OR 14 15-unit_1

Operations Research (III-2) 2014-15 Unit 1
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OPERATIONS RESEARCH
UNIT 1
Origin
 The term operations research (O.R.) was coined during World War II , when the British
military management called upon a group of scientists together to apply a scientific
approach in the study of military operations to win the battle.
 The main objective was to allocate scarce resources in an effective manner to various
military operations and to the activities within each operation.
 The effectiveness of operations research in military spread interest in it to other
government departments and industry.
 Due to the availability of faster and flexible computing facilities and the number of
qualified O.R. professionals, it is now widely used in military, business, industry,
transportation, public health, crime investigation, etc.
Application areas of OR in mechanical engineering
(1) Defence application:
Defence requires precision and accuracy. Hence it requires scientific decision making
techniques. Techniques like shortest path problems, scheduling algorithms, allocation
techniques can be used in defence forces.
(2) Industrial applications:
An industry or an organization has to manage production,
In production, the following OR techniques are used.
 Linear programming for overall production planning
 Integer programming for in-depth scheduling
 Network based techniques for event management
 Inventory control techniques for organizing material arrival and departure
 Replacement analysis for equipment condition study
 Queuing theory for deciding on buffer units
Phases and Processes of O.R.
(1) Formulate the problem:
 This is the most important process; it is generally lengthy and time consuming. The
activities that constitute this step are visits, observations, research, etc.
 With the help of such activities, the O.R. scientist gets sufficient information and
support to proceed and is better prepared to formulate the problem.

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 This process starts with understanding of the organizational climate, its objectives and
expectations. Further, the alternative courses of action are discovered in this step.
(2) Develop a model:
 Once a problem is formulated, the next step is to express the problem into a
mathematical model that represents systems, processes or environment in the form of
equations, relationships or formulas.
 We have to identify both the static and dynamic structural elements, and device
mathematical formulas to represent the interrelationships among elements.
 The proposed model may be field tested and modified in order to work under stated
environmental constraints.
 A model may also be modified if the management is not satisfied with the answer that it
gives.
(3) Select appropriate data input:
 No model will work appropriately if data input is not appropriate. The purpose of this
step is to have sufficient input to operate and test the model.
(4) Solution of the model:
 After selecting the appropriate data input, the next step is to find a solution.
 If the model is not behaving properly, then updating and modification is considered at
this stage.
(5) Validation of the model:
 A model is said to be valid if it can provide a reliable prediction of the system’s
performance.
 A model must be applicable for a longer time and can be updated from time to time
taking into consideration the past, present and future aspects of the problem.
(6) Implement the solution:
 The implementation of the solution involves so many behavioural issues and the
implementing authority is responsible for resolving these issues.
MODELS
 A model is a representation of a real object or situation using a set of simplifying
assumptions and relationships.
 Reality comprises a large number of variables, and a large number of often complex
interactions between them. In an effort to identify the most important variables, and
understand the most significant relationships, it is important to disregard the less

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important ones while building a model. Although this may render the approach
‘unrealistic’ it can provide insights into a problem, and a far greater predictive ability.
Models may be classified in various ways:
(1) Classification by Structure
(i) Iconic models are physical replicas of the real thing (for instance a toy car);
(ii) Analogue models provide a physical representation of the modelled activity
(for instance a thermometer); and
(iii) Mathematical models use symbols and mathematical relationships to evaluate
a situation.
(2) Classification by nature of environment
(i) Deterministic models, in which outcomes are precisely determined through
known relationships among states and events without any room for random
variation; and
(ii) Stochastic or probabilistic models where the inputs have a random element
and in which ranges of values for each variable (in the form of probability
distribution) are used
(3) Classification by Utility
(i) Descriptive models, which simply explain certain aspects of the problem or situation
or a system which the user can use for analysis;
(ii) Predictive models, which can predict the approximate result of the situation under
question; and
(iii) Prescriptive models, based on the approximate results obtained in the predictive
model) prescribe the courses of action to be taken by the manager to achieve the desired
goal.
(4) Classification by the behaviour of the problem variables
(i) Static models, which assume that there would be no changes in the values of variables
given in the problem for the given planning horizon due to any change in the
environment or conditions of the system. All the values given are independent of the
time; and
(ii) Dynamic models, in which the values of the given variablesgo on changing with time
or change in environment or change in the conditions of the given system.

