sem 4 mtma lp and gametheory part 1_new

7/29/2019 Sem 4 Mtma Lp and Gametheory Part 1_new

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LECTURE NOTES ON

LINEAR PROGRAMMING AND GAME THEORY

Pre-requisites: Matrices and Vectors

CHAPTER I

Mathematical formulation of Linear Programming Problem

Let us consider two real life situations to understand what we mean by a

programming problem. For any industry, the objective is to earn maximum profit

by selling products which are produced with limited available resources, keeping

the cost of production at a minimum. For a housewife the aim is to buy provisions

for the family at a minimum cost which will satisfy the needs of the family.

All these type of problems can be done mathematically by formulating a problem

which is known as a programming problem. Some restrictions or constraints are to

be adopted to formulate the problem. The function which is to be maximized or

minimized is called the objective function. If in a programming problem the

constraints and the objective function are of linear type then the problem is called a

linear programming problem. There are various types of linear programming

problems which we will consider through some examples.

Examples

1. (Production allocation problem) Four different type of metals, namely, iron,copper, zinc and manganese are required to produce commodities A, B and

C. To produce one unit of A, 40kg iron, 30kg copper, 7kg zinc and 4kg

manganese are needed. Similarly, to produce one unit of B, 70kg iron, 14kg

copper and 9kg manganese are needed and for producing one unit of C, 50kg

iron, 18kg copper and 8kg zinc are required. The total available quantities ofmetals are 1 metric ton iron, 5 quintals copper, 2 quintals of zinc and

manganese each. The profits are Rs 300, Rs 200 and Rs 100 by selling one

unit of A, B and C respectively. Formulate the problem mathematically.

Solution: Let z be the total profit and the problem is to maximize z(called

the objective function). We write below the given data in a tabular form:


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Iron Copper Zinc Manganese Profit

per unit

in RsA 40kg 30kg 7kg 4kg 300

B 70kg 14kg 0kg 9kg 200

C 60kg 18kg 8kg 0kg 100

Available

quantities 1000kg 500kg 200kg 200kgTo get maximum profit, suppose units of A, units of B and units ofC are to be produced. Then the total quantity of iron needed is

(40 +70 + 60 )kg. Similarly, the total quantity of copper, zinc andmanganese needed are (30 + 14 + 18 )kg , (7 + 0 + 8 )kgand (4 + 9 + 0 )kg respectively. From the conditions of the problemwe have, 40 + 70 + 60 1000

30 + 14 + 18 500

7 +0 +8 200

4 +9 +0 200The objective function is = 300 +200 +100 which is to be maximized.Hence the problem can be formulated as,

Maximize = 300 +200 +100 Subject to 40 + 70 + 60 1000

30 + 14 + 18 5007 +0 +8 200

4 +9 +0 200As none of the commodities produced can be negative, 0, 0, 0.


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All these inequalities are known as constraints or restrictions.

2.

(Diet problem) A patient needs daily 5mg, 20mg and 15mg of vitamins A, Band C respectively. The vitamins available from a mango are 0.5mg of A,

1mg of B, 1mg of C, that from an orange is 2mg of B, 3mg of C and that

from an apple is 0.5mg of A, 3mg of B, 1mg of C. If the cost of a mango, an

orange and an apple be Rs 0.50, Rs 0.25 and Rs 0.40 respectively, find the

minimum cost of buying the fruits so that the daily requirement of the

patient be met. Formulate the problem mathematically.

Solution: The problem is to find the minimum cost of buying the fruits. Let z be

the objective function. Let the number of mangoes, oranges and apples to bebought so that the cost is minimum and to get the minimum daily requirement

of the vitamins be , , respectively. Then the objective function is givenby

= 0.50 +0.25 +0.40 From the conditions of the problem

0.5 + 0 +0.5 5

+ 2 +3 20 +3 + 15 and

0, 0, 0Hence the problem is

Minimize = 0.50 +0.25 +0.40 .Subject to

0.5 + 0 +0.5 5

+ 2 +3 20 + 3 + 15

and 0, 0, 03. (Transportation problem) Three different types of vehicles A, B and C have

been used to transport 60 tons of solid and 35 tons of liquid substance. Type


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A vehicle can carry 7 tons solid and 3 tons liquid whereas B and C can carry

6 tons solid and 2 tons liquid and 3 tons solid and 4 tons liquid respectively.

The cost of transporting are Rs 500, Rs 400 and Rs 450 respectively pervehicle of type A, B and C respectively. Find the minimum cost of

transportation. Formulate the problem mathematically.

Solution: Let z be the objective function. Let the number of vehicles of type

A, B and C used to transport the materials so that the cost is minimum be

, , respectively. Then the objective function is = 500 +400 +450 . The quantities of solid and liquid transported by the vehicles are

7

+ 6

+ 3 tons and

3

+ 2

+ 4 tons respectively.

By the conditions of the problem, 7 + 6 + 3 60 and 3 + 2 +4 35 . Hence the problem isMinimize = 500 +400 +450 Subject to 7 + 6 + 3 603 + 2 + 4 35

And , , 04. An electronic company manufactures two radio models each on a separate

production line. The daily capacity of the first line is 60 radios and that ofthe second line is 75 radios. Each unit of the first model uses 10 pieces of a

certain electronic component, whereas each unit of the second model uses 8

pieces of the same component. The maximum daily availability of the

special component is 800 pieces. The profit per unit of models 1 and 2 are

Rs 500 and Rs 400 respectively. Determine the optimal daily production of

each model.

Solution: This is a maximization problem. Let , be the number of tworadio models each on a separate production line. Therefore the objectivefunction is = 500 +400 which is to be maximized. From theconditions of the problem we have 60 , 75 , 10 + 8 800.Hence the problem is

Maximize = 500 +400 Subject to 60


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75

10 + 8 800

And , 05. An agricultural firm has 180 tons of Nitrogen fertilizers, 50 tons of

Phosphate and 220 tons of Potash. It will be able to sell 3:3:4 mixtures of

these substances at a profit of Rs 15 per ton and 2:4:2 mixtures at a profit of

Rs 12 per ton respectively. Formulate a linear programming problem to

determine how many tons of these two mixtures should be prepares so as to

maximize profit.

Solution: Let the 3:3:4 mixture be called A and 2:4:2 mixture be called B.

Let , tons of these two mixtures be produced to get maximum profit.Thus the objective function is = 15 + 12 which is to bemaximized. Let us denote Nitrogen, Phosphate and Potash as N Ph and P

respectively.

Then in the mixture A , = = = (say).

= 3 , = 3 , = 4

= 10 .Similarly for the mixture B , = 2 , = 4 , = 2 = 8 .Thus the constraints are

+ 180 [since in A, amount of nitrogen

= = ] Similarly + 250 and + 220 . Hence the problem is

Maximize

= 15

+ 12

Subject to + 180 + 250

+ 220And , 0.


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6. A coin to be minted contains 40% silver, 50% copper, 10% nickel. The minthas available alloys A, B, C and D having the following composition and

costs, and availability of alloys:

%

silver

%

copper

%

nickel

Costs per

Kg

A 30 60 10 Rs 11

B 35 35 30 Rs 12

C 50 50 0 Rs 16

D 40 45 15 Rs 14

Availabil

ity ofalloys

Total 1000 Kgs

Present the problem of getting the alloys with specific composition at

minimum cost in the form of a L.P.P.

Solution: Let , , , Kg s be the quantities of alloys A, B, C, D usedfor the purpose. By the given condition + + + 1000 .The objective function is = 11 + 12 + 16 + 14 and the constraints are 0.3 +0.35 +0.5 +0.4 400 forsilver

0.6 +0.35 +0.5 +0.45 500 forcopper

0.1 +0.3 ++0.15 100 fornickel

Thus the L.P.P is Minimize = 11 + 12 + 16 + 14 Subject to 0.3 + 0.35 +0.5 +0.4 400

0.6

+0.35 +0.5 +0.45 500

0.1 +0.3 ++0.15 100 + + + 1000And , , 0

7. A hospital has the following minimum requirement for nurses.Period Clock time

(24 hours

day)

Minimum

number of

nurses


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required

1 6A.M-

10A.M

60

2 10A.M-

2P.M

70

3 2P.M-

6P.M

60

4 6P.M-

10P.M

50

5 10P.M-

2A.M

20

6 2A.M-

6A.M

30

Nurses report to the hospital wards at the beginning of each period and work for

eight consecutive hours. The hospital wants to determine the minimum number of

nurses so that there may be sufficient number of nurses available for each period.

Formulate this as a L.P.P.

Solution: This is a minimization problem. Let

, , , be the number of

nurses required for the period 1, 2, , 6. Then the objective function is

Minimize, = + ++ and the constraints can be written in thefollowing manner.

nurses work for the period 1 and 2 and nurses work for the period 2 and 3etc. Thus for the period 2,

+ 70.Similarly, for the periods 3, 4, 5, 6, 1 we have,

+ 60 + 50 + 20 + 30


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+ 60 , 0, = 1, 2, , 6

Mathematical formulation of a L.P.P.

From the discussion above, now we can mathematically formulate a general Linear

Programming Problem which can be stated as follows.

Find out a set of values , ,, which will optimize (either maximize orminimize) the linear function

= + ++ Subject to the restrictions + + + (=) + ++ (=)

..

+ ++ (=)

And the non-negative restrictions 0, = 1, 2, , where , , ( =1,2,, , = 1,2, . , ) are all constants and , ( = 1, 2, , ) arevariables. Each of the linear expressions on the left hand side connected to the

corresponding constants on the right side by only one of the signs , = and ,isknown as a constraint. A constraint is either an equation or an inequation.

