Convex Optimization and Machine Learning

Transcription

1 Convex Optimization and Machine Learning Mengliu Zhao Machine Learning Reading Group School of Computing Science Simon Fraser University March 12, 2014 Mengliu Zhao SFU-MLRG March 12, / 25

2 Introduction Formulation of binary SVM problem: Given training data set D = {(x i, y i ) x i R n, y i { 1, 1}, i = 1, 2,..., m} (1) We re trying to find the maximal-margin hyperplane, which can be described by its normal vector w which satisfies (b is some offset): minimize w 2 subject to y i (wx i b) 1 i = 1, 2,..., m (2) Comment We encounter a lot of constraint minimization problems in Machine Learning. Mengliu Zhao SFU-MLRG March 12, / 25

3 Why We Want Convex Problems? Mengliu Zhao SFU-MLRG March 12, / 25

4 Outline 1 Lagrange Dual Form 2 Dual Decomposition, Augmented Lagrangian and ADMM 3 SVM and Convex Optimization Mengliu Zhao SFU-MLRG March 12, / 25

5 Convex Optimization Problems General form of convex optimization problem is like following: minimize f 0 (x) subject to f i (x) 0, i = 1, 2,..., m h j (x) = 0, j = 1, 2,..., n (3) where f 0, f i are convex functions, h j are linear functions. Property The feasible set of a convex optimization problem is also convex. In other words, convex optimization problem is solving a convex function over a convex space. Mengliu Zhao SFU-MLRG March 12, / 25

6 General Constraint Problem with Lagrange Duality However, most constraint problems we optimize are not convex: minimize f 0 (x) subject to f i (x) 0, i = 1, 2,..., m h j (x) = 0, j = 1, 2,..., n (4) Lagrangian: L(x, λ, ν) = f 0 (x) + m λ i f i (x) + i=1 n ν j h j (x) (5) λ i (> 0), ν j are called Lagrangian multipliers or dual variables; the Lagrangian dual function is defined as: g(λ, µ) = inf x j=1 L(x, λ, ν) (6) Mengliu Zhao SFU-MLRG March 12, / 25

7 Geometric Explanation Primal Problem minimize x subject to x (7) (x ) (8) Mengliu Zhao SFU-MLRG March 12, / 25

8 Geometric Explanation Dual Problem x λ(x ) (9) λ =.1 :.1 : 2 (10) g(λ) = inf x { x λ(x ) λ =.1 :.1 : 2 (11) Mengliu Zhao SFU-MLRG March 12, / 25

9 Geometric Explanation Two Observations Observation (I) Dual function g(λ) is concave. maximize g(λ) subject to λ > 0 (12) is a convex optimization problem. Observation (II) Let p be the optimal value of the primal problem, then g(λ) p, λ (13) Mengliu Zhao SFU-MLRG March 12, / 25

10 Economic Explanation Company production cost f 0, with certain limits f i below a i (rules, resources): minimize f 0 (x) subject to f i (x) a i 0, i = 1, 2,..., m (14) However, if, the company can pay a fund rate of λ i > 0 to violate certain rules, which adds back to the total cost: g(λ) = inf λ {f 0(x) + i λ i (f i a i )} (15) In this case, the optimal value d for the company is the cost under the least favorable set of prices λ max g(λ). Mengliu Zhao SFU-MLRG March 12, / 25

11 Strong & Weak Duality How well does the dual problem approximate the original problem? 1 Weak Duality: optimal duality gap is always non-negative. 2 Strong Duality: duality gap is zero. Q: When does strong duality hold? Theorem (Slater s Theorem) p d 0 (16) p = d (17) D is feasible set. Assume the primal problem is convex: minimize f 0 (x) subject to f i (x) 0, i = 1, 2,..., m h j (x) = 0, j = 1, 2,..., n (18) If x relint D, and f i (x) < 0, i = 0, 1,..., m, then strong duality holds. Mengliu Zhao SFU-MLRG March 12, / 25

12 KKT Condition For constrained problem: minimize f 0 (x) subject to f i (x) 0, i = 1, 2,..., m h j (x) = 0, j = 1, 2,..., n (19) If x is the primal minimum, then it satisfies the following necessary condition: f 0 (x ) + m λ i f i (x ) + i=1 n ν j h j (x ) = 0 (20) j=1 Mengliu Zhao SFU-MLRG March 12, / 25

13 Outline 1 Lagrange Dual Form 2 Dual Decomposition, Augmented Lagrangian and ADMM 3 SVM and Convex Optimization Mengliu Zhao SFU-MLRG March 12, / 25

14 Dual Form, Then What? Once we get the dual problem, it s easy to solve, e.g., by gradient approach (dual ascent). Property If g(λ) is a convex (concave) function, then f (λ ) = 0 iff λ is the global minimizer (maximizer). Comment A lot of conditions need to be satisfied for a stable gradient method. Mengliu Zhao SFU-MLRG March 12, / 25

