The document describes the two phase method for solving linear programming problems. In phase I, artificial variables are introduced to obtain an initial basic feasible solution. The objective is to minimize the artificial variables subject to the original constraints. In phase II, the original objective function is optimized using the feasible solution from phase I as the starting point, without the artificial variables. Two examples are provided to illustrate the two phase method.
A problem is provided which is solved by using graphical and analytical method of linear programming method and then it is solved by using geometrical concept and algebraic concept of simplex method.
A problem is provided which is solved by using graphical and analytical method of linear programming method and then it is solved by using geometrical concept and algebraic concept of simplex method.
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Overview on Edible Vaccine: Pros & Cons with Mechanism
LINEAR PROGRAMMING
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
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
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
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
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
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