Linear Programming Problems {Operation Research}

LINEAR PROGRAMMING PROBLEM
INTRODUCTION
In a decision-making embroilment, model formulation is important
because it represents the essence of business decision problem. The
term formulation is used to mean the process of converting the verbal
description and numerical data into mathematical expressions which
represents the relevant relationship among decision factors, objectives
and restrictions on the use of resources. Linear Programming (LP) is a
particular type of technique used for economic allocation of 'scarce' or
'limited' resources, such as labour, material, machine, time, warehouse
space, capital, energy, etc. to several competing activities, such as
products, services, jobs, new equipment, projects, etc. on the basis of a
given criterion of optimally. The phrase scarce resources means
resources that are not in unlimited in availability during the planning
period. The criterion of optimality, generally is either performance, return
on investment, profit, cost, utilily, time, distance, etc.
George B Dantzing while working with US Air Force during World War
II, developed this technique, primarily for solving military logistics
problems. But now, it is being used extensively in all functional areas of
management, hospitals, airlines, agriculture, military operations, oil
refining, education, energy planning, pollution control, transportation
planning and scheduling, research and development, etc. Even though
these applications are diverse, all I.P models consist of certain common
properties and assumptions. Before applying linear programming to a
real-life decision problem, the decision-maker must be aware of all these
properties and assumptions, which are discussed later in this chapter.
Before discussing in detail the basic concepts and applications of
linear programming, let us be clear about the two words, linear and
programming. The word linear refers to linear relationship among
variables in a model. Thus, a given change in one variable will always
cause a resulting proportional change in another variable. For example,
doubling the investment on a certain project will exactly double the rate
of return. The word programming refers to modelling and solving a
problem mathematically that involves the economic allocation of limited

resources by choosing a particular course of action or strategy among
various alternative strategies to achieve the desired objective.
A large number of computer packages are available for solving a
mathematical LP model but there is no general package for building a
model. Model building is an art that improves with practice. To illustrate,
how to build I.P models, a variety of examples are given in this
chapter.
STRUCTURE OF LINEAR PROGRAMMING
General Structure of LP Model
The general structure of LP model consists of three components.
Decision variables (activities): We need to evaluate various
alternatives (courses of action) for arriving at the optimal value of
objective function. Obviously, if there are no alternatives to select from,
we would not need LP. The evaluation of various alternatives is guided
by the nature of objective function and availability of resources. For this,
we pursue certain activities usually denoted by x1, x2…xn. The value of
these activities represent the extent to which each of these is performed.
For example, in a product-mix manufacturing, the management may use
LP to decide how many units of each of the product to manufacture by
using its limited resources such as personnel, machinery, money,
material, etc.
These activities are also known as decision variables because they
arc under the decision-maker's control. These decision variables, usually
interrelated in terms of consumption of limited resources, require
simultaneous solutions. All decision variables are continuous,
controllable and non-negative. That is, x1>0, x2>0, ....xn>0.
The objective function: The objective function of each L.P problem is a
mathematical representation of the objective in terms of a measurable
quantity such as profit, cost, revenue, distance, etc. In its general form,
it is represented as:
Optimise (Maximise or Minimise) Z = c1x1 + c2X2. … cnxn

where Z is the mcasure-of-performance variable, which is a function of
x1, x2 ..., xn. Quantities c1, c2…cn are parameters that represent the
contribution of a unit of the respective variable x1, x2 ..., xn to the
measure-of-performance Z. The optimal value of the given objective
function is obtained by the graphical method or simplex method.
The constraints: There are always certain limitations (or constraints)
on the use of resources, e.g. labour, machine, raw material, space,
money, etc. that limit the degree to which objective can be achieved.
Such constraints must be expressed as linear equalities or inequalities in
terms of decision variables. The solution of an L.P model must satisfy
these constraints.
The linear programming method is a technique for choosing the best
alternative from a set of feasible alternatives, in situations in which the
objective function as well as the constraints can be expressed as linear
mathematical functions. In order to apply linear programming, there are
certain requirements to me met.
(a) There should be an objective which should be clearly identifiable
and measurable in quantitative terms. It could be, for example,
maximisation of sales, of profit, minimisation of cost, and so on.
(b) The activities to be included should be distinctly identifiable and
measurable in quantitative terms, for instance, the products included
in a production planning problem.
(c) The resources of the system which arc to be allocated for the
attainment of the goal should also be identifiable and measurable
quantitatively. They must be in limited supply. The technique would
involve allocation of these resources in a manner that would trade off
the returns on the investment of the resources for the attainment of
the objective.
(d) The relationships representing the objective as also the resource
limitation considerations, represented by the objective function and
the constraint equations or inequalities, respectively must be linear in
nature.
(e) There should be a series of feasible alternative courses of action
available to the decision makers, which are determined by the
resource constraints.

When these stated conditions are satisfied in a given situation, the
problem can be expressed in algebraic form, called the Linear
Programming Problem (LPP) and then solved for optimal decision. We
shall first illustrate the formulation of linear programming problems and
then consider the method of their solution.
ASSUMPTIONS OF LINEAR PROGRAMMING
The following four basic assumptions are necessary for all linear
programming models.
Certainty: In all LP models, it is assumed, that all model parameters
such as availability of resources, profit (or cost) contribution of a unit of
decision variable and consumption of resources by a unit of decision
variable must be known and is constant. In some cases, these may be
either random variables represented by a known distribution (general or
Additional info:
A large number of decision problems faced by a business manager involve allocation of
resources to various activities, with the objective of increasing profits or decreasing
costs, or both. When resources arc in abundance no difficulty is experienced. But such
cases are very rare. Practically in all situations, the management is confronted with the
problem of scarce resources. Normally, there are several activities to perform with
limitations of either of the resources or their use prevents each activity from being
performed to the desired level. Thus, the manager has to take a decision as to how
best the resources be allocated among the various activities.
The decision problem becomes complicated when a number of resources are required
to be allocated. There are several activities to perform. Rule of thumb, even for an
experienced manager, in all likelihood may not provide the right answer in such cases.
The decision problems can be formulated, and solved using mathematical
programming techniques.
Mathematical programming involves optimisation of a certain function, called the
objective functions subject to certain constraints. For example, a manager may be
faced with the problem of deciding the appropriate product mix of the four products.
With the profitability of the products along with the requirements of raw materials,
labour etc. known, his problem can be formulated as a mathematical programming
problem taking the objective function as the maximisation of profits obtainable from the
resources keeping in view the various constraints—the availability of raw materials,
labour supply, market and The methods of mathematical programming can be divided
into three groups: linear, integer, and non-programming.

may be statistical) or may tend to change, then the given problem can
be solved by a stochastic LP model or parametric programming. The
linear programming is obviously deterministic in nature.
Divisibility (or continuity): The solution values of decision variables
and resources are assumed to have either whole numbers (integers) or
mixed numbers (integer and fractional). However, if only integer variables
are desired, e.g. machines, employees, etc. the integer programming
method may be applied to get the desired values.
It is also an assumption of a linear programming model that the
decision variables are continuous. As a consequence, combinations of
output with fractional values, in the context of production problems, are
possible and obtained frequently. For example, the best solution to a
problem might be to produce 5 2/3 units of product A and 10 1/3 units
of product B per week.
Although in many situations we can have only integer values, but we can
deal with the fractional values, when they appear, in the following ways.
Firstly, when the decision is a one-shot decision, that is to say, it is not
repetitive in nature and has to be taken only once, we may round the
fractional values to the nearest integer values. However, when we do so,
we should evaluate the revised solution to determine whether the
solution represented by the rounded values is a feasible solution and
also whether the solution is the best integer solution. Secondly, if the
problem relates to a continuum of time and it is designed to determine
optimal solution for a given time period only, then the fractional values
may not be rounded. For instance in the context of a production
problem, a solution like the one given earlier to make 5 2/3 units of A
and 10 units of B per week, can be adopted without any difficulty. The
fractional amount of production would be taken to be the work-in-
progress and become a portion of the production of the following week.
In this case an output of 17 units of A and 31 units of B over a three-
week period would imply 5 2/3 units of A and 10 units of B per week.
Lastly, if we must insist on obtaining only integer values of the decision
variables, we may restate the problem as an integer programming
problem, forcing the solutions to be in integers only.

