Econ 508-A FINITE DIMENSIONAL OPTIMIZATION - NECESSARY CONDITIONS. Carmen Astorne-Figari Washington University in St. Louis.
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1 Econ 508-A FINITE DIMENSIONAL OPTIMIZATION - NECESSARY CONDITIONS Carmen Astorne-Figari Washington University in St. Louis August 12, 2010
2 INTRODUCTION General form of an optimization problem: max x f (x) s.t. x X x = (x 1, x 2,..., x N ): N-dimensional vector of choice variables or instruments. f (x) = f (x 1,..., x N ): real valued function called objective function. X R N : constraint set, feasible set or opportunity set.
3 INTRODUCTION An optimization problem in R N is one where, given function f : R N R, we choose the value of x over a given set X R N that maximizes or minimizes the value of f. Alternative notation: max x X f (x) min x X f (x) or max { f (x) x X} min { f (x) x X}
4 BASIC DEFINITIONS Given f and X as above, The set of attainable values of f on X, or image of X under f, denoted f (X), is defined by f (X) = {w R some x X such that f (x) = w} The interior of X is the set defined by int(x) = {x X some ε > 0 such that N ε (x) X} [pictures: R, R 2 ]
5 SOLUTIONS A solution to the problem max{f (x) x X} or maximizer of f on X is a point x s.t. f (x ) f (x) x X. A solution to the problem min{f (x) x X} or minimizer of f on X is a point x s.t. f (x ) f (x) x X.
6 LOCAL VS. GLOBAL SOLUTIONS x X is a global maximizer of f in X iff f (x ) f (x) x X. x X is a global minimizer of f in X iff f (x ) f (x) x X. x X is a local maximizer of f in X iff x N ε (x ) X. x X is a local minimizer of f in X iff x N ε (x ) X. f (x ) f (x) f (x ) f (x) [pictures]
7 Most of the definitions or results for maximization problems have an exact analog for minimization problems. From now on, the minimization analog will be omitted. A solution x X is interior iff there is an ε > 0 such that N ε (x ) X and f (x ) f (x) x N ε (x ). [picture]
8 SET OF SOLUTIONS The set of solutions to a max problem is denoted argmax (f (x) x X) = {x X f (x) f (x ) x X} Consider the following example: Let X = [ 1, 1], and f : R R be f (x) = x 2. [picture] Maximizing f on C has two solutions, x = 1 and x = 1. As we can see, the set arg max(f (x) x X) can have more than one element.
9 EXAMPLE 1 Let X = [0, 1], and f : R R be f (x) = x 2. [picture] The problem of maximizing f on X has exactly one solution, the point x = 1. The set arg max(f (x) x X) can be a singleton.
10 EXAMPLE 2 Let X = R +, and f : R + R be f (x) = x 2. [picture] The problem of maximizing f on X has no solution. The set arg max(f (x) x X) =.
11 EXISTENCE OF A SOLUTION A sufficient condition for existence of a solution is given by the Weierstrass theorem. THEOREM: (Weierstrass) If f is continuous and X is closed and bounded (hence compact) and nonempty, then there exist a global maximum and a global minimum. However, Weierstrass is sufficient but not necessary for existence of a solution. Example: X = (0, 2] and f : R R be f (x) = x 2 [picture]
12 OBJECTIVES OF OPTIMIZATION THEORY I. To identify a set of conditions on f and X that guarantee the existence of solutions to optimization problems. [Weierstrass theorem] II. To obtain a characterization of the set of solutions: A. Necessary conditions [picture] [differentiability] B. Sufficient conditions [picture] [convexity] C. Conditions that guarantee uniqueness of a solution D. A theory of parametric variation: [envelope theorem] Sometimes optimization problems are presented in parametric form: f and or X depend on parameters θ Θ, where Θ (set of feasible parameter values). We write: max{f (x, θ) x X(θ)}
13 MOTIVATION (1) Consumer s Utility Maximization Problem max u(x) s.t. p x m x 0 p R l ++ x R l Does a solution exist? The constraint set X = B(p, m) = {x R l : p x m} compact. [picture] is So if u is continuous, then a solution exists (by Weierstrass theorem).
