Duality Theory in Nonlinear Programming: Geometric Framework and Dual Problem, Slides of Computer Science

An introduction to duality theory in nonlinear programming, covering the geometrical framework, lagrange multipliers, and the dual problem. It includes examples of problems where a lagrange multiplier exists and does not exist, as well as the weak duality theorem and the relationship between the optimal primal and dual values.

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2012/2013

Uploaded on 03/27/2013

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NONLINEAR PROGRAMMING
LECTURE 18: DUALITY THEORY
LECTURE OUTLINE
Geometrical Framework for Duality
Lagrange Multipliers
The Dual Problem
Properties of the Dual Function
Consider the problem
minimize f (x)
subject to x X, gj
(x) 0, j =1,...,r,
assuming −∞ <f
< .
We assume that the problem is feasible and the
cost is bounded from below,
−∞ <f
= inf
xX
gj
(x)0,j=1,...,r
f(x) <
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NONLINEAR PROGRAMMING

LECTURE 18: DUALITY THEORY

LECTURE OUTLINE

  • Geometrical Framework for Duality
  • Lagrange Multipliers
  • The Dual Problem
  • Properties of the Dual Function
  • Consider the problem

minimize f (x)

subject to x ∈ X, gj (x) ≤ 0 , j = 1,... , r,

assuming −∞ < f ∗^ < ∞.

  • We assume that the problem is feasible and the

cost is bounded from below,

−∞ < f ∗^ = inf x∈X† gj†(x)≤ 0 , j=1,...,r†

f (x) < ∞

MIN COMMON POINT/MAX INTERCEPT POINT

  • Let S be a subset of n:
  • Min Common Point Problem: Among all points that

are common to both S and the nth axis,find the

one whose nth component is minimum.

  • Max Intercept Point Problem: Among all hyper-

planes that intersect the nth axis and support the

set S from “below”, find the hyperplane for which

point of intercept with the nth axis is maximum.

0 0

Min Common Point

Max Intercept Point Max Intercept Point

Min Common Point S S

(a) (b)

EXAMPLES: A L-MULTIPLIER EXISTS

0 (0,-1)

(μ*,1) (0,1)

(a)

(-1,0)

S = {(g(x),f(x)) | x ∈ X}

(-1,0)^0

(μ*,1)

(b)

S = {(g(x),f(x)) | x ∈ X}

0

(μ*,1)

(c)

(μ*,1)

(μ*,1)

S = {(g(x),f(x)) | x ∈ X}

min f(x) = x 1 - x 2 s.t. g(x) = x 1 + x 2 - 1 ≤ 0 x ∈ X = {(x 1 ,x 2 ) | x 1 ≥ 0, x 2 ≥ 0 }

min f(x) = (1/2) (x 12 + x 22 ) s.t. g(x) = x 1 - 1 ≤ 0 x ∈ X = R^2

min f(x) = |x 1 | + x 2 s.t. g(x) = x 1 ≤ 0 x ∈ X = {(x 1 ,x 2 ) | x 2 ≥ 0 }

EXAMPLES: A L-MULTIPLIER DOESN’T EXIST

(0,f*) = (0,0)

S = {(g(x),f(x)) | x ∈ X} min f(x) = x s.t. g(x) = x^2 ≤ 0 x ∈ X = R

(a)

(-1/2,0)

S = {(g(x),f(x)) | x ∈ X}

(b)

(0,f*) = (0,0)

(1/2,-1)

min f(x) = - x s.t. g(x) = x - 1/2 ≤ 0 x ∈ X = {0,1}

  • Proposition: Let μ∗^ be a Lagrange multiplier.

Then x∗^ is a global minimum of the primal problem

if and only if x∗^ is feasible and

x^ ∗ = arg min L(x, μ ∗^ ), μ∗^ ∗ x∈X j^ gj^ (x^ ) = 0,^ j^ = 1,... , r

WEAK DUALITY

  • The domain of q is

Dq = μ | q(μ) > −∞.

  • Proposition: The domain Dq is a convex set and

q is concave over Dq.

  • Proposition: (Weak Duality Theorem) We have

q ∗^ ≤ f ∗.

Proof: For all μ ≥ 0 , and x ∈ X with g(x) ≤ 0 , we

have

r q(μ) = inf L(z, μ) ≤ f (x) + μj gj (x) ≤ f (x), z∈X j=

so

q ∗^ = sup q(μ) ≤ inf f (x) = f ∗. μ≥ 0 x∈X, g(x)≤^0

DUAL OPTIMAL SOLUTIONS AND L-MULTIPLIERS

• Proposition: (a) If q ∗^ = f ∗^ , the set of Lagrange

multipliers is equal to the set of optimal dual solu-

tions. (b) If q ∗^ < f ∗^ , the set of Lagrange multipliers

is empty.

Proof: By definition, a vector μ∗^ ≥ 0 is a Lagrange

multiplier if and only if f ∗^ = q(μ∗) ≤ q ∗^ , which by the

weak duality theorem, holds if and only if there is

no duality gap and μ∗^ is a dual optimal solution.

Q.E.D.

μ

q(μ)

f* = 0

(a)

f* = 0 1 μ

q(μ)

  • 1
  • 1/

min f(x) = x s.t. g(x) = x^2 ≤ 0 x ∈ X = R

q(μ) = xmin ∈ R {x + μx^2 } ={- 1/- ∞( 4 if^ μ) μ^ ≤if 0 μ > 0

min f(x) = - x s.t. g(x) = x - 1/2 ≤ 0 x ∈ X = {0,1} q(μ) = min { - x + μ(x - 1/2)} = min{ - μ/2, μ/2 −1} x ∈ {0,1} (b)