Asteroide Santana, Santanu S. Dey. December 4, School of Industrial and Systems Engineering, Georgia Institute of Technology
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1 for Some for Asteroide Santana, Santanu S. Dey School of Industrial Systems Engineering, Georgia Institute of Technology December 4, / 38
2 1
3 1.1 Conic integer programs
4 for Conic integer programs Cutting planes: Cut-generating Stard integer program Consider the following integer program: min s.t. c x Ax b x Z n + where A R m n, b R m c R n. min c x s.t. Ax b 0 x Z n + R m + is a cone, which is closed, convex, pointed full dimensional 4 / 38
5 for Conic integer programs Cutting planes: Cut-generating Stard integer program Consider the following integer program: min s.t. c x Ax b x Z n + min c x s.t. Ax b 0 x Z n + where A R m n, b R m c R n. min c x s.t. Ax b R m + x Z n +, R m + is a cone, which is closed, convex, pointed full dimensional 4 / 38
6 for Conic integer programs Cutting planes: Cut-generating Stard conic integer program Consider the following conic integer program: inf s.t. c x Ax b K x Z n +, inf s.t. c x Ax K b x Z n +, K R m is a regular cone: closed, convex, pointed full dimensional. For instance, K is the cone: 5 / 38
7 for Conic integer programs Cutting planes: Cut-generating Conic integer program Consider the following conic integer program: inf s.t. c x Ax b K x Z n +, inf s.t. c x Ax K b x Z n + K R m is a regular cone: closed, convex, pointed full dimensional. For instance, K is the cone: 6 / 38
8 1.2 Cutting planes: Cut-generating
9 for Conic integer programs Cutting planes: Cut-generating Cutting planes approach Cutting planes are linear inequalities valid for: conv{x Z n + Ax K b}, in order to improve the dual bound over the stard convex relaxation. How can we generate these cuts in a systematic fashion? 8 / 38
10 for Conic integer programs Cutting planes: Cut-generating Cuts via Cut-generating Let m be the number of rows of the A matrix. Consider f : R m R such that 1. f (u) + f (v) f (u + v) for all u, v R m (subadditive); 2. u K v f (u) f (v) (non-decreasing w.r.t. K); 3. f (0) = 0. Denote the set of satisfying (1.), (2.), (3.) as F K. 9 / 38
11 for Conic integer programs Cutting planes: Cut-generating Cuts via Cut-generating Let m be the number of rows of the A matrix. Consider f : R m R such that 1. f (u) + f (v) f (u + v) for all u, v R m (subadditive); 2. u K v f (u) f (v) (non-decreasing w.r.t. K); 3. f (0) = 0. Denote the set of satisfying (1.), (2.), (3.) as F K. A valid inequality for the convex hull of the set of feasible solutions to is given by {x Z n + Ax K b}, n f (A j )x j f (b). j=1 9 / 38
12 for Conic integer programs Cutting planes: Cut-generating Cuts via Cut-generating Let m be the number of rows of the A matrix. Consider f : R m R such that 1. f (u) + f (v) f (u + v) for all u, v R m (subadditive); 2. u K v f (u) f (v) (non-decreasing w.r.t. K); 3. f (0) = 0. Denote the set of satisfying (1.), (2.), (3.) as F K. A valid inequality for the convex hull of the set of feasible solutions to is given by {x Z n + Ax K b}, n f (A j )x j f (b). j=1 n n Proof: ˆx is feasible = f (A j ) ˆx j f A j ˆx }{{} j j=1 j=1 (1.),(3.) }{{} (2.) f (b). 9 / 38
13 for Conic integer programs Cutting planes: Cut-generating Question: How good are? conv { x Z n + Ax K b }? = f F K {x R n + f (A j )x j f (b)} (1) 10 / 38
