Lecture 18: Optimization Programming

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1 Fall, 2016

2 Outline Unconstrained Optimization 1 Unconstrained Optimization 2 Equality-constrained Optimization Inequality-constrained Optimization Mixture-constrained Optimization 3 Quadratic Programming (QP), Linear Programming (LP)

3 Unconstrained Optimization Consider the minimization problem min x f (x), where x R d, and f is continuously differentiable. Necessary condition for local minimums If x is a local minimum of f (x), then f (x ) = 0. A point satisfying the necessary condition is called a stationary point, which can be a local minimum, local maximum, or saddle point. Sufficient conditions for x be a local minimum are: f (x ) = 0, 2 f (x ) is positive definite.

4 Equality- Inequality- Mixture- Consider the minimization problem min x f (x) subject to g i (x) 0, i = 1,, m. h j (x) = 0, j = 1,, k. Assume x R d. f : R d R is called the objective function g i : R d R, i = 1,, m are inequality constraints h j : R d R, j = 1,, k are equality constraints. Assume f, g i s, h j s are all continuously differentiable.

5 Single Equality Constraint Equality- Inequality- Mixture- Consider the minimization problem Introduce the Lagrange function: Lagrange extreme value theory: min x f (x) (1) subject to h(x) = 0. (2) L(x, λ) = f (x) + λh(x) If x is a local minimum under the constraint, then there exist some λ such that x L(x, λ ) = 0, λ L(x, λ ) = 0.

6 Interpretations Unconstrained Optimization Equality- Inequality- Mixture- Above conditions imply that (x, λ ) is a stationary point of L. The above equations are equivalent to x f (x ) + λ x h(x ) = 0, h(x ) = 0. Note the first condition gives d equations.

7 Multiple Equality Constraints Equality- Inequality- Mixture- Consider the minimization problem min x f (x) (3) subject to h j (x) = 0, j = 1,, k. (4) Introduce multiple Lagrange multipliers: λ = (λ 1,, λ k ) L(x, λ) = f (x) + k λ j h j (x) Lagrange extreme theory: If x is a local minimum under the constraint, then there exist some λ = (λ 1,, λ k ) such that j=1 x L(x, λ ) = 0, λ L(x, λ ) = 0. In other words, (x, λ ) is a stationary point of L.

8 Equality- Inequality- Mixture- Example: Entropy Maximization Problem Consider the minimization problem n min p i log(p i ), subject to p 1,,p n i=1 i=1 n p i = 1. i=1 The Lagrange function is n n L(p, λ) = p i log(p i ) + λ( p i 1). L p i = log(p i ) λ = 0, log(p i ) = 1 λ, i = 1,, n. Together with n i=1 p i = 1, we have p 1 = = p n = 1 n. This problem is known as maximizing information entropy. i=1

9 Generalized Lagrange Methods Consider the minimization problem Equality- Inequality- Mixture- min x f (x) subject to g i (x) 0, i = 1,, m. Introduce multiple non-negative Lagrange multipliers: µ 1 0,, µ m 0. m L(x, µ) = f (x) µ i g i (x). µ i s are also called dual variables or slack variables. Correspondingly, we call x as primal variables. We would like to minimize L with respect to x and maximize L with respect to µ. i=1

10 KKT Necessary Optimality Conditions Equality- Inequality- Mixture- If x is a local minimum under the above inequality constraints, then there exist some µ = (µ 1,, µ m) such that Stationarity: x L(x, µ ) = 0, i.e., x f (x ) m µ k xg i (x ) = 0. i=1 Primal feasibility: g i (x ) 0 for i = 1,, m. Dual feasibility: µ i 0 for i = 1,, m. Complementary slackness: KKT = Karush-Kuhn-Tucker µ i g i (x ) = 0, i = 1,, m.

