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LECTURE SLIDES ON CONVEX ANALYSIS AND OPTIMIZATION BASED ON 6.253 CLASS LECTURES AT THE MASS. INSTITUTE OF TECHNOLOGY CAMBRIDGE, MASS SPRING 2012 BY DIMITRI P. BERTSEKAS http://web.mit.edu/dimitrib/www/home.html Based on the book “Convex Optimization Theory,” Athena Scientific, 2009, including the on-line Chapter 6 and ...
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⇤ ⇤ x | • • which f and C are convex − − They are continuous problems They are nice, and have beautiful and intu- itive structure However, convexity permeates all of optimiza- • tion, including discrete problems Principal vehicle for continuous-discrete con- • nection is duality: − − The dual problem of a discrete prob...
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(y) Primal Description Values f (x) Dual Description Crossing points f ∗(y) 7 FENCHEL PRIMAL AND DUAL PROBLEMS f 1 (y) f 1 (y) +f 2 ( y) f 2 ( y) f1(x) Slope y f2(x) x x Primal Problem Description Vertical Distances Dual Problem Description Crossing Point Dierentials Primal problem: f1(x) +f 2(x) min x ⇤ ⌅ Dual problem...
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% Max Crossing %#0()1*22&'3(,*&'-(4/ Point q 0! 0 u 7 0 ! u 7 Max Crossing %#0()1*22&'3(,*&'-(4/ Point q (b) "5$ 6 M % 9 M % (a) "#$ w . Min Common %&'()*++*'(,*&'-(./ Point w Max Crossing Point q %#0()1*22&'3(,*&'-(4/ 0 ! (c) "8$ u 7 All of duality theory and all of (convex/concave) • minimax theory can be developed/e...
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). 13 MODERN VIEW OF CONVEX OPTIMIZATION • • Traditional view: Pre 1990s LPs are solved by simplex method NLPs are solved by gradient/Newton meth- ods Convex programs are special cases of NLPs − − − LP CONVEX NLP Simplex Duality Gradient/Newton Modern view: Post 1990s − − − LPs are often solved by nonsimplex/convex met...
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, 2009, including the on-line Chapter 6 and supplementary material at http://www.athenasc.com/convexduality.html Additional book references: − − − Rockafellar, “Convex Analysis,” 1970. Boyd and Vanderbergue, “Convex Optimiza- tion,” Cambridge U. Press, 2004. (On-line at http://www.stanford.edu/~boyd/cvxbook/) Bertseka...
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You can do your term paper on an applica- tion area 18 A NOTE ON THESE SLIDES These slides are a teaching aid, not a text Don’t expect a rigorous mathematical develop- • • ment The statements of theorems are fairly precise, • but the proofs are not Many proofs have been omitted or greatly ab- • breviated Figures are me...
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convex if � C, αx + (1 α)y  Operations that preserve convexity − ⌘ ⌘  x, y C, [0, 1] α ⌘ − Intersection, scalar multiplication, vector sum, closure, interior, linear transformations Special convex sets: − − Polyhedral sets: Nonempty sets of the form x { | a�jx ⌥ bj, j = 1, . . . , r } (always convex, closed, not alwa...
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-VALUED FUNCTIONS f (x) !"#$ Epigraph 01.23415 f (x) !"#$ Epigraph 01.23415 dom(f ) %&'()#*!+',-.&' Convex function x # x # dom(f ) /&',&'()#*!+',-.&' Nonconvex function The epigraph of a function f : X • the subset of n+1 given by � [ ] is , ⇣ −⇣ ◆→ epi(f ) = (x, w) x | ⌘ X, w ⌘ � , f (x) w ⌥ The effective domain of f ...
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) x | � ⇥ (ii) • (xk, wk) ✏ → (iii): Let (xk, wk) (x, w). Then f (xk) ⌦ wk, and epi(f ) with ⇤ ⌅ ⌥ w so (x, w) epi(f ) x. Then ⌘ V⇥ and xk → → (x, ⇤), so (x, ⇤) f (x) lim inf f (xk) k ⌥ ✏ ⌘ (iii) • (xk, ⇤) epi(f ), and x ⌥ ⌃ (i): Let xk} ⌦ { epi(f ) and (xk, ⇤) V⇥. ⌘ (ii): If xk → • xk}K → consider subsequence - contr...
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ˆ have the same epigraph, and both are not closed. But f is lower-semicon- tinuous while fˆ is not. Note that: • − If f is lower semicontinuous at all x it is not necessarily closed If f is closed, dom(f ) is not necessarily closed dom(f ), ⌘ − Proposition: Let f : X ] be a func- • tion. If dom(f ) is closed and f is l...
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= f (Ax) ] given by , ⇣ where A is an m ⇤ if f is convex (respectively, closed). n matrix is convex (or closed) (c) Consider fi : ⇣ is any index set. The function g : given by −⇣ → � ( , n ], i ⌘ n I, where I ] ( , −⇣ ⇣ → � g(x) = sup fi(x) i ⌦ is convex (or closed) if the fi are convex (respec- tively, closed). I 28◆ ...
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− x) f (x) − ⇥ f (x) + (z f (x) x)⇥ − x x + (z x) z − (b) 31 OPTIMALITY CONDITION n Let C be a nonempty convex subset of • let f : an open set that contains C. Then a vector x⇤ ⌘ minimizes f over C if and only if n and be convex and differentiable over C → � � � f (x⇤)�(x x⇤) 0, ≥ − ∇ x ⌘  C. Proof: If the condition ho...
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2 x � C (called the projection of z on − � • ⌘ � (a) For every z imum of over all x C). ⌘ (b) x⇤ is the projection of z if and only if (x − x⇤)�(z x⇤) 0, ⌥ − x ⌘  C Proof: (a) f is strictly convex and has compact level sets. (b) This is just the necessary and su⌅cient opti- mality condition f (x⇤)�(x x⇤) 0, ≥ − ∇ x ⌘ ...
