A Lower Bound on the Split Rank of Intersection Cuts
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1 A Lower Bound on the Split Rank of Intersection Cuts Santanu S. Dey H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology. 200
2 Outline Introduction: split rank, intersection cuts A lower bound on split rank of intersection cuts Outline of proof Application: mixing inequality Discussion
3 3 Split cut Let M := {(x, y) Z m R n Gx + Hy b} where G Q p m, H Q p n, and b Q p. Let M 0 := {(x, y) R m R n Gx + Hy b} denote the continuous relaxation of M.
4 Split cut Let M := {(x, y) Z m R n Gx + Hy b} where G Q p m, H Q p n, and b Q p. Let M 0 := {(x, y) R m R n Gx + Hy b} denote the continuous relaxation of M. Given a vector a Z m and c Z, any vector x Z m satisfies the split disjunction defined as (a T x c) (a T x c + ).
5 5 Split cut Let M := {(x, y) Z m R n Gx + Hy b} where G Q p m, H Q p n, and b Q p. Let M 0 := {(x, y) R m R n Gx + Hy b} denote the continuous relaxation of M. Given a vector a Z m and c Z, any vector x Z m satisfies the split disjunction defined as (a T x c) (a T x c + ). Let L 0 a,c := M 0 {(x, y) R m R n a T x c} and R 0 a,c := M 0 {(x, y) R m R n a T x c + } M 0 a,c := conv(l 0 a,c R 0 a,c) x <= 0 x >=
6 6 Split rank: A tool for comparing cuts with split cuts Definition (Split closure) The first split closure M is defined as a Z p,c ZM 0 a,c.
7 Split rank: A tool for comparing cuts with split cuts Definition (Split closure) The first split closure M is defined as a Z p,c ZM 0 a,c. [Cook et al.(988)] Split closure M is a polyhedron.
8 8 Split rank: A tool for comparing cuts with split cuts Definition (Split closure) The first split closure M is defined as a Z p,c ZM 0 a,c. [Cook et al.(988)] Split closure M is a polyhedron. Applying all possible split disjunctions to M, we obtain M 2,...
9 9 Split rank: A tool for comparing cuts with split cuts Definition (Split closure) The first split closure M is defined as a Z p,c ZM 0 a,c. [Cook et al.(988)] Split closure M is a polyhedron. Applying all possible split disjunctions to M, we obtain M 2,... Definition (Split rank) The split rank of a valid inequality α T x + β T y γ for M is defined as the smallest integer k such that α T x + β T y γ is valid for M k.
10 0 Some previous results. (Upper bound) Split rank of any inequality for a 0- MIP is at most n, when there are n binary variables: Balas (979), Nemhauser and Wolsey (990), Balas et. al. (993). 2. (Lower Bound) Split rank of certain class of cutting planes is not finite: Cook et. al. (988), Li and Richard (2008). 3. (Lower Bound) The upper bound on the split rank for 0- MIP is achieved: Cornuéjols and Li (2002). 4. (Upper Bound) Split rank of m row mixing inequalities is at most m: Dash and Günlük (2008). 5. (Upper Bound) Upper Bound on split rank for triangle and quadrilateral inequalities: D. and Louveaux (2009).
11 Basic MIP model for analysis: a relaxation of simplex tableau where. r i Q m 2. f Q m \ Z m. x = f + n i= r i y i x Z m y i 0 i {,..., n}
12 Basic MIP model for analysis: a relaxation of simplex tableau where. r i Q m 2. f Q m \ Z m. x = f + n i= r i y i x Z m y i 0 i {,..., n} Andersen et. al (2007), Borozan and Cornuéjols (2007), Cornuéjols and Margot (2008), D. and Wolsey (2008), Espinoza (2008), Zambelli (2008), Basu et. al (2008), Andersen et. al (2008).
