Cutting planes from two rows of a simplex tableau

Transcription

1 Cutting planes from two rows of a simplex tableau K. Andersen Q. Louveaux R. Weismantel L. Wolsey May, 6 Abstract

2 Introduction In this paper a basic geometric object is investigated that allows us to derive cutting planes for general mixed integer linear programs by considering two rows of a simplex tableau simultaneously. Throughout this paper the mixed integer linear program (MIP is given in equation form, i.e., the optimization problem is defined for a set I of integer variables, a set C of continuous variables, a rational matrix A, a rational vector b of right hand sides and a rational objective function vector c (of suitable dimensions (MIP max c T x subject to Ax = b, x, x i Z for i I. Let LP denote the linear programming relaxation of MIP, i.e., the linear program obtained by replacing the condition x i Z with the weaker condition x i R. In cutting plane theory, the goal is to determine linear inequalities i I C a ix i a that satisfy every feasible solution x to MIP satisfies i I C a ix i a, and a vertex x of LP is cut off by the linear inequality, i.e., i I C a ix i <a. From linear programming theory, it follows that a vertex x of LP corresponds to a basic feasible solution of a simplex tableau associated with subsets B and N of so-called basic and nonbasic variables x i + j N ā i,j x j = b i for i B. Any row associated with an index i B I such that b i Z gives rise to a set X(i := { x R N b i j N ā i,j x j Z, x j for all j N } whose analysis provides inequalities that are violated by x. Indeed, Gomory s mixed integer cuts [5] and mixed integer rounding cuts [7] are derived from such a basic set X(i, that is associated with exactly one index i B I such that b i Z. Interestingly, unlike in the pure integer case, no finite convergence proof of a cutting plane algorithm is known when Gomory s mixed integer cuts or mixed integer rounding cuts are applied only. More drastically, in [4], an interesting mixed integer program in three variables is presented, and it is shown that none of the known classes of general cutting planes contain the cut needed to solve this problem. Example : Consider the mixed integer set t x, t x, x + x + t, x Z and t R. The projection of this set onto the space of x and x variables is given by {(x,x R + : x + x } and is illustrated in Fig.. A simple analysis shows that the inequality x + x, or equivalently t, is valid. In [4] it is, however, shown that with the objective function z = max t, traditional mixed integer cutting plane algorithms do not converge finitely. In this paper we provide some insight into why cutting plane algorithms based on Gomory s mixed integer cuts, split cuts, lift-and-project cuts [3] and MIR cuts do not always converge in finite time. Roughly speaking, the reason is that all the traditional cutting planes arise from sets X(i associated with one integer variable x i.

3 r x x + x f =( 3. 3 r 3 x r Figure : The Instance in [4] We go one step further in this paper by considering two indices i,i B I simultaneously. It turns out that the underlying basic geometric object is significantly more complex than its one-variable counterpart. The set that we denote by X(i,i is described as X(i,i := { x R N b i j N ā i,j x j Z for i = i,i,x j for all j N }. Setting f := ( bi, b i R, and r j := ( ā i, ā i R, the set obtained from two rows of a simplex tableau can be represented as P I := {(x, s Z R n + : x = f + j N s j r j }, where f is fractional and r j R for all j N. Example (revisited: For the instance of Example, introduce slack variables, s,s and y in the three constraints. Then, solving as a linear program, the constraints of the optimal simplex tableau are t + 3 s + 3 s + 3 y = 3 x 3 s + 3 s + 3 y = 3 x + 3 s 3 s + 3 y = 3 Taking the last two rows, and rescaling s i = s i/3 for i =,, we obtain the set P I x s +s + 3 y = + 3 x +s s + 3 y = + 3 x Z,s R +,y R +. 3

4 Letting f =( 3, 3 T, r = (, T, r =(, T obtains the desired cut for conv(p I x + x + y or equivalently t, and r 3 =( 3, 3 (see Fig., one easily which, when used in a cutting plane algorithm, yields immediate termination. Our main contribution is to characterize geometrically all the facets of conv(p I. All facets are intersection cuts [], i.e., they can be obtained from a (two-dimensional convex body that does not contain any integer points in its interior. Our geometric approach is based on two important facts that we prove in this paper every facet is derivable from at most four nonbasic variables. with every facet F one can associate three or four particular vertices of conv(p I. The classification of F depends on how these k =3, 4 vertices can be partitioned into k sets of cardinality at most two. More precisely, the facets of conv(p I can be distinguished with respect to the number of sets that contain two integer points. Since k = 3 or k = 4, the following interesting situations occur no sets with cardinality two: all the k {3, 4} sets contain exactly one tight integer point. We call cuts of this type disection cuts. exactly one set has cardinality two: in this case we show that the inequality can be derived from lifting a cut associated with a two-variable subproblem to k variables. We call these cuts lifted two-variable cuts. two sets have cardinality two. In this case we show that the corresponding cuts are split cuts. Furthermore, we show that inequalities of the first two families are never split cuts. Our geometric approach allows us to generalize the cut introduced in Example. More specifically, the cut of Example is a degenerate case in the sense that it is almost a disection cut and almost a lifted two-variable cut: by perturbing the vectors r, r and r 3 slightly, the cut in Example can become both a disection cut and a lifted two-variable cut. We review some basic facts about the structure of conv(p I in Section. In Section 3 we explore the geometry of all the feasible points that are tight for a given facet of conv(p I. Section 4 explains our main result and presents the classification of all the facets of conv(p I. Finally, Section 5 presents a combinatorial polynomial-time algorithm to compute all the vertices of conv(p I. This algorithm can be used to construct a polynomial time separation algorithm. Basic structure of conv(p I The basic mixed-integer set considered in this paper is P I := {(x, s Z R n + : x = f + j N s j r j }, ( where N := {,,..., n}, f Q \Z and r j Q for all j N. The set P LP := {(x, s R R n + : x = f + j N s jr j } denotes the LP relaxation of P I. The j th unit vector in R n is denoted e j. In this section, we describe some basic properties of conv(p I. The vectors {r j } j N are called rays, and we assume r j for all j N. We first characterize the case when P I is empty. 4