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(5) Classification by the method of getting the solution
(i) Analytical models, which have a well-defined mathematical structure and can be
solved by the application of mathematical techniques. Examples include linear
Programming models, Transportation Problems, Assignment model etc.
(ii) Simulation models, which have a mathematical structure but can not be solved by
using mathematical techniques. They need experimental analyses to be done to be solved.
Popular Techniques of OR
(1) Linear Programming. Linear Programming (LP) is a mathematical technique of assigning a
fixed amount of resources to satisfy a number of demands in such a way that some objective is
optimized and other defined conditions are also satisfied.
(2) Transportation Problem. The transportation problem is a special type of linear
programming problem, where the objective is to minimize the cost of distributing a product from
a number of sources to a number of destinations.
(3) Assignment Problem. When the problem involves the allocation of n different facilities to n
different tasks, it is often termed as an assignment problem.
(4) Queuing Theory. The queuing problem is identified by the presence of a group of customers
who arrive randomly to receive some service. This theory helps in calculating the expected
number of people in the queue, expected waiting time in the queue, expected idle time for the
server, etc. Thus, this theory can be applied in such situations where decisions have to be taken
to minimize the extent and duration of the queue with minimum investment cost.
(5) Game Theory. It is used for decision making under conflicting situations where there are one
or more opponents (i.e., players). In the game theory, we consider two or more persons with
different objectives, each of whose actions influence the outcomes of the game. The game theory
provides solutions to such games, assuming that each of the players wants to maximize his
profits and minimize his losses.
(6) Inventory Control Models. It is concerned with the acquisition, storage, handling of
inventories so as to ensure the availability of inventory whenever needed and minimize wastage
and losses. It help managers to decide reordering time, reordering level and optimal ordering
quantity.
(7) Simulation. It is a technique that involves setting up a model of real situation and then
performing experiments. Simulation is used where it is very risky, cumbersome, or time
consuming to conduct real study or experiment to know more about a situation.
(8) Dynamic Programming. Dynamic programming is a methodology useful for solving
problems that involve taking decisions over several stages in a sequence. One thing common to
all problems in this category is that current decisions influence both present & future periods.
(9) Sequencing Theory. It is related to Waiting Line Theory. It is applicable when the facilities
are fixed, but the order of servicing may be controlled. The scheduling of service or sequencing
of jobs is done to minimize the relevant costs. For example, patients waiting for a series of tests
in a hospital, aircrafts waiting for landing clearances, etc.
(10) Replacement Models. These models are concerned with the problem of replacement of
machines, individuals, capital assets, etc. due to their deteriorating efficiency, failure, or
breakdown.

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Advantages & Limitations of Operations Research
Advantages
 Better Control: The management of large organizations recognize that it is a difficult
and costly affair to provide continuous executive supervision to every routine work. An
O.R. approach may provide the executive with an analytical and quantitative basis to
identify the problem area. The most frequently adopted applications in this category deal
with production scheduling and inventory replenishment.
 Better Systems: Often, an O.R. approach is initiated to analyze a particular problem of
decision making such as best location for factories, whether to open a new warehouse,
etc. It also helps in selecting economical means of transportation, jobs sequencing,
production scheduling, replacement of old machinery, etc.
 Better Decisions: O.R. models help in improved decision making and reduce the risk of
making erroneous decisions. O.R. approach gives the executive an improved insight into
how he makes his decisions.
 Better Co-ordination: An operations-research-oriented planning model helps in
coordinating different divisions of a company.
Limitations
 Dependence on an Electronic Computer: O.R. techniques try to find out an optimal
solution taking into account all the factors. In the modern society, these factors are
enormous and expressing them in quantity and establishing relationships among these
require voluminous calculations that can only be handled by computers.
 Non-Quantifiable Factors: O.R. techniques provide a solution only when all the
elements related to a problem can be quantified. All relevant variables do not lend
themselves to quantification. Factors that cannot be quantified find no place in O.R.
models.
 Distance between Manager and Operations Researcher: O.R. being specialist's job
requires a mathematician or a statistician, who might not be aware of the business
problems. Similarly, a manager fails to understand the complex working of O.R. Thus,
there is a gap between the two.
 Money and Time Costs: When the basic data are subjected to frequent changes,
incorporating them into the O.R. models is a costly affair. Moreover, a fairly good
solution at present may be more desirable than a perfect O.R. solution available after
sometime.
 Implementation: Implementation of decisions is a delicate task. It must take into
account the complexities of human relations and behaviour.

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LINEAR PROGRAMMING
INTRODUCTION
 A large number of decision problems faced by a manager involve allocation of resources
to various activities, with the objective of increasing profit, or decreasing cost.
 Normally, the resources are scarce (= limited) and the performance of a number of
activities within the constraints (= limitations) of limited resources is a challenge.
 A manager is, therefore required to decide as to how best to allocate resources among the
various activities.
 The linear programming method is a popular mathematical technique used to choose
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.
Definition: The general linear programming problem calls for optimizing (maximizing /
minimizing) a linear function of variables called the ‘OBJECTIVE FUNCTION’, subject to a set
of linear equations and / or inequalities called the ‘CONSTRAINTS’, or ‘RESTRICTIONS’.
REQUIREMENTS FOR APPLICATION OF LINEAR PROGRAMMING
(1) The aim or object should be clearly identifiable and definable in mathematical terms.
Example: Maximization of profit, Minimization of cost, Minimization of time etc.
(2) The activities involved should be distinct and measurable in quantitative terms. Example:
How many products of a particular type should be made in a time period? How many waiters are
to be employed during a time period? Etc.
(3) The resources to be allocated should be measurable quantitatively. Example: The availability
of material (in kgs), the availability of machine time (in hours); the availability of labour (in
man-hours), the demand for a product in the market (in units, litres) etc.
(4) The relationships representing the objective function and the constraints must be linear in
nature.
(5) There should be a series of feasible alternative courses of action available to the decision
maker, which is determined by the resource constraints.