The linear function = + ++ is known as the objectivefunction.

By using the matrix and vector notation the problem can be expressed in a compact

form as

Optimize = subject to the restrictions = , 0,where = is a m x n coefficient matrix.,


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= ( , , , ) is a n-component column vector, which is known as a costor price vector,

= ( , , , ) is a n-component column vector, which is known asdecision variable vector or legitimate variable vector and

= ( , , , ) is a m-component column vector, which is known asrequirement vector.

In all practical discussions, 0 . If some of them are negative, we make themby multiplying both sides of the inequality by (-1).

If all the constraints are equalities, then the L.P.P is reduced to

Optimize = subject to = , 0 .This form is called the standard form.

Feasible solution to a L.P.P: A set of values of the variables, which satisfy all the

constraints and all the non-negative restrictions of the variables, is known as the

feasible solution (F.S.) to the L.P.P.

Optimal solution to a L.P.P: A feasible solution to a L.P.P which makes theobjective function optimal is known as the optimal solution to the L.P.P

There are two ways of solving a linear programming problem: (1) Geometrical

method and (2) Algebraic method.

A particular L.P.P is either a minimization or a maximization problem. The

problem of minimization of the objective function is nothing but the problem of

maximization of the function

( )and vice versa and

min = max( )with

the same set of constraints and the same solution set.

Graphical or Geometrical Method of Solving a Linear Programming Problem

We will illustrate the method by giving examples.

Examples

Solve the following problems graphically.


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1. Maximize = 150 + 100 Subject to

8 + 5 60

4 + 5 40 , , 0 .

The constraints are treated as equations along with the non negativity relation. We

confine ourselves to the first quadrant of the xy plane and draw the lines given bythose equations. Then the directions of the inequalities indicate that the striped

region in the graph is the feasible region. For any particular value of z, the graph of

the objective function regarded as an equation is a straight line (called the profit

line in a maximization problem) and as z varies, a family of parallel lines is

generated. We have drawn the line corresponding to z=450. We see that the profit

z is proportional to the perpendicular distance of this straight line from the origin.

Hence the profit increases as this line moves away from the origin. Our aim is to

find a point in the feasible region which will give the maximum value of z. In orderto find that point we move the profit line away from origin keeping it parallel to

itself. By doing this we find that (5,4) is the last point in the feasible region which

the moving line encounters. Hence we get the optimal solution = 1150 for= 5, = 4 .Note: If we have a function to minimize, then the line corresponding to a particular

value of the objective function (called the cost line in a minimization problem) is

moved towards the origin.

0

3

7 10

Z=450Z=1150

4x+5y=40

8x+5y=60

12

8


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2. Solve graphically:Minimize

= 3 + 5

Subject to 2 + 3 12 + 3 4 3

Here the striped area is the feasible region. We have drawn the cost line

corresponding to z=30. As this is a minimization problem the cost line is moved

towards the origin and the cost function takes its minimum at = 19.5 for

= 1.5, = 3.

In both the problems above the L.P.P. has a unique solution.

3. Solve graphically:Minimize = + Subject to 5 + 9 45 + 2 4 , , 0

-3 0 4 6 10

3

4

6

2x+3y=12

-x+y=3

Z=30


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Here the striped area is the feasible region. We have drawn the cost line

corresponding to z=4. As this is a minimization problem the cost line

when moved towards the origin coincides with the boundary line+ = 2 and the optimum value is attained at all points lying on theline segment joining (2,0) and (0,2) including the end points. Hence there

are an infinite number of solutions. In this case we say that alternative

optimal solution exists.

4. Solve graphicallyMaximize = 3 + 4 Subject to 0+ 1 + 3 3 , , 0

x+y=2

5x+9y=45

y=4

z=4

0-3 4

z=12

x-y=0

-x+3y=3

x+y=1


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The striped region in the graph is the feasible region which is unbounded.. For any

particular value of z, the graph of the objective function regarded as an equation is

a straight line (called the profit line in a maximization problem) and as z varies, afamily of parallel lines is generated. We have drawn the line corresponding to

z=12. We see that the profit z is proportional to the perpendicular distance of this

straight line from the origin. Hence the profit increases as this line moves away

from the origin. As we move the profit line away from origin keeping it parallel to

itself we see that there is no finite maximum value of z.

Ex: Keeping everything else unaltered try solving the problem as a minimization

problem.

5. Solve graphicallyMaximize = 2 3 Subject to + 2 + 4, 0

It is clear that there is no feasible region.

In algebraic method, the problem can be solved only when all constraints are

equations. We now show how the constraints can be converted into equations.

Slack and Surplus Variables

When the constraints are inequations connected by the sign , in each

inequation a variable is added on the left hand side of it to convert ind sidet into an

equation. For example, the constraint

0

x+y=4x+y=2


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2 + 7 4is connected by the sign

. Then a variable is added to the left hand side and itis converted into an equation

2 + 7 + = 4From the above it is clear that the slack variables are non-negative quantities.

If the constraints are connected by ,in each inequation a variable is subtracted

from the left hand side to convert it into an equation. These variables are known as

surplus variables. For example,

2 + 7 4is converted into an equation by subtracting a variable from the left hand side.

2 + 7 = 4The surplus variables are also non-negative quantities.

Let a general L.P.P containing r variables and m constraints be

Optimize = + ++ subject to + ++ = , = 1,2, , , 0, =1,2, , ,where one and only one of the signs ,=, holds for each constraint, but the signsmay vary from one constraint to another. Let constraints out of the be

inequations (0 ). Then introducing k slack or surplus variables , , , , = +

, one to each of the inequations, all constraints can

be converted into equations containing n variables. We further assume that .The objective function is similarly accommodated with k slack or surplus variables , , , , the cost components of these variables are assumed to be

zero. Then the adjusted objective function is

= + + + + 0 + 0 ++0 , and then theproblem can be written as

Optimize

= subject to

= , 0,


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where is an matrix , known as coefficient matrix given by

= ( , , , ),where = ( , , , ) is a column vector associated with the vector, = 1,2, , .= ( , , , ,0,0,,0) is a n-component column vector,= ( , , , , , , , ) is a n-component column vector, and

= (

,

, ,

) is a m-component column vector.

The components of can be made positive by proper adjustments.

It is worth noting that the column vectors associated with the slack variables are all

unit vectors. As the cost components of the slack and surplus variables are all zero,

it can be verified easily that the solution set which optimizes also optimizes .Hence to solve the original L.P.P it is sufficient to solve the standard form of the

L.P.P. So, for further discussions we shall use the same notation for and .Problems

1. Transform the following Linear Programming Problems to the standardform:

(i) Maximize = 2 + 3 4 Subject to 4 + 2 43 + 2 + 3 6

+ 3 = 8 , 0, = 1,2,3 .Solution: First constraint is type and the second one is a type, soadding a slack and a surplus variable respectively, the two constraintsare converted into equations. Hence the transformed problem can be

written as

Maximize = 2 + 3 4 + 0 + 0 Subject to 4 + 2 + = 43 + 2 + 3 = 6

+ 3 = 8 , 0, = 1,2,3,4,5 .(ii) Maximize = +


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Subject to + 3 4

2 4 + 5

+ 2 2 3 , 0, = 1,2,3 .Solution: The problem can be transformed as

Maximize = + + 0 + 0 + 0 Subject to + 3 + = 42 4 + = 5

+ 2 2 = 3 ,

0, = 1,2,3,4,5,6.

, are surplus and is a slack variable. Making the secondcomponent of vector positive , the second equation can be written as2 + 4 + = 5andinthatcasethesurplusvariableischangedintoaslackvariable.

2. Express the following minimization problem as a standard maximizationproblem by introducing slack and surplus variables.

Minimize

= 4 + 2

Subject to 4 + 72 3 + 12 + + = 84 + 7 16 , 0, = 1,2,3 .Solution: After introducing slack variables in the first two constraints and a surplus

in the fourth, the converted problem is,

Minimize

= ( ) = 4 + 2 + 0 + 0 + 0

Subject to 4 + + = 72 3 + + = 12 + + = 84 + 7 = 16 , 0, =1,2,,6 .Writingtheaboveproblemasastandardmaximizationproblem


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Maximize = ( ) = 4 + 2 + 0 + 0 + 0 Subject to

4 + + = 7

2 3 + + = 12 + + = 84 + 7 = 16 , 0, =1,2,,6 .Variable unrestricted in sign

If a variable is unrestricted in sign, then it can be expressed as a difference oftwo non-negative variables, say,

/

, //

as

= /

//

,

/

0, //

0. If

/ > //, then > 0 , if / = // , then = 0 and if / < //, then < 0 .Hence is unrestricted in sign.

3. Write down the following L.P.P in the standard form.Maximize = 2 + 3

Subject to 4 + + 47 + 4 25

,

0, = 1,3,

unrestricted

in sign .

Solution: Introducing slack and surplus variables and writing = / //,where / 0, // 0,

the problem in the standard form is

Maximize = 2 + 3 / 3 // + 0 + 0 Subject to 4 + / // + = 47 + 4 / 4 // + = 25 ,. /, //, 0 .


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CHAPTER II

Basic Solutions of a set of Linear Simultaneous Equations

Let us consider linear equations with variables ( > ) and let the set ofequations be

+ ++ = + ++ =

+ ++ = This set of equations can be written in a compact form as

= , where,= is the coefficient matrix of order m x n,= ( , , , ) is a n-component column vector,

= ( , , , ) is a m-component column vector.We further assume that ( ) = , which indicates that all equations are linearlyindependent and none of them are redundant.

