Discrete Optimization 3 (006) 95 0 www.elsevier.com/locate/disopt Description of -integer continuous knapsack polyhedra A. Agra a,, M. Constantino b a Department of Mathematics and CEOC, University of Aveiro, Campus Universitário de Santiago, 380-93 Aveiro, Portugal b DEIO and CIO, University of Lisbon, Edifício C6, Campo Grande, 749-06 Lisboa, Portugal Received 0 January 004; received in revised form December 004; accepted October 005 Available online 7 April 006 Abstract In this paper we discuss the polyhedral structure of several mixed integer sets involving two integer variables. We show that the number of the corresponding facet-defining inequalities is polynomial on the size of the input data and their coefficients can also be computed in polynomial time using a known algorithm [D. Hirschberg, C. Wong, A polynomial-time algorithm for the knapsack problem with two variables, Journal of the Association for Computing Machinery 3 () (976) 47 54] for the two integer knapsack problem. These mixed integer sets may arise as substructures of more complex mixed integer sets that model the feasible solutions of real application problems. c 006 Elsevier B.V. All rights reserved. Keywords: Mixed integer programming; Polyhedral characterization; Single node flow problem. Introduction The description of the convex hull of elementary mixed integer sets has been useful in the generation of strong valid inequalities for general mixed integer problems. The so-called MIR inequalities are probably the most successful example in this area (see []). Recently, several elementary sets resulting from the aggregation of general MIP inequalities have been studied in order to generate strong cutting planes for those general models. Marchand and Wolsey [3] study the case of binary knapsack sets with a continuous variable and Günlük and Pochet [4] consider models with multiple constraints with a continuous variable and a single integer variable per inequality. In this paper we describe the convex hull of some mixed integer sets involving only two integer variables. First we consider sets of the form X = {(y, y, s) : y + a y D + δs, y, y, s 0, y, y integer}, where, a, D Z and δ {, }. Then we consider the integer single node flow set with two arcs: Z = {(x, x, y, y, s) : x + x D + s, 0 x y, 0 x a y, s 0, y, y integer} Corresponding author. E-mail addresses: aagra@mat.ua.pt (A. Agra), miguel.constantino@fc.ul.pt (M. Constantino). 57-586/$ - see front matter c 006 Elsevier B.V. All rights reserved. doi:0.06/j.disopt.005.0.008
96 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 and discuss the relationship between conv(z) and conv(z s=0 ) where Z s=0 = {(x, x, y, y ) : x + x D, 0 x y, 0 x a y, y, y integer}. Independently, Rajan [5] derived an equivalent description of conv(x). That description is based on an algorithm that uses an algorithm from Kannan [6] and then solves a sequence of diophantine approximations. Next we present the two main motivations to study the polyhedral structure of the convex hulls of these sets. The first motivation is based on the observation that most of the known valid inequalities for general mixed integer problems are generalizations of the basic MIR inequalities. These inequalities can be seen as the unique non-trivial facet-defining inequalities for the convex hull of sets involving one integer and one continuous variable. If we consider the set {(y, w) N 0 R : ay D + w} the basic MIR inequality for this set is given by y D/a + w a D/a D where D, a N and a does not divide D. In order to obtain new families of valid inequalities we study a model which has two integer and one continuous variables. The introduction of an extra integer variable increases considerably the complexity of the model. The reason is that in the -integer case, if the continuous variable is fixed we obtain a set whose convex hull has only one non-trivial facet-defining inequality, while in the -integer case we obtain a knapsack set with two integer variables which requires an algorithm to compute all coefficients necessary to obtain the polyhedral description. Nevertheless, this model allows a polynomial description and the models studied here inherit this property. A second motivation to study this kind of sets involving two integer and one continuous variable is the fact that they can be obtained from the relaxation and aggregation of more complex mixed integer sets that model real application problems. In particular, several such applications to lot-sizing models can be easily found. Sets of the form X and Z may arise when we consider machine capacity constraints in lot-sizing problems involving two types of batches, one small and one large. Other situations can occur from the aggregation of the well known flow conservation constraints and the elimination of the stock variables. For instance, Constantino [7] generates valid inequalities for the General Single Item (GSI) lot-sizing model with lower and upper bounds on production and start-up times by studying two - integer continuous sets obtained from relaxation and aggregation of the original model. Other applications also occur in telecommunication problems such as the capacitated expansion problem (see [8]). In this problem X occurs when we consider the flow capacity constraint on each arc. Typically, the flow capacity is equal to the installed capacity plus an integral multiple of several (in this case two) modularity sizes. In Section we describe the convex hull of the set {(y, y, s) N 0 R : y + a y D + s, s 0} with, a, D positive. This description is obtained using the following procedure: (i) eliminate the continuous variable by considering the restriction of the polyhedra to one of the two faces s = 0 or s = y + a y D; (ii) describe the convex hull of the resulting -integer knapsack sets using a polynomial algorithm from Hirschberg and Wong []; (iii) for each -integer facet-defining inequality find in polynomial time the lifting coefficient associated with the continuous variable. In Section 3 we consider other -integer continuous sets of the form X which we classify accordingly to the polyhedral description and give some results on the polyhedral structure of those models. Finally, in Section 4, using the same approach as in Section, we describe the convex hull of the integer single node flow set with two arcs and of the set Z s=0.. Polyhedral description of a -integer continuous set In this section we consider the set Y = {(y, y, s) : y + a y D + s, y, y, s 0, y, y integer} and assume that, a, D are positive integers and D > max{, a }. Let Q denote conv(y ). Consider the following sets (see Fig. ): Y = {(y, y ) N 0 : y + a y D}, Y = {(y, y ) N 0 : y + a y D} ()
