Lifting 2-integer knapsack inequalities

Transcription

1 Lifting 2-integer knapsack inequalities A. Agra University of Aveiro and C.I.O. M.F. Constantino D.E.I.O., University of Lisbon and C.I.O. October 1, 2003 Abstract In this paper we discuss the generation of strong valid inequalities for (mixed) integer knapsack sets based on lifting of valid inequalities for basic knapsack sets with two integer variables (and one continuous variable). The description of the basic polyhedra can be made in polynomial time. We use superadditive valid functions in order to obtain sequence independent lifting. 1 Introduction The idea of lifting was introduced by Gomory (1969). An important contribution was given by Wolsey (1977) when the link between superadditivity and the lifting procedure was established. The use of superadditive functions in the lifting procedure allows the introduction of several variables simultaneously which reduces the computational burden of the generation of valid inequalities. The procedure of lifting has been explored more intensively in the last few years for the generation of strong valid inequalities for (mixed) integer sets. Usually these inequalities are obtained from the lifting of facet defining inequalities of elementary polyhedra. We mention the works of Gu, Nemhauser & Savelsbergh (1999) and Gu, Nemhauser & Savelsbergh (2000) on the single node flow sets and on the use of superadditivity, the papers of Marchand & Wolsey (1999) and Marchand & Wolsey (2001) based on the lifting of the well known MIR inequalities. More recently Atamtürk has produced important work (2001a, 2001b, 2002) on the lifting of valid inequalities for mixed integer sets. We consider the integer set Y = y N N 0 : j N a jy j D} and the mixed integer set X = (y, s) N N 0 R : j N a jy j D + s, s 0} where a j, j N and D are positive integers. These sets may arise as aggregation of more general (mixed) integer sets or as subsets of more complex sets. First we consider the restriction of both sets obtained by setting all but two integer variables to zero. Suppose y 1, y 2 are those variables, thus the restricted sets are, respectively, Y = (y 1, y 2 ) N 2 0 : y 1 + a 2 y 2 D} and R = (y 1, y 2, s) N 2 0 R : y 1 + a 2 y 2 D + s, s 0}. Then, for each facet defining inequality of the convex hull of the restricted sets we construct valid inequalities for X and Y using superadditive valid lifting functions. In Section 2 we review some concepts on the lifting procedure. In Section 3 we summarize some results on the description of the 2-integer knapsack polyhedra. In Section 4 a polinomial superadditive valid lifting function for tight valid inequalities for the 2-integer knapsack set is given. In Section 5 1

2 we consider facet defining inequalities and characterize the lifting function for those inequalities. In Section 6 we construct several superadditive valid lifting functions for the facet defining inequalities. In Section 7 the results obtained in the previous sections are adapted to the integer knapsack set with a continuous variable. The computational experience is reported in Section 8. 2 Lifting and superadditivity In this section we introduce the basic concepts of the lifting procedure applied to the generation of valid inequalities for Y. These concepts are well known for the binary case (see, for instance, Wolsey 1977, Gu et al. 2000) and recently have been extended for the integer case (Atamtürk 2001b). Consider the restriction of this set obtained by setting y j = 0, j N \ M : Y M = y N M 0 : j M a jy j D}. Suppose α j y j α, (2.1) j M is a valid inequality for Y M. We assume that (2.1) is tight for Y M. The lifting problem consists of computing the lifting coefficients α j, for j N \ M in order to obtain a valid inequality α j y j + α j y j α j M j N\M for Y. To compute these coefficients we use the lifting function associated with (2.1), φ M (z) = min α j M α j y j s.t. a j y j D z, j M y j 0, and integer, j M. One way to lift variables is to introduce them one at a time (sequential lifting). Suppose y k, k N \ M is the first variable to be introduced. In order to obtain a valid inequality for Y M k} the lifting coefficient α k must satisfy α k y k φ M (a k y k ) for y k = 1, 2,..., D/a k, thus α k min n=1,..., D/a k φm (a k n) For z ]0, D], we denote by φ I M (z) = min φm (z n) n=1,..., D/z n } the function that gives the upper bound on the lifting coefficients. Notice that if y k is binary then α k φ M (a k ). Formally, Proposition 2.1 If (2.1) is valid for Y M and α k φ I M (a k) then α j y j + α k y k α (2.2) j M is valid for Y M k}. If (2.1) defines a facet of conv(y M ) and α k = φ I M (a k) then (2.2) defines a facet of conv(y M k} ). The lifting function associated with (2.2) is n φ M k} (z) = min n=0,..., D/a k φ M(z + na k ) nα k }. (2.3) }. 2

3 If y k is binary then n = 0, 1. The value of the lifting function may decrease as the set of variables is increased because φ M k} (z) φ M (z) (consider n = 0 in (2.3)). Thus the lifting coefficients may depend on the order that the variables are lifted. However there are cases where the order is not important. This happens when the lifting function has a nice property. Definition 2.2 A function f : A R R is superadditive on A if f(x 1 ) + f(x 2 ) f(x 1 + x 2 ) for all x 1, x 2, x 1 + x 2 A. Theorem 2.3 If φ M is superadditive on [0, D] R then, for all z [0, D] (i) φ M (z) = φ M k} (z) for all k N \ M; (ii) φ M (z) = φ I M (z). In general the lifting function is not superadditive. In that cases if we want to lift simultaneously all variables we may use other functions. Definition 2.4 A function ψ is said a valid lifting function if ψ(z) φ M (z) for all z [0, D]. Proposition 2.5 If ψ is a superadditive valid lifting function then ψ(z) φ I M (z) for all z [0, D]. Thus, if a valid lifting function is supperadditive we can use it to compute the lifting coefficients simultaneously. The following definition will be useful to compare several valid lifting functions. Definition 2.6 A valid lifting function ψ on A is said non-dominated if there is no other valid lifting function χ such that ψ(z) χ(z) φ M (z), for all z [0, D] and χ(z ) > ψ(z ) for some z [0, D]. Next we consider another property of the superadditive valid lifting functions. Definition 2.7 Let E = z [0, D] : φ M S (z) = φ M (z)} for all S N \ M, S. If ψ(z) = φ M (z), for all z E, ψ is said a maximal superadditive valid lifting function. In order to prove the existence of a superadditive valid lifting function Gu et al. (2000) constructed the following function: ζ(z) = min u [0,D z] φ M (z + u) φ M (u)}. And, using this function, they prove the following result. Lemma 2.8 If φ M (z) is integer for all z and if ψ is a superadditive valid lifting function such that ψ(z) > φ M (z) 1 for all z and ψ(z ) = φ M (z ) then z E. 3 Results on polyhedral 2-integer knapsack polyhedra In this section we summarize some results on the polyhedral description of the convex hull of the 2-integer knapsack sets: Y = (y 1, y 2 ) N 2 0 : y 1 + a 2 y 2 D} and Y = (y 1, y 2 ) N 2 0 : y 1 + a 2 y 2 D}. In order to obtain the extreme points of conv(y ), denoted by (a j, b j ), that maximize functions f 1 y 1 + f 2 y 2 with f 2 f 1 a 2 we can use the following algorithm given by Hirschberg & Wong (1976). 3

