Download presentation
Presentation is loading. Please wait.
Published byKerry Flynn Modified over 9 years ago
1
KKT Practice and Second Order Conditions from Nash and Sofer
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 pN(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
Close to Equality Case Equality FONC: x*
a2x = b a2x = b -a2 Polyhedron Ax>=b a3x = b a4x = b -a1 a1x = b x* contour set of function unconstrained minimum Which i are 0? What is the sign of I?
27
Close to Equality Case Equality FONC: x*
a2x = b a2x = b -a2 Polyhedron Ax>=b a3x = b a4x = b -a1 a1x = b x* Which i are 0? What is the sign of I?
28
Inequality Case Inequality FONC: x*
a2x = b a2x = b -a2 Polyhedron Ax>=b a3x = b a4x = b -a1 a1x = b x* 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
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
51
Nonlinear Inequality Constraints - V
52
Nonlinear Inequality Constraints - VI
53
Theorem 14.1 Necessary Conditions- Equality
If x* is a local min of f over {x|g(x)=0}, Z is a null-space matrix of the Jacobian g(x*)’, and x* is a regular point then
54
Theorem 14.3 Necessary Conditions
If x* is a local min of f over {x|g(x)>=0}, Z is a null-space matrix of the Jacobian g(x*)’, and x* is a regular point then
55
Regular point If x* is a regular point with respect to the constraints g(x*) if the gradient of the active constraints are linearly independent. For equality constraints, all constraints are active so should have linearly independent rows.
56
Necessary Example Show optimal solution x*=[1,0]’
is regular and find KKT point
57
Constraint Qualifications
Regularity is an example of a constraint qualification CQ. The KKT conditions are based on linearizations of the constraints. CQ guarantees that this linearization is not getting us into trouble. Problem is KKT point might not exist. There are many other CQ,e.g., for inequalities Slater is there exists g(x)<0. Note CQ not needed for linear constraints.
58
KKT Summary X* is global min Convex f Convex constraints
X* is local min CQ SOSC KKT Satisfied
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.