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 From the preceding result, f is convex. f (x), x)�∇ x, y C ⌘ (b) Similar to (a), we have f (y) > f (x) + (y f (x) for all x, y x)�∇ the preceding result. − C with x = y, and we use ⌘ (c) By contradiction ... similar. 34◆ ✓ CONVEX AND AFFINE HULLS Given a set X n: ⌦ � • • vector of the form m 0, and m i=1 αi = 1. � A...
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) Every x = 0 in cone(X) can be represented as a positive combination of vectors x1, . . . , xm from X that are linearly independent (so m n). ⌥ (b) Every x / X that belongs to conv(X) can ⌘ be represented as a convex combination of vectors x1, . . . , xm from X with m n + 1. ⌥ 36✓ PROOF OF CARATHEODORY’S THEOREM (a) ...
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� xi X � ⌘ dory, a sequence n+1 αkxk i i i=1 n+1 i=1 αk , and in conv(X) can , where for all k and i = 1. Since the � (αk 1, . . . , αk n+1, x1, . . . , xn+1) k k is bounded, it has a limit point ⇤ ⌅ (α1, . . . , αn+1, x1, . . . , xn+1) , ⇤ which must satisfy X for all i. xi The vector ⌘ ⌅ n+1 i=1 α = 1, and i αi 0, ≥ ...
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case where x C with xk → xk} ⌦ • • { C: See the figure. / C: Take sequence ⌘ x. Argue as in the figure. 40 ADDITIONAL MAJOR RESULTS • Let C be a nonempty convex set. (a) ri(C) is a nonempty convex set, and has the same a⌅ne hull as C. (b) Prolongation Lemma: x ri(C) if and only if every line segment in C having x as one ...
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) ≥ αf (x) + (1 − α)f (x), and since f (x) > f (x⇤), we must have f (x⇤) > f (x) - a contradiction. Q.E.D. Corollary: A nonconstant linear function can- • not attain a minimum at an interior point of a convex set. 42◆ CALCULUS OF REL. INTERIORS: SUMMARY The ri(C) and cl(C) of a convex set C “differ • very little.” − − ...
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(C) . ⌘ � ⇥ x x C The proof of ri(C) = ri cl(C) is similar. � ⇥ 44 LINEAR TRANSFORMATIONS Let C be a nonempty convex subset of n matrix. • let A be an m n and � ⇤ (a) We have A · ri(C) = ri(A C). (b) We have A cl(A · if C is bounded, then A cl(C) ⌦ C). Furthermore, · C). cl(C) = cl(A · · · Proof: (a) Intuition: Spheres...
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) ⌦ If one of C1 and C2 is bounded, then cl(C1) + cl(C2) = cl(C1 + C2) (b) We have ri(C1) ri(C2) ⌫ ri(C1 ⌫ ⌦ C2), cl(C1 C2) ⌫ ⌦ cl(C1) ⌫ cl(C2) If ri(C1) ⌫ ri(C2) = Ø, then ri(C1 ⌫ C2) = ri(C1) ⌫ ri(C2), cl(C1 C2) = cl(C1) cl(C2) ⌫ ⌫ Proof of (a): C1 + C2 is the result of the linear transformation (x1, x2) x1 + x2. ◆→ ...
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ri(C) , � . For every x ⌘ Mx ⌫ ri(C) = ri(Mx C) = (x, y) y | ⌘ ri(Cx) . ⌫ Combine the preceding two equations. ⇤ Q.E.D.⌅ 47✓ CONTINUITY OF CONVEX FUNCTIONS If f : n � → � • e4 = ( is convex, then it is continuous. 1, 1) yk e1 = (1, 1) xk xk+1 0 e3 = ( 1, 1) zk e2 = (1, 1) Proof: We will show that f is continuous at 0....
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] is , ⇣ −⇣ n � → [ , ⇣ −⇣ epi(cl f ) = cl epi(f ) The convex closure of f is� the function ⇥ epi(clˇ f ) = cl conv epi(f ) clˇ f with Proposition: For any�f : X� ◆→ [ ⇥⇥, −⇣ ] ⇣ • • inf f (x) = inf (cl f )(x) = inf (clˇ f )(x). x ⌦ ⌦� ⌦� X x x n n Also, any vector that attains the infimum of f over X also attains the i...
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finitely along d, we never cross the relative boundary of C to points outside C: x + αd C, ⌘ x C,  ⌘ α  ≥ 0 Recession Cone RC C d x + d 0 x Recession cone of C (denoted by RC): The set • of all directions of recession. RC is a cone containing the origin. • 51 RECESSION CONE THEOREM • Let C be a nonempty closed convex ...
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., let zk = x + kd, (z dk = k − zk − � x) x � � d � We have � + = d d dk d � � x zk − � x zk � − � � � d and x + dk → so dk → and closedness of C to conclude that x + d x x ⌅ � x + d. Use the convexity x zk − � � x ⌅ zk � − � x − zk � − � x − zk − x x 1, � , C. ⌘ 0, 53✓ LINEALITY SPACE The lineality space of a convex ...
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�� ⇤ ⇤ ⌅ These are the directions of recession of f . epi(f) ! “Slice” {(x,!) | f(x) " !} 0 Recession Cone of f Level Set V! = {x | f(x) " !} 55 RECESSION CONE OF LEVEL SETS Proposition: Let f : convex function ( n ] be a closed ⇣ � and consider the level sets −⇣ → , f (x) ⇤ , where ⇤ is a scalar. Then: • proper V⇥ = x...
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(a)-(d). • This behavior is independent of the starting dom(f ). • point x, as long as x ⌘ 57 RECESSION CONE OF A CONVEX FUNCTION For a closed proper convex function f : • , ( −⇣ level sets V⇥ = x | cession cone of f , and is denoted by Rf . ], the (common) recession cone of the nonempty ⇤ , ⇤ , is the re- f (x) ⌘ � ⇣ ...