A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 97 Fig.. Sets Y, Y, Y. and the corresponding convex hulls Q = conv(y ) and Q = conv(y ). The sets Y, Y are obtained when the continuous variable is eliminated by setting s = 0 and s = y + a y D, respectively. Let Q = {(y, y, s) Q : s = 0} and Q = {(y, y, s) Q : s = y + a y D}. We define the restriction of Q to s = 0 as the set Proj (y,y ) (Q ) and the restriction of Q to s = y + a y D as the set Proj (y,y ) (Q ). Remark. The convex hull of the restriction of a set S to the supporting hyperplane H of a valid inequality for S is equal to the restriction of the convex hull of S to H. This fact implies that the equalities Proj (y,y ) (Q ) = conv(y ) = Q and Proj (y,y ) (Q ) = conv(y ) = Q hold. For a given polyhedra P we denote by V (P) its set of extreme points. Lemma. If (y, y, s ) V (Q) then either s = 0 or s = y + a y D. Proof. Suppose not. Considering 0 < ɛ min{s, s y a y + D} and P = (y, y, s ɛ), P = (y, y, s + ɛ) then P, P Y and (y, y, s ) = /P + /P. Thus V (Q) V (Q ) V (Q ). Since Q and Q are faces of Q the converse is also true, so V (Q) = V (Q ) V (Q ). Lemma 3. (y, y, s ) V (Q) if and only if one of the following conditions holds: (i) s = 0 and (y, y ) V (Q ). (ii) s = y + a y D and (y, y ) V (Q ). Proof. Suppose (y, y, s ) V (Q). Lemma implies s = 0 or s = y + a y D. Suppose s = 0 (the other case is similar). Thus (y, y, 0) V (Q ). Using Corollary 3.6 in [9] we have (y, y ) V (Q ). To prove the converse suppose (y, y ) V (Q ). If (y, y, 0) V (Q) then there exists (y, y, s ) and (y, y, s ) Q such that (y, y, 0) = /(y, y, s ) + /(y, y, s ). As s 0, s 0 and /s + /s = 0 then s = s = 0. Using Remark we have (y, y ), (y, y ) Q, contradicting the hypothesis (y, y ) V (Q ). The proof for case (ii) is similar. Lemma 4. Let C(Q) = {(y, y, s) : y + a y s, y, y, s 0} denote the characteristic cone of Q. Then C(Q) = {r R 3 : r = 3 j= λ j r j, λ j 0, j =,, 3} where r = (0, 0, ), r = (, 0, ) and r 3 = (0,, a ). In order to obtain V (Q ) we present two algorithms given in []. Those algorithms compute the points in V (Q ) that maximize a function f y + f y with f f a. In order to obtain the extreme points that maximize the function with f f > a it suffices to exchange y with y. The first algorithm is a non-polynomial algorithm that computes coefficients with important properties (see Lemma 5 below) that are essential to prove Propositions 7 and 9. The second algorithm is a polynomial version of the first and it is the algorithm that we may use in practice to compute the necessary information to describe Q.
98 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 Algorithm HW Step 0: j, (y j, y j ) ( Da (e l, f l ) Step : While y j ck 0 do Set (i) (ii) ( ) a,, r. ) ( ), 0, k, (c k, d k ) a,, l, γ j D y j a y j, ρk c k + a d k, ρ l e l a f l if γ j ρ k set j j +, (y j, y j j ) (y, y j ) + r( c k, d k ); if γ j < ρ k and ρk ρl set k k +, (c k, d k ) (c k, d k ) + r(e l, f l ); (iii) if γ j < ρ k and ρk < ρl set l l +, (e l, f l ) (e l, f l ) + r(c k, d k ). ( ) The algorithm starts with a trivial extreme point (y, y ) = Da, 0 and, iteratively, computes rational c approximations, d and e f, for a ( d c a e f ). Once a fraction d c with a remainder ρ = c + a d not greater than the gap γ j is obtained another point (y j+, y j+ ) can be obtained replacing c units of with d units of a. Then γ j+ = γ j ρ. This algorithm is not polynomial and some points (y, y ) generated may not be extreme. In order to obtain only the extreme points in polynomial time as a function of the input data instead of considering always r = it suffices to compute r in Step as follows: { } γ j (i) r = min ρ k, y j c k { } ρ k ρ k γ j (ii) r = min, ; ρ l ρ l (iii) r =. ρ k ρ l ; Notice that these computations avoid the occurrence in consecutive iterations of the same case in Step. Although we are interested in obtaining the extreme points in polynomial time, the coefficients obtained using the non-polynomial Algorithm HW have some important properties we will use later. However, we notice here that all the information needed to describe Q can be obtained in O(log(D/ min{, a })) elementary operations using the polynomial version of Algorithm HW (see []). Next we introduce the notation for the non-polynomial case. We denote by k(l) the index of the pair (c, d) used, in (iii) of Step, to obtain (e l, f l ), that is (e l, f l ) = (e l, f l ) + (c k(l), d k(l) ). Similarly, we use the notation l(k) to denote the index of the pair (e, f ) used, in (ii) of Step, to obtain (c, d). Let n and n denote the number of distinct pairs (c, d) and (e, f ) generated, respectively. Now we summarize some of the properties of these coefficients. Lemma 5. (i) e l d k f l c k = if l = l(k) or k = k(l). (ii) c cn d,..., n (resp. d n a en f n e l f l, l =,..., n ) are the best approximations from below (resp. from above) to a e f. (iii) ck d k, k = for that size denominator. (iv) (a) The set {(c, d ),..., (c k, d k )} is an integral Hilbert basis for Cone{(c, d ), (c k, d k )}. (b) The set {(e, f ),..., (e l, f l )} is an integral Hilbert basis for Cone{(e, f ), (e l, f l )}. The first three properties are well known since the coefficients (c, d), (e, f ) are the sequence of convergents of the reduced continuous fraction into which a is expanded (see [0,]). (iv) is essentially due to [] and follows from the non-polynomial version of the algorithm. These algorithms can be easily adapted to generate V (Q ). We notice that as the number of extreme points is polynomial and these points can be obtained in polynomial time it is possible to give an extended formulation of polynomial size for Q using the relation Q = conv(v (Q)) + C(Q).
