A non-standart approach to Duality

Transcription

1 Department of Industrial and Systems Engineering Lehigh University Seminar Series, /27/2005

2 Outline 1 Duality Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation 2 Continuous Case Discrete Case LP formulation Superadditive Dual

3 References: Jean B. Lasserre, Duality and a Farkas lemma for integer programs. in: Optimization: Structure and Applications (E. Hunt and C.E.M. Pearce, Editors), Kluwer Academic Publishers (2004) Jean B. Lasserre, Integer programming duality and superadditive functions. Contemporary Mathematics 374, pp (2005)

4 Formulation Duality Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation I : P : LP f (b, c) := max x Ω(b) IP c x P d : f d (b, c) := max x Θ(b) Integration Summation Z ˆf (b, c) := e c x ds I d : ˆfd (b, c) := X Ω(b) x Θ(b) c x e c x where Ω(b) := {x R n : Ax = b, x 0} and Θ(b) := Ω(b) Z n.

5 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation The simple relation is that: or equivalently e f (b,c) = lim r ˆf (b, rc) 1/r, 1 f (b, c) = lim r r lnˆf (b, rc), e f d (b,c) = lim r ˆfd (b, rc) 1/r 1 f d (b, c) = lim r r ln ˆf d (b, rc)

6 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Let P, P d be some dual problems of P, P d. The well-known strong dual P can be obtained by Legendre-Fenchel Duality formulation. Let f : R n R, then the Fenchel conjugate: f (y) := sup x R n {x y f (x)} y R n If f is convex and lower semi-continuous f := f. Then, noting that f (y, c) is concave, P : f (b, c) := inf b f (λ, c)} = min λ A λ c} λ R m{λ λ R m{b No available transformation to get a strong P d (except the subadditive formulation, see my previous talk...) Let I, I d be the similar transforms of I, I d using Laplace-transform and Z-transform, respectively. Lasserre calls I, I d the natural duals: we can get closed form equations for ˆf (b, c) and ˆfd (b, c).

7 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Laplace transform f : C m + C of f : R m + R is Z f (λ) := e λ y f (y)dy R n + Applying to ˆf (b, c): ˆF(λ, c) := := Z e λ yˆf (y, c)dy R m + ny 1 (A λ c) k k=1

8 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Note that, ˆF(λ, c) is well defined when R(A λ c) > 0. To get I and hence a closed form ˆf (b, c), we solve the inverse Laplace transform problem: I : ˆf (b, c) := = 1 (2iπ) m 1 (2iπ) m Z γ+i γ i Z γ+i γ i e λ b ˆF(λ, c)dλ e λ b dλ ny (A λ c) k k=1 where γ is fixed and satisfies (A γ c) > 0. This complex integral can be solved directly by Cauchy residual techniques.

9 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation In fact, the multidimensional poles of ˆF are real and solve p B A B = c B for all bases of Ω(b). Then: ˆf (b, c) = X x b.f.s of Ω(b) e c x det(b) Y ( c k + p B A k ) k NB where B is the basis of the corresponding b.f.s. In other words, ˆf (b, c) is a weighted summation over the vertices of Ω(b).

10 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Z-transform f : C m + C of f : Z m + R is f (z) := X z y f (y) y Z m + Applying to ˆf d (b, c): ˆF d (z, c) := ny 1 1 e c k z A k k=1 which is well-defined if z A k > e c k k = 1,..., n

11 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation To get I d and hence a closed form ˆf d (b, c), we solve the inverse Z-transform problem: Z I d : ˆf 1 d (b, c) := ˆF (2iπ) m d (z, c)z b 1 dz z =γ where γ satisfies γ A k > e c k k = 1,..., n. Cauchy residue technique can be used, however, we have complex poles! Bases of ω(b) provide these poles. Each basis B provides det(b) complex poles in the form of: v z(k) = e p B+2iπ det(b) for k = 1,..., det(b) where v {v Z m v B = 0 mod det(b)}

12 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Combining these poles with discrete Brian and Vergne s formula: X ˆfd (b, c) = e c x U B (b, c) x b.f.s of Ω(b)

13 Integration & Counting Duality for Integration Brion and Vergne s continuous formula Duality for Summation Then, 1 f d (b, c) = lim r r ln ˆf d (b, rc) = max {c 1 x + lim x b.f.s of Ω(b) r r ln U B(b, rc)} = max {c x + 1 x b.f.s of Ω(b) q (deg(p b) deg(q b ))} = c x + ρ where q is the l.c.m of det(b): B is a feasible basis, P b and Q b are some real valued polynomials. Note that, ρ is the value of the Gomory group/asymptotic problem!

