Optimization Methods. Lecture 23: Semidenite Optimization

Transcription

1 5.93 Optimization Methods Lecture 23: Semidenite Optimization

2 Outline. Preliminaries Slide 2. SDO 3. Duality 4. SDO Modeling Power 5. Barrier lgorithm for SDO 2 Preliminaries symmetric matrix is positive semidenite ( ) if and only if Slide 2 n u u 8 u 2 R if and only if all eigenvalues of are nonnegative Inner product B = ij B ij j= 2. The trace The trace of a matrix is dened trace() = j j j= Slide 3 trace(b) = trace(b ) B = trace( B) = trace(b ) 3 SDO C symmetric n n matrix Slide 4 i i = : : : m symmetric n n matrices b i i = : : : m scalars Semidenite optimization problem (SDO) (P ) : min C X s:t: i X = b i i = : : : m X

3 3. Example n = 3 and m = C Slide 5 b = b 2 = 9 x x 2 x 3 X x 2 x 22 x 23 x3 x 32 x 33 Slide 6 (P ) : min x + 4 x x x x 33 s:t: x + 2 x x x x 33 = 4x x x x 33 = 9 x x 2 x 3 x 2 x 22 x 23 x3 x 32 x 33 X 3.2 Convexity (P ) : min C X s:t: i X = b i i = : : : m X The feasible set is convex: Slide 7 X X 2 feasible =) X + ( ; )X 2 feasible.. {z } {z } b i b i i (X + ( ; )X 2 ) = i X +( ; ) i X 2 = b i u (X + ( ; )X 2 )u = u X u +( ; ) u X 2 u {z } {z } 2

4 : = : = 3.3 LO as SDO LO : min c x s:t: x = b x a i : : : c : : : a B i2 : : C c : : : B 2 i = =.. C : : : : a in c n C Slide 8 Slide 9 (P ) : min C X s:t: i X = b i = i : : : m X ij X B i x : : : x B 2 : : : X : : x n : : : n j = i + : : : n C 4 Duality Equivalently, (D) : ) : (D max X m y i b i X m s:t: y i i + S = C S X m max y i b i mx s:t: C ; y i i Slide 3

5 4 4. Example (D) max y + s:t: S 9 y y S Slide (D) max y + 9 y 2 ; y ; y 2 2 ; y ; 2y 2 3 ; y ; 8y 2 2 ; y ; 2y 2 9 ; 3y ; 6y 2 ; 7y ; y 2 3 ; y ; 8y 2 ; 7y ; y 2 7 ; 5y ; 4y 2 Slide 2 y y Optimal value u y u :4847 y 2 u :45 y 4.2 Weak Duality Theorem Given a feasible solution X of (P ) and a feasible solution (y S) of (D), mx C X ; y i b i = S X mx If C X ; y i b i =, then X and (y S) are each optimal solutions to (P ) and (D) and SX = Slide 3

6 4.3 Proof We m ust show that if S and X, t h e n S X Let S = P D P and X = QEQ where P Q are orthonormal matrices and D E are nonnegative diagonal matrices S X = trace(s X) = trace(s X ) = trace(dp QEQ = trace(p D P QEQ ) P ) = D j j (P j= QEQ P ) j j since D j j and the diagonal of P QEQ P must be nonnegative. Suppose that trace(s X ) =. Then D j j (P QEQ P ) j j = j= Slide 4 Then, for each j = : : : n, D j j = o r ( P QEQ P ) j j =. The latter case implies that the j th row o f P QEQ P is all zeros. Therefore, DP QEQ P =, a n d s o S X = P D P QEQ =. 4.4 Strong Duality (P ) o r ( D) might not attain their respective optima There might be a duality gap, unless certain regularity conditions hold Slide 5 Theorem If there exist feasible solutions ^X for (P ) a n d ( ^y ^S) for (D) s u c h that ^X, ^S Then, both (P ) and (D) attain their optimal values z P and z Furthermore, z P = z D D 5 SDO Modeling Power 5. Quadratically Constrained Problems min ( x + b ) ( x + b ) ; c x ; d Slide 6 s:t: ( i x + b i ) ( i x + b i ) ; c i x ; d i 5

