NLP

1. NLP KKT Practice and Second Order Conditions from Nash and Sofer

2. Unconstrained • First Order Necessary Condition • Second Order Necessary • Second Order Sufficient

3. Easiest Problem • Linear equality constraints

4. KKT Conditions Note for equality – multipliers are unconstrained Complementarity not an issue

5. Null Space Representation • Let x* be a feasible point, Ax*=b. • Any other feasible point can be written as x=x*+p where Ap=0 • The feasible region {x : x*+p pN(A)} where N(A) is null space of A

6. Null and Range Spaces See Section 3.2 of Nash and Sofer for example

7. Orthogonality

8. Null Space Review

9. Constrained to Unconstrained You can convert any linear equality constrained optimization problem to an equivalent unconstrained problem • Method 1 substitution • Method 2 using Null space representation and a feasible point.

10. Example • Solve by substitution becomes

11. Null Space Method • x*= [4 0 0]’ • x=x*+Zv becomes

12. General Method • There exists a Null Space Matrix • The feasible region is: • Equivalent “Reduced” Problem

13. Optimality Conditions • Assume feasible point and convert to null space formulation

14. Where is KKT? • KKT implies null space • Null Space implies KKT Gradient is not in Null(A), thus it must be in Range(A’)

15. Lemma 14.1 Necessary Conditions • If x* is a local min of f over {x|Ax=b}, and Z is a null matrix • Or equivalently use KKT Conditions

16. Lemma 14.2 Sufficient Conditions • If x* satisfies (where Z is a basis matrix for Null(A)) then x* is a strict local minimizer

17. Lemma 14.2 Sufficient Conditions (KKT form) • If (x*,*) satisfies (where Z is a basis matrix for Null(A)) then x* is a strict local minimizer

18. Lagrangian Multiplier • * is called the Lagrangian Multiplier • It represents the sensitivity of solution to small perturbations of constraints

19. Optimality conditions • Consider min (x2+4y2)/2 s.t. x-y=10

20. Optimality conditions • Find KKT point Check SOSC

21. In Class Practice • Find a KKT point • Verify SONC and SOSC

22. Linear Equality Constraints - I

23. Linear Equality Constraints - II

24. Linear Equality Constraints - III so SOSC satisfied, and x* is a strict local minimum Objective is convex, so KKT conditions are sufficient.

25. Next Easiest Problem • Linear equality constraints Constraints form a polyhedron

26. x* Close to Equality Case Equality FONC: a2x = b a2x = b -a2 Polyhedron Ax>=b a3x = b a4x = b -a1 a1x = b contour set of function unconstrained minimum Which i are 0? What is the sign of I?

27. x* Close to Equality Case Equality FONC: a2x = b a2x = b -a2 Polyhedron Ax>=b a3x = b a4x = b -a1 a1x = b Which i are 0? What is the sign of I?

28. x* Inequality Case Inequality FONC: a2x = b a2x = b -a2 Polyhedron Ax>=b a3x = b a4x = b -a1 a1x = b Nonnegative Multipliers imply gradient points to the less than Side of the constraint.

29. Lagrangian Multipliers

30. Lemma 14.3 Necessary Conditions • If x* is a local min of f over {x|Ax≤b}, and Z is a null-space matrix for active constraints then for some vector *

31. Lemma 14.5 Sufficient Conditions (KKT form) • If (x*,*) satisfies

32. Lemma 14.5 Sufficient Conditions (KKT form) where Z+ is a basis matrix for Null(A +) and A + corresponds to nondegenerate active constraints) i.e.

33. Sufficient Example • Find solution and verify SOSC

34. Linear Inequality Constraints - I

35. Linear Inequality Constraints - II

36. Linear Inequality Constraints - III

37. Linear Inequality Constraints - IV

38. Example • Problem

39. You Try • Solve the problem using above theorems:

40. Why Necessary and Sufficient? • Sufficient conditions are good for? • Way to confirm that a candidate point • is a minimum (local) • But…not every min satisifies any given SC • Necessary tells you: • If necessary conditions don’t hold then you know you don’t have a minimum. • Under appropriate assumptions, every point that is a min satisfies the necessary cond. • Good stopping criteria • Algorithms look for points that satisfy Necessary conditions

41. General Constraints

42. Lagrangian Function • Optimality conditions expressed using Lagrangian function and Jacobian matrix were each row is a gradient of a constraint

43. Theorem 14.2 Sufficient Conditions Equality (KKT form) • If (x*,*) satisfies

44. Theorem 14.4 Sufficient Conditions Inequality (KKT) • If (x*,*) satisfies

45. Lemma 14.4 Sufficient Conditions (KKT form) where Z+ is a basis matrix for Null(A +) and A + corresponds to Jacobian of nondegenerate active constraints) i.e.

46. Sufficient Example • Find solution and verify SOSC

47. Nonlinear Inequality Constraints - I

48. Nonlinear Inequality Constraints - II

49. Nonlinear Inequality Constraints - III

50. Sufficient Example • Find solution and verify SOSC