DUALITY AND INTEGER PROGRAMMING. Jean B. LASSERRE
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1 LABORATOIRE d ANALYSE et d ARCHITECTURE des SYSTEMES DUALITY AND INTEGER PROGRAMMING Jean B. LASSERRE 1
2 Current solvers (CPLEX, XPRESS-MP) are rather efficient and can solve many large size problems with thousands of variables. However, the very small 5-variables knapsack problem max 213x x x x x x x x x x 5 = x 1, x 2, x 3, x 4, x 5 nonnegative integers still resists all efficient solvers (takes HOURS on CPLEX 8.1 and XPRESS-MP!) Optimal solution x = (7334, 0, 0, 0, 0)...! Any insight on integer problems is welcome... 2
3 The integer program max{ c x Ax = b; x N n } with A = [A 1,..., A n ] Z m n, b Z m has a dual problem min { f(b) f(a j) c j, j = 1,..., n} f Γ where Γ is the space of functions f : R m R that are superadditive (f(a + b) f(a) + f(b)), and with f(0) = 0. (Jeroslow, Wolsey, etc...) Elegant but rather abstract and not very practical; however, used to derive valid inequalities. Moreover, the celebrated Gomory cuts used in Solvers (CPLEX, XPRESS-MP,..) can be interpreted as such superadditive functions. 3
4 CONTINUOUS OPTIM. DISCRETE OPTIM. f(b, c) := max c x f d (b, c) := max c x [ [ Ax b Ax b s.t. x R n s.t. + x N n INTEGRATION SUMMATION f(b, c) := Ω ec x dx f d (b, c) := e c x [ [ Ω Ax b Ax b Ω := x R n Ω := x N n + 4
5 e f(b,c) = lim f(b, rc) 1/r ; e fd(b,c) = lim f r r d (b, rc) 1/r. or, equivalently 1 f(b, c) = lim r r ln f(b, 1 rc); f d (b, c) = lim r r ln f d (b, rc). Note in passing that we have the mean-value theorem f(b, c) = e c x λ(ω) = e c x vol(ω). and a Maslov max mean-value theorem f(b, c) = c x = c x µ maslov (Ω) = c x. Maslov duality for linear functionals in the Max-Plus algebra 5
6 f L(f) a linear functional in the max + algebra, i.e., L(f h) = L(f) L(h) f, h L(f) = with f dµ maslov. µ maslov (A B) = µ maslov (A) µ maslov (B) A, B B. 6
7 CONTINUOUS OPTIM. DISCRETE OPTIM. Legendre-Fenchel Duality?? INTEGRATION SUMMATION Laplace-Transform Z-Transform Duality Duality 7
8 Legendre-Fenchel duality : f : R n R convex; f : R n R. λ f (λ) = F(f)(λ) := sup y {λ y f(y)}. (One-sided) Laplace-Transform: f : R n + R; F : Cn C. λ F (λ) = L(f)(λ) := R n + e λ y f(y) dy. (One-sided) Z-Transform: f : Z n + R; F : Cn C. λ F (z) = Z(f)(z) := m Z n + z m f(m). 8
9 Observe : and e F(f)(λ) = L(e f )(λ) = R n + sup (λ y f(y)) e y e (λ y f(y)) dy = sup y {e (λ y f(y)) } Maslov e(λ y f(y)) dy exponential (Fenchel (f)) = Laplace (exponential (f)) in max, algebra 9
10 with Ω := {x R n Ax b; x 0} Fenchel-duality: Laplace-duality f(b, c) := max x Ω c x f(b, c) := f (λ, c) := min y {λ y f(y, c)} F (λ, c) := = Ω ec x dx e λy f(y, c) dy [ ] x 0 ec x Ax y e λy dy dx = 1 mj=1 λ j nk=1 (A λ c) k with : A λ c 0; λ 0 with : R(A λ c) > 0; R(λ) > 0 10
