a decision node reflect
the alternative decisions possible at that point.
Activity
1. The owner of the Burger Doodle Restaurant is considering two ways to expand operations:
opening a drive-up window or serving breakfast. The increase in profits resulting from these
proposed expansions depends on whether a competitor opens a franchise down the street. The
possible Profits from each expansion in operations given both future competitive situations are
shown in the following payoff table.
Competitor
Decision open Not Open
Drive-up window $-6,000 $20,000
Breakfast 4,000 8,000
Select the best decision using the following decision criteria.
a) Maximax
b) Maximin
c) Equally likely
d) Minimax regret
2. Consider the following payoff table for three alternatives, A,B, and C, under two future states
ADU –Department of Business Administration Page 146 of 214
Operations Research
Economic Conditions
Investment Good Bad
A $ 70,000 $ 25,000
B 120,000 -60,000
C 40, 000 40,000
Determine the decision using the following decision criteria.
a) Maximax
b) Maximin
c) Minimax regret
d) Hurwicz (α = 0.3)
e) Equal likelihood
3. An investor is considering investing in stock, real estate, or bonds under uncertain economic
conditions. The payoff table of returns for the investor’s decision situation is shown below
Economic Conditions
Investment Good Stable Poor
Stocks $ 5,000 $ 7,000 $ 3,000
Real estate -2,000 10,000 6,000
Bonds 4,000 4,000 4,000
Determine the best investment using the following decision criteria.
a) Equal likelihood
b) Maximin
c) Maximax
d) Hurwicz ( α = 0.3)
e) Minimax regret
Unit Six
Net works and Project Management
Unit Objectives
After completing this chapter you should be able to:
Define what a project is.
Describe the different types of networks
Use CPM to evaluate project’s time
Discuss the use of PERT in solving problems of project
Describe the concept of crashing
6.1. Introduction
It is essential to identify the various activities involved in the execution of Projects. The large
and complex projects of any organization involve a number of interrelated activities, which
might be performed independently, simultaneously, or one after the other. Modern management
has designed a network models approach to solve the problem associated with the allocation of
scarce resources of manpower, material, money and time to these interrelated activities.
? Dear learner, can you define the word project from your previous experiences and what you
heard about it
_______________________________________________________________________________
ADU –Department of Business Administration Page 148 of 214
_______________________________________________________________________________
_______________________________________________________________________________
_
Operations Research
A project can be defined as being a series of activities designed to achieve a specific objective,
and which has a definite beginning and a definite end. For network analysis to be of use, the
project must be capable of being split into a number of discrete activities, which relate together
in a logical and well-defined manner.
Network analysis involves the breaking down of a project into its constituent activities, and the
presentation of these activities in diagrammatic form.
Networks are one of the important tools of management science which easily solve problem by
presenting in visual formats. The shortest route problem, minimum spanning tree and maximal
flow models are useful for solving problems associated with allocation of scarce resources,
time, cost, and material consumption. Moreover, CPM and PERT, which we will discuss later
on also contributed a lot for complex projects management under different situations.
? Dear learner, do you know what a Gant chart is? Have you ever used it for any
purpose?
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
It 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, first identify the major activities that would be required. Next,
time estimates for each activity were made, and the sequence of activities was determined.
Once they were to occur.
The obvious advantage of Gantt chart is its simplicity, and this accounts for its popularity.
However, it fails to reveal certain relationships among activities that can be crucial to effective
project management.
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 charts can be useful for initial project planning, which then gives way to the use of
network diagrams.
? Dear learner, could you discuss about importance of network diagrams with your
colleagues?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_______________
Drawing a logic diagram is a skill requiring practice and ingenuity, and for major projects may
require two or three attempts before a satisfactory network or diagram is completed correctly.
Computer packages are often used to carry out this process, cutting drastically the time taken.
The straight lines connecting the circles represent a task that takes time or resources; these lines
are called activities. The arrow heads show the direction of the activity.
The beginning and end points of an activity are called events or nodes. Event is a point in the
line and does not consume any resource. A numbered circle represents it. The head event has
always a number higher than the tail event.
Activity
Tail Head
Merge and burst events
It is not necessary an event to be the ending event of the only one activity but can be the ending
event of two or more activities. Such event is defined as a Merge event.
Merge event
If the event happened to be the beginning event of two or more activities, it is defined as a
Burst event.
Burst event
Certain activities, which neither consume time nor resources but are used simply to represent a
connection or a link between the events, are known as dummies. It is shown in the network by a
dotted (broken) line.
The purpose of introducing dummy activity is:
To maintain uniqueness in the numbering system as every activity may have
distinct set of events by which the activity can be identified.
To maintain a proper logic in the network.
D
A Dummy
B
C
v) Numbering the Events
After the network is drawn in a logical sequence, every event is assigned a number.
The number sequence must be in such a way that it should reflect the flow of the network. In
event numbering, the following rules should be observed:
i. Event numbers should be unique.
ii) Event numbering should be carried out a sequential basis from
left to right.
iii) The initial event, which has all outgoing arrows with no incoming arrow, is
numbered 0 or 1.
iv) The head of an arrow should always bear a number higher than the one
assigned at the tail of the arrow.
v) Gaps should be left in the sequence of event numbering to accommodate
subsequent inclusion of activities, if necessary.
6.2.2. Rules for drawing a network
A complete network should have only one point of entry -the start event, and one point
of exit -the finish event.
Every activity must have one preceding event -the tail, and one following event - each
activity has one head.
Several activities may use the same tail event, and the same head event, but no two
activities may share the same head and tail events.
In addition to the rules described above, certain conventions are followed for the sake of clarity
and uniformity. There are two slightly different conventions for constructing the net work
diagrams. Under one convention, the arrows are used to designate activities, where as under the
other convention, the nodes are used to designate activities. These conventions are referred to
as activity- on-arrow (A-O-A) and activity – on- node (A-O-N) respectively. In order to avoid
confusion, the discussion here focuses primarily on activity- on- arrow convention. When we
use this convention:
Networks proceed from left to right -the start event is at the left hand side of the
diagram and the end event at the right hand side
A closed loop would produce an endless cycle in computer programmers with a built- in routine
for detection or identification of the cycle. Thus one property of a correctly constructed
network diagram is that it is non-cyclic.
B
A
b) Dangling: No activity should end without being joined to the end event. If it is not
so, a dummy activity is introduced in order to maintain the continuity of the
system. Such end-events other than the end of the project as a whole are called
dangling events. All activities must contribute to the progression of the network
or be discarded as irrelevant.
B
A E
D
C
F
c) Redundancy: Danglin If a dummy activity is the only activity
emanating from an event, it can be eliminated.
Dummy D F
A
C
E
B
? Dear learner, can you mention and discuss on the common mistakes made in drawing
net work diagrams.
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
____________
6.3. TYPES OF NETWORK MODELS
Section objective:
Up on completion of this section, the learner will be able to:
The objective of this network model is to obtain the shortest path in which one can minimize
distance, time or costs involved form the origin to destination.
If we have “n” location in the network, the total steps required to solve the shortest route
problem will be n-1. The determination of shortest route involves labeling procedure in which
each node is assigned with two numbers where by the 1 st label represent distance from the
original (source) node and the 2nd number refers the node that immediately precede the labeled
node. The labeling procedure begins with the original node where the label will be (O, S) to
indicate the distance is zero and it is starting point.