A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 99 In order to give an explicit linear description of Q we generate two families of inequalities. Restricting Y to s = 0 we obtain Y. Then generating a valid inequality to Y and lifting s we obtain a valid inequality to Y. Restricting Y to s = y + a y D we obtain Y. Generating a valid inequality to Y and lifting s we obtain a valid inequality to Y. Let us define Y = = {(y, y ) N 0 : y + a y = D}, Y > = Y \ Y = and Y < = Y \ Y =. Proposition 6. If α y + α y α defines a non-trivial facet of Q, then α y + α y α + βs defines a facet of Q, where β = max { α y +α y α y +a y D : (y, y ) Y > }. Proof. Consider a point (y, y, s ) Y with y + a y D (y, y ) Y. Since () is valid for all (y, y ) Y, β > 0 (in Proposition 7 below we determine this value, however by exhibiting points in Y > it is easy to verify that β > 0) and s 0 it follows that α y + α y α + βs. If y + a y > D then (y, y ) Y > and s > 0. Therefore, β α y + α y α y + a y D α y + α y α s α y + α y α + βs. Notice that since () defines a facet of Q and β is the smallest value in order for (3) to be valid for Q thus (3) must define a facet of Q. Now we address the question of how to compute β. We need to consider the coefficients computed by Algorithm HW. Notice that every facet of Q contains two points (y j, y j j+ ), (y, y j+ ) V (Q ) with (y j+, y j+ ) = (y j, y j ) + r( ck, d k ) for some k {,..., n } and some positive integer r. Thus α = d k κ, α = c k κ and α = κ(d k y j + ck y j ) for some κ > 0. We assume the coefficients are normalized, that is, κ =. Let us also assume α α a (the other case is similar, it suffices to exchange with a ) and assume that D is not divided by (in that case there would be no facet with α α < a since there would be only one extreme point that maximizes every function f y + f y with f f a ). We will use the notation γ (y, y ) = D y a y. Notice that γ j = γ (y j, y j ). The following proposition states that β depends on the gap γ j and on one of the remainders ρ l(k) = e l(k) a f l(k), ρ k = c k + a d k and, as we will see, it also depends explicitly on the position of the point (y j, y j ) in the plane. Notice that the coefficients (y j, y j ), (el(k), f l(k) ) and (c k, d k ) can be obtained using the polynomial version of Algorithm HW, and (c k, d k ) = (c k, d k ) (e l(k), f l(k) ). Thus, all the coefficients necessary to obtain β can be computed in polynomial time. Proposition 7. Let () define a facet of Q containing the points (y j, y j j+ ), (y, y j+ ) V (Q ) with (y j+, y j+ ) = (y j, y j ) + r( ck, d k ) for some k {,..., n } and some positive integer r. Assuming that α = d k, α = c k, then β = where η j = a η j (D/ ) D if k = and { l(k) ρ η j γ j, if y j = f l(k), ρ k γ j, if y j < f l(k), otherwise. Proof. We will use properties from Lemma 5. First consider the case k >. Let (y, y ) = { (y j, y j ) + (el(k), f l(k) ), if y j f l(k), (y j, y j ) + ( ck, d k ), if y j < f l(k). () (3)
00 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 Next we show that (y, y ) is the optimal solution to the maximization problem associated with the computation of β. Let τ(y, y ) denote τ(y, y ) = α y + α y α. So () can be written as τ(y, y ) 0. Notice that η j = γ { (y, y ) and, from Lemma 5, τ(y, y ) = which implies (y, y ) Y >. Using this notation we have β = max τ(y,y ) γ (y,y ) : (y τ(y, y ) Y > }. Hence, we need to show that,y ) γ (y,y ) τ(y,y ) γ (y,y ) for all (y, y ) Y >. Since the left-hand side is positive and (y, y ) Y > γ (y, y ) < 0 we only need to consider those points such that τ(y, y ) (notice that α, α, α are integer). Claim. (a) α η j 0; (b) a α η j 0. Proof of Claim : Let w = c k + d k a and w = e l(k) f l(k) a. Using (a) in Lemma 5 we obtain d k w + f l(k) w =. There is a positive integer r such that γ (y j, y j j+ ) = γ (y, y j+ ) + r( c k + d k a ) = γ (y j+, y j+ ) + r(w w ) w w 0. Since { w γ j, if y j f l(k), η j = w γ j, if y j < f l(k), we have w η j and w η j. Thus = d k w + f l(k) w (d k + f l(k) )η j = α η j. The proof of (b) is similar. Claim. (a) e l a f l η j (d k e l c k f l ) for all l l(k); (b) c t + a d t > η j ( d k c t + c k d t ) for all t k. Proof of Claim : For l = l(k) we have d k e l c k f l = and as we saw above e l a f l = w η j. For t = k we have d k c t + c k d t = 0 and w = c t + a d t > 0. The proof for all l < l(k) and for all t < k can be made by induction using simultaneously (a) and (b) and noticing that (c t, d t ) = (c t, d t ) (e l(t), f l(t) ) and (e l, f l ) = (e l, f l ) (c k(l), d k(l) ). Now consider (y, y ) (y, y ) such that (y, y ) Y > and τ(y, y ). Three cases may occur: Case. y y, y y. Let (c, d) = (y y, y y ). Thus τ(y, y ) γ (y, y ) = τ(y, y ) + α c + α d γ (y, y ) + c + a d. From Claim comes c( α η j ) + d(a α η j ) 0 α c+α d c+a d η j τ(y,y ) γ (y,y ) η j. Case. y y, y y. Let (c, d) = (y y, y y ). It cannot happen d c > ck because that would imply d k τ(y, y ) = τ(y, y ) ck c + d k d < τ(y, y ) τ(y, y ) 0. So d c ck. If c d k d < c let p = d ck c, d d k otherwise take p = 0. Thus (c + p, d) Cone{(c, d ), (c k, d k )}. Therefore, using (iv) in Lemma 5 (c, d) can be written as (c, d) = k t= β t (c t, d t ) p(, 0) where β t are non-negative integers. Thus τ(y, y ) γ (y, y ) = τ(y, y ) + k β t ( α c t + α d t ) + pα t=. γ (y, y ) + k β t ( c t + a d t ) + p t= From Claim, for t =,..., k we obtain α c t +α d t c t +a d t η j and, if p 0 from Claim, pα p η j thus τ(y,y ) γ (y,y ) η j. Case 3. y y, y y. First we prove that (e, f ) = (y y, y y ) satisfies e f Notice that τ(y, y ) implies e f ck d k. If a el(k) f l(k). Suppose e f < el(k) f l(k). e f ck d k, as ck d k is the best approximation from below to a for that size denominator, we conclude f d k. Similarly, if a < e f < el(k) f l(k) then f f l(k)+ f l(k) + d k > d k. In both cases we have y j y dk = y + f dk d k d k = f l(k). Thus (y, y ) = (y j, y j ) + (el(k), f l(k) ).
