Or Introduction

7/30/2019 Or Introduction

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OPERATIONS RESEARCH

Introduction

If you aspire to become a successful manager, one of the talents you must develop is

decision-making in business as the stakes are high therein because the managers

decisions affect many people.It may be noted that

1. todays business activities are very complex;

2. high cost of new technology, materials and labour;3. increasing number of competitors; and

4. limited time available during which many important decisions have to be made.

Operations Research (also known as quantitative methods, management science, decision

science and a few other names) can be viewed as a scientific method that has evolved asan aid to the decision-maker in all types of organizations.

Need For

i. Decision problems of modern management are so complex that only a systematicand scientifically based analysis can yield realistic solutions.

ii. Different types of quantitative models for solving these complex managerialproblems are availability.

iii. Availability of high-speed computers has made it possible both in terms of time and

cost to apply quantitative models to all real-life problems in all types oforganizations such as business, industry, military, government, and health and so

on.

Operations Research is not a fixed formula which can be applied to all types of problems.

This requires that the problems be defined, analyzed and solved in a rational, logical,systematic and scientific manner based on data, facts, information and logic and not on

intuition and subjective judgment. Operations Research, however, does not totallyeliminate the scope of qualitative or judgment ability of the decision-maker.

Nature of Operation Research

Operations Research helps to seek the optimal solution to a problem and not merely onewhich gives better solutions than the one currently in use.

Definitions of Operations Research

It is not possible to give a uniformly acceptable definition of Operations Research.However, few definitions which are commonly used and widely accepted are given

below:

Operations Research is the application of scientific methods, techniques and toolsto problems involving the Operations of a System so as to provide those in control

of the system with optimum solution to the problem.

Operations Research is the art of giving bad answers to problems which other wisehave worse answers.

Operations Research is a scientific method of providing executive departments with

a quantitative basis for decisions regarding the operations under their control.


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Operations Research is a management activity pursued in two complementary ways

one half by the free and bold exercise of commonsense untrammeled (notrestricted or constrained) by any routine, and the other half by the application of a

repertoire of well established pre-created methods and techniques.

Operations Research is a scientific approach to problems-solving for executive

management. Operations Research is concerned with scientifically deciding how best to design

and operate man-machine systems, usually under conditions requiring the

allocation of scarce resources.

Methodology of Operations Research

Formulate the problem

Build model

Acquire the required data

Solve the model

Validate the model

Implement the solution

Models

A. Physical All forms of diagrammes, drawings, graphs, maps, figures, charts

Iconic Scaled up/down version

Analogue one set up properties to represents another

(e.g. electric system/network to represent movement of trains/plains)B. Symbiotic or Mathematical: Abstract or general type

(Validation)

Define the Problem

Develop Model

Obtain Data

Solve Model

Test the solution

Implement


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used in Operations Research

easy to manipulate

yield more accurate results

Deterministic: no uncertainty about values of parameters

Probabilistic : one or more random variables exist

Components

o Decision variables : controllable/ manipulates/ where choice is made

o Result variables : Output/ dependent variables/ uncontrollable

o Exogenous variables : e.g. competitors strategies

Solutions

A. Feasible and infeasibleB. Optimal and non-optimal

C. Unique and multiple

Sensitivity Analysis or Post-optimality analysis: Determination of the behaviour of thesystem to changes in the system inputs and specification.

It is what if analysis

Operations Research Techniques

1. Allocation models (Programming Models, Transportation Model, and AssignmentModel etc.)

2. Inventory models

3. Waiting line (queuing) models4. Markovian models

5. Network models

6. Sequencing models7. Replacement models8. Simulation models

9. Decision and Game Theory models

Application Areas of Operations Research in Management

1. Finance, Budgeting and Investment (cash flow analysis, investment portfolio, credit

policies etc.)2. Marketing (Product selection, media mix, supply chain, logistic management, sales

management)

3. Materials management (Purchasing, procurement etc.)

4. Production management (Project scheduling, line balancing, maintenance, etc.)5. Personnel Management (Labour management, manpower management, planning

compensation, recruitment planning)

6. Research & Development. ( Project selection, reliability and alternative design)


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LINEAR PROGRAMMING

INTRODUCTION

The development of linear programming has been ranked among the most scientific

advances of the mid- 20th Century, and we must agree with this assessment. Its impact

since just 1950 has been extraordinary. Today it is a standard tool that has saved manythousands or millions of dollars for many companies and businesses even of moderate

size in the various industrialized countries of the world; and its use in other sectors has

been spreading rapidly.

The resource allocation problem affects all of us. For instance, each day you must

determine how to allocate your time among activities such as: study, work and recreation.Similarly, most of us work with a limited financial budget and must make difficult

decisions as to how these funds can best be used for family living. If you are a manager,

then the problem may arise in allocating limited resources (such as men, machines,

capital equipment, materials, etc.) among competing activities. Linear Programming (LP)

has been successfully applied to solve such type of problems.

Linear Programming is one of the most widely used and best understood OperationsResearch techniques. The LP is concerned with the problem of allocating limited

resources among competing activities in an optimal manner. This type of problem arises

in a number of situations such as manufacturing an item at a minimum cost, blending ofchemicals, allocating salesmen to sales territories, selection of various media for

advertising campaign, scheduling production, etc. In each of these situations the common

ingredient is the requirement that some type of scares/limited resource must be allocated

to some specific activity. Since the resources employed generally involve costs andproduce profits, LP problems generally concern either minimizing the cost or

maximizing the profit.

TERMINOLOGY

The word linear is used to describe the relationship among two or more variableswhich are directly proportional. For example, if doubling (or tripling) the production ofa product will exactly double (or triple) the profit and required resources, then it is a

linear relationship. The word programming means planning of activities in a manner

that achieves some optimal result with restricted resources.

A programme is optimal if it maximizes or minimizes some measure or criterion ofeffectiveness, such as profit, cost or sales.

REQUIREMENTS

Regardless of the way one defines Linear Programming, certain basic requirements whichare given below are necessary before it can be used for optimization problems.

Decision variables and their relationship: The decision (activity) variables refer tocandidates (products, services, projects etc.) that are competing with one another for


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sharing the given limited resources. These variables are usually inter-related in terms of

utilization of resources and need simultaneous solutions. The relationship among these

variables should be linear.

Objective function: The Linear Programming problem must have a well-defined

(explicit) objective function to optimize. For example, maximization of profit,minimization of costs or elapsed time of the system being studied is the objective

function. It should be expressed as linear function of decision variables.

Constraints: There must be limitations on resources which are to be allocated among

various competing activities. These resources may be production capacity, manpower,

time, space or machinery. These must be capable of being expressed as linear equalities

or inequalities in terms of decision variables.

Alternative courses of action: There must be alternative courses of action. For example,

there may be many processes open to a firm for producing a commodity and one process

can be substituted for another.

Non-negativity restriction: All variables must assume non-negative values, that is, allvariables must take on values equal to or greater than zero. Therefore, the problem should

not result in negative values for the variables.