In the shortest route algorithm nodes are labeled either permanently to indicate the final
labeling or temporary labeling to indicate that the labeling might be revised. Labels remain
temporary until it can be ascertained that no shorter route to a node exists.
Step 1: Start at node 1 and find the distance from node 1 to other nodes that can be directly
reached form node 1. Temporarily label each of these nodes with their distance from node 1
followed by a comma and a number 1. Then select the node that has smallest distance from
node 1, and make its label permanent. This can be done by shading the node.
Step 2: Find the distance from the new permanent node to each non permanent nodes that can
reached directly form this node. Temporarily label each of these nodes with the cumulative
distance from node 1 if a node has no label. Or, change the earlier assigned temporary label of a
node if its cumulative distance from node 1 through the new permanent node is less than the
previous temporary label. Then permanently label (shade) the node that has the smallest
cumulative distance form node 1.
Step 3: Repeat the preceding steps until all nodes have permanent labels.
Step 4: Identify the shortest route to each node from node 1 by working backward through the
tree according to the nodes label specified on the node.
Example :
Find the shortest route of the following network starting form node one. Travel between nodes
can be in either direction.
13
2
5
4 5
7
8 6
3
1
7
1
16 6
4
1st step
Shade node 1 and temporarily label node 2, 3 and 4, which can be directly reached from node 1.
The label should show the distance from node 1.
(5.1)
2 13
5 4
(O,
( S) (8, 1)
8 7
3 5 6
1
7
1
(7, 1)
ADU –Department of Business Administration Page 156 of 214
Operations Research
16 6
4
2nd Step
Select the node, which has the smallest distance from node 1, and label it permanent and shade
the node (i.e. node 2), then temporarily label nodes that can be directly reached from node 2.
Therefore node 5 is labeled as (18, 2.)
(5, 1)
2
13
5
(O, S) 4 (18,2)
(8,1)
8
3 5
7 7 6
7, 1
1
1 16
6
3rd Step
Identify the smallest distance from node 1 i.e. from node 3, 4, and 5, which have values of 8, 1,
7,1 and 18,2 respectively. Permanently label and shade node 4, which has the smallest distance
of 7. Identify non-permanent labels; directly reached from this node and temporarily label it by
the distance from node 1.
(5,1)
13
2
5 (18, 2)
(O, S) 4
(8, 1)
8 7 5
3
1
6
7
(7,1) (23, 4)
1
4 16 6
4th Step
The node with the smallest temporary label is node 3. Thus, its label becomes permanent and
the node is shaded. Node 3 can be reached directly from node 1 through node 4 with the same
distance of 8 (i.e. tie exists). Next, find each non-Permanent label that can be reached directly
from node 3. Node 5 is the only node with a value of (18, 2). However, the cumulative distance
of node 5 from node 1 through node 3 is 15 (i.e. 8+7=15). Because the route through node 3 is
shorter, we update the temporary label of node 5 to reflect this shorter route.
(5,1)
2
13
5 (15, 3)
(O, S)
(18,2)
(8,1) tie
7
1 8 5
3
7 1 6
(7, 1)
(23,4)
4 16
6
5th Step
Identify the node, which have smallest label from the un-shaded nodes or temporary labels.
Node 5 with a distance of 15,3 is selected and becomes permanent label.
(5,1)
2
13
5 4 (15,3)
(8,1) 18,2
(O,S) tie
8 3 7 5
1 6
7 (7,1) 1 (23,4)
16
4 6
6th Step
The only node which is not shaded and can be reached directly from the new permanent node is
node 6. Using this route, its cumulative distance from node 1 would be 15+6=21. Because this
is less than its current table update label to 21, 5 and make permanent & shade it.
2
13 (15, 3)
5 18,2
4
(O,S)
8, 1 tie
8 7 5
1 3 6
(21,5)
7 (23,4)
(7,1)
1
16 6
4
Pe
rmanent label of the final network indicates the shortest distance of each node from node 1 (the
starting node). In order to identify the route that yields the shortest distance to a particular node,
it is necessary to work backtracking. The shortest route of the nodes from node 1 to each node
is summarized below.
The minimum spanning tree problem involves in connecting all of the nodes of a network using
as little of the connecting material as possible. For example, the nodes might represent oil
storage tanks, and lines represent pipeline that are used to carry the oil between tanks. The cost
of the pipeline would be proportional to the length of the pipeline used. Hence, the objective
would be to connect all of the tanks using as little pipeline as possible. Similar activities include
designing communication systems by using minimum amount of wiring, designing highway
networks by using minimal amounts of materials etc.
Algorithm of Minimum Spanning Tree
Step 1. Start at any node, (usually, node 1 is used as the starting point), identify the node that
has shortest distance from node 1 and connect it to the node 1 using a line. If a tie occurs, break
it arbitrarily
Step 2. Find the shortest distance form the existing portion of the tree (i.e. the connected nodes)
to a node that is not yet connected. Make the connecting line from previously connected tree to
the new node which has shortest distance.
Step 3. Continue until all nodes have been connected to the tree.
Step 4. To find the total (minimal) length of the connecting distance, sum up their values.
Example:
Consider the following network where the nodes represent fuel storage tanks and the
connecting lines represent possible pipeline connections. The numbers on the lines represent the
distance in meters for a particular pipeline connection. Determine the spanning tree for the
network storage tanks and the amount of pipe that will be needed to make the connections.
26
22 6
2 4
24
13 18
27 33
5
1 20 30
32
18
49 7
3
Arbitrarily choose node 1 as a starting node consider all branches incident on it. They are 1-2
and 1-3 with distance of 13 and 18 respectively. Since 1-2 is the shortest, select this line and
connect it. Now node 1 and 2 are connected.
Next consider all arcs incident either on node 1 or node 2 that connect to other nodes. Such
lines are 1-3, 2-3 and 2-4 with a distance of 18, 20 and 22 respectively. Node 1-3 with a
distance of 18 of selected. Now the connected nodes are 1, 2, and 3.
Next consider all braches incident to Node 1, 2 and 3. These are 2-4, 3-4, 3-5 and 3-7 with
distance of 22, 27, 30 and 49. The shortest distance is 22 i.e. 2-4. Make 2-4 part of the network
2 2
2 4 6 13 2
1
5 18 2
1
7 3 2
3
A B
22
22
13 2 4
2 4 6
13
6 18
1
1 5 5 7
18
3
3 7
18
C D
22
2 22
13 4
6 2 4
18 13
18 6
1
5 24 1 5 24
18
3
7 32 7
18 3
E
F
The next step will be to identify the shortest line that is incident to node 3 and 4. These are 3-5,
3-7, 4-5 and 4-6 with a distance of 30, 49, 18 and 26. Since 4-5 is the shortest distance, include
it in the network.
Continue in this manner until all nodes are connected with their shortest distance. The network
solution for the above problem is shown on the following figure
The length of pipe that will be needed for this system can be found by summing the line lengths
(distance): i.e. 18+13+22+18+24+32= 127 meters.
It is assumed that there is a single input node (called a source) and a single output node
(called a sink).
There is flow conservation, i.e. the flow out of any node is equal to the flow into that
node.
Junctions or nodes cannot serve as a store, i.e. any flow arriving at a node is transferred
immediately to another location.