4 Algorithm HW D Step 0: j 1, (a j, b j ) (, 0), k 1, (c k, d k ) ( a2, 1), l 1, (e l, f l ) ( a2, 1), r 1 Step 1: While a j c k 0 do Set γ(a j, b j ) D a j a 2 b j, R (c k, d k ) c k + a 2 d k, R (e l, f l ) e l a 2 f l (i) if γ(a j, b j ) R (c k, d k ) set j j + 1, (a j, b j ) (a j 1, b j 1 ) + r( c k, d k ); (ii) if γ(a j, b j ) < R (c k, d k ) and R (c k, d k ) R (e l, f l ) set k k + 1, (c k, d k ) (c k 1, d k 1 ) + r(e l, f l ); (iii) if γ(a j, b j ) < R (c k, d k ) and R (c k, d k ) < R (e l, f l ) set l l + 1, (e l, f l ) (e l 1, f l 1 ) + r(c k, d k ). This algorithm is not polynomial and some points (a, b) generated may not be extreme. In order to obtain only the extreme points in polynomial time instead of computing, in Step 1, r = 1 it suffices to compute r as follows: γ(a j, b j ) (i) r = min R (c j, d j ) R (c k, d k ) (ii) r = min R (e l, f l ) (iii) r =,, a j c k } ; R (c k, d k ) γ(a j, b j ) R (e l, f l ) R (c k, d k ) R (e l, f l ). Although we are interested in obtaining the extreme points in polynomial time, the non polynomial Algorithm HW has some properties that we will use later. We denote by k(l) the index of the pair (c, d) used in (iii) of Step 1 to obtain (e l, f l ), this is, (e l, f l ) = (e l 1, f l 1 ) + (c k(l), d k(l) ). Similarly, we denote by l(k) the index of the pair (e, f) used in (ii) of Step 1 to obtain (c, d), and by k(j) the index of the pair (c, d) used in (i) of Step 1 to obtain to obtain (a j, b j ). In this last case l(k(j)) denote the index of the pair (e, f) used to obtain (c k(j), d k(j) ). Let n 1 and n 2 denote the number of distinct pairs (c, d) and (e, f) generated, respectively. Now we summarize some of the properties of these coefficients. Lemma 3.1 (i) e l c k f l d k = 1 if either l = l(k) or k = k(l). (ii) c1 d 1 cn1 d n1 a 2 en2 f n2 e1 f 1. (iii) ck d k, for k = 1,..., n 1 ( el f l, for l = 1,..., n 2) are the best approximations from below (from above) to a 2 for that size denominator. (iv) (a) The set (c 1, d 1 ),..., (c k, d k )} is an integral Hilbert basis for Cone(c 1, d 1 ), (c k, d k )}. (b) The set (e 1, f 1 ),..., (e l, f l )} is an integral Hilbert basis for Cone(e 1, f 1 ), (e l, f l )}. The first three properties are well known since the coefficients (c, d), (e, f) are those obtained by the continuous fraction method to approximate a 2 (see Schrijver 1986). Property (iv) is essentially due to (Weismantel 1996). To obtain the extreme extreme points that maximize functions with f 2 > a 2 it suffices to exchange f 1 with a 2. This algorithm can be easily adapted to generate the extreme points of conv(y ). } ; 4

5 4 Lifting the 2-integer knapsack inequalities In this section we consider the lifting of the 2-integer knapsack facet defining inequalities in order to obtain valid inequalities for Y. First we consider the restriction of Y obtained by setting all but two variables to zero. W.l.o.g. we assume that those two variables are y 1 and y 2. Thus the restricted set is Y M = Y = (y 1, y 2 ) N 2 0 : y 1 + a 2 y 2 D}, where M = 1, 2}. If either /a 2 or a 2 / are integer conv(y M ) has only one non trivial facet and, in that case, we may use the well known superadditive mixed integer rounding function as it is shown in Section 5. Thus we assume that /a 2 and a 2 / are not integer. Consider a valid inequality for Y. In this section we consider α2 y 1 + α 2 y 2 α (4.1) a2. The case α2 > a2 can be reduced to the previous one by exchanging with a 2. Let φ(z), 0 z D, be the lifting function associated to (4.1). For 0 z D, the value of φ(z) is obtained by solving the following knapsack problem, φ(z) = min α y 1 α 2 y 2 s.t. y 1 + a 2 y 2 D z y 1, y 2 N 0. For simplicity of notation we will omit the subscript M when M = 1, 2}. In general φ is not superadditive. Example 4.1 Consider the following set. 21y y 2 + 3y y , y j 0, and integer, j = 1, 2, 3, 4}. The inequality 5y y is valid for the restricted set with y 3 = y 4 = 0. The lifting function is: φ(z) = min 274 5y 1 18y 2 s.t. 21y y z, y 1, y 2 N φ φ I z Figure 1: Non superadditive lifting function of Example 4.1. The graph of this function and the graph of φ I, for 0 z 23, are shown in Figure 1. Considering z 1 = 3 and z 2 = 16 we have φ(3) + φ(16) = 5 > φ(19) = 4 and φ I (3) + φ I (16) = = 5

6 > φ I (19) = 4. Thus, functions φ and φ I are not superadditive. Introducing y 3 before y 4 we obtain 5y y y y 4 274, (5y y 2 + y 3 + 3y 4 274, if y 3, y 4 were binary) and introducing y 3 after y 4 we obtain, 5y 1 +18y y y (5y 1 +18y 2 +0y 3 +4y 4 274, in the binary case). Therefore, the lifting coefficients are sequence dependent. 4.1 A polynomial superadditive valid lifting function As φ, in general, is not superadditive, we will construct superadditive valid lifting functions. following proposition will help us to find such a function. The Proposition 4.2 Consider a function f : I = [0, D] R + 0 (i) f(0) = 0, (ii) f is convex. Then f is superadditive on I. R satisfying The proof is given in the Appendix. Next we construct a function, ψ 1, considering the convex lower envelope of the graph of φ. Given the knapsack coefficients,, a 2, D, let us define γ(a, b) = D a a 2 b and, for the inequality, y 1 + α 2 y 2 α define τ(a, b) = α a α 2 b. If (a, b) belongs to an ordered set E = (A t, B t ), t = 1,..., j}, then we use the notation γ t = γ(a t, B t ) and τ t = τ(a t, B t ). The function ψ 1 is a piecewise linear function whose graph contains the points (0, 0), (γ j, τ j ),..., (γ 1, τ 1 ). Although we are constructing ψ 1 for a general ordered set E, in particular, we are concerned with a subset of the set of the extreme points of conv(y ) (γ j 3, τ j 3 ) (γ j 2, τ j 2 ) (γ j, τ j ) (γ j 1, τ j 1 ) ψ 1 z Figure 2: Function ψ 1 for Example 4.1. Consider the function ψ 1, depending on the set E (set of ordered points),, α 2, α (coefficients of the valid inequality) and, a 2, D (coefficients of the knapsack constraint), defined by 0, 0 z γ j, ψ 1 (z) = τ t + τ t 1 τ t γ t 1 γ (z γ t ), γ t < z γ t 1, t = 2,..., j, t τ 1 + α1 (z γ 1 ), γ 1 < z D, The following propositions are proved in the Appendix. 6

7 Proposition 4.3 Let E = (A t, B t ), t = 1,..., j} be a set of points satisfying 0 A j A 1, B j > > B 1 0, γ t > γ t+1 and τ t > τ t+1 for t = 1,..., j 1. If A1 A 2 B 2 B 1 < < Aj 1 A j B j B j 1 α 2 a 2, then ψ 1 is convex on [0, D]. We are considering inequalities (4.1) defining proper faces of conv(y ) so, containing at least one extreme point of conv(y ). In this case, the set E is the subset of extreme points (, b 1 ),..., (a j, b j )}, of conv(y ) that maximizes functions π 1 y 1 + π 2 y 2 with ck(j) d π 2 α 2 a 2. It can be shown that k(j) π 1 γ t > γ t+1 and τ t > τ t+1 for t = 1,..., j 1. The following result ensures that ψ 1 is a valid supperadditive lifting function on [0, D]. In order to use this result notice that for t = 1,..., j 1, d k(t) y 1 + c k(t) y 2 d k(t) a t + c k(t) b t is a valid inequality for Y and that b 1 = 0. Notice that the first extreme point of conv(y ) is (, b 1 ) = ( D, 0). Proposition 4.4 Let E,, α 2, α,, a 2, D be the parameters satisfying the conditions of Proposition 4.3 and suppose that they satisfy, additionally, the following conditions: (i) B 1 = 0, (ii) for all t = 2,..., j, the inequality α t 1y 1 + α t 2y 2 α t 1A t + α t 2B t, where α t 1 = B t B t 1, α t 2 = A t 1 A t, is valid for Y. Then, the function ψ 1 is a superadditive valid lifting function on [0, D]. As the number of extreme points of conv(y ) is polynomial and, as those points can be generated in polynomial time using the polynomial version of Algorithm HW (see Hirschberg & Wong 1976), it follows that ψ 1 can be constructed in polynomial time. 5 Characterization of the lifting function for a facet defining inequality In order to construct better superadditive valid lifting functions, this means, functions ψ satisfying ψ 1 (z) ψ(z) φ(z) for all z [0, D] we need to characterize the lifting function φ. In this section we impose the hypotheses that (4.1) defines a facet of conv(y ). Therefore, we assume the facet includes two points (a j 1, b j 1 ) and (a j, b j ) generated by the Algorithm HW and so we assume = d k(j), α 2 = c k(j) and α = d k(j) a j + c k(j) b j. For the particular case of the facet defining inequality with coefficients = d 1 = 1, α 2 = c 1 we may consider the well known mixed integer rounding function (see Nemhauser & Wolsey 1988) k, if ka1 < z k + γ j 1, k 0,..., }, ψ(z) = k + z ka1 γj 1 γ j 1, if k + γ j 1 < z (k + 1), k 0,..., }, which is known to be superadditive. Therefore we will exclude this case from now on. Thus we assume c 1 d 1 < α 2 = ck(j) d a 2. k(j) Figure 1 shows that φ is a stepwise function. The length of the first step is γ(a j 1, b j 1 ) = D a j 1 a 2 b j 1 = 2 and the following steps have length either 3 or 5. Initially, the hight of each step is increased by one unit which follows from the properties of the coefficients of the facet defining inequalities (see (i) in Lemma 3.1). 7

