Operations ResearchCHAPTER 07 - THE TWO PHASE SIMPLEX METHOD

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.

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

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

Solution Write the problem in canonical form (add artificial variable to the 2nd and to the 3rd constraints)

Min Z= -3x1 + x2 + x3 S.T. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 - x5 + x6 = 3 -2x1 + x3 + x7 = 1

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

Solution Iteration 1: Basic: x4=11, x6=3, x7=1. W=4

0 0 0 0 0 1 1x1 x2 x3 x4 x5 x6 x7

0 x4 1 -2 1 1 0 0 0 111 x6 -4 1 2 0 -1 1 0 31 x7 -2 0 1 0 0 0 1 1

CJ 0 0 0

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

x3 is entering, x3= min{11,3/2,1}=1, x7 is leaving.

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Solution Iteration 2: Basic: x4=11-1=10, x6=3-2=1, x3=1. W=1

0 0 0 0 0 1 1x1 x2 x3 x4 x5 x6 x7

0 x4 3 -2 0 1 0 0 -1 101 x6 0 1 0 0 -1 1 -2 10 x3 -2 0 1 0 0 0 1 1

CJ 0 0 0 0

Solution C1= 0 – (0,1,1) = 0

C2= 0 – (0,1,1) = -1

C5= 0 – (0,1,1) = 1

x2 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].

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Solution Phase 2: -Change W to Z and remove the artificial variables

-The same last table but remove the artificial variables columns

Solution-3 1 1 0 0x1 x2 x3 x4 x5

0 x4 3 -2 0 1 0 101 x2 0 1 0 0 -1 10 x3 -2 0 1 0 0 1

CJ 0 0 0

Solution-3 1 1 0 0x1 x2 x3 x4 x5

0 x4 3 0 0 1 -2 121 x2 0 1 0 0 -1 10 x3 -2 0 1 0 0 1

CJ 0 0 0

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.

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Solution-3 1 1 0 0x1 x2 x3 x4 x5

-3 x1 1 0 0 1/3 -2/3 41 x2 0 1 0 0 -1 11 x3 0 0 1 2/3 -4/3 9

CJ 0 0 0

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.

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