Algorithm of Maximal flow model
Rule for Maximal Flow
1. Arbitrarily select a continuous path from the source node to the sink node that
has a positive flow capacity. Flow capacity of a path is determined by the smallest
branch capacity along the path chosen in the direction of flow.
2. In this continuous path determine the arc with the minimum flow capacity.
3. Reduce all the quantities along this path by the amount transported (i.e.
minimum of the path)
4. Repeat step 1- step 3 until all paths are used.
5. To get the maximum flow of the network, add all flows of the path.
Example:
Determine the maximum flow through the system of pipeline. The flow is from node 1 to node
4.
10 2 6
2
1 4
3 6
Step 1. Choose any path that will allow a positive flow from the source (node 1) to the sink
(node
4). The possible paths are 1-2-4, 1-2-3-4, 1-3-4 and 1-3-2-4. Suppose arbitrary select
path 1-2-4.
The maximum flow of 10 (Node 1-2) is limited by the 6 capacity of node 2-4. Therefore adjust
the capacity of the total path of 1-2-4 by flow of 6. (i.e. reduce by 6)
ADU –Department of Business Administration Page 163 of 214
Operations Research
4 2 6
2
1 4 6
6 6
3
Here, branch 2-4 is fully loaded and no additional flow to this path. But, path 1-2 still accept
additional flow of 4. Therefore, we can use path 1-2-3-4 to pass 2 units through path 1-2-3-4.
2
2 0
0
2+6 1
4 6+2
3
4
The flow capacity of 1-2-3-4 is fully utilized. i.e. no further positive flow is allowed through
this path. But there is a positive flow through path 1-3-4. Therefore a flow of 3 can be assigned
to this path. The resulting network will be as shown below.
2 0
2
2
ADU –Department of Business Administration Page 164 of 214
Operations Research
3+2+6 4 6+2+3
1
0 1
At this point, no additional flow is possible because no path has a positive flow capacity. Hence
the situation is optimal and maximal flow rate is 11 (6+2+3 = 11).
? Dear learner, please discuss on the different type of network models and their application areas.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
__
Being able to create an effective specification, supported by the analysis, design and
implementation stages of the project development, and to deliver a system to the required
standards and within the time and cost constraints imposed depends on the effective
management of the entire development process.
The primary objectives underlying any process of project management -whether related to
investments in information systems or to some other capital expenditure project -may be
encapsulated under two headings:
Meets the required needs of the users or the other objectives that have been established
Is produced to specified quality standards to satisfy user needs
Can be integrated within existing organizational information systems, structures and
processes
Is sufficiently flexible to respond to changes in the environment in which the system
will operate, or to the changing requirements of users -which may only become
articulated and refined during the development process itself
Provides appropriate support to decision makers at all levels: operational, tactical and
strategic.
Efficiency: ensuring that the development of the project -including development, delivery,
installation and final implementation -is
Undertaken within the manpower resources, costs and time targets or constraints
specified at the outset
Efficient in exploiting the resources of the project team members' and the users' time as
fully as possible, avoiding unnecessary idle time, delays or time wasted in undertaking
unnecessary tasks or activities
Effective in integrating the activities of those within the project team, and also those
interactions and dependencies with other parties outside the project team, including
users, suppliers, consultants, or managers
Capable of delivering the resources on time -neither too late nor too early to cause
problems of storage, loss of value due to deterioration, unexpected fluctuations in
planned cash flows, or loss of benefits from training due to excessive time required for
the user to apply the knowledge or skills acquired.
In practice, the project management process may involve the management of either a small
team of one or two internal systems staff, to cover all aspects of the project development, or, at
the other extreme, a large group of both internal and external staff, many of whom will be
specialists in limited features of the system during its development. Irrespective of the scale or
complexity of the project, the issues concerned are broadly similar and the practices used to
ensure improved effectiveness and efficiency may be applied equally well in all situations.
6.4.1. Project Management Elements
The terms effectiveness and efficiency, as defined above, may be expressed in terms of the key
elements of any project, namely:
Time
Resources
Costs
Quality.
The term 'efficiency' relates primarily to the elements of time, resource utilization and costs
(measuring the value of the output produced against the costs and time taken to produce the
system).
'Effectiveness' is concerned more with the quality of the system in terms of the required
performance standards and objectives. Let us look at these project elements in detail.
6.4.1.1. Time
The time factor represents a key parameter in the project management process: Typically, the
objectives of the project will stipulate either a target time for completing the development of the
project, or the date by which the system should be fully operational. Other intermediate dates or
targets may also be specified for completing particular component elements or phases in the
development process. Failure to meet these prescribed target completion dates may have both
direct and indirect consequences if the organization, some of which may be highly significant.
For example:
Losses of potential financial returns reduced operational efficiency and lost competitive
advantage from the system operation itself.
Reduced benefits and potential opportunities if other organizational development is tied
to the project -for example, the launch of a new product involving (expensive
promotional campaign).
Resources remaining idle: specific staff recruited and not utilized, and environment
facilities all arriving ahead of requirement and increasing the organization’s risk of loss,
damage or deterioration.
Delays that may require either the extension of contracts negotiated with specific
programmers or analysts hired only for the duration of the project (possibly increased
rates), or the use of overtime working to recover lost time (at higher rates)
The retention of existing computing equipment beyond the anticipated time
necessitating the extension of maintenance contracts and often incurring high
maintenance costs than the new equipment. This may also apply to staffing costs
particularly if significant reductions are anticipated as a result of the developments.
Unplanned cash flow effects from either early or delayed purchasing or delivery).
The impact on other project developments, either concurrent with the delayed project or
utilizing part of the current system project or the staff employed in its development.
6.4.1.2. Resources
For these purposes, the term resources relates to both material and human resources. Material
resources will include consumable materials and support services. Although the acquisition and
management of material resources improve demanding, the management of human resources is
usually the key element. The staffs involved are likely to differ in terms of their specialization
skills and experience (There will also be differences in personality and attitudes. Generally, the
management of resources involves:
Management priorities for both the time and resource elements will be reflected in the project
costs. A key performance indicator used by senior management to assess the project during its
development phase is the level of expenditure incurred relative to the budgeted expenditure.
Managing the costs of a project involves the following:
Forecasting and estimating the costs of the project in total so that the level of
anticipated expenditure can be established for each month or quarter. These costs will
normally be subdivided into a variety of sub-headings relating to capital and revenue
expenditures, and costs associated with staff, materials and overheads. (Total
expenditure approach )—Engineers estimates.
Monitoring the actual expenditure in total and under each of the sub-headings within
the budget. This will allow the project team to assess the degree to which actual
expenditure matches budgeted expenditure. In cases where there are significant
variances, the project team will be required to establish the cause and the consequences
for the total project costs.-- Variance Analysis
Significant overspending on a project may require the project manager and the
organization to review either the subsequent stages of the development process or the
initial system objectives and performance standards, with the intention of modifying the
more expendable elements to meet budget limits.—Feedback and feed forward
decisions.