Linearity and divisibility: All relationships (objective function and constraints) must

exhibit linearity, that is, relationship among decision variables must be directly

proportional. For example, if our resource increases by some percentage, then it should

increase the outcome by the same percentage. Divisibility means that the variables arenot limited to integers. It is assumed that decision variables are continuous, i.e.,

fractional values of these variables must be permissible in obtaining an optimal solution.

Deterministic: In LP model (objective function and constraints), it is assumed that all the

model coefficients are completely known (deterministic), e.g., profit per unit of each

product, and amount of resource available are assumed to be fixed during the planningperiod.

ASSUMPTIONS

A linear Programming model is based on the assumptions ofproportionality, additively,continuity, certainty, and finite choices.

1. Proportionality: A basic assumption of Linear Programming is that proportionality

exists in the objective function and the constraint inequalities. For example, if one

unit of a product is assumed to contribute Rs 10 toward profit, then the total

contribution would be equal to 10 x1, where X1 is the number of units of theproduct. For 4 units, it would equal Rs 40 and for 8 units it would be Rs 80, thus if

the output (and sales) is doubled, the profit would also be doubled. Similarly, if oneunit takes 2 hours of labour of a certain type, 10 units would require 20 hours, 20


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units would require 40 hoursand so on. In effect, then, proportionality

means that there are constant returns to scale----and there are no economics of

scale.

2. Additivity: Another assumption underlying the liner programming model is that in

the objective function and constraint inequalities both, the total of all the activities

is given by the sum total of each activity conducted separately. Thus, the totalprofit in the objective function is determined by the sum of the profit

contributed by each of the products separately. Similarly, the total amount of aresource used is equal to the sum of the resource values used by various

activities. 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).

3. Continuity: It is an assumption of a linear programming model that the decisionvariables 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 withthe 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 value.However, when we do so, we should valuate the revised solution to determine

whether the solution represented by the round value 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 giventime period only, then the fractional values many 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 1/3 units of B per week, can be adopted without anydifficulty. 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 B over a three week period wouldimply 5 2/3 units of A and 10 1/3 units of B per week. Lastly, if we must insist on

obtaining only integer value of the decision variables, we may restate the problem

as an integer programming problem, forcing the solutions to be in integers only.

4. Certainty: A further assumption underlying a Linear Programming model is thatthe various parameters, namely, the objective function coefficients, thecoefficients of the inequality/equality constraints and the constraint (resource)

values are known with certainty. Thus, the profit per unit of the product,requirements of materials and labour per unit, availability of materials and labour,

etc. are given or known in a problem involving these. The linear programming is

obviously deterministic in nature.5. 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


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problems, the output cannot obviously be negative, because a negative production

implies that we should be able to reverse the production and convert the finished

output back into the raw materials.

ADVANTAGES OF LINEAR PROGRAMMING

1. Linear Programming can be used to solve allocation type problems. Such problems

are very common and extremely important in organization. Their solution is

difficult due it the fact that an infinite number of possible solutions may exist. AnLP Model not only provides the optimal solution but does so in a very efficient

manner. Further, it provides additional information concerning the value of the

resources to be allocated.

2. Linear Programming gives possible and practical solution as there might be otherconstraints operating outside the problem which must be taken into account. Just

because we can produce so many units does not mean that they can be sold. It

allows modification of its mathematical solution for the sake of convenience to the

decision-maker.3. Linear Programming improves the quality of decisions. It makes decisions more

objective rather then subjective.4. Linear Programming helps in highlighting the bottlenecks in the production

processes.

5. Linear Programming helps in attaining optimum use of productive factors. It alsoindicates the significance and utility of these factors more effectively.

APPLICATION AREAS OF LINEAR PROGRAMMING

Linear Programming is the most widely used technique of decision-making in business,

industry and in various other fields. Few broad application areas of L.P are as under.

1. Production management: Linear Programming can be applied for determining the

optional product-mix to make best use of machine available. It is also used for

product smoothing and assemble line-balancing.2. Marketing management: Linear Programming helps in analyzing the audience

coverage if an advertising campaign based on the available advertising media and

budget. It also helps traveling salesman in finding the shortest route for this tour. It

is also used to determine the optimal distribution schedule for transporting theproduct form several warehouses to various market locations in such a way that the

total transportation cost in minimum.

3. Personnel management: Linear Programming enables the Personnel Manager toanalyze problems relating to recruitment, selection, training, and development of

manpower. It is also used to determine the minimum number of employees needed

to work in various shifts to meet a time varying workload.4. Financial management: Linear Programming helps in selection of specific

investments for among a wide variety of alternatives. Using LP model, decision can

be made with regard to how much production is to be supported by internally and

externally generated funds.


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5. Agriculture: Agricultural applications fall into two categories, 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. Thestudy of farm economics deals with inter-regional competition and optimum spatial

allocation of crop production. Efficient production patterns can be specified by LP

model under regional land resources and national demands constraints. An LP canbe applied in agricultural planning, e.g., allocation of limited resources such as

acreage, labour water supply and working capital, etc., in such a way as to

maximize net revenue.6. Military: Military applications include the problem of selecting an air weapons

system against the enemy so as to keep them pinned down and at the same time

minimize the amount of aviation gasoline used. A variation of transportation issue

is the problem of community defence against disaster, the solution of which yieldsthe 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.

Other applications of LP include the areas of administration, education, inventory control,fleet utilization, hospitals, awarding contracts, capital budgeting etc.

LIMITATIONS OF LINEAR PROGRAMMING

In spite of having wide field of applications, there are some limitations associated withLinear Programming which are given below:

1. The Linear Programming problem assumes the linearity of objective function and

constraints. But, generally, the objective function and the constraints in real-lifesituation concerning business cannot be solved by LP technique.

2. There is no guarantee that the solution by LP technique will give integer valued

solution. For example, in finding out many men and machines would be required toperform a particular job, the solution may be in fractions, and rounding off the

solution to the nearest integer may not yield an optimal solution. In such cases, LP

cannot be used.3. Linear Programming model does not take into consideration the effect of time and

uncertainty.

4. Parameters appearing in the model are assumed to be constant. But in real-life

situation, they are frequently neither known nor are they constants.5. Linear Programming deals with only a single objective, whereas in real-life

situations we may come across more than one objective. When above situations

(limitations) arise, other techniques such as integer programming can be utilized.


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FORMULATION OF LINEAR PROGRAMMING PROBLEMS

Here the formulation means writing or expressing a problem in a convenientmathematical form. Let us consider the following example:

Problem 1A furniture firm manufactures tables and chairs. Data given below shows the resources

consumed and unit profit in manufacturing a table and a chair. Here it is assumed that

wood and labour are the only two resources which are consumed in manufacturingfurniture. The manager of the firm wishes to determine how many tables and chairs

should be made to maximize the total profit. Let us formulate the problem as a linear

programming problem.