A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 0 This implies y j = y + f l(k) = y + f + f l(k) > d k. Now considering (y, y ) = (y j + ck, y j dk ) we have y, y 0 and γ (y, y ) 0. Thus (y, y ) Y and (y j, y j ) can be obtained as a linear convex combination of (y, y ) and (y j+, y j+ ) which contradicts the hypotheses that (y j, y j ) V (Q ). Let p = f el(k) e if e f l(k) f > e and p = 0 otherwise. Thus (e p, f ) Cone{(e, f ), (e l(k), f l(k) )}. Now we f can use (iv) in Lemma 5 and the proof is similar to the proof of Case. If k = set (y, y ) = ( D/, 0). Notice that k = implies j =, thus (y j, y j ) = ( D/, 0) and so τ(y, y ) =. For all other (y, y ) Y > such that τ(y, y ) only Case or Case considered above can occur. Claim also holds because c = a /, d = and η j < imply d η j 0 and a a / c η j. Thus, the proofs given for Case and Case are valid in this situation. Notice that in this case it is not necessary to use Claim in the proof of Case. Considering the restriction of Y to s = y + a y D we obtain similar results. Proposition 8. If α y + α y α defines a non-trivial facet of Q then ( β α )y + (a β α )y Dβ α + βs (5) defines a facet of Q, where β = max { α α y α y D y a y : (y, y ) Y < }. Proof. Notice that (5) can be written as β(s+d y a y )+α y +α y α. The validity for (y, y, s) Y with (y, y ) Y follows from the validity of (4) and β 0, s + D y a y 0. If (y, y, s) Y with (y, y ) Y <, then β α α y α y D y a y (D y a y )β α α y α y β(s + D y a y )+α y +α y α. Since () defines a facet of Q and β is the smallest value in order for (5) to be valid for Q thus (5) must define a facet of Q. Now we obtain β considering α suffices to exchange with a. α a and assuming that D is not divided by. The case α α a (4) is similar. It Proposition 9. Let (4) define a facet of Q containing the points (y j, y j j+ ), (y, y j+ ) V (Q ) with (y j+, y j+ ) = (y j, y j ) + r( el, f l ) for some l {,..., n } and some positive integer r. Assuming that α = f l, α = e l, then β = where η j = D a η j (D/ ) if l = and { γ η j j + ρ k(l), if y j = dk(l), γ j + ρ l, if y j < dk(l), otherwise. Since the proof of Proposition 9 is similar to the proof of Proposition 7 it will be omitted. Theorem 0. The families of inequality (3) and (5) with the trivial set of inequalities y 0, y 0, s 0, y + a y D + s suffice to describe Q. Proof. Considering a facet F defined by a non-trivial inequality µ y + µ y µ + νs ( f ) we show that this inequality either belongs to family (5) or it belongs to family (3). First, we show that (i) µ, µ, µ 0, (ii) ν > 0, (iii) ν > µ, νa > µ.
0 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 Proof of (i). Suppose µ < 0. There must exist a point (y, y, s ) F satisfying y > 0 since otherwise F {(y, y, s) : y = 0}. Thus (0, y, s ) Y and violates ( f ). Similarly we have µ 0. As (0, 0, 0) Y it follows that µ 0. Proof of (ii). Since (, 0, ) and (0,, a ) are directions of Q and µ and µ cannot be simultaneously zero (otherwise F {(y, y, s) : s = 0}) we have ν > 0. Proof of (iii). We only prove ν > µ since the other case is similar. The existence of the direction (, 0, ) implies ν µ. Suppose µ = ν. Let (y, y, s ) F Y that satisfies y + a y < D + s (there must exist such a point since, otherwise, F {(y, y, s) : y + a y = D + s}). From this inequality and from µ ν y + µ ν y = µ ν + s we obtain (a µ ν )y < (D µ ν ). Now, consider a point (y, y, s ) Y such that s = y + a y D (for instance, setting y = D a y if D a y or setting y = y, otherwise). This point must satisfy µ ν y + µ ν y µ ν + s because ( f ) is valid for Y. From this inequality and from s = y + a y D we have (a µ ν )y (D µ ν ) contradicting the previous inequality. Now we are ready to prove that ( f ) either belongs to (3) or to (5). As dim(f) = there are three affinely independent points lying in F. First we consider the case where F is bounded. Thus we may consider extreme points in Q. From Lemma 3 every extreme point of Q lies in one of the planes defined by s = 0 and y + a y = D + s. Noticing that the three extreme points cannot all lie in the same plane simultaneously, we first suppose two of them satisfy s = 0. If (y, y, s) is one of these two points, from Lemma 3, it must satisfy (y, y ) Q and, since s = 0, it also satisfies µ y + µ y = µ. From the validity of ( f ) for Y it follows that µ y + µ y µ is valid for Y and therefore it defines a facet of Q. Since µ and µ are non-negative and they cannot be simultaneously zero we know that this inequality defines a non-trivial facet for Q. Now, consider a third extreme point (y, y, s ) satisfying y +a y = D+s with s > 0 which implies (y, y ) Y >. Thus, every point