8 Γ 2 z = 10 Γ5 Γ z = 23 z = 2 (7, 2) Γ 3 z = 15 ( 11, 3) Γ 4 z = Γ 1 z = 7 Figure 3: Computing function φ given in Example 4.1. z = 0 For simplicity we refer to solutions of φ(z) instead of solutions to the problem associated to φ(z). Figure 3 shows how the optimal solution to φ(z) can be obtained. Notice that starting from (44, 3) (optimal solution to φ(2) = 0) the optimal solutions to the corresponding problems when we increase z can be obtained using two possible moves: (2, 7) and ( 11, 3). This last move is only used when the first one can not be used because the second coordinate becomes negative. So, for a given integer v we aim to find the solution associated to the greatest z such that φ(z) = v. To find this solution only these two moves have to be considered. Notice that after some iterations the value of the first coordinate may become negative and, in that case, we may conclude that for the corresponding v 1,..., α} there is no z such that φ(z) = v. Next we formalize these ideas and establish some results on the properties of function φ. First we characterize the solutions that are obtained using these two moves and prove that these moves are cyclical where each cycle has length. We also prove that in the first cycle all the solutions obtained using these two moves are feasible. Then, in Theorem 5.11 we characterize the function φ. For v = 1,..., α, we define the set Z(v) = 0 z D : φ(z) = v}. If Z(v) then Z(v) has a maximum. Below we show that this set is not empty for v = 1,...,. We denote by Γ v = (p v 1, p v 2) the v th point generated using those two moves. Formally, for v = 0, 1,... α 1, where Γ 0 = (a j 1, b j 1 ). Γ v+1 = (p v+1 1, p v+1 2 ) = (p v 1, p v 2) + (c k(j) 1, d k(j) 1 ) if p v 2 d k(j) 1 0, (p v 1, p v 2) (e l(k(j)), f l(k(j)) ) if p v 2 d k(j) 1 < 0, Example 1 (Cont.). See Figure 3. For = 21, a 2 = 76, we have (c 1, d 1 ) = (3, 1), (c 2, d 2 ) = (7, 2), (c 3, d 3 ) = (18, 5) and (e 1, f 1 ) = (4, 1), (e 2, f 2 ) = (11, 3). Thus = d 3 = 5, α 2 = c 3 = 18. As (a j 1, b j 1 ) = (44, 3), (a j, b j ) = (26, 8), then Γ 0 = (44, 3); p 0 2 = 3 d 2 = 2 Γ 1 = (44, 3) + (7, 2) = (51, 1); p 1 2 < d 2 Γ 2 = (51, 1) + ( 11, 3) = (40, 4); p 2 2 > d 2 Γ 3 = (47, 2); p 3 2 > d 2 Γ 4 = (54, 0); p 4 2 < d 2 Γ 5 = (43, 3). Lemma 5.1 τ(γ v ) = v, v 0,..., α}. 8

9 Proof: We prove this result by induction. As the point (a j 1, b j 1 ) belongs to the facet defined by the valid inequality (4.1), we have τ(γ 0 ) = 0. Suppose that τ(γ v ) = v. Using Lemma 3.1, d k(j) c k(j) 1 + c k(j) d k(j) 1 = 1. Thus, if p v 2 d k(j) 1 0, we have τ(γ v+1 ) = α p v+1 1 α 2 p v+1 2 = α p v 1 α 2 p v 2 c k(j) 1 + α 2 d k(j) 1 = τ(γ v ) d k(j) c k(j) 1 + c k(j) d k(j) 1 = v + 1. Similarly, since d k(j) e l(k(j)) c k(j) f l(k(j)) = 1, we conclude that τ(γ v+1 ) = v + 1 if p v 2 d k(j) 1 < 0. Lemma 5.2 p v 2 < = d k(j), v 0,..., α}. Proof: Notice that d k(j) = f l(k(j)) + d k(j) 1. We use induction. For v = 0 we have p 0 2 = b j 1 < d k(j), otherwise (a j 1, b j 1 ) would not be an extreme point. Suppose that p v 2 < f l(k(j)) + d k(j) 1. If p v 2 < d k(j) 1 then p v+1 2 = p v 2 + f l(k(j)) < d k(j) 1 + f l(k(j)). If d k(j) 1 p v 2 < d k(j) 1 + f l(k(j)) then 0 p v+1 2 = p v 2 d k(j) 1 < d k(j) 1 + f l(k(j)). Now we introduce parameters that indicate the move to take in each iteration. For v = 1,..., α define: 1 if Γ v = Γ v 1 + (c k(j) 1, d k(j) 1 ), δ v = 0 if Γ v = Γ v 1 (e l(j) 1, f l(k(j)) ). Let us denote q p = q j=p δ j, for q p and q p = 0, otherwise. q p indicates the number of moves using the direction (c k(j) 1, d k(j) 1 ) to go from Γ p 1 to Γ q. The following results are useful to compute the values of these parameters. b Lemma 5.3 v j 1 + v f l(k(j)) 1 =, v 1... α}. Proof: Let n = v 1. We know that p v 2 = b j 1 nd k(j) 1 +(v n)f l(k(j)) = b j 1 n(d k(j) 1 +f l(k(j)) )+ vf l(k(j)) = b j 1 nd k(j) + vf l(k(j)). As p v 2 0 and, from Lemma 5.2, p v 2 < d k(j) we have, Thus, n = b j 1 + v f l(k(j)) 1 < n bj 1 + v f l(k(j)). d k(j) d k(j) b j 1 + v f l(k(j)). d k(j) Next we stat that the sequence of parameters δ v is cyclical. Corollary 5.4 Let v = t +r with r 1,..., } and t 1,..., α/ }. Then v 1 = r 1+tf l(k(j)). Proof: Considering in Lemma 5.3 v = t +r we have v 1 = bj 1 +rf l(k(j)) +tf l(k(j)) = r 1 +tf l(k(j)). Using Corollary 5.4 to compute the points Γ v we obtain the following result. Proposition 5.5 Let v = t + r with r 1,..., } and t 1,..., α/ }. Then, Γ v = Γ r + t( 1, 0) 9

