3-Variable Problem by Simplex Method

Minimize: z = 51 2 83 subject to: 2x1 + 5x2 3 1

-2x1 - 122 + 3 9

-3x1 - 82 + 23 4

xi 0, i = 1 , ... , 3.

Starting with Basic Feasible Solution

Introduce slack variables as basic variables and the actual variables

as non-basic (set to 0), then the right-hand side of each constraint

represents the basic solution..

Canonical Form2x1 + 5x2 - x3 + x4 = 1

-2x1 - 12x2 + 3x3 + x5 = 9

-3x1 - 8x2 + 2x3 + x6 = 4

Treating the slack variables as basic and the others as non-basic,

the starting basic feasible solution is

Basic: x4 = 1, x5 = 9, x6 = 4

Non-basic: x1 = x2 = x3 = 0

z=0 is a basic feasible solution

We have the following situation:

Basic: x4 = 1, x5 = 9, x6 = 4 ;

Non basic: x1 = x2 = x3 = 0 ;

z= 0: 5x1 3x2 8x3 = 0.

The largest negative coefficient in the f equation is that of x3.

Thus, our next basic feasible solution should use x3 as one of

its basic variables.

The variable x3 is to be brought into the basic set as entering

variable implying that it will have a value greater than or equal to

zero in the next basic feasible solution. Therefore the columncontaining x3 is now pivotal column. For constraint 1, the

coefficient of x3 is negative and therefore, the choice of Row 1 aspivotal row is ruled out. constraint will remain positive as aresult of making x3 basic.

Basic x1 x2 x3 x4 x5 x6 zRHS (bi)

x4 2 5 -1 1 0 0 0 1

x5 -2 -12 3 0 1 0 0 9

x6 -3 -8 2 0 0 1 0 4

Obj. 5 -4 -8 0 0 0 -1 0

Basic x1 x2 x3 x4 x5 x6 zRHS (bi)

x4 1/2 1 0 1 0 0 3 R1+1/2R3x5 5/2 0 0 0 1 -3/2 0 3 R2-3/2R3x6 -3/2 -4 1 0 0 1/2 0 2

Obj. -7 -35 0 0 0 4 -1 16

Basic x1 x2 x3 x4 x5 x6 zRHS (bi)

x4 2 5 -1 1 0 0 0 1

x5 -2 -12 3 0 1 0 0 9

x6 -3 -8 2 0 0 1 0 4

Obj. 5 -4 -8 0 0 0 -1 0

Basic x1 x2 x3 x4 x5 x6 zRHS (bi)

x4 1/2 1 0 1 0 0 3

x5 5/2 0 0 0 1 -3/2 0 3

x6 -3/2 -4 1 0 0 1/2 0 2

Obj. -7 -35 0 0 0 4 -1 16

Basic x1 x2 x3 x4 x5 x6 zRHS (bi)

x4 1/2 1 0 1 0 0 3 Already 1

x5 5/2 0 0 0 1 -3/2 0 3 Already 0

x6 0 1 4 0 3/2 0 14 R3+4R1Obj. 21/2 0 0 35 0 43/2 -1 121

General Notation and Computer input

Matrix form

Input A B and CT

Now let us see what are

Surplus Variables

1 1

n m

ji i j

i j

a x b

1 1

1.n m

ji i n j j

i j

a x x b

But then another limitation arises when the coefficient of slack variables becomes

negative. To take care of such situation, Artificial Variables are added to the

system. The LPP Transforms to minimizing

Minimize '

1

n

i

i

z y

Subject to

1 1

n m

ji i i j

i j

a x y b

,i ix y 0

Try Examples Two phase Method on Page 137, 141, 145

DEGENERACY AND CYCLING

Let a LPP in Canonical form be given by

If there are several candidates for departing variables (tie among many) by criteria.

Then the value of basic variable not selected turns out to be zero.

A basic feasible solution in which some basic variables are zero is called

degenerate.

Example

2

4

6

1 3

1 2

1 2

1 2

1 2

z = 5x +3x

x - x

2x + x

-3x +2x

x 0 x 0

Examples

CHECK EXAMPLE PAGE 127

DUALITY

Maximize Minimize

Tc

z = x

Ax b

x 0 (13)

Primal

T

T

z' = b w

A x c

w 0(13)

Dual

2

5

1

1 2

1 2

1 2

1 2

1 2

z = 2x +3x

3x +2x

-x +2x

4x + x

x 0 x 0

4 2

3

1 2 3

1 2 3

1 2 3

1 2 3

z' = 2w +5w w

3w - w w

2w +2w w

w 0, w 0, w 0

Important Result: What is dual of the Dual problem? Prove it.

Check Statements of Theorems 3.4, 3.5, 3.6

Imp: Duality Theorem

Dual Simplex Method

With initial Tableau

Basic x1=x x2 =y x3=u x4 =v z =f bj

x3 2 2 1 0 0 8

x4 5 3 0 1 0 15

Obj.R -120 -100 0 0 1 0

Basic x1=x x2 =y x3=u x4 =v z =f bj

x3 0 4/5 1 -1/5 0 2

x1 1 3/5 0 1/5 0 3

Obj.R 0 -28 0 24 1 360

Basic x1=x x2 =y x3=u x4 =v z =f Xb

x2 0 1 5/4 -1/2 0 5/2

x1 1 0 -3/4 1/2 0 3/2

Obj.R 0 0 35 10 1 430