Lecture 19 Subgradient Methods. November 5, 2008

Transcription

1 Subgradient Methods November 5, 2008

2 Outline Lecture 19 Subgradients and Level Sets Subgradient Method Convergence and Convergence Rate Convex Optimization 1

3 Subgradients and Level Sets A vector s is a subgradient of a convex function f : R n R at ˆx dom f when subgradient inequality holds f(z) f(ˆx) + s T (z ˆx) for all z dom f We have interpreted the subgradient inequality in terms of a hyperplane in R n+1 supporting the epigraph epi f at (ˆx, f(ˆx)) Convex Optimization 2

4 Projecting the hyperplane and the epigraph on the set {(x, w) x R n, w = f(ˆx)} The subgradient inequality can be interpreted in terms of a hyperplane in R n supporting the level set L γ (f) at ˆx for γ = f(ˆx) For z in the level set L f(ˆx) (f) = {z f(z) f(ˆx)}, the subgradient inequality implies s T (z ˆx) 0 for all z L f(ˆx) (f) and the inequality is tight at z = ˆx. Convex Optimization 3

5 Another Interpretation of Subgradients A subgradient of f at any ˆx defines a hyperplane that cuts R n in two regions and one of them contains the level sets L γ (f) for all γ f(ˆx) The optimal set X lies entirely on one side of each such hyperplane Convex Optimization 4

6 Subgradient Method Constrained Convex Minimization Assumption 1 Assume that minimize subject to f(x) x X The function f(x) is convex and domf = R n The set X is nonempty closed and convex Subgradient Method Having the current iterate x k, a subgradient s k of f at x k, and a stepsize value α k > 0, the next iterate is given by x k+1 = P X [x k α k s k ] Polyak 1969, Ermoliev 1969, Shor 1985 Convex Optimization 5

7 Property ensuring that subgradient methods work Along the negative subgradient direction s k, there is a range of stepsize (0, ã), such that at every point x k αs k for α (0, α) one of the following two relations hold: The function value is decreased f(x k αs k ) < f(x k ) The distance to the optimal set is decreased dist(x k αs k, X ) < dist(x k, X ) Convex Optimization 6

8 Stepsize Choices Constant stepsize: α k = α for all k Diminishing stepsize: α k 0 and k=0 α k = + Dynamic stepsize for known f (relaxation step, Polyak s rule) Dynamic stepsize for unknown f where f lev k For example: α k = f(x k) f s k 2 α k = f(x k) fk lev s k 2 is an estimate of f based on the iterates available f lev k = min 0 i k f(x i ) δ with δ > 0 Convex Optimization 7

9 Lemma method Basic Iterate Relation Assumption 1 hold, and let {x k } be generated by a subgradient x k+1 = P X [x k α k s k ] with any stepsize rule. Then, for any x X and any k 0, Proof x k+1 x 2 x k x 2 2α k (f(x k ) f(x)) + α 2 k s k 2 Let x X and k be arbitrary. From the definition of the method: x k+1 x 2 = P[x k α k s k ] x 2 x k α k s k x 2 where the inequality follows from the nonexpansive property of the projection operation. Therefore, x k+1 x 2 x k x 2 2α k s T k (x k x) + α 2 k s k 2 By the subgradient inequality, we have s T k (x k x) f(x k ) f(x) for any x X, and thus the given relation follows Convex Optimization 8

10 Convergence for a Constant Stepsize Lecture 19 Theorem Let Assumption 1 hold and let f be finite. Assume that the subgradients {s k } are uniformly bounded, i.e., there is L > 0 such that s k L for all k Then, for the subgradient method with a constant stepsize α, we have lim inf k f(x k) f + αl2 2 Convex Optimization 9

11 Proof Barely different analysis than the one used for gradient projection method under assumption of bounded gradients. To arrive at a contradiction, assume that the relation does not hold, i.e., lim inf k f(x k) > f + αl2 2 Let f denote the limit inferior of f(x k ). Then, in view of the preceding inequality, there exists ɛ > 0 and K 1 such that f(x k ) 2ɛ + f + αl2 for all k K 2 Let y X be such that f(y) f + ɛ. Thus, f(x k ) f(y) ɛ + αl2 2 for all k K Convex Optimization 10