in F satisfies µ y +µ y = µ+νs where ν = µ y +µ y µ y +a y D. In order for ( f ) to define a valid inequality ν must satisfy ν = max{ µ y +µ y µ y +a y D : (y, y ) Y > }. Therefore ( f ) belongs to family (3) where α = µ, α = µ, α = µ and β = ν. Now we consider the case where there are two extreme points in the facet satisfying y + a y = D + s. Using Lemma 3, (y, y ) Q. Such points must satisfy µ y +µ y = µ+ν( y +a y D) (µ ν )y +(µ νa )y = µ ν D. Notice that from (iii), (µ ν ) < 0 and (µ νa ) < 0 and the validity of ( f ) for Y implies that (ν µ )y + (νa µ )y ν D µ is valid for Y and therefore it defines a non-trivial facet of Q. Let α = ν µ, α = ν µ, α = ν D µ. Now, considering a third extreme point (y, y, s ) with s = 0 and (y, y ) Y < we have µ y + µ y = µ (ν α )y + (νa α )y = ν D α α α y α y = ν(d y a y ). In order for ( f ) to define a valid inequality (y, y ) must satisfy (y, y ) arg max{ α α y α y D y a y : (y, y ) Y < }. Therefore ( f ) belongs to family (5) with α = ν µ, α = ν µ, α = ν D µ, β = ν. It remains to consider the case in which F is not bounded. Let (v, v, v 3 ) be a direction of F. Thus it must satisfy µ v + µ v = νv 3. From (iii) and noticing that v, v 0 and v and v cannot be simultaneously zero, we have ν v + νa v > νv 3 v + a v > v 3. But since (v, v, v 3 ) is also a direction of Q we obtain v + a v v 3 which contradicts the previous result. Thus there are no unbounded non-trivial facets. Example. Consider Y = {(y, y, s) : y + 76y 54 + s, y, y, s 0, y, y integer}. Setting s = 0 we obtain Y = {(y, y ) : y + 76y 54, y, y 0 and integer}. Using Algorithm HW (see Table ) we can obtain Q = {(y, y ) : y + 3y 54, y + 7y 09, 5y + 8y 74, 3y + y 66, y + 4y 60, y 0, y 0}. Based on the description of Q we obtain the following set of facet-defining inequalities for Q: y + 3y 54 + s, y + 7y 09 + s, 5y + 8y 74 + s, 3y + y 66 + s, y + 4y 60 + s 7. For instance, exchanging the coefficients of y and y (see the second table in Table ) we obtain the inequality y + 3y 66 containing (y j, y j j+ ) = (4, 4), (y, y j+ ) = (8, 6). Thus (c k, d k ) = (3, ),
A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 03 Table Coefficients obtained using the polynomial version of Algorithm HW a b γ c d R e f R a b γ c d R e f R 54 0 0 3 3 4 8 5 0 4 0 55 5 7 3 3 4 8 5 0 4 0 3 3 5 7 7 5 4 8 5 0 4 4 8 3 3 44 3 7 5 4 8 4 4 6 4 8 3 3 44 3 7 5 3 3 4 4 6 4 8 7 5 44 3 8 5 3 3 4 4 6 3 3 7 5 6 8 0 8 5 3 3 8 6 0 3 3 7 5 6 8 0 8 5 9 8 8 6 0 3 3 5 8 6 8 0 76 0 9 8 8 6 0 8 9 5 8 8 6 0 8 9 3 47 8 6 0 76 0 3 47 In the left table considering =, a = 76 and in the right table exchanging with a (corresponding to exchanging y with y ). (c k, d k ) = (, 4) and (e l, f l ) = (, 7). As b j < f l, then consider (4, 4) + (, 4) = (3, 8) Y. So η j = γ (y j, y j ) c k +a d k = 6 76 + 4 =. Hence β = /. Exchanging again the coefficients of y and y we obtain the inequality 3y + y 66 + s. Now, setting s = y + 76y 54, from s 0 we have Y = {(y, y ) : y + 76y 54, y, y 0 and integer}. Again, computing the coefficients, Q = {(y, y ) : 8y + 9y 440, 5y + 8y 74, y + 7y 07, y + y 6, y 0, y 0}. Based on the description of Q we derive the following set of inequalities: 5y + 8y 74 + s, 6y + y 33 + s, 9y + 34y 5 + s, 7y + 6y 930 + s. These two sets of inequalities with the trivial facet-defining inequalities y 0, y 0, s 0, y + 76y 54 + s suffice to describe Q. 3. Other -integer continuous sets In this section we briefly discuss other -integer continuous sets that may have practical importance. Since the polyhedral description for those cases is similar to the description of Q we focus on the differences between Q and the other polyhedra. First, consider X = {(y, y, s) N 0 R + : y a y D + s}, where, a, D > 0. Let P = conv(x). The major differences between P and Q are the corresponding characteristic cones and the fact that P may have one non-trivial unbounded facet. Next we consider these two issues. Lemm. Let C(P) = {(v, v, v 3 ) : v a v v 3, v, v, v 3 0} be the characteristic cone of P. Then C(P) = {r : r = 4 i= µ i r i, µ i 0, i =,, 3, 4} where r = (0, 0, ), r = (0,, 0), r 3 = (, 0, ), r 4 = (a,, 0). The bounded facets of P can be obtained in a similar fashion to the facets of Q, by considering the restriction of X to s = 0 and the restriction of X to s = y a y D. However P may have one more unbounded facet. Proposition 3. The unbounded facets of P are defined by y 0, y 0, s 0, y a y D + s and g y a y D m + s g m (6) where g = g.c.d.{, a } and m = D g D g. Notice that if m = 0, i.e. g divides D (the equation y a y = D, has integer solutions), the last two inequalities become the same inequality. Thus we will assume m 0.