14 MOTIVATION (2) Firm s Cost Minimization Problem min p x s.t. f (x) y x 0 p R l ++, y R ++ x R l Does a solution exist? The constraint set is not compact. [picture] If the constraint set is not empty, x s.t. f (x ) y. Compactify the constraint set: {x : p x p x and f (x) y}. Since the objective function is continuous, there exists a solution.
15 USEFUL THEOREMS (1) THEOREM: Let f denote the function whose value at any x is f (x). Then x is a maximum of f on X iff it is a minimum of f on X; and z is a minimum of f on X iff z is a maximum of f on X. PROOF: HW.
16 USEFUL THEOREMS (2) THEOREM: Let ϕ : R R be a strictly increasing function, that is, a function such that y > y ϕ(y) > ϕ(y ) Then x is a maximum of f on X iff x is a maximum of the composition of ϕ and f on X; and z is a minimum of f on X iff z is a minimum of ϕ f on X. REMARK: it suffices that ϕ be a strictly increasing function on just the set f (X).That is, that ϕ only satisfies ϕ(y) > ϕ(y ) for all y, y f (X) with y > y. PROOF: HW.
17 CONSTRAINED OPTIMIZATION Consider the problem (P): max f 0 (x) s.t. x S; f 1 (x) 0,... f m (x) 0 Where: S R N is convex f i : S R for i = 0, 1,...m are differentiable and concave
18 EXAMPLE Consumer s Utility Maximization Problem max u(x) s.t. x R l + m p x 0 where p R l ++
19 SLATER CONDITION Slater Condition: ˆx int(s) s.t. f i (ˆx) > 0 for i = 1,...m. EXAMPLE: f 0 (x) = x, f 1 (x) = x 2 Let S = R So the maximization problem is: max x s.t. x x R -x 2 0 There is no x R s.t. x 2 > 0 so Slater is violated.
20 THE LAGRANGEAN Associated with (P), we can define a function L : S R m + R given by: L(x, λ) = f 0 (x) + m i=1 λ i f i (x) Notice that, for a given value of λ, L(x, ) is concave in x, and for a given value of x, L(, λ) is convex in λ. A function with these properties is also called saddle function.
21 SADDLE POINT DEFINITION: a point (x, λ ) is a saddle point of L(x, λ) if L(x, λ ) L(x, λ ) L(x, λ) for all (x, λ) S R m +
22 SADDLE POINT: CHARACTERIZATION If x int(s), (x, λ ) is a saddle point iff 1. Df 0 (x ) + m i=1 λ i Df i (x ) = 0 2. f i (x ) 0, λ i 0 for i = 1,...m 3. m i=1 λ i f i (x ) = 0 Given condition 2, condition 3 can be replaced by 3. λ i f i (x ) = 0
23 KUHN TUCKER UNDER CONCAVITY THEOREM: (Kuhn-Tucker I) Assume f 0, f 1,..., f M are concave, continuous functions from a convex set S R N into R. Let the problem (P) and L(x, λ) be as described above. Then: (i) (ii) If (x, λ ) S R M is a saddle point of L(x, λ), then x solves (P). Assume that the Slater condition holds. Then if x S is a solution to (P), λ R M such that (x, λ ) is a saddle point of L(x, λ). This version of Kuhn Tucker doesn t require differentiability.
24 EFFECTIVE CONSTRAINTS DEFINITION: an inequality constraint is effective or binding at a certain point x if f i (x ) = 0 i.e. if the constraint holds with equality at x.
25 KUHN TUCKER WITHOUT CONCAVITY THEOREM: (Kuhn Tucker II) Let f 0 : S R be a C 1 function on a certain open set S R N, and let f i : R N R, i = 1,..., M be C 1 functions. Suppose that x is a local maximum of f 0 on the set D = S {x R N f i (x) 0, i = 1,...M}. Let E {1,..., M} denote the set of effective constraints at x. Suppose that the derivatives {Df i (x ) i E} form an independent set of vectors. Then there exist λ i R, i = 1,..., M s.t. (i) λ i 0, i = 1,..., M; (ii) λ i f i (x ) = 0, i = 1,..., M; (iii) Df 0 (x ) + M λ i f i (x ) = 0 i=1 NOTE: these conditions are only necessary.
