A geometric perspective on lifting

Transcription

1 A geometric perspective on lifting Michele Conforti Università di Padova, Gérard Cornuéjols Carnegie Mellon University and Université d Aix-Marseille, gc0v@andrew.cmu.edu Giacomo Zambelli Università di Padova, giacomo@math.unipd.it May 2009 Abstract Recently, it has been shown that minimal inequalities for a continuous relaxation of mixed integer linear programs are associated with maximal lattice-free convex sets. In this paper we show how to lift these inequalities for integral nonbasic variables by considering maximal lattice-free convex sets in a higher-dimensional space. We apply this approach to several examples. In particular we identify cases where the lifting is sequence-independent, and therefore unique. Introduction Let S be the set of integral points in some rational polyhedron in R n such that dim(s = n. We consider the following semi-infinite relaxation to a general MILP x = f + rs r + ry r r R n r R n x S s r 0, r R n ( y r 0, y r Z, r R n s, y have finite support. Given two functions ψ and π from R n to R, the inequality ψ(rs r + π(ry r r R n r R n (2 Supported by NSF grant CMMI065349, ONR grant N and ANR grant BLAN

2 is valid for ( if it holds for every (x, s, y satisfying (. If (2 is valid, we say that the function (ψ, π is valid for (. A valid function (ψ, π is minimal if there is no valid function (ψ, π distinct from (ψ, π such that ψ (r ψ(r, π (r π(r for all r R n. The following simpler model has been studied recently [9] x = f + rs r r R n x S s r 0, r R n (3 s has finite support. We refer to this model as the continuous semi-infinite relaxation relative to f. Given a valid function ψ for (3, the function π is a lifting of ψ if (ψ, π is valid for (. Minimal valid inequalities for (3 are well understood in terms of maximal S-free convex sets. We are interested in characterizing liftings of minimal valid inequalities for (3. If ψ is a minimal valid function for (3 and π is a lifting of ψ such that (ψ, π is minimal, we say that π is a minimal lifting of ψ. We remark that, given any valid function ψ for (3 and a lifting π of ψ, the function π defined by π (r = min{ψ(r, π(r} is also a lifting for ψ. Indeed, given ( s, ȳ satisfying (, we show that r R n ψ(r s r + r R n π (rȳ r. Let ( s, ỹ be defined by s r = s r, ỹ r = ȳ r for every r R n such that π(r ψ(r, and s r = s r +ȳ r, ỹ r = 0 for every r R n such that ψ(r < π(r. One can readily verify that ( s, ỹ satisfies (, hence r R n ψ(r s r + r R n π(rỹ r. Furthermore, r R n ψ(r s r + r R n π (rȳ r = r R n ψ(r s r + r R n π(rỹ r In particular, if ψ is a minimal valid function for (3 and π is a minimal lifting of ψ, then π ψ. We first concentrate on deriving the best possible lifting coefficient of one single integer variable. Namely, given d R n, we consider the model x = f + rs r + dz r R n x S s r 0, r R n (4 z 0, z Z, s has finite support. Given a minimal valid function ψ for (3, we want to determine the minimum scalar λ such that the inequality r R n ψ(rs r + λz is valid for (4. Given d R n, let π l (d be such minimum λ. By definition, π l π for every lifting π of ψ. In general, the function (ψ, π l is not valid for (. However, when (ψ, π l is valid, π l is the unique minimal lifting for ψ. In this paper we give a geometric characterization of the function π l, and use this characterization to analyze specific functions ψ in which π l is the unique minimal lifting. 2