04 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 Proof. First we show that (6) defines an unbounded facet. To prove validity we show that (6) can be seen as a basic MIR inequality. Considering a = g, y = ( /g)y (a /g)y and w = s, then from () we obtain s ( /g)y (a /g)y D/g + g D/g D a g y a y D m + s g m. To prove that (6) defines a facet consider the following three affinely independent points: (a, b, 0), (a, b, 0) + (a,, 0), (a, b, g m) where (a, b) is a solution to y a y = D m with y, y Z + and (a, b ) is a solution to y a y = D m + g with y, y Z +. Next we show that if an inequality defines an unbounded non-trivial facet then it is defined up to the multiple of a constant by (6). Let α y α y α + βs (7) define such facet of P denoted by F. As (0, 0, 0) P we have α 0. If α < 0 then F {(y, y, s) : y = 0}. So, α 0. The existence of the directions (a,, 0) and (, 0, ) implies α a α and α β, respectively. Let (v, v, v 3 ) be a direction of F. From s 0 it follows that v 3 0. Suppose v 3 > 0. There exists one (y, y, s ) F with y a y < D + s. Thus, there exists a λ > 0 such that (y, y, s ) = (y, y, s ) + λ(v, v, v 3 ) and it satisfies y a y < D + s and s > 0. As β > 0 (the case β = 0 implies α = 0 and hence F {(y, y, s) : y = 0}) then the point (y, y, s+ ) P with s + = max{0, y a y D} violates (7). Thus v 3 = 0 and so α v = α v. Using this equation and the inequality α a α we conclude that a v v. To prove the equality a v = v we suppose a v > v. Let (y 0, y0, s0 ) F with y 0 a y 0 < D +s0. Setting λ = max{0, ( (D + s 0 y 0 a y 0))/( v + a v )}, the point (y 0 + λv +, y 0 + λv, s 0 ) P violates (7). Since a v = v and α v = α v the inequality (7) can be written as y a y α + β s where α = α α and β = α β. As β > 0, every point lying in F X either satisfies s = 0 or s = y a y D (otherwise we could decrease the value of s). Let us consider a point (y, y, s ) satisfying s = 0. Therefore y a y = α. Since (7) is a valid inequality we have α = max{ y a y : y a y D, y, y N 0 } = D m. Therefore β > (otherwise every point (y, y, s ) X satisfying y a y = D + s would violate (7)). Now consider (y, y, s ) F satisfying s = y a y D. Thus y a y = D m + β ( y a y D). So β m = + y a y D. Since (7) is a valid inequality we have (y, y ) = arg min{y a y : y a y D, y, y N 0 }. Therefore y a y D = g m β = g g m. Now, we consider all mixed integer sets of the form: X = {(y, y, s) N 0 R + : y + a y D + δs} where, a, D Z, δ {, }. The description of conv(x) is trivial for sets X where and a have the same sign and D has the opposite sign. If δ < 0 then all non-trivial facets of conv(x) are obtained from the description of the convex hull of the restriction of X to s = 0. For these cases α y + α y α + βs defines a non-trivial facet of conv(x) if and only if β = 0 and α y + α y α defines a non-trivial facet of conv(x s=0 ) where X s=0 = {(y, y ) N 0 : y + a y D}. Notice that lifting each non-trivial facet-defining inequality for conv(x s=0 ) we obtain β = 0 and, conversely, each non-trivial facet-defining inequality for conv(x) satisfies β = 0. It remains to consider four cases with δ > 0. The case > 0, a > 0 and D > 0 was considered in Section. The case < 0, a < 0, D < 0 is similar to the previous one. The case where > 0 and a < 0 was considered above and the case < 0 and a > 0 is similar to that one. Other important -integer continuous sets arise when upper and lower bounds on the integer variables are considered. In those cases a polyhedral description can be obtained in a way similar to that given in Section. However, for each particular situation it is necessary to adapt Algorithm HW to generate the extreme points of the pure -integer polyhedra. To compute the lifting coefficient β of the continuous variable, some minor modifications must be made in order to ensure that the point chosen to compute β (the point (y, y ) considered in the proof of Proposition 7) satisfies these new constraints.
4. The integer single node flow set with two arcs A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 05 In this section we consider the mixed integer set Z defined by: x + x D + s, x y, x a y, x, x, s 0, y, y integers. We assume that, a, D are positive integers with D > max{, a }. It is important to notice that there are only two integer variables involved in this model and so, for this particular structure, all the information needed to describe conv(z) will also be obtained from the -integer knapsack sets that result from the elimination of the continuous variables. Since there are several similarities between the study of conv(z) and the study of Q in Section we will omit some details. First consider the characteristic cone of conv(z), denoted by C(Z). Lemm4. C(Z) = {(x, x, y, y, s) : x + x s, x y, x a y, x, x, s 0} = {r R 5 : r = 5 j= λ j r j, λ j 0, j =,..., 5} where r = (0, 0, 0, 0, ), r = (0, 0,, 0, 0), r 3 = (0, 0, 0,, 0), r 4 = (, 0,, 0, ) and r 5 = (0, a, 0,, a ). All the extreme points of conv(z) lie in the intersection of three of the following four hyperplanes defined by x = y, x = a y, x + x = D + s and s = 0. Thus, every extreme point of conv(z) has to satisfy one of the following set of conditions: (i) x = y, x = a y, s = 0, (ii) x = D x + s, x = a y, s = 0, (iii) x = y, x = D x + s, s = 0, (iv) x = y, x = a y, s = x + x D. In case (i) we have (y, y ) Y. In case (ii), noticing that 0 x y implies that 0 D a y y, we have (y, y ) Y where Y = {(y, y ) N 0 : y + a y D, y D/a }. Note that Y differs from Y because it includes the additional constraint y D/a that is implied by the nonnegativity constraint x 0. Similarly, in case (iii) we have (y, y ) Y where Y = {(y, y ) N 0 : y + a y D, y D/ }. Finally, in case (iv), we have (y, y ) Y. Let us define Y > = Y \ Y = and Y > = Y \ Y =. First we consider the valid inequalities obtained from the lifting of facet-defining inequalities for Q = conv(y ) (corresponding to case (i)). Proposition 5. If α y + α y α is a valid facet-defining valid for Q then the inequality β (x y ) + β (x a y ) + α y + α y α + βs (4) is a valid facet-defining inequality for conv(z), where α y + α α β = max y + a y D : (y, y ) Y >, α y + α y α β = max y + a y D : (y, y ) Y > α y + α y α and β = max y + a y D : (y, y ) Y >. (8) (9) (0) () () (3)