5 Lemma The set P I is empty if and only if (i All rays {r j } j N are parallel. (ii The lines {f + s j r j : s j R} for j N do not contain any integer points. Proof: Clearly (i and (ii are sufficient. Suppose P I =. If there are j, k N such that r j and r k are not parallel, the set f + cone({r j,r k } contains integer points. Hence all rays {r j } j N are parallel, and the sets {f + s j r j : s j R} for j N do not contain any integer points. In the remainder of the paper we assume P I. The next lemma gives a characterization of conv(p I in terms of vertices and extreme rays. Lemma (i The dimension of conv(p I is n. (ii The extreme rays of conv(p I are (r j,e j for j N. (iii The vertices (x I,s I of conv(p I take the following two forms: (a (x I,s I = (x I,s I j e j, where x I = f + s I j rj Z and j N (an integer point on the ray {f + s j r j : s j }. (b (x I,s I = (x I,s I j e j + s I k e k, where x I = f + s I j rj + s I k rk Z and j, k N (an integer point in the set f + cone({r j,r k }. Proof: Let ( x, s P I be arbitrary. (ii: For any j N, since r j Q, there exists a positive integer q j such that q j r j is integer. Hence we have f + j N s jr j + q j r j Z. This proves (r j,e j is a ray of conv(p I for all j N. In addition, every other ray of conv(p I can be expressed as a conic combination of these rays. (i: The (n + points ( x, s and {( x + q j r j, s + q j e j } j N are in P I and affinely independent. (iii: If ( x, s is a vertex of conv(p I, then x is integer, and s is a basic solution to the system x = f + j N s jr j and s j for all j N. Using Lemma, we now give a simple form for the valid inequalities for conv(p I considered in the remainder of the paper. Corollary Every non-trivial valid inequality for P I that is tight at a point ( x, s P I can be written in the form α j s j, ( where α j for all j N. j N Proof: Let j N α js j β be a non-trivial valid inequality for conv(p I, and let ( x, s P I be a tight feasible point, i.e., j N α j s j = β. Suppose there exists j N such that α i <. From Lemma.(ii, we have ( x+r j, s+e j conv(p I, which contradicts the validity of j N α js j β for conv(p I. Hence α j for all j N. Since j N α j s j, we can not have β<. If β =, j N α js j β is a non-negative combination of the inequalities s j for j N, which contradicts the assumption that j N α js j β is non-trivial. For an inequality j N α js j of the form (, let Nα := {j N : α j =} denote the variables with coefficient zero, and let Nα := N \ N α denote the remainder of the variables. We now introduce an object that is associated with the inequality j N α js j. We will use this object to obtain a two dimensional representation of the facets of conv(p I. 5

6 Lemma 3 Let j N α js j be a valid inequality for conv(p I of the form (. Define v j := f + α j r j for j N. Consider the convex polyhedron in R α L α = {x R : there exists s R n + s.t. (x, s P LP and j N α j s j }. (i L α = conv({f} {v j } j N α + cone({rj } j N α. (ii interior(l α does not contain any integer points. (iii If interior(l α, then f interior(l α. Proof: (i First suppose x L α. This means there exists s R n + such that j N α j s j and x = f + j N s jr j. Therefore, we may write x = λ f + j N λ jv j + α j Nα s jr j, where λ j = s j α j for j Nα and λ = j N λ j, which shows x conv({f} {v j } j N + α cone({r j } j N α. Conversely, if x conv({f} {v j } j N + cone({rj } α j N α, then x = λ f + j Nα λ jv j + j N µ α j r j, where {µ j } j N R +, λ, {λ j } α N R + and λ α + j N λ j =. Hence, we may write x = f + j N s jr j, where s j = λj α j for j Nα and s j = µ j for j Nα, and therefore x L α. (ii If x interior(l α, then there exists s R n + such that ( x, s P LP and j N α j s j <. Since j N α js j is valid for P I, we can not have that x is integer. (iii This is clear from the facts that f can be represented using s = and interior(l α can be written in the form {x R : s R n + s.t. (x, s P LP and j N α js j < }. Example : Consider the set P I = {(x, s Z R 5 + : x = f + ( where f = ( 4 ( s +, and consider the inequality ( 3 s + ( s 3 + ( s 4 + s 5 }, s +s +4s 3 + s s 5. (3 The corresponding set L α is shown in Fig.. As can be seen from the figure, L α does not contain any integer points in its interior. It follows that (3 is valid for conv(p I. Note that, conversely, the coefficients α j for j =,,..., 5 can be obtained from the polygon L α as follows: α j is the ratio between the length of r j and the distance between f and v j. In particular, if the length of r j is, then α j is the inverse of the distance from f to v j. The interior of L α gives a two-dimensional representation of the points x R that are affected by the addition of the inequality j N α js j to the LP relaxation P LP of P I. In other words, for any (x, s P LP that satisfies j N α js j <, we have x interior(l α. Furthermore, for a facet defining inequality j N α js j of conv(p I, there exists n affinely independent points (x i,s i P I, i =,,..., n, such that j N α js i j =. The integer points {xi } i N are on the boundary of L α, i.e., they belong to the following integer set: X α := {x Z : s R n + s.t. (x, s P LP and j N α j s j =}. We have X α = L α Z, and X α whenever j N α js j defines a facet of conv(p I. In Sect. 3 we characterize conv(x α. This characterization is then used in Sect. 4 to characterize L α. We first characterize the facets of conv(p I that have zero coefficients. 6

7 r 3 r r v 3 v v v 4 v 5 r 4 r 5 Figure : The set L α for a valid inequality for conv(p I 7