13 3 Basic MIP model for analysis: a relaxation of simplex tableau where. r i Q m 2. f Q m \ Z m. x = f + n i= r i y i x Z m y i 0 i {,..., n} Andersen et. al (2007), Borozan and Cornuéjols (2007), Cornuéjols and Margot (2008), D. and Wolsey (2008), Espinoza (2008), Zambelli (2008), Basu et. al (2008), Andersen et. al (2008). Another Example: Mixing Set x i + y 0 f i i {,..., m} x Z m, y 0 0
14 Basic MIP model for analysis: a relaxation of simplex tableau where. r i Q m 2. f Q m \ Z m. x = f + n i= r i y i x Z m y i 0 i {,..., n} Andersen et. al (2007), Borozan and Cornuéjols (2007), Cornuéjols and Margot (2008), D. and Wolsey (2008), Espinoza (2008), Zambelli (2008), Basu et. al (2008), Andersen et. al (2008). Another Example: Mixing Set x i + y 0 f i i {,..., m} Introducing non-negative slack variable: x Z m, y 0 0 x i = f i y 0 + y i i {,..., m} x Z m, y 0 0, y i 0 i {,..., m}
15 5 Basic MIP model for analysis: a relaxation of simplex tableau where. r i Q m 2. f Q m \ Z m. x = f + n i= r i y i x Z m, Ax b y i 0 i {,..., n} Andersen et. al (2007), Borozan and Cornuéjols (2007), Cornuéjols and Margot (2008), D. and Wolsey (2008), Espinoza (2008), Zambelli (2008), Basu et. al (2008), Andersen et. al (2008). Another Example: Mixing Set x i + y 0 f i i {,..., m} Introducing non-negative slack variable: x Z m, y 0 0 x i = f i y 0 + y i i {,..., m} x Z m, y 0 0, y i 0 i {,..., m}
16 6 Intersection cut: lattice-free convex sets to valid cuts Let P R m be a convex set and Lattice-free, i.e., int(p) Z m = and f int(p).
17 7 Intersection cut: lattice-free convex sets to valid cuts Let P R m be a convex set and Lattice-free, i.e., int(p) Z m = and f int(p). Intersection Cut: [Balas (97)] π P (r i )y i, () i= where π P (r i ) = { λ if λ 0, s.t. f + r i boundary(p) λ 0 if r i is a ray for P. (2)
18 8 Example of intersection cut Projection on x-space 2.5 r 3 x = 0.7.5y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y x 0 2 r f r 2-0 x 2
19 9 Example of intersection cut Projection on x-space x r f r 2 r 3 x = 0.7.5y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y 3 0 Lattice-free convex set P: Convex Hull of (.32,.2) ( , 0.5) (0.7,.05) - 0 x 2
20 20 Example of intersection cut Projection on x-space 2.5 u 0.5 x r 0 2 v f r r 2-0 x 2 x = 0.7.5y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y 3 0 Lattice-free convex set P: Convex Hull of (.32,.2) ( , 0.5) (0.7,.05) f + r (u+v)/v y in cut is u+v v. Bnd(P) Coefficient of.456y y y 3
21 2 Problem statement To obtain an insight into the (lower bounds on) split rank of intersection cuts.
22 Main result: split rank depends on orientation of integer points satisfied at equality "As the number of integer points lying on distinct facets of P increases, the split rank increases."
23 23 Main result: split rank depends on orientation of integer points satisfied at equality "As the number of integer points lying on distinct facets of P increases, the split rank increases." More precisely: Step : Under a assumption on the columns of our basic relaxation of simplex tableau, construct a polyhedral subset of P (P is the lattice-free convex set used to generate the intersection cut) : restricted lattice-free set.