ResourceUnit requirements

Amount availableTable Chair

Wood (sq. ft.) 30 20 300

Labour (hours) 5 10 110Unit profit (Rs.) 600 800

Formulation: We will begin by treating the number of tables and chairs to bemanufactured as unknown quantities or decision variables. Let X1 and X2 be the

number of tables and chairs to be manufactured, respectively.

Objective function: Since total profit consists of the profit derived from selling tables at

Rs. 600 each plus the profit derived from selling chairs at Rs. 800 each. thus, 600X 1 is the

profit earned by selling tables and 800X2 is the corresponding profit by selling chairs, Asthe manager wants to achieves the greatest possible profit (say Z) it can be stated

algebraically by writing profit equation as:Maximize Z = 600X1 + 800X2

In this form, the profit expression provides the objective function.

Constraints: Constraints are limitations or restrictions placed on availability ofresources:

i) Wood used for tables + wood used for chairs < Available Wood (sq. ft.)

i.e., 30X1 + 20X2 < 300

ii) Labour hours for table + hours for chair < Available Labour (hours)

i.e., 5X1 + 10X2 < 110

Further we cannot have negative production, i.e., either we manufacture or do not

manufacture.i.e., X1 > 0 and X2 > 0

These are called non-negativity restrictions.


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We can now state the problem, also referred to as a Linear Programming problem in full.

Max Z= 600X1 + 800X230X1 + 20X2 < 300

5X1 + 10X2 < 110

X1, X2 > 0ProblemGiven the data below for products A and B

ResourceProduct Resource

availableA B

R1 60 20 1200

R2 40 50 2000

Contribution per Unit Rs. 3 Rs. 2

Formulate the given problem as a linear programming problem.

FormulationMaximise Z = 3A+ 2B

Subject to 60A+ 20B < 1200

40A+50B < 2000

A, B > 0

Problem 3

The vitamins A and B are found in two different foods F 1 and F2. One unit of food F1contains two units of vitamin A and five units of vitamin B. One unit of food F 2 contains

four units of vitamin A and two units of vitamin B. One unit of food F 1 and F2 cost Rs. 3and Rs. 2.50, respectively. The minimum daily requirement for a person of vitamin A and

B is 40 and 50 units, respectively. Find the optimal mix of food F1 and F2 at theminimum cost which meets of daily minimum requirement of vitamin A and B. Assumethat anything in excess of daily minimum requirement of vitamin A and B is not harmful.

Let us, formulate the given problem the given problem as a linear programming problem.

Formulation: The data of the given problem can conveniently be summarized in thefollowing tabular form:

Food

Vitamin F1 F2 Daily requirement

A 2 4 40

B 5 2 50Cost (Rs.) per unit 3 2.5

Let X1 and X2be the number of units needed of food F1 and F2 respectively to meet the

daily requirement of vitamin A and B

Since the daily minimum requirement of vitamins A and B is 40 and 50 units

respectively, the constraints are given by


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2X1 + 4X2 > 40

5X1 + 2X2 > 50

As cost per unit of food F1 and F2 are Rs. 3 and Rs. 2.50, respectively, then the objectivefunction will become

Minimize Z = 3X1 + 2.5X2

Also either we need some units of food F1 and F2 or at worst we may not need any unit of

the food. Therefore, the non-negativity restriction is given as:

X1 > 0 and X2 > 0

The given problem then can be expressed in LP model as:

Min Z= 3X1 + 2.5X2

2X1 + 4X2 >40

5X1 + 2X2 > 110

and, X1, X2 > 0

Problem 4

Suppose you are trying to select the cheapest combination of two foods X and Y to meet

daily vitamin needs. The vitamin needs call for at least 40 units of vitamin A, 50 units ofvitamin B, and 49 units of vitamin C. Each gram of food X provides 4 units of vitamin A,

10 units of vitamin B, and 7 units of vitamin C. Each gram of food Y provides 10 units of

vitamin A, 5 units of vitamin B, and 7 units of vitamin C. Food X costs Rs. 5 per gram,

and food Y costs Rs. 8 per gram. The objective is to find the least expensive way tosatisfy your vitamin needs. Formulate this problem as a linear programming problem.

Formulation

Minimize Z = 5X +8Y

Subject to 4X+10Y> 40

10X+5Y> 507X+7Y > 49

Problem 5

A company is making two products A and B. The cost of producing one unit of product Aand B is Rs. 60 and Rs. 80, respectively. As per the agreement, the company has to

supply at least 200 units of product B to its regular customers. One unit of product A

requires one machine hour whereas product B has machine hours available abundantlywithin the company. Total machine hours available for product A are 400 hours. One unit

of each product A and B requires one labour hour each and total of 500 labour hours are

available.

The company wants to minimize the cost of production by satisfying the given

requirements. Let us formulate this problem as a liner programming problem.


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Formulation: Let X1 and X2 be the number of units of product A and B to be produced,

respectively, then LP model is given by:

Min Z= 60X1 + 80X2

X2 > 200 (agreement constraint)

X1 < 400 (machine hours constraint for product A)X1 + X2 < 500 (Labour hours constraint)

and X1, X2 > 0

MATHEMATICAL FORMULATION

Instead of two decision variables and few constraints, if we have n decision variables and

m constraints in the problem, then we would have the following type of mathematicalformulation of LP problem.

Optimize (maximize or minimize)

Z = C1X1 + C2X2 ++ CnXn

Subject to a11X1 + a12X2 + .. + a1nXn b1

a21X1 + a22X2 + .. + a2nXn b1 . . . .

. . . .

. . . .

. . . .and am1X1 + am2X2 + .. + amnXn bm

X1,X2,, Xn > 0

The above formulation may be put in the following compact form by using the ()summation notation.

Optimize (max. or min.) Z

Subject to

And Xj > 0; j = 1,2..,n (non-negativity restrictions)

>


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Where,

Xj = decision variables, i.e., quantity of the jth variable of interest to the decision-maker.

Cj = constant representing per unit contribution to the objective function (profit or cost)of the jth decision variable.

aij = constant, representing exchange coefficients of the jth decision variable in the ith

constraint.bi = constant, representing ith constraint or availability.

And the expression ( ) means that each constraint may take only one of the three

possible forms:i) less than or equal to ( < )

ii) equal to (=)

iii) greater than or equal to ( > )

The expression Xj > 0 simply means that the Xjs must be non-negative

The basic difference between a maximization and minimization problem in linear

programming is found in the signs of the inequalities of constraints. The constraints are

generally expressed by less than or equal to ( < ) sign in the maximization problem,whereas those of the minimization problem are expressed by greater than or equal to

( > ) sign.

