8. Optimization Daisuke Oyama Mathematics II May 8, 2020 - PowerPoint PPT Presentation

8. Optimization Daisuke Oyama Mathematics II May 8, 2020

Unconstrained Maximization Problem Let X ⊂ R N be a nonempty set. Definition 8.1 For a function f : X → R , ▶ ¯ x ∈ X is a (strict) local maximizer of f if there exists an open neighborhood A ⊂ X of ¯ x relative to X such that f (¯ x ) ≥ f ( x ) for all x ∈ A ( f (¯ x ) > f ( x ) for all x ∈ A with x ̸ = ¯ x ); ▶ ¯ x ∈ X is a maximizer (or global maximizer ) of f if f (¯ x ) ≥ f ( x ) for all x ∈ X . (Local and global minimizers are defined analogously.) 1 / 30

First-Order Condition for Optimality Let X ⊂ R N be a nonempty set. Proposition 8.1 For f : X → R , if ▶ ¯ x ∈ X is a local maximizer or local minimizer of f , ▶ ¯ x ∈ Int X , and ▶ f is differentiable at ¯ x , then ∇ f (¯ x ) = 0 . Proof Apply the FOC for the one variable case to f ( x i , ¯ x − i ) for each i = 1 , . . . , N . 2 / 30

Second-Order Condition for Optimality Let X ⊂ R N be a nonempty set. Proposition 8.2 For f : X → R , suppose that ¯ x ∈ Int X and that f is differentiable on Int X and ∇ f is differentiable at ¯ x . 1. If ¯ x is a local maximizer of f , then D 2 f (¯ x ) is negative semi-definite. x ) = 0 and D 2 f (¯ 2. If ∇ f (¯ x ) is negative definite, then ¯ x is a strict local maximizer of f . 3 / 30

Proof 1. ▶ Fix any z ∈ R N , z ̸ = 0 . Let h ( α ) = f (¯ x + αz ) − f (¯ x ) (where α ∈ R is sufficiently close to 0 ). Note that h is differentiable and h ′ is differentiable at α = 0 . ▶ Recall that h ′′ ( α ) = z · D 2 f (¯ x + αz ) z . ▶ If ¯ x is a local maximizer of f , then α = 0 is a local maximizer of h . ▶ If h ′′ (0) > 0 , then α = 0 would be a strict local minimizer. ▶ Thus, h ′′ (0) ≤ 0 , or z · D 2 f (¯ x ) z ≤ 0 . 4 / 30

2. ▶ Suppose that ∇ f (¯ x ) = 0 and D 2 f (¯ x ) is negative definite. ▶ Since z · D 2 f (¯ x ) z is continuos in z and since { z ∈ R N | ∥ z ∥ = 1 } is compact, it follows from the assumption of negative definiteness and the Extreme Value Theorem that there is some ε > 0 such that 1 x ) u + ε < 0 for all u ∈ R N such that ∥ u ∥ = 1 . 2 u · D 2 f (¯ ▶ Since ∇ f (¯ x ) = 0 , by Taylor’s Theorem we can take a sufficiently small δ > 0 such that 1 2 z · D 2 f (¯ 0 < ∥ z ∥ < δ ⇒ f (¯ x + z ) − f (¯ x ) x ) z − ≤ ε. ∥ z ∥ 2 ∥ z ∥ 2 5 / 30

▶ Now take any x ∈ B δ (¯ x ) , x ̸ = ¯ x . Then, f ( x ) − f (¯ x ) ≤ 1 x − ¯ x x ) x − ¯ x x ∥ · D 2 f (¯ x ∥ + ε < 0 , x ∥ 2 ∥ x − ¯ 2 ∥ x − ¯ ∥ x − ¯ where the last inequality follows from the choice of ε . Thus, f ( x ) < f (¯ x ) . 6 / 30

Concave Functions Let X ⊂ R N be a nonempty convex set. Proposition 8.3 For f : X → R , suppose that ¯ x ∈ Int X and f is differentiable at ¯ x . ▶ Suppose that f is concave. If ∇ f (¯ x ) = 0 , then ¯ x is a global maximizer of f . ▶ Suppose that f is strictly concave. If ∇ f (¯ x ) = 0 , then ¯ x is a unique global maximizer of f . 7 / 30