11 Fenchel-duality: Laplace-duality f(b, c) = min{λ b f (λ, c)} f(b, c) = 1 λ Γ eλ b F (λ, c) dλ = min{b λ A (2iπ) m e λ b λ c; λ 0} = λ mj=1 λ nk=1 j (A dλ λ c) k Γ (2iπ) m where Γ C m Γ = u1 +i um and +i u 1 i u m i with u R m ; A u > c; u > 0. Polytope volume: > Brion, Brion and Vergnes, Barvinok and Pommersheim, Lasserre and Zeron,... Multidimensional integration over a simplex, ellipsoid > Lasserre and Zeron. 11
12 Remark: Logarithmic Barrier function of min{λ b A λ c) 0; λ 0} λ ϕ(r, λ) := rλ b n k=1 ln (A λ c) k m j=1 ln(λ j ). f(b, rc) = 1 r n f(rb, c) = Hence, lim r e min λ ϕ(r,λ) = = 1 r n (2iπ) m 1 r n (2iπ) m Γ e rλ b mj=1 λ j nk=1 (A λ c) k dλ Γ eϕ(r,λ) dλ lim min e ϕ(r,λ) 1 = lim r λ r r n (2iπ) m Γ eϕ(r,λ) dλ 12
13 Fenchel-duality: Laplace-duality f(b, c) = min{λ b f (λ, c)} f(b, c) = 1 λ = min λ {b λ A λ c; λ 0} = (2iπ) m (2iπ) m e λ b mj=1 Γ λ nk=1 j (A dλ λ c) k Γ = u1 +i Γ eλ b F (λ, c) dλ um where +i u 1 i u m i with u R m ; A u > c; u > 0. Simplex Algorithm Vertices of the polyhedron {Ax b; x 0} Cauchy s Residue Technique Poles of the function 1 mj=1 λ nk=1 j (A λ c) k THE SAME! 13
14 Laplace-Duality algorithm: Cauchy s Residue Technique f(b, c) = 1 (2iπ) m Γ e λ b mj=1 λ j nk=1 (A λ c) k dλ. Integration w.r.t. λ 1, λ 2,..., λ m (one variable at a time) by repeated application of Cauchy s residue theorem. Yields a tree of depth m whose level k is integration w.r.t. λ k on the integration path {u k i, u k + i } while the other variables λ k+1, λ k+2,..., λ m are fixed, on their respective integration path {u j i, u j + i }. Application of Cauchy s Residue Theorem yields a rational fraction in the variables λ k+1,..., λ m, and so on, until level m 1 where one obtains a rational fraction in the single variable λ m. 14
15 Brion and Vergne s continuous formula Terminology of LP in standard form: Let A σ := [A σ1... A σm ] be a basis of max{c x Ax = b; x 0}, with x(σ) the corresponding vertex, π σ := c σa 1 σ the associated dual variable and the reduced cost vector c k π σ A k, k σ. Then : f(b, c) = x(σ): vertex of Ω(b) from which it easily follows that e c x(σ) det(a σ ) k σ( c k + π σ A k ) log [ lim r f(b, rc) r] 1/r = max vertex of Ω(b) c x(σ). 15
16 Let Ω(y) := {x R n Ax = y; x 0}. Continuous Laplace-duality Discrete Z-duality f(y, c) := F (λ, c) := Ω(y) ec x dµ f d (y, c) := R m + e λy f(y, c) dy F d (z, c) := e c x x Ω(y) Z n y Z m z y f d (y, c) = 1 nk=1 (A λ c) k = n k=1 1 1 e c kz A 1k 1 z A mk m with R((A λ) k ) > c k k with z A 1k 1 z A mk m > e c k k 16
17 Continuous Laplace-duality Discrete Z-duality I. F (λ, c) f(b, c) I. F d (z, c) f d (b, c) by Inverse Laplace Transform and Cauchy s Residue Th. II. Data appear as by Inverse Z-Transform and Cauchy s Residue Th. II. Data appear as COEFFICIENTS EXPONENTS! of dual variables λ in F (λ, c) of dual variables z in F d (z, c). III. The poles of F (λ, c) are REAL III. The poles of F d (z, c) are COMPLEX and much more numerous! 17