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+ αd) α − f (x) Thus rf (d) is the “asymptotic slope” of f in the • direction d. In fact, rf (d) = lim α ⌃ ∇ f (x + αd)�d, x, d n ⌘ �  if f is differentiable. Calculus of recession functions: • rf1+ ··· +fm(d) =r f1(d) + · · · + rfm (d), rsupi I fi(d) = sup rfi(d) i I ⌦ 2 59◆ LOCAL AND GLOBAL MINIMA Consider minimizin...
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• pact if the level sets of f are compact. (An Extension of the) Weierstrass’ Theo- • rem: The set of minima of f over X is nonempty and compact if X is closed, f is lower semicontin- uous over X, and one of the following conditions holds: (1) X is bounded. (2) Some set x and bounded. ⌘ ⇤ (3) For every sequence , we ha...
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the It follows recession cone of X ⇤, when X ⇤ = Ø. that X ⇤ is nonempty and compact if and only if RX . Q.E.D. Rf = ⌫ 0 ⌫ { } 62◆ ✓ ✓ ✓ EXISTENCE OF SOLUTION, SUM OF FNS Let fi : n ( , ], i = 1, . . . , m, be closed ⇣ • proper convex functions such that the function −⇣ → � f = f1 + + fm · · · is proper. Assume that a...
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follows: Does a function f : minimum over a set X? � n ( , ⇣ −⇣ → ] attain a This is true if and only if Intersection of nonempty x X | ⌘ f (x) ⇤k ⌥ is nonempty. ⇤ ⌅ Level Sets of f X Optimal Solution 65◆ ROLE OF CLOSED SET INTERSECTIONS II If C is closed and A is a matrix, closed? is A C Nk x Ck C y yk+1 yk AC • If C...
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poly- • hedral. To be shown later by a more refined method. 67 ROLE OF CLOSED SET INTERSECTIONS III Let F : n+m ( , ] be a closed proper ⇣ • convex function, and consider −⇣ → � f (x) = inf F (x, z) ⌦� m z If F (x, z) is closed, is f (x) closed? Critical question in duality theory. − 1st fact: If F is convex, then f is ...
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z) = ⌦� z ✏ if x >0, if x = 0, if x <0, 0 1 ⇣ is not closed. 69 PARTIAL MINIMIZATION THEOREM Let F : ( • convex function, and consider f (x) = inf z −⇣ → ⇣ � n+m , ] be a closed proper m F (x, z). ⌦� Every set intersection theorem yields a closed- • ness result. The simplest case is the following: Preservation of Close...
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sets in the sequence Once the appropriately refined set intersection • theory is developed, sharper results relating to the three questions can be obtained The remaining slides up to hyperplanes sum- • marize this development as an aid for self-study using Sections 1.4.2, 1.4.3, and Sections 3.2, 3.3 71 ASYMPTOTIC SEQUE...
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ractive Set Sequence %&'()*+,&-+./* %0'(123,*+,&-+./* (b) Nonretractive Set Sequence A closed halfspace (viewed as a sequence with • identical components) is retractive. Intersections and Cartesian products of retrac- • tive set sequences are retractive. A polyhedral set is retractive. Also the vec- • tor sum of a conv...
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k) { ⌫ corresponding to d Proof: The set of common directions of recession of C k is RX R. For any asymptotic sequence xk} (1) xk − (2) xk − So C k { } X (because X is retractive) Ck (because d is retractive. ⌫ L) RX R: ⌘ ⌘ ⌘ ⌘ d d 75 NEED TO ASSUME THAT X IS RETRACTIVE X X Ck+1 Ck Ck+1 Ck Consider ⌫ k=0 C k, with C k ...
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� Q.E.D. 77 CLOSURE UNDER LINEAR TRANSFORMATION Let C be a nonempty closed convex, and let A • be a matrix with nullspace N (A). (a) A C is closed if RC N (A) LC. ⌫ ⌦ C) is closed if X is a retractive set (b) A(X and ⌫ y � ⌅ Proof: (Outline) Let We prove RX RC N (A) LC, ⌫ ⌫ ⌦ A C with yk → { ⌫ k=0Ck = Ø, where Ck = C ⌫...
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, m. Then C1 + set. · · · · · · � ⌘ Special Case: If C1 and − C2 are closed convex RC2 = 0 . • sets, then C1 C2 is closed if RC1 ⌫ − } Proof: The Cartesian product C = C1 Cm is closed convex, and its recession cone is RC = RCm. Let A be defined by RC1 ⇤ · · · ⇤ ⇤ · · · ⇤ { A(x1, . . . , xm) = x1 + + xm · · · A C = C1 + ...
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2  ⌘ C2, or a�x2 b ⌥ ⌥ a�x1, x1  ⌘ C1, x2  ⌘ C2 If x belongs to the closure of a set C, a hyper- • plane that separates C and the singleton set is said be supporting C at x. x } { 81 VISUALIZATION Separating and supporting hyperplanes: • a C2 C1 (a) a C x (b) A separating x a�x = b • | C1 and C2 is called strictly s...
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.E.D. → ⇣ 83 SEPARATING HYPERPLANE THEOREM n. Let C1 and C2 be two nonempty convex subsets • If C1 and C2 are disjoint, there exists a of hyperplane that separates them, i.e., there exists a vector a = 0 such that � a�x1 ⌥ a�x2, x1  ⌘ C1,  x2 ⌘ C2. Proof: Consider the convex set C1 − C2 = x2 { − x1 | x1 ⌘ C1, x2 C2 }...
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C2 a strictly separating hyperplane without C1 being closed. − − 85 LECTURE 7 LECTURE OUTLINE Review of hyperplane separation Nonvertical hyperplanes Convex conjugate functions Conjugacy theorem Examples • • • • • Reading: Section 1.5, 1.6 86 ADDITIONAL THEOREMS Fundamental Characterization: The clo- • n is the sure of...
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, ⇥) is � It intersects the (n+1)st axis at ξ = (µ/⇥)�u+w, • where (u, w) is any vector on the hyperplane. w (µ, ) (u, w) µ u + w 0 Nonvertical Hyperplane (µ, 0) Vertical Hyperplane u A nonvertical hyperplane that contains the epi- •graph of a function in its “upper” halfspace, pro- vides lower bounds to the function v...