11 Sufficient Optimality Conditions Equality- Inequality- Mixture- The above necessary conditions are also sufficient for optimality, if f, g i s are continuously differentiable f and g i s are convex functions. Under this case, There exists a solution x satisfying the KKT condition, x is actually a global optimum. Examples: SVM problems

12 Alternative Constraints Equality- Inequality- Mixture- Consider the minimization problem min x f (x) subject to g i (x) 0, i = 1,, m. Introduce multiple non-negative Lagrange multipliers: µ 1 0,, µ m 0. m L(x, µ) = f (x) + µ i g i (x). µ i s are also called dual variables or slack variables. Correspondingly, we call x as primal variables. We would like to minimize L with respect to x and maximize L with respect to µ. i=1

13 KKT Necessary Optimality Conditions Equality- Inequality- Mixture- If x is a local minimum under the above inequality constraints, then there exist some µ = (µ 1,, µ m) such that Stationarity: x L(x, µ ) = 0, i.e., x f (x ) + m µ k xg i (x ) = 0. i=1 Primal feasibility: g i (x ) 0 for i = 1,, m. Dual feasibility: µ i 0 for i = 1,, m. Complementary slackness: µ i g i (x ) = 0, i = 1,, m.

14 Generalized Lagrange Methods Consider the minimization problem Equality- Inequality- Mixture- min x f (x) subject to g i (x) 0, i = 1,, m. h j (x) = 0, j = 1,, k. Introduce two sets of Lagrange multipliers: Define the Lagrange function: µ 1 0,, µ m 0, λ 1,, λ k. L(x, µ, λ) = f (x) m µ i g i (x) + i=1 k λ j h j (x) j=1 µ i s and λ j s are dual variables or slack variables.

15 KKT Necessary Optimality Conditions Equality- Inequality- Mixture- If x is a local minimum under the above inequality constraints, then there exist some (µ, λ ) such that Stationarity: x L(x, µ, λ ) = 0, i.e., x f (x ) Primal feasibility: m µ i x g i (x ) + i=1 k λ j x h j (x ) = 0. j=1 g i (x ) 0, i = 1,, m; h j (x ) = 0, j = 1,, k. Dual feasibility: µ i 0 for i = 1,, m. Complementary slackness: µ i g i (x ) = 0, i = 1,, m.

16 Sufficient Optimality Conditions Equality- Inequality- Mixture- The above necessary conditions are also sufficient for optimality, if f, g i s are continuously differentiable. f and g i s are convex functions, h j s are linear constraints. Under this case, There exists a solution x satisfying the KKT condition, x is actually a global optimum.

17 Quadratic Programming (QP) Quadratic Programming (QP) The objective function is quadratic in x The constraints are linear in x. Consider the minimization problem min x f (x) = 1 2 xt Qx + c T x, subject to Ax b, Ex = d. where Q is a symmetric matrix of d d, A is a matrix of m d, E is a matrix of k d, c R d, b R m, and d R k. Example: SVM

18 Properties of QP Unconstrained Optimization Quadratic Programming (QP) If Q is a non-negative definite matrix, then f (x) is convex. If Q is a non-negative definite, the above QP has a global minimizer if there exists at least one x satisfying the constraints and f (x) is bounded below in the feasible region. If Q is a positive definite, the above QP has a unique global minimizer. If Q = 0, the problem becomes a linear programming (LP). From optimization theory, If Q is non-negative definite, the necessary and sufficient condition for a point x to be a global minimizer is for it to satisfy the KKT condition.

19 Solving QP Unconstrained Optimization Quadratic Programming (QP) If there are only equality constraints, then the QP can be solved by a linear system. In general, a variety of methods for solving the QP are commonly used, including Methods: ellipsoid, interior point, active set, conjugate gradient, etc Software packages: Matlab, CPLEX, AMPL, GAMS, etc. For positive definite Q, the ellipsoid method solves the problem in polynomial time. In other words, the running time is upper bounded by a polynomial in the size of the input for the algorithm, i.e., T = O(d k ) for some k.

20 Dual Problem for QP Quadratic Programming (QP) Consider the minimization problem The Lagrange function. min f (x) = 1 x 2 xt Qx, subject to Ax b. L(x, λ) = f (x) + λ T (Ax b). We need to minimize L with respect to x and maximize L with respect to λ. The dual function is defined as a function of dual variables only by solving x first with any fixed λ: G(λ) = inf x L(x, λ).

21 Dual Problem Derivation for QP Quadratic Programming (QP) With fixed λ, we solve the equation L x (x, λ) = 0 and obtain x = Q 1 A T λ. Plug this into L, we get the dual function The dual problem of the QP is G(λ) = 1 2 λt AQ 1 A T b T λ. max G(λ) = 1 λ 2 λt AQ 1 A T b T λ, subject to λ 0. We first find λ = arg max λ G(λ), and then compute x = Q 1 A T λ. The idea is similar to the profile method.

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