The set of equations can also be written in the form

+ + .+ ++ = where = ( , , , ),an m component column vector and all are non-null vectors. These vectors are

called activity vectors. From the theory of linear algebra, we know that here

infinitely many solutions exist. We will now find a particular type of solutions of

the set of equations which are finite in number.

From the set of n column vectors we arbitrarily select m linearly independentvectors (there exists at least one such set of vectors since ( ) = , and > )which constitutes a basis B of the Euclidean space . The vectors which are notincluded in the selected set are called non-basic vectors. Assuming that all

variables associated with the non-basic vectors are zero, we get a set of m


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equations with m variables. The coefficient matrix here is the basis matrix and

hence is non-singular. Hence there exists a unique solution for the set of m

equations containing m variables. This solution is called a Basic Solution. Thevariables associated with the basis vectors are called basic variables. The number

of basic variables are m, and the number of non-basic variables(the ones associated

with the non-basic vectors) are whose values are assumed to be zero. Thenthe set of equations are reduced to

= ,Where is the basis matrix and

is the m component column vector consisting

of the basic variables. Using the matrix inversion method of finding the solution ofa set of equations

( ) = , or, = = , where is the m-componentcolumn vector written as = ( , ,...... , ).The general solution is written as = , 0 ,where 0 is a ( )component null vector.

Since out of n vectors, m vectors constitute a basis, then theoretically the

maximum number of basis matrices are nCm and hence the maximum number of

basic solutions arenCm . Hence the basic solutions are finite in number. We now

formally define a basic solution.

Basic Solution: Let us consider a system of simultaneous linear equations

containing variables ( > ) and write the set of equations as = , where( ) = . If any arbitrary non-singular sub-matrix(say ), be selected from, and we assume all ( ) variables not associated with the column vectors ofare zero, then the solution so obtained is called a basic solution. The variables

associated with the columns of the non-singular matrix are called basic variables

and the remaining variables whose values are assumed to be zero, are callednon-basic variables. The values of each of the basic variables can be positive,

negative or zero. From this we can conclude that a solution is said to be a basic

solution if the vectors associated with the non-zero vectors are linearlyindependent. This condition is necessary and sufficient.


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Non-Degenerate Basic Solution: If the values of all the basic variables are non-

zero then the basic solution is known as a Non-Degenerate Basic Solution.

Degenerate Basic Solution: If the value of at least one basic variable is zero then

the basic solution is known as a Non-Degenerate Basic Solution.

Basic Feasible Solution (B.F.S): The solution set of a L.P.P. which is feasible as

well as basic is known as a Basic Feasible Solution.

Non-degenerate B.F.S: The solution to a L.P.P. where all the components

corresponding to the basic variables are positive is called a Non-degenerate B.F.S.

Degenerate B.F.S: The solution to a L.P.P. where the value of at least one basicvariable is zero is called a Degenerate B.F.S.

Examples

1. Find the basic solutions of the system of equations given below and identifythe nature of the solution.2 + 4 2 = 10

10

+ 3

+ 7

= 33

2. Given that = 2, = 1, = 0 is a solution of a system of equations3 2 + = 89 6 + 4 = 24

Is this solution basic? Justify.


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CHAPTER III

N-Dimensional Euclidean Space and Convex Set

We will denote the N-Dimensional Euclidean Space by or or . The pointsin are all column vectors.Point Set: Point sets are sets whose elements are all points in .Line: If = ( , , , ) and = ( , , , ) be two pointsin , then the line joining the points and , ( ) is a set of pointsgiven by

= { : = + (1 ) , }Line segment: If = ( , , , ) and = ( , , , ) betwo points in , then the line segment joining the points and , ( ) isa set X of points given by

= { : = + (1 ) , 0 1, }Hyperplane: A hyperplane in is a set of points given by

= { : = } ,Where = ( , , , ), not all = 0 , is a fixed element of and= ( , , , ) is an element of .A hyperplane can be defined as a set of points which will satisfy + ++ = .A hyperplane divides the space into three mutually exclusive disjoint sets givenby = ( , , , ) = { : > } , = { : = } , = { : < } .

The sets and are called open half spaces.In a L.P.P. , the objective function and the constraints with equality sign are all

hyperplanes.


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Hypersphere: A hypersphere in with centre at = ( , , , ) andradius

> 0is defined to be the set of points given by

= { : | | = },

where = ( , , , ) .The equation can be written as

( ) + ( ) ++ ( ) = The hypersphere in two dimensions is a circle and in three dimensions is a sphere.

An - neighbourhood: An - neighbourhood about a point is defined as the set

of points lying inside the hypersphere with centre at and radius

> 0, i.e. , the -

neighbourhood about the point is a set of points given by = { : | | < }.An interior point of a set: A point is an interior point of the set if there exists an

- neighbourhood about the point which contains only points of the same set.

From the definition it is clear that an interior point of a set is an element of the

set .

Boundary point of a set: A point is a boundary point of a set if every -

neighbourhood about the point (

> 0)contains points which are in the set and

points which are not in the set . A boundary point may or may not be an elementof the set .

Open set: A set is said to be open if it contains only interior points.

Closed set: A set is said to be closed if it contains all its boundary points.

Bounded set: A set is said to be strictly bounded set if there exists a positive

number such that for any point belonging to , | | . For everybelonging to , if , then the set is bounded from below.Convex Combination and Convex SetsConvex Combination of a set of points: The convex combination of a set of

points , , , in a space is also a point in the same space, given by= + ++ where 0 and for all and = 1 .


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For different values of the scalar quantities , = 1,2, , satisfying =1

, and

0for all , a set of points will be obtained from the convex

combinations of the set of finite points which is a point set in . This point setis known as a convex polyhedron.

The point set , called the convex polyhedron is given by

= { : = , = 1 and 0 for all }.From the above definition it is also clear that a line segment is a convex

combination of two distinct points in the same vector space.

Convex Set: A point set is said to be a convex set if the convex combination of any

two points of the set is in the set. In other words, if the line segment joining any

two distinct points of the set is in the set then the set is known as a convex set.

Extreme points of a convex set: A point is an extreme point of the convex set if

it cannot be expressed as a convex combination of two other distinct points , of the set , i.e, cannot be expressed as

= + (1 ) , 0 < < 1.

From the definition, it is clear that all extreme points of a convex set are boundary

points but all boundary points are not necessarily extreme points. Every point of

the boundary of a circle is an extreme point of the convex set which includes the

boundary and interior of the circle. The extreme points of a square are its four

vertices.

Convex hull: If be a point set, then Convex hull of which is denoted by ( ),is the set of all convex combinations of set of points from . If the set consists of

a finite number of points then the convex hull ( ) is called a convex polyhedron.For a convex polyhedron, any point in the set can be expressed as a convex

combination of its extreme points.

Simplex: A simplex is an -dimensional convex polyhedron having exactly + 1vertices.

Theorem 1: Intersection of two convex sets is also a convex set.


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Proof: Let , be two convex sets and let = . It is required to provethat is also a convex set.

Let , be two distinct points of . Then , and , . Let bea point given by

= + (1 ) , 0 1 .As is a convex combination of the points , and , are convex sets, then is a point of as well as . Hence is a point of= . Hence is a

convex set.

Note1: Intersection of a finite number of convex sets is a convex set.

Note 2: Union of two convex sets may not be a convex set.

Theorem 2: A hyperplane is a convex set.

Proof: Let the point set be a hyperplane given by = { : = } . Let , be two distinct points of . Then = and = . Let be a pointgiven by = + (1 ) , 0 1 .Therefore, = + (1 ) = + (1 ) = which indicatesthat is also a point of = .But is a convex combination of two distinct points and of . Hence is aconvex set.

Note: Set is also a closed set.

Theorem 3: Convex polyhedron is a convex set.

Proof: Let be a point set consisting of a finite number of points , , . , in .We have to show that the convex polyhedron ( ) = = : = , 0, = 1 }.Let , be any two distinct points of given by

=

,

0,

= 1


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= , 0, = 1 Consider = + (1 ) , 0 1.Then

= + (1 ) = { + (1 ) } = where = + (1 ) .Now, = + (1 ) =1 and 0 as 0, 0 and0 1 .Hence is also a point of which is a convex combination of two distinct pointsof . Hence is a convex set.Theorem 4: The set of all feasible solutions to a L.P.P = , 0 is a closedconvex set.

Proof: Let be the point set of all feasible solutions of = , 0.If the set has only one point, then there is nothing to prove.

If has at least two distinct points and , then = , 0 and = , 0.

Consider a point such that = + (1 ) , 0 1 .Thus = + (1 ) = + (1 ) = .Again 0 as 0 and 0 and 0 1.Then

is also a feasible solution to the problem

= , 0.

But is a convex combination of two distinct points and of the set . Thusis a convex set.

Now the finite number of constraints represented by = are closed sets andalso the set of inequations (finite) represented by 0 are closed sets andtherefore the intersection of a finite number of closed sets which is the set of all

feasible solutions is a closed set.


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Note: If the L.P.P has at least two feasible solutions then it has an infinite number

of feasible solutions

Theorem 5; All B.F.S of the set of equations = , 0 are extreme points ofthe convex set of feasible solutions of the equations and conversely.

Proof: Let = ( , , , ) be the coefficient matrix of order , > and let us assume that be the basis matrix = ( , , , ) where, , , are the column vectors corresponding to the first variables, , , .