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spaces as per (a). 90✓ CONJUGATE CONVEX FUNCTIONS Consider a function f and its epigraph • Nonvertical hyperplanes supporting epi(f ) Crossing points of vertical axis ◆→ f (y) = sup x�y x n ⌦� ⇤ 0 f x) , ( − y n. ⌘ � ⌅ y, 1) ( f (x) Slope = y x inf x ⇥⇤ n{ f (x) x�y } = f (y) For any f : [ • function is defined by n � ...
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, − y n, ⌘ � ⌦� ⇤ ⌅ note that f is convex and closed . Reason: epi(f ) is the intersection of the epigraphs • of the linear functions of y as x ranges over x�y − f (x) n. � Consider the conjugate of the conjugate: • • f (x) = sup n y ⌦� ⇤ y�x − f (y) , x n. ⌘ � ⌅ f is convex and closed. Important fact/Conjugacy theorem...
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��� ⇤ (a) We have y�x − f (y) , x n ⌘ � ⌅ f (x) ≥ f (x), x  ⌘ � n (b) If f is convex, then properness of any one of f , f , and f implies properness of the other two. (c) If f is closed proper and convex, then f (x) = f (x), (d) If clˇ f (x) > for all x −⇣ clˇ f (x) = f (x), x  ⌘ � n n, then ⌘ � x n ⌘ �  95◆ PROOF ...
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are x�y f ing point of the strictly sep. hyperplane. Hence (x), and lie st�rictly ab⇥ove and� below the⇥ cross- − f (x) and x�y − the fact f x�y f (x) > x�y f . Q.E.D. − ≥ f (x) − 96 A COUNTEREXAMPLE A counterexample (with closed convex but im- • proper f ) showing the need to assume properness in order for f = f : f ...
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� y�x f (y) inf x ⇥⇤ n{ f (x) x�y } = f (y) 99 A FEW EXAMPLES • • lp and lq norm conjugacy, where 1 p + 1 q = 1 f (x) = 1 n p i=1 ⌧ xi p, | | f (y) = 1 q n i=1 ⌧ yi q | | Conjugate of a strictly convex quadratic f (x) = x�Qx + a�x + b, 1 2 f (y) = (y 1 2 a)�Q− 1(y a) b. − − − Conjugate of a function obtained by inverti...
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A CONE - POLAR CONE The conjugate of the indicator function ⌅C is • the support function, ↵C(y) = supx C y�x. ⌦ If C is a cone, • ↵C(y) = 0 ⇣ � y�x 0, if otherwise ⌥ x ⌘  C, i.e., ↵C is the indicator function ⌅C⇤ of the cone C ⇤ = y { | y�x ⌥ 0, x C } ⌘  This is called the polar cone of C. By the Conjugacy Theorem th...
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COMMON / MAX CROSSING PROBLEMS • • We introduce a pair of fundamental problems: Let M be a nonempty subset of n+1 � (a) Min Common Point Problem: Consider all vectors that are common to M and the (n + 1)st axis. Find one whose (n + 1)st compo- nent is minimum. (b) Max Crossing Point Problem: Consider non- vertical hype...
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� w + µ�u, (u, w) M ⌘  Max crossing problem is to maximize ξ subject to ξ n, or , µ inf (u,w) w + µ�u ⌥ M { ⌦ } ⌘ � maximize q(µ) =↵ inf (u,w) ⌦ M{ w + µ�u } subject to µ n. ⌘ � 105 GENERIC PROPERTIES – WEAK DUALITY Min common problem inf w M ⌦ Max crossing problem (0,w) • • maximize q(µ) =↵ inf (u,w) ⌦ M{ w + µ�u } s...
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inf p(u) | { w ⌅ } w+µ�u , } { and finally q(µ) = inf u m ⌦� (µ, 1) p(u) p(u) +µ �u ⇤ ⌅ M = epi(p) w = p(0) q = p(0) 0 u µ) q(µ) = p( Thus, q(µ) = p ( − − µ) and • q⇤ = sup q(µ) = sup 0 ( µ) p ( µ) = p (0) · − − − n µ ⌦� n µ ⌦� ⇤ ⌅ 107◆ GENERAL OPTIMIZATION DUALITY Consider minimizing a function f : n [ −⇣ � ] be a fun...
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q(µ) = r ⌦� µ inf F (0, r µ ⌦� − − µ) = inf F (0, µ), r µ ⌦� − and weak duality has the form w⇤ = inf F (x, 0) ⌦� x n ≥ − inf F (0, µ) = q r µ ⌦� ⇤ 108◆ ◆ CONSTRAINED OPTIMIZATION Minimize f : over the set n � C = → � X x ⌘ n and g : ⇤ | n � 0 , ⌥ r. ⌅ g(x) → � where X ⌦ � Introduce a “perturbed constraint set” • • Cu...
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). Then • • q(µ) = inf u ⌥ p(u ) + µ⇧u r r ⇤ , x u = = ⌥ ⌥ inf x ⌥ −∞ � inf ⌅f (x) + µ⇧u X, g(x) u ⇤ X L(x, µ)⇤ if µ 0, ⌅ ⌥ otherwise. 110 LINEAR PROGRAMMING DUALITY Consider the linear program • minimize c�x subject to a�jx ≥ bj, j = 1, . . . , r, where c For µ • ⌘ � ≥ n, aj n, and bj , j = 1, . . . , r. ⌘ � ⌘ � 0, ...
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Point w M 6 % . w Min Common Point w %&'()*++*'(,*&'-(./ M % M % Max Crossing %#0()1*22&'3(,*&'-(4/ Point q 0! 0 u 7 0 ! u 7 (a) "#$ w . Min Common %&'()*++*'(,*&'-(./ Point w Max Crossing Point q %#0()1*22&'3(,*&'-(4/ Max Crossing %#0()1*22&'3(,*&'-(4/ Point q (b) "5$ 6 M % 9 M % 0 ! (c) "8$ u 7 112 REVIEW OF THE MC/M...