3 A valid function (ψ, π is extreme for ( if there do not exist distinct valid functions (ψ, π, (ψ 2, π 2 such that (ψ, π = 2 (ψ, π + 2 (ψ2, π 2. Note that if ψ is extreme for (3, then ψ is minimal. Remark. If ψ is extreme for (3 and π l is a lifting for ψ, then (ψ, π l is extreme for (. Indeed, given valid functions (ψ, π, (ψ 2, π 2 such that (ψ, π = 2 (ψ, π + 2 (ψ2, π 2, then ψ = ψ 2 = ψ, since ψ is extreme for (3, and π = π 2 = π l since π π l and π 2 π l. 2 Lifting and S-free convex sets We observe that (4 is equivalent to the following ( ( x f = + x n+ 0 r R n ( r s 0 r + ( d z (x, x n+ S Z + s r 0, r R n (5 z 0, s has finite support. Indeed (x, s, z is a solution for (4 if and only if (x, x n+, s, z is a solution to (5 by setting x n+ = z. Note that the above is obtained from the continuous semi-infinite relaxation relative to ( f by setting to 0 all variables relative to rays with nonzero (n + -th component, except for ( d. Therefore, given any valid function ψ for the continuous semi-infinite relaxation relative to ( f, then if we let ψ(r = ψ( r 0 for r R n and λ = ψ ( d, the inequality r R n ψ(rs r + λz is valid for (5 and for (4. A convex set is S-free if it does not contain any point of S in its interior. Maximal S-free convex sets were characterized in [4], where it was also shown that there is a one-to-one correspondence between minimal valid functions for (3 and maximal S-free convex sets with f in their interior. Theorem 2. A full-dimensional convex set B is a maximal S-free convex set if and only if it is a polyhedron such that B does not contain any point of S in its interior and each facet of B contains a point of S in its relative interior. Furthermore if B conv(s has nonempty interior, lin(b contains rec(b conv(s. We explain how minimal valid inequalities for (3 arise from maximal S-free convex sets. Let B a polyhedron with f in its interior, and let a,..., a t R q such that B = {x R n a i (x f, i =..., t}. We define the function ψ B : R n R by ψ B (r = max i=,...,t a ir. Note that the function ψ B is subadditive, i.e. ψ B (r + ψ B (r ψ B (r + r, and positively homogeneous, i.e. ψ B (λr = λψ B (r for every λ 0. 3

4 We claim that, if B is a maximal S-free convex set, r R n ψ B (rs r is valid for (3. (6 Indeed, let (x, s be a solution of (3. Note that x S, thus x / int(b. Then ψ B (rs r = ψ B (rs r ψ B ( rs r = ψ B (x f, r R n r R n r R n where the first equation follows from positive homogeneity, the first inequality follows from subadditivity of ψ B and the last one follows from the fact that x / int(b. The above functions are minimal [4],[8]. It was proved in [4] that the converse is also true, namely that every minimal function valid for (3 is of the form ψ B where B is a maximal S-free convex set with f in its interior. Example. We consider problem ( when n =, 0 < f < and S = Z. In this case the only maximal S-free convex set containing f is the interval B = [0, ]. Thus B = {x R f (x f, ( f (x f } and ψ B (r = max{ f r, ( f r}. Let ψ be a minimal valid function for (3, and let B = {x R n a i (x f, i =,..., t} be a maximal S-free convex set with f in its interior such that ψ = ψ B. We define the set B(λ R n+ as follows B(λ = { ( x, x n+ R n+ a i (x f + (λ a i dx n+, i =,..., t}. (7 Theorem 3. The inequality r R n ψ(rs r + λz is valid for (4 if and only if B(λ is (S Z + -free. Proof. Let ψ = ψ B(λ. By construction, ψ ( r 0 = ψ(r for all r R n, while ψ ( d = λ. We show the if part of the statement. Given λ such that B(λ is (S Z + -free, it follows by claim (6 that the function ψ is valid for the continuous semi-infinite relaxation relative to ( f 0. This implies that r R n ψ(rs r + λz is valid for (4. We now prove the only if part. Let λ be such that r R n ψ(rs r + λz is valid for (4. Given a point ( x x n+ S Z+, we show that such point is not in the interior of B(λ. Indeed, let r = x x n+ d f, z = x n+, and ( s r r R n be defined by { if r = r, s r = 0 otherwise. Note that f + r R n r s r + d z = f + r + x n+ d = x. Since x S and r R n ψ(rs r + λz is valid for (4, we have ψ(r s r + λ z = ψ( r + λ x n+ = max a i r + λ x n+ r R n i=,...,t = max [a i( x f + (λ a i d x n+ ]. i=,...,t Thus there exists i {,..., t} such that a i ( x f + (λ a i d x n+. This shows that ( x x n+ is not in the interior of B(λ. 4