6.4.1.4. Quality
The task of managing quality differs from that of managing the elements of time, resources and
costs, which are concerned primarily with the inputs, processes and activities involved in
developing a system. Quality management is primarily focused on the output of the
development process as an operational system. It may be more difficult to measure quality than
the three other elements, because it may be assessed against more subjective criteria. Thus, the
perception of quality may differ, depending on the position or role of the individual in relation
to the project. While different perceptions of quality exist, the important assessment of quality
relates to the project’s ability to meet the users' requirements and expectations. Generally, the
management of quality Incorporates
Establishing quality and performance standards for the project at the outset and
developing methods for measuring these less tangible elements
Establishing procedures and methods of working which will assist in assuring the
achievement of the standards in the final system
Monitoring the project regularly against the desired standards and ensuring that the
necessary procedures are being observed.
? Dear learner, can you discuss how quality, cost, time and, resource determine the
effectiveness and efficiency of the project?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
________
The Project Initiation Document (PID), sometimes called a 'statement of work' or ‘project
charter', is a formal document listing the goals, constraints and success criteria for the project -
the rules of the game. The PID, once written, is subject to negotiation and modification by the
various stakeholders of the project. Once they formally agree its content, it becomes the
document that is referred to in the case of any disagreement later as to precisely what the
project was intended to achieve.
According to Eric Verzuh, a PID should contain at least the following sections:
Scope statement: This puts boundaries to the project by outlining the major activities of
the project. It is important to include this section in order to prevent 'scope creep',
where additional activities are added during the project, and making achievement of the
cost and time objectives totally impossible.
Deliverables: What are the main outcomes expected from the project? The focus on
what makes the success of the project easier to measure. Deliverables tend to be tangible
elements of the project, such as reports, assets and other outputs.
Cost and time estimates: Even at this early stage, it is a good idea for the project team
to have some feel for the organization’s expectations in terms of the project budget.
These estimates will necessarily be modified later in the project, but are necessary to
give a starting point for planning.
Objectives: A clear statement of the mission, critical success factors and milestones of
the project
Stakeholders: A list of the major stakeholders in the project and their interest in the
project.
? Dear learner, what are the basic sections of project initiation documents?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
______
1) Strategic. This level involves taking a broad view of the total project and of all the major
stages involved. The aim is to integrate and co-ordinate the contributions, both direct and
indirect, from each area of the organization and from outside it. While the decision making is
inevitably concerned with the issues of time, resources, costs and quality, the perspective
adopted is to view these issues in aggregate terms -say, the total capital expenditure or staffing
costs incurred on the project to date in relation to the budgeted levels. Reporting at this level
will generally be directed towards the organization’s senior management, and in addition to
being in a summarized form, it is also likely to be on a monthly or quarterly timescale to allow
the planning, monitoring and control activities to focus on broad issues of performance,
timescales or costs.
2) Tactical This level is concerned with viewing and managing the project on a short- to
medium-term timescale. The actual time horizon of planning, decision making and control will
depend on the total scale of the project in terms of resources and time, but it would be usual to
review both the work in hand and the work to be undertaken in each of the main activity areas
of the project over the coming one or two months. In larger projects, separate teams of
development staff may be given specific responsibility for monitoring and controlling the main
activity areas. While still concerned with the four basic elements of time, resources, costs and
quality, the information used for this purpose is likely to be more detailed and the reporting
period will probably be shorter. For example, the performance of a particular team of staff
during the previous week may be evaluated to measure the progress made and to identify any
problems and the appropriate corrective action.
3) Operational Management: activities at this level are concerned with the more detailed
aspect of supervising and controlling the ongoing operations of the development. This will tend
to focus on the problems associated with a particular activity and on the direct supervision of
staff. The main tasks are resolving any
Specific technical difficulties encountered, ensuring the maintenance of: appropriate standards,
motivating the staff and trouble-shooting. While the priority remains the maintenance of
effectiveness and efficiency, this is now directly related to specific activities and to individual
members of the project team. One of the more important consequences is that the management
process at this level will involve less formalized reporting systems; the style usually depends
upon a more direct and frequent method of monitoring performance and greater emphasis on
interpersonal relationships.
6.4.4. Duties of Project manager
The project manager may be required to interface with all three levels (strategic, tactical and
operational) at different times, although, in the larger projects, the team or group leaders would
be responsible for the operational level management of the activities and personnel within their
team. The project manager's primary responsibility is to ensure effectiveness and efficiency
across the complete project, which involves the following action at each level:
1) Strategic: At this level, the manager should:
ADU –Department of Business Administration Page 172 of 214
Operations Research
evaluate and analyze strategic impact of each proposed system on the organization’s
operations, in promoting competitive advantage, and on the other existing or proposed
information systems within the organization
report to senior management on the results of the strategic evaluation and quantify the
potential costs and benefits that may be derived from each system development
contribute to the strategic decision-making process of the organization in terms of the
information resource
review and evaluate the strategic consequences of the progress achieved in developing
the system at regular intervals.
convert the objectives, targets and performance standards agreed for the project into
operational plans
develop the outline and detailed plans for the project, identifying the time, resource and
cost parameters, and producing a schedule of activities for the project staff
monitor the performance of the project against the agreed plans, budgets and quality
performance standards, and report progress to both the steering committee and the
management sponsor
liaise with management and staff in the user departments, to ensure an effective
interface in terms of integration and co-operation between the users and the project
development staff
facilitate a similar level of co-operation with project staff and outside agencies - either
with suppliers or distributors for the company's main business, or with the specialist
agencies supporting the system development project itself
analyze deviations from planned performance, evaluating alternative solutions and
recommending appropriate control changes to future plans or targets
develop appropriate information systems to provide the necessary Information to
monitor and control the project.
manage and control the daily activities of the project team members, either directly or
through the project team leaders
conduct regular progress meetings with the project staff and team leaders to identify and
resolve existing and potential difficulties
ADU –Department of Business Administration Page 173 of 214
Operations Research
appraise the performance of individual members of the project development staff and
seek to assure the achievement of the required quality standards throughout the project
development.
The project manager plays a key role in the effective management of system development and
(although other participants may contribute significantly to this process) fulfils the major role of
integrating all the various contributions.
3. Estimation of the probabilities of the events occurring. It will be difficult to establish precise
probabilities, but it will be necessary to establish some type of prioritization: some risks will
be much more likely than others. Attention must first be paid to the more important threats.
4. Decide how the risks will be handled. Risks can be handled as follows:
Do nothing. This is appropriate where the effect is small or the chance of occurrence
very remote
Off-load the risk, for example by arranging for third parties to complete part of the
project
Investigate the risk further and try to protect against it. For example arrange to have
additional staff available in case of project overrun.
Thereat Identification and slippage reduction
a) Possible threats:
The following can threaten the success of a project. Suggestions are included as to how
minimize the slippage involved with those threats.
Poor management
Many project leaders will be from technical backgrounds and they may not have the proper
management skills for controlling large projects.
Project leaders should be properly trained so that they have managerial skills as well as
technical skills. They should not be given large critically important projects until they have
proved themselves on smaller exercises.
Poor planning
Managers have not made use of the various planning methods available: network analysis,
Gantt charts. They have not broken the project down into its various activities and estimated a
time and cost for each.
Unrealistic deadlines
There is often pressure from users for projects to be completed quickly. Project teams,
particularly if they have had to win the job competitively, may have suggested times that are
unrealistic. Project managers must look critically at the deadlines. They should identify the
critical activities of the project and ensure that these do not slip.