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113 MINIMAX PROBLEMS Given φ : X consider Z ◆→ � ⇤ minimize , where X n, Z m ⌦ � ⌦ � sup φ(x, z) z ⌦ subject to x X Z ⌘ or • • • maximize inf φ(x, z) x subject to z Z. X ⌦ ⌘ Some important contexts: Constrained optimization duality theory Zero sum game theory − − We always have sup inf φ(x, z) z x X Z ⌦ ⌦ ⌥ inf x ⌦ X z...
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gives aij • to the 2nd. Mixed strategies are allowed: The two players • select probability distributions x = (x1, . . . , xn), z = (z1, . . . , zm) over their possible choices. Probability of (i, j) is xizj, so the expected • amount to be paid by the 1st player x�Az = aijxizj i,j ⌧ where A is the n m matrix with elemen...
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, z) z x X Z ⌥ ⌥ By the minimax inequality, the above holds as an equality throughout, so the minimax equality and Eq. (*) hold. Conversely, if Eq. (*) holds, then sup inf ⌅(x, z) = inf ⌅(x, z⇥) z Z ⌥ x X ⌥ ⇤ ⌅(x⇥, z⇥) x X ⌥ sup ⌅(x⇥, z) = inf sup ⌅(x, z) x z X z Z Z ⌥ ⌥ ⇤ ⌥ Using the minimax equ., (x⇤, z⇤) is a saddle...
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, z), z z ⌦ ⌦� Z m w⇤ = p(0) = inf ⌦ The dual function can be shown to be sup (clˆ φ)(x, z). ⌦� X z x m • so • q(µ) = inf (clˆ φ)(x, µ), X x ⌦ µ  ⌘ � m so if φ(x, · ) is concave and closed, w⇤ = inf ⌦ x X z m ⌦� sup φ(x, z), q⇤ = sup inf φ(x, z) m x X z ⌦� ⌦ 119◆ PROOF OF FORM OF DUAL FUNCTION Write p(u) = inf x p xX...
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X p⇤ x( µ) ⌅ m X u ⌦� − = inf (clˆ φ)(x, µ) ⌅ − ⇤ ⌦ m m u ⌦� = inf u ⌦� inf x ⌦ = inf X x = x X ⌦ 120 DUALITY THEOREMS Assume that w⇤ < ⇣ • and that the set M = (u, w) | ⇤ is convex. there exists w with w w and (u, w) M ⌃ ⇤ ⌅ Min Common/Max Crossing Theorem I: • We have q⇤ = w⇤ if and only if for every sequence (uk, ...
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⌘ −⇣ ⌘ � D = u | there exists w ⇤ contains the origin in its relative interior. Then q⇤ = w⇤ and there exists µ such that q(µ) = q⇤. w (µ, 1) w∗ = q∗ M M 0 D u w w q∗ 0 M M u D Furthermore, the set is nonempty µ • and compact if and only if D contains the origin in its interior. q(µ) = q⇤} { | Min Common/Max Crossing T...
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w⇤ − ⇧) • / cl(M ) for any ⇧ > 0. ⌘ w w w∗ − ⇥ wk lim inf k ⇥⇤ 0 M (uk, wk) (uk+1, wk+1) (uk, wk) (uk+1, wk+1) M u 123 PROOF OF THEOREM I (CONTINUED) ⇧ ⌘ Step 2: M does not contain any vertical lines. 1) would be a direction cl(M ), ⇤) 0 belongs to • If this were not so, (0, − ). Because (0, w of recession of cl(M ⇧) (...
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through (0, w⇤), contains M in one of its closed halfspaces, but does not fully contain M , i.e., for some (µ, ⇥) = (0, 0) ⇥w⇤ ⌥ µ�u + ⇥w, ⇥w⇤ < sup (u,w) ⌦ { M M , (u, w)  µ�u + ⇥w ⌘ } Will show that the hyperplane is nonvertical. any (u, w) M , the set M contains the Since for • 0. If w halfline (u, w) | ≥ ri(D) µ�u ...
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≥ X . ⇤ Then Q⇤ is nonempty and compact if and only⌅ if X such that gj(x) < 0 there exists a vector x for all j = 1, . . . , r. ⌘ (g(x), f (x)) x | ⌅ X (g(x), f (x)) x | ⌅ X (g(x), f (x)) x | ⌅ X � g(x), f (x) � ⇥ 0 ⇥ � ⇥ � ⇥ (µ, 1) 0 (µ, 1) 0 0 } (a) (b) (c) The lemma asserts the existence of a nonverti- r+1, with nor...
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/MC Theorem II applies: we have • D = u | there exists w with (u, w) M ⌘ ⌘ � and 0 ⇤ ⌘ int(D), because g(x), f (x) M . ⌘ ⌅ � ⇥ 127 LECTURE 10 LECTURE OUTLINE Min Common/Max Crossing Th. III Nonlinear Farkas Lemma/Linear Constraints Linear Programming Duality Convex Programming Duality • • • • Optimality Conditions • Re...
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P = Ø, where ⌫ D˜ = u | there exists w with (u, w) ⌘ � ˜ M } ⌘ ⇤ ⌘ Then q⇤ = w⇤, there is a max crossing solution, 0 and all max crossing solutions µ satisfy µ�d for all d ⌥ RP . Comparison with Th. II: Since D = D˜ • the condition 0 ri(D) of Theorem II is P , − ⌘ ri(D˜ ) ri(P ) = Ø ⌫ 129✓ ✓ PROOF OF MC/MC TH. III Con...