In particular any minimization problem can be converted into equivalent maximization

problem by changing the sign of Cjs in the objective function. For example, the linear

objectives function. For example, the linear objective function

Minimize Z

is equivalent to Maximize Z* =

Thus, for example, maximizing the objective function

Z = 3X1 + 5X2

Is equivalent to minimizing its negative counterpart, i.e.,

Z* = -Z = -3X1-5X2

Equality Sign

So far, we have discussed only those constraints which are either expressed by ( < ) signor by (>) sign. If there exists a situation, when some of the constraints are expressed by

equations, each of the equations may be replaced by two inequalities. For example,

a11X1 + a12X2 = b1

>

b1or - a11X1 - a12X2 < - b1

Unrestricted Variables

In certain problems, situations may exist, when some decision variables are unrestricted

in sign (positive, negative or zero). In all such cases, the decision variables can always beexpressed as the difference in sign, then we define two non-negative variables X i

+ and Xi-

such that

Xi = Xi+ - Xi

-

Xi+, Xi

- > 0

SOLUTION TO LP PROBLEM

Several methods or algorithms have been developed to get the optimal solution to LP

problem like graphical method and Simplex method

Graphical Method

Graphical method is used for solving those LP problems which involve only twovariables. Consider example of furniture manufacturer mentioned above. On our graph,

let the horizontal axis represent the number of tables manufactured and the vertical axis

the number of chairs manufactured. We will plot a line for each of the two constraints

and the two non-negativity restrictions. We shall begin by plotting the linescorresponding to the non-negativity restrictions. This plotting suggests that the graphical

method always starts with the first quadrant of the graph.

Now in order to plot the constraints on the graph, temporarily we consider inequalities as

equations, i.e,

30X1 + 20X2 = 3005X1 + 10X2 = 110

When plotted on the graph, these will represent straight lines. All points on these straight

lines represent combinations of tables and chairs quantities that consumer the exactamount of resource that is available. In other worlds, points on these lines are satisfying

the equality and any point below the line will include the inequality part of the constraint.

A straight line is completely specified by knowing any point that falls on that line.

Therefore, to plot any constraint, we need only to specify two points on that line and then

draw the line by connecting these two points. To illustrate let X 1 =0 in the availabilitywood constraint, then we get:

30(X1 = 0) + 20X2 = 300

Or

X2 = 15


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Figure I Graphical Solution for Maximization Problem

X2

Chairs

(0, 15)

(0, 11) 30X1 + 20X2 = 300

B (4, 9)

Feasible

Solution 5X1 + 10X2 = 110Region

Tables X1 A(10,0) (22,0)

This represents the vertical interceptor on X2 axis and is written as (0, 15), i.e., o tables

and 15 chairs. Similarly, horizontal interceptor on X1 axis is obtained by putting X2 = 0we get

20X1+ 20(0) = 300

OrX1 = 10

It is also written as (10, 0) and represents 10 tables and 0 chairs. Join the two interceptorsto draw the constraint line. Thus, the points on or below the line represent the region

satisfied by this constraint. Any point which does not lie on or below the line gives

infeasible solution.

Similarly, the availability (labour) constraint can also be plotted on the graph as shown in

the Figure.

Any combination of value of X1 and X2 which satisfies the given constraints is called a

feasible solution. The OABC in the figure satisfied by the constraints is shown by

shaded area and is also called feasible solution region or space. The coordinates of thepoints on the corners of the region OABC can be obtained by solving equations

intersecting on these points.

100 5 15 20 22 25

10

0

5

11C


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The shaded region may contain an infinite number of points which would satisfy the

constraints of the given LP problem. However, it can be proved that the optimal solution

of any LP problem corresponds to one of the corner points (also called extreme points) ofthe feasible solution region. Thus we confine only to those points which correspond to

corners of feasible solution region to get an optimal value of the objective function of

given LP problem.

In the present example, value of objective function on the corner points can be evaluated

as follows:

Coordinates of corner point Objective function

Z= 6X1 + 8x2

Value

O (0, 0) 600 (0) + 800 (0) 0

A (10, 0) 600 (10) + 800 (0) 6000

B (4, 9) 600(4) + 800 (9) 9600

C (0, 11) 600 (0) + 800 (11) 8800

The maximum profit of Rs. 9600 is obtained at the corner point B (4,9), i.e. X 1 = 4 X2 = 9

which also satisfy the given constraints. Hence to maximize profit, the company must

produce 4 tables and 9 chairs.

The steps of graphical method can be summarized as follows:

1. Formulate the linear programming problem.2. Plot the constraint lines considering them as equations.

3. Identify the feasible solution region.

4. Locate the corner points of the feasible region.5. Calculate the value of the objective function on the corner points.

6. Choose the point where the objective function has optimal value.

USING THE GRAPHICAL METHOD FOR MINIMIZATION

The solution of minimization problem follows the same procedure as that of

maximization problem. The basic difference is that we now want the smallest possiblevalue for the objective function. The earlier discussed problem is reproduced below:

Min Z= 3X1 + 2.24 X2Subject to 2X1 + 4 X2 > 40

5X1 + 2 X2 > 50

and X1, X2 > 0

Because there are only two variables X1 and X2 we can construct a graph of the set of

feasible solution. Recall that the non-negativity restrictions reduce this feasible region to

the first quadrant of the graph. To this quadrant we first plot the constraint for vitamin Aas a straight line given by

2X1 + 4 X2 = 40


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We now obtain the two points necessary to plot this line. We first set X 1 = 0, which yield

X2 = 10 and set X2 =0 getting X1=20. We then plot this line as shown in the graph.

Similarly, we can plot the constraint for vitamin B. since the inequalities are of thegreater than or equal to type, the feasible region is formed by considering the area to the

upper right side of each equation (a way from the origin)

The area above CBD is satisfied by the two constraints and is shown by shaded areawhich is termed as feasible solution region. There is a significant difference between the

feasible region for this minimization problem and the one for the maximization problem.

The feasible region for this minimization problem is unbounded and unlimited becauseany combination of vitamins A and B will satisfy the constraints.

Since the optimal solution corresponds to one of the corner (extreme) points, we will

calculate the values of the objective function for each corner point. Viz., D (20,0); B (7.5,

6.25); and C (0.25). The calculations are shown below in the table.

Figure II Graphical Solution for Minimization Problem

5X1 + 2X2 > 50

FeasibleSolution

Region

B (7.5, 6.25

2X1 + 4 X2 > 40

D (20,0)

X1

FOOD F1

100 5 15 20

10

0

5

15

25

20

X2

C (0, 25)

FOOD

F1


18/35

Coordinates of corner

point

Objective function

Z= 3X1 + 2.25x2

Value of Z

D (20,0) 3 (20) + 2.25 (0) 60

B (7.5,6.25) 3 (10) + 2.25 (6.25) 36.5625

C(0,25) 3 (4) + 2.25 (25) 56.25

The minimum cost is obtained at the corner point B (7.5, 6.25), i.e., X 1 = 7.5 and X2 =

6.25. Hence to minimize cost, and to meet the daily vitamins requirement, one should

have 7.5 units of food F1 and 6.25 units of food F2.

Although the two examples so far discussed have contained constraints of one type either

all less than or equal to or all greater than or equal to, it is possible to have mixedconstraints.

SPECIAL CASES IN LINEAR PROGRAMMING

Each of the LP problems discussed had a unique optimal solution. However, it is possiblefor LP problems to have some special cases. Some of these special cases include:

1. Alternative optimal solution

2. Unbounded solution

3. Infeasible solution4. Redundant constraint

Each of these special cases will now be described briefly.