8 Lemma 4 Any facet defining inequality j N α js j for conv(p I of the form ( that has some zero coefficients is a split cut. In other words, if N α, there exists (π, π Z Z such that L α {(x,x :π π x + π x π +}. Proof: Let k Nα be arbitrary. Then the line {f + µr k : µ R} does not contain any integer points. Furthermore, if j Nα, j k, is such that rk and r j are not parallel, then f + cone({r k,r j } contains integer points. It follows that all rays {r j } j N α are parallel. By letting π := (r k =( rk and π r k := (π T f, we have {f + µr k : µ R} = {( x x :π x + π x = π }. Now define: π := max{π x + π x : π x + π x π and x Z }, and π := min{π x + π x : π x + π x π and x Z }. We have π <π <π, and the set S π := {x R : π π x + π x π } does not contain any integer points in its interior. We now show L α S π by showing that every vertex v m = f + α m r m of L α, where m Nα, satisfies vm S π. Suppose v m satisfies π vm +π vm <π (the case π vm + π vm >π is symmetric. By definition of π, there exists xi Z such that π xi + π xi = π, and xi = λv m + ( λ(f + δr k, where λ ], [, for some δ>. We then have x I = f + λ α m r m + δ( λr k. Inserting this representation of x I into the inequality j N α js j λ then gives α m α m + α k δ( λ =λ<, which contradicts the validity of j N α js j for P I. Hence L α S π. To finish the proof, we show that we may write S π = {x R : π π x + π x π +} for some (π, π Z Z. First observe that we can assume (by scaling that π, π and π are integers. Next observe that any common divisor of π and π also divides both π and π (this follows from the fact that there exists x,x Z such that π x + π x = π and π x + π x = π. Hence we can assume that π and π are relative prime. Now the Integral Farkas Lemma (see [9] implies that the set {x Z : π x + π x =} is non-empty. It follows that we must have π = π +, since otherwise the point ȳ := x + x Z, where x {x Z : π x + π x = } and x {x Z : π x + π x = π}, satisfies π <π ȳ + π ȳ <π, which contradicts that S π does not contain any integer points in its interior. 3 A characterization of conv(x α As a preliminary step of our analysis, we first characterize the set conv(x α. We assume α j > for all j N. Clearly conv(x α is a convex polygon with only integer vertices, and since X α L α, conv(x α does not have any integer points in its interior. We first limit the number of vertices of conv(x α to four. Lemma 5 Let P R be a convex polygon with integer vertices that has no integer points in its interior. (i P has at most four vertices (ii If P has four vertices, then at least two of its four facets are parallel. (iii If P is not a triangle with integer points in the interior of all three facets (see Fig. 4.(c, then there exists parallel lines πx = π and πx = π +, (π, π Z 3, such that P is contained in the corresponding split set, i.e., P {x R : π πx π +}. Proof: (i-(ii: Let v,v,...,v p denote the vertices of P. Wlog assume v =, and that v and v 3 are such that P cone(v,v 3. Observe that cone(v,v 3 can be written as the disjoint union of the following parallelograms 8

9 Π i,j := {z R : z =(i + λ v +(j + λ 3 v 3, where λ,λ 3 [, [}, where i, j. The parallelograms Π i,j are illustrated in Fig. 3. Note that the following translation property holds: for any integer point w i,j in parallelogram Π i,j, there is a corresponding integer point w i,j in parallelogram Π i,j at the same position. Specifically, if the point w i,j =(i+λ v + (j + λ 3 v 3 =(iv + jv 3 + (λ v + λ 3 v 3 Π i,j is integer, then λ v + λ 3 v 3 is integer, and therefore w i,j := (i v + j v 3 +(λ v + λ 3 v 3 = (i + λ v +(j + λ 3 v 3 Π i,j is integer (see Fig. 3. We first claim that every vertex v k, k 4, must be either on the line L = {v + λv 3 λ> }, or on the line L 3 = {λv + v 3 λ> }. First suppose v k Π i,j, where i, j. Then either v k L or v k L 3, since otherwise the integer point v + v 3 is an interior point of P. Now suppose v k Π,j, where j (the case when v k Π i,, i, is symmetric. This means there exists λ,λ 3 [, [ such that v k = λ v +(j + λ 3 v 3. If λ =, v 3 is on the line between v and v k, which is a contradiction, so we have λ >. From the translation property discussed above, we have w := λ v +(j +λ 3 v 3 Z. We claim w is in the interior of P. Indeed, we may write where µ := λ j+λ 3 w = j +λ 3 j + λ 3 v k + λ j + λ 3 v, ], [ and µ k := j +λ3 j+λ 3 ], [. Hence we have w = µ v +µ k v k +( µ µ k v, where µ + µ k = (j+λ3 ( λ j+λ 3 ], [, and this shows w is in the interior of P. Finally suppose v k Π,. We may therefore write v k = λ v + λ 3 v 3, where λ,λ 3 [, [. If λ =, v k is on the line between v and v 3, and if λ 3 =, v k is on the line between v and v, so we can assume λ,λ 3 >. If λ + λ 3 <, then v k is in the interior of conv({v,v,v 3 }, and if λ + λ 3 =, v k can not be a vertex. Finally, if λ + λ 3 >, the point w := (v + v 3 v k = ( λ v + ( λ 3 v 3 is integer and in the interior of conv({v,v,v 3 }. It follows that v k must be either in L or L 3. To finish the proof of (i and (ii, we only need to argue that L L 3 can only contain one vertex of P. Clearly L and L 3 can only contain one vertex each, so we may assume p = 5, v 4 = v + λ 3 v 3 L and v 5 = v 3 + λ v L 3. Furthermore, we have either λ,λ 3 > or λ,λ 3 <, since otherwise either v 4 is in the interior of conv({v,v,v 3,v 5 }, or v 5 is in the interior of conv({v,v,v 3,v 4 }. If λ,λ 3 >, the point v + v 3 is in the interior of P, and if λ,λ 3 <, the point w := λ v + λ 3 v 3 is integer and in the interior of P. (iii Consider again the parallelogram Π,. If det(v,v 3 =, it is well known that the vertex is the only integer point contained in Π,. Hence all parallelograms Π i,j do not have any integer point in their interior. In particular j Z Π,j and i Z Π i, do not have any integer point in their interior. Furthermore P is always contained in at least one of these two sets since a potential vertex v 4 would be either in L or in L 3. Now suppose det(v,v 3 >. Therefore Π, contains more than one integer point. If P is a quadrangle, observe that these points must be on the boundary of Π,, since otherwise P would contain integer points in its interior. If P is a triangle, we assume that the edge (v,v 3 does not contain an integer point in its interior. Otherwise we can re-index the vertices so that this holds. In this case, it is also true that the integer points of Π, are on its boundary. We claim that all integer points of Π, must be on the same side of Π,, i.e., either in S := {λv : λ< }, or in S 3 := {λv 3 : λ< }. Suppose we have w = λ v Z with <λ < and w 3 = λ 3 v 3 Z with <λ 3 <. Note that λ and λ 3 can be chosen in ], /] by possibly considering v w or v 3 w 3, which are also integer. Therefore w + w 3 = λ v + λ 3 v 3 is integer and in the interior of P if P is a quadrangle. This contradicts the hypothesis that the edge (v,v 3 does not contain an integer point if it is a triangle. Hence either S or S 3 does not contain any integer point. This implies that either j Z Π,j or i Z Π i, does not contain an integer point in its interior. From the proof of Lemma 5, it follows that the polygons in Fig. 4 include all possible polygons that can be included in the set L α in the case when L α is bounded and of dimension. The 9