12 By using Basic Iterate Relation with α k = α and x = y, we obtain x k+1 y 2 x k y 2α (f(x k ) f(y)) + α 2 s k 2 x k y 2α ( x k y 2 2αɛ ɛ + αl2 2 ) + α 2 s k 2 for all k K where the last inequality follows from the subgradient boundedness. Hence, for all k K, x k+1 y 2 x k y 2 2αɛ x K y 2 2αɛ(k + 1 K) For sufficiently large k, the right hand side becomes negative - a contradiction. Comments: With the constant step, the convergence to f is within an error that depends on the stepsize α and the bound on subgradient norms Convex Optimization 11

13 Subgradient Method with Momentum Term The subgradient method x k+1 = P X [x k α k s k ] can be slow when the level sets are prolonged (ill-conditioned) A popular modification proposed by Camerini, Fratta and Maffioli 1975 x k+1 = P X [x k α k s k + β k (x k x k 1 )] heavy-ball method The term β k (x k x k 1 ) is the momentum term Convex Optimization 12

14 Subgradient with Dynamic Stepsize with Known f x k+1 = P X [x k α k s k ] α k = f(x k) f s k 2 Primarily useful when solving convex feasibility problem: g j (x) 0, j = 1,..., m and x X Lecture 19 The problem can be reduced to minimize max 1 j m g j (x) subject to x X or minimize mj=1 max{0, g j (x)} subject to x X This method provides insights into the behavior (and analysis) of a modification of the stepsize when f is replaced by an estimate Convex Optimization 13

15 Insights for Dynamic Stepsize using f Assume that Assumption 1 holds and the optimal set X is not empty. Let x / X be arbitrary, and let s be a subgradient of f at x Consider the half-space in R n given by S x = {z R n s T (z x) + f( x) f } The optimal set X is contained in S x The point x does not belong to S x The distance from x to the half-space S x is the distance from x to the hyperplane H x defining S x : H x = {z R n s T (z x) + f( x) = f } Convex Optimization 14

16 The distance from x to the hyperplane H x is determined by finding the projection z of x on H x : Consider a ray from x in direction s: { x ts t 0} The projection z is the intersection of the ray with the hyperplane: ts T s + f( x) = f t = f( x) f Hence the projection is: z = x f( x) f s 2 The distance is x z = f( x) f s s s 2 Convex Optimization 15

17 The method with dynamic step: x k+1 = P X [x k α k s k ], α k = f(x k ) f Can be written as x k+1 = P [ x k f(x k) f s k ] s k s k s k 2 Uses the distance from x k to H k = {z s T (z x k ) + f(x k ) = f } as a stepsize, dist(x k, H k ) = f(x k ) f s Moves along a normalized subgradient s k s k Convex Optimization 16

18 Convergence Result Lecture 19 Theorem Assume that Assumption 1 holds and X is nonempty. Then, {x k } generated by the subgradient method with the dynamic stepsize using f converges to an optimal solution. Proof Using the Basic Iterate Relation, we obtain for any x X x k+1 x 2 x k x 2 2α k (f(x k ) f ) + α 2 k s k 2 ( = x k x 2 α k s k 2 2 f(x k) f α s k 2 k = x k x 2 (f(x k) f ) 2 s k 2 The sequence {x k } is bounded, thus subgradient norms s k are uniformly bounded by some scalar L. Therefore, for all k and any x X, x k+1 x 2 x k x 2 (f(x k) f ) 2 (1) L 2 ) Convex Optimization 17

19 Eq. (1) implies that (f(x k ) f ) 2 k=0 L 2 x k x 2 x k+1 x 2, and (f(x k ) f ) 2 L 2 x 0 x 2 Lecture 19 Hence f(x k ) f. Since {x k } X is bounded, it has limit points that belong to X by closedness of X. By continuity of f and f(x k ) f, it follows that all limit points are optimal. We now show that {x k } has a unique limit point. To arrive at a contradiction, suppose there are at least two: say x and ˆx with x ˆx. Let {x m } M and {x k } K be subsequences converging to x and ˆx, respectively. By Eq. (1), it follows that for all m, k with m k x m x < x k x for all x X Let x = ˆx, and m M and k K. Then by taking limits, we obtain lim x m m ˆx lim x k ˆx k m M k K implying that x ˆx = 0 - a contradiction. Convex Optimization 18