Alternative Optimal solution

All our examples so far have resulted in unique optimal solution. This may not always betrue. Any particular LP problem may have multiple solutions, all of which are optimal.

Unbounded Solution

When a LP solution is permitted to be infinitely large, it is known to be unbounded. For

example, in a maximization problem at least one of the constraints should be an equality

or less than or equal to ()type, then there will be no upper limit on the feasible region. Similarly, for minimization,

there must be equality or a greater than or equal to (>) type constraint if a solution is to be

found.

Infeasible Solution

If there is no solution that satisfies all the constraints, it is said to be infeasible. Such acondition generally indicates that the LP problem has been wrongly formulated. The only

way to arrive at a feasible solution is to reformulate one or more of the constraints.


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Redundant Constraint

In a properly formulated LP problem, each of the corner will define a portion of theboundary of the feasible solution region. Whenever, a constraint does not define a portion

of the boundary of the feasible solution region, it is called a redundant constraint. A

redundant constraint is unnecessary for the problem and may be omitted from theproblem formulation.

LIMITATIONS OF THE GRAPHICAL METHOD

The graphical method is widely used as an academic exercise because it enables the

student to visualize the LP solution process. The method is severely limited in application

by the fact that the number of variables should be restricted to two dimensions. For threevariables problem, it becomes very difficult as one must be familiar with three

dimensional space. Of course, the method breaks down completely for four or more

variables.

Other methods of solution which avoid the limitations of the graphical method have been

developed. The most widely used method is known as Simplex Method.

USE OF COMPUTER

Real-life LP problem are almost too large and complex for graphical or manual simplex

methods. Therefore, they are solved with the help of computers. Numerous LP packages

exist for solving linear programming problems, ranging from simplified procedures

designed for classroom use to sophisticated and complex programmes. Virtually allcomputer manufactures and software consultants offer these LP packages.

SUMING UP

Linear Programming is powerful technique of Operations Research designed to solve

allocation problems. It helps in choosing the best combinations of various activities tomeet various criteria in such a way that the solution obtained is optimal. LP may be

applicable if you have a problem which (a) is deterministic in nature, (b) involves

multiple variables which are inter-related, (c) has a set of criteria which must be met, and

(d) has a single objective. All LP problems can be formulated in a common format. Thegraphical method provides a good conceptual introduction to linear programming

problems. However, its usefulness in solving practical problems is quite limited. For real-

life problems, simplex method with the help of computer packages can be used.

KEY WORDS

Constraints: An upper limit on the availability of a resource or a lower limit on

necessary levels to achieve.

Corner (or extra) points: The points of feasible solution region formed by the

constraints and the non-negativity restrictions of a problem.


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Feasible solution region: The region defined by the constraints and the non-negativity

restrictions of the LP problem.

Linear Programming: A linear deterministic model used to solve the problem of

allocating limited resources among competing activities.

Non- negativity restrictions: The conditions that require the values of the decision

variables to be either zero or more than zero.

Objective function: An equation that specifies the dependent relationship between the

decision objective and the decision variables.


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Linear Programming: Simplex Method

Simplex Method: Mathematical iterative procedure known as simplex

method was developed in 1957 by George B Dantzig.

Basic Concepts\TermsThe optimal solution to a linear programming problem, if it exists,

always occurs at one of the corner points of the feasible solution space.

The evaluation of corner points always starts at the point of origin (initial

basic feasible solution)

Standard Form: A linear programme in which all the constraints are written

as equalities.

Slack Variable: A variable added to the LHS of a less than or equal to

constraint, to convert the constraint into an equality is called the slack

variable.Surplus Variable: A variable subtracted from the LHS of the greater than

or equal to constraint, to convert the constraint into an equality is called a

surplus variable.

Basic Solution: For a system of m simultaneous linear equations in n

variables (n > m), a solution obtained by setting (n-m) variables equals to zeroand solving for the remaining m variables is called as a basic solution. In

addition, the solution for the m variables must be unique. The (n-m)

variables set equal to zero in any solution are called non-basic variables. The

other m variables whose values are obtained by solving the remaining systemof equations are referred to as Basic Variables.

Basic Feasible Solution: A basic feasible solution to a linear programming

problem is a basic solution for which m variables are solved for, are all

greater than or equal to zero. In other words, a basic solution which happens

to be feasible is called a basic feasible solution.

Optimal Solution: Any basic feasible solution which optimizes the objective

function of a general linear programming problem is called an optimal basic

feasible solution to the linear programming problem.

Tableau Form: The form in which a linear programme must be written priorto setting up the initial simplex tableau.

Simplex Tableau: A table is used to keep track of the calculations made at

each iterations when the simplex solution method is employed.

Zj Row: The numbers in this row under each variable represents the total

contribution of outgoing profit when one unit of a non-basic variable is

introduced into the basis in place of basic variable.


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Cj-Zj (net evaluation or index) Row: The row containing the net profit (or

loss) that will result from introducing one unit of the variable indicated in

that column in the solution. Numbers in index row are also known as

SHADOW PRICES. Thus a positive (or negative) number in the index row

indicates an algebraic reduction (or increment) in the objective function ifone unit of the variable at the head of that column, were introduced in the

basis (solution)

Pivot (or Key) Column: The column with the largest positive number in the

net evaluation row of maximization problem, or the largest negative number

in the net evaluation row in a minimization problem. It indicates which

variable will enter the solution next.

Pivot (or Key) Row: The row corresponding to the variable that will leave

the basis in order to make room for the entering variable. The departing

variable will correspond to the smallest positive ratio found by dividing thequantity (solution) column values by pivot column values for each row.

Pivot (or Key) Element: The number at the intersection of the pivot row and

pivot column.


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LINEAR PROGRAMMING: THE SIMPLEX METHOD

INTRODUCTION

We can solve LP problem with two or more than two variables by using a systematic

procedure called the Simplex Method. The simplex method is the name given to the

solution algorithm for solving LP Problems developed by George B. Dantzig in 1947.By solution algorithm we refer to an iterative procedure having fixed computational

rules that leads to a solution to the problem in a finite number of steps. By using the

simplex method, one is capable of solving large LP problems efficiently. The simplexmethod allows evaluation of the corner points in such a way that each successive corner

point gives the same or better solution than the previous one. This search for examining

successive corner points will continue until no better solution can be found.