10 dashed lines in Fig. 4 indicate the possible split sets that include P. We excluded from Fig. 4 the cases when X α is of dimension. We note that Lemma 5.(iii (existence of split sets proves that there cannot be any triangles where two facets have interior integer points, and also that no quadrangle can have more than two facets that have integer points in the interior. L Π, Π, w, Π, w, w, Π, w, w, Π, w, Π, v 3 Π, Π, w, Π, w, w, L 3 v Figure 3: Partitioning of cone(v,v 3 into parallelograms 4 A characterization of the facets of conv(p I In this section we focus on the set L α. As in the previous section, we assume α j > for all j N. Due to the direct correspondence between the set L α and a facet defining inequality j N α js j for conv(p I, this gives a characterization of the facets of conv(p I. The main result in this section is that L α can have at most four vertices. In other words, we prove Theorem Let j N α js j be a facet defining inequality for conv(p I that satisfies α j > for all j N. Then L α is a polygon with at most four vertices. Theorem shows that there exists a set S N such that S 4 and j S α js j is facet defining for conv(p I (S, where P I (S := {(x, s Z R S + : x = f + j S s j r j }. Throughout this section we assume that no two rays point in the same direction. If two variables j,j N are such that j j and r j = δr j for some δ >, then the halflines {x R : x = f + s j r j,s j } and {x R : x = f + s j r j,s j } intersect the boundary of L α at the same point, and therefore L α = conv({f} {v j } j N = conv({f} {v j } j N\{j}, where v j := f + α j r j for j N. This assumption does therefore not effect the validity of Theorem. The proof of Theorem is based on characterizing the vertices conv(p I that are tight for j N α js j. Since conv(p I has dimension n, there exists n affinely independent vertices

11 (a A triangle: no facet has interior integer points (b A triangle: one facet has interior integer points (c A triangle: all facets have interior integer points (d A quadrangle: no facet has interior integer points (e A quadrangle: one facet has interior integer points (f A quadrangle: two facets have interior integer points Figure 4: All integer polygons that do not have interior integer points of conv(p I that satisfy j N α js j with equality. In the remainder of this section, the set {(y i,t i } i M P I denotes n such points, where M := {,,..., n}. In other words we have α j t i j =, for i M. (4 j N Since j N α js j is facet defining for conv(p I, the values {α j } j N are the unique solution to (4. It follows that every variable j N is involved in system (4. We will show that there exists a subset S N of variables and a set of S affinely independent vertices of conv(p I such that S 4 and {α j } j S is the unique solution to the equality system defined by these vertices. Observe that the set {y i } i M is the set of vertices of conv(x α. We can therefore partition the equalities (4 according to which vertex of conv(x α is involved in each equality. The following notation will be used intensively in the remainder of this section. Notation (i The number k 4 denotes the number of vertices of conv(x α. (ii The set {x v } v K denotes the vertices of conv(x α, where K := {,,..., k}. (iii For every v K, the set M v := {i M : y i = x v }, denotes the equalities of (4 for which x v is the vertex of conv(x α involved in the corresponding equality of (4. The system (4 can now be rewritten as α j t i j =, for every vertex x v of conv(x α and i M v. (5 j N

12 Lemma.(iii demonstrates that for a vertex ( x, s of conv(p I, s is positive on at most two coordinates j,j N, and x f + cone({r j,r j }. If for some i M, t i is positive on only one coordinate, then y i = f + t i j rj for some j N, and the system (5 includes the equality α j t i j =, which simply states α j =. A point x Z that satisfies x {x R : x = f + s t i j r j,s j } for j some j N will be called a ray point in the remainder of the paper. In order to characterize the equalities of (5 that correspond to vertices x v of conv(x α that are not ray points, we introduce the following concepts. Definition Let j N α js j be valid for conv(p I. Suppose x Z is not a ray point, and that x f + cone({r j,r j }, where j,j N. This implies x = f + s j r j + s j r j, where s j,s j > are unique. (a The pair (j,j is said to give a representation of x. (b If α j s j +α j s j =, (j,j is said to give a tight representation of x wrt. j N α js j. (c If (i,i N N satisfies cone({r i,r i } cone({r j,r j }, the pair (i,i is said to define a subcone of (j,j. Example (continued: Consider again the set ( P I = {(x, s Z R 5 + : x = f + where f = ( 4 ( s + ( 3 s + ( s 3 + ( s 4 +, and the valid inequality s +s +4s 3 + s s 5 for conv(p I. The point x = (, is on the boundary of L α (see Fig.. We have that x can be written in any of the following forms s 5 }, x =f+ 4 r + 4 r, x =f+ 3 7 r + 8 r3, x =f r + 4 r4. It follows that (,, (, 3 and (, 4 all give representations of x. Note that (, and (, 3 give tight representations of x wrt. the inequality s +s +4s 3 + s s 5, whereas (, 4 does not. Finally note that (, 5 defines a subcone of (, 4. Observe that, for a vertex x v of conv(x α and an equality i M v, if t i has support two, the equality j N α jt i j = of system (5 can be written as α j t i j + α j t i j =, where j,j N satisfy t i j,t i j >. Hence, the equality j N α jt i j = corresponds to the tight representation (j,j of x v wrt. j N α js j. Furthermore, if t i is positive on only one coordinate, then x v is a ray point. We now characterize the set of tight representations of an arbitrary integer point x Z, which is not a ray point T α ( x := {(j,j N N :(j,j gives a tight representation of x wrt. α j s j }. j N We will show (Lemma 6 and Lemma 7 that T α ( x contains a unique maximal representation (j x,j x T α( x that satisfies:

13 (i Every subcone (j,j of (j x,j x that gives a representation of x satisfies (j,j T α ( x. (ii Conversely, every (j,j T α ( x defines a subcone of (j x,j x. We then use the maximal representations of the vertices of conv(x α to rewrite (5 (Lemma 8. From this system, we construct a subset S of variables such that S = k and j S α js j is facet defining for conv(p I (S. To prove (i and (ii, there are two cases to consider. For two representations (i,i and (j,j of x, either one of the two cones (i,i and (j,j is contained in the other, or their intersection defines a subcone of both (i,i and (j,j (note that we cannot have that their intersection is empty, since they both give a representation of x. We first consider the case when one cone defines a subcone of another cone. Lemma 6 Let j N α js j be a facet defining inequality for conv(p I that satisfies α j > for all j N, and let x Z. Then (j,j T α ( x implies (i,i T α ( x for every subcone (i,i of (j,j that gives a representation of x. Proof: Suppose (j,j T α ( x. Observe that it suffices to prove the following: for any j 3 N such that r j3 cone({r j,r j } and (j,j 3 gives a representation of x, the representation (j,j 3 is tight wrt. j N α js j. The result for all remaining subcones of (j,j follows from repeated application of this result. For simplicity we assume j =, j = and j 3 = 3. Since x f + cone({r,r }, x f + cone({r,r 3 } and r 3 cone({r,r }, we may write x = f + u r + u r, x = f + v r + v 3 r 3 and r 3 = w r + w r, where u,u,v,v 3,w,w. Furthermore, since (, gives a tight representation of x wrt. j N α js j, we have α u + α u =. Finally we have α v + α 3 v 3, since j N α js j is valid for P I. If also α v + α 3 v 3 =, we are done, so suppose α v + α 3 v 3 >. We first argue that this implies α 3 >α w +α w. Since x = f +u r +u r = f +v r +v 3 r 3, it follows that (u v r = v 3 r 3 u r. Now, using the representation r 3 = w r + w r, we get (u v v 3 w r +(u v 3 w r =. Since r and r are linearly independent, we obtain: (u v =v 3 w and u = v 3 w. Now we have α v + α 3 v 3 > =α u + α u, which implies (v u α α u + α 3 v 3 >. Using the identities derived above, we get v 3 w α α v 3 w + α 3 v 3 >, or equivalently v 3 ( w α α w + α 3 >. It follows that α 3 >α w + α w. We now derive a contradiction to the identity α 3 >α w +α w. Since j N α js j defines a facet of conv(p I, there must exist x Z and k N such that (3,k gives a tight representation of x wrt. j N α js j. In other words, there exists x Z, k N and δ 3,δ k such that x = f + δ 3 r 3 + δ k r k and α 3 δ 3 + α k δ k =. Furthermore, we can choose x, δ 3 and δ k such that r 3 is used in the representation of x, i.e., we can assume δ 3 >. Now, using the representation r 3 = w r + w r then gives x = f + δ 3 r 3 + δ k r k = f + δ 3 w r + δ 3 w r + δ k r k. Since j N α js j is valid for P I, we have α δ 3 w + α δ 3 w + α k δ k =α 3 δ 3 +α k δ k. This implies δ 3 (α 3 α w α w, and therefore α 3 α w α w, which is a contradiction. To finish the proof of (i and (ii, we need to consider the case of two cones (j,j, (j 3,j 4 T α ( x, where neither of the two cones is a subcone of the other cone. This case is captured in the following lemma. Lemma 7 Let j N α js j be a facet defining inequality for conv(p I satisfying α j > for j N, and suppose x Z is not a ray point. Also suppose the intersection between the cones (j,j, (j 3,j 4 T α ( x is given by the subcone (j,j 3 of both (j,j and (j 3,j 4. Then (j,j 4 T α ( x, i.e., (j,j 4 also gives a tight representation of x. 3

14 Proof: For simplicity assume j =, j =, j 3 = 3 and j 4 = 4. Since the cones (, and (3, 4 intersect in the subcone (, 3, we have r 3 cone({r,r }, r cone({r 3,r 4 }, r 4 / cone({r,r } and r / cone({r 3,r 4 }. We first represent x in the translated cones in which we have a tight representation of x. In other words, we can write x = f + u r + u r, (6 x = f + v 3 r 3 + v 4 r 4 and (7 x = f + z r + z 3 r 3, (8 where u,u,v 3,v 4,z,z 3 >. Note that Lemma 6 proves that (8 gives a tight represention of x. Using (6-(8, we obtain the relation ( T, I T, I ( r ( u = r, (9 T, I T, I r 3 v 4 r 4 where T is the matrix T := ( T, T, ( (z = u z 3 and I is the T, T, z (z 3 v 3 identity matrix. On the other hand, the tightness of the representations (6-(8 leads to the following identities α u +α u =, ( α 3 v 3 +α 4 v 4 = and ( α z +α 4 z 3 =, ( where, again, the last identity follows from Lemma 6. Using (-(, we obtain the relation ( T, T, T, T, ( α α 3 = ( u α v 4 α 4. (3 We now argue that T is non-singular. Suppose, for a contradiction, that T, T, = T, T,. From (7 and (8 we obtain v 4 r 4 = (z 3 v 3 r 3 + z r, which implies z 3 < v 3, since r 4 / cone({r,r } cone({r,r 3 }. Multiplying the first equation of (3 with T, gives T, T, α + T, T, α 3 = u T, α, which implies T, (T, α + T, α 3 =u T, α. By using the definition of T, this can be rewritten as z 3 (α z +(z 3 v 3 α 3 =u α (z 3 v 3. Since z α + z 3 α 3 =, this implies z 3 ( v 3 α 3 =u α (z 3 v 3. However, from ( we have v 3 α 3 ], [, so z 3 ( v 3 α 3 > and u α (z 3 v 3 <, which is a contradiction. Hence T is non-singular. We now solve (9 for an expression of r and r 3 in terms of r and r 4. The inverse of the coefficient matrix on the left hand side of (9 is given by ( T, I T, I ( T, T, T, T, denotes the inverse of T. We therefore obtain T, I T, I, where T := r = λ r + λ 4 r 4 and (4 r 3 = µ r + µ 4 r 4, (5 where λ := u T,,λ 4 := v 4 T,,µ := u T, and µ 4 := v 4 T,. Similarly, solving (3 to express α and α 3 in terms of α and α 4 gives α = λ α + λ 4 α 4 and (6 α 3 = µ α + µ 4 α 4. (7 4

