Two Phase Method- Linear Programming

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 1

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 2

3. Phase I 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 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. 3

4. 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. 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. 4

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. 5

6. Example 1 Solve given LPP by Two-Phase Method 6

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 7 - Max

8. Basis Variable C B X B X 1 X 2 X 3 S 1 S 2 A 1 A 1 -1 20 2 1 -6 1 S 1 76 6 5 10 1 S 2 50 8 -3 6 1 8 C J -1

9. X1 is entering variable and S2 is leaving variable Basis Variable C B X B X 1 X 2 X 3 S 1 S 2 A 1 θ A 1 -1 20 2 1 -6 1 10 S 1 76 6 5 10 1 76/6 S 2 50 8 -3 6 1 50/8 Z J -2 -1 6 -1 C J - Z J 2 1 -6 9 C J -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 C B X B X 1 X 2 X 3 S 1 S 2 A 1 θ A 1 -1 7.5 1.75 -7.5 -1/4 1 4.28 S 1 77/2 29/4 11/2 1 -0.75 5.31 X 1 50/8 1 -3/8 6/8 1/8 --- Z J -1.75 7.5 1/4 -1 C J - Z J 1.75 -7.5 -1/4 10 C J -1 X2 is entering variable and A 1 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 C B X B X 1 X 2 X 3 S 1 S 2 X 2 4.28 1 -4.28 -0.14 S 1 7.47 36.53 0.765 X 1 7.85 1 -0.86 0.073 Z J C J -Z J 11

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. 12

13. Basis Variable C B X B X 1 X 2 X 3 S 1 S 2 X 2 -4 4.28 1 -4.28 -0.14 S 1 7.47 36.53 0.765 X 1 5 7.85 1 -0.86 0.073 Z J 5 -4 12.82 0.925 C J -Z J -9.82 -0.925 13 C J 5 -4 3

14. As the given problem is of maximization and all the values in C J -Z J row are either zero or negative, an optimal solution is reached and is given by X 1 =7.855 X 2 =4.28 and Z=5X 1 -4X 2 +3X 3 Z=5(7.855)-4(4.28)+3(0) = 22.15 14

15. Example 2 Solve by Two-Phase Simplex Method 15

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 16 Max - -

17. Phase I Basis Variable C B X B X 1 X 2 X 3 S 1 S 2 A 1 A 2 A 1 -1 15 2 4 6 -1 1 A 2 -1 12 6 1 6 -1 1 17 C J -1 -1

18. Row Calculations New Z*=R3+R1+R2 X3 is entering variable and A2 is leaving variable Basis Variable C B X B X 1 X 2 X 3 S 1 S 2 A 1 A 2 θ A 1 -1 15 2 4 6 -1 1 O 15/6 A 2 -1 12 6 1 6 -1 1 12/6 Z J -8 -5 -12 1 -1 -1 C J -Z J 8 5 12 -1 -1 18 C J -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 C B X B X 1 X 2 X 3 S 1 S 2 A 1 θ A 1 -1 3 -4 3 -1 1 1 1 X 3 2 1 1/6 1 -1/6 12 Z J 4 -3 1 -1 -1 C J -Z J -4 3 -1 1 19 C J -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 C B X B X 1 X 2 X 3 S 1 S 2 X 2 1 -4/3 1 -1/3 1/3 X 3 11/6 11/9 1 1/18 -2/9 Z J C J -Z J 20 C J

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. 21

22. Initial Basic Feasible Solution Basis Variable C B X B X 1 X 2 X 3 S 1 S 2 θ X 2 -3 1 -4/3 1 -1/3 1/3 9/4 X 3 -9 11/6 11/9 1 1/18 -2/9 1.5 Z J -7 -3 -9 -1/2 -1 C J -Z J 3 1/2 1 22 C J -4 -3 -9 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 C B X B X 1 X 2 X 3 S 1 S 2 X 2 -3 3 1 12/11 -3/11 13/33 X 1 -4 3/2 1 9/11 1/22 -2/11 Z J -4 -3 72/11 7/11 3/11 C J -Z J -27/11 -7/11 -3/11 23 C J -4 -3 -9

24. X 1 =3/2 X 3 =3 Hence Z = -4X 1 -3X 2 -9X 3 Z = - 4(1.5)-3(3)-9(0) Z = - 15 24

25. Thank You 25