THE SIMPLEX METHOD (MAXIMISATION CASE)It was pointed out during the graphical solution procedure that the search for an optimal

solution can be limited to only the corner points of the feasible solution region. This was

very easy to do manually for a problem with two variables and less number ofconstraints, but for larger problems a more efficient procedure is necessary for

identifying and evaluating corner points. This is one of the main objectives of the simplexmethod. The simplex method is a well defined systematic procedure called an algorithm.The simplex algorithm provides a more structured method for moving from one corner

point to another, always maintaining or improving the objective function, until an optimalsolution is obtained. To explain and develop the simplex method, let us consider the

following example

Max Z = 6X1 + 8 X2Subject to 30 X1+ 20 X2 300

5X1 + 10 X2 110

and X1, X2 0

The problem must first be stated in equation form so that the simplex method can be

used. Thus, every inequality constraint in the LP problem must first be converted into an

equality constraint so that the problem can be written in a standard form. We do this byadding a slack variable to each constraint. Each slack variable corresponds to the amount

of unused capacity (or resources) for the constraint to which it is added. We will use the

symbol S to represent slack variables. Slack variables are always added to the less

than type of constraint to make it an equality. These slack variables must be non-

negative; otherwise, the capacity utilized will exceed the total available capacity. The

constraints for the given problem can now be rewritten with slack variables to form the

equalities as given below:

30 X1+ 20 X2 + S1 = 300

5X1 + 10 X2 + S2 = 110

Since slack variables represent unused resources, their contribution in the objective

function is zero. Including these slack variables in the objective function, we get


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Max Z = 6X1 + 8 X2 + 0S1 + 0S2

The logic of the simplex method is based on the fact that only the corner points of thefeasible solution region can give unique optimal solution. The search starts with a

solution at the origin indicating nothing can be produced and therefore the values of

decision variables are zero, i.e., X1 = 0 and X2 = 0. When we are not producinganything, obviously we are left with unused capacity of S1 = 300 and S2 = 110. We note

that the current solution has two variables (slack variables S1 and S2) with non-zero

solution value and two variables (decision variables X1 and X2) with zero values.

Variables with non zero solution values are called basic variables . Variables with

zero values are called non basic variables. There will always be the same numberof basic variables as there are the numbers of constraints, provided the constraints are

not redundant and that a basic feasible solution exists. Solutions with basics variables

are called basic solutions. The basic solutions can further be divided into categories:

feasible and infeasible. The basic feasible solutions are those that satisfy all the

constraints. The simplex procedure searches for basic feasible solutions only at the

corner points of the feasible solution region. Introducing the slack variables, our problemcan be given below:

Max Z = 6X1 + 8 X2 + 0S1 + 0S2Subject to 30 X1 + 20 X2 + S1 = 300

5X1 + 10 X2 + S2 = 110

and X1, X2, S1, S2 0

To use simplex method, it is convenient to put the problem in a tabular form as shown

below:

Table 1

CONTRIBUTION PER UNIT 6 8 0 0MINIMUM

RATIOBASICVARIABLES

SOLUTIONVALUES

X1 X2 S1 S2

0

0

S1

S2

300

110

30

5

20

10

1

0

0

1

300/20= 15

110/10= 11

KEY ROW

CONTRIBUTION PER UNIT Zj 0 0 0 0

NET CONTRIBUTION PER UNIT Cj Zj 6 8 0 0

KEY ELEMENT KEY COLUMN

The first row of the table indicates the value of Cj, the coefficients of the objective

function, and indicates the contribution per unit to the objective function of each of thevariables. The second row of the table provides the column heading for the table. The

first column heading lists the coefficient of the objective function of the current basicvariables. The second column represents the basic variables in the current solution. The

next column, with heading solution values, provides the current solution. In our

example, when the solution is at the origin, the basic variables are the slack variables S 1and S2 and these are listed in the second column of the Table. Referring back to the first

column, the coefficients for these two basic variables in the objective functions are S 1 = 0

Cj


25/35

and S2 = 0, respectively. If the current solution at the origin is X 1 = 0 and X2 = 0, then the

solution values correspond to S1 = 300 and S2= 110 as shown in the third column of the

Table. The next four columns headed by X1, X2, S1, and S2 are the coefficients of theconstraint set.

The Zj row represents the decrease in the value of the objective function that will result ifone unit of the variable is brought into the solution. Therefore, the Zj values can be

thought of as the objective function contribution loss per unit and is determined by

summing the products of the coefficients of Cj column and the physical rates ofsubstitution (coefficients in the constraint set) associated with corresponding basic

variables. In our example, contribution loss per unit, (Zj) row values, are determined as

given below:

Contribution loss per unit (Zj) = Summation of (Coefficients of Cj column *

corresponding Coefficients of the constraint set)

For X1 column: Zj = 0 * 30 + 0*5 = 0

Similarly we can calculate Zj values for other columns as shown in the Table.

The last (Cj - Zj) row represents the net contribution per unit and its value is

determined by subtracting the appropriate Zj value from the corresponding coefficient Cjvalue in the objective function for that column. This value of (Cj - Zj) is the differencebetween the contribution Cj and the loss Zj that result from one unit of Xj being produced.

For determining the value of the last row (Cj Zj), the calculations are shown below:

Net contribution per unit (Cj - Zj) = contribution per unit contribution loss per unit

In our example, the values of net contribution per unit (Cj - Zj) for X 1 column are

determined as given below:

Net contribution per unit (Cj - Zj) = 6-0 = 6 and similarly we can calculate the other

values of (Cj - Zj) for other columns as shown in Table.

The positive value in the (Cj - Zj) row indicate improvement possibility in the existing

solution. The objective is to maximize profit. Therefore, consider that column where per

unit contribution is the largest. In our case, we notice that in X 2 column, contribution perunit (i.e., Rs. 8) is maximum. This criterion helps us to know the variable to be entered

into the solution basis. Thus X2 is the entering variable. The column corresponding to

entering variable is termed as key column. Since we have decided to enter one variableas the basic variable into the solution basis, therefore, we replace one of the existing basic

variables to depart from the solution basis. The variable to depart is identified by forming

the ratios of solution values to physical rates of substitution of entering variable. Thus, inour example, we have

For S1; ratio = 300/20 = 15

For S2 : ratio = 110/10 = 11


26/35

The one with minimum ratio (S2) represents the variable to depart from the solution basis.

Thus S2 is the departing variable. This procedure of selecting the departing variable

guarantees that no basic variable will ever be negative. The row corresponding to thedeparting variable is called key row. The element on the inter-section of key row and

key column is called the key element and is denoted by making a box ( ) in the

simplex table.

At this point, develop a new improved solution by revising Table I. Now have two tables

to carry out the calculation: the old table and a new table.

To revise the key row divide all values in the key row (S 2) by the value of the key

element (i.e.10) and replace departing variable (S2) by the entering variable (X2).

Also replace the new values in Cj column accordingly. Pull all other values so obtained at

the appropriate places. In our example, this new row becomes:

8 X2 11 5 /10 1 0 1 /10

For other non-key rows, new values can be obtained as given below:

New row value= old row value (corresponding to old value in the key column *

corresponding new value in the revised key row).

In our example, the values of other row (S1) can be obtained as follows:

For solution value column: = 300 20 11 = 80For X1 column: = 30 20 5/10 = 20

For X2 column: = 20 - 20 1 =0

For S1 column: = 1 - 20 0 = 1For S2 column: = 0 20 1 / 10= -2

Therefore, the value for the new row becomes

S1 80 20 0 1 - 2

The value of Cj and (Cj Zj) rows are calculated in the same way as discussed earlier.The value of the objective function is calculated by substituting the values of the

variables in the objective function. For example, in this improved solution

Z= 0 X 80 + 8 X 11 = 88

The new revised improved solution is shown in Table II.