14 Continuous Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Note that a polynomial Q R[λ 1,..., λ m] can be written Q(λ) = X α S Q α λ α = X Q α λ α λαm m α S where S N m and Q α are real coefficients α S. Let A R m n, b R m. Then the folllowing two statements are equivalent: (a) The linear system Ax = b has a nonnegative solution x R n. (b) The polynomial b λ can be written b λ = nx Q j (λ)(a jλ), λ R m j=1 for some polynomials Q j R[λ 1,..., λ m], j = 1,..., n, all with nonnegative coefficients. Proof:

15 Discrete Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Let, wlog (as long as the feasible region is compact), A N m n, b N m. Then the folllowing two statements are equivalent: (a) The linear system Ax = b has a solution x N n. (b) The polynomial z b 1 can be written z b 1 = nx Q j (z)(z A j 1), z R m j=1 for some polynomials Q j R[z 1,..., z m], j = 1,..., n, all with nonnegative coefficients. Moreover, the total degree of each polynomial Q j can be bounded by b = P mx m j=1 b j min A ik. k=1,...,n Proof: i=1

16 Discrete Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Let q 0 be the vector of nonnegative coefficients of all polynomials Q j s. If D := {Mq = r, q 0} then, there exist such polynomials. Note that Mq = r state that the nx polynomials z b 1 and Q j (z)(z A j 1) are identical by equating the respective coefficients. j=1

17 Discrete Case Duality Continuous Case Discrete Case LP formulation Superadditive Dual Each Q j may be restricted to contain only monomials {z α : α b A j, α N m } Hence, in D, q R s and M R p s where p = s = my (b i + 1) i=1 nx my s j with s j = (b i A ij + 1), j = 1,..., n j=1 i=1 Note that, p is the number of monomials z α with α b and s j is the number of monomials z α with α A j b. Example: Note that M is totally unimodular.

18 Continuous Case Discrete Case LP formulation Superadditive Dual An LP equivalent to P d Let e sj = (1,..., 1) R s j, j = 1,..., n and let E N n s be the n-block diagonal matrix, whose each diagonal block is a row vector e sj. Then, 1 f d (b, c) = max{c Eq Mq = r, q 0} 2 If q is an optimal solution, then x := Eq is the associated optimal solution of P d. Proof:

19 A class of superadditive functions Continuous Case Discrete Case LP formulation Superadditive Dual Let D N m be a finite set s.t 0 D and if α D β D, β α. D be the set of functions π : N m R {+ } s.t. π(0) = 0 and π(α) < + if α D, or π(α) = +, otherwise. Given π D, f π : N m R {+ } is defined as f π(x) := inf {π(α + x) π(α)}, x α D Nm Lemma: For every π D : 1 f π D 2 f π π, and f π is superadditive 3 if π is superadditive, then π = f π. Proof:

20 Superadditive Dual Duality Continuous Case Discrete Case LP formulation Superadditive Dual Let D := {α N m α b}. Then the dual of max{c Eq Mq = r, q 0} is min γ s.t γ(b) γ(0) γ(α + A j ) γ(α) c j, α + A j D, j = 1,..., n Letting π(α) = γ(α) γ(0), α D and extending π to N m, the dual becomes ρ 1 = min π D s.t π(b) π(α + A j ) π(α) c j, α D, j = 1,..., n

21 Superadditive Dual Duality Continuous Case Discrete Case LP formulation Superadditive Dual Now, assume that f d (b, c) > and consider the following problem ρ 2 = inf π D s.t f π(b) f π(a j ) c j, j = 1,..., n where f π : N m R defined as before for every π D. Then, f d (b, c) = ρ 1 = ρ 2 = f π (b) for some π D Proof: Example:

22 Superadditive Dual Duality Continuous Case Discrete Case LP formulation Superadditive Dual Note the similar dual formulation of Wolsey min π s.t π(b) π(λ) + π(µ) π(λ + µ), 0 λ + µ b π(a j ) c j, j = 1,..., n π(0) = 0 where the first constraint set state that π : D R is superadditive. The number of variables are the same, however, this one has O(p 2 ) constraints whereas the introduced one has O(np).