This presentation deals with the two phase method which is used to handle constraints with equality and greater or equal to inequality.

1. Linear Programming
Two Phase Method
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2. Two Phase Method
 For 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
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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 (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.
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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.
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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.
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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
  
  
  
  
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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
MinZ x x x A
MinZ x x x A
x x x A
x x x S
x x x S
   
    
   
   
   
7
-Max

8. Basis
Varia
ble
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
8
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 10
S1 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 -1
CJ- ZJ 2 1 -6 0 0 0
9
CJ 0 0 0 0 0 -1

10.  Row Calculations
 New R3=Old R3/8
 New R1=New R3*2-Old R1
 New 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
10
CJ 0 0 0 0 0 -1
X2 is entering variable and A1 is leaving variable

11.  Row Calculations
 New R1=Old R1/1.75
 New R2=New R1*29/4-Old R2
 New 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
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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
Max Z x x x S S
Max Z x x x S S
    
     
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13. Basis
Variable
CB XB X1 X2 X3 S1 S2
X2 -4 4.28 0 1 -4.28 0 -0.14
S1 0 7.47 0 0 36.53 0 0.765
X1 5 7.85 1 0 -0.86 0 0.073
ZJ 5 -4 12.82 0 0.925
CJ-ZJ 0 0 -9.82 0 -0.925
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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=7.855
 X2=4.28 and
 Z=5X1-4X2+3X3
 Z=5(7.855)-4(4.28)+3(0)
= 22.15
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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
   
  
  

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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
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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
Min Z x x x A A
Min Z A A
x x x S A
x 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
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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
Variabl
e
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
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CJ 0 0 0 0 0 -1 -1

19.  Row Calculations
 New R2=Old R2/6
 New 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
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CJ 0 0 0 0 0 -1

20.  Row Calculations
 New R1=Old R1/3
 New 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 0
CJ-ZJ 0 0 0 0 0
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CJ 0 0 0 0 0

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
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
     
    
   
   

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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
22
CJ -4 -3 -9 0 0
X1 is entering variable and X3 is leaving variable

23.  Row Calculations
 New 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
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CJ -4 -3 -9 0 0

24.  X1=3/2
 X3=3
 Hence Z= -4X1 -3X2 -9X3
Z= -4(1.5)-3(3)-9(0)
Z= -15
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25. Thank You
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