Insufficient budget
Too few people are employed on the project, inadequate materials is bought, the cheapest (not
the best) solutions are always sought. Of course, organizations cannot ignore costs and should
try to get good value for money. However, it is important to be objective about what a given
cost budget can produce by way of project outcomes. If money is tight, it might be better to do
a smaller project thoroughly than a larger one poorly.
Moving targets
The project specification keeps changing as the project progresses. This will certainly add costs
and delay to the project. User’s requirements should be thoroughly examined and the analyst
should check understanding before the project is started. Techniques such as structured
walkthroughs and prototyping will help here.
b) Project change procedure
One of the most effective methods of dealing with the need to amend projects is to have a
project change procedure. Although this procedure will not remove some of the risks discussed
above, It will enable some changes to be made to the project with minimal disruption and
slippage occurring.
The main issue to be aware of in changing a project is that the later into a project that a change
is made, then the more difficult it will be to accommodate that change and the greater will be
the expense of that change. A change management procedure for a project will normally
involve the following activities:
Identifying the need for change
This may arise from many sources including user input, technical difficulties with
implementing part of the project, time or cost efficiencies identified by the project team etc.
Any change will be discussed with the Project Manager initially.
A more formal explanation of the change is produced, stating clearly the need for the change,
what the change will be and the costs and benefits associated with the change.
Feasibility of change
The project manager and senior members of the project team will check that the change is
actually possible in terms of technical and social feasibility
When the case for a change has been checked, the change document will be placed before the
Steering committee for discussion, and approval if the change is accepted
Major changes will also require the authorization of the project sponsor and possibly the Board
or similar decision-making body of the organization
The project plan will be amended to take account of the change. Deadlines and costs will be
revised.
Make change
The change is actually carried out and, where necessary, tested to ensure that there are no
conflicts with other sections of the project.
? Dear learner, assume you are a project manager in your organization. What do you think
of your major duties? Discuss it with your friends.
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
The key elements of project management are planning and control. Deciding and specifying
what to do is the function of a project plan. Making sure it is done right is the function of
project control.
Planning is vital if control is required later. Planning is necessary to avoid wasting time,
resources and effort. It should also be flexible to take into consideration any change that may
occur through unforeseen circumstances during the development of the project.
Construct a plan -list tasks, timescales and dependencies, allocate tasks, decide on the
tools to be used and identify control systems.
Control systems must be capable of measuring progress, reporting deviations and taking
corrective action.
The criterion for a good control system includes the following:
The plan and its control should be viable
Corrective action taken if there are any discrepancies and the requirements, schedules
and budgets are updated.
Planning and control is very important as it enables people to understand what is needed to
meet the goals and reduces uncertainty of outcomes.
The following are the main advantages afforded by using critical path methods:
1. Easier visualization of relationships. The network diagram that is produced shows how the
different tasks and activities relate together, making it easier to understand a project intuitively,
improving planning and making it easier to communicate details of the project plan to interested
parties.
2. More effective planning. CPM forces management to think a project through thoroughly. It
requires careful and detailed planning, and the discipline imposed often justifies its use even
without the other benefits.
3. Better focusing on problem areas. The technique enables the manager to pinpoint likely
bottlenecks and problem areas before they occur
4. Improved resource allocations. Manpower and other resources can be directed to those
parts of the project where they will have the most effect in reducing cost and speeding up the
completion of the project. Overtime can be eliminated, or confined to those jobs where it will
do the most good.
5. Studying alternative options. Management can simulate the effect of alternative courses of
action. They can also gauge the effect of problems in carrying out particular tasks and can,
therefore, make contingency plans.
7. Improved project monitoring. By comparing the actual performance of each task with the
schedule, a manager can immediately recognize when problems are occurring, can identify
when those problems are important and can take the appropriate action in time to rescue the
project.
One of the most significant benefits of network analysis is that it enhances management's ability
to monitor the project as it is being carried out. After the project has been analyzed a series of
charts, diagrams, graphs, narratives and tables are drawn up. These can be used as a
reporting system that will identify immediately a critical activity that is out of control, or a non-
critical activity that has used up its float.
Once problems have been identified, extra resources can be brought in or transferred between
activities. Managers are thus in a better position to correct the situation before the problems
become insoluble.
6.4.6.3. Types of Project management tools
A project manager has a range of project management tools available to assist with the planning
and control of individual systems projects. These tools have been refined and developed over
the years, but they are all designed to improve the effectiveness of the project management
process.
The project management tools available include:
Work Breakdown Structure
Gantt chart
Resource Histogram
Budget.
A Work Breakdown Structure is a results-oriented family tree that captures all the work of a
project in an organized way. It is often portrayed graphically as a hierarchical tree; however, it
can also be a tabular list of "element" categories and tasks or the indented task list that appears
in the Gantt chart schedule. One of the main methods of breaking down work into manageable
units, and then allocating those units to members of the project team, is to use a work
breakdown structure. The basic idea of creating a WBS is to take the work involved in the
project and break it down into smaller and smaller parts, until the project consists of a series of
work packages. These can then be put into the project plan as activities.
Example one:
The CBE has started a project to upgrade the BPR performed on its functional units. To manage
this project, the Bank has established three teams each team consisting of four members. The
project was expected to be finalized by the end of July, 2007. In this case, the Bank can apply
the concept of Work Breakdown Structure as follows.
1. Top-level WBS
This breaks the project down into stages, often using the project life cycle stages as major sub-
divisions of the project. Mostly, BPR has five stages:
BPR
Recommendation Implementation
Documentation Analysis Problem
Identification
Documentation
2.2 Analysis
Analysis
Problem Identification
Compare what is with what should be Identify the causes, effects and
impact of the problem
2.4. Recommendation
Recommendation
2.5. Implementation
Implementation
Domestic
Credit Services Foreign Banking
Banking
Example Two:
It is derived from the earlier PRINCE technique, which was initially developed in 1989 by the
Central Computer and Telecommunications Agency (CCTA) as a UK Government standard for
information technology (IT) project management; however, it soon became regularly applied
outside the purely IT environment. PRINCE was released in 1996 as a generic project
management method. PRINCE has become increasingly popular and is now the de facto
standard for project management in the UK. Its use has spread beyond the UK to more than 50
other countries.
The PRINCE methodology uses a different approach to analyzing the work to be done. The
assumption made with PRINCE is that it is easier to see, in a complex project, what is to be
produced than it is to know how that will happen. PRINCE therefore uses a system of product
breakdown structures (PBS) to analyze the project.
Management products are those products associated with the planning and control of the
project. They include, for example, the project initiation document (PID), the project plan, etc.
Technical products are those things the project has been set up to create.
Quality products are things associated with the definition and control of quality.
PLANNING TOOLS
Having completed the work breakdown structure, it will be possible to construct a project
network.
6.4.6.2.2. Gantt chart
The Gantt chart is a timeline chart. It clearly shows when each task is to begin, the time it will
take to complete each task, and which tasks will be going on simultaneously. You may want to
use more than one level of Gantt chart.
A Gantt chart, named after its inventor, shows the activities of a project, the same as CPA.
However, these activities are presented as a bar chart with the start and finishing times clearly
identified.
This is a diagrammatic representation which shows the various activities in a project. The aim
of the CPA is to identify how those activities link together and to show the critical path or the
sequence of activities where a delay will result in the overall project being delayed. An activity
is said to be critical if a delay in its start will cause a further delay in the completion of the
entire project.