Let be a B.F.S and is given by

= , 0, where

= and

0is the

( ) component null vector.We have to show that is an extreme point of the convex set of feasible

solutions of the equation = , 0.Let be not an extreme point of the convex set . Then there exist two points/, //, / // in such that it is possible to express as

=/

+ (1 )//

, 0 < < 1, where /

,// are given by

/ = , , // = , where contains components of /,corresponding to the variables , , , and contains the remaining( ) components of /. Similarly and contains the first and theremaining ( ) components of //respectively.Thus, = , + (1 ) , = + (1 ) , + (1 ) .As = , 0 , then equating the components corresponding to the last ( )variables, we get + (1 ) = 0 which is possible only when = 0 and = 0 [as 0 , 0 and 0 < < 1].Thus / = , 0, // = , 0 . Hence and are the m components of thesolution set corresponding to the basic variables , , , for which the basismatrix is . Then = and = . Hence, = = . So the threepoints , /, // are not different and therefore cannot be expressed as a convexcombination of two distinct points. So a B.F.S is an extreme point.


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Conversely, let us assume that is an extreme point of the convex set of feasible

solutions of the equation

= , 0.

We have to show that is a B.F.S.

Let = , , , ,0,,0 , number of zero components are , 0 for= 1,2, , .If the column vectors , , , associated with the variables , , , respectively are L.I (which is possible only for ) then , the extreme point ofthe convex set, is a B.F.S and we have nothing to prove.

If , , , are not L.I then = and = 0 with at leastone 0 .Let > 0 , then from the above two equations we get ( ) = .Consider such that 0 < < , where = min , 0 .Then

0for

= 1,2, ,.

Hence the two points

/= 1+ 1, 2+ 2, , + ,0, ,0 and//

= , , , ,0,,0are points of the convex set .

Now,

/ +

// = , so can be expressed as = / + (1 ) // where= 1/2 .

Thus is being expressed as a convex combination of two distinct points of

which contradicts the assumption that is an extreme point. So the column vectors

, , , are L.I, and hence is a B.F.S.Note: There is a one to one correspondence between the extreme points and B.F.S

in case of non-degenerate B.F.S.


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Eamples

1.

In , prove that the set = {( , )| + 2 5} is a convex set.Solution: The set is non empty. Let ( , ) and ( , ) be two points of theset. Then + 2 5 and + 2 5 .The convex combination of the two points is a point given by + (1 ) , + (1 ) ,.Now + (1 ) + 2 + (1 ) = ( + 2 ) + (1 )( + 2 ) 5 + 5(1 ) = 5So the convex combination of the two points is 0 1 a point of the set.Thus the set is a convex set.

2. Prove that the set defined by = { : | | 2} is a convex set.Solution: The set is non empty. Let and be two points of the set . Then| | 2 and | | 2 .The convex combination of the two points is a point = + (1 ) , 0 1 .Now | + (1 ) | | | + |(1 ) | | || | + |(1 )|| | 2 + 2(1 ) 2 .Hence

. So the set is a convex set.

3. Prove that in , the set = {( , )| + 4} is a convex set.Solution: Let ( , ) and ( , ) be two points of the set .Then + 4,and + 4.

The convex combination of the two points is a point given by

+ (1 ) , + (1 ) ,.Now + (1 ) + + (1 )

= ( + ) + (1 )( + ) + 2 (1 )( + )

( + ) + (1 )( + )+ () 4 + 4(1 ) + 8 (1 ),since + + + 8= 4

Therefore the point . Hence the set is a convex set.


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CHAPTER IV

Fundamental Properties of Simplex MethodReduction of a F.S. to a B.F.S

Theorem: if a linear programming problem = , 0, where is thecoefficient matrix ( > ), ( ) = has one feasible solution, then it has atleast one basic feasible solution.

Proof: Let = ( , , , ) be a feasible solution to the set of equations

= , 0. Out of components of the feasible solution, let components be

positive and the remaining components be zero (1 ) and we alsomake an assumption that the first components are positive and the last components are zero.

Then = ( , , , ,0,,0), number of zeroes being .If , , , be the column vectors corresponding to the variables, , , , then + ++ = , = . (1)We will consider three cases

(i) and the column vectors , , , are linearly independent (L.I)(ii) > (iii) and the column vectors , , , are linearly dependent (L.D)Case(i)If

and the column vectors

, , , are L.I , then by definition

the F.S. is a B.F.S. If= , the solution is a non degenerate B.F.S and if < ,the solution is a degenerate B.F.S.Case(ii)If > and the columns , , , are L.D, the solution is not basic.By applying a technique given below, the number of positive components in the

solution can be reduced one by one till the corresponding column vectors are L.I.

( This will be possible as a set of one non-null vector is L.I.)


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Procedure: As the column vectors , , , are L.D, there exist scalars , = 1,2, ,

, not all zero, such that

+ ++ = 0 , = 0 .. (2)Now at least one is positive (if not, multiply both sides of equation (1) by 1 ).Let = max , = 1,2, , ,As all > 0 and > 0 , then is essentially a positive quantity.Multiplying equation (2) by

1/and subtracting from equation (1) we get

( ) = .. (3)which indicates that

/ = , , , ,0 ,,0is a solution set of the equations = .Now . That implies or, 0, = 0 for at least one j.Then / = 0, = 1,2, , , at least one of a them is equal to zero.Therefore / = ( /, /, , / ,0 ,,0) is also a feasible solution of = with maximum number of positive variables 1 . By applying this methodrepeatedly we ultimately get a basic feasible solution.

(iii)In this case, as the vectors are L.D, we use the above procedure to get a B.F.S.

We state another theorem without proof.

Theorem (statement only) The necessary and sufficient condition that all basic

solutions will exist and will be non-degenerate is that , every set of column

vectors of the augmented matrix is linearly independent.


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Problems

1.

= 1, = 3, = 2 is a feasible solution of the equations2 + 4 2 = 10,10 + 3 + 7 = 33Reduce the above F.S to a B.F.S.

Solution: The given equations can be written as = where= = 2 4 210 3 7 and ( ) = 2 . Hence the two equations

are L.I. , but , , are L.D. Hence there exist three constants , , ,(not all zero) such that + + = 0 ,or,

210 + 43 + 27 = 0which gives

2 + 4 2 = 0 and 10 + 3 + 7 = 0By cross multiplication,

= = = = (say).Then we get = 1, = 1, = 1 .Hence + + = 0.Therefore = max , > 0 = , = .Hence a feasible solution is given by

/ = , , = 1 + 2, 3 2, 2 2 = 3,1,0 which is aB.F.S.

2. Given (1,1,2) is a feasible solution of the equations + 2 + 3 = 9,2 + = 3Reduce the above F.S to one or more B.F.S.

Solution: The given equations can be written as = where= = 1 2 32 1 1 and ( ) = 2 . The equations = can be

written as + + = . As (1,1,2) is a solution of = , we have


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+ + 2 = . . (1)Hence the two equations are L.I. , but

, , are L.D. So there exist three

constants , , , (not all zero) such that + + = 0 ,or, 12 + 21 + 31 = 0 which gives + 2 + 3 = 0 and 2 + = 0

By cross multiplication,

= = = = (say).Then we get = 1, = 1, = 1 .Hence + = 0. . (2)Therefore = max , > 0 = , = 1 which occurs at = 1,2.Thus we shall have to eliminate either or from the set of vectors , , to get a basis and hence a basic solution. Subtracting (2) from (1) we get,

0 + 0 + 3 = which shows that (0,0,3) is a feasible solution and as, and , are L.I , the solution is a B.F.S.Again taking

= = = / = , we get + = 0 which givesanother B.F.S. as (3,3,0).

Fundamental Theorem of Linear Programming:

Statement: If a L.P.P. , optimize = subject to = , 0 , where isthe coefficient matrix ( > ) , ( ) = , has an optimal solution thenthere exists at least one B.F.S. for which the objective function will be optimal.Proof: It is sufficient to consider a maximization problem as a minimization

problem can be converted into a maximization problem.

Let = , , , be an optimal feasible solution to the problem whichmakes the objective function maximum. Out of , , , let k components(1 ) are positive and the remaining ( ) components are zero. We


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further make an assumption that the first k components are positive. Thus the

optimal solution is

= , , , ,0,,0 ,

( )zero components, and

= = .If , , , be the column vectors associated with the variables , , , then the optimal solution will be a B.F.S provided the vectors , ,, areL.I. This is possible only if .We know = , 0, = 1,2, , . (1)Let us assume that

, , , are L.D.

Then = 0 with at least one > 0. . (2)Taking = max which is a positive quantity, and the solution set/ = /, /, , / ,0,,0 where / = 0, = 1,2, , which

contains maximum 1 positive components. Let / = 0, then/

= /

, /

, , /

,0,,0,

( + 1)zero components. The value of the

objective function for this solution set is

= / = / (as / = 0)= = = .

If =0 (which will be proved at the end) , then = and the solution set/ is also an optimal solution. If the column vectors corresponding to/, /, , / are L.I. then the optimal solution / is a B.F.S. If the column

vectors are L.I then repeating the above procedure at most a finite number of times

we will finally get a B.F.S (as a single non-null vector is L.I.) which is also an

optimal solution.