5 Example (continued. In the previous example, let d R and λ R. If λ 0, then the set B(λ is the 2-dimensional polyhedron with two facets, containing the points ( ( 0 and 0 respectively and with one vertex, namely ( f 0 + λ ( d. If λ = 0, then B(λ is the split [0, ] + ( d. It is immediate to verify that, for λ < 0, the interior of B(λ contains one of the integral points ( ( d or d. For example, let f = 4. For d = 3 2, ψ B(d = 2. One can readily verify that B(λ is Z Z + -free if and only if λ 2 3, otherwise it contains the point ( 2. Hence πl (d = 2 3. For d =, ψ B (d = 4 3. It is immediate that B(λ is Z Z +-free if and only if λ 0, hence π l (d = 0. Figure : Example: f = 4. Left: d = 3 2. Right: d =. Theorem 4. Let ψ be a minimal valid function for (3 and π be a minimal lifting of ψ. Then there exists ε > 0 such that ψ, π and π l coincide on the ball of radius ε centered at the origin. Proof. Since ψ is a minimal valid function for (3, there exists a maximal S-free convex set B = {x R n a i (x f, i =,..., t} such that ψ = ψ B. Let α = max max (a i a j r i,j t r = Since B is a maximal S-free convex set, every facet of B contains a point of S in its relative interior. Hence, for i =,..., t, there exists x i S such that a i (x i f = and a j (x i f γ i, j i, for some positive γ i. Let ε > 0 such that εα γ i for i =,..., t. Let d R n such that d ε. We will show that, for every λ < ψ(d, B(λ contains a point of S Z + in its interior. By Theorem 3, this implies that π l (d ψ(d. Since π l π ψ, this implies π l (d = π(d = ψ(d. Let i, i t, such that ψ(d = a i d. Let λ = ψ(d δ for some δ > 0. We show that B(λ contains the point ( x i in its interior. Indeed, by (7, B(λ is the set of points in R n+ satisfying the inequalities a j (x f + [(a i a j d δ]x n+, j =,..., t. 5

6 Substituting ( x i we obtain a i (x i f δ <, a j (x i f + (a i a j d δ <, j =,..., t, j i, where the first inequality follows from a i (x i f =, while the second follows from a j (x i f γ i, d ε, and (a i a j (d/ d α by our choice of α. Thus ( x i is in the interior of B(λ. Example (continued. From the previous example where n =, 0 < f < and S = Z, note that π l (d = ψ B (d for every d [ f, f]. Indeed, if d < 0, then B(λ contains ( 0 for all λ < ψ B (d, while if d 0 then B(λ contains ( for all λ < ψb (d. Furthermore, for λ = ψ B (d, if d < 0 the facet of B(λ containing ( ( 0 is vertical and contains the point 0, if d 0 then the facet of B(λ containing ( ( is vertical and contains the point. Theorem 4 implies that, for every minimal valid function ψ for (3, there exists a region R ψ R n containing the origin in its interior such that ψ and π coincide in R ψ for every minimal lifting π of ψ for (. Lemma 5. Let ψ be a minimal valid function, and π be a minimal lifting of ψ. Then i For every r R n and w Z n lin(conv(s, π(r = π(r + w. ii For every r R n such that r + w R ψ for some w Z n lin(conv(s, π(r = ψ(r + w. Proof. i Let r R n and w Z n lin(conv(s. Suppose π( r π( r + w. Since w Z n lin(conv(s, we may assume π( r > π( r + w. Since w Z n lin(conv(s, then a point x R n is in S if and only if x + w S. Thus a point ( x, s, ȳ satisfies ( if and only if ( x + wȳ r, s, ỹ satisfies (, where ỹ r = 0, ỹ r+w = ȳ r+w + ȳ r, and ỹ r = ȳ r for every r R n \ { r, r + w}. This shows that the function π defined by π ( r = π( r + w, π (r = π(r for every r R n \ { r} is a lifting of ψ, contradicting the minimality of π. ii It follows from i that π(r = π(r + w. By definition of R ψ, π(r + w = ψ(r + w. The above lemma implies the following result. Theorem 6. If for every r R n there exists w r Z n lin(conv(s such that r+w R ψ, then there exists a unique minimal lifting for ψ, namely the function π defined by π(r = ψ(r+w r. Furthermore π = π l. Note that, if for some r R ψ there exists w Z n lin(conv(s such that r + w R ψ, then ψ(r + w = ψ(r. Example (continued. From the previous example where n =, 0 < f < and S = Z, we have shown that ψ(r = π l (r for every r [ f, f]. Note that, for every r R, r r + f [ f, f]. Thus π l (r = ψ(r r + f for all r R, and π l is the unique minimal lifting for ψ. Thus π l (r = max{ f (r r + f, ( f (r r + f }. More explicitly, if r r < f, then π l (r = r r f, while if r r f, π l(r = r r f. 6