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ASSUMPTIONS UNDERLYING LINEAR PROGRAMMING
(1) Proportionality
 A primary requirement of a linear programming problem is that the objective function
and every constraint function must be linear.
 As an example, if 1 kg of a product costs Rs 2, then 10 kg will cost Rs 20. Similarly, if a
steel mill can produce 200 tonnes in an hour, it can produce 1000 tonnes in 5 hours.
 This assumption, thus, ignores bulk discounts and idle times
(2) Additivity
 Additivity indicates that the total of all activities is given by the sum total of each activity
conducted separately.
 For example, if it takes T1hours on machine G to make a unit of product A and T2 hours
to make a unit of product B, then time required on machine G to produce X1 units of A
and X2 units of product B is
T1X1 + T2X2. Thus, the time required to change the set up on the machine from product
A to product B is neglected.
(3) Continuity
 It is assumed that the decision variables are continuous. As a consequence, combinations
of output with fractional values (especially in the context of production problems) are
possible. For example, the solution to an LP problem might yield a solution X1 = 2.54
tonnes; X2 = 3.18 tonnes etc.
 Normally, we deal with integer values, but even fractional values are permissible.
(4) Deterministic
 Various parameters, namely the objective function coefficients and the coefficients of the
variables in the constraints are known with certainty.
 Hence, linear programming is deterministic in nature.
(5) Finite Choices
 A linear programming model also assumes that a limited number of choices are available
to the decision maker

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APPLICATIONS OF LINEAR PROGRAMMING
The following are the various applications of LP problems in productions:
Production
(1) Product Mix and product proportioning; (2) Production planning; (3) Assembly line
balancing; (4) Oil refining; (5) Paper trimming; (6) Hospital scheduling; (7) Agricultural
operations
FORMULATIONS OF LINEAR PROGRAMMING PROBLEMS
Step 1: DECISION VARIABLES: Identify the decision variables. These variables are generally
written as X1, X2,… etc. as the number of units to be produced (or recruited) and contributing
towards the objective function.
Step 2: OBJECTIVE FUNCTION: Identify the aim (object) of the LP problem in terms of
profit, time, cost etc. Thus the objective function is to be written as maximization of profit or
minimization of cost or minimization of time etc.
Step 3: CONSTRAINTS: Constraints are inequalities (such as machine time, availability of
materials, market demand etc) representing restrictions on the value of the objective function .
These constraints have to be written in the linear form.
Step 4: NON-NEGATIVITY CONSTRAINTS: Since the decision variables cannot take
negative values (except in the case of certain unrestricted variables), the non-negative constraints
have to be incorporated in the LP problem.
ADVANTAGES OF LINEAR PROGRAMMING
(1) Linear Programming is a useful technique to obtain the optimum use of productive resources.
It helps a decision maker to ensure effective use of scarce resources by their proper deployment.
(2) Due to its structured form, a linear programming technique improves the quality of decision
making.
(3) The LP model generates a large number of alternate solutions and hence helps in reaching
practical solutions at optimum working level. It also permits in modifications of the
mathematical model to suit the decision maker’s requirements.
(4) The LP technique also indicates ideal capacity of machines or materials in a production
process. In fact it helps the decision maker to decide whether his resources can intentionally be
kept idle in order to work on optimal level of the objective, if certain constraints demand so.
(5) The LP technique can also cater to changing situations. The changed conditions can be used
to readjust the plan decided for execution.

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LIMITATIONS OF LINEAR PROGRAMMING
(1) There is no guarantee that linear programming will give an integer valued solution. For
example, a solution may result in producing 8.291 cars. In such a situation, the decision maker
will examine the possibility of producing 8 cars as well as 9 cars and will take a decision on
which ensures higher profits subject to given constraints. But this rounding does not always
ensure a good answer. Only integer programming techniques, which are much more complicated,
can handle such cases.
(2) Under LP approach, uncertainty is not allowed. The linear programming model operates only
when values for costs, objective function coefficients and the coefficients of the variables in the
constraints are known with certainty. But, in real life, these values are always not known with
certainty.
(3) The assumption of linearity is another formidable limitation of the LP model. A primary
requirement of a linear programming problem is that the objective function and every constraint
function must be linear. We are thus dealing with a system where that has constant returns to
scale. In many situations, the input-output rate for an activity varies with the activity level. The
constraints in real-life situations covering business and industrial problems are not linearly
related to the variables. In most economic situations, sooner or later, the law of diminishing
marginal returns begins to operate.
(4) Linear programming will fail to give a solution if the management has conflicting multiple
goals because in the LP model, there is only one goal, which is expressed in the objective
function, e.g., maximizing profit or minimizing cost etc.
GRAPHICAL METHOD OF SOLVING A LINEAR PROGRAMMING PROBLEM
(EXTREME POINT APPROACH)
This method can be used to solve an LP problem only if there are two decision variables.
This method operates based on the fundamental theorem of linear programming which states that
“the collection of all feasible solutions to LP problems constitutes a convex set whose extreme
(corner) points correspond to the basic feasible solutions”.
Steps in the graphical method:
(1) Formulate the problem as an LP with two decision variables.
(2) Consider each inequality constraint as an equation.
(3) Plot each equation on the graph sheet, as each one will geometrically represent a straight line.

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(4) The inequality conditions (as per the original formulation) are now used to get the feasible
region satisfying all constraints. If the inequality is of the form ≤ , the region below the line in
the first quadrant is shaded.. Similarly, for the constraint of the form ≥ the region above the
corresponding line falling in the first quadrant is shaded. The points lying in the common region
will satisfy all the constraints simultaneously. The common region thus obtained is called the
feasible region.
(5) The extreme points of the feasible region (the convex set) would cover the solution area and
these extreme points of the feasibility region correspond to the basic feasible solutions.
(6) Calculate the value of the objective function at each of these extreme points and the desired
value (either maximum or minimum as per the objective function) would be the optimal solution
of the problem.