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To prove that = 0, if possible let 0.Then

> 0 < 0

. We multiply

by a quantity

( 0), such that

> 0.Hence + > , or, ( + ) > . . (3)Multiplying equation (2) by and adding to (1) we get ( + ) = ,which shows that + , = 1,2, , is a solution set of the system = .Value of is given by the relation

max , > 0 min , < 0 .As for > 0, + > 0 gives and for < 0, + > 0 gives .Hence for particular values of it is always possible to get + 0 for all .So the solution set + , = 1,2, , is a feasible solution of the system

= . From (3) it is clear that this solution set gives the value of the objectivefunction greater than which contradicts the fact that is the maximum value ofthe objective function.

Hence = 0 .


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CHAPTER V

Simplex MethodAfter introduction of the slack and surplus variables and by proper adjustment of z,

let us consider the L.P.P. as

Maximize = subject to = , 0,where is the coefficient matrix given by

= ( , , , ),where = ( , , , ) is a column vector associated with the vector, = 1,2, , .= ( , , , ,0,0,,0) is a n-component column vector,= ( , , , , , , , ) is a n-component column vector,

where , , , are either slack or surplus variables and

= (

,

, ,

) is a -component column vector.

We make two assumptions: components of are non negative by proper

adjustments and < (this assumption is non restrictive).As none of the converted equations are redundant then there exists at least one

set of column vectors, say, , , , of the coefficient matrix which arelinearly independent. Then one basis matrix which is a submatrix of is givenby = .Let

, , , be the variables associated with the basic vectors

, , , respectively. Then the basic variable vector is = .The solution set corresponding to the basic variables is = .We assume that 0 , i.e. the solution is a B.F.S.


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Let , , , be the coefficients of , , , respectively in theobjective function

= , then

= is an m component

column vector known as the associated cost vector.

Now a value is defined as = + ++ = . is the value of the objective function corresponding to the B.F.S, where the

basis matrix is .

Now, , , , are L.I and so is a basis of . Therefore all the vectors canbe expressed as a linear combination of , , , .Let = + ++ = where = ,Therefore = .Net evaluation: Evaluation is defined as which is usually denoted by . So is given by = = = + ++ and is called the net evaluation.

If the coefficient matrix contains unit column vectors which are L.I, then

this set of vectors constitute a basis matrix. Let , , , , , beindependent unit vectors of the coefficient matrix, all of which may not be placedin the ascending order of ( = 1,2, , ). For example, , , may occur atthe 5

th,7

th, 3

rdcolumn of respectively. But the basis matrix is the identity

matrix. Hence the components of the solution set corresponding to the basic

variables are = , = 1,2, , and = = , that is the vectors are nothing but the column vectors due to this transformation.Note: In the simplex method all equations are adjusted so that the basis matrix is

the identity matrix and 0 for all .Optimality test: For a maximization problem, if at any stage, 0 for allthen the current solution is optimal. If < 0 for at least one and for this atleast one > 0 , then the value of the objective function can be improved further.If any < 0 and 0 for all then the problem has no finite optimal valueand the problem is said to have an unbounded solution.


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Selection of a vector to enter the next basis and a vector to leave the previous

basis : If

< 0for at least one and for this at least one

> 0, then we

shall have to select a new basis. Thus one new vector is to be selected from (which is not in the previous basis) to replace a vector in the previous basis to form

a new basis.

If = min{ , < 0} , then is the vector to enter in the newbasis and the k

thcolumn of the simplex table is called the key column or the pivot

column. If the minimum occurs for more than one value of then the selection is

arbitrary.

Let be the vector to enter in the new basis.If min{ , > 0} = , then the vector in the rth position of the current basiswill be replaced by . The rth row of the table is called the key row and iscalled the key element. If the value of r is not unique, then again the choice is

arbitrary.

We will illustrate the simplex method in details through examples.Before going

into the examples, we state and prove one more theorem.

Theorem: Minimum value of is the negative of the maximum value of( ) withthe same solution set. In other words, = ( ) with the same solutionset.

Proof: Let = attain its minimum at = then = .Hence or, .Therefore,max( ) = or, = max( ) or, = ( ) ,with the same solution set. Similarly, = ( ).


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Examples

1.

Solve the L.P.P.Maximize = 5 + 2 + 2 Subject to + 2 2 30

+ 3 + 36 , , , 0 .Solution: This is a maximization problem , 0, = 1,2 and the constraints areboth " " type. So introducing two slack variables , , one to each constraint,we get the following converted equations.

+ 2 2 + = 30

+ 3 + + = 36 , , , , , 0 .The adjusted objective function is

= 5 + 2 + 2 + 0 + 0 . In notations, the new problem isMax = subject to = , 0, where = ,

= 11 ,

= 23 ,

= 21 ,

= 10 ,

= 01 , =

3036,

= = = = 3036 , = = = 00 = 0, = = + = 0, = = = that is = .

With the above information we now proceed to construct the initial simplex table.

Initial simplex table

c 5 2 2 0 0

Basis b ( ) ( ) Min ratio= , > 0 0 30 1 2 -2 1 0 = 30 0 36 1 3 1 0 1 361 = 36 = 0 5 -2 -2 0 0


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Rule of construction of the second table: The new basis is and and thereforethey must be the columns of the identity matrix. We make the necessary row

operations as follows:

is the key row, is the key column, is the key element./ = new key row = / = /

The same notations will be used in all the tables but the entries will keep changing.

Second simplex table (1st iteration)

c 5 2 2 0 0

Basis b Min ratio = , > 0 5 30 1 2 -2 1 0

0 6 0 1

3 -1 1

= 2

= 0 8 12 5 0= 150

is the key row, is the key column, is the key element./ = new key row =

/ = /


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Third simplex table (2nd iteration)

C 5 2 2 0 0Basis B

5 34 1 8/3 0 1/3 2/3 2 2 0 1/3 1 -1/3 1/3

0 12

0 1 4

Z=174

Here 0 for all . Hence the solution is optimal. = 174 for = 34, = 0, = 2 .

2. Solve the L.P.P. by simplex methodMaximize = 4 + 7 Subject to

2 + 1000

+ 600 2 1000 , , 0 .This is a maximization problem. Multiplying the third constraint by (-1) we

get + 2 1000 . Hence all 0 and all constraints are type.Introducing three slack variables , , , one to each constraint we get thefollowing converted equations2 + + = 1000

+ ++ = 600

+ 2 ++ = 1000 , , , , , 0 .The adjusted objective function is = 4 + 7 + 0 + 0 + 0 .In notations, the new problem is

Max = subject to = , 0, where = , = 211 , =

112, = 100, =

010 , = 001 , =

10006001000,


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=

=

=

=

1000

6001000, =

=

=

0

00,

= = + + = 0, = = = thatis = . With the above information we now proceed to construct theinitial simplex table.

C 4 7 0 0 0

Basis b Min ratio= , > 0

0 1000 2 1 1 0 0

=1000 0 600 1 1 0 1 0 6001 = 600 0 1000 1 2 0 0 1 10002 = 500 = 0 -4 7 0 0 0

is the key element. / = new key row = , / = / , = 1,2. 0 500 3/2 0 1 0 -1/2 /=1000/3 0 100 12

1 0 1 -1/2 1001/2 = 200 7 500 1/2 0 0 0 1/2 / =1000 3500 12

0 0 0 7/2

is the key element. / = new key row = , / = / , = 1,3. 0 200 0 0 1 -3 1 4 200 1 0 0 2 -1 7 400 0 1 0 -1 1 3600 0 0 0 1 3

Here 0 for all . Hence the solution is optimal.

= 3600for

= 200,

= 400.


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3. Solve the L.P.P.Minimize

= 2 + 3

Subject to 2 5 74 + 87 + 2 16, , 0 .Solution: This is a minimization problem. Let / = .Then min =max( ) = /. We solve the problem for / and the requiredmin = /. Introducing two slack variables the converted equationsare 2 5 + = 7

4 + + = 8

7 + 2 + = 16, , , , , 0 .The adjusted objective function is/ = 2 + 3 + 0 + 0 + 0 . The initial basis = ( , , ) = and we start the simplex table and solve the problem. We solve the

problem in a compact manner as shown below.

c -2 3 0 0 0

Basis b Min ratio= , > 0 0 7 2 -5 1 0 0 72 0 8 4 1 0 1 0 72 0 16 7 2 0 0 1 167 / = 0 2 3 0 0 0

is the key element. / = new key row = , / = / , = 1,2. 0 3 0 -11/2 1 -1/2 0 2 2 1 0 0 0 2 0 0 -7/4 1 / =4 0 7/2 0 0


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Here 0 for allj. Therefore the solution is optimal. Hence / = 4 . Now

min = /

= 4. Hence the minimum value of is

4corresponding to

the optimal basic feasible solution. = = 322, i.e., for =2, = 0, the objective function of the original problem attains its minimum value.This solution is a degenerate B.F.S.

4. Use simplex method to solve the L.P.PMaximize = 2 +

Subject to + 2 73 + + 2 3 , , , 0 .Solution: Adding two slack variables , , one to each constraint, theconverted equations are + 2 + = 73 + + 2 + = 3, 0 for = 1, ,5.

The adjusted objective function is = 0 + 2 + + 0 + 0 .= 73 0, = = is the initial unit basis matrix.

C 0 2 1 0 0

Basis B ( ) ( ) Min ratio= , > 0 0 7 1 1 -2 1 0 71 = 7 0 3 -3 1 2 0 1 31 = 3 = 0 0 2 -1 0 0

is the key element.

/

=new key row

=

,

/

= /

.

0 4 4 0 -4 1 -1 44 = 1 2 3 -3 1 2 0 1 ---

= 6 6 0 3 0 2 is the key element. / = new key row = , / = / .

