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# LINEAR PROGRAMMING

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TWO PHASE METHOD

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### LINEAR PROGRAMMING

1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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