Additivity: The value of the objective function for the given values of
decision variables and the total sum of resources used, must be equal to
the sum of the contributions (profit or cost) earned from each decision
variable and the sum of the resources used by each decision variable,
respectively. For example, the total profit earned by 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
by A and B must be equal to the sum of resources used for A and B
individually.
This assumption implies that there is no interaction among the
decision variables (interaction is possible when, for example, some
product is a by-product of another one).
Finite choices A linear programming model also assumes that a
limited number of choices are available to a decision-maker and the
decision variables do not assume negative values. Thus, only non-
negative levels of activity are considered feasible. This assumption is
indeed a realistic one. For instance, in the production problems, the
output cannot obviously be negative, because a negative production
implies that we should be able to reverse the production process and
convert the finished output back into the raw materials!
Linearity (or proportionality): All relationships in the LP model (i.e. in
both objective function and constraints) must be linear. In other words,
for any decision variable j, the amount of particular resource say i used
and its contribution to the cost one in objective function must be
proportional to its amount. For example, if production of one unit of a
product uses 5 hours of a particular resource, then making 3 units of that
product uses 3 x 5 = 15 hours of that resource.
ADVANTAGES OF LINEAR PROGRAMMING
Following are certain advantages of linear programming.
1.Linear programming helps in attaining the optimum use of
productive resources. It also indicates how a decision-maker can
employ his productive factors effectively by selecting and
distributing (allocating) these resources.
2.Linear programming techniques improve the quality of decisions.
The decision-making approach of the user of this technique
becomes more objective and less subjective.
3.Linear programming techniques provide possible and practical

solutions since there might be other constraints operating outside
the problem which must be taken into account. Just because we
can produce
so many units docs not mean that they can be sold. Thus,
necessary modification of its mathematical solution is required for
the sake of convenience to the decision-maker.
4.Highlighting of bottlenecks in the production processes is the most
significant advantage of this technique. For example, when a
bottleneck occurs, some machines cannot meet demand while other
remains idle for some of the time.
5.Linear programming also helps in re-evaluation of a basic plan for
changing conditions. If conditions change when the plan is partly
carried out, they can be determined so as to adjust the remainder of
the plan for best results.
LIMITATIONS OF LINEAR PROGRAMMING
In spite of having many advantages and wide areas of applications,
there arc some limitations associated with this technique. These are
given below. Linear programming treats all relationships among decision
variables as linear. However, generally, neither the objective functions
nor the constraints in real-life situations concerning business and
industrial
problems are linearly related to the variables.
1.While solving an LP model, there is no guarantee that we will get
integer valued solutions. For example, in finding out how many men
and machines would be required lo perform a particular job, a non-
integer valued solution will be meaningless. Rounding off the
solution to the nearest integer will not yield an optimal solution. In
such cases, integer programming is used to ensure integer value to
the decision variables.
2.Linear programming model does not take into consideration the
effect of time and uncertainty. Thus, the LP model should be
defined in such a way that any change due to internal as well as
external factors can be incorporated.
3.Sometimes large-scale problems can be solved with linear
programming techniques even when assistance of computer is
available. For it, the main problem can be fragmented into several
small problems and solving each one separately.
4.Parameters appearing in the model are assumed to be constant but
in real-life situations, they are frequently neither known nor
constant.
It deals with only single objective, whereas in real-life situations we
may come across conflicting multi-objective problems. In such cases,
instead of the LP model, a goal programming model is used to get
satisfactory values of these objectives.
APPLICATION AREAS OF LINEAR PROGRAMMING
Linear programming is the most widely used technique of decision-
making in business and Industry and in various other fields. In this
section, we will discuss a few of the broad application areas of linear
programming.

Agricultural Applications
These applications fall into categories of farm economics and farm
management. The former deals with agricultural economy of a nation or
region, while the latter is concerned with the problems of the individual
farm.
The study of farm economics deals with inter-regional competition and
optimum allocation of crop production. Efficient production patterns can
be specified by a linear programming model under regional land
resources and national demand constraints.
Linear programming can be applied in agricultural planning, e.g.
allocation of limited resources such as acreage, labour, water supply
and working capital, etc. in a way so as to maximise net revenue.
Military Applications
Military applications include the problem of selecting an air weapon
system against enemy so as to keep them pinned down and at the
same time minimising the amount of aviation gasoline used. A variation
of the transportation problem that maximises the total tonnage of bombs
dropped on a set of targets and the problem of community defence
against disaster, the solution of which yields the number of defence
units that should be used in a given attack in order to provide the
required level of protection at the lowest possible cost.
Production Management
(i) Product mix A company can produce several different products,
each of which requires the use of limited production resources. In
such cases, it is essential to determine the quantity of each
product to be produced knowing its marginal contribution and
amount of available resource used by it. The objective is to
maximise the total contribution, subject to all constraints.
(ii) Production planning This deals with the determination of
minimum cost production plan over planning period of an item with
a fluctuating demand, considering the initial number of units in
inventory, production capacity, constraints on production,
manpower and all relevant cost factors. The objective is to
minimise total operation costs.
(iii) Assembly-line balancing This problem is likely to arise when an
item can be made by assembling different components. The
process of assembling requires some specified sequcnce(s). The
objective is to minimise the total elapse time.
(iv) Blending problems These problems arise when a product can be
made from a variety of available raw materials, each of which has
a particular composition and price. The objective here is to
determine the minimum cost blend, subject to availability of the raw
materials, and minimum and maximum constraints on certain
product constituents.
(v) Trim loss When an item is made to a standard size (e.g. glass,
paper sheet), the problem that arises is to determine which

combination of requirements should be produced from standard
materials in order to minimise the trim loss.
Financial Management
(i) Portfolio selection This deals with the selection of specific
investment activity among several other activities. The objective
is to find the allocation which maximises the total expected return
or minimises risk under certain limitations.
(ii) Profit planning This deals with the maximisation of the profit
margin from investment in plant facilities and equipment, cash in
hand and inventory.
Marketing Management
(i) Media selection Linear programming technique helps in
determining the advertising media mix so as to maximise the
effective exposure, subject to limitation of budget, specified
exposure rates to different market segments, specified minimum
and maximum number of advertisements in various media.
(if) Travelling salesman problem The problem of salesman is to find
the shortest route from a given city, visiting each of the specified
cities and then returning to the original point of departure,
provided no city shall be visited twice during the tour. Such type of
problems can be solved with the help of the modified assignment
technique.
Cm) Physical distribution Linear programming determines the most
economic and efficient manner of locating manufacturing plants
and distribution centres for physical distribution.
Personnel Management
a) Staffing problem Linear programming is used to allocate
optimum manpower to a particular job so as to minimise the total
overtime cost or total manpower.
b) Determination of equitable salaries Linear programming
technique has been used in determining equitable salaries and sales
incentives.
c) Job evaluation and selection Selection of suitable person for a
specified job and evaluation of job in organisations has been done
with the help of linear programming technique.
Other applications of linear programming lie in the area of
administration, education, fleet utilisation, awarding contracts, hospital
administration and capital budgeting, etc.