26 ANOTHER APPROACH State problems in standard form : MAX PROBLEM MIN PROBLEM max f (x) min f (x) s.t. g 1 (x) 0 s.t. g 1 (x) 0.. g K (x) 0 g K (x) 0
27 EXAMPLE (i) Consumer: (ii) Firm: max u(x) min p x s.t. p x m 0 s.t. f (x) y 0 x 1 0 x x N 0 x N 0
28 CONSTRAINT QUALIFICATION Recall we defined the set E which contains only the indices of the binding (effective) constraints. DEFINITION: constraint qualification (CQ) holds at x iff { g i (x ) : i E} is independent.
29 KUHN TUCKER II RESTATED THEOREM (KT): Consider a MAX problem in the standard form. Let f, g k be C 1. Let x be a local maximum. Suppose that CQ holds at x. Then λ k 0 s.t. (1) f (x ) = K λ k g k(x ) k=1 (2) λ k g k(x ) = 0 k (complementary slackness) NOTE: (2) says that if g k (x ) < 0 (kth constraint not binding), then λ k = 0. We can rewrite (1): f (x ) = k E λ k g k(x )
30 EXAMPLE Let f (x 1, x 2 ) = x 2. Let the constraints be g 1 = (x 1 1) 2 + x [a disk centered at (1, 0)] g 2 = (x 1 + 1) 2 + x [a disk centered at ( 1, 0)] The only feasible point is (0, 0), so it is the solution. f (ˆx ) = (0, 1) g 1 (x ) = ( 2, 0) g 2 (x ) = (2, 0)
31 EXAMPLE (2) There s no way of writing f (x ) as a linear combination of g 1 (x ) and g 2 (x )! Condition KT(1) fails. What is wrong with this example? Constraint qualification (CQ) fails at (0, 0). But CQ isn t necessary. There are other restrictions on the MAX problem sufficient to guarantee KT conditions hold under other assumptions (see previous section, Kuhn Tucker under concavity)
32 BINDING VS. ACTIVE CONSTRAINTS EXAMPLE: max f (x) = x 2 s.t. g(x) = x 0 [picture] The solution is x = 0. The constraint is binding, but λ = 0 because f (x ) = 0. Relaxing the constraint does not change the solution. Call constraint k active if λ k > 0. Then g in the previous example is binding, but not active. If a constraint is active, then it is binding (by condition KT (2)). Most of the times, binding constraints will be active, but not always.
33 SLACK CONSTRAINTS CAN AFFECT THE GLOBAL SOLUTIONS EXAMPLE: max f (x) = x 2 + x 4 s.t. g 1 (x) = x 1 0, g 2 (x) = x 1 0 [picture: W ] We have three constrained maxima at 1, 0 and 1. At x = 0, none of the constraints is binding: g 1 (x ) < 0 and g 2 (x ) < 0. However, if, say, we relax g 2 unique maximum at x = 3. to x 3 0, then there would be a KT is a result about local rather than global maximization. Even if we relax the constraint, x = 0 remains a local maximum.