0 1 1 0 -1 -1/4 2 6 0 1 -1 1/4

= 12

0 0 -3 3/2 1/2


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In the third table < 0 for = 3 . But < 0 for all . Hence the problem hasunbounded solution.5. Use simplex method to solve the L.P.P

Maximize = 5 + 2 Subject to 6 + 10 3010 + 4 20 , , 0 .

Show that the solution is not unique. Write down a general form of all the optimal

solutions.

Solution: Adding two slack variables , , one to each constraint, the convertedequations are 6 + 10 + = 30

10 + 4 + = 20, 0 for = 1,2,3,4.The adjusted objective function is = 5 + 2 + 0 + 0 .= 3020 0, = = is the initial unit basis matrix.

Simplextables

C 5 2 0 0

Basis B ( ) Min ratio= , > 0 0 30 6 10 1 0 306 = 5 0 20 10 4 0 1 2010 = 2 = 0 5 -2 0 0 is the key element. / = new key row = , / = / 0 18 0 385

1 -3/5 1838/5 = 4519

5 2 1 2/5 0 1/10 22/5 = 5 = 10 0 0 0


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is the key element. / = new key row = , / = / 2 45/19 0 1 5/38 -3/38 5 20/19 1 0 -2/19 5/38

= 10 0 0 0 In the second table, 0 for allj. Therefore the solution is optimal and = 10 at = 2, = 0. But = 0 corresponding to a non-basicvector . Thus the solution is not unique. Using to enter in the next basis, thethird table gives the same value of z but for = 20/19, = 45/19. We knowthat if there exists more than one optimal solution, then there exist an infinite

number of optimal solutions, given bythe convex combination of the optimal

solutions / = 20 and // = 20/1945/19. Hence all the optimal solutions are givenby / + (1 ) //,0 1,

= 20 + (1 ) 20/1945/19These solutions are called alternative optima.

Artificial variables

To solve a problem by simplex method, we rewrite all the constraints as equations

by introducing slack/surplus variables. We consider the following example.

Maximize = 2 + 3 4 Subject to

4 + 2 4

3 + 2 + 3 6 + 3 = 8 , 0, = 1,2,3 .

First constraint is type and the second one is a type, so adding a slack and asurplus variable respectively, the two constraints are converted into equations.

Hence the transformed problem can be written as


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Maximize = 2 + 3 4 + 0 + 0 Subject to

4 + 2 + = 4

3 + 2 + 3 = 6 + 3 = 8 , 0, = 1,2,3,4,5 .To get the initial B.F.S for using the simplex method, we require an identity matrix

as a sub-matrix of the coefficient matrix. To get that, we need to introduce some

more variables which will be called the artificial variables. Even if a constraint is

given as an equation, we still add an artificial variable (A.V) to get an initial B.F.S.

So after introducing artificial variables the above problem is written as4 +2 + = 4

3 + 2 + 3 + = 6 + 3 + = 8 , 0, = 1, ,7 .

Then the basis matrix is = .In an attempt to solve a problem involving artificial variables by using simplex

method, the following three cases may arise.

1. No artificial variables are present in the basis at the optimal stage indicatesthat all A.V.s are at the zero level and hence the solution obtained is optimal.

2. At the optimal stage, some artificial variables are present in the basis at thepositive level indicates that there does not exist a F.S to the problem.

3. At the optimal stage, some artificial variables are present in the basis but atzero level indicates that some constraints are redundant.

Problems involving artificial variables can be solved in two ways:

(a)Charnes method of penalties or Big M-Method(b)Two-phase Method

Charnes method of penalties or Big M-Method

In this method, after rewriting the constraints by introducing slack, surplus and

artificial variables, we adjust the objective function by assigning a large negative

cost, say to each artificial variable. In the example given above, the objectivefunction becomes Maximize

= 2 + 3 4 + 0 + 0 .


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We then solve the problem using simplex method as explained earlier, the only

point to remember is that once an artificial variable leaves a basis we drop the

column corresponding to the vector associated with that A.V.

6. Solve the L.P.PMaximize = 2 3 Subject to + 25 + 4 467 + 2 32 , 0, = 1,2 .Solution: In the first constraint = 2 < 0, so making it positive, 2

. Introducing slack and surplus variables the converted equations are

+ = 25 + 4 + = 467 + 2 = 32 , 0, = 1, ,5 .The coefficient matrixdoes not contain a unit basis matrix, so we introduce an A.V in the third

constraint and the set of equations are + = 25 + 4 + = 467 + 2 + = 32

,

0, = 1, ,6

The adjusted objective function is = 2 3 + 0 + 0 + 0 assigning very large negative price to the A.V.c 2 -3 0 0 0 -M

Basis b Min ratio= , > 0 0 2 1 -1 1 0 0 0 2

1= 2

0 46 5 4 0 1 0 0 465 = 915 -M 32 7 2 0 0 -1 1 327 = 447 7 2 -2M+3 0 0 M 0 is the key element. / = new key row = , / = / , = 2,3


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Table 2

c 2 -3 0 0 0 -M

Basis b Min ratio= , > 0 2 2 1 -1 1 0 0 0 0 36 0 9 -5 1 0 0 369 = 4

-M 18

0

9-7 0 -1 1

189 = 2 0 9M+ 1 7M+2 0 M 0 is the key element. / = new key row = , / = / , = 1,2

Table 3

c 2 -3 0 0 0

Basis b

2 4 1 0 2/9 0 -1/9 0 18 0 0 2 1 1

-3 2 0 1 -7/9 0 -1/9 z=2 0 0 25/9 0 1/9

As 0 for allj, optimality condition is reached. The artificial vector isnot present in the final basis. Therefore the A.V is zero at the final stage. Hencethe optimal solution obtained is a B.F.S and the maximum value of z is 2 for

= 4, = 2 .7. Solve the L.P.P


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Maximize = + 2 Subject to

5 10

2 2 + = 10 , 0, = 1,2 .Soln: Introducing slack, surplus and artificial variables, the converted problem is

Maximize = 2 3 + 0 + 0 subject to

5 + = 10

2 + = 2 + + = 10 , 0, = 1, ,6 is a slack variable, is a surplus variable and , are artificial variables. We

now construct the simplex tables and solve the problem.

c 2 -3 0 0 -M -M

Basis b Min ratio= , > 0 0 10 1 -5 1 0 0 0 =10 -M 2 2 -1 0 -1 1 0 22 = 1 -M 10 1 1 0 0 0 1 =10 3 1 -2 0 M 0 0

is the key element. / = new key row = , / = / , = 1,3 0 9 0 -9/2 1 1/2 0 ---

1 1 1 -1/2 0 -1/2 0 ---


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-M 9 0 3

2

0 1/2 1 9

3/2= 6

0 32 52

0 0

is the key element. / = new key row = , / = / , = 1,2 0 36 0 0 1 2

1 4 1 0 0 -1/3 2 6 0 1 0 1/3

16 0 0 0 1/3As 0 for all j, optimality condition is reached. The artificial vectors areall driven out from the final basis. Hence the optimal solution obtained is a B.F.S

and the maximum value of z is 16 for = 4, = 6 .8. Solving by Big M method prove that the following L.P.P. has no F.S.

Maximize = 2 + 5 Subject to + 2 + 2 2

+ 3 + 4 = 124 + 3 + 2 24 , 0, = 1,2,3 .Solution: Introducing slack, surplus and artificial variables the converted equationsand the adjusted objective functions are

Maximize = 2 + 5 + 0 + 0 + 2 + 2 + = 2

+ 3 + 4 + = 12


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4 + 3 + 2 + = 24 , 0, = 1, ,7 .We now construct the simplex tables.

c 7 -1 5 0 -M 0 -M

Basis b min = , > 0 0 2 1 -1 1 0 0 0 0 21 = 2 -M 12 5/2 9 -5 1 0 0 0 12

5/2 =245

-M 24 4 9 -7 0 -1 1 1 244 = 6 132 M 2

-6M

+1

-6M

-5

0 0 M 0

is the key element. / = new key row = , / = / , = 2,3

7 2 1 -1 1 0 0 0 0

-M 7 0 1 0 0

-M 16 0 0 1 1

0 7M+5

7M

-1

132 M

+ 2

0 M 0

As 0 for allj, optimality condition is reached. The artificial variables and are present at a positive level in the optimal solution. Hence the problemhas no feasible solution.

Note: We need not complete the table if the optimality condition is reached.

Two phase method


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In this method the problems are to be solved in two phases.

In the first phase, we first rewrite the constraints as equations by introducing slack,surplus and artificial variables. Then we consider a new objective function = , where is an artificial variable. We will first solve the auxiliary problem :Minimize = subject to the constraints.Here three cases may arise.

(i) Min = 0 and all artificial vectors are removed from the basis. Then wego into the second phase by removing the columns corresponding to the

artificial vectors from the table, and then solving the original problem as

usual with the changed values of and .(ii) Min = 0 but some artificial vectors are present at zero level in the

basis. This shows that some constraints may be redundant. Taking the

final simplex table of the auxiliary problem as the initial simplex table of

the original problem (with the changed values of and ), onlyconsidering the cost of the artificial variables which are basic as zero.

(iii) Min > 0 (this can happen only when one or more artificial variableremains in the basis at a positive level) and the optimality conditions are

satisfied in the auxiliary L.P.P. In that case the problem has no feasible

solution.

We illustrate the method through some examples.