GRAPHICAL SOLUTION
Extreme point enumeration approach
Convex Polyhedron
TYPES OF SOLUTION
(a) Solution. Values of decision variables xj (j = 1, 2, 3, ….n) which satisfy
the constraints of the general L. P. P., is called the solution to that L.
P. P.
(b) Feasible solution. Any solution that also satisfies the nonnegative
restrictions of the general L. P. P. is called a feasible solution.
(c) Basic Solution. For a set of m simultaneous equations in n unknowns
(n> m). a
solution obtained by setting (n - m) of the variables equal to zero and
solving the remaining m equations in m unknowns is called a basic
solution. Zero variables (n - m) are called non-basic variables and
remaining m are called basic variables and constitute a basic solution.
(d) Basic Feasible Solution. A feasible solution to a general L.P.P. which
is also basic
solution is called a basic feasible solution.
(e) Optimum Feasible Solution. Any basic feasible solution which
optimizes (maximizes or minimizes) the objective function of a general
L.P.P. is known as an optimum feasible solution to that L.P.P.
(f) Degenerate Solution. A basic solution to the system of equations is
called degenerate if one or more of the basic variables become equal
to zero.
SPECIAL CASES IN LINEAR PROGRAMMING
Alternative (or Multiple) Optimal Solutions
So far we have seen that the optimal solution of any linear programming
problem occurs at an extreme point of the feasible region and the
solution is unique, i.e. no other solution yields the same value of the
objective function. However, in certain cases, a given LP problem may
have more than one solution yielding the same optimal objective
function value. Each of such optimal solutions is termed as alternative
optimal Solution.
There are two conditions that should be satisfied for an alternative
optimal solution to exist:
(i)The given objective function is parallel to a constraint that
forms the boundary (or edge) of the feasible solutions region. In
other words, the slope of the objective function is same as that
of the constraint forming the boundary of the feasible solutions
region, and

(ii) The constraint should
form a boundary on the feasible region in the direction of optimal
movement of the objective function. In other words, the
constraint should be an active constraint.
Remark: The constraint is said to be active or binding or tight, if at
optimality, the left-hand side of a constraint equals the right-hand side.
In other words, an equality constraint is always active. An inequality
constraint may or may not be active.
Geometrically, an active constraint is one that passes through one of
the extreme points of the feasible solution space.
Unbounded Solution
Sometimes an LP problem will not have a finite solution. This means
when one or more decision variable values and the value of the
objective function (maximisation case) are permitted to increase
infinitely without violating the feasibility conditions, then the solution is
said to be unbounded. It is important to note that there is the difference
between a feasible region being unbounded and an LP problem being
unbounded. It is possible for a feasible region to be unbounded but LP
problem not to be unbounded, that is an unbounded feasible region may
yield some definite value of the objective function. The general cause
for an unbounded LP problem is a mistake in mathematical model
formulation.
Infeasible solution
It has already been stated that a solution is called feasible if it satisfies
all the constraints and the non-negativity conditions. Sometimes it is
possible that the constraints may be inconsistent so that there is no
feasible solution to the problem. Such a situation is called infeasibility.
This means, there is no unique (single) feasible region. Such a problem
arises due to wrong model formulation with conflicting constraints.
Infeasibility depends solely on the constraints as has nothing to do with
the objective function.
In the graphic approach to the solution to an LPP, the infeasibility is
evident if its feasible region is empty so that there is no feasible region
in which all the constraints may be satisfied simultaneously.
Infeasibility vs Unboundedness
Both infeasibility and unboundedness have a similarity in that there is no
optimal solution in either case. But there is a striking difference between
the two, while in infeasibility there is not a single feasible solution, in
unboundedness there are infinite feasible solutions but none of them can
be termed as the optimal.

Redundancy
A redundant constraint is one that does not affect the feasible solutions
region (or space) and thus redundancy of any constraint does not cause
any difficulty in solving an LP problem graphically. In other words
constraint appears redundant when it may be more binding than
another.
SIMPLEX
Define l.p.p. (mathematical)
Explain the summary procedure for the maximization case of the
simplex method.
Step 1 Formulate the problem
 Translate the technical specifications of the problems into
inequalities, and make a precise statement of the objective function.
 Convert the inequalities into equalities by the addition of nonnegative
slack variables. These inequalities should be symmetric or balanced
so that each slack variable appears in each equation with a proper
coefficient.
 Modify the objective function to include the slack variables.
Step 2 Design an initial program (A basic feasible solution)
 Calculate the net evaluation row: To get a number in the net
evaluation row under a column, multiply the entries in that column by
the corresponding numbers in the objective column, and add the
products. Then subtract this sum from the number listed in the
objective row at the top of the column. Enter the result in the net
evaluation row under the column.
 Test : Examine the entries in the net evaluation row for the given
simplex tableau. If all the zero or negative, the optimal solution has
been obtained. Otherwise, the presence of any positive entry in the
net-evaluation row indicates that a better program can be obtained.
Step 3 Revise the program
 Find the key column. The column under which falls the largest
positive net-evaluation- row entry is the key column.
 Find the key row and the key number. Divide the entries in the
“quantity” column by the corresponding nonnegative entries of the key
column to form replacement ratios, and compare these ratios. The
row in which falls the smallest replacement ratio is the key row. The
number which lies at the intersection of the key row and the key
column is the key number.

 Transform the key row. Divide all the numbers in the key row (starting
with and to right of the “quantity” column) by the key number. The
resulting numbers form the corresponding row of the next tableau.
 Transform the non-key rows. Subtract from the old row number of a
given key row (in each column ) the product of the corresponding key-
row number and the corresponding fixed ratio formed by dividing the
old row number in the key column by the key number. The result will
give the corresponding new row number. Make this transformation for
all the non-key rows.
 Enter the results of (3) and (4) above in a tableau representing the
revised program.
Step 4 Obtain the optimal program
Repeat steps 3 & 4 until a program has been derived.
[ Linear-programming problems involving the minimization of an
objective function usually contain structural of the “greater than equal to”
type. They can also be solved by the simplex method. The simplex
procedure for solving a linear-programming problem in which the
objective is to minimize rather than maximize a given function, although
basically the same as above, requires sufficient modifications to deserve
the listing of a separate summary. ]
9. Summary procedure for the simplex method (minimization case)
Step 1 Formulate the problem
 Translate the technical specification of the problem into inequalities,
and make a precise statement of the objectivity function.
 Convert the equalities into inequalities by the subtraction of
nonnegative slack variables. Then modify these equations by the
addition of nonnegative artificial slack variables. These equations
should be asymmetric or balanced so that each slack and artificial
slack variable appears in each equation with a proper coefficient.
 Modify the objective functions to include all the slack artificial slack
variables.
Step 2 Design an initial program (a basic feasible solution).
 Design the first program so that only the artificial slack variables are
included in the solution. Place the program in a simplex
tableau. In the objective row, above each column variable, place the
corresponding coefficient of that variable from step 1.c.In particular,
place a zero above each column containing an artificial slack variable.