24 Main result: split rank depends on orientation of integer points satisfied at equality "As the number of integer points lying on distinct facets of P increases, the split rank increases." More precisely: Step : Under a assumption on the columns of our basic relaxation of simplex tableau, construct a polyhedral subset of P (P is the lattice-free convex set used to generate the intersection cut) : restricted lattice-free set. Step 2: Then the following result is proven: log 2 (t) is a lower bound on the split rank of intersection cut where {x, x 2,..., x t } is a subset of integer points on the boundary of the restricted lattice-free set such that no two points lie on the same facet of the restricted lattice-free set.
25 25 Assumption on columns Some notation: The lattice-free set P R m that is used to generate the intersection cut may not be a bounded set. If P is unbounded, some of the columns r i may be rays of P.
26 26 Assumption on columns Some notation: The lattice-free set P R m that is used to generate the intersection cut may not be a bounded set. If P is unbounded, some of the columns r i may be rays of P. Let r i be a ray for P if i {n +,..., n}.
27 27 Assumption on columns Some notation: The lattice-free set P R m that is used to generate the intersection cut may not be a bounded set. If P is unbounded, some of the columns r i may be rays of P. Let r i be a ray for P if i {n +,..., n}. Assumption: There exist subsets S v {,..., n } and S r {n +,..., n} of the columns such that. Q := conv i S v f + r i π P (r i ) + cone i Sr { }{{} r i } has a }{{} ray of P Bnd(P) dimensions of S v + S r 2. f affine.hull( Q). Q: Restricted lattice-free set.
28 28 Construction of Q r -2 r 4 r 3 r 5 r r -2 r 4 r 3 r 2 r
29 29 The lower bound result Theorem Let {x, x 2,..., x t } be a subset of integer points on the boundary of the restricted lattice-free set Q such that no two points lie on the same facet of Q. Then a lower bound on the split rank of the intersection cut is log 2 (t).
30 Outline of proof. 30
31 3 How to prove a lower bound on split rank? Set: x = f + r i y i x Z m, y R n + i= Intersection Cut: π P (r i )y i i=
32 32 How to prove a lower bound on split rank? Set: x = f + r i y i x Z m, y R n + i= Intersection Cut: π P (r i )y i. There exists ( x, ỹ) M k that does not satisfy the intersection cut, i.e., n i= πp (r i )ỹ i < i=
33 33 How to prove a lower bound on split rank? Set: x = f + r i y i x Z m, y R n + i= Intersection Cut: π P (r i )y i. There exists ( x, ỹ) M k that does not satisfy the intersection cut, i.e., n i= πp (r i )ỹ i < Intersection Cut has a split rank of at least k +. i=
34 34 How to prove a lower bound on split rank? Set: x = f + r i y i x Z m, y R n + i= Intersection Cut: π P (r i )y i. There exists ( x, ỹ) M k that does not satisfy the intersection cut, i.e., n i= πp (r i )ỹ i < Intersection Cut has a split rank of at least k We will show that points of the form (x I, y I δλ Sv,Sr ) M log2( I ), but does not satisfy intersection cut; where (x I, y I ) belongs to convex hull of M, where I {,..., t}. λ S v,s r R n is a fixed vector. i=
35 35 How to prove a lower bound on split rank? Set: x = f + r i y i x Z m, y R n + i= Intersection Cut: π P (r i )y i. There exists ( x, ỹ) M k that does not satisfy the intersection cut, i.e., n i= πp (r i )ỹ i < Intersection Cut has a split rank of at least k We will show that points of the form (x I, y I δλ Sv,Sr ) M log2( I ), but does not satisfy intersection cut; where (x I, y I ) belongs to convex hull of M, where I {,..., t}. λ S v,s r R n is a fixed vector. If δ > 0, then (x I, y I δλ Sv,Sr ) does not satisfy the intersection cut. i=
36 36 How to prove a lower bound on split rank? Set: x = f + r i y i x Z m, y R n + i= Intersection Cut: π P (r i )y i. There exists ( x, ỹ) M k that does not satisfy the intersection cut, i.e., n i= πp (r i )ỹ i < Intersection Cut has a split rank of at least k We will show that points of the form (x I, y I δλ Sv,Sr ) M log2( I ), but does not satisfy intersection cut; where (x I, y I ) belongs to convex hull of M, where I {,..., t}. λ S v,s r R n is a fixed vector. If δ > 0, then (x I, y I δλ Sv,Sr ) does not satisfy the intersection cut. We do the following next:. Construction of λ Sv,Sr R n and (x I, y I ) R m R n +. i= 2. Show that (x I, y I δλ Sv,Sr ) M log 2( I ). 3. Show that (x I, y I δλ Sv,Sr ) is cut off by the intersection cut for δ > 0.