7 i = : : : m (x + b) (x + b) ; c x ; d, " # I x + b (x + b) c x + d Slide 7 min t s:t: ( x + b ) ( x + b ) ; c x ; d ; t, min s:t: ( i x + b i ) ( i x + b i ) ; c i x ; d i 8 i t " # I x + b ( x + b ) c x + d + t " # I i x + b i ( i x + b i ) c i x + d i 8 i Slide Eigenvalue Problems X: symmetric n n matrix Slide 9 max (X) = largest eigenvalue of X (X) 2 (X) m (X) eigenvalues of X i (X + t I) = i (X) + t Theorem: max (X) t, t I ; X Sum of k largest eigenvalues: kx i (X) t, t ; k s ; trace(z) Follows from the characterization: kx Z Z ; X + s I i (X) = maxfx V : trace(v ) = k V Ig 6

8 j= We w ant to bound the fundamental frequency y: score of a random student o n k tests 5.3 Optimizing Structural Dynamics Select x i, cross-sectional area of structure i, i = : : : n P P M (x) = M + x i M i, mass matrix i K(x) = K + x i K i, stiness matrix Structure weight w = w + Dynamics i P d(t) v ector of displacements d i (t) = ij cos(! j t ; j ) i x i w i (x) M d + K(x)d = Slide 2 Slide 2 det(k(x) ; M (x)! 2 ) =!! 2! n =2 Fundamental frequency:! = min (M K(x)) (x)! () M (x) 2 ; K(x) Minimize weight Problem: Minimize weight subject to Fundamental frequency! Limits on cross-sectional areas Formulation P min w + i x i w i s:t: M (x) 2 ; K(x) li x i u i 5.4 Measurements with Noise x: ability of a random student o n k tests Slide 22 Slide 23 E[x] = x, E[(x ; x)(x ; x) ] = v: testing error of k tests, independent o f x E[v] =, E[vv ] = D, diagonal (unknown) 7

9 y = x + v E[y] = x b E[(y ; x)(y ; x) ] = = + D Objective: Estimate reliably x and Take samples of y from which w e can estimate b x, Slide 24 e x: total ability o n tests e y: total test score Reliability of test:= Var[e x] e e e De = = ; Var[ey] e e b e e b We can nd a lower bound on the reliability of the test Slide 25 min e e s:t: + D = b D D diagonal Equivalently, max e D e s:t: D b D diagonal 5.5 Further Tricks Slide 26 If B, B C = C D () D ; CB ; C x x + 2 b x + c 8 x () c b b 8

10 5.6 MXCUT Given G = ( N E ) undirected graph, weights w ij o n e d g e ( i j) 2 E Find a subset S N : P i2s j2 S w ij is maximized x j = for j 2 S and x j = ; for j 2 S 5.6. Reformulation M X C U T : max Let Y = xx, i.e., Y ij = x i x j Let W = [ w ij ] Equivalent F ormulation Relaxation Y = xx Relaxation M X C U T : max RELX : max n n X X 4 j= w ij ( ; x i x j ) s:t: x j 2 f ; g j = : : : n n n X X 4 j= w ij ; W Y s:t: x j 2 f ; g j = : : : n Y j j = j = : : : n Y = xx n n X X 4 j= w ij ; W Y s:t: Y j j = j = : : : n Slide 27 Slide 28 Slide 29 It turns out that: Y M X C U T RELX Slide 3 :87856 RELX M X C U T RELX The value of the SDO relaxation is guaranteed to be no more than 2% higher than the value of the very dicult to solve problem MXCUT 9

11 Logarithmic barrier problem SDO represents the present and future in continuous optimization Barrier lgorithm is very powerful Many good solvers available: SeDuMi, SDPT3, SDP, etc. 6 Barrier lgorithm for SDO X, (X) : : : n(x) Slide 3 Natural barrier to repel X from the boundary (X) > : : : n(x) > : ; log( i (X)) = j= ny ; log( i (X)) = ; log(det(x)) j= Slide 32 Barrier algorithm needs O from to log(det(x)) min B (X) = C X ; s:t: i X = b i i = : : : m X p n log iterations to reduce duality g a p 7 Conclusions SDO is a very powerful modeling tool Slide 33 Pointers to literature and solvers: www-user.tu-chemnitz.de/~helmberg/semidef.html