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NLP

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NLP

KKT Practice and Second Order Conditions from Nash and Sofer

- First Order Necessary Condition
- Second Order Necessary
- Second Order Sufficient

- Linear equality constraints

Note for equality – multipliers are unconstrained

Complementarity not an issue

- 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

See Section 3.2 of Nash and Sofer for example

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.

- Solve by substitution
becomes

- x*= [4 0 0]’
- x=x*+Zv
becomes

- There exists a Null Space Matrix
- The feasible region is:
- Equivalent “Reduced” Problem

- Assume feasible point and convert to null space formulation

- KKT implies null space
- Null Space implies KKT
Gradient is not in Null(A), thus it must be in Range(A’)

- If x* is a local min of f over {x|Ax=b}, and Z is a null matrix
- Or equivalently use KKT Conditions

- If x* satisfies (where Z is a basis matrix for Null(A))
then x* is a strict local minimizer

- If (x*,*) satisfies (where Z is a basis matrix for Null(A))
then x* is a strict local minimizer

- * is called the Lagrangian Multiplier
- It represents the sensitivity of solution to small perturbations of constraints

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

- Find KKT point Check SOSC

- Find a KKT point
- Verify SONC and
SOSC

so SOSC satisfied, and x* is a strict local minimum

Objective is convex, so KKT conditions are sufficient.

- Linear equality constraints
Constraints form a polyhedron

x*

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?

x*

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?

x*

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.

- 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 *

- If (x*,*) satisfies

where Z+ is a basis matrix for Null(A +) and A + corresponds to nondegenerate active constraints)

i.e.

- Find solution and verify SOSC

- Problem

- Solve the problem using above theorems:

- 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

- Optimality conditions expressed using
Lagrangian function

and Jacobian matrix

were each row is a gradient of a constraint

- If (x*,*) satisfies

- If (x*,*) satisfies

where Z+ is a basis matrix for Null(A +) and A + corresponds to Jacobian of nondegenerate active constraints)

i.e.

- Find solution and verify SOSC

- Find solution and verify SOSC

- 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

- 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

- 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.

- Show optimal solution x*=[1,0]’
is regular and find KKT point

- 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.

X* is global min

Convex f

Convex constraints

X* is local min

SOSC

CQ

KKT Satisfied