Step 3 Test and revise the program.
 Calculate the net evaluation row. Toto get a number in the net
evaluation row under a column, multiplying the entries in that column
by the corresponding number in the objective column, and add the
products. Then subtract the sum from the number listed in the
objective row above the column. Enter the result in the net-evaluation
row under the column.
 Test. Examine the entries in the net-evaluation row for the given
simplex tableau. If all the entries are zero or positive, the optimum
solution has been obtained. Otherwise, the presence of any negative
entry in the net-evaluation row indicates that a better program can be
obtained.
 Revise the program.
 Find the key column. The under which falls the largest negative net-
evaluation-entry is the key column.
 Find the key row and the key number. Divide the entries in the
“Quantity” column by the corresponding nonnegative entries in the
key column to form replacement ratios, and compare these ratios.
The row in which the smallest replacement ratio falls is the key row.
The number which lies at the intersection of the key row and the key
column is the key number.
 Transform the key row. Divide all the numbers in the key row (staring
with and to the right to the of the “Quantity” column by the key
number. The resulting numbers form the corresponding row of the
next tableau.
 Transform the non-key rows. Subtract from the old row the number of
a given non-key row (in each column) the product of the
corresponding fixed ratio formed by dividing the old row number in the
key column by the number. The result will give the corresponding new
row number. Make the transformation for all rows.
 Enter the results of (3) and (4) above in a tableau representing the
revised program.
Step 4 Obtain the optimal program.
Repeat steps 3and 4 until an optimal program has been derived.
We repeat the following comments comparing the maximization
and minimization problems as solved by the simplex method.
The procedure for calculating the net-evaluation row is the same in
both cases. However, whereas the largest positive value is chosen to
identify the incoming product in a maximization problem, the most

negative value is chosen in a minimization problem. The rest of the
mechanics, namely, the transformation of the key and the non-key rows,
is exactly the same. The decision rule identifying the optimal solution is
the absence of any positive value in the non-evaluation row in the
maximization problem, and the absence of any negative value in the
minimization problem.
Algorithm, degeneracy
DUAL SIMPLEX
In ordinary simplex method we start with feasible but non-optimal
solution while in Dual Simplex Method, we start with infeasible but
optimal solution. Successive iterations will maintain optimality to remove
infeasibility of the solution. The following steps are followed to arrive at
optimal feasible solution.
(1) Write down objective function as maximization and all constraints
as ≤ or =.
(2) Construct first dual simplex table from the given problem in usual
manner.
(3) The leaving variable is the basic variable having the most negative
value (Break ties, if any, arbitrarily). If all the basic variables are
non-negative, the process ends and the feasible (optimal) solution
is reached.
(4) To determine the entering variable take ratios of the coefficients of
non-basic variables in the objective function to the corresponding
coefficients in the row associated with the leaving variable. Ignore
the ratios with positive or zero denominators. The entering variable
is the one with the smallest absolute value of the ratio. (Break ties,
if any, arbitrarily). If all the denominators are zero or positive, the
problem has no feasible solution.
After selecting the entering and leaving variables, row operations
are applied as usual to obtain the next table.
Application of this Dual Simplex Method is useful in Sensitivity
Analysis. For example, suppose a new constraint is added to the
problem after the optimal solution is reached. If this constraint is not
satisfied by the optimal solution, the problem remains optimal but it
becomes infeasible. The Dual Simplex Method is then used to clear the
infeasibility in the problem.

Example : Minimize Z = 2x1 + x2 subject to 3x1 + x2 ≥ 3, 4xx +
3x2 ≥ 6,
x1 + 2x2 ≤ 3, x1, x2 ≥ 0.
CPM and PERT
CRITICAL PATH ANALYSIS
Advantages of critical path analysis
1. It allows for a comprehensive view of the entire project. Because
of the sequential and concurrent relationships, time scheduling
becomes very effective. Identifying the critical activities keeps the
executive alert and in a state of preparedness, with alternative
plans ready in case these are needed. Breaking down the project
into the smaller components permits better and closed control.
2. Critical path analysis offers economic and an effective system of
control based on the principle of management by exception. That
is need for corrective action arises only in exceptional situations
and in most other cases; performance is in conformity with the
plans.
3. It is a dynamic pool of management which calls for a constant
review, early formulation of the network, and finding the current
path of relevance and optimum resources allocation.
Events
The beginning and ending points of an activity or a group of activities are
called events, aka nodes and connectors. It is often graphically
represented by a numbered circle. All activities in the network must
commence from some event. Such events are called the tail event.
Similarly all activities in the network must our terminal point called the
head event.
In the network, symbol "i" is used for the tail event, also called the
preceding event and "j" for the head event also called the succeeding
event of an activity.
The activity is then denoted by "i-j".
Conventions adopted in drawing networks

There are two conventions normally adopted while drawing networks.
They are
Time flows from left to right.
Head events always have a number higher than that of tail events.
Graphical representation of events and activities
Events are represented by numbers within circles. Activities are
represented by arrows; the arrowheads represent the completing of the
activities. Lengths and orientation of the arrow are of no significance and
are only chosen for convenience.
Fundamental properties governing the representation of events and
activities
The representation of events and activities is governed by one simple
dependency rule which requires that an activity which depends on
another activity is shown to emerge from the head event of the activity
upon which depends and that only dependent activities are drawn in this
way.
Errors and logical sequence: two types of errors in logic may arise when
drawing network, particularly when it is a complicated one. These are
known as looping and dangling.
Looping: Normally in the network, the arrow points from left to right. This
convention is to be strictly adhered, as this would avoid the illogical
looping, as shown wrongly below.
Dangling: The situation represented by the following diagram is also at
fault, since the activity represented by the dangling arrow 9-11 is
undertaken with no result.
1 2 3 4

To overcome problems arising due to dangling arrows, we must make
sure that
All events except the first and the last must have at least one
activity entering and one activity leaving them.
All activities must start to finish within event.
Duplicate activities: consider the following figure
Here A and B are known as duplicate activities because they have the
same head and tail events. A remedy the situation is introduction of a
dummy activity.
Dummy activity: it is a hypothetical activity which consumes known
source or time. It is represented by a dotted line and inserted in the
network to clarify activity that turn on the following situations
(i) it is created to make equities with, starting and finishing events
distinguishable,
(ii) to identify and maintain the proper precedence relationship between
activities that are not connected by events
(iii) to bring all "loose ends" to single initial and a single terminal event in
each network using dummies if necessary.
6 7
8
9
6 7 8
A
B

For example the problem of duplicate activities may be circumvented as
shown:
Numbering the events
The event numbers in the network should in some respects reflects the
logical sequences. When a complicated network has been drawn when
the problem of assigning numbers to the events involved in the network
arises. A rule devised by D. R. Fulkerson, involving the following steps
may be followed to resolve the problem of numbering events.
1. And initial event is one which has arrow/arrows coming out of it
and none of the arrow entering it. In a network there will be only
one such event. Call it 1.
2. Delete all arrows coming of from the event 1. This will give us at
least one more initial event.
3. Numbers these events as 2,3…
4. delete all emerging arrows from these numbered events which will
create new initial events. Then follow steps 3.
5. Continue the above steps the last event is obtained which has no
arrows coming out of it.
Consider the numbering of events in the following figure:
6 8
A
B7
1 2 4
5
3
6 7 8
X
Y