7 Given a tableau row x = f + h i= pi s i + k j= qj y j, where s i 0, i =,..., h, and y j 0 and integer, j =,..., h, the inequality h i= ψ(pi s i + k j= π l(q j y j is h i= p i 0 p i f s i + h i= p i <0 pi f s i + k j= q j q j < f q j q j y j + f k j= q j q j f which is the Gomory Mixed Integer Cut associated with the tableau row. 3 Applications 3. Wedge inequalities q j q j y j, f We consider the problem ( where n = 2 and S = Z Z +. We focus on inequalities arising from maximal S-free convex sets with 2 sides and one vertex. We call such sets wedges. Figure 2: Wedges and corresponding region R shaded in gray. The inequality corresponding to the wedge on the right has a unique minimal lifting. Let B = {x R 2 a i (x f, i =, 2} be such a maximal S-free convex set. Since B is S-free, its only vertex must be in the interior of conv(s, rec(b has dimension 2 and for every nonzero element r rec(b, r 2 < 0. Note that rec(conv(s = R R + and B has empty lineality space. By Theorem 2, lin(b rec(b conv(s, hence rec(b conv(s ( =. In particular, (R {0} rec(b =, thus by symmetry we may assume a ( < 0 and a2 > 0, that is a < 0 and a 2 > 0. Let ˆr be a nonzero vector such that a ˆr = a 2ˆr. Note that any point x R 2 can be uniquely written as x = f + α xˆr + β x( where α x, β x R. Let x S be a point in the relative interior of one of the two facets of B, say a h ( x f =, a k ( x f <. Note that 0 > (a k a h ( x f = β x (a k a h, hence β x < 0 if h = and β x > 0 if h = 2. Let x be a point of S in the relative interior of the facet defined by a (x f such that β x is largest possible, and x 2 be a point of S in the relative interior of the facet defined by a 2 (x f such that β x2 is smallest possible. Let β i = β xi. Note that β < 0 < β 2. We define the region R = [β, β 2 ] + ˆr. (See Figure 2. 7