34 INTUITION: ONE BINDING CONSTRAINT Mountain example. By (1), f (x ) = λ g(x ) g(x ) = 0: g cannot be increased. g(x ) level set g(x) (the fence). (we can only move along the level set g(x )). f (x ) is also level set g(x) (any movement that we re allowed to make does not increase f ). [pictures]
35 EXAMPLE 1 max f (x) = x s.t. x 0, x 1 [picture] g 1 {}}{ λ 1 ( x 0) λ 2 (x }{{ 1 } 0) g 2 At x = 1 : f (x) = Df (x) = 1 2 x f (x ) = 1 2 λ 1 = 0 by KT (2). λ 2 solves f (x ) = λ 2 g 2(x ) 1 2 = λ 2(1) λ 2 = 1 2 [picture of the gradients]
36 EXAMPLE 2 max f (x) = e x s.t. x 0 ( x 0) [picture] At x = 0 : By (2), λ = 1 f (x) = e x f (x ) = 1 g(x ) = 1 [picture of gradients]
37 EXAMPLE 3 For N = 2, suppose (1) doesn t hold ( f (x ) and g(x )are not collinear) Then it s feasible to move and increase f simultaneously. [picture]
38 INTUITION: TWO BINDING CONSTRAINTS [picture] Mountain example. By (1), f (x ) = λ 1 g 1(x ) + λ 2 g 2(x ). The gradient of the objective function lies in the cone spanned by the gradients of each binding constraint. [picture]
39 PROOF OF KUHN TUCKER Define W = {x R N : λ k s.t. x = λ k g k (x ) x E } WTS: if x is a local max, then f (x ) W. By contraposition: if f (x ) / W, then x can t be a local max. Suppose f (x ) / W. Since W is convex and closed, and f (x ) is compact and convex (a point), by the (strict) Separating Hyperplane Theorem, v 0 R N and c R such that f (x ) v > c > w v w W
40 PROOF OF KUHN TUCKER (CONTINUED 1) (i) Since 0 W (set all λ k = 0), c > 0, so f (x ) v > 0. Given f (x ) v > 0, D v f (x ) > 0 (there is a movement in direction v that increases f ) (ii) Since λ k g k (x ) W, then c > λ k g k (x ) v For any λ k > 0, c/λ k > g k (x ) v Taking limits as λ, 0 g k (x ) v ( k E) The movement in direction v is feasible.
41 PROOF OF KUHN TUCKER (CONTINUED 2) Define J = {k E : g k (x ) v = 0} If J =, then g k (x ) v < 0 k E α sufficiently small, g k (x + αv) < g k (x ) = 0 (g k no longer binds); for k / E, g k (x + αv) < 0, so by continuity of g, g k (x + αv) < 0 α sufficiently small. So k, g k (x + αv) < 0 α sufficiently small, so the movement is feasible. Also, given that f (x ) v > 0, then f (x + αv) > f (x ) for α sufficiently small, so x cannot be a local maximum.
42 PROOF OF KUHN TUCKER (CONTINUED 3) If J, then at least one k E for which g k (x )v = 0, so the point x + αv might not be feasible. Use Implicit Function Theorem to argue that points x that are (a) feasible, (b) arbitrarily close to x, and (c) the movement from x to x is arbitrarily close to v. Let K = J. Let S = { g(x ) : k J}. By CQ, S is independent. By CQ and Implicit Fcn Thm, equations g k (x ) = 0 k J implicitly define a C 1 function ψ : R (N K) R N s.t. k J and z R (N K), g k (ψ(z)) = 0, Dψ(z) has full rank, and ψ(0) = x That is, ψ gives a N K-dimensional surface consisting of all points near x for which the constraints in J hold with equality.
43 PROOF OF KUHN TUCKER (CONTINUED 4) Given that g k (ψ(z)) = 0 z R N K and all k J, by the Chain Rule, Dg k (x )Dψ(0)z = 0 z R N K and all k J. Let A = [D k ] k J, (K N). By CQ, A has full rank (K), so its null space has dimension N K. Since Dψ has full rank (N K), it maps onto the null space of A. Since v lies in the null space of A, z v R (N K) s.t. v = Dψ(0)z v. For sufficiently small α > 0, ψ(αz v ) gives such point x that fulfills (a), (b) and (c).
44 PROOF OF KUHN TUCKER (CONTINUED 5) We already know that k J, g k (ψ(αz v ) = 0 α > 0. For k E\J, by Chain Rule, Dg k (x )Dψ(0)z v = Dg k (x )v < 0, which implies that g k (ψ(αz v )) < g k (x ) = 0 for α sufficiently small. For k / E, g k (x ) < 0. By continuity, g k (ψ(αz v )) < 0 for α sufficiently small. Thus, x = ψ(αz) is feasible. By Chain Rule, Df (x )Dψ(0)z v = Df (x )v. Since f (x ) v > 0, Df (x )v > 0 Hence, for α sufficiently small, f (x) > f (x ).
