1

Linear ProgrammingTwo Phase Method

2

Two Phase MethodFor greater than or equal to

constraint, the slack variable has a negative co efficient

Equality constraints do not have slack variables

If either of constraint is part of the model, there is no convenient IBFS and hence two phase method is used

3

Phase I1. In this phase, we find an IBFS to the original

problem, for this all artificial variable are to be driven to zero. To do this an artificial objective function (Z*) is created which is the sum of all artificial variables. The new objective function is then subjected to the constraints of the given original problem using the simplex method. At the end of Phase I, three cases arises

A. If the minimum value of Z*=0, and no artificial variable appears in the basis at a positive level then the given problem has no feasible solution and procedure terminates.

4

B. If the minimum value of Z*=0, and no artificial variable appears in the basis, then a basic feasible solution to the given problem is obtained.

C. If the minimum value of the Z*=0 and one or more artificial variable appears in the basis at zero level, then a feasible solution to the original problem is obtained. However, we must take care of this artificial variable and see that it never become positive during Phase II computations.

5

Phase IIWhen Phase I results in (B) or (C), we go on

for Phase II to find optimum solution to the given LP problem. The basic feasible solution found at the end of Phase I now used as a starting solution for the original LP problem. Mean that find table of Phase I becomes initial table for Phase II in which artificial (auxiliary) objective function is replaced by the original objective function. Simplex method is then applied to arrive at optimum solution.

6

Example 1Solve given LPP by Two-Phase Method

1 2 3

1 2 3

1 2 3

1 2 3

5 4 3Subject to 2 6 20

6 5 10 768 3 6 50

MaxZ x x xx x x

x x xx x x

7

Add artificial variable to the first constraint and slack variable to second and third constraints.

Phase IAssigning a cost 1 to artificial variable and

cost o to other variables, the objective function of the auxiliary LPP is 1 2 3 1

1 2 3 1

1 2 3 1

1 2 3 1

1 2 3 2

* 0 0 0* 0 0 0 0

Subject to 2 6 206 5 10 76

8 3 6 50

MinZ x x x AMinZ x x x A

x x x Ax x x Sx x x S

-Max

8

Basis Variable

CB XB X1 X2 X3 S1 S2 A1

A1 -1 20 2 1 -6 0 0 1

S1 0 76 6 5 10 1 0 0

S2 0 50 8 -3 6 0 1 0

CJ 0 0 0 0 0 -1

9

X1 is entering variable and S2 is leaving variable

Basis Variable

CB XB X1 X2 X3 S1 S2 A1 θ

A1 -1 20 2 1 -6 0 0 1 10S1 0 76 6 5 10 1 0 0 76/6

S2 0 50 8 -3 6 0 1 0 50/8

ZJ -2 -1 6 0 0 -1CJ- ZJ 2 1 -6 0 0 0

CJ 0 0 0 0 0 -1

10

Row CalculationsNew R3=Old R3/8New R1=New R3*2-Old R1New R2=NewR3*6-Old R2

X2 is entering variable and A1 is leaving variable

Basis Variable

CB XB X1 X2 X3 S1 S2 A1 θ

A1 -1 7.5 0 1.75 -7.5 0 -1/4 1 4.28

S1 0 77/2 0 29/4 11/2 1 -0.75 0 5.31

X1 0 50/8 1 -3/8 6/8 0 1/8 0 ---

ZJ 0 -1.75 7.5 0 1/4 -1

CJ- ZJ 0 1.75 -7.5 0 -1/4 0

CJ 0 0 0 0 0 -1

X2 is entering variable and A1 is leaving variable

11

Row CalculationsNew R1=Old R1/1.75New R2=New R1*29/4-Old R2New R3=NewR1*(3/8)+Old R3

As there is no artificial variable in the basis go to Phase II

Basis Variable

CB XB X1 X2 X3 S1 S2

X2 0 4.28 0 1 -4.28 0 -0.14

S1 0 7.47 0 0 36.53 0 0.765

X1 0 7.85 1 0 -0.86 0 0.073

ZJ 0 0 0 0 0

CJ-ZJ 0 0 0 0 0

12

Phase IIConsider the final Simplex table of Phase I,

consider the actual cost associated with the original variables. Delete the artificial variable A1 column from the table as it is eliminated in Phase II.