27/35

Table IICONTRIBUTION PER UNIT 6 8 0 0 MINIMUM

RATIOBASIC

VARIABLES

SOLUTION

VALUES

X1 X2 S1 S2

0

8

S1

X2

80

11

20

5/10

0

1

1

0

-2

1/10

80/20= 4

KEY ROW

11 5/10 =22

CONTRIBUTION PER UNIT Zj 4 8 0 8/10Z = 88

NET CONTRIBUTION PER UNIT Cj Zj2 0 0 -8/10


Proceeding in the same manner, we notice that variable X1 will enter into the solution and

variable S1 will depart. The key element is 20. The next improved solution is given below

in Table III.

Table III

CONTRIBUTION PER UNIT 6 8 0 0

BASIC VARIABLES SOLUTION VALUES X1 X2 S1 S2

6

8

X1

X2

4

9

1

0

0

1

1/20

-1/40

-1/10

3/20

Z = 96CONTRIBUTION PER UNIT Zj 6 8 1/10 6/10

NET CONTRIBUTION PER UNIT Cj Zj

0 0 -1/10 -6/10

Since all entries in the net contribution per unit (Cj Zj) row are negative, therefore this

indicates no sign of further improvement in the contribution; therefore, we have toterminate the procedure. Hence Table III provides the optimal solution i.e.,

X1 = 4; X2 = 9; S1=0; S2 =0 and Z = 96

These optimum values suggest that the furniture manufacture should manufacture four

tables and nine chairs and the maximum profit is Rs. 96.

You are requested to compare the results so obtained by simplex method with that ofgraphical method. What do you observe?

Problem 1: Use simplex method to solve the following LP problem:

Max Z= 4X + 5Y + 8Z

Subject to X + Y + Z < 1003X + 2Y + 4Z < 500

and X, Y, Z > 0

Cj

Cj


28/35

Problem 2: A manufacture produces four products A, B, C, and D each of which is

processed on three machines X, Y and Z. The time required to manufacture one of each

of the four products and the capacity of each of the three machines in indicated in the

following tables:

Processing Time (hrs)

Product Machine X Machine Y Machine Z

A 1.5 4 2B 2 1 3

C 4 2 1

D 3 1 2

Capacity (hrs) 550 700 200

The profit contribution per unit of the four products A, B, C and D is Rs. 4, 6, 3 and 1

respectively. The manufacture wants to determine its optional product-mix:

a. Formulate the above as a LPP

b. Solve it with the simplex method

c. Find out the optional product-mix and the total maximum profit contribution.

d. Which constraints are binding?

e. Which machine (s) has (have) excess capacity available? How much?

f. If the profit contribution from product B increases by Rs. 2 per unit, will the

optimal product-mix change?

g. What are the shadow prices of the machine hours on the three machines?

h. If machine A is to be shut down for 50 hours due to repairs, what will be the

effect on profits?

i. If the company whishes to expand the production capacity, which of the three

machines should be given priority?

j. If a customer is prepared to pay higher prices for product A, how much should the

price be increased so that the companys profit remains unchanged?

MINIMIZATION CASEIn problems involving minimization of the objective function one simple way is to

convert the problem by multiplying the objective function by 1. This yield negative

solution values whose sign must be reversed for application. The method however is notrecommended since a direct solution is more convenient to use.


29/35

Let us take the example earlier solved through graphical approach. The formulated

problem is:

Minimize Z= 3X1 + 2.5 X2Subject to 2X1 + 4X2 > 40

5X1 + 2X2 > 50and X1, X2 > 0

The first step in the simplex method is to convert the inequality constraint into equalityconstraint. In minimization problems the slack variables are actually referred to as

surplus variables (negative slack). Rather than representing unused capacity, they

represent the excess amount by which a particular requirement is met. Converting

inequalities into equalities and denoting surplus variables by S1 and S2, we get

2X1 + 4 X2 S1 = 40

5X1 + 2 X2 S2 = 50

The surplus variables take negative values, which violates the non-negativity restrictions.

To overcome this problem, we need to add an additional variable to each constraint thathas a positive value with a surplus variable. These new variables are called artificial

variables because they are used to convert the origin artificially from infeasible to

feasible. It is emphasized that the use of an artificial variable is not limited tominimization problems. An artificial variable may be used in maximization problems

as well. In general, as long as at least one greaterthan or equalto constraint isinvolved, regardless of minimization or maximization problem, artificial variable is

used. In our example, introducing artificial variables, the constraints become;

2X1 + 4 X2 S1 + A1 = 40

5X1 + 2 X2 S2 + A2 = 50

The simplex method then selects the artificial variables as the initial basic variables .

Therefore, the decision and surplus variables are non-basic variables and can be setequal to zero. It is noted that with the addition of the artificial variable the origin stands

converted from infeasible point to feasible one. To correct this problem, we must add

each artificial variable to the objective function. To ensure that the artificial variables

are not basic variables in the optimal solution, we assign them very high costs . Oneconvenient way of doing this is to assign each artificial variable a cost ofM, whereMis

defined to be a very large number. It is emphasized that the artificial variables are

used only as a mathematical convenience to obtain an initial basic feasible solution .This is the reason why the name artificial has been given to these variables since they

are fictitious and have no physical meaning for the original problem. Introducing

artificial variables in the objective function, we get

Min Z = 3X1 + 2.25X2 + 0S1 + 0S2 + MA1 + MA2

The modified problem is now to


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Min Z = 3X1 + 2.25X2 + 0S1 + 0S2 + MA1 + MA2Subject to 2X1 + 4X2 - S1 + A1 = 40

5X1 + 2X2 S2 + A2 = 50

And X1, X2, S1, S2, A1, A2 > 0

Now set up the initial simplex table exactly the way it was done for maximization

problem. This is shown below:

Table I

CONTRIBUTION

PER UNIT3 2.25 0 0 M M MINIMUM

RATIOBASIC

VARIABLES

SOLUTION

VALUESX1 X2 S1 S2 A1 A2

M

M

A1

A2

40

50

2

5

4

2

-1

0

0

-1

1

0

0

1

40/2 = 20

50/5 =10KEY ROW

Zj 7M 6M -M -M M MZ = 90M

Cj Zj 3-7 M 2.25-

6M

M M 0 0


Table II

CONTRIBUTION

PER UNIT3 2.25 0 0 M M MINIMUM

RATIOBASIC

VARIABLES

SOLUTION

VALUESX1 X2 S1 S2 A1 A2

M

3

A1

X1

20

10

0

1

16/5

2/5

-1

0

2/5

-1/5

1

0

-2/5

1/5

20 x 5/16=6.25KEY ROW

40/2 = 20

Zj 3 16M/5

+6/5

-M 2M/5

-3/5

M -2M/5

+ 3/5 Z= 30+20M

Cj Zj 0 -16M/5

+ 21/20

M -2M/5

+3/5

0 7M/5

3/5


The entering variable will be selected for that variable for which the value of (Cj Zj) is the largest negative. In the case, variable X1 has the largest negative value. Thus