10 Proof: Γ v can be written as Γ v = Γ 0 + = Γ 0 + v i=t+1 v δ i (d k(j) 1, c k(j) 1 ) i=1 δ i (c k(j) 1, d k(j) 1 ) v i=t+1 v (1 δ i )(e l(k(j)), f l(k(j)) ) i=1 (1 δ i )(e l(k(j)), f l(k(j)) ) t t + δ i (c k(j) 1, d k(j) 1 ) (1 δ i )(e l(k(j)), f l(k(j)) ). i=1 i=1 Corollary 5.4 implies (i) δ tα1+k = δ k for k 1,..., } and (ii) t i=1 (1 δ i) = t tα1 1 = t( f l(k(j)) ) = td k(j) 1. Using (i), (ii) and part (i) of Lemma 3.1, it follows that, Γ v = Γ r + tf l(k(j)) (c k(j) 1, d k(j) 1 ) td k(j) 1 (e l(k(j)), f l(k(j)) ) = Γ r + t( 1, 0). Next we present a technical result we will use later. The proof can be found in the Appendix. Lemma 5.6 s 1 r+s r+1 1 for all s, r 1,..., α}, such that s + r 1,..., α}. It is important to notice that there is no guarantee that p v 1 0. Therefore Γ v may not be a feasible solution to φ(z). Next consider, for each z [0, D], the following relaxation of the problem associated to the lifting function where y 1 may assume negative values: φ(z) = minα y 1 α 2 y 2 : y 1 + a 2 y 2 D z, y 1 Z, y 2 N 0 }. Let z v = max0 z D : φ(z) = v}, for v = 0,..., α. Notice that φ(γ v ) = v so, there always exists a z [0, D] such that φ(z) = v. For v = 0 we have z 0 = γ(a j 1, b j 1 ) = γ j 1. As φ is a non decreasing function, z v+1 z v for all v = 0,..., α 1. Next we explain how to obtain the optimal solution to φ(z v ) for all v = 1,..., α. Notice that if φ(z v ) has more than one optimal solution then α 2 = ck(j) d = a 2. k(j) Proposition 5.7 Γ v = (p v 1, p v 2) is an optimal solution to φ(z v ) for v = 1,..., α. Proof: We prove by induction. Γ 0 = (a j 1, b j 1 ) is an optimal solution to φ(z 0 ). Suppose Γ v = (p v 1, p v 2) is an optimal solution to φ(z v ) and that (p 1, p 2 ) is an optimal solution to φ(z v+1 ). We are assuming c 1 d 1 < α 2 = ck(j) d a 2. Next we consider two cases: (i) p k(j) 1 p v 1 and p 2 p v a 2; (ii) p 1 p v 1 and p 2 p v 2. 1 The proof that it can not occur p 1 > p v 1, p 2 > p v 2 and p 1 < p v 1, p 2 < p v 2, is trivial. Consider case (i). Let (a, b) = (p 1 p v 1, p v 2 p 2 ). Now we prove that (a, b) Cone(a 2, ), ( a2, 1)}. Notice that (c 1, d 1 ) = ( a2, 1). Let (a, b) = λ 1 (a 2, ) + λ 2 (c 1, d 1 ). We must show that λ 1, λ 2 0. The case λ 1, λ 2 < 0 can not occur because a 0, b 0. If λ 1 0 and λ 2 < 0 we obtain z v+1 = D p 1 a 2 p 2 = D (p v 1 + a) a 2 (p v 2 b) = D p v 1 a 2 p v 2 (λ 1 a 2 + λ 2 c 1 ) + a 2 (λ 1 + λ 2 d 1 ) = z v +λ 1 ( a 2 +a 2 )+λ 2 ( c 1 +a 2 d 1 ) = z v +λ 2 ( c 1 +a 2 d 1 ) < z v because, from (ii) of Lemma 3.1, c 1 + a 2 d 1 > 0. This is absurd since z v+1 z v. Consider λ 1 < 0 and λ 2 0. First assume b > 0. Notice that inequalities λ 1 < 0, λ 2 0 and a 2 > c1 d 1 imply a b = λ 1a 2 + λ 2 c 1 λ 1 + λ 2 d 1 < c1 d 1. Thus, since a b < c1 d 1 < α 2 10 we have α 2 b > c1 d 1 b > a and

11 as d 1 = 1 and all the coefficients are non negative integers, it follows that α 2 b c 1 b + 1 a + 2. Therefore v + 1 = α p 1 α 2 p 2 = α (p v 1 + a) α 2 (p v 2 b) = α p v 1 α 2 p v 2 a + α 2 b v + 2 which is absurd. If b = 0 then v + 1 = α p 1 α 2 p 2 = α (p v 1 + a) α 2 (p v 2 b) = v a < v. From Lemma 3.1, part (iv), we know that (a, b) can be obtained as non negative integer combination of the vectors in G 1 = (c 1, d 1 ),..., (c n1, d n1 )}. Thus, (a, b) = n 1 k=1 ρ k(c k, d k ) where ρ k 0 and integer for all k = 1,..., n 1. Since v + 1 = α p 1 α 2 p 2 = α p v 1 α 2 p v 2 a + α 2 b = v a + α 2 b a + α 2 b = 1. Then, 1 = a + α 2 b = n 1 k=1 ρ ku k where u k = c k + α 2 d k. Using Lemma 3.1 it can be shown that: u k < 0 for k > k(j), u k = 0 for k = k(j), u k = 1 for k = k(j) 1, u k > 1 for k < k(j) 1 (notice that = d k(j), α 2 = c k(j) ). From Lemma 5.2 we have, for all v 1,..., α}, p v 2 < d k(j). Thus ρ k = 0 for k > k(j) 1, which implies ρ k = 1 for k = k(j) 1 and ρ k = 0 for k < k(j) 1. Thus (a, b) = (c k(j) 1, d k(j) 1 ). The proof of case (ii) is similar. Consider (a, b) = (p v 1 p 1, p 2 p v 2). Again, proving that (a, b) belongs to Cone( a2, 1), (a 2, )} then, from Lemma 3.1, (a, b) can be written as a non negative linear combination of the vectors in G 2 = (e 1, f 1 ),..., (e n2, f n2 )}, this is, (a, b) = n 2 l=1 ρ l(e l, f l ) where ρ l 0 and integer, for all l = 1,..., n 2. Similarly we conclude that 1 = a α 2 b = n 2 l=1 ρ lν l where ν l = e l α 2 f l. Since α2 a2 el f l, we have ν l > 0 for all l and, in particular, ν l = 1 for l = l(k(j)),..., l(k(j) + 1) and ν l > 1 for all other values of l. Thus, ρ l = 1 where l l(k(j)),..., l(k(j) + 1)} and ρ l = 0 for l l. We must prove that l = l(k(j)). Consider the solution Γ v (e l(k(j)), f l(k(j)) ). Let z = D (p v 1 e l(k(j)) ) a 2 (p v 2 + f l(k(j)) ). As z > z v and as τ(γ v (e l(k(j)), f l(k(j)) )) = v + 1 it follows that φ(z ) = v + 1. By definition of z v+1 we have z v+1 z. Suppose ρ l = 1 for l l(k(j)) + 1,..., l(k(j) + 1)}. Thus (a, b) = (e l, f l ). Noticing that, for l > l(k(j)), e l a 2 f l < e l(k(j)) a 2 f l(k(j)), it follows that z v+1 = D p 1 a 2 p 2 = D p v 1 a 2 p v 2 + e l a 2 f l = z v + e l a 2 f l < z v + e l(k(j)) a 2 f l(k(j)) = z z v+1, which is absurd. Thus ρ l = 0 for l = l(k(j)) + 1,..., l(k(j) + 1) and ρ l(k(j)) = 1. If p v 2 d k(j) 1 < 0 then Γ v ( c k(j) 1, d k(j) 1 ) is not feasible. In this case (i) can not occur because, as we saw in the proof of (i), Γ v ( c k(j) 1, d k(j) 1 ) is the unique point, (p 1, p 2 ), satisfying p 1 p v 1, p 2 p v 2 and τ(p 1, p 2 ) = v + 1. If p v 2 d k(j) 1 0 then τ(γ v ( c k(j) 1, d k(j) 1 )) = τ(γ v (e l(k(j)), f l(k(j)) )) = v + 1. Noticing that c k(j) 1 +a 2 d k(j) 1 > e l(k(j)) a 2 f l(k(j)) we can prove, as we did above to prove l = l(k(j)), that z v+1 is obtained at the point Γ v ( c k(j) 1, d k(j) 1 ). Thus Γ v+1 = Γ v +(c k(j) 1, d k(j) 1 ) is optimal to φ(z v+1 ). Starting at Γ 0 = (a j 1, b j 1 ), we may compute the optimal solution to φ(z v ) for all v = 1,..., α, iteratively. As the set of feasible solutions to φ(z) is a subset of the set of feasible solution to φ(z) we have the following consequence. Corollary 5.8 If p v 1 0 then Γ v is the optimal solution to φ(z) for all z v 1 < z z v and, in this case φ(z) = v. Lemma 5.9 Z(v) if and only if Γ v is a feasible solution to φ(z), for all z v 1 < z z v. 11