15 Now, using for instance (6 and (4, we obtain x = f +(u + u λ r +(u λ 4 r 4, and: (u + u λ α +(u λ 4 α 4 = (using ( ( u α +u λ α +(u λ 4 α 4 = +u (λ α + λ 4 α 4 α =. (using (6 To finish the proof, we only need to argue that we indeed have x f + cone({r,r 4 }, i.e., that x = f + δ r + δ 4 r 4 with δ = u + u λ and δ 4 = u λ 4 satisfying δ,δ 4. If δ and δ 4 >, we have x = f + δ r + δ 4 r 4 = f + u r + u r, which means δ 4 r 4 =(u δ r + u r cone({r,r }, which is a contradiction. Similarly, if δ > and δ 4, we have x = f + δ r + δ 4 r 4 = f + v 3 r 3 + v 4 r 4, which implies δ r = v 3 r 3 +(v 4 δ 4 r 4 cone({r 3,r 4 }, which is also a contradiction. Hence we can assume δ,δ 4. However, since δ = u + u λ and δ 4 = u λ 4, this implies λ,λ 4, and this contradicts what was shown above, namely that the representation x = f + δ r + δ 4 r 4 satisfies α δ + α 4 δ 4 =. By using Lemma 6 and Lemma 7, i.e., the characterization of the tight representations of an integer point, we now rewrite the subsystems of (5 that correspond to vertices x v of conv(x α that satisfy M v >. Lemma 8 Let j N α js j be a facet defining inequality for conv(p I s.t. α j > for j N. Consider a vertex x v of conv(x α that satisfies M v >, and the subsystem of (5 corresponding to x v α j t i j =, for every i M v. (8 j N Let J v := {j N : t i j > for some i M v } denote the variables that appear in (8. There exist two variables j v,j v J v such that (i (j v,jv gives a tight representation of xv wrt. j N α js j, i.e., there exists w j v,w j v > such that x v = f + w j v r jv + wj v r jv and wj v α j v + w j v α j v =. (ii For every j J v \{j v,jv }, we have rj cone(r jv,r j v, i.e., there exists w j,wj > such that r j = w j rjv + w j rjv. (iii {α j } j N is the unique solution to the system (5 obtained from (5 by replacing (8 with w j v α j v + w j v α j v =, (9 α j = w j α j v + wj α j v, for every j J v \{j v,j v }. ( Proof: Observe that, since M v >, there exists i M v such that t i is non-zero on two coordinates. Wlog suppose this is true for all i M v. For every i M v, let j,j i i J v be such that t i,t i >. Then (j i j i j i,ji gives a tight representation of xv wrt. j N α js j for every i M v. It follows from Lemma 6 and Lemma 7 that there exists j v,j v J v such that j v j v,(j v,j v gives a tight representation of x v wrt. j N α js j, and (j i,ji defines a subcone of (jv,jv for all i M v. This shows (i and (ii. We now show (iii. Clearly equality (9 is implied by (5, since (j v,j v gives a tight representation of x v wrt. j N α js j (note that (9 might not be part of the system (5. Now consider a variable j J v \{j v,j v }, and define the point x := f + r j on the w j α j v +wj α j v halfline {f + s j r j : s j }. By using (ii, we obtain x w = f+ j r jv w w j α j v +wj α + j r jv j v w j α j v +wj α. j v 5

16 Inserting this representation of x into j N α js j then gives w j α j v w j α j v +wj α j v + w j α j v w j α j v +wj α j v This shows x is on the boundary of L α, and since the intersection point between the halfline {f + s j r j : s j } and the boundary of L α is unique, we have x = v j = f + α j r j, which implies α j = w j α j v + wj α j v. Hence the system (9-( is implied by (5. We now show the converse, i.e., that {α j } j N is the unique solution to the system (5 obtained from (5 by replacing (8 with (9-(. Therefore let i M v be arbitrary, and let (j,j := (j,j i i for simplicity. We need to show t i j α j + t i j α j =, where x v = f + t i j r j + t i j r j. From (9-( we know α j = w j α j v +wj α j v, α j = w j α j v +wj α j v and α j v w j v +α j v w j v r j =. =, where = w j rjv +w j rjv, r j = w j rjv +w j rjv and x v = f+w j v r jv +wj v r jv. Inserting the expressions for r j and r j into the identity x v = f + t i j r j + t i j r j then gives x v = f +(t i j w j + ti j w j rjv + (t i j w j +ti j w j rjv. Since the representation of x v in the set f +cone(r jv,r j v is unique, it follows that w j v = t i j w j +ti j w j and w j v = ti j w j +ti j w j. Now, from the equality α j vw j v +α j v w j v =, we get = α j v (t i j w j +ti j w j + α j v (ti j w j +ti j w j =ti j (w j α j v +wj α j v + ti j (w j α j v +wj α j v. Since α j = w j α j v + wj α j v and α j = w j α j v + wj α j v, we obtain ti j α j + t i j α j =. By using Lemma 8, we are now able to finish the proof of Theorem. Lemma 9 Suppose j N α js j is facet defining for conv(p I and α j > for j N. Choose a set S N of variables as follows. Initialize S =, and for every vertex x v of conv(x α : (i If M v = {ī}, update S := S J v, where J v := {j N : tīj variables appearing in equality ī of system (5. > } denotes the (at most two (ii If M v >, update S := S {j v,j v }, where (j v,j v denotes the tight representation of x v given in Lemma 8. Then the inequality j S α js j is facet defining for P I (S, S = k and we have L α = conv({f} {v j } j S, where v j = f + α j for j S. Proof: From Lemma 8 and the fact that j N α js j is facet defining for conv(p I, we know that {α j } j N is the unique solution to the system α j tīj =, for all v K s.t. M v = {ī}, ( j N w j v α j v + w j v α j v =, for all v K s.t. M v >, ( w j α j v + wj α j v = α j, for all v K s.t. M v > and j J v \{j v,j v }, (3 Observe that, for a vertex x v of conv(x α such that M v > and a variable j J v \{j v,j v }, α j only appears in system (3. It follows that {α j } j S is the unique solution to the subsystem (-( of (-(3. Now, since the system (-( consists of exactly K = k equalities, it follows that S = k. We note that, from a facet defining inequality j S α js j for conv(p I (S, where S = k, the coefficients on the variables j N \ S can be simultaneously lifted from equality ( of Lemma 8, i.e., by computing the intersection point between the halfline {f + s j r j : s j } and the boundary of L α. We now use Theorem to categorize the facet defining inequalities j N α js j for conv(p I. For simplicity, we only consider the most general case, namely when none of the vertices of conv(x α are ray points. Furthermore, we only consider k = 3 and k = 4. When k =, j N α js j is a facet defining inequality for a cone defined by two rays. We divide the facets of conv(p I into the following three main categories. 6

17 (i Disection cuts (Fig. 5: Every vertex of conv(x α belongs to a different facet of L α. (ii Lifted two-variable cuts (Fig. 6: Exactly one facet of L α contains two vertices of conv(x α. Observe that this implies that there is a set S S, S =, such that j S α js j is facet defining for conv(p I (S. (iii Split cuts: Two facets of L α each contain two vertices of conv(x α. (a Disection cut from a triangle (b Disection cut from a quadrangle Figure 5: Disection cuts (a Lifted two-variable cut from triangle (b Lifted two-variable cut from quadrangle Figure 6: Lifted two-variable cuts An example of a cut that is not a split cut was given in [4] (see Fig.. This cut is the only cut when conv(x α is the triangle of Fig. 4.(c, and, necessarily, L α = conv(x α in this case. Hence, all three rays that define this triangle are ray points. As mentioned in the introduction, the cut in [4] can be viewed as being on the boundary between disection cuts and lifted two-variable cuts. Indeed, by perturbing the three rays that define the triangle slightly, both a disection cut and a lifted two-variable cut can be obtained. 7