37 37 Sv,Sr Construction of λ x 0 2 r f r 3 x = 0.7 2y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y 3 0 The cut is:.93y +.29y y r 2-0 x 2
38 38 Sv,Sr Construction of λ x r f r 2 r 3 x = 0.7 2y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y 3 0 The cut is:.93y +.29y y 3 Consider λ R 3 : (,, 2). Then ( ) ( ) ( ) 2 0 λ +λ 0 2 +λ 2 3 = 0 }{{}}{{}}{{} r r 2 r 3-0 x 2
39 39 Sv,Sr Construction of λ x r r 2-0 x 2 f r 3 x = 0.7 2y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y 3 0 The cut is:.93y +.29y y 3 Consider λ R 3 : (,, 2). Then ( ) ( ) ( ) 2 0 λ +λ 0 2 +λ 2 3 = 0 }{{}}{{}}{{} r r 2 r 3 λ.93 }{{} π P (r ) +λ 2.29 }{{} π P (r 2 ) +λ }{{} π P (r 3 ) > 0
40 40 Sv,Sr Construction of λ x r x 2 Proposition Sv,Sr If Assumption holds, then λ. n i= f r 2 r 3 x = 0.7 2y +0y 2 +y 3 x 2 = y 2y 2 +y 3 x, x 2 Z y, y 2, y 3 0 The cut is:.93y +.29y y 3 Consider λ R 3 : (,, 2). Then ( ) ( ) ( ) 2 0 λ +λ 0 2 +λ 2 3 = 0 }{{}}{{}}{{} r r 2 r 3 λ.93 }{{} π P (r ),Sr λsv i r i = 0 ( 0 is the origin in R m ) and +λ 2.29 }{{} π P (r 2 ) +λ }{{} π P (r 3 ) > 0 R n Sv,Sr (λi = 0 i / S v S r ) such that, 2. n i= πp (r i Sv,Sr )λi > 0.
41 4 Construction of (x I, y I ) x = (, 0) = f + 0r + 0.4r r 3 π P (r )(0) + π P (r 2 )(0.4) + π P (r 3 )(0.3) = 2.5 r 3 x 2 x {2,3} 0.5 r x {,2} f 0 x {3,} -0.5 x 2 x 3 x r 2-0 x 2
42 Construction of (x I, y I ) 2.5 r 3 x 2 x {2,3} 0.5 r x {,2} f 0 x {3,} -0.5 x 2 x 3 x x = (, 0) = f + 0r + 0.4r r 3 π P (r )(0) + π P (r 2 )(0.4) + π P (r 3 )(0.3) = x = (, 0); y := (0, 0.4, 0.3) x 2 = (, ); y 2 := (0., 0, 0.5) x 3 = (0, 0); y 3 := (0.3, 0.25, 0) r 2-0 x 2
43 43 Construction of (x I, y I ) 2.5 r 3 x 2 x {2,3} 0.5 r x {,2} f 0 x {3,} -0.5 x 2 x 3 x x = (, 0) = f + 0r + 0.4r r 3 π P (r )(0) + π P (r 2 )(0.4) + π P (r 3 )(0.3) = x = (, 0); y := (0, 0.4, 0.3) x 2 = (, ); y 2 := (0., 0, 0.5) x 3 = (0, 0); y 3 := (0.3, 0.25, 0) Note (x, y ), (x 2, y 2 ), (x 3, y 3 ) belong to convex hull r 2-0 x 2