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SIMPLEX METHOD
Introduction
When a large number of variables (more than 2) are involved in a linear programming problem,
the solution by graphical method is not possible. The simplex method provides an efficient
technique for solving LPP problems of any magnitude, involving two or more variables. In this
method, the objective function is used to control the development and evaluation of each feasible
solution of the problem.
The simplex method was formulated by G.B.Dantzig in 1947. This algorithm is an iterative
procedure for finding, in a systematic manner, the optimal solution that comes from the corner
points of the feasible region. The simplex algorithm considers only those feasible solutions
which are provided by the corner points and that too not all of them. It is a very efficient
algorithm as it has the merit to indicate whether a given solution is optimal or not.
Method of Application of Simplex Algorithm
(1) First of all, an appropriately selected set of variables is introduced into the problem.
(2) An iterative process is started by assigning values only to those variables so introduced and
the primary decision variables of the problem are all set equal to zero.
(3) The algorithm then replaces one of the initial variables by another variable – the variable
which contributes most to the desired optimal solution enters in, while the variable creating the
bottleneck to the optimal solution goes out. This improves the value of the objective function.
(4) The procedure of substitution of variables is repeated until no further improvement in the
objective function value is possible. The algorithm terminates thus indicating that the optimal
solution has been obtained.
For application of simplex method, the following conditions must be satisfied:
(a) Right Hand Side (RHS) of each constraint should be non-negative. In case of negative RHS,
the whole inequality is to be multiplied by -1.
(b) Each of the decision variables of the problem should be non-negative. In case of unrestricted
variables, each of them should be treated as the difference of two non-negative variables.

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Basic Definitions in Simplex Algorithm
(1) Basic Solution: There may be n variables and m constraints in a linear programming
problem. When we evaluate the solution of this problem by setting (n-m) of the variables to zero
and solve the other m equations, we obtain a unique solution. This solution is called the “Basic
Solution”.
(2) Basic variables and non-basic variables: The variables which are not set equal to zero and
appear in the basic solution are called the basic variables. The variables which are set equal to
zero and do not appear in the basic solution are called the non-basic variables.
(3) Basic Feasible Solution: When the basic solution satisfies even the non-negativity
constraints, it is called as the “Basic Feasible Solution”.
(4) Cj row: It is the row containing the coefficients of all the variables (decision variables, slack
variables, surplus variables or artificial variables).
(5) Cj-Zj = ∆j row or Index row: It is the row containing net profit or loss resulting from
introducing one unit of the variable in that column in the solution. A positive number in that row
would indicate improvement in the objective function value if that variable enters the basis. A
negative number, on the other hand, would indicate reduction in the value of the objective
function if that variable enters the basis.
(6) Constraints: Restrictions on the problem solution arising from limited resources, such as
availability of material, manpower, machine time etc.
(7) Infeasible solution: The solution violating at least one constraint.
(8) Iterations: In the simplex method, the procedure is that of constant improvement type from
one basic feasible solution to another. These steps of moving from one solution to a better
solution are called iterations.
(9) Pivot Column (Key Column): The column with the largest positive number in Cj-Zj row in
a maximization problem (or the smallest number in a minimization problem) is called the key
column (or the pivot column). This column indicates the variable entering the solution in the next
iteration by replacing an appropriate variable.
(10) Pivot Row (Key Row): This is the row containing the outgoing variable. The outgoing
variable, to be replaced by the incoming variable, is decided as the variable with the minimum
positive ratio (= Xb/Cb).
(11) Pivot Element (Key Element): The element at the point of intersection of the key column
and the key row is called the key element (or Pivot Element).

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(12) Linear Function: A mathematical function in which a linear relation exists among the
variables is called a linear function.
Ex: 3X1 + 2 X2 ≤ 300; 8X1 + 5X2 ≥ 250 etc.
(13) Objective Function: The function representing the aim (object) of the LP problem in terms
of profit, time, cost etc. is called the objective function. Thus the objective function is to be
written as maximization of profit or minimization of cost or minimization of time etc.
(14) Optimal Solution: The Optimal solution is the best of all basic feasible solutions. The value
of the objective function is the highest for this basic feasible solution.
(15) Multiple Optimal Solutions: Sometimes, alternative solutions to the values of decision
variables may exist. The value of the objective function would be the same in all these
alternative optimal solutions.
(16) Slack Variable: A less than or equal to (≤) inequality constraint is converted into an
equation by introducing a slack variable. In a physical sense, the slack variables represent unused
resources. The slack variables do not contribute anything to the value of the objective function.
Hence their coefficients in the objective function are to be zeros.
(17) Surplus Variables: A greater than or equal to (≥) inequality constraint is converted into an
equation by introducing a surplus variable. In a physical sense, the surplus variables represent the
amount of resources over and above the minimum required level. The surplus variables do not
contribute anything to the value of the objective function. Hence their coefficients in the
objective function are to be zeros.
(18) Artificial Variables: When the constraints are of the type ≥, then equality conversion
would need surplus variables subtracted. But as per the standard form, this cannot be an initial
basic feasible solution because simplex method does not accept negative variables in the starting
basic feasible solution. Hence, just to convert it into initial basic feasible solution, artificial
variables (dummy variables) are added to convert the constraints into standard form with these
variables as the basic variables.
By adding these additional variables, we apply the simplex method not to the objective function
of the original problem, but to a new function containing artificial variables. This goes contrary
to the problem definition and hence it becomes imperative to remove these variables from the
solution and not to permit their re-entry. It is for this reason that a very large arbitrary penalty
(called Big M) is assigned to each of the artificial variables as its coefficient value in the
objective function.
(19) Degeneracy: An LP is degenerate if in a basic feasible solution, one of the basic variables
takes on a zero value. Degeneracy is caused by redundant constraint(s) and could cost simplex
method extra iterations