12 Proof: If Γ v is a feasible solution to φ(z), then Z(v) because γ(γ v ) = γ(p v 1, p v 2) belongs to Z(v). To prove the implication in the other direction suppose that Γ v is not a feasible solution and that Z(v). Then there exists z Z(v) such that φ(z ) = v. Let (a, b) be an optimal solution to φ(z ). As v = τ(γ v ) = τ(a, b) then α p v 1 α 2 p v 2 = α a α 2 b (a p v 1) = α 2 (p v 2 b) a pv 1 p v 2 b = α 2 a 2. Notice that a 0 > p v a 1, thus b < p v 2. The case α 2 = a 2 can not occur because as α 2 1 is the best approximation to a 2 for that size denominator, we would have p v 2 b d k(j), contra- a p v 1 dicting Lemma 5.2. So, p v 2 b < a 2 a p v 1 < a 2 p v 2 a 2 b p v 1 a 2 p v 2 < a a 2 b D p v 1 a 2 p v 2 < D a a 2 b z v = γ(p v 1, p v 2) < γ(a, b). Setting z = γ(a, b) > z v we have φ(z) = φ(z) = v, obtained at (a, b), which contradicts the definition of z v. It remains to determine the value of φ(z), z v 1 < z z v, when Z(v) =. Proposition 5.10 Let z r < z z r+1 with r 1,..., α 1}. Then φ(z) = v where v = minj r + 1,..., α} : p j 1 0}. Proof: Notice that φ(z) is integer for all z. For z > z r we have φ(z) > r. From Lemma 5.9, Z(k) = for all k = r + 1,..., v 1, i.e., there is no z z v 1 such that φ(z) = k, for k = r + 1,..., v 1, which implies φ(z) v, for z > z r. On the other hand, as Γ v is a feasible solution to φ(z) with z r < z z r+1, then φ(z) v. So φ(z) = v. Finally, from this discussion we can characterize function φ. Theorem 5.11 Let v k denote the k th value in 1,..., α} such that Z(v k ) and define v 0 = 0. Thus 0, if 0 z z 0, φ(z) = v k+1, if z vk < z z vk+1, k 0,..., n }, where n denotes the number of values v 1,..., α} for which Z(v). Lemma 5.12 Z(v) for all v = 0,...,. Proof: Suppose that there exists v 1,..., } such that Z(v) = and Z(v 1). Notice that Z(0). Thus p2 v 1 d k(j) 1 < 0 and p1 v 1 e l(k(j)) < 0 p1 v 1 e l(k(j)) 1. Since b j = b j 1 + rd k(j) for some positive integer r, then b j d k(j) = d k(j) 1 + f l(k(j)). So, as p2 v 1 d k(j) 1 < 0 we have p2 v 1 < b j f l(k(j)). Hence φ(z v 1 ) = α p1 v 1 α 2 p2 v 1 = (a j p1 v 1 )+α 2 (b j p2 v 1 ) > (a j e l(k(j)) +1)+α 2 (b j b j +f l(k(j)) ) (0 e l(k(j)) +1)+α 2 f l(k(j)) = d k(j) d k(j) e l(k(j)) + c k(j) f l(k(j)) = d k(j) 1. Therefore, v 1 = φ(z v 1 ) > 1 v >. Lemma 5.12 states that for all v = 0,...,, Γ v is a feasible solution and therefore the optimal solution to φ(z). Let w 1 = d k(j) 1 a 2 c k(j) 1 and w 2 = f l(k(j)) a 2 + e l(k(j)). Thus, w 1 is the length of the steps corresponding to moves based in the vector (c k(j) 1, d k(j) 1 ) and w 2 is the length of the steps corresponding to moves based in the vector ( e l(k(j)), f l(k(j)) ). Using the notation of Algorithm HW we have w 1 = R (c k(j) 1, d k(j) 1 ) and w 2 = R (e l(k(j)), f l(k(j)) ). Hence w 1 w 2. Notice that w 1 w 2 = (c k(j) 1 + e l(k(j)) ) + a 2 (d k(j) 1 + f l(k(j)) )) = c k(j) + a 2 d k(j) = R (c k(j), d k(j) ). As γ j 1 12

13 γ j c k(j) + a 2 d k(j) (observe that (a j, b j ) = (a j 1, b j 1 ) + r( c k(j), d k(j) ) for some integer r > 0) and as γ j 0 we have γ j 1 w 1 w 2. Next we state these conclusions as a result. Lemma 5.13 (i) w 1 w 2 0; (ii) γ j 1 w 1 w 2. As a corollary of Proposition 5.5 and noticing that w 2 + f l(k(j)) (w 1 w 2 ) = ( f l(k(j)) )w 2 + f l(k(j)) w 1 = d k(j) 1 w 2 + f l(k(j)) w 1 = (last equality follows from part (i) of Lemma 3.1) we obtain the following result. Corollary 5.14 Let v = t + r with r 0,..., 1} and v 1,..., α}. Then z v = γ j 1 + vw 2 + (w 1 w 2 ) v 1 = γ j 1 + (t + r)w 2 + (tf l(k(j)) + r 1)(w 1 w 2 ) = γ j 1 + t + rw 2 + r 1(w 1 w 2 ). It can be verified that z vn = z α = D. Remark 5.15 Although we are considering z [0, D], it is important for the study in Section 7, to describe φ on ], 0[. In this case, for v N, we can obtain an optimal solution, Γ v = (p v 1, p v 2 ), for φ(z v ) iteratively, Γ v = Γ v+1 (e l(k(j)), f l(k(j)) ) if p2 v+1 f l(k(j)) 0, + ( c k(j) 1, d k(j) 1 ) if p2 v+1 f l(k(j)) < 0, where Γ 0 = (a j 1, b j 1 ). Again we define, for v N, 1 if Γ v = Γ v+1 + (e l(k(j)), f l(k(j)) ), δ v = 0 if Γ v = Γ v+1 + ( c k(j) 1, d k(j) 1 ). For v N and for z v 1 < z z v we have φ(z) = v, where z v = γ j 1 vw 2 (w 1 w 2 ) v i=1 δ i and Γ v is an optimal solution to φ(z). This characterization, for z < 0, can be proven as we proved the corresponding properties for φ, with z 0. Notice that for all v N, z 0 : φ(z) = v}. 6 Superadditive valid lifting functions for facet defining inequalities In this section we construct new superadditive valid lifting functions, depending on the parameters γ j 1, w 1, w 2, δ 1,..., δ α, used to characterize the lifting function in the previous section. We start by considering functions with a similar analytical expression for intervals with the same length. To do that consider the family C of functions on [0, D] such that χ C if and only if χ is a superadditive valid lifting function and can be written as: 0, if 0 z z 0, χ(z) = v + χ 1 (z z v ), if z v < z z v+1 and δ v+1 = 1, v = 0,..., α 1, v + χ 2 (z z v ), if z v < z z v+1 and δ v+1 = 0, v = 0,..., α 1, r ki where, for i 1, 2}, χ i (r) =, if r k i, and 0 < k 1 < w 1, 0 < k 2 < w 2. This family will 1, if r > k i, also be very useful since it can easily be extended for the knapsack inequalities with one continuous variable as we will see in Section 7. 13