Here we proceed from left to right. The event with the least x coordinate
is assigned the smallest integer, say 1. Other events are assigned
progressively higher integers with regard to x coordinate. If two were
more events such as 4 & 5 have the same x coordinate the one towards
the arrow should have a higher number.
It is better to number of the events as 10, 20, 30.... to afford insertions of
more activities and events omitted by oversight.
Concurrent activities
Activities may not always be discreet that is, they may be done in part
allowing the subsequent activities to commence before the preceding
activity is fully completed. Activities of this kind are to be frequently
encountered in batch production. If, for example, a batch of 50 spindles
is to be processed onto machines obviously it is not necessary to
process all the items of the batch on the first machines and then transfer
these to the next machines. A few items processed on the first machines
may be transferred to the second machine before completion of the
entire batch on the first machine. Concurrent activities are to be
encountered in sewage work, trenching, laying pipe, welding pipe, and
backfilling all going on simultaneously with suitable lags on construction
work.
Forward pass computations
As stated above, the purpose of the forward pass is to compute the
earliest start (EST) and finish time (EFT) for each activity. The EST time
indicates the earliest time that a given activity can be scheduled.
Earliest finish time for an activity indicates the time by which the activity
can be completed, at the earliest. To compute these time estimates, we
will first of all compute the earliest allowable occurrence time for various
events of the network.
It is a convention to keep the earliest allowable occurrence time of the
START event as zero.
To understand, how this time estimate for other events is computed, let
us consider the following network diagram.

Earliest allowable occurrence time
In the network shown above, event 1 stands for the beginning of the
activity 1 – 2 and we can say that it occurs at the time zero i.e. E1 = 0.
Event 2 stands for the finish of the activity 1 – 2 , thus event 2 can occur
at the earliest time E2 which is computed as
E2 = 0 + D 12 = 0 + 6 = 6
Where D12 stands for the duration of activity 1 – 2
Event 3 stands for finish of the activity 2 –3 and its earliest time is
E3 = E2 + D23 = 6 + 8 = 14
The event 4 can occur either at the end of the activity 3-4 or at the
finish of activity 2 – 4. In this case, there will be two time estimates as
follows:
E4 = E2 + D34 = 14 + 0 = 14
E4 = E2 + D24 = 6 + 10 = 16
In case, two or more time estimates exist for a particular
event, then the time estimate with maximum value is retained as the
earliest event time and other values are discarded. This maximum
value represents the completion of all the activities ending at the event
under consideration. In the above example, the earliest event time for
event 4 will be 16.
1 2
3
4
5 6
E1= 0 E2= 6
E3= 14
E4= 16
E5= 36 E6= 52
6
8
10
20
6
0
16
L1= 0 L2= 6
L3= 16
L4= 16
L5= 36 L6= 52

A general rule can also be given here for determining the earliest
event time as below:
E1 = Max (Ei + Dii)
When E1 is the earliest time for even j, Ei is the earliest time for
event I and Dii is the duration of the activity i-j.
Earliest start and finish times of an activity
After computing the earliest event time of various events, one can easily
compute the earliest start and finish times of all the activities on the
network. The earliest start time of an activity is given by the
earliest allowable occurrence time of the tail even of that activity.
Thus, in our example, the earliest start time of the activity 1-2 will be
given by the earliest time of the event 1 i.e. it will be 0. The earliest start
time for the activities 2-3 and 2-4 will be given by the earliest time of
event 2 which is equal to 6. The earliest time for the activities 3-4 and 3-
5 will be 14 which is the earliest time for the event 3 and so on.
The earliest finish time of an activity will be simply equal to the
earliest start time of the activity plus the duration of that activity.
Hence, in our example, earliest finish time of activity 1-2 will be 0 + 6 =
6, for activity 2-3, it will be 6 + 8 = 14 and for activity 2-4, it will be 6 + 10
= 16 and so on.
The complete computations for all the activities are shown in columns
(3) and (4) of Table 1.
1.4.3 Backward pass computations
The purpose of the backward pass is to compute latest start and finish
times for each activity. These computations are precisely a “mirror
image” of the forward pass computations. The term “latest allowable
occurrence” of an even (denoted by L1) is used in the sense that the
project terminal event must occur on or before some arbitrary scheduled
time. Thus, the backward pass computations are started rolling
back by arbitrarily specifying the latest allowable occurrence time
for the project terminal event. If no schedule date for the completion

of the project is specified, then the convention of setting the latest
allowable time for the terminal event equal to its earliest time,
determined in the forward pass, is usually followed i.e. L = E for the
terminal event of the project. This convention is called the zero slack
convention. Following this, one can also interpret the latest allowable
activity finish time (LFT) as the time to which the completion of an
activity can be delayed without directly causing any increase in the
total time to complete the project.
To explain the computation, let us again consider the network diagram in
figure 24. The terminal even is 6 so we set L6 = E6 = 52 and we start
rolling back. The latest allowable occurrence time for the events 5 and 4
are L5 = 52 – 16 = 36 and L4 = 36 – 20 = 16 respectively. It may be
noted here that can roll back to event 3 via activity 3 –5 as well as
activity 3 – 4. So there are two latest allowable occurrence times for the
event 3 as given below:
L3 = L4 – D34 = 16 – 0 = 16
L3 = L5 – D35 = 36 – 6 = 30
We retain the minimum value as the latest occurrence time for the
event 3 and ignore other values. Therefore, the latest allowable
occurrence time for the event 3 is 16. Similarly
L2 = L3 – D23 = 16 – 8 = 8
L2 = L4 – D24 = 16 – 10 = 6
The latest occurrence time for the event 2 is thus 6 and the latest
occurrence time for the event 1 is equal to its earliest time i.e. zero.
In general, latest allowable occurrence time of an event can be
calculated by selecting an appropriate formula among the following two:
Li = Li – Dii or
Li = minimum (Li – Dii)
The second formula is used for the event having two or more latest
allowable occurrence time estimates.
Latest start and finish times of an activity

After computing latest allowable occurrence time for various
events, one can compute the latest start and finish times of an activity.
The latest finish time of an activity is equal to the latest allowable
occurrence time of the head event of that activity.
i.e. LFT (i-j) = Lj
The latest start time of an activity is equal to its latest finish
time minus its duration.
i.e. LST (i-j) = LFT (i-j) – Dij
These computations are shown in column (4) and column (6) of
Table 1, given below:-
Table 1
Start Finish
Activity Duration Earliest
time
Latest
time
Earliest
time
Latest
time
(1) (2) (3) (4) (5) (6)
1-2 6 0 0 6 6
2-3 8 6 8 14 16
2-4 10 6 6 16 16
3-4 0 14 16 14 16
3-5 6 14 30 20 36
4-5 20 16 16 36 36
5-6 16 36 36 52 52
The critical path determination
After having computed various time estimates, we are now interested in
finding the critical Path of the network. A network will consist of a
number of parts. A path is a continuous series of activities through the

network that leads from the initial event (or node) of the network to its
terminal event. For finding the critical Path, we list out all possible paths
through a network along with their duration. In the network under
consideration, various paths have been listed as follows:
Path length in days
1-2-3-5-6 36
1-2-4-5-6 52
1-2-3-4-5-650
Critical Path: a path in a project network is called critical if it is the
longest Path. The activities lying on the critical Path are called the critical
activities.
In the above example, the Path 1-2-4-5-6 with the longest duration of 52
days is the critical Path and the activities 1-2,2-4, 4-5 and 5-6 are the
critical activities.
Calculation of floats
It may be observed that for every critical activities in a network, the
earliest start and latest start times are the same. This is so since the
critical activities cannot be scheduled later than the earliest scheduled
time without delaying the total project duration, they do not have any
flexibility in scheduling. However, noncritical activities do have some
flexibility. That is these activities can be delayed for sometime without
affecting the project duration. This flexibility is termed as slack in case of
an event and as floats in case of an activity.
Some people do not make any distinction between a slack and a float.
Slack time for an event
The slack time or slack of an event in a network is the difference
between the latest event time and the earliest event time.
Mathematically it may be calculated using the formula Li – Ei where Li is
the latest allowable occurrence time and Ei is the earliest allowable
occurrence time of an event i.