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(20) Simplex Table: A table used for calculations during various iterations of the simplex
iterations.
(21) Canonical form of an LPP: An LP is said to be in the canonical form if the constraints are
in the inequalities form.
(21) Standard form of an LPP: An LP is said to be in the standard form if the constraints are in
the equation form.
SIMPLEX ALGORITHM (MAXIMIZATION CASE)
Step 1: Formulation of the standard form:
(a) Formulate the given problem as an LP model.
Maximize Z = 4 X1 + 10 X2
Subject to 2X1 + X2 ≤ 50
2X1 + 5X2≤ 100
2X1 + 3X2 ≤ 90
X1, X2 ≥ 0
(b) The standard form of the simplex method is obtained by ensuring that all inequalities are of
the type ≤. If not, the relevant constraint must be multiplied by -1 on both sides.
(c) Addition of slack variables to the inequalities will ensure their conversion into equalities.
Also we assign a zero-cost coefficient to each of these slack variables in the objective function.
Maximize Z = 4 X1 + 10 X2 + 0 S1 + 0S2 + 0 S3
Subject to 2X1 + X2 +S1 = 50
2X1 + 5X2 + S2 = 100
2X1 + 3X2 + S3 = 90
Step 2: Set up initial solution:
The initial solution of the simplex method is obtained by assigning all decision variables values
as zero in order to begin the solution from zero level i.e., setting the value of objective function
as zero and gradually improving it by iteration process.

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Cj 4 10 0 0 0
Basic
Variables
CB X1 X2 S1 S2 S3 Value
XB
Ratio
XB/Xi
S1 0 2 1 1 0 0 50
S2 0 2 5 0 1 0 100
S3 0 2 3 0 0 1 90
Zj 0 0 0 0 0 0
Cj-Zj 4 10 0 0 0
Step 3: Testing the solution for optimality:
On careful examination of the index row (Cj-Zj row), if all the elements of the Cj-Zj row are
negative or zero, the basic feasible solution obtained is optimal.
If the solution is not optimal, it can be improved in the next iteration by replacing one of the
basic variables by the entering variable as decided in Step 4.
Step 4: (a) Identify entering variable: The improvement is brought about by selecting the
column with the highest Cj-Zj value. This column is called the pivot (or key) column. The
corresponding variable becomes the entering variable.
Cj 4 10 0 0 0
Basic
Variables
XB
XB/Xj
Ratio
S1 0 2 1 1 0 0 50 50/1 =
50
S2 0 2 5 0 1 0 100 100/5 =
20
S3 0 2 3 0 0 1 90 90/3 =
30
Zj 0 0 0 0 0 0
Cj-Zj ---- 4 10 0 0 0
Here, X2 becomes the entering variable as Cj- Zj = 10 for X2
(b) Identify leaving variable: Having identified the pivot column, all values of XB are divided
by the corresponding elements of the pivot column for each row to obtain the XB/Xj ratios. The
ratio where the least positive ratio occurs is called the pivot row. The variable corresponding to
this row is the variable leaving the basis. In this problem, S2 is the variable leaving the basis..
Step 5: Identify pivot element:
The value at the intersection of the pivot column and the pivot row is called the pivot element (or
key element). In this problem, the pivot element is 5 as it is at the intersection of the X2 column
(the pivot column) and the S2 row (the pivot row).

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Step 6: Iteration for improved solution:
(a) Replace outgoing variable with the entering variable and enter relevant coefficients in Basic
Variables column.
(b) Compute the pivot row with reference to the newly entered variable by dividing the old row
quantities by the pivot element.
(c) Modify the values for other rows. The elements are recalculated as follows:
New row element = Old row element- corresponding element in the pivot column
multiplied by the corresponding new element of the revised pivot row.
Cj 4 10 0 0 0
Basic
Variables
XB
XB/Xj
Ratio
S1 0 8/5 0 1 -1/5 0 30
X2 10 2/5 1 0 1/5 0 20
S3 0 4/5 0 0 -3/5 1 30
Zj 200 4 10 0 2 0 Zj
Cj-Zj ---- 0 0 0 -2 0 Cj- Zj
On careful examination of the index row (Cj-Zj row), if all the elements of the Cj-Zj row are
negative or zero, the basic feasible solution obtained is optimal.
The values of XB for the corresponding variables in the basis will indicate the values of the
variables contributing towards the objective function.
In this problem, since all the Cj- Zj s are either zero or negative, the optimal solution is reached.
The optimal solution is X2 = 30 (X1= 0 ). The value of Zmax = 4 X1 + 10 X2 = 200.
If the solution is not optimal, proceed to Step 8.
Step 8: Revise or improve the solution:
The steps from Step 4 to Step 7 are repeated till optimality conditions are fulfilled and the
solution is obtained.
Note: Rule for breaking ties for deciding entering variable and leaving variable
Whenever two equal values are encountered, we can break ties arbitrarily. However to reduce the
computation effort, the following can be helpful:

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(a) For pivot column, select the left most element. (b) For pivot row, select the one nearest to the
top.
SIMPLEX ALGORITHM (MINIMIZATION CASE)
When any of the constraints is of the ≥ type, we have to add negative slack variables (i.e.,
subtract surplus variables) to convert inequalities to equations.
For ex: the constraint 2X1+ 3 X2 ≥ 20 is converted as 2X1+ 3 X2 - S1 = 20.
This results in getting solutions such as - S1 = 20 or S1 = - 20. But as per the standard form, this
cannot be an initial basic feasible solution because simplex method does not accept negative
variables in the starting basic feasible solution. Hence, just to convert it into initial basic feasible
solution, artificial variables (dummy variables) are added to convert the constraints into standard
form with these variables as the basic variables.
By adding these additional variables, we apply the simplex method not to the objective function
of the original problem, but to a new function containing artificial variables. This goes contrary
to the problem definition and hence it becomes imperative to remove these variables from the
solution and not to permit their re-entry.
I. Two-Phase Method
Step 1:
(a) If all the constraints in the given LP are of the ≤ type, then we should add only slack variables
and should not add artificial variables and can go directly to Phase II.
(b) Otherwise, the necessary number of surplus variables and artificial variables are added to the
inequality constraint to convert it to an equality constraint.
(c) The minimization problem should be converted to a maximization problem by multiplying
the objective function by (-1).
Step 2:
(a) Assign zero coefficients to decision variables (X1, X2, X3, ,,,,,,,,,,,,,) and the surplus variables
(s1, s2, s3 etc.)
(b) Assign (-1) coefficient to each of the artificial variables.
This yields the following auxiliary problem:
Max Z*
= -A1 –A2- A3 – A4…………….-Am
Subject to Ʃ aiXi = bi

18
Step 3:
(a) Apply the Simplex algorithm to solve the auxiliary LP problem.
The following two cases may arise:
(i) Max Z*
= 0 and no artificial variable is present in the basis. This means that the basis consists
of only decision variables (and slack variables) and hence we may move to Phase II to obtain an
optimal basic feasible solution to the original LP problem.
(ii) Max Z*
= 0 and at least one artificial variable is present in the basis with a positive value.
This means that no feasible solution exists for the original LP problem.
Step 4:
(a) The last Simplex table of Phase I can be used as the initial Simplex table for Phase II.
(b) Assign the actual coefficients to the decision variables and zero coefficients to the surplus
variables in the objective function.
(c) Then apply the usual Simplex algorithm to the modified Simplex table in order to get the
optimal solution to the original problem.
Example of Two-phase Simplex Method
Minimize Z = X1 + X2
Subject to 2 X1 + X2 ≥ 4
X1 + 7 X2 ≥ 7; X1, X2 ≥ 0
Converting the given problem into auxiliary problem by converting the minimization problem
into maximization type and adding surplus and artificial variables,
Maximize Z*
= - X1 - X2 + 0S1+ 0S2
Subject to 2 X1 + X2 – S1 + A1 = 4
X1 + 7 X2 -S2 + A2 = 7
; X1, X2, S1, S2, A1, A2 ≥ 0
Phase I
This phase starts by considering the following auxiliary LP problem
Maximize Z*
= - A1 - A2

19
Subject to 2 X1 + X2 – S1 + A1 = 4
X1 + 7 X2 -S2 + A2 = 7
X1, X2, S1, S2, A1, A2 ≥ 0
Putting X1 = 0, X2 = 0, S1 = 0 , S2 = 0, the initial solution is A1 = 4 and A2 = 7
PHASE I SIMPLEX TABLE 1
Cj 0 0 0 0 -1 -1
Basis Coefft X1 X2 S1 S2 A1 A2 Value Ratio
A1 -1 2 1 -1 0 1 0 4 4
A2 -1 1 7 0 -1 0 1 7 1
Zj* -11 - 3 - 8 1 1 -1 - 1
Cj-Zj* ----- 3 8 -1 -1 0 0
Since all Cj-Zj* are not equal to zero or negative, this is NOT the optimal solution.
The entering variable is X2 and the leaving variable is A2. The pivot element = 7.
Since A2 is the leaving variable, A2 column is deleted from the next table.
Cj 0 0 0 0 -1
Basis Coefft X1 X2 S1 S2 A1 Value Ratio
A1 -1 13/7 0 -1 1/7 1 3 21/13
X2 0 1/7 1 0 -1/7 0 1 7
Zj* -3 -13/7 0 1 -1/7 -1
Cj-Zj* ---- 13/7 0 -1 1/7 0
The entering variable is X1 and the leaving variable is A1. The pivot element = 13/7.