X2 is selected as the entering variable. The selection of the departing variable is made

exactly the same way as in the case of maximization. Thus artificial variable A2 ischosen as the departing variable. Therefore, the element which corresponds to the

intersection of key row and key column, i.e., 5 becomes the key element. Having selected

the key element, the computation procedure is the same as for the maximization case. The

Cj

Cj


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revised improved solution is shown in Table II. Proceeding in the same manner, Table III

as shown below is obtained:

Table III

CONTRIBUTION PER UNIT 3 2.25 0 0 M M

BASIC

VARIABL

ES

SOLUTI

ON

VALUES

X1 X2 S1 S2 A1 A2

2.25

3

X1

X1

6.25

7.50

0

1

1

0

-5/16

1/8

1/8

-1/4

5/16

-1/8

-1/8

1/4

Z=

36.5625

Zj 3 2.25 -21/64 -15/32 21/64 15/32

Cj Zj 0 0 21/64 15/32 M

-21/64

M

- 15/32

In the (Cj Zj) row of Table III, there are no negative values. Thus the solution is

optimal. Hence to meet the daily vitamins requirement, one should have 7.5 units of food

F1 and 6.25 units of food F2 and the minimum cost is Rs. 36.5625.

Problem: Use simplex method to solve the following LP problem:

Min Z = 80 X1 + 100 X2Subject to 80 X1 + 60 X2 > 1500

20 X1 + 90 X2 > 1200

and X1, X2 > 0

THE SIMPLEX METHOD (MIXED TYPE CONSTRAINTS)

Let us now consider a situation when the constraints are of mixed type i.e. not merely of

either > type or < type. The LP problem as given below is of this type.

Min Z = 60 X1 + 80X2X2 > 200

X1 < 400X1+ X2 = 500

And X1, X2 > 0

The problem can be converted into the standard form by adding slack, surplus and

artificial variables in the set of constraints and assigning appropriate cost to these

variables in the objective function. Thus the problem can be stated as follows:Minimize Z = 60X1 + 80X2 + 0S1 + 0S2 + MA1 + MA2Subject to X1 - S1 + A1 = 200

X1 + S2 = 400X1 + X2 + A2 = 400

And X1, X2, S1, S2, A1, A2 > 0

Cj


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The initial simplex table for this problem is shown below in Table I

Table I

Contribution

per unit

60 80 0 0 M M Minimum Ratios

Basic

variables

Solution

values

X1 X2 S1 S2 A1 A2

M

0

M

A1

S2

A2

200

400

500

0

1

1

1

0

1

-1

0

0

0

1

0

1

0

0

0

0

1

200/1

KEY ROW

----

500/1 = 500

Zj M 2M -M 0 M MZ= 700M

Cj Zj 60-M 80-2M M 0 0 0


Since our problem is of minimization, therefore, instead of selecting the largest positivevalue in (Cj Zj) row, we shall consider the largest negative value. Since both X 1 and X2have negative values in (Cj Zj) row, therefore we select the largest negative value

corresponding to X2 column as key column. The key row is obtained by choosing theminimum ratio which corresponds to A4. The intersection of these two elements gives us

the key element, i.e. I. Continuing as before; we get the improved solution as shown in

Table II.

Table II

60 80 0 0 M M Minimum

RatiosBasic

variables

Solution

values

X1 X2 S1 S2 A1 A2

80

0

M

X2

S2

A2

200

400

300

0

1

1

1

0

0

-1

0

1

0

1

0

1

0

-1

0

0

1

-

400/1 = 400

300/1= 300

KEY ROW

Zj M 80 M-80 0 -M+80 MZ=16000

+ 300MCj Zj 60-M 0 80-2M 0 2M -80 0


In the improved solution, variable X1 has the negative value and therefore this solution is

not optimal. As shown in the Table, variable X1 will be the entering variable and A2 will

be the departing variable. Proceeding as before, we get the next improved solution asshown below in Table III.

Cj

Cj


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Table III

60 80 0 0 M M

Basic

variables

Solution

values

X1 X2 S1 S2 A1 A2

800

60

X2

S2

X1

200100

300

00

1

10

0

-1-1

1

01

0

11

-1

0-1

1

Z=3400

0

Zj 60 80 -20 0 20 60

Cj Zj 0 0 20 0 M -20 M-60

Since all entries in (Cj Zj) row are either positive or zero, therefore, further

improvement in the objective function is not possible. Hence, the optimal solution is:

X1 = 300, X2 = 200, S1 = 0, S2 = 100 and Z=34,000

Problem: Use simplex method to solve the following LP problem:

Objective Function: Min Z = 3X1 + 8 X2Subject to X1 + X2 = 200

X1 < 80

X2 > 60

and X1, X2, 0.

SPECIAL CASES IN APPLYING THE SIMPLEX METHOD

There are several special situations which are confronted while applying the simplex

method manually. Some of these cases are presented hereunder.

Tie for the Key Column

In simplex method, one proceeds from one basic feasible solution to another improved

solution until an optimum solution (if any) is arrived at. To go from one iteration to the

next a key column is selected based on the most positive (or negative) value in (Cj Zj)row in the case of maximization (or minimization) problem. But a situation may arise

when at any intersection, two or more columns have exactly the same positive (or

negative) value in (Cj Zj) row. In the situation, selection for column can be made

arbitrarily.

Tie for the Key Row (Degeneracy)

If the number of positive variables in the solution is less than the number of constraints,the solution is said to degenerate. In the simplex method, degeneracy occurs when there

is a tie for the minimum ratio for choosing the departing variable. Again the choice for

selecting the departing variable may be made arbitrarily.

Cj


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simplex method to both the maximization and the minimization problems, having less

than equal to and greater than constraints have been discussed. We have also

reviewed the special situations that commonly arise in applying the simplex method.Although the simplex method is efficient, large problems are not easy to solve manually

but rather require computers. Numerous computer programmes have been developed for

this purpose and are available on most computers.

KEY WORDS

Algorithm: A formalized systematic procedure for solving problems.

Artificial Variable: A variable that has no physical meaning and is used to obtain an

initial basic feasible solution to a LP problem.

Basic Variable: The variables that normally take non-zero values to obtain a solution.

Degeneracy: A situation in the simplex method where a basic variable takes on the value

zero.

Redundant Constraint: A constraint that does not affect the feasible solution region.Simplex Method: An algorithm for solving LP problems investigates feasible corner

points only, always maintaining or improving the objective function, until an optimalsolution is obtained.

Slack Variable: A variable used to convert a less than or equal to constraint into an

equality constraint by adding it to the left hand side of the constraint.

Surplus Variable: A variable used to convert a greater than or equal to constraint into an

equality constraint by subtracting it for the left hand side of the constraint.

Unbounded Solution: A solution involving the infinite use of some resource.

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