Proof ▶ Take any x ∈ X , x ̸ = ¯ x . ▶ If f is concave, then we have f ( x ) ≤ f (¯ x ) + ∇ f (¯ x ) · ( x − ¯ x ) , with a strict inequality if f is strictly concave. ▶ Thus, if ∇ f (¯ x ) = 0 , we have f ( x ) ≤ f (¯ x ) if f is concave, and f ( x ) < f (¯ x ) if f is strictly concave. 8 / 30

Equality Constrained Maximization Problem Let X ⊂ R N be a nonempty open set, and f, g 1 , . . . , g M : X → R , where M < N . Consider the maximization problem: max f ( x ) (P) x s. t. g 1 ( x ) = 0 . . . g M ( x ) = 0 . ▶ Write g : X → R M , x �→ ( g 1 ( x ) , . . . , g M ( x )) , and C = { x ∈ X | g ( x ) = 0 } . ▶ ¯ x ∈ C is a local (global, resp.) constrained maximizer of (P) if it is a local (global, resp.) maximizer of f | C . 9 / 30

First-Order Condition for Optimality Proposition 8.4 Suppose that ▶ f, g 1 , . . . , g M are of C 1 class; ▶ ¯ x ∈ C is a local constrained maximizer of (P) ; and ▶ rank Dg (¯ x ) = M (“constraint qualification”). λ M ) ∈ R M (Lagrange multipliers) Then there exist unique (¯ λ 1 , . . . , ¯ such that M ¯ ∑ ∇ f (¯ x ) = λ m ∇ g m (¯ x ) . m =1 10 / 30

Expression with Lagrangian ▶ Let L : X × R M → R be defined by M ∑ L ( x, λ ) = f ( x ) − λ m g m ( x ) . m =1 ▶ Then the FOC is: λ ∈ R M such that there exists ¯ ∂L x, ¯ (¯ λ ) = 0 , n = 1 , . . . , N, ∂x n ∂L x, ¯ (¯ λ ) = 0 , m = 1 , . . . , M, ∂λ m or x, ¯ ∇ L (¯ λ ) = 0 . 11 / 30

Proof ▶ Let ¯ x ∈ C be a local constrained maximizer. x ) ∈ R M × N has rank M . By assumption Dg (¯ ▶ Without loss of generality, assume that the first M columns of Dg (¯ x ) are linearly independent. Write x = ( p, q ) , where p ∈ R M and q ∈ R N − M . ▶ By the Implicit Function Theorem, the equation g ( p, q ) = 0 is locally solved as p = η ( q ) , where q )] − 1 D q g (¯ Dη (¯ q ) = − [ D p g (¯ p, ¯ p, ¯ q ) . ▶ Consider the unconstrained maximization problem F ( q ) = f ( η ( q ) , q ) , where ¯ q is a local maximizer. 12 / 30

▶ By the FOC DF (¯ q ) = 0 , we have 0 = D q f ( η ( q ) , q ) | q =¯ q = D p f (¯ x ) Dη (¯ q ) + D q f (¯ x ) x )] − 1 D q g (¯ = − D p f (¯ x )[ D p g (¯ x ) + D q f (¯ x ) . λ T = D p f (¯ ▶ Let ¯ x )] − 1 , where ¯ λ ∈ R M . x )[ D p g (¯ ▶ Then we have x ) = ¯ λ T D p g (¯ x ) = ¯ λ T D q g (¯ D p f (¯ x ) , D q f (¯ x ) , or M x ) T ¯ ∑ ¯ ∇ f (¯ x ) = Dg (¯ λ = λ m ∇ g m (¯ x ) . m =1 13 / 30

Second-Order Condition for Optimality Proposition 8.5 Suppose that f, g 1 , . . . , g M are of C 2 class, ¯ x ∈ C , and rank Dg (¯ x ) = M . Denote W = { z ∈ R N | Dg (¯ x ) z = 0 } . 1. If ¯ x is a local constrained maximizer of (P) , x, ¯ then D 2 x L (¯ λ ) is negative semi-definite on W , λ ∈ R M is such that ∇ L (¯ where ¯ x, ¯ λ ) = 0 . λ ∈ R M such that 2. If there exists ¯ x, ¯ x, ¯ λ ) = 0 and D 2 ∇ L (¯ x L (¯ λ ) is negative definite on W , then ¯ x is a strict local constrained maximizer of (P) . 14 / 30