18
19 Let σ := [A σ1 A σm ] be a feasible basis of {Ax = b, x 0}, with µ(σ) := det(a σ ), and dual variable π σ A σ = c σ. Continuous Laplace-duality basis σ poles : A σ λ = c σ a single real pole λ in R m Discrete Z-duality basis σ poles : z A σ = e c σ µ(σ) complex poles z = e λ in C m λ = π σ λ = π σ + i 2π v µ(σ) ; v V σ Z m V σ = {v Z m v A σ = 0 mod µ(σ)} 18
20 So both f d (b, c) and f(b, c) are computed with the same Cauchy s residue technique but the resulting contribution of each vertex for f d (b, c) is more complicated to evaluate because of the additional complex zeros. 19
21 Brion and Vergne s discrete formula Let σ := [A σ1 A σm ] be a feasible basis of {Ax = b, x 0}, with µ(σ) := det(a σ ), and µ(σ) dual variables z = e λ e 2iπv/µ(σ) C m with λ = π σ + i 2π v µ(σ) ; v V σ := {v Z m v A σ = 0 mod µ(σ)}. Re-interpreted with these data, Brion and Vergne s original discrete formula reads (Lasserre) f d (b, c) = e c x(σ) x(σ): vertex of Ω(b) 1 µ(σ) v V σ e2iπv b/µ(σ) k σ(1 e (2iπv A k /µ(σ)) e (c k π σ A k ) 20
22 Back to optimization : f d (b, c) = max{c x Ax = b; x N n }. Theorem max σ e c x(σ) lim r (Lasserre).Assume that v V σ k σ e2iπv b/µ(σ) (1 e (2iπv A k /µ(σ)) e r(c k π σ A k ) ) } is attained at a unique basis σ. Then : f d (b, c) = c x(σ ) + (c k π σ Ak )x k = k σ σ is an optimal basis of the linear program, and j c j x j. 1/r x(σ ) (resp. x ) is an optimal solution of the linear (resp. integer) program. 21
23 In this case : f d (b, c) = c x(σ ) + max A σ u k σ (c k π σ Ak )x k + k σ A k x k = b u Z m ; x k N k σ f d (b, c) = c x(σ ) + opt. value of GOMORY relaxation! 22
24 As a consequence, Corollary For t N sufficiently large, the gap f(tb, c) f d (tb, c) between the continuous and discrete optimal values is a (constant) periodic function with period µ(σ ) = det(a σ ), where σ is the optimal basis of the continous LP. The periodicity is due to the complex poles z = e λ with λ = π σ + i2π v µ(σ ), v V σ Zm ; v A σ = 0 mod µ(σ ). 23
25 KNAPSACK REVISITED Let a N n+1, 0 c R n+1, b N, and s := i a i. f d (b, c) = max x N n{c x a x = b}; f d (b, c) = x N n {e c x a x = b}. F d (z, c) = z s nj=1 (z a j e c j) ; f d (b, c) = 1 2iπ with γ R + and ln γ > max n i=1 c i/a i. z =γ zb 1 F d (z, c) dz The poles {z jk } of F satisfy : ln z jk = c j /a j + 2iπk/a j ; k = 1,..., a j ; j = 1,..., n 24
26 Poles of F(z) Integration path z = r 0 z = c_1/a_1 z = c_2/a_2 z =c_3/a_3 25 z = c_4/a_4
27 When the ratios {c k /a k } are close to each other, the integration of z b 1 F d (z) on the circle z = γ becomes more difficult because nearly all the poles of F d (z, c) contribute. Similarly, it is also known that the corresponding knapsack problems are difficult to solve... 26