14 Consider the function ψ 2 that belongs to C with k 1 = w 1 and k 2 = w 2, defined by, 0, if 0 z z0, ψ 2 (z) = v + z zv z v+1 z v, if z v < z z v+1, v = 0,..., α 1. An example of the graph of such function is shown in Figure ψ 2 φ z 0 = 2 z 1 = 7 z 2 = 10 z 3 = 15 z 4 = 20 z 5 = 23 z Figure 4: Example of ψ 2. Proposition 6.1 Function ψ 2 is a valid supperadditive valid function on [0, D]. We omit the proof because below we construct a new function, ψ 3, dominating ψ 2 and the proof of superadditivity of ψ 2 is similar to the proof of superadditivity of ψ 3. Lemma 6.2 ψ 1 (z) ψ 2 (z), z [0, D]. Proof: As ψ 1 is convex on [0, D] then its epigraph, Epi(ψ 1 ) = (y, z) R [0, D] : y ψ 1 (z)}, is a convex set. Proposition 4.4 ensures (z, φ(z)) Epi(ψ 1 ) for all z [0, D]. For all z [z 0, D], the point (z, ψ 2 (z)) can be written as linear convex combination of two points: (z v, φ(z v )) and (z v+1, φ(z v+1 )) in Epi(ψ 1 ), where z v z z v+1 (for z z 0 we have ψ 2 (z) = φ(z)). So (z, ψ 2 (z)) Epi(ψ 1 ) which implies ψ 1 (z) ψ 2 (z), z [0, D]. Although ψ 1 is dominated by ψ 2, it is important to notice that ψ 1 (z) can be computed in a polynomial number of steps while ψ 2 (z) may require a non polynomial number of elementary operations because we need to compute δ 1,..., δ α1. Notice also that the function ψ 2 requires more restricted conditions than ψ 1, namely, it requires that (4.1) defines a facet of conv(y ). Next we construct a better superadditive valid lifting function. Consider ψ 3 (see Figure 5) defined by 0, if 0 z z 0, v + z zv µ, if z v < z z v + µ and δ v+1 = 0, v = 0,..., α 1, ψ 3 (z) = v + 1, if z v + µ < z z v+1 and δ v+1 = 0, v = 0,..., α 1, v + z zv λ, if z v < z z v + λ and δ v+1 = 1, v = 0,..., α 1, v + 1, if z v + λ < z z v+1 and δ v+1 = 1, v = 0,..., α 1, where µ = w 1 γ j 1 and λ = w 1 + (w 1 w 2 ) γ j 1. Notice that as γ j 1 w 1 w 2 we have µ w 2 and λ w 1. Therefore, when γ j 1 > w 1 w 2, ψ 3 dominates ψ 2. Proposition 6.3 The function ψ 3 is a superadditive valid lifting function. 14

15 ψ 3 φ knapsack constraint: 21y y valid inequality: 3y y z 0 = 6 z 1 = 14 z 2 = 19 z 3 = 27 z 4 = 32 z 5 = 40 z 6 = 48 z 7 = 53 z 8 = 61 z 9 = 69 Figure 5: Example of ψ 3 with µ = 2 and λ = 5. The proof is left to the Appendix. In order to find better superadditive valid lifting functions we must consider more restrictive conditions on the sequence of the values of the parameters δ. Next we construct several valid lifting functions satisfying ψ 3 (z) ψ 4 (z) ψ 5 (z) ψ 6 (z) φ(z), z [0, D], and indicate sufficient conditions on the parameters δ under which superadditivity holds. In that case these functions belong to C. Proposition 6.4 If, for each n and m one of the following conditions hold: (i) m 1 m 1 < n+1 n+m and δ n+m+1 = 0, then the function 0, if 0 z z 0, ψ 4 (z) = v + z zv µ, if z v < z z v + µ, v = 0,..., α 1, v + 1, if z v + µ < z z v+1, v = 0,..., α 1, n+m n+1 ; (ii) where µ = w 1 γ j 1, is a superadditive valid lifting function. The proof is similar to the proof of case δ n+1 = δ m+1 = 0 in Proposition 6.3. Noticing that the sequence of the parameters is cyclical it is only necessary to check conditions (i) and (ii) for n = 1,..., 1 and m = 1,..., ψ 4 φ z 0 = 2 z 1 = 7 z 2 = 10 z 3 = 15 z 4 = 20 z 5 = 23 z Figure 6: Example of ψ 4 with µ = 3. If more restrictive conditions on the sequence of the values of the parameters δ are verified we can use a better superadditive valid lifting function. 15

16 Proposition 6.5 If, for each n and m, one of the following conditions hold: (i) m 1 > n+m n+1 ; (ii) m 1 = n+1 n+m, δ n+1 = δ m+1 = δ n+m+1 = 1; (iii) m 1 = n+1 n+m, δ n+m+1 = 0, then 0, if 0 z z 0, v + z zv θ, if z v < z z v + θ and δ v+1 = 0, v = 0,..., α 1, ψ 5 (z) = v + 1, if z v + θ < z z v+1 and δ v+1 = 0, v = 0,..., α 1, v + z zv µ, if z v < z z v + µ and δ v+1 = 1, v = 0,..., α 1, v + 1, if z v + µ < z z v+1 and δ v+1 = 1, v = 0,..., α 1, where µ = w 1 γ j 1 and θ = w 2 γ j 1, is a superadditive valid lifting function. The proof is similar to the proof of Proposition 6.3. However, it is important to notice that if γ j 1 w 2, then it can be proven that there are n and m such that m 1 < n+1 n+m and hence the hypothesis of Proposition 6.5 do not hold. Under further more restrict conditions superadditivity of ψ 6 holds. Proposition 6.6 If, for each n and m one of the following conditions hold: (i) m 1 m 1 = n+1 n+m and δ n+m+1 = 0, then 0, if 0 z z 0, ψ 6 (z) = v + z zv θ, if z v < z z v + θ, v = 0,..., α 1, v + 1, if z v + θ < z z v+1, v = 0,..., α 1, > n+m n+1 ; (ii) where θ = w 2 γ j 1, is a superadditive valid lifting function. The proof is similar to the proof of case δ n+1 = δ m+1 = 0 in Proposition 6.3, so it will be omitted. Consider v = maxv 1,..., α} : φ(z) = φ(z) for all 0 z z v }. Next we analise some properties of these valid lifting functions on [0, z v ]. On [z v, D], φ may increase faster in some intervals since solutions Γ v with v > z v may be unfeasible. Notice that ψ 3 is a superadditive function while superadditivity of ψ 4, ψ 5, ψ 6, requires more restrict conditions. All these four functions, when superadditivity holds, belong to C. Next we prove that for each specific sequence of parameters δ one of these functions is the best function in C, this means, there is no function in C dominating the corresponding function. Proposition 6.7 Consider the conditions for superadditivity of ψ 6, ψ 5, ψ 4, respectively: (C6) For each n and m one of the following conditions hold: (i) m 1 δ n+m+1 = 0. (C5) For each n and m one of the following conditions hold: (i) m 1 δ n+m+1 = 0; (iii) m 1 = n+1 n+m and δ n+1 = δ m+1 = δ n+m+1 = 1. (C4) For each n and m one of the following conditions hold: (i) m 1 δ n+m+1 = 0. > n+m n+1 ; (ii) m 1 = n+m n+1 and > n+m n+1 ; (ii) m 1 = n+m n+1 and n+m n+1 ; (ii) m 1 < n+m n+1 and Let χ C be a valid lifting function for φ. (a) If (C6) hold on [0, z v ], then χ(z) ψ 6 (z) for all z [0, z v ]. (b) If (C5) hold and (C6) do not hold on [0, z v ], then χ(z) ψ 5 (z) for all z [0, z v ]. (c) If (C4) hold and (C5) do not hold on [0, z v ], then χ(z) ψ 4 (z) for all z [0, z v ]. (d) If (C4) do not hold [0, z v ], then χ(z) ψ 3 (z) for all z [0, z v ]. 16

