1. ENGM 435/535 Optimization
Adapting to Non-standard forms

2. Standard Form for LP Max Z c X s t a b = + £ × ³ . , n m mn 1 2 11 1221 22 . ,

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

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

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

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

7. New Feasible Region
Now, only feasible solutions on the blue line segment 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)

8. Equality Constraint Max Z = 3X1 + 5X2 s.t. X1 +Xs1 = 8,000
3X1 + 4X 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?

9. Equality Constraint Max Z = 3X1 + 5X2 - MXa1 s.t. X1 +Xs1 = 8,000
3X1 + 4X 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?

10. 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

11. Greater Than Constraints
0.6X1+ 0.4X2 > 6
0.6X1+ 0.4X Xs2 = 6
0.6X1+ 0.4X Xs2 +Xa2 = 6

12. 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

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

14. 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.

15. Final Note (row 0)