28 Let r N, c 1 /a 1 < c 2 /a 2 < < c n /a n and c j N for all j. F d (z, rc) z = z s 1 nj=1 (z a j e rc j) = n j=1 Q j (z) (z a j e rc j), for some polynomials z Q j (z) := a j 1 k=0 Q jk z k = a j 1 k=0 [ Mjk (e r ) R jk (e r ) for some polynomials {M jk, R jk } of the variable y := e r. ] z k, In principle they can be obtained by symbolic calculation. Therefore, with b = s j mod a j, for all j = 1,..., n, f(b, rc) = n j=1 Q j(aj s j 1) y (b s j)c j a j = n j=1 M j(aj s j 1) (y) (b s j )c j R j(aj s j 1) (y) y a j 27
29 Hence e f d(b,c) = lim f r d (b, rc) 1/r = lim r n j=1 Thus, for sufficiently large b, e f d(b,c) = max j=1,...,n lim r M j(aj s j 1) (er (b s ) j )c j R j(aj s j 1) (er ) er a j M j(a j s j 1) (er ) R j(aj s j 1) (er ) er 1/r (b s j )c j a j 1/r f d (b, c) = c 1b a 1 c 1s 1 a 1 + deg M 1(a1 s 1 1) (y) deg R 1(a 1 s 1 1) (y) 28
30 Example : Consider the problem f(b, c) := max x N 2{3x 1 + 5x 2 2x 1 + 3x 2 b}. Let y = e r. We obtain F (z) z = z 5 (z 1)(z 2 y 3 )(z 3 y 5 ) = (y9 + y 7 + y 6 ) + z (y 7 + y 6 + y 4 ) + z 2 (y 6 + y 4 + y 2 ) (y 5 1)(y 1)(z 3 y 5 ) (y5 + y 3 ) + z (y 3 + y 2 ) (y 3 1)(y 1)(z 2 y 3 ) + 1 (y 5 1)(y 3 1)(z 1) Hence, f d (b, c) = f(b, c) if b = 0 mod 3 and for large b, f d (b, c) = f d (b, c) 5/3 + 1 whenever b = 1 mod 3 f d (b, c) = f d (b, c) 10/3 + 3 whenever b = 2 mod 3 29
31 A Discrete Farkas Lemma Let A N m n, b N m and consider the problem deciding whether or not Ax = b has a solution x N n. Theorem: (i) Ax = b has a solution x N n if and only if the polynomial b z b 1 in R[z 1,..., z m ] can be written z b 1 1 z b m 1 = n j=1 Q j (z)(z A j 1) = n j=1 Q j (z)(z A 1j 1 z A mj m 1) for some polynomials Q j (z), all with nonnegative coefficients. (ii) The degree of the Q j s is bounded by b := m j=1 b j max k m j=1 A jk. 30
32 A single LP to solve with n ( b ) ( +m b variables, b ) +m m constraints and a (sparse) matrix of coefficients in {0, ±1} One also retrieves the classical Farkas Lemma in R n, that is, {x R n Ax = b; x 0} = A u 0 b u 0. Indeed, if Ax = b has a solution x N n, then with u = ln z, Therefore, e b u 1 = n j=1 Q j (e u 1,..., e um )(e (A u) j 1). A u 0 e (A u) j 1 0 e b u 1 0 b u 0, and one retrieves (*), i.e., Ax = b has a solution x R n. 31
33 The general case A Z m n, b Z m. Let Ω := {x R n Ax = b; x 0} be a polytope. Let α N n be such that for every column A j of A, A kj +α j 0 k = 1,..., m; let N β ρ(α) := max{α x x Ω}. Theorem: (i) Ax = b has a solution x N n if and only if the polynomial z z b (zy) β 1 in R[z 1,..., z m ] can be written z b (zy) β 1 = Q 0 (z, y)(zy 1) + n j=1 Q j (z, y)(z A j(zy) α j 1), for some polynomials Q j (z), all with nonnegative coefficients. (ii) The degree of the Q j s is bounded by b := (m+1)β + m j=1 32 b j.