8 Lemma 7. For every d R, π l (d = ψ B (d. Proof. Let d R, that is d = αˆr + β (, for some α R and β [β, β 2 ]. We consider the case β 0. The case β 0 is similar. Note that (a a 2 d = α(a a 2 ˆr + β(a a 2 0 since (a a 2 ˆr = 0, β 0, a < 0 and a 2 > 0. Hence ψ B (d = max{a d, a 2 d} = a d. We will show that, for every λ < ψ B (d, the set B(λ defined in (7 contains the point ( x in its interior. By Theorem 3, this will imply πl (d ψ B (d, and thus π l (d = ψ B (d. Let λ = ψ B (d δ for some δ > 0. Then B(λ is the set of x R 3 satisfying a (x f δx 3, a 2 (x f + (a a 2 dx 3 δx 3. Substituting ( x in the first inequality, we obtain a (x f δ = δ <. Substituting in the second inequality, we obtain a 2 (x f + (a a 2 d δ = α x a 2ˆr + β a 2 + α(a a 2 ˆr + β(a a 2 δ = α x a ˆr + β a + (β β (a a 2 δ a (x f δ = δ < where the first inequality in the last row follows from β β, a < 0, a 2 > 0. Thus ( x is in the interior of B(λ. Let y and y 2 be the intersection of the facets defined by a (x f and a 2 (x f, respectively, with the axis x 2 = 0. That is a (y f =, y 2 = 0, and a 2(y 2 f =, y 2 2 = 0. Since B is S-free, y2 y, where equality holds if and only if y, y 2 are integral. Furthermore, it is not difficult to show that β 2 β y 2 y. Thus β 2 β = if and only if y, y 2 are integral vectors. In this case, for every r R 2 there exists w r Z {0} such that r + w r R. Since lin(conv(s = R {0}, by Theorem 6, π l (r is the unique minimal lifting of ψ B, and π l (r = ψ B (r + w r for every r R 2. Dey and Wolsey [9] show that ψ B is extreme for (3 if and only if B contains at least three points of S. Thus Remark implies the following: Theorem 8. If B contains at least three points of S and B (R {0} is an interval of length one, then (ψ B, π l is a valid extreme inequality for (. 3.2 Simplicial polytopes In this section we focus on valid inequalities for (3 arising from maximal lattice-free simplicial polytopes, in the case where S = Z n. Recall that a polytope is simplicial if each of its facets is a simplex. Let B = {x R n a i (x f, i =,..., t} be an n-dimensional maximal lattice-free simplicial polytope and let v,..., v p be its vertices. For i =,..., t, let V i {,..., p} be the set of indices of vertices of the facet defined by a i (x f, that is a i v j = for all j V i. Let r i = v i f, i =,..., p. Note that, since B is simplicial, {r j j F i } consists of 8

9 n linearly independent vectors, for i =,..., t, and a i r j = for all j F i, while a i r j < for all j / F i. Let x be an integral point in the interior of the facet defined by a i (x f, that is a i ( x f =, a j ( x f <, j i. Then x can be uniquely written as x = f + j F i ᾱ j r j, where j F i ᾱ j =, ᾱ j 0, j F i. Let R( x = { j F i α j r j 0 α j ᾱ j, j F i }. Let us denote by I the set of all points x in Z n such that x is contained in the relative interior of some facet of B. Let R = x I R( x. Lemma 9. For every d R, π l (d = ψ B (d. Proof. We only need to show that, given x I and d R( x, π l (d = ψ B (d. By symmetry we may assume that x is in the relative interior of the facet defined by a ( x f, and that F = {,..., n}. Let ᾱ,..., ᾱ n nonnegative such that n j= ᾱj = and x = f + n j= ᾱjr j. Since d R( x, there exist α,..., α n such that d = n j= α jr j and 0 α j ᾱ j, j =,..., n. Note that, for i =,..., t, (a a i d = n j= α j(a a i r j 0. Thus ψ B (d = a d. We will show that, for every λ < ψ B (d, the set B(λ defined as in (7 contains the point ( x in its interior. By Theorem 3, this will imply πl (d ψ B (d, and thus π l (d = ψ B (d. Let λ = ψ B (d δ for some δ > 0. Then B(λ is the set of x R n+ satisfying a (x f δx n+, a i (x f + (a a i dx n+ δx n+, i = 2,..., t. Substituting ( x in the first inequality, we obtain a ( x f δ = δ <. Substituting in the ith inequality, i = 2,..., n +, we obtain a i ( x f + (a a i d δ = = ᾱ j a i r j + j= ᾱ j j= = α j (a a i r j δ j= ᾱ j ( a i r j + j= α j ( a i r j δ j= (ᾱ j α j ( a i r j δ j= δ < where the equality in the second line follows from a i r j = for j =,..., n, the equality on the third line follows from n j= ᾱj =, while the first inequality on the last line follows from α j ᾱ j and a i r j. In light of Theorem 6, we are interested in cases where for every r R n there exists w r Z n such that r + w r R, since in this case π l is the unique minimal lifting. Dey and Wolsey [8] studied the case n = 2. In this case maximal lattice free polytopes are either triangles or quadrilaterals [0]. Dey and Wolsey show that the above property holds if and only if B is a triangle containing at least four integral points (see Figure 3, while it does not hold if B is a triangle containing exactly three integral points or if B is a quadrilateral. They also show that, when B is a triangle with at least four integral points, 9

