4 - 1 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 4. Algebra and Duality Example: non-convex polynomial optimization ² Weak duality and duality gap ² The dual is not intrinsic ² The cone of valid inequalities ² Algebraic geometry ² The cone generated by a set of polynomials ² An algebraic approach to duality ² Example: feasibility ² Searching the cone ² Interpretation as formal proof ² Example: linear inequalities and Farkas lemma ² 4 - 2 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example minimize x1x2 subject to x 0 1 ¸ 0.6 x 0 2 ¸ 1 0.4 2 2 0.75 x1 + x2 1 0.2 · 0.5 0 0.25 1 0.75 The objective is not convex. 0.5 0 0.25 ² 0 The Lagrange dual function is ² g(¸) = inf x x ¸ x ¸ x + ¸ (x2 + x2 1) x 1 2 ¡ 1 1 ¡ 2 2 3 1 2 ¡ ³ ´ T 1 ¡ 1 ¸1 2¸3 1 ¸1 1 ¸3 2 if ¸3 > 2 = 8¡ ¡ "¸2# " 1 2¸3# "¸2# > <> otherwise, except bdry ¡1 > :> 4 - 3 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example, continued The dual problem is maximize g(¸) −0.5 −0.75 subject to ¸1 0 ¸ ) λ −1 ¸2 0 g( ¸ 0 −1.25 1 0.2 ¸3 0.4 ¸ 2 −1.5 0.6 0.5 0.6 0.7 0.8 0.8 λ 0.9 1 1 1 λ 3 By symmetry, if the maximum g(¸) is attained, then ¸1 = ¸2 at ² optimality The optimal g(¸?) = 1 at ¸? = (0; 0; 1) ² ¡2 2 Here we see an example of a duality gap; the primal optimal is strictly ² greater than the dual optimal 4 - 4 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example, continued It turns out that, using the Schur complement, the dual problem may be written as maximize 2 2¸ ¸ ¸ ¡ ¡ 3 1 2 subject to ¸ 2¸ 1 > 0 2 1 3 3 ¸2 1 2¸3 ¸41 > 0 5 ¸2 > 0 In this workshop we'll see a systematic way to convert a dual problem to an SDP, whenever the objective and constraint functions are polynomials. 4 - 5 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 The Dual is Not Intrinsic The dual problem, and its corresponding optimal value, are not prop- ² erties of the primal feasible set and objective function alone. Instead, they depend on the particular equations and inequalities used ² To construct equivalent primal optimization problems with di®erent duals: replace the objective f (x) by h(f (x)) where h is increasing ² 0 0 introduce new variables and associated constraints, e.g. ² minimize (x x )2 + (x x )2 1 ¡ 2 2 ¡ 3 is replaced by minimize (x x )2 + (x x )2 1 ¡ 2 4 ¡ 3 subject to x2 = x4 add redundant constraints ² 4 - 6 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example Adding the redundant constraint x1x2 0 to the previous example gives ¸ 0.6 minimize x1x2 1 0.4 0.75 subject to x 0 0.2 1 0.5 ¸ 0 x2 0 0.25 1 ¸ 0.75 2 2 0.5 x + x 1 0 0.25 1 2 · 0 x x 0 1 2 ¸ Clearly, this has the same primal feasible set and same optimal value as before 4 - 7 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example Continued The Lagrange dual function is g(¸) = inf x x ¸ x ¸ x + ¸ (x2 + x2 1) ¸ x x x 1 2 ¡ 1 1 ¡ 2 2 3 1 2 ¡ ¡ 4 1 2 ³ ´ T 1 ¡ 1 ¸1 2¸3 1 ¸4 ¸1 ¸3 2 ¡ if 2¸3 > 1 ¸4 = 8¡ ¡ "¸2# "1 ¸4 2¸3 # "¸2# ¡ > ¡ > < otherwise, except bdry ¡1 > :> Again, this problem may also be written as an SDP. The optimal value ² is g(¸?) = 0 at ¸? = (0; 0; 0; 1) Adding redundant constraints makes the dual bound tighter ² This always happens! Such redundant constraints are called valid ² inequalities. 4 - 8 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Constructing Valid Inequalities The function f : Rn R is called a valid inequality if ! f(x) 0 for all feasible x ¸ Given a set of inequality constraints, we can generate others as follows. (i) If f1 and f2 de¯ne valid inequalities, then so does h(x) = f1(x)+f2(x) (ii) If f1 and f2 de¯ne valid inequalities, then so does h(x) = f1(x)f2(x) (iii) For any f, the function h(x) = f(x)2 de¯nes a valid inequality We view these as inference rules Now we can use algebra to generate valid inequalities. 