The sequence of critical activities in a network is called the critical path. It is the longest path in
the network from the starting event to the ending event and defines the minimum time required
to complete the project. In the network it is denoted by double line. This path identifies all the
critical activities of the project. Hence, for the activity (i,j) to lie on the critical path, following
condition must be satisfied.
a) ESi= LSi
b) EFi = LFj
c) ESj- ESi = LFj- LFi = tij.
ES i, EF j, are the earliest start, and finish time of the event j and i. LS i, LF j, are the latest
start, finish time of the event j and i.
Required:
a) Draw an arrow diagram representing the project.
b) Find the total float for each activity.
c) Find the critical path and the total project duration.
Solution:
5 5
15
2
1 1 3
3 4 7
14
15 14
8
3 12
6
Earliest Latest
Normal Total float LFj-
Activity Start Finisgh Start Finish
time ESj or LFi-ESi
ESi ESj LFi LFj
1-2 15 0 15 0 15 0
1-3 15 0 15 3 18 3
2-3 3 15 18 15 18 0
2-5 5 15 20 32 37 17
3-4 8 18 26 18 26 0
3-6 12 18 30 28 40 10
4-5 1 26 40 26 40 0
5-6 3 27 30 37 40 10
6-7 14 40 54 40 54 0
PERT
PERT is a further development of the CPA approach which deals with the influence of changes
in time on the cost of a project. In summary, the approach involves four key activities:
1. Estimating the normal duration for an activity and the normal cost associated with this
time period.
2. Estimating the extent to which this normal duration may be reduced and the additional
costs of doing so. The additional costs may arise from the hiring of additional systems
staff or the overtime worked by existing staff. However, the additional costs do not
always relate to human resources and may, for example, involve additional payments to
secure faster methods of delivering the supplies required. Further, not all activities are
capable of being reduced in terms of time, and for those where reductions may be
achieved, the costs may rise more steeply as successively more time is saved on the
activity. In this, the reductions usually plateau at a lower limit in terms of duration and it
proves impossible to secure further reductions.
3. Addressing those activities on the critical path, so that each activity is ranked in terms
of the cost of saving one week of time. Starting with the activity that costs least to save
a week, and moving down the list in terms of the ranking, produces an assessment of the
time that it is possible to save on the total project time and the consequent increase in
total costs for each incremental week saved. The process is complicated by the fact that
initial changes in the duration of activities on the critical path may result in a change in
the direction of the critical path requiring the consideration of the time/cost trade-off of
other activities.
4. The project managers assessing the potential benefits of saving time on the project
against the incremental costs of doing so. The application of the time/cost trade-off is
not only useful before the start of a project, but can become a useful aid to the project
manager in controlling the project, as it will indicate the costs associated with proposed
changes in future activities and the time that may be saved.
Computing Algorithm for PERT (Network and Activity Slack Time, Earliest and
Latest Activity Times)
In managing the activities of a project, it is some times useful to know how soon or how late an
individual activity can be started or finished without affecting the scheduled completion date of
the total project.
Four symbols are commonly used to designate the earliest and latest activity times.
ES= The earliest start time for an activity. The assumption is that all predecessor activities are
started at their earliest starting time.
EF= The earliest Finish the time for an activity. The assumption is that the activity starts on its
ES and takes its expected time, T.
Therefore,
EF= Es + t
LF= The latest finish in time for an activity without delaying the project. The assumption is
that successive activities take their expected time.
LS= The latest start time for an activity without delaying the project.
LS= LF - t
The process of calculating ES and EF times involves calculations in sequence from left to right
( in the network) and is some times referred to as a forward pass of calculations. Thus, the ES
of an activity can be determined by summing the times of all preceding activities where two
paths converge at node the, longest path (time wise) govern.
The process of calculating ES and EF times involves calculations in sequence from left to right
(in the network) and is some times referred to as a forward pass of calculations. Thus, the ES
of an activity can be determined by summing the times of all preceding activities where two
paths converge at node, the longest path (time wise) governs.
Example 1: Computation of earliest start and latest start times for the activities in the net
work model for 2 simple data collection project.
Sequence Activity
1-2 Design questionnaire
2-3 Prepare questionnaire
3-4 Prepare Time Table
3-5 Select interviewers
3-6 Select target respondents
4-5 Arrange facilities
4-6 Notify the respondents
5-6 Train the interviewers
6-7 Undertake the interview
7-8 Organize the collected data
6 6 5
7
ADU –Department of Business Administration Page7192 of 214 8
Operations Research
5 8 4
1 2 3 4
4
6
5
Required:
a) Determine the critical path
b) How much slack time is available in the path containing notifying the respondents
Solution Time
a. 1-2 -3-5-6-7-8 36 days
b. 1-2-3-4-5-6-7-8 38 days
c. 1-2-3-4-6-7-8 32 days
d. 1-2-3-6-7-8 31 days
The length of any path can be determined by summing the expected times of the activities on
that path. The path with the longest time is of particular interest because it determines the
projects completion time. If there are any delays along the longest path, here will be
corresponding delays in project completion time. So to shorten the project completion time one
must focus on the longest path. This longest path is so called the critical path and its activities
are referred to as critical activities.
Thus, one the above example, the critical path corresponds to path B with a time requirement of
38 days.
A path slack is the difference between the length of a given path and the length of the critical
path. Thus, the slack corresponding to path C is;
Slack = Time for critical path - Time for = 38 - 32 = 6 days
Path Slack
A 38-36 2
B 38-38 0
C 38-32 6
D 38-31 7
The computation of earliest times involves "forward pass" through the network and the
computation of the latest times involves a "back-ward pass" through the network. Hence for
finding the LS and LF, we must begin with the earliest finish time of the last activities (i.e. the
project length) and use that time as latest finish time for the last activity.
Having been provided with this information, let us try to find the following facts;
a) Find the expected duration and variance of each activity
b) Find the expected duration of the project.
c) Calculate the variance and standard deviation of project length and what is the
probability that the project will be completed:
1. at least 4 days earlier than expected?
2. no more than 4 days later than expected?
ADU –Department of Business Administration Page 195 of 214
Operations Research
d) If the project due date is 41 days what is the probability of meeting the due date.
Solution
Ze Expected path
Paths Activity to tm tp To+4tm+tp length ( E the S2
6 path)
4
1-2-3-5-6-7-8 1-2 3 5 7 5 37 /9
16/9
2-3 4 8 12 8
3-5 3 6 9 6 1
5-6 2 5 8 5 1
6-7 2 5 14 6 4
7-8 4 7 10 7 1
4
1-2-3-4-5-6-7-8 1-2 3 5 7 5 39 /9
2-3 4 8 12 8 Critical path 16
/9
4
3-4 2 4 6 4 /9
4
4-5 2 4 6 4 /9
5-6 2 5 8 5 1
Path Activity (2 path) Path
1-2-3-5-6-7-8 1-2
83
2-3 /9
3-5
5-6
6-7
7-8
1-2-3-4-5-6-7-8 1-2
2-3
3-4
82
4-5 /9
5-6 Project variance
6-7
7-8
1-2-3-4-6-7-8 1-2
2-3
285/
3-4 36
4-6
6-7
7-8
1-2-3-6-7-8 1-2 74/9
ADU –Department of Business Administration Page 196 of 214
2-3
3-6
6-7
7-8
Operations Research
Thus, project variance is 82/9 and standard deviation of . To assist our calculations, let us
approximate the values as: Variance = 9 and standard deviation as 3.