10 Figure 3: is shaded. Lattice free triangles giving inequalities with a unique minimal lifting. Region R (ψ B, π l is extreme for (. This fact also follows from Remark and from the fact that ψ B is extreme for (3 whenever B is a maximal lattice-free triangle [7]. We next show that the above property holds when B is the n-dimensional simplex conv{0, ne,..., ne n }, where e i denotes the ith unit vector. We assume that f is in the interior of B. The picture on the left in Figure 3 shows the case n = 2. Note that B = {x R n n i= x i n, x i 0, i =,..., n}. The point e e i, where e denotes the vector of all ones, is the unique integral point in the relative interior of the facet of B defined by x i 0 and e is the unique integral point in the relative interior of the facet of B defined by n i= x i n. Thus I = {e, e e,..., e e n } where e denotes the vector of all ones. Let d,..., d n+ be defined as follows: d i = e i n f, i =,..., n and dn+ = n f. Then R(e = { n j= α jd j 0 α i, i =,..., n} and R(e e i = { n+ j= α jd j 0 α k, k =,..., n +, α i = 0}. Therefore R = { n+ j= α jd j 0 α i, i =,..., n +, α i = 0 for some i, i n + } Lemma 0. Let B = conv{0, ne,..., ne n }. For every r R n, there exists w Z n such that r + w R. Proof. Note that, for i, j n +, d i d j Z n. Let C i = cone{d j j i, j n + }, i =,..., n +. Note that n+ i= C i = R n and C i C k = cone{d j j i, k, j n + }. Furthermore, d i C i for i =,..., n +. Claim: Let r R n and let i such that r C i. There exists a unique α R n+ such that r = n+ j= α jd j and α i = 0. Furthermore, α is nonnegative and α j α j for every nonnegative α R n+ such that r = n+ j= α j dj. We prove the claim. Since C i is generated by n linearly independent vectors, r can be uniquely written as r = n+ j= α jd j such that α i = 0, and α must be nonnegative since r C i. 0

