linear programming

04/12/2023 Two-Phase Method 1

Linear Programming:Two Phase Method

By

Er. ASHISH BANSODE M.E. Civil-Water Res. Engg.

DEPARTMENT OF CIVIL ENGINEERING.GOVERNMENT COLLEGE OF ENGINEERING

AURANGABAD-431 005


Two Phase MethodIn the Big M Method, we observed

that it was frequently necessary to add artificial variables to the constraints to obtain an initial basic feasible solution to an LPP. If problem is to be solved, the artificial variable must be driven to zero.

The two phase method is another method to handle these artificial variable. Here the LP problem is solved in two phase.


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 (w) is created which is the sum of all artificial variables. The new objective function is then minimized, 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 w=0, and no artificial variable appears in the basis at a positive level then the given problem has no feasible solution and procedure terminates.


B. If the minimum value of w=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 w=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.


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.

Note that the new objective function is always of minimization type regardless of whether the original problem of maximization or minimization type.


Example 1Solve given LPP by Two-Phase Method

1 2 3

1 2 3

1 2 3

1 2 3

5 4 3

Subject to 2 6 20

6 5 10 76

8 3 6 50

Max Z x x x

x x x

x x x

x x x


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 20

6 5 10 76

8 3 6 50

MinZ x x x A

MinZ x x x A

x x x A

x x x S

x x x S


Basis Variable

Coefficients of RHS Ratio

X1 X2 X3 S1 S2 A1

A1 2 1 -6 0 0 1 20

S1 6 5 10 1 0 0 76

S2 8 -3 6 0 1 0 50

Z* 0 0 0 0 0 -1 0


Row Calculations New Z=Old Z+R1

X1 is entering variable and S2 is leaving variable

Basis Variable


X1 X2 X3 S1 S2 A1

A1 2 1 -6 0 0 1 20 10

S1 6 5 10 1 0 0 76 76/6

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

Z* 2 1 -6 0 0 0 20


Row CalculationsNew R3=Old R3/8New R1=New R3*2-Old R1New R2=NewR3*6-Old R2New Z*=New R3*2-Old Z*

X2 is entering variable and A1 is leaving variable

Basis Variable


X1 X2 X3 S1 S2 A1

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

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

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

Z* 0 1.75 -7.5 0 -1/4 0 7.5


Row CalculationsNew R1=Old R1/1.75New R2=New R1*29/4-Old R2New R3=NewR1*(3/8)+Old R3New Z*=New R1-Old Z*

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

Basis Variable


X1 X2 X3 S1 S2 A1

X2 0 1 -4.28 0 -0.14 0.57 4.28

S1 0 0 36.53 0 0.765 -4.13 7.47

X1 1 0 -8.86 0 0.073 0.041 7.85

Z* 0 0 0 0 0 0.08 0.01


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 0

5 4 3 0 0 0

MaxZ x x x S S

Max Z x x x S S


Row calculation: New Z=Old z+5(R3)-4(R1)

Basis Variable


X1 X2 X3 S1 S2

X2 0 1 -4.28 0 -0.14 4.28 0

S1 0 0 36.53 0 0.765 7.47 0

X1 1 0 -0.86 0 0.073 7.85 7.85

Z* -5 4 -3 0 0

Basis Variable


X1 X2 X3 S1 S2

X2 0 1 -4.28 0 -0.14 4.28

S1 0 0 36.53 0 0.765 7.47

X1 1 0 -0.86 0 0.073 7.855

Z* 0 0 9.82 0 0.925 22.148


As the given problem is of maximization and all the values in Z row are either zero or positive, an optimal solution is reached and is given by

X1=7.855X2=4.28 and Z=5X1-4X2+3X3Z=5(7.855)-4(4.28)+3(0) = 22.15


Example 2Solve by Two-Phase Simplex Method

1 2 3

1 2 3

1 2 3

1 2 3

4 3 9

Subject to 2 4 6 15

6 6 12

, , 0

Max Z x x x

x x x

x x x

x x x


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 15

6 6 12

MinZ x x x A A

MinZ A A

x x x S A

x x x S A


Phase IBasis Variable

Coefficients of RHS

X1 X2 X3 S1 S2 A1 A1

A1 2 4 6 -1 0 1 O 15

A2 6 1 6 0 -1 0 1 12

Z* 0 0 0 0 0 -1 -1 0


Row Calculations New Z*=R3+R1+R2


Basis Variable


X1 X2 X3 S1 S2 A1 A1

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

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

Z* 8 5 12 -1 -1 0 0 27


Row CalculationsNew R2=Old R2/6New R1= New R2*6-Old R1New Z*=New R2*12-Old Z*


Basis Variable


X1 X2 X3 S1 S2 A1 A2

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

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

Z* -4 3 0 -1 1 0 -2 3


Row CalculationsNew R1=Old R1/3New R2= New R1*(1/6)-Old R2New Z*=New R1*3-Old Z*

Optimality condition is satisfied as Z* is having zero value

Basis Variable

Coefficients of RHS

X1 X2 X3 S1 S2 A1 A2

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

X3 11/9 0 1 1/18 -2/9 -2/27 2/9 11/6

Z* 0 0 0 0 0 -1 -1 0


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 0

4 3 9 0 0

Subject to 2 4 6 0 15

6 6 0 12

, , 0

Max Z x x x S S

Max Z x x x S S

x x x S

x x x S

x x x


Initial Basic Feasible Solution

Row calculations New Z=OldZ-3R1-9R2

Basis Variable

Coefficients of RHS

X1 X2 X3 S1 S2

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

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

Z 4 3 9 0 0 0


X1 is entering variable and X3 is leaving variable

Basis Variable


X1 X2 X3 S1 S2

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

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

Z -3 0 0 1/2 1 -19/5


Row CalculationsNew R2=Old R2/(11/9)New R1=New R2+Old R1New Z= New R2*3+Old Z

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

Basis Variable

Coefficients of RHS

Ratio

X1 X2 X3 S1 S2

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

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

Z 0 0 27/11 7/11 3/11 -15


X1=3/2X3=3Hence Z=-4x1-3x2-9x3

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

Z=-15


Exercise 1.

2.

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

2

Subject to 4 6 3 8

3 6 4 1

2 3 5 4

, , 0

Max Z x x x

x x x

x x x

x x x

x x x

1 2

1 2

1 2

1 2

2

Subject to 2

4

, 0

MinZ x x

x x

x x

x x


3.1 2 3

1 2 3

1 2

2 3

1 2 3

5 2 3

Subject to 2 2 2

3 4 3

3 5

, , 0

Max Z x x x

x x x

x x

x x

x x x

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