18 Since the cut presented in [4] is not a split cut, and this cut can be viewed as being a boundary cut between disection cuts and lifted two-variable cuts, a natural question is whether or not disection cuts and lifted two-variable cuts are split cuts. We finish this section by answering this question. Lemma Let j N α js j be a facet defining inequality for conv(p I satisfying α j > for j N, and suppose none of the vertices of conv(x α are ray points. If j N α js j is a disection cut or a lifted two-variable cut, then j N α js j is not a split cut. Proof: Observe that, if j N α js j is a split cut, then there exists (π, π Z Z such that L α is contained in the split set S π := {x R : π π x + π x π +}. Furthermore, all points x X α and all vertices of L α must be either on the line π T x = π, or on the line π T x = π +. However, this implies that there must be two facets of L α that do not contain any integer points. Example 3: Consider the set P I = {(x, s Z R 5 + : x = f + ( 5 where f = ( 3 ( 5 s + 3 ( 4 s + 3 ( s ( s 4 +. We computed all the facets of conv(p I for this example. In Table, we report how many of the 5 facets belong to each of the categories defined above. Disection Disection Lifted -var Lifted -var split (triangle (quadrangle (triangle (quadrangle cut # facets percentage Table : A categorization of the facets for an example s 5 }, 5 An algorithm to compute the vertices of conv(p I In sections 3 and 4 we described the structure of the facets of conv(p I. In particular, we demonstrated that every facet of conv(p I can be obtained from a choice of at most four vertices of conv(p I. Therefore, to compute the facets of conv(p I, a method for generating the vertices of conv(p I is needed, and this is the topic of this section. It is well known that the facets of any mixed integer set can be computed from its polar. The polar of conv(p I can be represented as a polyhedron by including one inequality for each extreme point and each extreme ray of conv(p I. In our case, every extreme point (x v,s v of conv(p I can be represented by two rays r jv and r kv, i.e., x v = s jv r jv + s kv r kv, where s jv,s kv (Lemma.(iii. Therefore, the polar of conv(p I is the set of u R n that satisfy s v j v u jv + s v k v u kv, for all vertices (x v,s v of conv(p I, u j, for all j N, where non-negativity follows from the extreme rays of conv(p I. To set up and optimize over such a linear system, the vertices of conv(p I must be computed. In this section, we present a combinatorial algorithm for achieving this. Since the number of vertices of conv(p I is polynomial, 8

19 an overall algorithm that is based on optimizing over the polar of conv(p I runs in polynomial time. Consider two arbitrary rays r i and r i, where i,i N. We now show how to compute all vertices of conv(p I that can be obtained from r i and r i. For simplicity assume i =,i =, and let P I (f, R ={x Z : x = f + r s + r s,s,s }, (4 be the integer set generated from r and r, where R := cone({r,r }. Observe that (4 can be reformulated by using a unimodular transformation. A unimodular transformation maps integer vertices to integer vertices and therefore preserves integrality. Observation Let T Q be a unimodular matrix, and let {x v } v V denote the vertices of conv(p I (f, R. Define f := Tf, R := cone({tr, T r } and x v := Tx v for v V. Then conv(p I (f, R = conv({x v } v V +R, and conv(p I ( f, R = conv({ x v } v V + R. To simplify the exposition, we use a unimodular transformation to represent the problem in a standard form. This is achieved with a variant of the Hermite normal form (see [9]. Observation Let C Z be nonsingular. There exists a unimodular matrix T Q such that ( p (i TC =, where q q p. (ii The size of all entries in TC is polynomially bounded by the encoding length of C. ( ( From Observations and it follows that we can assume r = and r p =. q We now describe how to successively construct the vertices of conv(p I (f, R. The algorithm constructs a new vertex in every iteration, where the first vertex is trivially obtained as follows. Lemma The point w := ( f, f T is a vertex of conv(p I (f, R. Furthermore, the inequality x f is facet defining for conv(p I (f, R. Proof: This follows directly from the fact that R = cone({(, T, (p,q T }, where p, q. The remaining vertices of conv(p I (f, R are constructed in a particular order, which we now discuss. If there exists a vertex w of conv(p I (f, R with the same first coordinate as w, then we consider w as the next vertex of conv(p I (f, R. Observe that there can only be one such vertex, and that this vertex, if it exists, can be constructed by simple rounding. In the following, we therefore assume for simplicity that no such vertex exists. Now consider the set R := w + R with vertex w and extreme rays r and r (see Fig. 7.(a. Observe that w is the only vertex of conv(p I (f, R contained in R. Therefore all the remaining vertices of conv(p I (f, R are contained in the set (f + R \ R (the hatched area of Fig. 7.(a. Furthermore, observe that the slope of the line between w and any remaining vertex w of conv(p I (f, R is larger than the slope q p of r (see Fig. 7.(b. The next vertex w is then defined to be the vertex w of conv(p I (f, R for which this slope is as large as possible. Similarily, by induction, for any t {, 3,..., r}, w (t is the only vertex of conv(p I (f, R in the set R t := w (t + R, and the next vertex w t is defined to be the vertex w of conv(p I (f, R, among the vertices that have not been chosen so far, for which the slope of the line between w (t and w is as large as possible. 9

20 w f w f w (a Initial vertex w and the set R (b The vertex w and the corresponding facet Figure 7: Construction the vertex w We now introduce the following notation. For any k {, 3,..., r}, let (p k,q k T := w k w (k. Then the slope of the line between w (k and w k is given by q k p k, and by the choice of the ordering w,w,..., w r of the vertices of conv(p I (f, R, we have that the slopes are ordered in decreasing order q p > q 3 p 3 >... > q r p r. Also note that all slopes are upper approximations to the slope q p of r, i.e., p k > q p for all k =, 3,..., r, and that the approximation becomes better as k increases. The next lemma demonstrates that vertex w k can be obtained from the previous vertices w,w,...,w (k by solving an optimization problem. Lemma Let k r. The pair (p k,q k is an optimal solution to the optimization problem: q k max q p s.t. q p q (5 p ( p w (k + f + R q p, q Z +. Furthermore, there exists an optimal solution (p h,q h to (5 and an integer δ k Z + such that (p k,q k =δ k (p h,q h, and the set {(p, q Z :(p, q =δ(p h,q h,δ Z + and δ δ k } describes all optimal solutions to (5. We now relate problem (5 to the directed simultaneous diophantine approximation problem. The directed simultaneous diophantine approximation problem D( ˆqˆp,N is the problem of finding