14 for Conic integer programs Cutting planes: Cut-generating Question: How good are? conv { x Z n + Ax K b }? = K R m + (stard IP) f F K {x R n + f (A j )x j f (b)} (1) Table: Strong Duality Conditions A is rational data matrix Johnson (1973, 1979), Jeroslow (1978, 1979) 10 / 38
15 for Conic integer programs Cutting planes: Cut-generating Question: How good are? conv { x Z n + Ax K b }? = K R m + (stard IP) K is arbitrary regular cone f F K {x R n + f (A j )x j f (b)} (1) Table: Strong Duality Conditions A is rational data matrix Discrete Slater condition: ˆx Z n + s.t. Aˆx b int(k) Johnson (1973, 1979), Jeroslow (1978, 1979) Morán, D., Vielma (2012) 10 / 38
16 for Conic integer programs Cutting planes: Cut-generating More structure known on F K when K = R m + 1 We only need a subset C m F R m + (Blair, Jeroslow (1982)): conv { x Z n + Ax R m + b } = where C m is a set of : {x R n + f (A j )x j f (b)}, f F R m C m + C m contains (non-decreasing) linear. f, g C m, then αf + βg C m, where α, β 0. f CG m, then f C M. 11 / 38
17 for Conic integer programs Cutting planes: Cut-generating More structure known on F K when K = R m + 1 We only need a subset C m F R m + (Blair, Jeroslow (1982)): conv { x Z n + Ax R m + b } = where C m is a set of : {x R n + f (A j )x j f (b)}, f F R m C m + C m contains (non-decreasing) linear. f, g C m, then αf + βg C m, where α, β 0. f CG m, then f C M. 2 The simplest function in this family are the famous Chvátal-Gomory cuts: where λ R m +. f λ (u) = λu, 11 / 38
18 Question: Can we identify structured sub-family of CGFs that already provide the integer hull for more general K?
19 2
20 2.1 : Definition
21 for Definition LCF bounded cut-off region LCFs are not sufficient (Dual cone K := {y x, y 0 x K}.) 15 / 38
22 for Definition LCF bounded cut-off region LCFs are not sufficient For w 1, w 2,, w p K define f : R m R as f (v) = g((w 1 ) v, (w 2 ) v,..., (w p ) v), where g F R p +. Then, f F K. 15 / 38
23 for Definition LCF bounded cut-off region LCFs are not sufficient For w 1, w 2,, w p K define f : R m R as f (v) = g((w 1 ) v, (w 2 ) v,..., (w p ) v), where g F R p +. Then, f F K. 15 / 38
24 for Definition LCF bounded cut-off region LCFs are not sufficient For w 1, w 2,, w p K define f : R m R as f (v) = g((w 1 ) v, (w 2 ) v,..., (w p ) v), where g F R p +. Then, f F K. 15 / 38
25 2.2 bounded cut-off region
26 for Definition LCF bounded cut-off region LCFs are not sufficient Which cuts can we obtain using LCFs Theorem Suppose T = {x R n Ax K b} has non-empty interior. Let π x π 0 be a valid inequality for the integer hull of T. Assume B := {x T π x π 0 } is bounded. Then, there exist w 1, w 2,..., w p K, such that 1 π x π 0 is a valid inequality for the integer hull of Q = {x R n (w i ) Ax (w i ) b, i [p]}, with 2 (w i ) A rational 3 p 2 n. 17 / 38
27 for Definition LCF bounded cut-off region LCFs are not sufficient Which cuts can we obtain using LCFs Theorem Suppose T = {x R n Ax K b} has non-empty interior. Let π x π 0 be a valid inequality for the integer hull of T. Assume B := {x T π x π 0 } is bounded. Then, there exist w 1, w 2,..., w p K, such that 1 π x π 0 is a valid inequality for the integer hull of Q = {x R n (w i ) Ax (w i ) b, i [p]}, with 2 (w i ) A rational 3 p 2 n. 17 / 38