45 USING THE KT CONDITIONS Bad news! There is no easy way of finding points x and multipliers λ that satisfy the KT conditions. Cookbook? Try solving the unconstrained problem first. If the solution satisfies the constraints, you re done. If it doesn t, make an educated guess of which of the constraints might be binding.
46 EXAMPLE 1 max f (x) = x 1/2 1 x 1/3 3 s.t. 4x 1 + 8x 2 + 3x 3 9 x 0 2 x 1/6 In the standard form: max f (x) = x 1/2 1 x 1/3 2 x 1/6 3 s.t. 4x 1 + 8x 2 + 3x x 0
47 EXAMPLE 1 i. None of the non-negativity constraints will bind. How do we know? If any of the x n = 0, then f (x) = 0. Take any feasible point, for instance, ( 1 4, 1 8, 1 3 ) plug it into the constraint: 3 9 = 6 0, which yields f (x) = ( 1 2 )2 ( 1 3 )1/6 > 0. ii. The constraint 4x 1 + 8x 2 + 3x 3 9 = 0 is binding. How do we know? Df (x) >> 0. From KT(2), we know that λ 2 = λ 3 = λ 4 = 0. Use KT(1) to find λ 1.
48 EXAMPLE 1 iii. Want to avoid messy calculations? Use the second useful theorem : apply a strictly increasing transformation to f (x), and solve the max problem. Let ˆf (x) = Ln(f (x)). This transformation is strictly increasing: D(Ln(y)) = 1 y > 0 (we already know that f (x) > 0) So solving the problem: max ˆf (x) = 1 2 ln x ln x ln x 3 s.t. 4x 1 + 8x 2 + 3x x 0 Yields the same solution as the original problem.
49 EXAMPLE 1 Use KT(1): 1 2x 1 1 3x 2 1 6x 3 4 = λ or 1 2λ 1 1 3λ 1 1 6λ 1 = 4x 1 8x 2 3x 3 Substituting in the constraint, we get 1 2λ λ λ 1 = 9, λ 1 = Substituting back, x = However, I haven t shown yet that x is a solution. So far, we only know that it satisfies the KT necessary conditions. To show that x is a solution, I need sufficient conditions.
50 EXAMPLE 2 max f (x) = x x x s.t. 4x 1 + 8x 2 + 3x 3 9 x 0 In standard form: max f (x) = x x x s.t. g 1 = 4x 1 + 8x 2 + 3x x 0 Now it s no longer obvious that the solution has x >> 0. What do we know now? (i) Df (x) >> 0, so the first constraint will bind: 4x 1 + 8x 2 + 3x 3 9 = 0.
51 EXAMPLE 2 Now start guessing! Guess 1: x >> 0? If only the first constraint binds, then by KT(2), λ 2 = λ 3 = λ 4 = x And by KT(1), x 2 +1 = λ x 3 +1 After some calculations we get x 1 x 2 = x 3 But this point is not in the feasible set! and λ 1 = 10 16
52 EXAMPLE 2 This bad guess can give us a clue of what the solution looks like. Guess 2: x 1 = x 2 = 0 Then, since the first constraint binds, x 3 = 3, so x = (0, 0, 3) 1 f (x 2 ) = 1 g 1 (x ) = 4 8, g 2 (x ) = 1 0, g 3 (x ) = Write f (x ) = λ 1 g 1 (x ) + λ 2 g 2 (x ) + λ 3 g 3 (x ) in matrix form: = λ 1 λ λ 4 3 and λ λ 2 = 1 > 0 2 λ 3 1 Set λ 4 = 0, and both KT(1) and KT(2) hold.
53 EXAMPLE 2 What would have happened if I had guessed that x 2 = x 3 = 0? x = ( 9 4, 0, 0), By KT(2), λ 2 = 0 By KT(1), λ 1 λ λ 4 So λ k 0 is violated. which is feasible, so we would expect KT to fail. [The sign of KT multipliers is important]. KT can catch bad guesses.
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