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assumption, ⇥ = 0, so we may assume that ⇥ = 1. a direction of recession of C ≥ ⇤ 130✓ ✓ PROOF (CONTINUED) Hence, • w⇤ + µ�z so that ⌥ inf (u,v) ⌦ C1 { v + µ�u , } z P, ⌘  w⇤ ⌥ = (u,v) inf C1, z ⌦ inf ˜M ⌦ P { v + µ�(u z) − P ⇤ v + µ�u ⌅ } (u,v) ⌦ = inf (u,v) ⌦ = q(µ) − v + µ�u } { M Using q⇤ ⌥ q⇤ = w⇤. w⇤ (weak dual...
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= M˜ + Positive Orthant, where ⌅ M˜ = (Ax b, w) | − (x, w) ⌘ epi(f ) ⇤ w w epi(f ) (x⇥, w⇥) (x, w) 0 } Ax b ⇥ (Ax − b, w) ⇧⌅ x ˜M w⇥ 0 } p(u) = inf b ⇤ − Ax ⌅ u f (x) (µ, 1) M w⇥ q(µ) 0 } p(u) < w u (u, w) | � M ⇤ ⇥ u ⇤ epi(p) (2) There is an x ri(dom(f )) s. t. Ax b 0. Then q⇤ = w⇤ and there is a µ ⌘ − 0 with q(µ) = q...
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. ⇤ Assume that there exists a vector x that Ax 0. Then Q⇤ is nonempty. ⌘ b − ⌥ ⌅ ri(X) such Proof: As before, apply special case of MC/MC Th. III of preceding slide, using the fact w⇤ ≥ 0, implied by the assumption. w (0, w∗) 0 M = (u, w) Ax b − ⇥ | u, for some (x, w) epi(f ) ⌅ ⇤ (Ax − b, f (x)) x | ⌅ X ⇤ ⌅ ⌅ u (µ, 1)...
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, ≥ − x  ⌘ � m, or Aµ = c. 134 LINEAR PROGRAMMING DUALITY Consider the linear program • minimize c�x subject to a�jx ≥ bj, j = 1, . . . , r, where c n, aj n, and bj ⌘ � ⌘ � The dual problem is , j = 1, . . . , r. ⌘ � • • maximize b�µ r subject to ajµj = c, µ 0. ≥ j=1 ⌧ Linear Programming Duality Theorem: (a) If either...
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 } 0, c can c = r j=1 ⌧ µ⇤j aj, µ⇤j ≥ 0,  j ⌘ J, µ⇤j = 0, j / ⌘  J. Taking inner product with x⇤, we obtain c�x⇤ = f ⇤, shows that q⇤ = f ⇤ b�µ⇤, which in view of q⇤ ⌥ and that µ⇤ is optimal. 136 LINEAR PROGRAMMING OPT. CONDITIONS A pair of vectors (x⇤, µ⇤) form a primal and dual optimal solution pair if and only if...
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b�µ⇤ = c�x⇤. From Eq. (**), we obtain Eq. (*). 137 CONVEX PROGRAMMING Consider the problem minimize f (x) subject to x X, gj(x) ⌘ ⌥ 0, j = 1, . . . , r, ⌦ � ◆→ � where X gj : X n is convex, and f : X are convex. Assume f ⇤: finite. Recall the connection with the max crossing • problem in the MC/MC framework where M = ep...
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X w/ g(x) − ⌥ r f ⇤ ⌥ f (x) + µ⇤j gj(x), x ⌘  X j=1 ⌧ It follows that • f ⇤ ⌥ inf X x ⌦ ⇤ f (x)+µ⇤�g(x) ⌅ inf X, g(x) 0 ⌅ ⌥ x ⌦ f (x) = f ⇤. Thus equality holds throughout, and we have f ⇤ = inf ◆ x X ⌦ ⌫ f (x) + r j=1 ⌧ µ⇤j gj(x)  ⇠ = q(µ⇤)   139 QUADRATIC PROGRAMMING DUALITY • Consider the quadratic program 2 x�Q...
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have q⇤ = f ⇤, and the vectors x⇤ and µ⇤ are • optimal solutions of the primal and dual problems, respectively, iff x⇤ is feasible, µ⇤ ≥ x⇤ ⌘ arg min L(x, µ⇤), µ⇤j gj(x⇤) = 0, 0, and (1) Proof: If q⇤ = f ⇤, and x⇤, µ⇤ are optimal, then j.  X ⌦ x f ⇤ = q⇤ = q(µ⇤) = inf L(x, µ⇤) x X ⌦ r L(x⇤, µ⇤) ⌥ = f (x⇤) + µ⇤j gj(x⇤) ...
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: • Ax⇤ ⌥ b, µ⇤ ≥ 0 Lagrangian optimality holds [x⇤ minimizes L(x, µ⇤) • over x n]. This yields ⌘ � x⇤ = − Q− 1(c + A�µ⇤) Complementary slackness holds [(Ax⇤ − • 0]. It can be written as b)�µ⇤ = µ⇤j > 0 ✏ a�jx⇤ = bj,  j = 1, . . . , r, where a�j is the jth row of A, and bj is the jth component of b. 142 LINEAR EQUALIT...
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m. ⌘ � 143 DUALITY AND OPTIMALITY COND. • Pure equality constraints: (a) Assume that f ⇤: finite and there exists x ⌘ ri(X) such that Ax = b. Then f ⇤ = q⇤ and there exists a dual optimal solution. (b) f ⇤ = q⇤, and (x⇤, ⌃⇤) are a primal and dual optimal solution pair if and only if x⇤ is fea- sible, and x⇤ ⌘ arg min L(...
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Farkas’ Lemma Linear programming (duality, opt. conditions) Convex programming • • • minimize subject to x f (x) X, g(x) ⌘ ⌥ 0, Ax = b, g1(x), . . . , gr(x) �, f : where X is convex, g(x) = vex. and gj : X X (Nonlin. Farkas’ Lemma, duality, opt. conditions) , j = 1, . . . , r, are con ⇥ ◆→ � ◆→ � � 145 DUALITY AND OPTI...