34 Back to standard Farkas lemma {x R n Ax = b, x 0} = [ A λ 0 ] b λ 0. But, equivalently {x R n Ax = b, x 0} = if and only if the polynomial λ b λ can be written b λ = n j=1 Q j (λ)(a λ) j, for some polynomials {Q j } R[λ 1,..., λ m ], all with nonnegative coefficients. In this case, each Q j is necessarily a constant, that is, Q j Q j (0) = x j 0, and Ax = b! 33
35 P = {x R n Ax = b, x 0} P Z n x P x = Q(0,..., 0) with x integer hull (P) x = Q(1,..., 1) with Q R[λ 1,..., λ m ] Q R[e λ 1,..., e λ m] b λ = Q, A λ e b λ 1 = Q, e A λ 1 n Q 0 Q 0 Comparing continuous and discrete Farkas lemma
36 An equivalent Linear program Let 0 q = {q jα } R ns be the coefficients of the Q j s in z b 1 = n j=1 Q j (z)(z A j 1) They are solutions of a linear system M q = r, q 0 for some matrix M and vector r, both with 0, ±1 coefficients. ** M and r are easily obtained from A, b with no computation Write q = (q 1, q 2,..., q n ) with each q j ĉ jα := c j for all α = {q jα } R s, and let 35
37 Theorem : Let A N m n, b N m, c R n. (i) The integer program P max{c x Ax = b, x N n } has same value as the linear program Q max{ n j=1 ĉ j q j M q = r; q 0}. (ii) Let q be an optimal vertex, and let x j := α q jα j = 1,..., n. Then x N n and x is an optimal solution of P. 36
38 The link with superadditive functions The LP-dual Q of the linear program Q reads Q min π {π r M π ĉ}. More precisely, with D := n j=1 {0, 1,..., b j } N m, Q min π π(b) π(0) s.t. π(α + A j ) π(α) c j, α D, j = 1,..., n Let Π := {π : N m R {+ } π(x) = if and only if x D}. For every π Π, let f π : N m R {+ } be the function f π (x) := inf α D π(α + x) π(α), x Nm 37
39 For every π Π, the function f π is superadditive and f π (0) = 0. The LP dual Q reads Q min π Π f π(b) s.t. f π (A j ) c j, j = 1,..., n. Thus, Q is a simplified and explicit form of the abstract dual of Jeroslow, Wolsey, stated in terms of superadditive functions. In the abstract dual one may restrict to the subclass of superadditive functions derived from the representation z b 1 = n j=1 Q j (z A j 1). 38
40 And, with P = {x R n + Ax = b} the integer hull co(p Zn ) reads co(p Z n ) = {x R n n j=1 f π (A j )x j f π (b)}, for finitely many π, generators of the convex cone (π, λ) : π(α + A j ) π(α) + λ j 0, α + A j D, j = 1,... n 39
41 CONCLUSION Generating functions permit to exhibit a natural duality for integer programming, an IP-analogue of LP duality. This duality also shows which kind of superadditive functions are useful in the abstract dual of Jeroslow, Wolsey. This might help providing efficient Gomory cuts in MIP solvers like CPLEX, or XPRESS-MP. 40
42 Another dual problem Let A Z m n, b Z m, c R n. Let y f(y, c) = max{c x Ax = y; x 0}. The Fenchel transform of the convex function f(., c) is the convex function λ ( f) (λ, c) = sup y R m λ y + f(y, c). The dual problem of the linear program is obtained from Fenchel duality as f(b, c) = inf λ R mb λ + ( f) ( λ, c) = inf λ R m b λ + sup x R m(c A λ) x = min {b λ A λ c} Equivalently e f(b,c) = inf λ R m sup e(b Ax) λ e c x x R m 41
43 Define Hence, ρ := inf z C m f d (z, c) = R z b sup R ( z b Ax ) e c x = inf f x N n z C m d (z, c). n j=1 (z A je c j) x j < if z A j e c j j (that is, A ln z c). Next, (writing z C as e λ e iθ ) ρ inf sup R ( z b Ax ) e c x = inf sup z R m x R n λ R m e(b Ax) λ e c x = e f(b,c) x R n Finally, with z C m arbitrary fixed sup R ( z b Ax ) e c x e c x = e f d(b,c) x N n Hence f d (b, c) ln ρ f(b, c). 42
44 Let σ be an optimal basis of the linear program. Under uniqueness of the max σ in Brion and Vergne s formula, and an additional technical condition e f d(b,c) = ρ = max x N n R ( ẑ b Ax e c x ) = f d (ẑ, c) where ẑ A j = γ e c j j σ for some real γ > 1. ẑ is an optimal solution of the dual problem inf f z C m d (z, c) 43
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