KKT conditions I Lecture 14 ME EN 575 Andrew Ning aning@byu.edu - - PDF document

kkt conditions i
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KKT conditions I Lecture 14 ME EN 575 Andrew Ning aning@byu.edu - - PDF document

KKT conditions I Lecture 14 ME EN 575 Andrew Ning aning@byu.edu Outline Equality Constraints minimize f ( x ) x R n with respect to subject to lb < x < ub c j ( x ) = 0 , j = 1 , . . . , m c k ( x ) 0 , k = 1 , . . .


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KKT conditions I

Lecture 14

ME EN 575 Andrew Ning aning@byu.edu

Outline

Equality Constraints

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minimize f(x) with respect to x ∈ Rn subject to lb < x < ub ˆ cj(x) = 0, j = 1, . . . , ˆ m ck(x) ≤ 0, k = 1, . . . , m

Equality Constraints

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Motivating Problem:

minimize x1 + x2 subject to x2

1 + x2 2 = 8

Small group exercise

  • What is the unconstrained optimum?
  • Draw the direction of the function gradient:

∇f.

  • Rewrite the constraint in our convention.
  • Draw the direction(s) for the constraint

gradients: ∇c.

  • What is the constrained optimal solution and

where is it located?

  • What do you notice about ∇f and ∇c at the
  • ptimum? Does that make sense?
  • How could you define the optimality criteria

mathematically?

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A More Formal Motivation

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Define the Lagrangian L(x, λ) = f(x) + λˆ c(x)

Extend to m constraints

∂L ∂xi = ∂f ∂xi +

ˆ m

  • j=1

ˆ λj ∂ˆ cj ∂xi = 0, (i = 1, . . . , n) ∂L ∂ˆ λj = ˆ cj = 0, (j = 1, . . . , ˆ m).

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λj is called a Lagrange multiplier, and there is a separate one for each constraint.