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�� j  146 COUNTEREXAMPLE I Strong Duality Counterexample: Consider • minimize f (x) = e− subject to x1 = 0,  x1x2 X = x ⌘ x x { | ≥ 0 } Here f ⇤ = 1 and f is convex (its Hessian is > 0 in the interior of X). The dual function is q(⌃) = inf 0 ⇧ x e−  x1x2 + ⌃x1 = 0 0, if ⌃ ≥ otherwise, � −⇣ (when ⌃ ≥ ative for x x1 ≥...
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Solutions Counterexample: , f (x) = x, g(x) = x2. Then x⇤ = 0 is • Let X = the only feasible/optimal solution, and we have � q(µ) = inf { ⌦� x x + µx2 = } − 1 4µ , µ > 0,  for µ and q(µ) = −⇣ However, there is no µ⇤ ≥ q⇤ = 0. ⌥ 0, so that q⇤ = f ⇤ = 0. 0 such that q(µ⇤) = The perturbation function is • p(u) = inf x = ...
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� Dual problem: max ⌅ {− min⌅{− q(⌃) or } • − f (⌃) 1 ⇤ − f 2 ( ⌃) } − ⌅ = 1 (⌃) +f f 2 ( minimize n, subject to ⌃ ⌘ � ⌃) − where f 1 and f 2 are the conjugates. 150◆ ◆ • FENCHEL DUALITY THEOREM Consider the Fenchel framework: (a) If f ⇤ is finite and ri = dom(f1) , then f ⇤ = q⇤ and there exists at least one dom(...
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RIC INTERPRETATION f 1 () q() f 2 ( ) f = q f1(x) Slope Slope f2(x) x x When dom(f1) = dom(f2) = n, and f1 and • f2 are differentiable, the optimality condition is equivalent to � ⌃⇤ = ∇ f1(x⇤) = −∇ f2(x⇤) By reversing the roles of the (symmetric) primal • and dual problems, we obtain alternative criteria for strong...
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f ⌥(⇤) = sup 1 n x ⌥ ⇤ where C ⇤ = ⇤⇧x − f (x) ⌅ , f ⌥ 2 (⇤) = sup ⇤⇧x = x C ⌥ 0 ⇧ if ⇤ if ⇤ C⇥, ⌃ / C , ⇥ ⌃ � ⌃ ⌃�x ⌥ { The dual problem is | 0, x C . } ⌘  • • minimize subject to ⌃ f (⌃) ˆ C, ⌘ where f is the conjugate of f and Cˆ = ⌃ { | ⌃�x 0, x C . } ⌘  ≥ Cˆ and − ˆ C are called the dual and polar cones. 154◆ ...
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• • • Primal problem is minimize c�x subject to x We have b − ⌘ S, x C. ⌘ f (⌃) = sup (⌃ x S − b c)�x = sup(⌃ c)�(y + b) = ⌦ − (⌃ � ⇣ c)�b − if ⌃ if ⌃ y S − ⌦ c ⌘ c / ⌘ S⊥, S. − − Dual problem is equivalent to minimize b�⌃ subject to ⌃ c S , − ⌘ ⊥ ⌃ C.ˆ ⌘ If X ri(C) =Ø, there is no dualit y gap an d • there exists a du...
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ization: ⇣ � p(u) = inf F (x, u). ⌦� Under the given assumption, p is closed convex. x n 157 LECTURE 12 LECTURE OUTLINE Subgradients Fenchel inequality Sensitivity in constrained optimization Subdifferential calculus • • • • Optimality conditions • Reading: Section 5.4 158 SUBGRADIENTS f (z) 0 g, 1) ( x, f (x) � ⇥z Let ...
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:3)(cid:1)(cid:10)(cid:1)(cid:13)(cid:11)(cid:14)(cid:15)(cid:7)(cid:4)(cid:1)(cid:2)(cid:8)(cid:6)(cid:9)(cid:3)(cid:2)(cid:14)(cid:9)(cid:1)(cid:5)(cid:1)(cid:8)(cid:3)(cid:17) 1) f (x) = max 0, (1/2)(x2 � -1 (cid:5)(cid:1)(cid:8) 0 (cid:7) 1 (cid:8) ⇥ x (cid:14) (cid:2)(cid:12)(cid:2)(cid:14)(cid:3) f (x) (cid:8) 1 ...
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proper convex. • • M = epi(fx), fx(z) = f (x + z) f (x) − f (z) 0 Epigraph of f g, 1) ( x, f (x) � ⇥z fx(z) Translated Epigraph of f g, 1) ( 0 z By 2nd MC/MC Duality Theorem, ◆f (x) is • nonempty and compact if and only if x is in the interior of dom(f ). More generally: for every x ri dom(f )), ⌘ • ◆f (x) = S⊥ + G, � ...
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} , ⌥ we have NC(x) = 0 { } cone � ai { | a�ix = bi } if x if x int(C), ⌘ / int(C). ⌘ � ⇥ Proof: Given x, disregard inequalities with • a�ix < bi, and translate C to move x to 0, so it becomes a cone. The polar cone is NC(x). 163 FENCHEL INEQUALITY n � Let f : ] be proper convex and ⇣ • let f be its conjugate. Using th...
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) We have x⇥ � n. So all x ⌘ � ⇤ ⌘ X ⇤ iff f (x) f (x⇤) for ≥ X ⇤ x⇤ ⌘ where: iff 0 ⌘ ◆f (x⇤) iff x⇤ ⌘ ◆f (0) − − 1st relation follows from the subgradient in- equality 2nd relation follows from the conjugate sub- gradient theorem (b) ◆f (0) is nonempty if 0 ri dom(f ) . ⌘ (c) ◆f (0) is nonempty and compact int dom(f ) . ...
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∇ − p(0) is • ⌅ Q⇤ p(u) = x inf X, g(x) f (x), u ⌦ ⌅ If p is convex and differentiable, µ⇤j = ◆p(0) ◆uj , − j = 1, . . . , r. 166 EXAMPLE: SUBDIFF. OF SUPPORT FUNCTION Consider the support function ↵X (y) of a set • X. To calculate ◆↵X (y) at some y, we introduce r(y) = ↵X (y + y), y n. ⌘ � We have ◆↵X (y) = ◆r(0) = arg...