06 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 Notice that the value of β can be computed as indicated in Proposition 7 and the computation of β and β is similar with small differences. If k > it is necessary to check if the point (y j, y j ) + (el(k), f l(k) ) satisfies the additional condition, y D/a or y D/, according to the coefficient we are computing, β or β, respectively. If k = with α =, α = a / then β = /( (D a )/ + a D). If α = /a, α = then β = /(a (D )/a + D). Now we are ready to prove Proposition 5. Proof. Consider (x, x, y, y, s) Z. If (y, y ) Y, from validity of (3) for Y and noticing that β (x y ) 0, β (x a y ) 0 and βs 0 we conclude that (x, x, y, y, s) satisfies (4). Now suppose (y, y ) Y >. We may assume that only one of the three cases may occur: s > 0, x < y or x < a y. In fact we are checking all the extreme points. Note that this verification is enough since β > 0 and, as in Claim, we have α j β j a j 0, j {, } and βa j α j, j {, } which ensures that moving along any direction of C(Z) from a feasible point we obtain only feasible solutions. Assume x < y. If (y, y ) Y > then from the definition of β we have Thus, β ( y + a y D) α y + α y α β (D y a y ) + α y + α y α. β (x y ) + β (x a y ) + α y + α y β (D y a y ) + α y + α y α α + βs. The case (y, y ) Y > \ Y > cannot occur because it would imply y > D/a and, as we are assuming x = a y, s = 0 we would have x + x > D + s. The cases x < a y and s > 0 are similar. Since (3) defines a facet of Q and β, β, β take the smallest values in order for (4) to be valid for conv(z) then (4) must define a facet of conv(z). Example 6. Consider the set, Z = {(x, x, y, y, s) : x + x 54 + s, 0 x y, 0 x 76y, s 0, y, y integer}. Based on the description of Q we obtain the following set of facet-defining inequalities for conv(z): y + 3y + (x y ) + 4 (x 76y ) 54 + s, y + 7y + (x y ) + 6 (x 76y ) 09 + s, 5y + 8y + (x y ) + 3 (x 76y ) 74 + s, 3y + y + (x y ) + (x 76y ) 66 + s, y + 4y + 7 (x y ) + 7 (x 76y ) 60 + 7 s. For instance, considering the facet-defining inequality y + 3y 54 for Q we have (c, d ) = (, 3), thus k =. So, β = β = /( 54/ 54) = /. That is, β and β are obtained for (y, y ) = (55, 0). However, β = /( (54 76)/ + 76 54) = /4. We obtain the inequality y + 3y + (x y ) + 4 (x 76y ) 54 + s. Next we state similar results for the remaining cases without proof. Proposition 7. If α y + α y α (5) is a valid facet-defining inequality for conv(y ) containing only points in Y > then the inequality βs + α y + α y α + β (x + x D s) + β (x a y ) (6) is valid for Z and defines a facet of conv(z), where
A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 07 α α y α y β = max y + a y D : (y, y ) Y >, α α y α y β = max : (y, y ) Y < D y a y α α y α y and β = max y + a y D : (y, y ) Y >. Note that β and β are zero if (5) is valid for Y > and Y >, respectively, and they are strictly positive otherwise. In this last case it can be easily checked that β and β are obtained for (y, y ) = ( D/, 0). So β = β. The restriction in Proposition 7 that the valid inequality (5) for Y must contain only points in Y > ensures that β and β are nonnegative. The facet-defining inequalities excluded by this restriction belong also to the family of valid inequalities introduced in Proposition 5. In Example 6, conv(y ) = {(y, y ) : 8y +9y 440, 5y +8y 74, y +7y 07, y 0, y 0}. Only y + 7y 07 defines a non-trivial facet that includes only points in Y >. Based on that inequality we obtain: 0s + y + 7y 07 + 6 (x + x D s) + 0(x 76y ). Now we have a similar result considering Y. Proposition 8. If α y + α y α is a valid facet-defining inequality for conv(y ) containing only points in Y > the inequality βs + α y + α y α + β (x y ) + β (x + x D s) (8) is valid for Z and defines a facet of conv(z), where α α y α y β = max y + a y D : (y, y ) Y >, α α y α y β = max y + a y D : (y, y ) Y > α α y α y and β = max : (y, y ) Y <. D y a y Continuing Example 6, conv(y ) = {(y, y ) : y, y + 4y 56, 3y + y 66, 5y + 8y 74, y + 7y 07, y + y 6, y 0, y 0}. Based on these inequalities we derive the following set of facet-defining inequalities for conv(z): s + y + 0 (x + x 54 s) + (x y ), s + y + 4y 56 + 7 (x + x 54 s) + (x y ), 0s + y + 7y 07 + 6 (x + x 54 s) + 0(x y ), 0s + y + y 6 + 4 (x + x 54 s) + 0(x y ). Finally, we consider the last set of inequalities. Proposition 9. If α y + α y α is a valid facet-defining inequality for conv(y ) containing only points in Y > then the inequality β(s + D x x ) + α y + α y α + β (x y ) + β (x a y ) (0) (7) (9)
08 A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 is valid for Z and defines a facet of conv(z), where α α y α y β = max : (y, y ) Y <, D y a y α α y α y β = max y + a y D : (y, y ) Y > α α y α y and β = max y + a y D : (y, y ) Y >. Continuing Example 6 it can be checked that every facet of the form (0) can also be obtained either from (6) or from (8). Note that the convex hulls of Y, Y and Y are very similar and, in general, they share a set of facets. Thus, several facet-defining inequalities of conv(z) may belong to more than one of the families (6), (8), (0). In fact, all but the facet in the border of Q adjacent to the facet defined by x = 0 (resp. x = 0) are also facets of conv(y ) (resp. conv(y )). Hence, conv(y ) can only contribute a new facet if it has only one non-trivial facet. Now we consider two unbounded facet-defining inequalities that can be obtained by the MIR procedure. Proposition 0. The inequality x i γ i y i (a i γ i ) D/a i + s where γ i = D a i D/a i, and i {, }, is valid for Z. Proof. Consider the basic MIR inequality () with y = y i, a = a i, w = s + a i y i x i. In Example 6 we have two unbounded non-trivial facet-defining inequalities: x 0y 53+s, x 69y 98+s. In this example the inequalities presented so far with the trivial facet-defining inequalities suffice to describe conv(z). Proposition. Inequalities () are