28 for Definition LCF bounded cut-off region LCFs are not sufficient Proof Sketch 1 Because B is bounded, it is possible to find dual multipliers w 1,..., w q such that G := { x (w i ) Ax (w i ) b } } {{ } Outer approx of T {x π x π 0 } }{{} Cut-off region is bounded. 2 If int(g) Z n, add additional finite separating hyperplanes update G. 3 Stard parity argument to say that we need only 2 n these inequalities to obtain Q. Lemma Let C be a closed convex set. If int ((rec.cone(c)) ) z C, then there exists π Q n such that π z < π 0 π x x C. 18 / 38
29 for Definition LCF bounded cut-off region LCFs are not sufficient are sufficient for compact sets In particular, if T is compact has non-empty interior, then linear describe the integer hull of T, using no more than 2 n dual multipliers at a time. 19 / 38
30 for Definition LCF bounded cut-off region LCFs are not sufficient are sufficient for compact sets In particular, if T is compact has non-empty interior, then linear describe the integer hull of T, using no more than 2 n dual multipliers at a time. 19 / 38
31 2.2 are not enough
32 for Definition LCF bounded cut-off region LCFs are not sufficient Let Epigraph of a stard hyperbola T = {(x 1, x 2 ) R 2 + : x 1 x 2 1}. This set is conic representable. 21 / 38
33 for Definition LCF bounded cut-off region LCFs are not sufficient L m := If K = L 3, then where Second-order cone { } x R m x1 2 + x x m 1 2 x m T = {(x 1, x 2 ) R 2 + : x 1 x 2 1} = {x Z 2 + : Ax K b}, A = , b = / 38
34 for Epigraph of a stard hyperbola (cont.) Clearly, x 1 1 is valid inequality for the integer hull of T. Definition LCF bounded cut-off region LCFs are not sufficient However, every linear outer approximation for T contains integer points of the form (0, k). Thus, linear will never produce the cut x 1 1! 23 / 38
35 3 cuts
36 3.1 cuts: Description
37 for Description of function fγ in R 2 Theorem Consider the function f γ : R m R defined as: f γ (v) = The new CGF { γ v + 1 if v j 0 γ v Z, γ v otherwise, where γ Γ j interior (L m ) with j {1, 2,, m 1} m 1 Γ j := {γ R m γ m γ i, γ m > γ j }. i=1 Then, f γ F L m. 26 / 38
38 for Description of function fγ in R 2 Theorem Consider the function f γ : R m R defined as: f γ (v) = The new CGF { γ v + 1 if v j 0 γ v Z, γ v otherwise, where γ Γ j interior (L m ) with j {1, 2,, m 1} m 1 Γ j := {γ R m γ m γ i, γ m > γ j }. i=1 Then, f γ F L m. 26 / 38
39 for Description of function fγ in R 2 Example Consider again the hyperbola T = {x Z 2 + : Ax K b}, where A = 1 1, b = Choose j = 1 γ = (0, 0.5, 0.5). Then, { 0.5v v if v v v 3 Z, f γ (v) = 0.5v v 3 otherwise. Thus, f γ (A 1 ) = 1, f γ (A 2 ) = 0, f γ (b) = 1, which yields f γ (A 1 )x 1 + f γ (A 2 )x 2 f γ (b) x / 38
40 for Description of function fγ in R 2 Subadditivity Proposition For all j {1, 2,, m 1} γ Γ j interior (L m ), f γ (v) = { γ v + 1 if v j 0 γ v Z, γ v is subbaditive, i.e., u, v R m otherwise, f γ (u + v) f γ (u) + f γ (v). Proof: If u or v fits the first clause, then we have f γ (u+v) γ (u + v) +1 γ u + γ v +1 f γ (u)+f γ (v). Suppose u v do not satisfy the first clause. Two cases: 1. u + v does not fit in the first clause: it follows from subadditivity of ; 28 / 38