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xed x n, consider ⌘ � • Ax = j | a�jx + bj = f (x) and the function r(x) = max a� ⇤ jx j | ⌅ ⌘ Ax . It can be seen that ◆f (x) =⇤ ◆r(0). Since r is the support function of the finite set ⌅ • • aj { j Ax} ⌘ | , we see that ◆f (x) = ◆r(0) = conv aj { | j ⌘ Ax} ⇥ � 168 CHAIN RULE m ( � Let f : ] be convex, and A be , ⇣ • a...
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⌃ �(Az y) − Since the min over z is unconstrained, we hav⌅e ⇤ d = A�⌃, so Ax arg miny f (y) ⌃�y , or m ⌘ ⌦� − f (y) ≥ ⇤ f (Ax) +⌃ �(y Ax), − y ⌅ m. ⌘ �  ◆f (Ax), so that d = A�⌃ Hence ⌃ ⌘ It follows that ◆F (x) dral case, dom(f ) is polyhedral. Q.E.D. A�◆f (Ax). ⌘ A�◆f (Ax). In the polyhe- ⌦ 169◆ ✓ ✓ SUM OF FUNCTIONS...
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�� ⌥ ⌫ ⌫ � ⌥ m i=k+1 ri dom(fi) = Ø. � ⇥� 170◆ ✓ ◆ ✓ CONSTRAINED OPTIMALITY CONDITION n ( Let f : , −⇣ • � be a convex subset of the following four conditions holds: ] be proper convex, let X n, and assume that one of ⇣ � → (i) ri dom(f ) ri(X) = Ø. ⌫ f (ii) f is polyhedral and dom( ) ⇥ � ri( X ) = Ø . ⌫ ⌫ (iii) X is ...
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�) ⌘ −∇ NC(x⇤), which is equivalent to f (x⇤)�(x x⇤) 0, ≥ − ∇ x ⌘  X. In the figure on the right, f is nondifferentiable, • and the condition is that g − ⌘ NC(x⇤) for some g ◆f (x⇤). ⌘ 172 LECTURE 13 LECTURE OUTLINE • • Problem Structures Separable problems Integer/discrete problems – Branch-and-bound Large sum problems...
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It is still useful in a branch-and-bound scheme. 0, 1 { 174◆ ◆ LARGE SUM PROBLEMS Consider cost function of the form m • f (x) = fi(x), m is very large, i=1 ⌧ → � n � where fi : are convex. Some examples: Dual cost of a separable problem. • • • Data analysis/machine learning: x is pa- • rameter vector of a model; each...
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0, and c is a Examples: • 0, t The quadratic penalty P (t) = max − } The nondifferentiable penalty P�(t) = max⇥{ − Another possibility: Initially discard some of • the constraints, solve a less constrained problem, and later reintroduce constraints that seem to be violated at the optimum (outer approximation). { 2 . 0, ...
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uality with the definitions • f1(x) = f (x), f2(x) = 0 ⇣ � if x if x C, ⌘ / C. ⌘ The conjugates are f ⌥ 1 (⇤) = sup n ⌥ x ⇤⇧x − ) f (x ⇤ ⌅ , f ⌥ 2 (⇤) = sup ⇤⇧x = x C ⌥ 0 ⇧ if ⇤ if ⇤ C⇥ , ⌃ / C⇥, ⌃ � where C ⇤ = cone of C. ⌃ { | ⌃�x 0, x C } ⌘  ⌥ is the polar The dual problem is • minimize subject to ⌃ f (⌃) ˆ C, ⌘ wh...
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⌃ y S − ⌦ ⌘ ⊥, c S S⊥, c / ⌘ − − so the dual problem can be written as minimize b�⌃ subject to ⌃ c − ⌘ S⊥, ˆ C. ⌃ ⌘ The primal and dual have the same form. If C is closed, the dual of the dual yields the • • primal. 180 SPECIAL LINEAR-CONIC FORMS min Ax=b, x C ⌦ c�x ⇐✏ min c�x C ⌦ − b Ax ⇐✏ A c max b�⌃, ˆ C A0⌅ − ⌦ max...
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the dual as minimize x�(c subject to c − A�⌃) ˆ C − A�⌃ ⌘ discard the constant x�c, use the fact Ax = b, and change from min to max. 181 SOME EXAMPLES Nonnegative Orthant: C = { The Second Order Cone: Let x x 0 . } ≥ | • • C = (x1, . . . , xn) � xn | ≥ x3 ! x2 1 + · · · + x2 n 1 − � x1 x2 The Positive Semidefinite Cone:...
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m i=1 ⌧ m i=1 ⌧ b�i⌃i A�i⌃i = c, ⌃i ⌘ Ci, i = 1, . . . , m, where ⌃ = (⌃1, . . . , ⌃m). The duality theory is no more favorable than • the one for linear-conic problems. There is no duality gap if there exists a feasible • solution in the interior of the 2nd order cones Ci. Generally, 2nd order cone problems can be • r...
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� · · · ⇤ • • maximize subject to m i=1 ⌧ m i=1 ⌧ b�i⌃i A�i⌃i = c, ⌃i ⌘ Ci, i = 1, . . . , m, where ⌃ = (⌃1, . . . , ⌃m). 186 EXAMPLE: ROBUST LINEAR PROGRAMMING minimize c�x subject to a�jx bj, ⌥  (aj, bj) ⌘ Tj, j = 1, . . . , r, where c n, and Tj is a given subset of n+1. ⌘ � � We convert the problem to the equivalen...
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− ⌘ Cj, where 187 SEMIDEFINITE PROGRAMMING Consider the symmetric n • product < X, Y >= trace(XY ) = ⇤ n matrices. Inner n i,j=1 x y . ij ij • Let C be the cone of pos. semidefinite matrices. � C is self-dual, and its interior is the set of pos- • itive definite matrices. Fix symmetric matrices D, A1, . . . , Am, and • v...
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