Total float of an activity
The total activity float is equal to the difference between the earliest and
latest allowable start or finish times for the activity in question. Thus, for
an activity (i-j), the total float is given by:
TFij = LST – EST or TFij = LFT – EFT
In other words, it is the difference between the maximum time available
for the activity and the actual time it takes to complete. Thus, total float
indicates the amount of time by which the actual completion of an
activity can exceed its earliest expected completion time without causing
any delay in the project duration.
Free float
It is defined as that portion of the total float within which an activity can
be manipulated without affecting the float of the succeeding activities. It
can be determined by subtracting the head event slack from the total
float of an activity.
i.e. FFij = TFij – (slack of event j)
The free float indicates the value by which an activity in question can be
delayed beyond the earliest starting point without affecting the earliest
start, and therefore the total float of the activities following it.
Independent float
It is defined as that portion of the total float within which an activity can
be delayed for start without affecting float of the preceding activities. It is
computed by subtracting the tail event slack from the free float.
i.e. IFij = FFij – (slack of event i)
The independent float is always either equal to a less than the free float
of an activity. If a negative value is obtained, the independent float is
taken to be 0.
Interfering float

Utilisation of the float of an activity can affect the float of subsequent
activities in the network. Thus, interfering float can be defined as that
part of the total float which causes a reduction in the float of the
successor activities. In other words, it can be defined as the difference
between the latest finish time of the activity under consideration and the
earliest start time of the following activity, or 0, whichever is larger. Thus,
interfering float refers to that portion of the activity float which cannot be
consumed without affecting adversely the float of the subsequent activity
or activities.
Example
Activity Duration
1-2 4 days
1-3 12 days
1-4 10 days
2-4 8 days
2-5 6 days
3-6 8 days
4-6 10 days
5-7 10 days
6-7 0 days
6-8 8 days
7-8 10 days
8-9 6 days
With the help of the activities given above draw a network. Determine its
critical path, earliest start time, earliest finish tine, latest start time, latest
finish time, total float, free float and independent float.
PROGRAMME EVALUATION AND REVIEW TECHNIQUE
(a) The optimistic time estimate: this is the estimate of the shortest
possible time in which an activity can be completed on the ideal
conditions. For this estimate, no provision for delays of setbacks are
made. We shall denote this estimate by to.
(b) The pessimistic time estimate: this is the maximum possible time
equity to accomplish the job. If everything went long and normal
situations prevailed, this would be the time estimate. It is denoted by tp.

(c) The most likely time estimate: this is the time which lies between the
optimistic and pessimistic time estimates. It assumes that things go in a
normal way with few setbacks. It is represented by tm.
Beta distribution is found to give fairly satisfactory results for most of the
activities. For distribution of this type the standard deviation is
approximately 1/6 of the range.
i.e
6
op
t
tt
S


The variance, therefore is
2
2
6 




 
 op
t
tt
S
Expected time:
The expected time (te) is the average time taken for the completion of
the job. By using beta distribution, the expected time can be obtained
from the following formula.





 

6
4 pmo
e
ttt
t
Probability of achieving completion date
Suppose we wish to find out the probability that the project will be
completed within the scheduled completion time. The time te as
determined by beta distribution after taking into account the time
estimates viz. to, tp, tm only represents a 50% chance that the activity will
be completed within time te.
In general are project consisting of several activities will have a normal
distribution, that is for the project as a whole, the distribution curve will
be a normal curve and the probability of competing project in time equal
to the mean value Te which is ½.

A standardised normal curve has an area equal to 1 and the standard
deviation of 1. Further, it is symmetrical about the mean value te. Hence
the area under the curve AC is 50% of the total area under the curve
ACB. The area under the curve ACD depends on the location of Ts along
the time axis. The point te can be taken as the reference point and the
distance (te- Ts) can be expressed in terms of standard deviation. For
example, if Ts is on the right of te at a distance of one standard deviation
in the enclosed by ACD is 84.1%. If Ts is on the left of te at a distance of
one standard deviation then the area enclosed is 15.9%.
A distance of +1 corresponds to 84.1% probability and a distance of -1
corresponds to 15.9% probability.
We calculate the value of the standard normal variate (Z) as follows:
..
1
DS
TT
Z cp

Where T1 denotes the duration in which we wish to complete the project
and Tcp represents the duration of the critical path, S.D. stands for
standard deviation of the earliest finish of a network.
Computed variance Vt (= 2
tS ) of all the activity durations of the critical
path. Sum up these and take the square root. This yields the S.D. of the
earliest finish time of a network. Let the critical path duration the
designated by Tcp assuming normal distribution for the total duration, you
should be in a position to find the confidence interval for Tcp. Look at the
standard normal probability distribution tables for the probability of
competing project within the given duration of T1.
Examples 10. PERT calculations yield a project length of 60 weeks with
the variance of 9. Within how many weeks we expect the project to be
completed with the probability of 0.99? (That is the project length that he
would expect to be exceeded only by 1% of the time if the project were
repeated many times in an identical manner).
Solution
Tcp = 60 S.D. = 9 = 3
60 + (3 x 2.3) = 67

A few comments on assumptions of PERT AND CPM
1. Beta distribution may not always be applicable.
2. The formula for the expected duration and S. D. are simplifications.
Maccrinnon and Ryavec reached the conclusion that in certain
cases the errors, because of these assumptions may even be to
the tune of 33 percent.
3. The errors owing to the aforesaid simplification and assumptions
may be compounded or may cancel each other to an extent.
4. In computing the S. D. of the critical path, independence of
activities is implied. Limitations of resources may invalidate the
independence which exists by the very definition of an activity.
5. It may not always be possible to sort of completely identifiable
activities and state where they begin and where they end.
6. In projects fraught with uncertainty it is natural that the existing
alternatives with differing outcomes. For example for particular
hardness is not obtained in a metal, an alloy might have to be used
that is more expensive and also inferior on certain technical
considerations. There have been theoretical developments in this
regard, and it may be worthwhile to incorporate the concept of
decision tree analysis depending on the situation.
7. Time estimates have an element of subjectiveness and, to that
extent, the techniques could be weak. The contractors reacts with
this weakness shrewdly whilst bidding. If there are cost plus
contracts they would deliberately underestimate the time for
chances of being awarded the contract. Incentive type contracts
might lead to an opposite bias.
8. Cost-time trade-offs, for deriving the cost curve slopes, are
subjective again and call for ability of expertise of the technology
as well as genuine effort to estimate. Often the engineers tend to
be lax here; occasionally with the honest deliberation even, their
guesses may be wide off the mark.
Distinctions between PERT and CPM
The PERT and CPM models are similar in terms of their basic structure,
rationale and mode of analysis. However, there are certain distinctions
between pert and CPM networks which are ennumerated below.
1. CPM is activity oriented that is CPM network is built on the basis of
activities. Also results of various calculations are considered in