17 Proof: Consider case (a). Suppose that for t = z n + r and r < θ, χ(t) > ψ 6 (t) (for r θ and n v we have ψ 6 (t) = φ(t)). Let t = γ j 1 + θ r. Thus χ(t + t ) = χ(z n + r + γ j 1 + θ r) = χ(z n + w 2 ) φ(z n+1 ) = n + 1. On the other hand, χ(t + t ) χ(t) + χ(t ) > ψ 6 (t) + ψ 6 (t ) = n + r/θ + (θ r)/θ = n + 1 which is absurd. Now we prove (b). First we prove that for all n and m with δ n+1 = δ m+1 = 0 it can not always occur m 1 > n+1 n+m and δ n+m+1 = 0. If δ 1 = 0 consider n = m = 0 which implies m 1 = n+1 n+m = 0. If δ 1 = 1 consider m + 1 = mink 1,..., } : δ k = 0} and n + 1 = maxk 1,..., } : δ k = 0} thus, m 1 > n+1 n+m and δ n+m+1 = 1. Therefore, as there must exist n and m with δ n+1 = δ m+1 = 0 such that either m 1 = n+1 n+m or m 1 > n+1 n+m and δ n+m+1 = 1 we can prove, as we did in case case (a) considering t = z m + θ r, that for all n such that δ n+1 = 0, χ(z) ψ 5 (z), z [z n, z n+1 ]. Now suppose that there is a r, 0 < r < µ, such that for all v with δ v+1 = 1 and for all z = z v + r we have χ(z) = χ 1 (z) > ψ 5 (z). Case (b) implies that there are n and m with n + m + 1 v such that m 1 = n+1 n+m and δ n+1 = δ m+1 = δ n+m+1 = 1. Consider t = z n + r and t = z m + µ r. Thus, χ(t + t ) = χ(z n + z m + r + µ r) = χ(z n+m + γ j 1 + µ) = χ(z n+m+1 ) φ(z n+m+1 ) = n + m + 1 and χ(t + t ) χ(t) + χ(t ) > ψ 5 (t) + ψ 5 (t ) = n + r/µ + m + (µ r)/µ) = n + m + 1, which is absurd. Proof of (c). This case implies that there are n and m such that one of the following conditions hold (see Lemma 5.6): (1) m 1 = n+1 n+m 1 and δ n+m+1 = 0. (2) m 1 = n+1 n+m and δ m+1 = 0, δ n+1 = 1, δ n+m+1 = 1. (3) m 1 = n+1 n+m and δ m+1 = 1, δ n+1 = 0, δ n+m+1 = 1. Notice that it can not happen m 1 = n+1 n+m and δ n+1 = 0, δ m+1 = 0, δ n+m+1 = 1 because it would imply that m+1 1 = n+2 n+m+1 2 which contradicts Lemma 5.6. Also notice that (1) imply (2). In order to (1) occur it is necessary that there is m, m < m, such that m 1 = m n+1 and δ m +1 = 0, δ n+m +1 = 1. Since it can not occur 1 m = n+m n+1 and δ n+1 = 0, δ m +1 = 0, δ n+m +1 = 1 we also have δ n+1 = 0. Suppose that there is r, 0 < r < µ, such that for all v with δ v+1 = δ, δ 0, 1} and for all z = z v + r, we have χ(z) > ψ 4 (z). Consider case (2) (the case (3) is similar) and suppose χ(t) > ψ 4 (t) for t = z n + r. Let t = z m + µ r. Thus χ(t + t ) = χ(z n+m + γ j 1 + µ) = χ(z n+m+1 ) = n + m + 1 and χ(t + t ) χ(t) + χ(t ) > ψ 4 (t) + ψ 4 (t ) = n + m + 1. Finally consider case (d). This case implies that there is n and m such m 1 = n+1 n+m 1 and δ n+m+1 = 1. And these conditions can only occur (see proof of Proposition 6.3) with δ n+1 = δ m+1 = 1. Therefore, if χ 1 (t) > ψ 3 (t) for t = z n + r and δ n+1 = 1 then we obtain again the contradiction χ(t + t ) > n + m + 1 = χ(t + t ) with t = z m + r and r = λ r. Now we consider χ 2. Notice that if there is n and m such that m 1 = n+1 n+m 1 and δ n+1 = δ m+1 = δ n+m+1 = 1 then there must exist n and m such that m 1 = n+m n+1 and δ n+1 = 1, δ m +1 = 0, δ n+m +1 = δ n+m +2 = 1. In this case if χ 2 (z) > ψ 3 (z) for z = z v + r with δ v = 0 and r < µ, then χ 2 (t) > ψ 3 (t) for t = z m + k 2, δ m +1 = 0 and k 2 < µ. Considering t = z n + λ we have, χ(t+t ) = χ(z n+m +γ j 1 +λ+k 2 ) = χ(z n+m +w 1 +(w 1 w 2 )+k 2 ) = χ 1 (z n+m +1 +(w 1 w 2 )+k 2 ) = n + m (w1 w2)+k2 λ superadditive. < n + m (w1 w2)+µ λ = n + m + 2 = χ 1 (t ) + χ 2 (t). Thus, χ is not Notice that we proved stronger results than the results stated in Proposition 6.7. In fact we proved that ψ 6 can not be dominated by other superadditive valid lifting function. Only for the case of ψ 3 with 17

18 z [z n, z n+1 ] and δ n+1 = 0 we used explicitly the expression of χ 2. In all other cases the proofs are valid considering χ 1 and χ 2 as generic functions. However, if there is n and m with δ n+1 = δ m+1 = 0 and m+1 1 < n+1 n+m + δ n+m+1 then the expression of χ 2 is not important to prove that ψ 3 can not be dominated by other valid lifting function in C. Otherwise, if for all n and m with δ n+1 = δ m+1 = 0 we have m+1 1 n+m ψ 3 (z) = v + 1 µ λ + z zv λ n+1 + δ n+m+1 then we can improve ψ 3 for z [z v, z v+1 ] with δ v+1 = 0, by setting, if z v z < z v + µ and δ v+1 = 0, v = 0,..., α 1. The function with this modification is not in C and can not be useful for Section 7. However, if conditions (C4) do not hold it can be proven, as we did above, that this function dominates all other functions χ C where χ 1 and χ 2 are generic functions provided that χ is a superadditive valid lifting function. Next we analyze maximality of the superadditive valid lifting functions given above. Lemma 2.8 implies that for all z [0, z v ] for which the value of the functions ψ i, i 2, 3, 4, 5, 6}, coincides with the value of φ is in the maximality set E. Remember that if φ M S (z) = φ M (z), for all S, S N \M, then z E. Proposition 6.8 [t, t + γ j 1 ] E for t = 1,..., D. Proof: Consider S such that S N \ M. As φ M S (z) φ(z) for all z, we must show that the strict inequality can not occur for z [t, t + γ j 1 ]. Suppose that there exists z [t, t + γ j 1 ] such that φ M S (z) < φ(z) = t. Let y N M S be the optimal solution to φ M S (z). Then φ M S (z) = α j M S α jyj < t = φ(z) and M S a jyj D z D t. Setting y 1 = y1 +t and y j = y j for j S (M \1}) then j M S a jy j D and α j M S α jy j = α M S α jyj t < 0, i.e. φ M S (0) < 0 which is absurd because it implies that M S α jy j α is not valid for Y M S. Noticing that ψ i (k ), for i 2, 3, 4, 5, 6}, may not assume the value φ(k ) = k we will improve these functions. Lemma 6.9 Let ψ be a valid superadditive lifting function such that ψ(k + r) = k + ψ(r) for all k = 0,..., D/ and γ j 1 r <, then the function ψ defined by ψ k, if z A, (z) = ψ(z), otherwise, where A = k=0,..., D/ [k, k + γ j 1 ], is a superadditive valid lifting function. Proof: From the definition of ψ and from Proposition 6.8 it follows that φ(z) ψ (z) ψ(z) for all z [0, D]. Next we prove superadditivity of ψ. Case t 1 A, t 2 A. Suppose t 1 = k 1 + r 1, t 2 = k 2 + r 2, with 0 r 1 γ j 1, 0 r 2 γ j 1. Then ψ (t 1 + t 2 ) ψ ((k 1 + k 2 ) ) = (k 1 + k 2 ) = ψ (t 1 ) + ψ (t 2 ). Case t 1 A, t 2 A (similarly for t 1 A, t 2 A). Let t 1 = k 1 + r 1, t 2 = k 2 + r 2, with 0 r 1 γ j 1 and γ j 1 < r 2 <. If r 1 + r 2 < then t 1 + t 2 A and r 1 + r 2 A which implies ψ (t 1 +t 2 ) = ψ(t 1 +t 2 ) = (k 1 +k 2 ) +ψ(r 1 +r 2 ) k 1 +ψ(r 1 )+k 2 +ψ(r 2 ) = k 1 +0+k 2 +ψ(r 2 ) = ψ (t 1 )+ψ (t 2 ). If r 1 +r 2 then t 1 +t 2 A and r 1 +r 2 A because r 1 +r 2 = +r with 0 r < γ j 1. Thus ψ (t 1 + t 2 ) = (k 1 + k 2 + 1) k 1 + k 2 + ψ(r 2 ) = ψ (t 1 ) + ψ (t 2 ). 18

