ENGM 435/535 Optimization Adapting to Non-standard forms

Standard Form for LP Max Z c X s t a b = + £ × ³ . , n m mn 1 2 11 12 21 22 . ,

Wyndor Glass Max Z = 3X1 + 5X2 s.t. X1 < 8,000 X2 < 6,000

Equality Constraint Max Z = 3X1 + 5X2 s.t. X1 < 8,000 X2 < 6,000 Suppose constraint 3 is a hard constraint

New Feasible Region 2 4 6 8 10 12 8 6 4 2 Now, only feasible solutions on the blue line segment (0,6) (4,6) (8,3) (8,0) (0,0)

Equality Constraint Max Z = 3X1 + 5X2 s.t. X1 +Xs1 = 8,000 Add Slack Variables

New Feasible Region Now, only feasible solutions on the blue line segment 2 4 6 8 10 12 8 6 4 2 (0,6) (4,6) In past, we start at origin which is no longer feasible (8,3) (8,0) (0,0)

Equality Constraint Max Z = 3X1 + 5X2 s.t. X1 +Xs1 = 8,000 3X1 + 4X2 + Xa1 = 36,000 X1 > 0 X2 > 0 Add an artificial variable Now, I can easily find an initial feasible, but how do I keep Xa1 out of final solution?

Equality Constraint Max Z = 3X1 + 5X2 - MXa1 s.t. X1 +Xs1 = 8,000 3X1 + 4X2 + Xa1 = 36,000 X1 > 0 X2 > 0 Add a large penalty Now, I can easily find an initial feasible, but how do I keep Xa1 out of final solution?

Summary for Equality Constraint Max Z = 3X1 + 5X2 s.t. X1 < 8 X2 < 3X1 + 4X2 = 36 X1 > 0 X2 > 0 Max Z = 3X1 + 5X2 – MXa1 s.t. X1 < 8 X2 < 3X1 + 4X2 < 36 X1 > 0 X2 > 0

Greater Than Constraints 0.6X1+ 0.4X2 > 6 0.6X1+ 0.4X2 - Xs2 = 6 0.6X1+ 0.4X2 - Xs2 +Xa2 = 6

Minimization Problems Min Z = .4X1 + .5X2 s.t. 0.3X1+ 0.1X2 < 2.7 0.5X1+ 0.5X2 = 6 0.6X1+ 0.4X2 > 6 X1 > 0 X2 > 0 Max -Z = -.4X1 - .5X2 s.t. 0.3X1+ 0.1X2 < 2.7 0.5X1+ 0.5X2 = 6 0.6X1+ 0.4X2 > 6 X1 > 0 X2 > 0

Adapt to Std Form Min Z = .4X1 + .5X2 Max -Z = -.4X1 - .5X2 – MXa1 –MXa2 s.t. 0.3X1+ 0.1X2 + Xs1 =2.7 0.5X1+ 0.5X2 + Xa1 = 6 0.6X1+ 0.4X2 -Xs2 + Xa2 = 6 X1 > 0 X2 > 0

Final Note (row 0) Row 0 has no identifiable basics. We must remove M in row 0 by multiplying rows 2 and 3 by –M and adding to row 0.

Final Note (row 0)