44 Construction of (x I, y I ) 2.5 r 3 x 2 x {2,3} 0.5 r x {,2} f 0 x {3,} r 2-0 x 2 x 2 x 3 x x = (, 0) = f + 0r + 0.4r r 3 π P (r )(0) + π P (r 2 )(0.4) + π P (r 3 )(0.3) = x = (, 0); y := (0, 0.4, 0.3) x 2 = (, ); y 2 := (0., 0, 0.5) x 3 = (0, 0); y 3 := (0.3, 0.25, 0) Note (x, y ), (x 2, y 2 ), (x 3, y 3 ) belong to convex hull. For I {,..., t}, let x I = I Similarly y I = I x i. i I i I The points (x I, y I ) are satisfied at equality by π P, y i i= π P (r i )y I i =
45 45 A key lemma r x {2,3} x x 3 x {3,} x f r 2 r 3 x 2 x {,2} x = (, 0); y := (0, 0.4, 0.3) x 2 = (, ); y 2 := (0., 0, 0.5) x 3 = (0, 0); y 3 := (0.3, 0.25, 0) y {,2} := (0.05, 0.2, 0.4) y {2,3} := (0.2, 0.25, 0.25) y {3,} := (0.5, 0.325, 0.5) y {,2,3} := (0., 0.20, 0.) x Lemma If I >, then y I i > 0 i (S v S r ).
46 46 Points belonging to the LP relaxation Remember (x I, y I ) belongs to the LP relaxation for all I, i.e., x I = f + yi I r i. (3) i= There exists a vector λ Sv,Sr such that Sv,Sr λi r i = 0 (4) i= If I >, then y I i > 0 i(s v S r ) δ > 0 s.t.y I δλ Sv,Sr 0
47 47 Points belonging to the LP relaxation Remember (x I, y I ) belongs to the LP relaxation for all I, i.e., x I = f + yi I r i. (3) i= There exists a vector λ Sv,Sr such that Sv,Sr λi r i = 0 (4) Also, i= If I >, then y I i > 0 i(s v S r ) δ > 0 s.t.y I δλ Sv,Sr 0 δ > 0 s.t.(x I, y I δλ Sv,Sr ) belongs to the LP relaxation. i= π P (r i Sv,Sr )λi > 0 and i= π P (r i )y I i = (5)
48 Points belonging to the LP relaxation Remember (x I, y I ) belongs to the LP relaxation for all I, i.e., x I = f + yi I r i. (3) i= There exists a vector λ Sv,Sr such that Sv,Sr λi r i = 0 (4) Also, i= If I >, then y I i > 0 i(s v S r ) δ > 0 s.t.y I δλ Sv,Sr 0 δ > 0 s.t.(x I, y I δλ Sv,Sr ) belongs to the LP relaxation. i= π P (r i Sv,Sr )λi > 0 and i= π P (r i )y I i = π P (r i )(y I δλ Sv,Sr ) i <. i= If I > 2 0, then the point (x I, y I δλ Sv,Sr ) M 0 and is cut off by the intersection cut. 48 (5)
49 49 Proof by induction Let γ I,k = max{δ (x I, y I δλ Sv,Sr ) M k }. We just proved: If I >, then γ I,0 > 0. If I {,..., t} and I > 2 z, then γ I,z > 0
50 50 Proof by induction Let γ I,k = max{δ (x I, y I δλ Sv,Sr ) M k }. We just proved: If I >, then γ I,0 > 0. If I {,..., t} and I > 2 z, then γ I,z > 0 (x I, y I δλ Sv,Sr ) M log 2( I ).