1 3 1 2

1 3 1 2

5 4 3 0 05 4 3 0 0 0

MaxZ x x x S SMax Z x x x S S

13

Basis Variable

CB XB X1 X2 X3 S1 S2

X2 -4 4.28 0 1 -4.28 0 -0.14S1 0 7.47 0 0 36.53 0 0.765X1 5 7.85 1 0 -0.86 0 0.073

ZJ 5 -4 12.82 0 0.925CJ-ZJ 0 0 -9.82 0 -0.925

CJ 5 -4 3 0 0

14

As the given problem is of maximization and all the values in CJ-ZJ row are either zero or negative, an optimal solution is reached and is given by

X1=7.855X2=4.28 and Z=5X1-4X2+3X3

Z=5(7.855)-4(4.28)+3(0) = 22.15

15

Example 2Solve by Two-Phase Simplex Method

1 2 3

1 2 3

1 2 3

1 2 3

4 3 9Subject to 2 4 6 15

6 6 12, , 0

MaxZ x x xx x xx x xx x x

16

Add artificial variable to the first constraint and slack variable to second and third constraints.

Phase IAssigning a cost 1 to artificial variable and

cost o to other variables, the objective function of the auxiliary LPP is

A new auxiliary linear programming problem1 2 3 1 2

1 2

1 2 3 1 1

1 2 3 2 2

* 0 0 0* 0

2 4 6 156 6 12

MinZ x x x A AMinZ A Ax x x S Ax x x S A

Max

- -

17

Phase I

Basis Variable

CB XB X1 X2 X3 S1 S2 A1 A2

A1 -1 15 2 4 6 -1 0 1 0

A2 -1 12 6 1 6 0 -1 0 1

CJ 0 0 0 0 0 -1 -1

18

Row Calculations New Z*=R3+R1+R2

X3 is entering variable and A2 is leaving variable

Basis Variable

CB XB X1 X2 X3 S1 S2 A1 A2 θ

A1 -1 15 2 4 6 -1 0 1 O 15/6

A2 -1 12 6 1 6 0 -1 0 1 12/6

ZJ -8 -5 -12 1 0 -1 -1

CJ-ZJ 8 5 12 -1 -1 0 0

CJ 0 0 0 0 0 -1 -1

19

Row CalculationsNew R2=Old R2/6New R1= New R2*6-Old R1

X2 is entering variable and A1 is leaving variable

Basis Variable

CB XB X1 X2 X3 S1 S2 A1 θ

A1 -1 3 -4 3 0 -1 1 1 1

X3 0 2 1 1/6 1 0 -1/6 0 12

ZJ 4 -3 0 1 -1 -1

CJ-ZJ -4 3 0 -1 1 0

CJ 0 0 0 0 0 -1

20

Row CalculationsNew R1=Old R1/3New R2= New R1*(1/6)-Old R2

Optimality condition is satisfied as Z* is having zero value

Basis Variable

CB XB X1 X2 X3 S1 S2

X2 0 1 -4/3 1 0 -1/3 1/3

X3 0 11/6 11/9 0 1 1/18 -2/9

ZJ 0 0 0 0 0CJ-ZJ 0 0 0 0 0

CJ 0 0 0 0 0

21

Phase IIOriginal objective function is given asConsider the final Simplex table of Phase I,

consider the actual cost associated with the original variables. Delete the artificial variable A1 column from the table as it is eliminated in Phase II.

1 2 3 1 2

1 2 3 1 2

1 2 3 1

1 2 3 2

1 2 3

4 3 9 0 04 3 9 0 0

Subject to 2 4 6 0 156 6 0 12

, , 0

MaxZ x x x S SMax Z x x x S S

x x x Sx x x Sx x x

22

Initial Basic Feasible Solution

Basis Variable

CB XB X1 X2 X3 S1 S2 θ

X2 -3 1 -4/3 1 0 -1/3 1/3 9/4

X3 -9 11/6 11/9 0 1 1/18 -2/9 1.5

ZJ -7 -3 -9 -1/2 -1

CJ-ZJ 3 0 0 1/2 1

CJ -4 -3 -9 0 0

X1 is entering variable and X3 is leaving variable

23

Row CalculationsNew R2=Old R2/(11/9)New R1=New R2+Old R1

As all the values in Z row are zero or negative, the condition of optimality is reached.

Basis Variable

CB XB X1 X2 X3 S1 S2

X2 -3 3 0 1 12/11 -3/11 13/33

X1 -4 3/2 1 0 9/11 1/22 -2/11

ZJ -4 -3 72/11 7/11 3/11

CJ-ZJ 0 0 -27/11 -7/11 -3/11

CJ -4 -3 -9 0 0

24

X1=3/2X3=3Hence Z= -4X1 -3X2 -9X3

Z= -4(1.5)-3(3)-9(0) Z= -15

25

Thank You