41 for Description of function fγ in R 2 Subadditivity (cont) 2. u + v satisfies the first clause: u j + v j 0, γ (u + v) = γ u + γ v Z. (a) Then, u j 0 or v j 0. WLOG assume u j 0. Hence γ u / Z, (b) since u doesn t fit in the first clause. It follows from (a) (b) γ v / Z. (c) Combining (a), (b) (c) we conclude f γ (u) + f γ (v) = γ u + γ v = γ u + γ v + 1 = f γ (u + v). 29 / 38
42 for Description of function fγ in R 2 Non-decreasing w.r.t. L m Proposition For all j {1, 2,, m 1} γ Γ j interior (L m ), f γ (v) = { γ v + 1 if v j 0 γ v Z, γ v otherwise, is non-decreasing w.r.t. L m, i.e., u L m v f γ (v) f γ (u). Proof: Let w L m j [m 1]. If γ L m, then γ w 0. If, in addition, γ Γ j interior (L m ) w j 0, then one can prove that γ w > 0. Suppose u L m v. For w := u v we find γ u γ v, ( ) which holds strictly whenever u j v j 0. Two cases: 30 / 38
43 for Description of function fγ in R 2 Non-decreasing w.r.t. L m (cont) 1. u fits in the first clause: using ( ) we obtain f γ (v) γ v + 1 γ u + 1 = f γ (u). 2. u does not fit in the first clause: (i) v does not fit in the first clause: using ( ) we obtain f γ (v) = γ v γ u = f γ (u). (ii) v fits in the first clause: so v j 0 γ v Z. - If u j = 0, then u j v j 0 hence ( ) holds strictly. Thus, f γ (v) = γ v + 1 γ u = f γ (u). - If u j 0, then γ u / Z (since u does not satisfy the first clause), using ( ) we obtain γ v γ u < γ u f γ (v) = γ v+1 γ u = f γ (u). 31 / 38
44 for Description of function fγ in R 2 f γ does not belong to F R m + Note that f γ is not necessarily non-decreasing with respect to R 3 +. Indeed, let j = 1 γ = (0, ρ, ρ) where ρ is a positive scalar. Then { ρ(v2 + v 3 ) + 1 if v 1 0 ρ(v 2 + v 3 ) Z, f γ (v 1, v 2, v 3 ) = ρ(v 2 + v 3 ) otherwise. Consider the vectors u = (0, 0, 1/ρ) v = ( 1, 0, 1/ρ). Then u R 3 + v, however f γ (u) = 1 < 2 = f γ (v). 32 / 38
45 3. f γ in R 2
46 for Description of function fγ in R 2 Consider where Conic sections in R 2 W = m W i, i=1 W i = {x R 2 A i x L m i b i }, where A i R m i 2, b i R m i L m i is the cone in R m i. (W i is a parabola, ellipse, branch of hyperbola, half-space) 34 / 38
47 for Description of function fγ in R 2 Theorem Assume interior W each constraint A i x L m i b i in the description of W is either a half-space or a single conic section. Then the following statements hold: (i) If W Z 2 =, then this fact can be certified with the application of at most two inequalities generated from linear or some f γ ; (ii) Assume interior(w ) Z 2. Every face π x π 0 of the integer hull of W, where π Z 2 is non-zero, can be obtained with exactly one function or one f γ. 35 / 38
48 for Description of function fγ in R 2 Conclusions Given a conic set with non-empty interior a valid inequality for its integer hull, if the set cut-off is bounded, then the valid inequality can be obtained via ; If the conic set is compact, then its integer hull can be described by ; For sets that are conic representable, we introduced a new interesting family of, f γ ; In R 2, the family f γ combined with linear are enough to describe the integer hull of the underlying conic set, under minor assumptions. 36 / 38
49 for Research questions Description of function fγ in R 2 1 Is the closure wrt f γ locally polyhedral? 2 Is the rank wrt f γ finite? 3 Is there any sort of natural generalization of f γ to other cones? 37 / 38
50 for Thank you! Description of function fγ in R 2 38 / 38
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