terms of activities of the project. On the other hand, PERT is event
oriented.
2. CPM is a deterministic model that is it does not take into account
the uncertainties involved in the estimation of time for execution of
a job or an activity. It completely ignores the probabilistic element
of the problem. Pert, however is the probabilistic model. It uses
three estimates of the activity time; optimistic, pessimistic and
most likely; with a view to take into account time uncertainty. Thus,
the expected duration of each activity is probabilistic and expected
duration indicates that there is 50% probability of getting the job
done within that time.
3. CPM places dual emphasis on time and cost and evaluates the
trade-off between project cost and project time. By deploying
additional resources, it allows the critical path project manager to
manipulate project duration within certain limits so that project
duration can be shortened at an optimal cost. On the other hand,
pert is primarily concerned with time. It helps the manager to
schedule and coordinate various activities so that the project can
be completed on schedule time.
4. CPM is commonly used for those projects which are repetitive in
nature and where one has prior experience of handling similar
projects. What is generally used for those projects with time
required to complete various activities are not known before hand.
Thus, pert is widely used for planning and scheduling research and
development projects.
PROJECT CRASHING
In some cases, there are compelling reasons to complete a project
earlier than the originally estimated time duration of the critical path
computed on the basis of normal activity times, by employing extra
resources. An example would be introduction of a new project. The
motives in hastening the project might be to ensure that the competitors
do not steal a march. Increase or decrease in the total duration of the
completion time for project is closely associated with cost
considerations. In such cases when the total time duration is reduced,
the project cost increases, but in some exceptional cases project cost is
reduced as well. Production cost occurs in the cases of those projects

which make use of a certain type of resources for example a machine
and whose time is more valuable than the operator’s time.
Some definitions:
Activity cost: it is defined as the cost of performing and completing a
particular activity or task.
Crash cost, Cc: this is the direct cost that is anticipated in completing an
activity within the crash time.
Crash time, Ct: This is the minimum time required to complete an
activity.
Normal cost Nc: this is the lowest possible direct cost required to
complete an activity.
Normal time Nt: this is the minimum time required to complete an activity
at normal cost.
Activity costs slope: the costs slope indicates the additional cost incurred
per unit of time saved in reducing the duration of an
activity.

Let OA represent the normal duration of completing a job and OC the
normal cost involved to complete the job. Assume that the management
wish to reduce the time of completing the job to OB from normal time
OA. Therefore under such a situation the cost of the project increases
and it goes upto say OD (Crash Cost). This only amounts to saving that
by reducing the time period by BA the cost has increased by the amount
CD. The rate of increase in the cost of activity per unit decrease in time
is known as cost slope and is described as follows.
Activity cost slope =
OBOA
OCOD
AB
CD



=
CrashtimeNormaltime
NormalCosttCrash

cos
Optimum duration: the total project cost is the sum of the direct and
indirect costs. In case the direct cost varies with the project duration
time, the total cost would have the shape as indicated in the above
figure.
t Point A, the cost will be minimum. The time corresponding to this point
Point A is called the optimum duration and the cost as optimum cost for
the project.
Crash
cost
Normal
cost
Crash
Time
Normal
Time
COST
DURATION FOR THE JOB
D E
C
F
ABO

TRANSPORTATION
What is the transportation problem?
The transportation problem consists of three components. First we
can formulate a linear objective function, which is to be minimized.
This function will represent the total shipping cost of all the goods to
be sent to all the destinations. Second we can write a table of
constraints. Of the seven constraints of this problem, three (one for
each row) will give the relationships between the origin capacities and
the goods to be received by different destinations. These are called
capacity constraints. The other four constraints (one for each column)
will relationships between destination requirements and the goods to
be shipped from different origins. These are called as requirement
constraints. Third we can specify a set of non-negativity constraints
for the structural variables xij. They will state that no negative
shipments be permitted. The general correspondence between a
typical linear programming problem and the transportation problem is
thus complete.
Minimize
COST
A
O
TOTAL PROJECT COST
DIRECT COST
INDIRECT COST
Crash NormalOptimal
TIME

General Transportation Tableau
Destination (Di)
Origin
(oj)
D 1 D 2 _ _ _ _ _
_
Dn Ai
O1
C11
X11
C12
X12
C1n
X1n
A1
O2
C21
X21
C22
X22
C2n
X2n
A2
|
|
|
Om
Cm1
Xm1
Cm2
Xm2
Cmn
Xmn
Am
Req Bj B1 B2 -------- Bn
 

n
j
j
m
i
i BA
11

What are the methods to solve a transportation problem?
A. Procedure summary for the modified distribution method
(minimization case)
Step 1 Obtain a basic feasible solution
An initial basic feasible solution for a given transportation may be
obtained by following the northwest corner rule, by the application of
Vogel’s approximation method, or by simple inspection.
Test for step 1: A basic feasible solution must include shipments
covering m+n-1 cells. That is number of occupied cells is 1 less than
the number of rows and columns in the transportation matrix.
If number is more then recheck the data. If the number of occupied
cells is less than m+n-1 then this is a degenerate solution. To resolve
the degeneracy, add one or more epsilons to some “suitable” empty
cells so that the number of occupied cells becomes equal to m+n-1.
Step 2 Determine the opportunity costs of the empty cells
(Opportunity costs = Implied costs –Actual costs)
a) Determine a complete set of row and column numbers (values).
When in a given program, the number of occupied cells equals
m+n-1, proceed to assign row and column numbers in such a
manner that for each occupied cell the relationship cij =ui+vj holds.
To start, a value of zero can be assigned to any row having an
occupied cell. For each occupied cell, the actual shipping cost per
unit should equal the sum of its row and column values.
b) Calculate the implied cost of empty cells. Once all the row and
column values have been assigned, the implied cost of a given
empty cell can be calculated as follows :
Implied cost = row value + column value.
c) Determine the opportunity costs of empty cells. The opportunity
cost of an empty cell is determined by subtracting the actual cost
of the empty cell from the implied cost. In other words, opportunity
cost, for each cell is given by
Opportunity cost = ui+ vj –cij.