the unique inequalities defining unbounded facets of conv(z). Proof. Suppose F is an unbounded facet of conv(z) defined by ζ x + ζ x µ y µ y µ + νs. () First notice that ζ, ζ, µ, µ, µ, ν 0 and ν max{ζ, ζ }. The direction r cannot belong to the characteristic cone of F (denoted by C(F)) because, otherwise, we would have ν = 0 which implies ζ = ζ = 0 and, therefore, µ = µ = 0. Now we consider r 4. If r 4 C(F) then ν = ζ µ. Consider a point (x, x, y, y, s) F Z satisfying x + x < D + s. As ν > 0 and µ 0 then ζ > 0 x = y. Thus y + x < D +s. On the other hand as (x, x, y, y, s) F we have (ζ µ )y +ζ x µ y = µ+νs. Using ν = ζ µ we obtain y + (ζ /ν)x (µ /ν)y = µ/ν + s. From this equation and from the inequality y + x < D + s we have (ζ /ν)x + (µ /ν)y + x < D µ/ν. Now, considering the point (x, x, y, y, s ) Z with y = (D x s)/, x = y and s = x + x D we have y + x = D + s. But, as () is valid for Z we also have (ζ µ )y + ζ x µ y µ + νs. Again, since ν = ζ µ it follows that y + (ζ /ν)x (µ /ν)y µ/ν + s. Using the equation y + x = D + s we obtain (ζ /ν)x + (µ /ν)y + x D µ/ν, contradicting the previous strict inequality. Hence r 4 C(F). Similarly r 5 C(F). Hence, only r and r 3 may belong to C(F). Suppose r C(F). Thus µ = 0 which implies ζ = 0 because, otherwise, we would have x + x = D + s for every point in F. Thus r 3 cannot belong to C(F) because we would obtain a valid inequality dominated by s 0. Since ζ > 0 we can write () as x µ y µ + ν s. We know that ν = ν/ζ. Suppose ν > then s = 0 for every point in the facet because if s > 0 we could decrease x and s in the same amount obtaining a point violating (). Thus ν =. Now we prove that µ = (a µ ) D/a. Considering the point (x, x, y, y, s) = (0, a D/a, 0, D/a, 0) Z we have µ (a µ ) D/a. Suppose µ > (a µ ) D/a. As x a y x µ y (a µ )y we conclude that if (x, x, y, y, s) F then, as µ < a, we have y > D/a and, therefore, x + x = D + s. Thus, µ = (a µ ) D/a. Finally we show that µ = (D a D/a ). Considering the point (0, D, 0, D/a +, 0) it follows that D µ ( D/a + ) (a µ ) D/a. Thus µ D a D/a. Suppose µ > D a D/a. From x D + s we have x µ y (D a D/a ) + a D/a µ y + s. If (x, x, y, y, s) F then ()
A. Agra, M. Constantino / Discrete Optimization 3 (006) 95 0 09 µ +s (D a D/a )+a D/a µ y +s µ D/a (D a D/a ) µ y µ (y D/a ) (D a D/a ). As we are assuming µ > D a D/a we must have y D/a which implies x = a y. Hence µ = D a D/a. Theorem. conv(z) is completely described by the trivial facet-defining inequalities and the families (4), (6), (8), (0) and (). Proof. Since the proof is similar to the proof of Theorem 0 we omit some technical details. Consider a facet F defined by a non-trivial inequality ζ x + ζ x µ y µ y µ + νs ( f ). In the case where ( f ) defines an unbounded facet it was proven in Proposition that ( f ) is of type (). Thus, suppose ( f ) defines a bounded facet and therefore it must include five affinely independent extreme points. Notice that every extreme point satisfies one of the four types of conditions (i) (iv). There must exist one extreme point satisfying each one of those conditions otherwise ( f ) would not define a facet. For instance, if there were no points satisfying conditions (i) then we would have x + x = D + s for every point in F. At least one set of conditions must be satisfied by two extreme points. If that set is (i), (ii), (iii) or (iv) then, as in the proof of Theorem 0, we may conclude that ( f ) can be written as (4), (6), (8) or (0), respectively. Observe that those cases excluded in the hypothesis of Propositions 7 9 correspond to situations where (i) is satisfied by two extreme points and, therefore, the corresponding facets are also of type (4). Now we consider the set Z s=0 = {(x, x, y, y ) : x + x D, 0 x y, 0 x a y, y, y integer}. Obviously, restricting conv(z) to s = 0 we obtain conv(z s=0 ). However the relation is stronger. In the description of the bounded facets of conv(z) only the family of inequalities (0) was obtained from the lifting of -integer valid inequalities considering s 0 (the sets of conditions (i), (ii), (iii) consider s = 0). However, as is illustrated in Example 6, the set of facets of conv(y ) is usually contained in the set of facets of conv(y ) and conv(y ). In that case the set of inequalities (0) is contained in the sets of inequalities (6) and (8). In fact, it can be checked that the unique exception is the case where conv(y ) has only one non-trivial facet. Thus if conv(y ) has more than one non-trivial facet then a valid inequality for Z, ζ x + ζ x µ y µ y µ + νs defines a non-trivial facet of conv(z) if and only if ν = max{ζ, ζ } and ζ x + ζ x µ y µ y µ is valid for Z s=0 and defines a non-trivial facet of conv(z s=0 ). Acknowledgements We thank Laurence Wolsey for his helpful suggestions. This work was partly carried out within the framework of ADONET, a European network in Algorithmic Discrete Optimization, contract no. MRTN-CT-003-504438. References [] D. Hirschberg, C. Wong, A polynomial-time algorithm for the knapsack problem with two variables, Journal of the Association for Computing Machinery 3 () (976) 47 54. [] H. Marchand, L. Wolsey, Aggregation and mixed integer rounding to solve mips, Operations Research 49 (3) (00) 363 37. [3] H. Marchand, L. Wolsey, The 0- knapsack problem with a single continuous variable, Mathematical Programming 85 () (999) 5 33. [4] O. Günlük, Y. Pochet, Mixing mixed-integer inequalities, Mathematical Programming 90 (3) (00) 49 457. [5] D. Rajan, Designing capacitated survivable networks: Polyhedral analysis and algorithms, Ph.D. Thesis, University of California, Berkeley, 004. [6] K. Kannan, A polynomial algorithm for the two-variable integer programming problem, Journal of the Association for Computing Machinery 7 (980) 8. [7] M. Constantino, A polyhedral approach to a production planning problem, Annals of Operations Research 96 (000) 75 95. [8] D. Bienstock, O. Günlük, Capacitated network design polyhedral structure and computation, INFORMS Journal on Computing 8 (996) 43 59.
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