20
Cj 0 0 0 0
Basis Coefft X1 X2 S1 S2 Value Ratio
X1 0 1 0 -7/13 1/13 21/13
X2 0 0 1 1/13 -2/13 10/13
Zj* 0 0 0 0 0
Cj-Zj* ---- 0 0 0 0
Since all Cj-Zj* are equal to zero, this is the optimal solution for Phase I LP problem. So, now,
we move to Phase II.
Phase II
The last Simplex table of Phase I can be used as the initial Simplex table for Phase II.
(b) Assign the actual coefficients to the decision variables and zero coefficients to the surplus
variables in the objective function.
(c) Then apply the usual Simplex algorithm to the modified Simplex table in order to get the
optimal solution to the original problem.
PHASE II SIMPLEX TABLE 1
Cj -1 -1 0 0
X1 -1 1 0 -7/13 1/13 21/13
X2 -1 0 1 1/13 -2/13 10/13
Zj* -31/13 -1 -1 6/13 1/13
Cj-Zj* ---- 0 0 - 6/13 -1/13
Since all Cj – Zj* s are either zero or negative, the optimal solution to Phase II problem is
reached.
The optimal solution is:
X1 = 21/13
X2 = 10/13
Z* = -31/13
But Z = - Z*
Therefore, Zmin = 31/13; X1 = 21/13; X2 = 10/13

21
II. Big M Method
The Big-M method is another method of removing artificial variables from the basis.
In this method, we assign large undesirable (unacceptable) penalty coefficients to artificial
variables from the point of view of the objective function,
If the objective function is to be maximized, then a very large negative price is assigned to each
artificial variable.
The penalty to be assigned is designated as – M for the maximization problems.
Step 1: The standard simplex table is obtained by adding slack, surplus and artificial variables.
The slack and surplus variables are assigned zero coefficients and artificial variables assigned
(- M ) coefficients in the objective function.
Step 2: Initial basic feasible solution is obtained by assigning zero values to the decision
variables and slack variables.
On careful examination of the index row (Cj-Zj row), if all the elements of the Cj-Zj row are ≤ 0,
the basic feasible solution obtained is optimal. If the solution is not optimal (i.e., any of the Cj-Zj
values are ≥ 0) , it can be improved in the next iteration by replacing one of the basic variables
by the entering variable as decided in Step 4.
Step 4: (a) Identify entering variable: The improvement is brought about by selecting the
column with the largest positive (Cj-Zj) value. This column is called the pivot (or key) column.
The corresponding variable becomes the entering variable.
(b) Identify leaving variable: Having identified the pivot column, all values of XB are divided
by the corresponding elements of the pivot column for each row to obtain the XB/Xj ratios. The
ratio where the least positive ratio occurs is called the pivot row. The variable corresponding to
this row is the variable leaving the basis.
Step 5: Repeat Steps 3 and 4 to ensure optimal solution with no artificial variable in the solution.

22
Example for Big-M method
Minimize Z = X1 + X2
Subject to 2 X1 + X2 ≥ 4
X1 + 7 X2 ≥ 7; X1, X2 ≥ 0
Converting the given problem into standard form by converting the minimization problem into
maximization type and adding surplus and artificial variables,
Maximize Z*
= - X1 - X2- + 0 S1+ 0S2 - MA1-MA2
Subject to 2 X1 + X2 – S1 + A1 = 4
X1 + 7 X2 -S2 + A2 = 7
X1, X2, S1, S2, A1, A2 ≥ 0
Putting X1 = 0, X2 = 0, S1 = 0, S2 = 0, the initial solution is A1 = 4 and A2 = 7
SIMPLEX TABLE 1
Cj -1 -1 0 0 -M -M
Basis Coefft X1 X2 S1 S2 A1 A2 Value Ratio
A1 -M 2 1 -1 0 1 0 4 4
A2 -M 1 7 0 -1 0 1 7 1
Zj* -11M - 3M - 8M M M -M - M
Cj-Zj* ----- 3M-1 8M-1 -M -M 0 0
The entering variable is X2 and the leaving variable is A2. The pivot element = 7.
SIMPLEX TABLE 2
Cj -1 -1 0 0 -M
Basis Coefft X1 X2 S1 S2 A1 Value Ratio
A1 -M 13/7 0 -1 1/7 1 3 21/13
X2 -1 1/7 1 0 -1/7 0 1 7
Zj* -3M-1 -13/7M-
1/7
0 M -1/7M
+1/7
-1
Cj-Zj* ---- 13/7 M -
6/7
0 -M 1/7M-1/7 0

23
The entering variable is X1 and the leaving variable is A1. The pivot element = 13/7.
SIMPLEX TABLE 3
Cj -1 -1 0 0
X1 -1 1 0 -7/13 1/13 21/13
X2 -1 0 1 1/13 -2/13 10/13
Zj* -31/13 -1 -1 6/13 1/13
Cj-Zj* ---- 0 0 -6/13 -1/13
Since all Cj-Zj* are equal to either zero or negative, this is the optimal solution for the LP
problem.
The optimal solution is:
X1 = 21/13
X2 = 10/13
Z* = -31/13
But Z = - Z*
Therefore, Zmin = 31/13; X1 = 21/13; X2 = 10/13