Inequality Constrained Maximization Problem Let X ⊂ R N be a nonempty open set, and f, g 1 , . . . , g M , h 1 , . . . , h K : X → R , where M < N . Consider the maximization problem: max f ( x ) (P) x s. t. g 1 ( x ) = 0 . . . g M ( x ) = 0 h 1 ( x ) ≤ 0 . . . h K ( x ) ≤ 0 . ▶ Write C = { x ∈ X | g ( x ) = 0 , h ( x ) ≤ 0 } . ▶ ¯ x ∈ C is a local (global, resp.) constrained maximizer of (P) if it is a local (global, resp.) maximizer of f | C . 15 / 30

First-Order Condition for Optimality (KKT Conditions) For x ∈ C , write I ( x ) = { k | h k ( x ) = 0 } . Proposition 8.6 Suppose that ▶ f, g 1 , . . . , g M , h 1 , . . . , h K are of C 1 class; ▶ ¯ x ∈ C is a local constrained maximizer of (P) ; and ▶ ∇ g 1 (¯ x ) , . . . , ∇ g M (¯ x ) and ∇ h k (¯ x ) , k ∈ I (¯ x ) , are linearly independent (“constraint qualification”). µ M ∈ R and ¯ λ 1 , . . . , ¯ Then there exist ¯ µ 1 , . . . , ¯ λ K ∈ R such that M K ∑ ∑ ¯ (i) ∇ f (¯ x ) = µ m ∇ g m (¯ ¯ x ) + λ k ∇ h k (¯ x ) , and m =1 k =1 (ii) ¯ λ k ≥ 0 and ¯ λ k h k (¯ x ) = 0 for each k = 1 , . . . , K . 16 / 30

▶ “ ¯ λ k h k (¯ x ) = 0 ” is called the complementarity condition . ▶ It says: ¯ λ k = 0 for all k / ∈ I (¯ x ) , where I ( x ) = { k | h k ( x ) = 0 } . 17 / 30

Example 1 Let X = R . Consider x ∈ [0 , 1] f ( x ) , max or max f ( x ) x s. t. h 1 ( x ) = − x ≤ 0 h 2 ( x ) = x − 1 ≤ 0 . ▶ If ¯ x ∈ [0 , 1] is a local constrained maximizer, then clearly we have: 1. if ¯ x ∈ (0 , 1) , then f ′ (¯ x ) = 0 , 2. if ¯ x = 0 , then f ′ (¯ x ) ≤ 0 , 3. if ¯ x = 1 , then f ′ (¯ x ) ≥ 0 . 18 / 30

Example 1 ▶ Let L ( x, λ ) = f ( x ) − λ 1 ( − x ) − λ 2 ( x − 1) . ▶ The KKT conditions are: L x ( x, λ ) = f ′ ( x ) + λ 1 − λ 2 = 0 ⇐ ⇒ f ′ ( x ) = − λ 1 + λ 2 , λ 1 ≥ 0 , λ 1 ( − x ) = 0 , λ 2 ≥ 0 , λ 2 ( x − 1) = 0 . ▶ By these, 1. if − ¯ x < 0 and ¯ x − 1 < 0 , then λ 1 = λ 2 = 0 , so f ′ (¯ x ) = 0 , 2. if − ¯ x = 0 and ¯ x − 1 < 0 , then λ 2 = 0 , so f ′ (¯ x ) = − λ 1 ≤ 0 , 3. if − ¯ x < 0 and ¯ x − 1 = 0 , then λ 1 = 0 , so f ′ (¯ x ) = λ 2 ≥ 0 . 19 / 30

Example 1 ▶ To see why we have λ k ≥ 0 , suppose that ¯ x satisfies the constraint h k ( x ) ≤ 0 with “ = ” (i.e., h k (¯ x ) = 0 ). ▶ For z ≈ 0 , f (¯ x ) + f ′ (¯ x + z ) ≈ h ′ x + z ) ≈ f (¯ x ) z and h k (¯ k (¯ x ) z . ▶ If f ′ (¯ x ) > 0 , then for small ε > 0 , ¯ x + ε has to violate the constraint, for which we have to have h ′ k (¯ x ) ≥ 0 . (Constraint qualification implies that h ′ k (¯ x ) ̸ = 0 .) ▶ If f ′ (¯ x ) < 0 , then for small ε > 0 , ¯ x − ε has to violate the constraint, for which we have to have h ′ k (¯ x ) ≤ 0 . ▶ In these cases, we have f ′ (¯ x ) = λ k h ′ k (¯ x ) with λ k > 0 . ▶ It is possible that f ′ (¯ x ) = 0 , so it may be the case that λ k = 0 . 20 / 30