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
10/3/2013 Two-Phase Method 1

2. Two Phase Method
 In 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.
10/3/2013 2Two-Phase Method

3. Phase I
1. 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.
10/3/2013 3Two-Phase Method

4. 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.
10/3/2013 4Two-Phase Method

5. Phase II
 When 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.
10/3/2013 5Two-Phase Method

6. Example 1
 Solve 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
10/3/2013 6Two-Phase Method

7.  Add artificial variable to the first constraint and slack
variable to second and third constraints.
 Phase I
 Assigning 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
Min Z x x x A
Min Z x x x A
x x x A
x x x S
x x x S
10/3/2013 7Two-Phase Method

8. 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
10/3/2013 8Two-Phase Method

9.  Row Calculations
 New Z=Old Z+R1
 X1 is entering variable and S2 is leaving variable
Basis
Variable
Coefficients of RHS Ratio
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
10/3/2013 9Two-Phase Method

10.  Row Calculations
 New R3=Old R3/8
 New R1=New R3*2-Old R1
 New R2=NewR3*6-Old R2
 New Z*=New R3*2-Old Z*
 X2 is entering variable and A1 is leaving variable
Basis
Variable
Coefficients of RHS Ratio
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
10/3/2013 10Two-Phase Method

11.  Row Calculations
 New R1=Old R1/1.75
 New R2=New R1*29/4-Old R2
 New R3=NewR1*(3/8)+Old R3
 New Z*=New R1-Old Z*
 As there is no artificial variable in the basis go to Phase
II
Basis
Variable
Coefficients of RHS Ratio
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
10/3/2013 11Two-Phase Method

12. Phase II
 Consider 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
M ax Z x x x S S
M ax Z x x x S S
10/3/2013 12Two-Phase Method

13.  Row calculation: New Z=Old z+5(R3)-4(R1)
Basis
Variable
Coefficients of RHS Ratio
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
Coefficients of RHS Ratio
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 22.14810/3/2013 13Two-Phase Method

14.  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=7.855
 X2=4.28 and
 Z=5X1-4X2+3X3
 Z=5(7.855)-4(4.28)+3(0)
 = 22.15
10/3/2013 14Two-Phase Method

15. Example 2
 Solve 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
10/3/2013 15Two-Phase Method

16.  Add artificial variable to the first constraint and slack
variable to second and third constraints.
 Phase I
 Assigning 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 problem
10/3/2013 Two-Phase Method 16
1 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
M in Z x x x A A
M in Z A A
x x x S A
x x x S A

17. Phase I
Basis
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
10/3/2013 17Two-Phase Method

18.  Row Calculations
 New Z*=R3+R1+R2
 X3 is entering variable and A2 is leaving variable
Basis
Variabl
e
Coefficients of RHS Ratio
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
10/3/2013 18Two-Phase Method

19.  Row Calculations
 New R2=Old R2/6
 New R1= New R2*6-Old R1
 New Z*=New R2*12-Old Z*
 X2 is entering variable and A1 is leaving variable
Basis
Variabl
e
Coefficients of RHS Ratio
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
10/3/2013 19Two-Phase Method

20.  Row Calculations
 New R1=Old R1/3
 New R2= New R1*(1/6)-Old R2
 New 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
10/3/2013 20Two-Phase Method

21. Phase II
 Original objective function is given as
 Consider 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
M ax Z x x x S S
M ax Z x x x S S
x x x S
x x x S
x x x
10/3/2013 21Two-Phase Method

22. 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
10/3/2013 22Two-Phase Method

23.  X1 is entering variable and X3 is leaving variable
Basis
Variable
Coefficients of RHS Ratio
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
10/3/2013 23Two-Phase Method

24.  Row Calculations
 New R2=Old R2/(11/9)
 New R1=New R2+Old R1
 New 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 RH
S
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
10/3/2013 24Two-Phase Method

25.  X1=3/2
 X3=3
 Hence Z=-4x1-3x2-9x3
Z=-4(1.5)-3(3)-9(0)
Z=-15
10/3/2013 25Two-Phase Method

26. 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
M ax 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
Min Z x x
x x
x x
x x
10/3/2013 26Two-Phase Method

27. 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
M ax Z x x x
x x x
x x
x x
x x x
10/3/2013 27Two-Phase Method