51 5 Proof by induction Let γ I,k = max{δ (x I, y I δλ Sv,Sr ) M k }. We just proved: If I >, then γ I,0 > 0. If I {,..., t} and I > 2 z, then γ I,z > 0 (x I, y I δλ Sv,Sr ) M log 2( I ). 2.5 x 2 x {2,3} 0.5 x {,2} {, 2,3} x 0 x x 3 x {3,} x x - 0 2
52 52 Proof by induction Let γ I,k = max{δ (x I, y I δλ Sv,Sr ) M k }. We just proved: If I >, then γ I,0 > 0. If I {,..., t} and I > 2 z, then γ I,z > 0 (x I, y I δλ Sv,Sr ) M log 2( I ). 2.5 x 2 x {2,3} 0.5 x {,2} {, 2,3} x 0 x x 3 x {3,} x. Any disjunction π T x π 0 π T x π 0 + contains at least two of the integer points on one side, i.e., π T x π 0 and π T x 2 π So the point (x {,2}, y {,2} γ {,2},0 λ Sv,Sr ) M 0 {(x, y) π T x π 0 } x - 0 2
53 53 Proof by induction Let γ I,k = max{δ (x I, y I δλ Sv,Sr ) M k }. We just proved: If I >, then γ I,0 > 0. If I {,..., t} and I > 2 z, then γ I,z > 0 (x I, y I δλ Sv,Sr ) M log 2( I ). 2.5 x {2,3} 0.5 x {,2} {, 2,3} x 0 x x 3 x {3,} x x x Any disjunction π T x π 0 π T x π 0 + contains at least two of the integer points on one side, i.e., π T x π 0 and π T x 2 π So the point (x {,2}, y {,2} γ {,2},0 λ Sv,Sr ) M 0 {(x, y) π T x π 0 }. 3. So a point of the form (x {,2,3}, y {,2,3} δλ Sv,Sr ) conv(m 0 {(x, y) π T x π 0 }) (M 0 {(x, y) π T x π 0 + }) where δ > 0.
54 54 An useful extension Corollary Let { M 0 x = f + n := i= r i y i, x R m, y R n +, Ax b. and M = M 0 x Z m. Let P R m be a lattice-free convex set containing f in its interior. Let the inequality π P (r i )y i (6) i= be an intersection cut. Let {x, x 2,..., x t } be a subset of integer points on the boundary of the restricted lattice-free set such that no two points lie on the same facet of Q and Ax j b j l. Then a lower bound on the split rank of intersection cut is log 2 (t).
55 Application 55
56 56 Mixing inequalities The mixing set: x i = f i y 0 + y i i {,..., m} x i Z i {,..., m}, y 0 0, y i 0 i {,..., m} The mixing inequality (type ) y 0 f n f x (f i f i )x i, i=2
57 57 Mixing inequalities The mixing set: x i = f i y 0 + y i i {,..., m} x i Z i {,..., m}, y 0 0, y i 0 i {,..., m} The mixing inequality (type ) y 0 f n f x ( f n) y 0 + f D D y + i=2 where D = n i= f i(f i+ f i ) + f n( f n). (f i f i )x i, i=2 (f i f i ) y i, D
58 58 Mixing inequalities The mixing set: x i = f i y 0 + y i i {,..., m} x i Z i {,..., m}, y 0 0, y i 0 i {,..., m} The mixing inequality (type ) y 0 f n f x ( f n) y 0 + f D D y + i=2 (f i f i )x i, i=2 (f i f i ) y i, D where D = n i= f i(f i+ f i ) + f n( f n). The mixing inequality is an intersection cut generated using a maximal lattice-free simplex.
59 59 Mixing inequalities The mixing set: x i = f i y 0 + y i i {,..., m} x i Z i {,..., m}, y 0 0, y i 0 i {,..., m} The mixing inequality (type ) y 0 f n f x ( f n) y 0 + f D D y + i=2 (f i f i )x i, i=2 (f i f i ) y i, D where D = n i= f i(f i+ f i ) + f n( f n). The mixing inequality is an intersection cut generated using a maximal lattice-free simplex. If 0 < f < f 2 <... < f m <, then each facet of the mixing lattice-free set contains at least one integer point in its relative interior of the form: (0, 0, 0,..0), (0, 0, 0,..., ), (0, 0, 0,...,, ),..., (,,,..., ).