4 - 9 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 The Cone of Valid Inequalities The set of polynomial functions on Rn with real coe±cients is denoted ² R[x1; : : : ; xn] Computationally, they are easy to parametrize. We will consider poly- ² nomial constraint functions. A set of polynomials P R[x1; : : : ; xn] is called a cone if ½ (i) f P and f P implies f f P 1 2 2 2 1 2 2 (ii) f P and f P implies f + f P 1 2 2 2 1 2 2 2 (iii) f R[x1; : : : ; xn] implies f P 2 2 It is called a proper cone if 1 P ¡ 62 By applying the above rules to the inequality constraint functions, we can generate a cone of valid inequalities 4 - 10 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Algebraic Geometry There is a correspondence between the geometric object (the feasible ² subset of Rn) and the algebraic object (the cone of valid inequalities) This is a dual relationship; we'll see more of this later. ² The dual problem is constructed from the cone. ² For equality constraints, there is another algebraic object; the ideal ² generated by the equality constraints. For optimization, we need to look both at the geometric objects (for ² the primal) and the algebraic objects (for the dual problem) 4 - 11 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Cones For S Rn, the cone de¯ned by S is ² ½ (S) = f R[x1; : : : ; xn] f(x) 0 for all x S C 2 ¸ 2 n ¯ o ¯ If P and P are cones, then so is P P ² 1 2 1 \ 2 A polynomial f R[x1; : : : ; xn] is called a sum-of-squares (SOS) if ² 2 r 2 f(x) = si(x) Xi=1 for some polynomials s1; : : : ; sr and some r 0. The set of SOS polynomials § is a cone. ¸ Every cone contains §. ² 4 - 12 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Cones The set monoid f1; : : : ; fm R[x1; : : : ; xn] is the set of all ¯nite f g ½ products of polynomials fi, together with 1. The smallest cone containing the polynomials f1; : : : ; fm is r cone f1; : : : ; fm = sigi s0; : : : ; sr §; f g ( j 2 Xi=1 gi monoid f1; : : : ; fm 2 f g ) cone f ; : : : ; fm is called the cone generated by f ; : : : ; fm f 1 g 1 4 - 13 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Explicit Parametrization of the Cone If f ; : : : ; fm are valid inequalities, then so is every polynomial ² 1 in cone f ; : : : ; fm f 1 g The polynomial h is an element of cone f ; : : : ; fm if and only if ² f 1 g m h(x) = s0(x) + si(x)fi(x) + rij(x)fi(x)fj(x) + : : : i=1 i=j X X6 where the si and rij are sums-of-squares. 4 - 14 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 An Algebraic Approach to Duality Suppose f1; : : : ; fm are polynomials, and consider the feasibility problem n does there exist x R such that 2 f (x) 0 for all i = 1; : : : ; m i ¸ Every polynomial in cone f ; : : : ; fm is non-negative on the feasible set. f 1 g So if there is a polynomial q cone f ; : : : ; fm which satis¯es 2 f 1 g n q(x) " < 0 for all x R · ¡ 2 then the primal problem is infeasible. 4 - 15 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example Let's look at the feasibility version of the previous problem. Given t R, 2 does there exist x R2 such that 2 x x t 1 2 · x2 + x2 1 1 2 · x 0 1 ¸ x 0 2 ¸ Equivalently, is the set S nonempty, where n S = x R fi(x) 0 for all i = 1; : : : ; m 2 j ¸ n o where f (x) = t x x f (x) = 1 x2 x2 1 ¡ 1 2 2 ¡ 1 ¡ 2 f3(x) = x1 f4(x) = x2 4 - 16 Algebra and Duality P. Parrilo and S. Lall, CDC 2003 2003.12.07.01 Example Continued Let q(x) = f (x) + 1f (x). Then clearly q cone f ; f ; f ; f and 1 2 2 2 f 1 2 3 4g q(x) = t x x + 1(1 x2 x2) ¡ 1 2 2 ¡ 1 ¡ 2 = t + 1 1(x + x )2 2 ¡ 2 1 2 t + 1 · 2 1 So for any t < 2, the primal problem is infeasible. ¡ 1 Corresponds to Lagrange multipliers (1; 0; 0; 2) for the thm.