the two phase method - operations research

the two phase method - operations research

1. Operations Research Chapter 07 - The Two Phase Simplex Method

2. The Two Phase Simplex Method Phase I: We create an artificial objective function as the sum of all the artificial variables, and we minimize this objective function using the tableau simplex method. If the minimum value of this artificial objective function is zero, then this means that all the artificial variables have been reduced to zero, and we have a basic feasible solution to the original problem, and we proceed to phase II. If this minimum value is positive, then this means that the original problem is infeasible, so we terminate. Phase II: The final tableau of phase I becomes the first tableau of phase II using the original objective function now. Use the tableau method again to obtain an optimal solution.

3. Example Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 <= 11 -4x1 + x2 +2x3 >= 3 -2x1 + x3 = 1

4. Solution Write the problem in standard form Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 - x5 = 3 -2x1 + x3 = 1

5. Solution Write the problem in canonical form (add artificial variable to the 2 nd and to the 3 rd constraints) Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 - x5 + x6 = 3 -2x1 + x3 + x7 = 1

6. Solution Phase 1: -Change Z to W and add to it the artificial variables -W is always minimized -Keep the constraints Min W= x6 + x7 S.T. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 - x5 + x6 = 3 -2x1 + x3 + x7 = 1

7. Solution Iteration 1: Basic: x4=11, x6=3, x7=1. W=4 1 1 x1 x2 x3 x4 x5 x6 x7 x4 1 -2 1 1 11 1 x6 -4 1 2 -1 1 3 1 x7 -2 1 1 1 CJ

8. Solution C1= 0 – (0,1,1) = 6 C2= 0 – (0,1,1) = -1 C3= 0 – (0,1,1) = -3 C5= 0 – (0,1,1) = 1 x 3 is entering, x3= min{11,3/2,1}=1, x7 is leaving. 1 -4 -2 -2 1 1 2 1 -1

9. Solution Iteration 2: Basic: x4=11-1=10, x6=3-2=1, x3=1. W=1 1 1 x1 x2 x3 x4 x5 x6 x7 x4 3 -2 1 -1 10 1 x6 1 -1 1 -2 1 x3 -2 1 1 1 CJ

10. Solution C1= 0 – (0,1,1) = 0 C2= 0 – (0,1,1) = -1 C5= 0 – (0,1,1) = 1 x 2 is entering, x2= min{-5,1,inf.}=1, x6 is leaving. Basic: x4=12, x2=1, x3=1. W=0 [End of phase 1 because W is zero ]. 3 -2 -2 1 -1

11. Solution Phase 2: -Change W to Z and remove the artificial variables -The same last table but remove the artificial variables columns

12. Solution -3 1 1 x1 x2 x3 x4 x5 x4 3 -2 1 10 1 x2 1 -1 1 x3 -2 1 1 CJ

13. Solution -3 1 1 x1 x2 x3 x4 x5 x4 3 1 -2 12 1 x2 1 -1 1 x3 -2 1 1 CJ

14. Solution C1= -3 – (0,1,1) = -1 C5= 0 – (0,1,1) = 1 x1 is entering, x1= min{4,inf.,-1/2}=4, x4 is leaving. Basic: x1=4, x2=1, x3=9. Z=-12+1+9=-2. 3 -2 -2 -1

15. Solution -3 1 1 x1 x2 x3 x4 x5 -3 x1 1 1/3 -2/3 4 1 x2 1 -1 1 1 x3 1 2/3 -4/3 9 CJ

16. Solution C4= 0 – (-3,1,1) = 1/3 > 0 C5= 0 – (-3,1,1) = 1/3 > 0 No entering variables, and the current solution is optimal. Basic: x1=4, x2=1, x3=9. Z=-12+1+9=-2. 1/3 2/3 -2/3 -1 -4/3