24
DUALITY
 In the context of linear programming, duality implies that each linear programming
problem can be analyzed in two different ways but would have equivalent solutions.
 Any LP problem (either maximization or minimization) can be stated in another
equivalent form based on the same data.
 The new LP problem is called dual linear programming problem or in short dual.
 In general, it is immaterial which of the two problems is called primal or dual, since the
dual of the dual is the primal.
If Primal Then Dual
(1) Objective is to maximize (1) Objective is to minimize
(2) Objective is to minimize (2) Objective is to maximize
(3) j th (For example 2nd
) primal variable (3) j th (2nd
) dual constraint
(4) k th (example 3 rd) primal constraint (4) k th (3 rd) dual variable
(5) Primal Constraints ≤ type (5) Dual Constraints ≥ type
(6) Primal Constraints ≥ type (6) Dual Constraints ≤ type
(7) Primal Constraint is = type (7) Dual variable (corresponding to this
constraint) is unrestricted in sign
(8) Primal variable is unrestricted in sign (8) Dual Constraint (corresponding to this
variable) is = type
Rules for Constructing the Dual from Primal
(1) A dual variable is defined for each constraint in the primal LP problem and vice-
versa. Thus, for a given primal LP problem with “n” variables and “m” constraints, there
exists a dual LP problem with “m” variables and “n” constraints and vice-versa.
(2) The Right-Hand Side constants b1, b2, ………, bm of the constraints of the primal LP
problem become the coefficients of the dual variables Y1, Y2, ………Ym in the dual
objective function Zy.
(3) The coefficients C1, C2, …….Cn of the primal variables X1, X2, ….Xn in the objective
function become the Right-Hand side constants in the dual LP problem.
(4) For a maximization primal LP problem with all ≤ type constraints, there exists a
minimization dual LP problem with all ≥ type constraints and vice-versa.
Thus, the inequality sign is reversed in all the constraints (except the non-negativity
constraints),
(5) The matrix of the coefficients of variables in the constraints of a dual is the transpose
of the matrix of coefficients of variables in the constraints of the primal and vice-versa,

25
i.e., coefficients of the primal variables X1, X2, …….Xn in the constraints of a primal LP
problem are the coefficients of dual variables in first, second,…. n th constraints for the
dual problem respectively.
(5) If the objective function of a primal LP problem is to be maximized, the objective
function of the dual is to be minimized and vice-versa.
(6) If the j th primal constraint is = type, then the jth dual variable is unrestricted in sign
and vice-versa.
(7) The maximum values of the objective function of the primal LP problem and the dual
LP problems are equal.
Example for Primal, Dual and Dual of the Dual (=Primal)
PRIMAL DUAL DUAL OF DUAL
Maximize
Zx = X1-X2+3X3
s.t. X1+ X2 + X3 ≤ 10
2 X1 – X2 – X3 ≤ 2
2 X1 – 2 X2 – 3 X3 ≤ 6
X1, X2, X3 ≥ 0
Minimize
Zy = 10Y1 + 2 Y2 + 6 Y3
s.t. Y1 + 2 Y2 + 2 Y3 ≥ 1
Y1 – Y2 – 2 Y3 ≥ -1
Y1 – Y2 – 3 Y3 ≥ 3
Y1, Y2, Y3 ≥ 0
Maximize
Zx = X1-X2+3X3
s.t. X1+ X2 + X3 ≤ 10
2 X1 – X2 – X3 ≤ 2
2 X1 – 2 X2 – 3 X3 ≤ 6
X1, X2, X3 ≥ 0
UNIVERSITY QUESTIONS FROM UNIT 1
OR MODELS & LINEAR PROGRAMMING
Q (1) (a) Discuss the Operations Research models.
(b) Explain the graphical solution with an example.
Q (2) Use the two-phase method to solve the following linear programming problem.
Min Z = X1 + X2
Subject to the constraints
2 X1 + X2 ≥ 4
X1 + 7 X2 ≥ 7
X1, X2 ≥ 0
Q (3) (a) Explain the terms: feasible solution, optimum solution, and basic feasible solution as
applied to LPP.
(b) Use Big-M method to minimize Z = 3 X + 2.5 Y
Subject to 2 X + 4 Y ≥ 40
3 X + 2 Y ≥ 50
X, Y ≥ 0.

26
Q 4 (a) Define Operations Research.
(b) Give at least five applications of Operations Research.
(c) Explain characteristics of Operations Research.
Q (5) (a) Solve the following LPP using graphical method.
Maximize Z = 2 x1 + 3 x2
Subject to the constraints x1 + x2 ≤ 400; 2 x1 + x2 ≤ 600 x1 ≥ 0; x2 ≥ 0.
(b) What are artificial variables? State their importance.
(c) Explain Simplex method in detail.
Q (6) (a) Discuss scientific methods of OR.
(b) Maximize Z = 3 x1 + 2 x2
Subject to the constraints 2x1 + x2 ≤ 1; 3 x1 + 4x2 ≥ 4 x1 ≥ 0; x2 ≥ 0.
Q (7) (a) What is Operations Research? Explain the concept, and its scope.
b) Explain the models of Operations Research as applicable to business and industry.
Q (8) (a) Explain the use of artificial variable and its use in linear programming.
b) Food X contains 6 units of vitamin A per gram and 7 units of vitamin B per gram and cost is
12 paise/gm. Food Y contains 8 units of vitamin A per gram and 12 units of vitamin B per gram
and cost is20 paise/gm. The daily minimum requirement of vitamin A and B are 100 and 120
units respectively. Use graphical method to find the cost of product min.

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