11 Given a nonnegative α R n+ such that r = n+ j= α j dj distinct from α, then α i > 0. Hence n+ d i = (α i (α j α j d j j= j i thus α j α j 0 since d i C i, hence by the above argument d i can be uniquely written as a linear combination of the extreme rays of C i, and such combination is nonnegative. This proves the claim. Let us now consider r R n. Let i be such that r C i, i n +. Let α R n+ such that r = n+ j= α jd j and α i = 0. By the above claim α is nonnegative. Let ᾱ = max j=,...,n+ α j. If ᾱ, then r R. If not, α k = ᾱ > for some k n +. Let r = r + (e i e k = r + (d i d k. Then r = j i,k α jd j + d i + (α k d k. Let h be such that r C h, h n + and let α R n+ be the unique vector such that r = n+ j= α j dj and α h = 0. By the previous claim, α satisfies the following properties r r Z n and α h = 0, 0 α j α j, j i, j n +, 0 α i, 0 α k α k. Thus, either max j=,...,n+ α j ᾱ, or the number of indices j such that α j = ᾱ is smaller than the number of indices j such that α j = ᾱ. This implies the statement of the lemma. It can be shown that, in this case, R is a polytope with ( n+ 2 pairs of parallel facets, and that R has volume. Thus, by Lemma 0, all possible translations of R by integral vectors form a tiling of R n. Therefore for every d R n, there exists w d Z n such that d + w d R. By Theorem 6, the function π l defined by π l (d = ψ B (d + w d is the unique minimal lifting of ψ B. Whenever B is a maximal lattice-free simplex, ψ B is extreme for (3. Indeed, if v,..., v n+ are the vertices of B and we define r j = v j f, j =,..., n +, ψ B is extreme for (3 if and only if n+ j= ψ B(r j s j is extreme for the convex hull of the set R f (r,..., r n+ defined as the set of all s R n+ such that f + n+ j= rj s j Z n and s 0 (see [9]. In this case, since each facet of B contains an integral point, for i =,..., n + there exists s i R n+ such that s i j > 0 for all j i, j n +, si i = 0 and n+ j= si j rj Z n. Hence s,..., s n+ are linearly independent points of R f (r,..., r n+, and n+ j= ψ B(r j s i j = for i =,..., n +. This shows that n+ j= ψ B(r j s j defines a facet of conv(r f (r,..., r n+, and thus it is extreme for conv(r f (r,..., r n+. Therefore ψ B is extreme for (3. The above statement and Remark imply the following. Theorem. If B = conv(0, ne,..., ne n, (ψ B, π l is extreme for ( with S = Z n.

12 3.3 Simple cones We consider the case were S = Z n Z + and the maximal S-free convex set B is the translation of a simple cone. That is, B has a unique vertex v, and B v is a simple cone. Recall that a polyhedral cone in R n is simple if it is generated by n linearly independent vectors, and therefore it has n facets. This case extends the wedge inequalities of Section 3.. Let B = {x R n a i (x f, i =..., n}. By Theorem 2, rec(b rec(conv(s is contained in the lineality space of B, which is empty. Therefore B conv(s is bounded. Therefore the polytope B (R n {0} is an an (n -dimensional simplex P. Let v,..., v n be the vertices of P, and let r j = v j f, j =,..., n. By symmetry, we may assume that a i r j = for i, j n, i j, and a i r i <. Let ˆr = v f. Note that, for i =,..., n, a iˆr =. Let x be a point of S in the relative interior of one of the facets of B, say the facet defined by a h (x f. Then x can be uniquely written as x = f + ᾱˆr + n j= ᾱjr j such that 0 ᾱ j, j =,..., n, and ᾱ h = 0. Let R( x = { n j= α jr j 0 α j ᾱ j, j =,..., n} + ˆr. Let us denote by I the set of all points x in S such that x is contained in the relative interior of some facet of B. Let R = x I R( x. Lemma 2. For every d R, π l (d = ψ B (d. Proof. We only need to show that, given x I and d R( x, π l (d = ψ B (d. By symmetry we may assume that x is in the relative interior of the facet defined by a (x f. Let ᾱ R and ᾱ 2,..., ᾱ n nonnegative such that x = f + ᾱˆr + n j=2 ᾱjr j. Since d R( x, there exist α R and α,..., α n such that d = αˆr + n j=2 α jr j and 0 α j ᾱ j, j = 2,..., n. Note that, for i = 2,..., t, (a a i d = α(a a i ˆr + n j=2 α j(a a i r j 0, since (a a i ˆr = 0 and (a a i r j 0. Thus ψ B (d = a d. We will show that, for every λ < ψ B (d, the set B(λ defined in (7 contains the point ( x in its interior. By Theorem 3, this will imply πl (d ψ B (d, and thus π l (d = ψ B (d. Let λ = ψ B (d δ for some δ > 0. Then B(λ is the set of x R n+ satisfying a (x f δx n+, a i (x f + (a a i dx n+ δx n+ i = 2,..., t. Substituting ( x in the first inequality, we obtain a ( x f δ = δ <. Substituting in the ith inequality, i = 2,..., n +, we obtain a i ( x f + (a a i d δ = ᾱa iˆr + = ᾱa ˆr + ᾱ j a i r j + α(a a i ˆr + j=2 α j (a a i r j δ j=2 ᾱ j a r j ᾱ i (a a i r i + α i (a a i r i δ j=2 = a ( x f (ᾱ i α i (a a i r i δ δ < where the equality in the second line follows from a iˆr = a ˆr and a r j = a i r j for all 2 j n such that i j, while the first inequality on the last line follows from α i ᾱ i and a i r i < = a r i. 2