µ = expected time
δ = standard time
Therefore, to find the probability that the project will be completed at least 4 days earlier than
expected;
So, p (Z<-1.33)
= p(X< 35)
= 0.0918
= 9.18%
Z -1.33 0
And the probability that the project will be completed at a no more than 4 days later than
expected is ;
P= ( X<43) = P(Z<1.33)
= 0.9082
ADU –Department of Business Administration Page 197 of 214
Operations Research
= 90.82%
N 1.33
6.4.6.2.4. Budget
The project budget shows for each month of the project, the proposed expenditure. This will be
updated as the project progresses with the actual expenditure and the reason for any variances.
Almost all of these tools can be in some form of computer software, For example, a budget or
Gantt chart can be presented on a spreadsheet. However, there is also. Specialist project
management software available to produce critical path analysis and Gantt charts directly from
the data entered into the software.
The information analysis syllabus requires detailed knowledge of the project networks (network
analysis) and Gantt charts.
6.4.6.2.5. Resource Histogram (Resource allocation)
This is a stacked bar chart showing the number and mix of staff required over the duration of a
project. The network diagram gives the logical sequence of activities in a project. The analysis
we have discussed so far takes no account of any limitations imposed by availability of
resources. The initial activity schedule was drawn up on the implied assumption that all of the
resources which are called for will be available. This is not necessarily so, and even if it is,
using the resources as required by this first schedule may be uneconomic.
The way in which activities are scheduled when resource considerations are taken into account,
depends on the objectives of the project controllers. It may be of greatest importance that the
At the outset of the project, the project manager should determine the objective for the
allocation of resources. The objective may be:
1. To maximize the utilization of the resources. The utilization can be assessed in terms of a
utilization factor:
Utilization factor = Total resource employed
Total resource provided
Manpower
required
Available
5 10 15 20
Time, weeks
Whenever a resource limit is exceeded, either more resources must be put into the project, or
activities must be re-scheduled in some way. It may even be necessary to delay the project
completion. Some activities, although they do not have a logical sequential relationship in the
project, cannot go on at the same time, because of resource limitations. This restriction can be
allowed for in the resource profile by limiting the resource available. This prevents these
activities being scheduled for the same time periods.
SUR Construction Share Co. have been awarded a contract to re-surface a car park. The project
Manager has identified eight essential activities in this work. The details are listed below
A - 3 1
B - 6 1
C - 7 2
D A 8 2
E C 4 1
F B,E 3 2
G C 10 2
H F,G 3 1
Unfortunately, due to the pressure of work on other jobs, SUR can spare only 4 men for the car
park job. We will determine how long the job will takes and how the men should be allocated.
Assume that all of the men can do all of the activities.
Solution: We will build up a resource profile, starting all activities at the earliest times. We can
then re-schedule the activities in order to meet the limitations on the number of men available.
We first draw the network and identify the critical path.
3 12
A 4 D
3
14
11
B 8 H
0 20
1 3 5 6
6 3
0 20 7
7
C G
17
17
7 10
2 The overall completion time, without
resource considerations, is 20 days.
The critical path is C-G-H.
We will construct a Gantt chart and build up a resource profile, assuming that all jobs are
started as soon as possible. The chart shows the float allocated at the end of each activity. This
diagram helps us to identify which activities are concurrent and which activities it is possible to
re-schedule, without causing an overall delay in the project.
A D
(2)
3 11
Activities
B
(1) 6
F
12
11
C G H
5 10 15
20
Resource profile for earliest start times for the example given
5 -- 4 Men available
4
D
D D
+
3 + +
A
B C E
+ F
+ +
B +
C G
2 + G
C
H G
ADU –Department of Business Administration Page 202 of 214
Operations Research
5 10 15 20
Days
We can see from the resource profile that the limit of 4 men available is exceeded when activity
D is in progress between days 3 and 11. It is not possible to re-schedule activities in order to
accommodate D entirely. The critical activities C and G both require 2 men, so that D cannot be
started during days 0-17, unless no other non-critical jobs are being carried out. If D is moved
to its latest start time of 12 days, there will still be days 12 to 14, when we require more than 4
men. G(2 men), F(2 men) and D(2 men) would be in progress during these 2 days. We either
hire one more man for this period, or we must start activity D when F has finished in day 14.
This will delay the project by 2 days, extending the duration from 20 to 22 days.
There are many project management software tools available. We will look at a sample of
products that are available. These product descriptions are intended as a basic introduction. Due
to the ever changing software products, it might be useful to visit the Web sites of the
individual software manufacturers to get up-to-date product information. Many sites provide
downloads and free trial usage of their products.
Microsoft
Microsoft Project (MS Project) dominates project management software systems. It must be
installed on every user's computer and is compatible with Microsoft Office. This allows team
members to save to the database documents created in any Office application. Users become
familiarized more quickly due to the same tool bars being used in MS Office. There are no
limits as to the number of tasks it can handle.
One very good feature of MS Project is that it allows multiple projects to share a common
resource pool; use existing e-mail infrastructure to communicate with other team members;
"flag" certain tasks as reminders; split tasks; analyze project using "worst" and "most-likely"
cases and many more.
Most project management software packages will generate Critical Path networks, Gantt
Charts, Resource histograms, billing information, cost estimates, actual times and costs
incurred, reports for project managers and the Steering Committee. As with most packages, it
achieves what traditional, human, project management has always achieved, but with greater
efficiency and speed, and with the ability to handle much greater complexity.
The use of project management software allows the manager to achieve the more
tedious, clerical tasks quickly and efficiently. The challenge for the project manager is
not to draw the diagrams, but to identify dependencies, and manage the resources
available
The software is able to perform the calculations for deriving critical paths and floats
available much faster than humans can
The software can help to track resources used across multiple projects
Having the plans available on software makes maintenance and adjustments easier.
Billing complications, arising from cross-project allocation, can be handled more simply
by the software
Document management: all documents produced for the project can be stored, and their
quality criteria set and checked, using the software.
The software is able to generate reports on the progress of the project, and automatically
flag any delays to the critical path, with recommended action.
Estimates of activity times for projects are usually made for some given level of resources. In
many situations it is possible to reduce the length of a project by injecting additional resources.
The impetus to shorten projects may reflect effects to avoid late penalties, to take advantage of
monetary incentives for timely completion of project, or to free resources for use on other
projects. In many cases however, the desire to shorten the length of a project merely reflects an
attempt to reduce the indirect costs associated with running the project, such as facilities and
equipment costs, supervision, and labor and presumed costs. Managers often have certain
options at their disposal that allow them to shorten, or crash, certain options at their disposal
that options are using additional personnel or more efficient equipment and relaxing work
specifications. Hence, a project manager may be able to shorten a project, there by realizing a
savings on indirect
project costs by increasing direct expenses to speed up the project. The goal in evaluating time-
and trade-offs is to identify a plan that will minimize the sum of the indirect and direct project
costs.