21 the best upper approximation q p of a rational number ˆqˆp under the constraint that p N min q p s.t. q, (6 p ˆqˆp p N, p, q Z +. The following theorem gives the relationship between the directed simultaneous diophantine approximation problem and problem (5. Theorem 3 Assuming conv(p I (f, R does not have a vertex with the same first coordinate as w, the problem (5 has the same set of optimal solutions as the directed simultaneous diophantine approximation problem D( q p,n with N = p k. Proof: Consider the optimal solution (p k,q k to (5. Also consider an optimal solution ( p, q to D( q p,p k. We will prove that q k p k = q p. Note that this implies that every optimal solution (p,q to (5 must be optimal for D( q p,p k (since (p,q is feasible for D( q p,p k and attains the same objective value as ( p, q. Conversely every optimal solution (p,q to D( q p,p k satisfies q p = q k p k, and since p p k,(p,q is feasible for (5. We now prove q k p k since ( p, q is feasible for (6, and we also have q k q, since q p < q k = q p. Suppose, for a contradiction, that q p < q k p k. Clearly we have p k p, p k. We first argue that ( w (k pk p + q k q f + R. (7 Suppose (7 is not satisfied, i.e., we have the situation in Fig. 8. Observe that the slope of the line between the points w (k and w (k + ( p, q is identical to the slope of the line between w (k +(p k p, q k q and w (k +(p k,q k, and that this slope is given by q p. Hence, if (7 q is not satisfied, then we have q p < p. However, this contradicts the fact that ( p, q is feasible for D( q p,p k. Hence (7 is satisfied Next suppose p = p k. If also q = q k, we are done, and if q k > q, (7 implies that the vertical line through w (k contains integer points different from w (k that are contained f + R. However, this contradicts the assumption that conv(p I (f, R does not have a vertex with the same first coordinate as w. Finally suppose p k > p and q k > q. Since q p < q k p k, we have Indeed (8 can be written as q k q p k p > q k p k. (8 p k q k p k q > p k q k q k p. (9 It now follows from (8 and (7 that (p k p, q k q is a feasible solution to (5 with a better objective value than (p k,q k, which is a contradiction. The optimal solutions of the directed simultaneous diophantine approximation problem D( q p,n (( ( can be characterized from a minimal Hilbert basis of the cone R p = cone,, where q a Hilbert basis of a rational polyhedral cone is defined as follows.

22 w (k + ( pk p q k q ( pk w (k + ( q k w (k + p q w (k Figure 8: Existence of ( p, q Definition Let C R n be a rational polyhedral cone. A finite set H(C ={h,..., h t } C Z n is called a Hilbert basis of C if every z C Z n has a representation of the form z = t λ i h i, i= where λ i Z +. Furthermore, if C is pointed, there exists a unique Hilbert basis H (C of C of minimum cardinality. The following result is due to Henk and Weismantel [6]. Theorem 4 Let ˆp, ˆq Z +. For every N Z +, there exists an optimal solution to the problem D( ˆqˆp,N defined by (6 which is a member of the Hilbert basis H (R of the cone R = (( ( cone,. Conversely, for every member h of the Hilbert basis H ˆpˆq (C of R, there exists N Z + such that h is an optimal solution of D( ˆqˆp,N. An algorithm to construct the vertices of conv(p I (f, R now follows directly from Theorem 3 and Theorem 4. We start from the vertex w =( f, f. To (( construct ( the next vertex, we consider the Hilbert basis elements of H (R, where R p = cone,. We sort q the elements of H (R in terms of decreasing slopes. The first element h H (R for which w + h f + R leads to the next vertex. Notice that, if there is a vertex of conv(p I (f, R with the same first coordinate as w, then this vertex can be found using the element (, of H (R, which is always contained in H (R. Also note that a similar approach has been used recently by Agra and Constantino [] to determine a complete description for multidimensional knapsack problems in two variables. The cones analyzed in [], however, slightly differ from the family of cones studied here.

23 The algorithm above, however, does not run in polynomial time. It is well known that a Hilbert basis of a two-dimensional cone can be of exponential size. An algorithm, which is based on checking every element of H (R, can therefore have exponential running time. Fortunately, we can overcome this difficulty by exploiting the special structure of Hilbert bases of two-dimensional cones. A minimal Hilbert basis of a two-dimensional cone can be described by a polynomial number of vertices and edges (see [8]. The following theorem gives the structure of a minimal Hilbert basis of a two-dimensional cone in terms of vertices and edges. Theorem 5 Let C = cone(r,r R be a rational polyhedral cone of dimension two. There exists a polynomial number of vectors e,...,e s Z, called the edges, and numbers k,..., k s Z + such that, if we denote the partial sums σ i = i j= k je j, H (R ={r,r + e,..., r + k e = r + σ, r + σ + e,..., r + σ + k e = r + σ,. r + σ (s + e s,...,r + σ s = r }. We can now modify the initial algorithm as follows. Instead of checking every element of H (R in turn, we sequentially check every edge of H (R. This can be done in polynomial time using ( ( p p q binary search. To do this, consider the inverse of the matrix given by. q q To check whether the elements e + je of the Hilbert basis H (R of R leads to a new vertex of conv(p I (f, R from a previous vertex w (k of conv(p I (f, R, where j Z + and e, e Z + are edges of H (R, one needs to check the following condition ( p q q ( (w (k + e + je f. (3 The second inequality of (3 is trivially satisfied since w (k >f. Hence the first inequality of (3 determines whether an element of the Hilbert basis H (R along the edge e + je provides a new vertex of conv(p I (f, R. Example 4 Consider the cone (( R = cone ( 37, 33 and f = (/3, 5/7. The edges of the Hilbert basis H (R =H (cone((,, (33, 37 are given by (,, (, 4, (3, 3, (4, 6, (53, 3, (45, 63,, with k =,k =,k 3 =3,k 4 =,k 5 = k 6 =. The algorithm takes the following steps. Step Construct the initial vertex w =( f, f = (,. Step Check the edge e = (,. Find j such that (, +j f + R. We obtain j., i.e., w = (,. Step Check the edge e = (, 4, i.e., Hilbert basis elements (j, + 4j. Find minimal j such that (, + (j, + 4j f + R. We obtain j.4, i.e., we do not find a new vertex because j must be smaller than k =. Step 3 Check the edge e 3 = (3, 3, i.e., Hilbert basis elements ( + 3j, 9 + 3j. Find minimal j such that (, + ( + 3j, 9 + 3j f + R. We obtain j.5. Hence j =, and w 3 = (9, 37 is the next vertex. 3