13 Note that P is an n -dimensional simplex in R n {0} and P does not contain any point of Z n {0} in its interior. Suppose that P is maximal lattice free in R n {0}. In this case we can apply the results of Section 3.2 to identify cases where π l is a lifting of ψ B. Let f be the intersection of the line f + ˆr with R n {0}, and let r j = v j f. For every point x Z n {0} in the relative interior of one of the facets of P, say the facet defined by a h (x f, x can be uniquely written as x = f + n j= ᾱj r j such that 0 ᾱ j, j =,..., n, and ᾱ h = 0. Let R( x = { n j= α j r j 0 α j ᾱ j, j =,..., n}. Note that R( x = R( x (R n {0}. Let Ī be the set of all points in x Zn {0} in the relative interior of some of the facets of P. We define R = x R( x. Ī Then R R + ˆr. Hence, if for every r R n {0} there exists w Z n {0} such that r + w R, it also holds that for every r R n there exists w r Z n {0} such that r + w r R. Since R n {0} is the lineality space of conv(s, Theorem 6 implies that π l is the unique minimal lifting of ψ B, and π l (r = ψ(r + w r. The above property holds, for example, when n = 2 and P is an interval of length one (as seen in Section 3., when n = 3 and P is a maximal lattice-free triangle containing at least four points in Z 2 {0}, or for general n when P is a unimodular transformation of conv(0, (n e,..., (n e n. References [] K. Andersen, Q. Louveaux, R. Weismantel, L. A. Wolsey, Cutting Planes from Two Rows of a Simplex Tableau, Proceedings of IPCO XII, Ithaca, New York (June 2007, Lecture Notes in Computer Science 453, -5. [2] K. Andersen, Q. Louveaux, R. Weismantel, An Analysis of Mixed Integer Linear Sets Based on Lattice Point Free Convex Sets, manuscript, [3] A. Basu, M. Conforti, G. Cornuéjols, G. Zambelli, Maximal lattice-free convex sets in linear subspaces, manuscript (March [4] A. Basu, M. Conforti, G. Cornuéjols, G. Zambelli, Minimal inequalities for an infinite relaxation of integer programs, manuscript (April [5] A. Basu, G. Cornuéjols, G. Zambelli, Convex Sets and Minimal Sublinear Functions, manuscript (March [6] V. Borozan, G. Cornuéjols, Minimal Valid Inequalities for Integer Constraints, technical report (July 2007, to appear in Mathematics of Operations Research. [7] G. Cornuéjols, F. Margot, On the Facets of Mixed Integer Programs with Two Integer Variables and Two Constraints (September 2007, to appear in Mathematical Programming. [8] S.S. Dey, L.A. Wolsey, Lifting Integer Variables in Minimal Inequalities Corresponding to Lattice-Free Triangles, IPCO 2008, Bertinoro, Italy (May 2008, Lecture Notes in Computer Science 5035,

14 [9] S.S. Dey, L.A. Wolsey, Constrained Infinite Group Relaxations of MIPs, manuscript (March [0] L. Lovász, Geometry of Numbers and Integer Programming, Mathematical Programming: Recent Developements and Applications, M. Iri and K. Tanabe eds., Kluwer (989,