15 Dual Ascent for Solving Dual Problem Let s look at a simplified version of the constrained problem: Its dual form: Update x, λ at each iteration: minimize subject to f (x) Ax = b (21) max g(λ) = max {min L(x, λ)} (22) λ x L(λ, x) = f (x) + λ(ax b) (23) x k+1 = min L(x, λ k ) x (24) λ k+1 = λ k + α k+1 g(x k+1, λ k ) (25) Question What if we have a much more complex situation? Mengliu Zhao SFU-MLRG March 12, / 25

16 Dual Decomposition Suppose the problem is of high dimension, ˆx = (x, z), and f (ˆx) is separable: f (ˆx) = f 1 (x) + f 2 (z) (26) Aˆx b = (A 1 x b 1 ) + (A 2 z b 2 ) (27) Then we can do dual ascent on each dimension separately: Comment L 1 (x) = f 1 (x) + λ 1 (A 1 x b 1 ) (28) L 2 (x) = f 2 (x) + λ 2 (A 2 x b 2 ) (29) x k+1 = min L 1 (x, z k, λ k ) x (30) z k+1 = min L 2 (x k+1, z, λ k ) z (31) λ k+1 = λ k + α k+1 g(x k+1, z k+1, λ k ) (32) 1 Simple dual ascent is usually slow; Mengliu Zhao SFU-MLRG March 12, / 25

17 Alternative to Dual Ascent Augmented Lagrangian Primal problem: minimize subject to f (x) Ax = b (33) Dual problem: L(x, λ, θ) = f (x) + λ T (Ax b) + θ 2 Ax b 2 2 (34) Update by method of multipliers (fixed step): x k+1 := min L(x, λ k, θ) x (35) λ k+1 := λ k + θ(ax k+1 b) (36) Mengliu Zhao SFU-MLRG March 12, / 25

18 Method of Multipliers Comparing to dual ascent: 1 Good news: convergence under more relaxed conditions; 2 Bad news: dual decomposition no longer works (now we have quadratic terms)! Comment ADMM can help! Mengliu Zhao SFU-MLRG March 12, / 25

19 ADMM Alternating Direction Method of Multipliers minimize subject to f (x) + g(z) Ax + Bz = b (37) Its Lagrangian is: L θ (x, λ, z) = f (x) + g(z) + λ T (Ax + Bz b) + θ 2 Ax + Bz b 2 2 (38) ADMM scheme: x k+1 := min x L θ (x, z k, λ k ) z k+1 := min z L θ (x k+1, z, y k ) λ k+1 := λ k + θ(ax k+1 + Bz k+1 b) (39) Mengliu Zhao SFU-MLRG March 12, / 25

20 A Closer Look at ADMM Comment We need more convincing evidence that the scheme will work! The thing unnatual here is the new variable z. We ll check the KKT condition with the constraint problem above: g(z) + B T λ = 0 (40) We ll check if this could be satisfied by the ADMM scheme. Since z k+1 minimized L θ (x k+1, z, λ k ), then 0 = g(z k+1 + B T λ k + θb T (Ax k+1 + Bz k+1 b)) (41) = g(z k+1 + B T λ k (42) Which means the KKT condition is satisfied. Mengliu Zhao SFU-MLRG March 12, / 25

21 Outline 1 Lagrange Dual Form 2 Dual Decomposition, Augmented Lagrangian and ADMM 3 SVM and Convex Optimization Mengliu Zhao SFU-MLRG March 12, / 25

22 Dual Form of SVM Now let s come back to the constrained version of SVM model: minimize w 2 subject to y i (wx i b) 1 i = 1, 2,..., m (43) It s easy to convert it to Lagrangian dual form as following: max {min λ w,b { w λ i [1 y i (wx i b)]}} (44) Comment The formulation is too complex! We can do further to simplify it! Mengliu Zhao SFU-MLRG March 12, / 25

23 Dual Form of SVM Check KKT condition, taking 1-order derivative of w and b on Lagrangian function w λ i [1 y i (wx i b)]: w = i λ i y i x i (45) Replace them back in (44), we have: 0 = λ i y i (46) max g(λ) = max { λ i 1 λ λ 2 yi y j λ i λ j (x i ) T x j } s.t. λ i 0, i = 1, 2,..., m λi y i = 0 (47) Mengliu Zhao SFU-MLRG March 12, / 25

24 Summary 1 Lagrangian Duality, KKT condition 2 Dual Decomposition, Augmented Lagrangian, ADMM 3 Example using Lagrangian Duality on SVM Mengliu Zhao SFU-MLRG March 12, / 25

25 Thank You! Mengliu Zhao SFU-MLRG March 12, / 25