I.3. LMI DUALITY. Didier HENRION EECI Graduate School on Control Supélec - Spring 2010

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1 I.3. LMI DUALITY Didier HENRION EECI Graduate School on Control Supélec - Spring 2010

2 Primal and dual For primal problem p = inf x g 0 (x) s.t. g i (x) 0 define Lagrangian L(x, z) = g 0 (x) + i z i g i (x) = [g 0 (x) g 1 (x) g 2 (x) ][1 z 1 z 2 ] T and Lagrange dual function f(z) = inf x where z is a dual multiplier L(x, z) Function f is always concave even if primal problem is nonconvex

3 Weak duality Define dual problem d = sup z f(z) s.t. z 0 which is always convex since f is concave Weak duality always holds: p d because f(z) g 0 (x) + i z i g i (x) }{{} 0 for any primal feasible x and dual feasible z g 0 (x) The difference p d 0 is called duality gap

4 Strong duality Sometimes, assumptions ensure that strong duality holds: p = d An example is Slater s constraint qualification assuming a strictly feasible convex primal (or dual) problem

5 Geometric interpretation of duality Consider the primal optimization problem p = inf x g 0 (x) s.t. g 1 (x) 0 with Lagrangian L(x, z) = g 0 (x) + zg 1 (x) dual function f(z) = inf x L(x, z) and dual problem d = sup z f(z) s.t. z 0

6 Geometric duality g (x) 0 L(x,z) f(z) g (x) 1 (z,1) Lagrangian L(x, z) = g 0 (x) + zg 1 (x) is a supporting line with (negative) slope z, whose intersection with g 1 (x) = 0 axis gives dual function f(z) = inf x L(x, z)

7 Geometric duality g (x) 0 p* d* g (x) 1 Three supporting lines, including the optimum z yielding d < p (duality gap = no strong duality here)

8 Complementary slackness Suppose that strong duality holds, let x be primal optimal and z be dual optimal, then g 0 (x ) = f(z ) = inf x ( g0 (x) + i z i g i(x) ) g 0 (x ) + i z i g i(x ) g 0 (x ) from which it follows that z i g i(x ) = 0 This is complementary slackness: z i > 0 = g i(x ) = 0 or equivalently g i (x ) < 0 = z i = 0 In words, the ith optimal multiplier is zero unless the ith constraint is active at the optimum

9 KKT optimality conditions Assuming that functions g i are differentiable and that strong duality holds, then the gradient of Lagrangian L(x, z ) over x vanishes at x : g i (x ) 0 (primal feasible) zi 0 (dual feasible) zi g i(x ) = 0 (complementary) g 0 (x ) + i zi g i(x ) = 0 Necessary Karush-Kuhn-Tucker conditions satisfied by any primal and dual optimal pair For convex problems, KKT conditions are also sufficient

10 Equality constraints Multipliers corresponding to equality constraints are unconstrained: p = inf x g 0 (x) s.t. h j (x) = 0 g i (x) 0 Lagrangian L(x, y, z) = g 0 (x) + j y j h j (x) + i z i g i (x) dual function f(y, z) = inf x L(x, y, z) dual problem d = sup y,z f(y, z) s.t. z 0 no constraint on multiplier vector y

11 LP duality Primal LP dual function Dual LP p = inf x c T x s.t. Ax = b x 0 f(y, z) = { inf x (c T x + y T (b Ax) z T x) b = T y if c A T y z = 0 otherwise d = sup y b T y s.t. z = c A T y 0

12 SDP duality Primal SDP dual function p = inf X trace CX s.t. trace A i X = b i X 0 f(y, Z) = inf X (trace CX + { i y i (b i trace A i X) trace ZX) b = T y if C Z i y i A i = 0 otherwise Dual SDP d = sup y b T y s.t. Z = C i y i A i 0

13 Example of SDP duality gap Example Consider the primal semidefinite program with dual inf x 1 s.t. 0 x 1 0 x 1 x x 1 0 sup y 1 y 2 (1 + y 1 )/2 y 3 s.t. (1 + y 1 )/2 0 y 4 y 3 y 4 y 1 0 In the primal necessarily x 1 = 0 (x 1 appears in a row with zero diagonal entry) so the primal optimum is x 1 = 0 Similarly, in the dual necessarily (1+y 1 )/2 = 0 so the dual optimum is y 1 = 1 There is a nonzero duality gap here

14 Theorem of the alternatives Consider primal feasibility problem and dual feasibility problem g i (x) 0 f(y) < 0, y 0 with dual function f(y) = sup x i y i g i (x) Dual feasible implies primal infeasible Proof: if x is primal feasible then f(y) = sup x i y i g i (x) i y i g i (x ) and hence f(y) 0 for all y 0 Separating hyperplanes in convex analysis Can be generalized in the context of convex conic programming..

15 Farkas lemma When solving a primal/dual conic problem inf c T x s.t. Ax = b x K sup s.t. b T y c A T y K in the absence of a duality gap, then either x is optimal and y certifies optimality, i.e. b T y = c T x, or y is optimal and x certifies optimality, i.e. c T x = b T y, or there is no x K with Ax = b and this is certified by y, i.e. b T y > 0 and A T y K, or there is no y such that c A T y K and this is certified by x, i.e. c T x < 0, Ax = 0, x K Either we find a feasible point or we certify that no such point exists

16 LMI duality In the LMI formulation, the primal problem is actually in dual SDP form (confusing indeed..) p = inf x c T x s.t. F (x) = F 0 + i x i F i 0 with dual LMI in primal SDP form d = sup Z trace F 0 Z s.t. trace F i Z = c i Z 0 Primal (resp. dual) not strictly feasible iff there exists a certificate of infeasibility provided by the dual (resp. primal)

17 Theorem of alternatives for LMIs For the LMI mapping F (x) = F 0 + i x i F i Exactly one statement is true there exists x s.t. F (x) 0 there exists a nonzero Z = Z T 0 s.t. trace F 0 Z 0 and trace F i Z = 0 for i > 0 Useful for detecting infeasibility of LMIs Rich literature on theorems of alternatives for generalized inequalities, e.g. nonpolyhedral convex cones

18 S-procedure S-procedure: frequently useful in robust and nonlinear control, is an outcome of the theorem of alternatives There exists no nonzero complex vector x such that x A i x 0, i = 1,..., p if there exist real numbers y i 0 such that p i=1 y ia i 0 If there exists x 0 such that x 0 A ix 0 > 0 for some i, the converse also holds when p = 2 for real quadratic forms when p = 3 for complex quadratic forms

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