Crashing -is accelerating project of those critical activities that have the lowest ratio of
incremental cost to incremental time saved
Crash time -is the minimum time in which activity can be completed in case additional
remoras are inducted.
Crash cost -is the total cost of completing an activity in crash time.
The objective of project crash cost analysis is to reduce the total projected completion time (to
avoid late penalties, to take advantage of monetary incentives for timely completion of a
project, or to free resources for we on other the projects), while minimizing the cost of crashing.
Project completion time can be shortened only by crashing critical activities, which follows that
not all project activities should be crashed. However, when activities are crashed, the critical
path may change, requiring further crashing of previously non-critical activates in order to
further reduce the project completion time. In a nutshell, crashing means adding extra
resources, and managers are usually interested in speeding up project at the least additional
cost.
In order to make a rational decision about which activities (if any) to crash and the extent of
crashing desirable, a manager needs the following information:
1. Regular time and crash time estimates for each activity (normal)
2. Regular (normal) cost and crash cost estimates for “
3. A list of activates that are on the critical path
Note: Activities on critical path are potential candidates for crashing between
shortening non critical activities would not have an impact on total project
duration and activities are crashed according to crashing costs. Crash those
activities with the lowest cost first. Moreover, crashing should continue as long
as the cost to crash is less than the benefit received from crashing. These benefits
might take the form of incentive payments for early completion of the project as
part of a government contract, or they might reflect savings in the indirect
project costs, or both
Note that two or more paths may become critical as the original critical path becomes
shorter, so that subsequent improvements will require simultaneous shortening of two or
more
paths. In some cases it will be most economical to shorten an activity that is on two or more
of the critical paths. This is true whenever the crashing cost for a joint activity is less than
the sum of crashing one activity on each separate path.
Crash Normal
Time
Example:
Using information below develop an optimum time – cost solution. Assume that indirect project
costs are birr 1000 per day.
10
6 b 2 End
Start a f
5 9 e
C 4
d
Solution
Step 1. Determine activities on critical path and non critical path and their lengths.
Path Length
1-2-5-6 18
1-3-4-5-6 20 Critical path
2. Rank the critical path activates in order of lowest crashing cost and determine the no of
days each can be crashed.
3. Begin shortening the project, one day at a time, and check after each reduction to see
which path is critical.
4. 1. Shorten activity C by one day – length becomes 19 days
2. Shorten e by one day since shortening c is not possible. – length 18 days.
3. Since there are two critical paths, further improvement will necessitate
shortening one activity on each.
SUMMARY
18 18 18 17
1-2-5-6
20 19 18 17
1-3-4-5-6
Activity to be
- c d f
crashed
Note: The next activities to be crashed were ‘e’ and‘d’. However, if the two activities were
crashed, the total cost will be greater than (600+ 700 = 1300) the benefit (i.e., 1000/day).
Therefore, we stop the crashing at this point.
Total cost of crashing will be the normal cost of the activities plus the crash cost.
? Dear learner, can you compute the total cost of the above problem?
__________________________________________________________
__________________________________________________________
_
Summary
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 coordination the work.
PERT and CPM are the two commonly used techniques for developing and
monitoring projects. Although each technique is developed independently and
expressly different purposes, time and practice has erased most of the original
differences so that now little distinction can be made between the two. Either one
provides the manager with a rational approach to project planning along with a
graphical display of project activities. Both depict the sequential relationships that
exist among activities and reveal to managers which activity must be completed on
time in order to achieve timely completion of the project. Managers can use that
information to direct their attention towards the most critical activities.
In some cases it may be possible to shorten the length of a project by shortening one
or more of the project activities. Typically such gains are achieved by the use of
additional resources, although in some cases it may be possible to transfer resources
among project activities. Generally, projects are shortened either to the point where
the cost of additional reduction would exceed the benefit of additional reduction or
to the point where further improvements, although desirable, would be physically
impossible.
Activity:
Determine:
a)Critical path and Critical activities
b) Duration of the project
c) Slack time of activities
Estimat
(months
Activit a m b
y
1--->2 4 8 12
1--->3 6 10 15
1--->4 2 10 14
2--->5 3 6 9
2--->6 1 4 13
3--->5 3 6 18
4--->5 0 0 0
4--->7 2 8 12
5--->8 9 15 22
5--->7 5 12 21
7--->8 5 6 12
6--->9 7 20 25
8--->9 3 8 20
e. Critical path
f. Expected project duration and variance
3. The Stone River Textile Mill was inspected by OSHA and was found to be in violation of a
number of safety regulations. The OSHA inspectors ordered the mill to alter some existing
machinery to make it safer (Le., add safety guards, etc.); purchase some new machinery to
replace older, dangerous machinery; and relocate some machinery to make safer passages and
unobstructed entrances and exits. OSHA gave the mill only 35 weeks to make the changes; if
the changes w re not made by then, the mill would be fined $300,000. The mill determined the
activities in a PERT network that would have to be completed and then estimated the indicated
activity times, as shown below. Construct the PERT network for this
project and determine the following:
Time
Estimates
1.
4.The following table provides the information necessary to construct a
project network and project crash (weeks)
Activity
data. Description a m b
1-->2 Activity
Order new machinery Time 1 2 3
(weeks) Activity Cost ($)
Activit
1-->3 (i,j) PlanPredecessor Normal
new physical layoutCrash Normal 2 5 8 Crash
ya (1. 2) - 16 8 $2,00 $4,400
1-->4 Determine safety changes in 1 3 5
b (1,3) - 14 9 1,0000 1.800
c (2.4)existing
a machinery 8 6 500 700
d (2,5) a 5 4 600 1,300
e
2-->6 (3,5)
Receivebequipment 4 2 1,500 4 10 25 3.000
f (3.6) b 6 4 800 1,600
g
2-->5 (4,6)
Hire newc employees10 7 3,000 3 7 12 4.500
h (5,6) d.e 15 10 5,000 8.000
3-->7 Make plant alterations 10 15 25
4-->8 5 9 14 a. Constr
Make changes in existing uct the
machinery project
networ
5-->6 Dummy 0 0 0 k.
b. Manual
6-->7 Dummy 0 0 0
ly
7-->8 Dummy 0 0 0 crash
6-->9 Train new employees 2 3 7 the
networ
7-->9 Install new machinery 1 4 6
k to 28
8-->9 Relocate old machinery 2 5 10 weeks.
c. Formul
9-->10 Conduct employee safety orientation 2 2 2 ate the
general
linear programming model for this network.
d. Formulate the linear programming crashing model that would crash this model by the
maximum amount
5. Indirect costs for a project are $12,000 per week for as long as the project lasts. The
project manager has supplied the cost and time information shown below.
a 3 11,000
b 3 3,000 first week, 4000 others
c 2 6000
d 1 1000
e 3 6000
f 1 2000
10 14
a b
13 6
d
15 c
e 8
f
6. The following table provides the information necessary to construct a project network and project crash data.
Activity Time Activity Cost ($)
(weeks)
Activit
(i,j) Predecessor Normal Crash Normal Crash
y
a (1,2) 20 8 $1,000 $1,480
b (1,4) - 24 1,400
20 1,200
c
(1,3) - 14 7 700 1,190
d
(2,4) a 10 6 500 820
8. Given the following network with the indicated distances between nodes (in miles),
determine the shortest route from node 1 to each of the other four nodes (2, 3, 4, and