9. Solve the L.P.P by two phase method.Minimize = 3 + 5

Subject to + 2 8

3 + 2 12

5 + 6 60 , 0, = 1,2 .Solution: Introducing slack, surplus and artificial variables the convertedequations are

+ 2 + = 83 + 2 + = 125 + 6 + = 60 , 0, = 1, ,7 .


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In the first phase the objective function is Minimize = + or, Maximize

= . Hence the auxiliary L.P.P is

Maximize / = Subject to + 2 + = 8

3 + 2 + = 125 + 6 + = 60, 0, = 1, ,7 .

Therefore the simplex table for the first phase is

Simplex table for the auxiliary L.P.P

0 0 0 0 0 -1 -1Basis b min

-1 8 1 2 -1 0 0 1 0 82 = 4 -1 12 3 2 0 -1 0 0 1 12

2 = 6

0 60 5 6 0 0 1 0 0 606 = 10 -20 4 -4 1 1 0 0 0

is the key element. / = new key row = , / = / , = 2,3 0 4 1/2

1 -1/2 0 0 1/2 0

41/2 = 8

-1 4 2 0 1 -1 0 -1 1 42 = 2 0 36 2 0 3 0 1 -3 0 362 = 18 -4 2 0 -1 1 0 2 0


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is the key element. / = new key row = , / = / , = 1,3 0 3 0 1 -3/4 1/4 0 3/4 -1/4 0 2 1 0 -1/2 0 -1/2 1/2

0 32 0 0 2 1 1 -2 -1 0 0 0 0 0 0 1 1

0 . Hence the optimality conditions are satisfied and Min = / = 0 = 0 . All A.V s are driven out from the basis and Min = 0 . So weproceed to solve the original problem after removing the columns corresponding to

the artificial vectors.

2nd

Phase

The objective function of the original problem is min = 3 + 5 , that is,max / = = 3 5 + 0 + 0 + 0 .Initial simplex table of the original L.P.P.

-3 -5 0 0 0

Basis b

-5 3 0

1 -3/4 0

-3 2 1 0 -1/2 0

0 32 0 0 2 1 1 -21 0 0 9/4 0


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As 0 for allj, optimality condition is reached. Hence the optimal solutionis

= /

= (21) = 21for

= 2, = 3.

10.Solve the L.P.PMaximize = 5 + 11 Subject to 2 + 43 + 4 242 3 6 , 0, = 1,2by two phase method and prove that the problem has no feasible solution.

Soln: Introducing slack, surplus and artificial variables the converted equations are

2 + + = 43 + 4 + = 242 3 + = 6 , 0, = 1, ,7 .In the first phase the objective function is Minimize = + or, Maximize = . Hence the auxiliary L.P.P is

Maximize / = Subject to 2 + + = 4

3 + 4 + = 242 3 + = 6 , 0, = 1, ,7 .

Therefore the simplex table for the first phase is

Simplex table for the auxiliary L.P.P

0 0 0 0 0 -1 -1Basis b min

0 4 2 1 1 0 0 0 0 42 = 2


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-1 24 3 4 0 -1 0 1 0 24

3= 8

-1 6 2 3 0 0 -1 0 1 62 = 3 -30 5 -1 0 1 1 0 0

is the key element. / = new key row = , / = / , = 2,3 2 1 -1/2 0 0 1/2 0 18 2 0 1 -1 0 -1 1 2 2 0 3 0 1 -3 0 -20 0 3/2 5/2 1 1 0 0

Here the optimality condition is reached but artificial variables are present at a

positive level in the final basis. Hence the problem has no feasible solution.

Problems with variables unrestricted in sign

11.Solve the L.P.PMaximize = 2 + 5 Subject to 2 + 12

+ 4 , 0, is unrestricted in sign.

Solution: As is unrestricted in sign, may be written as = / //, where/, // 0 . Hence the problem can be written asMaximize = 2 + 5 / 5 //Subject to 2 + / // 12

+ / // 4 , , /, // 0


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Now we solve the problem as usual and finally replace / // by .


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CHAPTER VI

Duality Theory

Associated with every L.P.P there exists a corresponding L.P.P. The original

problem is called the primal problem and the corresponding problem as the

dual problem.

We will first introduce the concept of duality through an example.

Food ( ) ( ) Requirement 3 2 20 units 4 3 30 unitscost Rs. 7 Rs. 5 Per unit

Let units of and units of be required to get the minimum amountof vitamins. This is a problem of minimization. The L.P.P. is

Minimize = 7 + 5 Subject to 3 + 2 204 + 3 30, , 0 .Let us now consider the corresponding problem.

A dealer sells the above mentioned vitamins and separately. Hisproblem is to fix the cost per unit of and in such a way that the price of and do not exceed the amount mentioned above. His problem is also to

get a maximum amount by selling the vitamins.

Let and be the price per unit of and respectively.Therefore the problem is

Maximize = 20 + 30 Subject to

3 + 4 7

2 + 3 5, , 0 .Now if we take = 3 24 3 , = 2030 , = , = 75,The initial problem can be written as

Minimize = subject to , 0.Now the corresponding problem is

Maximize = subject to , 0.


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The above is an example of primal-dual problem. Generally the initial

problem is called the primal problem and the corresponding problem as the

dual problem.Standard form of primal

A L.P.P is said to be in standard form if

(i) All constraints involve the sign inaproblemofmaximization,or

(ii) All constraints involve the sign inaproblemofminimization.Given a L.P.P, we write it in the standard form as follows

Maximize = + ++ Subject to + ++

+ ++

+ + + , 0 for all .Here the constants , = 1, , and , = 1, , are unrestricted in sign.The corresponding dual problem isMinimize = + ++ Subject to + ++

+ + +

+ ++ , 0 for all , where= is an m component dual variable vector.Putting = ( ), = ( ), = ( ), = ( ) , the aboveprimal and the dual problem can be written as

Maximize = subject to , 0.Now the corresponding problem is


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Minimize = subject to , 0.Theorem: Dual of the dual ts the primal itself.

Proof: Let the primal problem be Max ( ) subject to , 0 ...(1)The dual of (1) is Min ( ) subject to , 0 . (2)(2) is equivalent to Max ( ) subject to ( ) , 0 .(3)where Min ( ) = Max( ) ..(4)The dual of (3) is Min ( ) subject to ( ) , 0 .(5)(5) is equivalent to Max ( ) subject to , 0 which is exactlythe original problem.Hence the theorem.

From this we conclude that if either problem is considered as a primal then

the other will be its dual.

Example: Write down the dual of the problem

Maximize

= 2 3

Subject to 4 10 + 3 3 4 , 0, = 1,2 .Solution: Rewriting the problem with all type constraints we haveMaximize = 2 3 Subject to 4 10 + 3

+ 3 4,

0, = 1,2

which in the standard form is

Max 2 3 subject to 1 41 11 3

1034 , 0, = 1,2 .Therefore the dual of the problem is

Min = 10 3 4 subject to 1 1 1 4 1 3

23,


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0, = 1,2,3Or the dual problem is Min

= 10 + 3 4

Subject to + 24 + + 3 3, 0, = 1,2,3Weak Duality Theorem

Theorem: If be any F.S to the primal Max = subject to , 0, and be any F.S to the dual problem

Min

= subject to

, 0, then

.

Proof: We have for any F.S of the dual, or, ( ) or, . .. (1) be any F.S to the primal. Post multiplying (1) by we have( ) , [since 0]

or, , [since ]or,

(as

is a scalar)

Hence the theorem is proved.

Note: If and be the optimal feasible solutions of the primal and the dualrespectively then min

Theorem: If and be any two feasible solutions of the primal max = subject to , 0, and the corresponding dual, min = subject to , 0

respectively and

= , then

and

are the optimal

feasible solutions of the primal and the dual respectively.

Proof: From the previous theorem, for any two F.S. and of the primal andthe dual [as is a F.S. of the dual]

or, = ,from which we get max( ) = which implies that is an optimal feasiblesolution of the primal.


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In the same way we can prove that min = min = , that is, is anoptimal feasible solution of the dual.

Fundamental Duality Theorem

Theorem: (a) If either the primal, Max = subject to , 0, or thedual, Min = subject to , 0 has a finite optimal solution, thenthe other problem will also have a finite optimal solution. Also the optimal values

of the objective functions in both the problems will be the same, that is

Max

= Min .

Proof: We first assume that the primal has an optimal feasible solution which has

been obtained by simplex method. Let us convert the constraints of the primal in

the following form

+ = , 0, 0,Where is a set of m slack variables and is a unit matrix of order m, b isunrestricted in sign as in the original problem. We assume that an optimal solution

is obtained without having to make each component of the requirement vector bnon negative.

Let be the optimal feasible solution of the primal problem corresponding to thefinal basis B and let be the associated cost vector. Therefore = And the corresponding optimal value of the objective function is

= max( ) = ( ) .Since

is optimal, we have

0(in a maximization problem) for all in

the final table. Thus

0, [ is the jth column vector in the final table]or,

or, ( ) ( 000)(as , , , are slack vectors and cost component of each one is 0)


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or, ( , ) ( , 0)Equating we get, and 0 . .. (1)Putting = 0 where = ( ) , we get from (1), and = ( ) ( ) = which indicates that is a feasible

solution to the dual problem.

Now we have to show that is also an optimal solution to the dual problem. = = ( ) (as is a scalar)= = ( ) = ( ) = = max Hence is an optimal solution to the dual problem and

= min = max .Similarly, starting with the finite optimal value of the dual problem, if it exists, we

can prove that primal also has an optimal value of the objective function andmax = min .The above theorem can be stated in an alternative way as follows:

Theorem: A feasible solution to a primal maximization problem with objectivefunction , will be optimal, if and only if, there exists a feasible solution tothe dual minimization problem with the objective function such that = .The proof is almost exactly the same as the above theorem.