60 Mixing inequalities The mixing set: x i = f i y 0 + y i i {,..., m} x i Z i {,..., m}, y 0 0, y i 0 i {,..., m} The mixing inequality (type ) y 0 f n f x ( f n) y 0 + f D D y + i=2 (f i f i )x i, i=2 (f i f i ) y i, D where D = n i= f i(f i+ f i ) + f n( f n). The mixing inequality is an intersection cut generated using a maximal lattice-free simplex. If 0 < f < f 2 <... < f m <, then each facet of the mixing lattice-free set contains at least one integer point in its relative interior of the form: (0, 0, 0,..0), (0, 0, 0,..., ), (0, 0, 0,...,, ),..., (,,,..., ). If 0 < f < f 2 <... < f m <, then the split rank of the mixing inequality is at least log 2 (m + ). (Note: Dash and Günlük (2008) prove an upper bound on the split rank of mixing inequalities is m.) 60
61 6 Discrete lot-sizing problem with initial stock A formulation for single item discrete lot-sizing problem with initial stock s 0, with binary variables v u representing the decision to produce in the period u, and with constant capacity C is s 0 C + t u= v u d t C s 0 0, v u {0, } u {,..., n}, where d t is the sum of the demands from periods to t. t n (7)
62 62 Discrete lot-sizing problem with initial stock A formulation for single item discrete lot-sizing problem with initial stock s 0, with binary variables v u representing the decision to produce in the period u, and with constant capacity C is s 0 C + t u= v u d t C s 0 0, v u {0, } u {,..., n}, where d t is the sum of the demands from periods to t. Setting y 0 = s 0 C and f t = d t, we can re-write (7) as C t n (7) y 0 + x t f t t n (8) 0 x t x t t n (9) t x t = v u t n (0) y 0 0, x t Z + t n, v u {0, } u n. () If 0 < f < f 2 <... < f n <, then the split rank of facet-defining inequalities is at least log 2 (n + ). u=
63 63 Discussion Upshot: This is a non-trivial lower bound on the split rank of intersection cuts. The main insight: the effect of the orientation of integer feasible points satisfied at equality by intersection cuts on the split rank of the inequality. We also used this result to derive a lower bound on the split rank of mixing inequalities.
64 64 Discussion Upshot: This is a non-trivial lower bound on the split rank of intersection cuts. The main insight: the effect of the orientation of integer feasible points satisfied at equality by intersection cuts on the split rank of the inequality. We also used this result to derive a lower bound on the split rank of mixing inequalities. Possibilities of strengthening the result: It is however not clear whether the split rank of an inequality is a function of the structure of Q (for the best choice of S v and S r ) alone.
65 65 Discussion Upshot: This is a non-trivial lower bound on the split rank of intersection cuts. The main insight: the effect of the orientation of integer feasible points satisfied at equality by intersection cuts on the split rank of the inequality. We also used this result to derive a lower bound on the split rank of mixing inequalities. Possibilities of strengthening the result: It is however not clear whether the split rank of an inequality is a function of the structure of Q (for the best choice of S v and S r ) alone. This lower bound seems to be weak when the vertices of the lattice-free set Q are all integral and each facet contains an integer point in its relative interior. Theorem (Li and Richard (2008)) If Q R m is the lattice-free simplex defined by the following vertices: (0, 0, 0,...0), (m, 0, 0,...0), (0, m, 0,...0),..., (0, 0, 0..., m), then a lower bound to the split rank of the inequality is infinite. A more general result that combines the result of above Theorem?
66 Thank You. 66
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