19 Finally, if t 1 A, t 2 A then ψ (t 1 + t 2 ) ψ(t 1 + t 2 ) ψ(t 1 ) + ψ(t 2 ) = ψ (t 1 ) + ψ (t 2 ). The fact that ψ i, with i 2, 3, 4, 5, 6}, satisfies ψ(k + r) = k + ψ(r) for all k = 1,..., D/ and γ j 1 r < is a consequence of Corollary Using this procedure to improve ψ 1 we may obtain a non superadditive function. However, for each facet defining inequality with k(j) > 1, we may use ψ 1 to lift all the coefficients since ψ 1(z) ψ 2(z) for all z [0, D]. Notice that, in general, the interval [0, z v ] is not too much restrictive. Remember that z v z α1 = + γ j 1. In Example 4.1, z v = 850. We will not analyze these functions on [z v, D]. Remark 6.10 In this section we constructed several superaddive valid lifting functions based on a non polynomial number of parameters. However, it is not clear whether is it possible to compute the coefficients using these functions in polynomial time. 7 Lifting the 2-integer continuous knapsack inequalities In this section we consider mixed integer knapsack sets of the form: X = (y, s) N N 0 R : j N a jy j D + s, s 0} where a j, j N and D are positive integers. To generate strong valid inequalities for X we restrict X by setting all integer variables to zero except two of them. W.l.o.g. we assume that those two variables are y 1 and y 2. Then, using a superadditive valid lifting function, we lift each facet defining inequality of the restricted set R = (y 1, y 2, s) N 2 0 R : y 1 +a 2 y 2 D+s, s 0}. The description of conv(r) is given in (Agra & Constantino 2003). Each non trivial facet defining inequality of conv(r), y 1 + α 2 y 2 α + βs. (7.1) belongs to one of two families. First we consider the family that is obtained from the lifting of a facet defining inequality for conv(y ) after s has been set to zero. Proposition 7.1 Consider a facet defining inequality for conv(y ), containing the extreme points of conv(y ), (a j 1, b j 1 ) and (a j, b j ) with (a j, b j ) = (a j 1, b j 1 ) + r( c k(j), d k(j) ) for some positive integer r, d k(j) y 1 + c k(j) y 2 d k(j) a j + c k(j) b j (7.2) then, if ck(j) a2 d k(j), the following inequality defines a facet of conv(r), d k(j) y 1 + c k(j) y 2 d k(j) a j + c k(j) b j + 1 j s (7.3) where j = (D/ ) D if k = 1 and j γ(a j 1, b j 1 ) + e l(k(j)) a 2 f l(k(j)), if b j 1 f l(k(j)), = γ(a j 1, b j 1 ) c k(j) 1 + a 2 d k(j) 1, if b j 1 < f l(k(j)), otherwise. Coefficients (a j 1, b j 1 ), (e l(k(j)), f l(k(j)) ) and (c k(j), d k(j) ) can be obtained using the polynomial version of Algorithm HW and (c k(j) 1, d k(j) 1 ) can be obtained as (c k(j) 1, d k(j) 1 ) = (c k(j), d k(j) ) r(e l(k(j)), f l(k(j)) ) where r = R (e l(k(j)),f l(k(j)) ). The case ck(j) a2 R (c k(j),d k(j) ) d k(j) is similar. It suffices to exchange 19

20 with a 2. Considering w 1 = a 2 d k(j) 1 c k(j) 1 and w 2 = a 2 f l(k(j)) + e l(k(j)), then j can be written as j w 2 γ j 1, if b j 1 f l(k(j)), = w 1 γ j 1, if b j 1 < f l(k(j)). Since w 1 w 2 and γ j 1 w 1 w 2, we have j w 2 w 1. The lifting function associated to (7.3) is given by: φ c (z) = min α y 1 α 2 y 2 + βs s. t. y 1 + a 2 y 2 D z + s, s 0, y 1, y 2 N 0. where = d k(j), α 2 = c k(j), α = d k(j) a j + c k(j) b j, β = 1 j. In the following consider z v, v = 0,..., α, and v k, k = 0,..., n, as defined in Section 6. Proposition 7.2 For z 0, 0, if z z 0, φ c v k + z zv k, if z (z) = j vk z < z vk + j (v k+1 v k ), for k = 0,..., n, v k+1, if z vk + j (v k+1 v k ) z z vk+1, for k = 0,..., n, α + z D, if z > D. j Proof: First consider 0 z D. Hence φ c can be written as φ c (z) = min s>0 φ(z), φ(z s) + 1 s}. j Suppose z vk z < z vk + j (v k+1 v k ). Considering t = z z vk > 0 we have 0 t < j (v k+1 v k ), thus φ c (z) φ(z t) + t = v j k + t. In order to prove the other direction we consider the following j cases. If 0 < s < t then as z s > z vk, it follows that φ(z s) + s = v j k+1 + s v j k + t because, j as we are assuming t < j (v k+1 v k ) and as s > 0, we have t j < s j + (v j k+1 v k ). If s > t and z vp < z s z vp+1 with p 0,..., k 1}, then s z vk z vp+1 +t = (v k v p+1 )w 2 +(w 1 w 2 ) v k v p+1 +t (v k v p+1 )w 2 + t (v k v p+1 ) j + t. Therefore φ(z s) + s j v p+1 + v k v p+1 + t = v j k + t = v k w 2 + (w 1 w 2 ) v k 1 + pw 2 + (w 1 w 2 ) p = v p+1 + s v j p+1 + (v k v p+1) j +t j. If s > t and z j p 1 < z s z p for p N 0, then s z vk z p + t i=1 δ i + t (v k + p)w 2 + t (v k + p) j + t. Therefore φ(z s)+ s = p+ s j p+ (v k+p) j +t j v j k + t. Hence φ(z s)+ s j φ(z t)+ t j = v j k + t for j all s > 0. Now suppose z > D. Again, φ c (z) = min s>z D φ(z (z D)) + 1 (z D), φ(z s) + 1 j s}. j Similarly, it can be checked that for s > z D, φ(z s) + 1 s φ(z (z D)) + z D j = α + z D j. j The case z vk + j (v k+1 v k ) z z vk+1, is similar to the previous one. It is important to notice that, in general, φ c is not superadditive (in the example of Figure 7, φ c (3) + φ c (3) > φ c (6)). In order to extend the superadditive functions obtained in Section 4, ψ i, i 2, 3, 4, 5, 6}, for [D, + [ we define z v, for v > α, v N, as z v = γ j 1 +vw 2 +(w 2 w 1 ) v i=1 δ i, where, for i > α, δ i = δ r with r = i (i 1)/. The proofs of superadditivity of ψ i, i 2, 3, 4, 5, 6}, are also valid replacing v = 0,..., α 1 with v N 0. Next we prove that the extended functions of ψ i, i 2, 3, 4}, are also valid considering φc. Lemma 7.3 ψ 4(z) φ c (z), for all z [0, + [. 20