5.7 Initialization Revisited

1. 5.7 Initialization Revisited • Motivation: a solution for the transformed system is feasible for the original system if and only if all the artificial variables are equal to zero. • Two methods are available for this purpose: Big M Two-Phase • They look quite different but are essentially equivalent.

2. Big M • Associate with each artificial variable a very unattractive cost coefficient (cj). • Since we try to optimize the objective function, the optimal solution generated by such a model will make the artificial variables as small as possible. • If the problem is feasible, the smallest feasible value of any artificial variable is zero.

3. Thus, if the problems is feasible, this approach will yield an optimal solution with all the artificial variables equal to zero.

4. Example (5.41)

5. Initial tableau isnotin canonical form: • Have to set the reduced costs of the artificial variables to zero. • We use (legal)row operations for this purpose.

8. Note that, as expected, all the artificial variables are nonbasic (thus equal to zero). • The optimal solution is x=(6,6,0,4,0,0)

9. Remark • Once an artificial variable is out of the basis, we never put it back into the basis. • Thus, once an artificial variable is out of the basis, we can “ignore” its column. • If you want to handle M numerically (i.e set it to a given value) make sure that it is not too large, but also not too small • The Big M method is “not nice”. • What happens if opt=min ?

10. 2-Phase Method • Phase 1: Find a basic feasible solution to the original problem (i.e. take the artificial variables out of the basis). • Phase 2: Find an optimal solution to the original problem, ignoring the artificial variables.

11. Phase 1 • Let w := sum of the artificial variables w* := minimumvalue of w subject to the constraints. • Because the artificial variables must satisfy the nonnegativity constraint, w*=0 if and only if all the artificial variables are equal to zero. • Thus, the goal in Phase 1 is tominimize w (regardless of what is the value of opt in the original problem)

12. Case 1: w*>0 The problem is not feasible!!! (why!) • Case 2: w*=0 and all the artificial variables are non-basic A basic feasible solution to the original problem has been generated. Continue with Phase 2. • Case 3: w*=0, but at least one artificial variable is in the basis. Using pivot operations, take all the artificial variables out of the basis.

13. 5.7.3 Example

14. Phase 1

15. We have to restore the canonical form (by legal row operations)

16. 0 0 corrections!!!

17. End of Phase 1: All the artificial variables are out of the basis.

18. Phase 2 • We now have to restore the original objective function: z’ = -3x1- 5x2 3 5 0 0

19. 3 5 0 0 • This is not in canonical form, so we use legal row operations to restore the canonical form.

20. Remark • Read the material in the lecture notes concerning the relationship between the Big M method and the 2-Phase Method and make sure you understand why there are “equivalent” and why the 2 Phase Method is better. (end of section 5.7)

21. 5.8 Algorithm Complexity • Worst case is very bad • In practice : surprisingly well!!!