Theorem: (b) If either of the primal or the dual has unbounded solution then the

other will have no feasible solution.

Proof: Let us assume that the primal has an unbounded solution. If the dual

problem has a finite optimal solution, then the primal will also have a finite

optimal solution, which is a contradiction. We now prove that the dual has no

feasible solution.


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When the primal objective function is unbounded,

we have max = max and since for any feasible solution of the dual, max = max for all feasible solutions of the dual, whichindicates that there is no feasible whose components are finite. Hence we can

conclude that the dual has no feasible solution.

Note: Converse of this theorem is not necessarily true.

Examples

1. Solve the following problem by solving its dual using simplex method.Min = 3 + Subject to 2 + 14

4 , , 0Solution: The dual of the problem is

Max = 14 + 4 Subject to 2 + 3

1,

, 0

Now we solve the dual problem by the simplex method.

Simplex tables

C 5 2 2 0

Basis B Min ratio= , > 0

0 3 2 1 1 0

32

0 1 1 1 0 1 11 = 1 0 14 4 0 0

is the key element. / = new key row = , / = / 0 1 0 3 1 2 1

3


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14 1 1 1 0 1

14

0

18 0 14

is the key element. / = new key row = , / = / 4 1/3 0 1 1/3 -2/3 14 4/3 1 0 1/3 1/3

20 0 0 6 2Here all 0. So the solution is optimal.Max = 20 at =

, = 1/3.

Now = 6 and = 2 corresponding to the slack vectors and atthe optimal stage.

Hence the primal optimal solution is = 6, = 2, so min =Max = 20 at = 6, = 2 .

Note: Advantage of solving the dual problem is that we are able to solve the primal

without using artificial variables.

2. Solve the problem by solving its dual using simplex method.Max = 3 + 4

Subject to + 102 + 3 18 8

6 , , 0Solution: The dual of the problem is

Min = 10 + 18 + 8 + 6 Subject to + 2 + 3 + 3 + 4 , , , , 0

Now we solve the dual problem by the simplex method.


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After introducing surplus variables (as the constraints are type) we see thatidentity matrix is already present in the coefficient matrix. So we do not need to

add artificial variables. Then we change the problem to a maximization problem as

Max ( ) = 10 18 8 6 Subject to + 2 + = 3

+ 3 + = 4 , , , 0Simplex tables

c -10 -18 -8 -6 0 0

Basis b Min ratio= , > 0 -8 3 1 2 1 0 -1 0 =10 -6 4 1 3 0 1 0 -1 43

-48 -4 16 0 0 8 6

is the key element.

/

=new key row

= ,

/

= /

-8 1/3

-18 4/3

-80/3 4/3 0 0 16/3 8 2/3Here all 0. So the solution is optimal.Min = Max( ) = 80/3 at = 0, = , = , = 0. Thus usingthe property of duality theory, Max = Min = 80/3 at = 8, = whichare the values corresponding to the surplus vectors ( ) and ( )respectively.

3. Solve the following primal problem by solving its dual.Min = 10 + 2


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Subject to + 2 + 2 1 2 1

+ 3 3 , , , 0Solution: The dual of the above problem is Max = + 3

Subject to + + 102 22 2 + 3 0, , , 0

After introducing slack variables the converted problem is

Max = + 3 + 0. + 0. + 0. Subject to + + + = 102 + = 22 2 +3 + = 0 , 0, = 1, ,6Simplex tables

c 1 -1 3 0 0 0Basis b Min ratio= , > 0

0 10 1 1 1 1 0 0 =10 0 2 2 0 -1 0 1 0 0 0 2 -2 3 0 0 1 0

3 = 0

0 1 1 -3 0 0 0 is the key element. / = new key row = , / = / , = 1,2 0 10 1/3 53

0 1 0 -1/3 105/3 = 6


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0 2 8/3 -2/3 0 0 1 1/3 ---

3 0 2/3 -2/3 1 0 0 1/3 ---

0 1 1 0 0 0 1 is the key element. / = new key row = , / = / , = 2,3 6

6

4

6 6/5 0 0 3/5 0 4/5As 0 for allj, optimality condition is reached. Hence the optimal solutionobtained is a B.F.S and the maximum value of

is 6 for

= 0, = 6, = 4.

Therefore min = 6. Now = , = 0, = 4/5.Thus = 6 for = 3/5, = 0, = 4/5 .


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CHAPTER VIITRANSPORTATION AND ASSIGNMENT PROBLEM

Transportation problem (T.P) is a particular type of linear programming problem.

Here, a particular commodity which is stored at different warehouses (origins) is to

be transported to different distribution centres (destinations) in such a way that the

transportation cost is minimum.

Let us consider an example where there are origins

with the quantity

available at each be , = 1,2, , and destinations with the quantitiesrequired, i.e. the demand at each be , = 1,2, , .We make an assumption = = . This assumption is notrestrictive.

If in a particular problem this condition is satisfied, it is called a balanced

transportation problem.

If the condition is , then it is called an unbalancedtransportation problem.We shall first discuss balanced transportation problems.

Destinations

origin

capacities

demands

In the above table, denotes the number of units transported from the ith originto the jth destination.


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Let denote the cost of transporting each unit from the ith origin to the jthdestination. In general,

0for all

,.

The problem is to determine the quantities , = 1,2, , , = 1,2, , , whichis to be transported from the ith origin to the jth destination such that the

transportation cost is minimum subject to the condition = .Mathematically the problem can be written as

min = subject to the constraints = , = 1,2, , = , = 1,2, ,

and = .Clearly 0 for all , .We state a few theorems without proof.

Theorem: There exists a feasible solution to each T.P. which is given by = ,= 1,2, , , = 1,2, , , where = = .Theorem: In each T.P. there exists at least one B.F.S. which makes the objective

function a minimum.

Theorem: In a balanced T.P. having origins and destinations ( , 2) theexact number of basic variables is + 1 .T.P. is a special case of a L.P.P. , therefore it can be solved by using simplex

method. But the method is not suitable for solving a T.P. A specially designed

table called the transportation table is used to solve a T.P.

In this table there are squares or rectangles arranged in rows and columns.

Each squares or rectangle is called a cell. The cell which is in the ith row and the

jth column is called the ( , ) cell. Each cost component is displayed at thebottom right corner of the corresponding cell. A component (if 0) of afeasible solution is displayed inside a square at the top left hand corner of the cell


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( , ) . The capacities of the origins and demands of the destinations are listed inthe outer column and the outer row respectively as given in the table below.

Transportation Table

Destinations

Origin

Capacities

Demands

Determination of an initial B.F.S

We will discuss two methods of obtaining an initial B.F.S , (i) North-West corner

rule and (ii) Vogels Approximation method (VAM) .

(i) North-West corner ruleStep 1: Compute min( , ) . Select = min( , ) and allocate the value of in the cell (1,1), i.e. the cell in the North-West corner of the transportation

table.

Step 2: If < , the capacity of the origin will be exhausted , so all othercells in the first row will be empty, but some demand remains in the destination .Compute min( , ) . Select = min( , ) and allocate the valueof in the cell (2,1).Step 3: If < , then the demand of is satisfied, so all the remainingcells in the first column will be empty, but some capacity remains in the origin .Compute min( , ) . Select = min( + , ) and allocate thevalue of in the cell (2,2).


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Repeat the above process till the capacities of all the origins are exhausted and the

demands of all the destinations are satisfied.

The feasible solution obtained by this method is always a B.F.S.

Problem: Determine an initial B.F.S. of the following transportation problem by

North-West corner rule.

destinations

origin

capacities

4

6

9

516

2 6 4 1 12 5 7 2 9 15

12 14 9 8demands

Solution:

4 6 9 5 16 2 6 4 1 12 5 7 2 9 15

12 14 9 8(ii) Vogels Approximation method (VAM)

Step 1: Determine the difference between the lowest and next to lowest cost for

each row and each column and display them within first brackets against the

respective rows and columns.

Step 2: Find the row or column for which this difference is maximum. Let this

occur at the ith row. Select the lowest cost in the ith row. Let it be . Allocate

12 4

10 2

7 8


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= min( , ) in the cell ( , ). If the maximum difference is not unique thenselect arbitrarily.

Step 3: If < , cross out the ith row and diminish by . If < , crossout the j

thcolumn and diminish by . If = , delete only one of the ith row

or the jth

column.

Step 4: Compute the row and the column differences for the reduced transportation

table and repeat the procedure discussed above till the capacities of all the origins

are exhausted and the demands of all the destinations are satisfied.

Problem: Determine an initial B.F.S. of the following transportation problem byVAM.

destinations

origin

capacities

19 30 50 10 7 70 30 40 60 9 40 8

70 20

185 8 7 14

demands

Solution:

19 30 50 10 7(9) 7(9) 2(40) 2(40) 70 30 40 60 9(10)

9(20)

9(20)

9(20)

9 40 8 70 20 18(12) 10(20) 10(50)

5(21) 8(22) 7(10) 14(10)5(21) 7(10) 14(10)

7(10) 14(10)7(10) 4(50)7 2

8

5

10

2

7 2


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Loops in a Transportation problem

In a transportation problem, an ordered set of four or more cells are said to form aloop (i) if and only if two consecut

sem 4 mtma lp and gametheory part 1_new

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