If the opportunity costs of all the cells are non-positive, an optimal
solution has been obtained else a better program can be
designed. Thus step 2 serves as a test for optimality.
Step 3 Designed an improved program
Design a new program such that the empty cell having the largest
opportunity cost is included in the solution. This is accomplished in the
following manner:
a) draw a loop of horizontal and vertical arrows in such a manner
that it starts from the cell to be filled, passes to an occupied cell
in the same row or column as the empty cell, and then, making
a series of alternate horizontal and vertical turns through
occupied cells returns to the empty cell.
b) Place the plus (+) sign in the empty cell to be filled. Then
alternately, place minus signs and plus signs at the beginnings
and ends of the connecting ends of the loops.
c) Examine those occupied cells in which the minus signs have
been placed. Of these, the cell having the least number is
vacated by transferring these units to the empty cell. This is
accomplished by adding the same amount to all cells having
plus signs and subtracting it from all cells having minus signs.
The improved program should have the same number of
occupied cells as the preceding program. If the number is less
then the problem becomes degenerate. In such a case , add
epsilons to some recently vacated cells such that the number of
occupied cells again equals m+n-1.
Step 4: Repeat steps 2 and 3 until a program is achieved in which
each empty cell has a n opportunity cost value which is either zero
or negative. This program will be an optimal program.
B. Modified distribution method (maximization case)
Except for one transformation, a transportation objective is to
maximize a given function can be solved by the MODI algorithm
as presented above. The transformation is made by subtracting all
the cij’s from the highest cij (profit) of the given transportation
matrix. The transformed cij’s give us the relative costs, and the
problem then becomes a minimization problem. Once an optimal
solution to this minimization problem has been found, the value of
the objective function can calculated by inserting the original

values of the cij’s for those routes, which form the basis (occupied
cells) in the optimal solution.
What is degeneracy? How can it be resolved?
The solution for non-degenerate basic feasible solution with exactly
m+n-1 strictly positive allocations in independent positions has been
discussed so far. However, sometimes it is not possible to get such
initial feasible solution to start with. Thus degeneracy occurs in the
transport problem whenever the number of occupied cells is less than
m+n-1.
Degeneracy in transportation can occur in two ways:
Basic feasible solutions may be degenerate from the initial stage
onward.
They may become degenerate at any intermediate stage.
Resolution of degeneracy during the initial stage.
To resolve degeneracy, allocate an extremely small amount of goods
(close to zero) to one or more of the empty cells so that the number of
occupied cells becomes m+n-1. The containing this small allocation, is of
course, considered to be an occupied cell.
Rule: The extremely small quantity usually denoted by the Greek letter
(delta) is introduced in the least cost independent cell subject to
following rule. If the necessary, two or more ’s can be introduced in the
least and second least cost independent cells.
Resolution of degeneracy during the solution stages
The transportation problem may also become degenerate during
solution stages. This happens when most favorable quantity is allocated
to the empty cell having the largest negative cell evaluation resulting in
simultaneous vacation of two or more of the currently occupied cells. To
solve degeneracy allocate (delta) to one or more of recently vacated
cells so that the no. of occupied cells is m+n-1 in the new solution.

What is an unbalanced Transportation problem? Explain the
procedure for balancing an unbalanced TP.
If in a TP the sum of all available quantities is not equal to the sum of
requirements, that is, m n
 ai <>  bj
i=1 j=1 then such a problem is called an unbalanced
transportation problem. An unbalanced transportation problem can be
modified to a balanced problem by introducing a fictitious sink in the first
case and a fictitious source in the second. The inflow from the fictitious
source to the sink represents the unfilled demand at that sink. Similarly
the inflow from the source to the fictitious sink represents the surplus at
that source. For convenience the cost of transportation of a unit from the
fictitious source or to the fictitious sink as the case maybe is assumed
zero. Following is the procedure to balance a TP.
Balancing the given transportation problem
To solve a given transportation problem we must establish equality
between the total capacities of the origins and the total requirements of
the destinations. Three cases can arise.
Case 1 : bi = dj In this case, the total capacity of the origins equals
the total requirements of the destinations. The problem can be arranged
in the form of a matrix, along with relevant cost data, and the
transportation algorithm may be applied directly to obtain a solution.
Case 2 : bi > dj In this case, the total capacity of the origins exceeds
the total requirement of the destinations. A “dummy” destination can be
added to the matrix to absorb the excess capacity. The cost of shipping
from each origin to this dummy destination is assumed to be zero. The
adding of a dummy destination establishes equality between the total
origin capacities and the total destination requirements. The problem is
then amenable to solution by the transportation algorithm.
ASSIGNMENT
Define assignment problem- The name assignment problem originates
from the classical problems where the objective is to assign a number of
origins (jobs) to the equal number of destinations (persons) t a minimum

cost or at maximum profit. To explain the nature of assignment problem,
suppose there are n jobs to be performed and n persons are available
for doing these jobs. Assume that each person can do each job at a
time, though with varying degree of efficiency. Let cij be the cost
(payment) if the ith person is assigned the jth job, the problem is to find
the assignment (which job should be assigned to which person) so that
the total cost for performing all jobs is minimum. Problems of this kind
are known as assignment problems.
Further these types of problems ma consist of assigning men to offices,
classes to rooms, drivers to trucks, trucks to delivery routes etc. The
assignment problem can be stated in the form of a n x n matrix [cij] of
real numbers as given in the table below.
JOBS
1 2 ……………………. j
……………… n
1 C11 C12 ….. C1j
…… C1n
2 C21 C22 ….. C2j
…… C2n
. . .
. . . . .
. PERSONS i Ci1 Ci2
…... Cij …… Cin
. .
. . . . .
. .
n Cn1 Cn2 …… Cnj
…… Cnn

Explain the mathematical formation of an assignment problem.
Mathematically the assignment problem can be stated as:
Minimize the total cost
n m n
Z=   cij xij , i= 1,2,3,…..  xij = 1 for all I
(workers available)
i=1j=1 j = 1
n
subject to restrictions of the form:  xij = 1
for all j (job req)
xij = 1 : if ith person is assigned the jth job .
= 0 : if not
n
 xij +1 (one job is done by the ith person , j=
1,2,……. n )
j=1
n
and  xij =1 ( only one person should be assigned the jth
job , j =1,2,3,……n )
i=1
where xij denotes that the jth job is to assign to the ith person.
 MINIMISE ( TOTAL COST ) m m
Z =   cij xij
i= 1 j = 1
Subject to

n
 xij = ai j i= 1,2 …… m (capacity constraints)
j = 1
m
 xij = bj ; j = 1,2 ….. n (req. constraint)
and xij  0 for all i & j
Explain the procedure summary for the assignment method for (a)
minimization case & (b) maximization case.
(a) The minimasation case:
Step 1 Determine the total opportunity cost matrix
a) Arrive at a column opportunity cost matrix by subtracting the lowest
entry of each column of the given payoff matrix from all the entries in
the column.
b) Then subtract the lowest entry of each row of the matrix obtained
in (a) from all the entries in its row.
The result of step 1b gives the total-opportunity-cost matrix.
Step 2 Determine whether an optimal assignment can be made
a) Cover all the zeroes of the current total-opportunity-cost matrix with
the minimum possible
number of horizontal and vertical lines.
b) If the number of lines in step 2a equals the number of rows (or
columns) of the matrix, the problem can be solved. Make a complete
assignment so that the total opportunity cost involved in the
assignment is zero.
c) if the number of lines drawn in step 2a is less than the number of
rows (or columns ) of the matrix, proceed to step 3.

Step 3 Revise the total opportunity cost matrix
a) Subtract the lowest entry in the uncovered cells of the current total
opportunity cost matrix from all the uncovered cells.
b) Add the same lowest entry to only those cells in which the covering
lines of step 2 cross.
The result of steps 3a and 3b is a revised total opportunity cost
matrix.
Step 4
Repeat steps 2 and 3 until an optimal assignment having a total
opportunity cost of zero can be made.
Maximisation case
Except for one transformation, an assignment problem in which the
objective is to maximize the total payoff measure can be solved by the
assignment algorithm presented above. The transformation involves
subtracting all the entries of the original payoff matrix from the highest
entry of the original payoff matrix. The transformed entries give us the “
relative costs “and the problem then becomes a minimization problem.
Once the optimal assignment for this transformed minimization problem
has been identified, the total value of the original payoff measure can be
found by adding the individual original entries for those cells to which the
assignments have been made.

Linear Programming Problems {Operation Research}

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