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The continuous knapsack set
Sanjeeb DashIBM Research
Oktay GunlukIBM Research
Laurence WolseyCore
January 31, 2014
Abstract
We study the convex hull of the continuous knapsack set which consists of a single inequality constraintwith n non-negative integer and m non-negative bounded continuous variables. When n = 1, this set isa slight generalization of the single arc flow set studied by Magnanti, Mirchandani, and Vachani (1993).We first show that in any facet-defining inequality, the number of distinct non-zero coefficients of thecontinuous variables is bounded by 2n − n. Our next result is to show that when n = 2, this upperbound is actually 1. This implies that when n = 2, the coefficients of the continuous variables in anyfacet-defining inequality are either 0 or 1 after scaling, and that all the facets can be obtained from facetsof continuous knapsack sets with m = 1. The convex hull of the sets with n = 2 and m = 1 is thenshown to be given by facets of either two-variable pure-integer knapsack sets or continuous knapsacksets with n = 2 and m = 1 in which the continuous variable is unbounded. The convex hull of these twosets has been completely described by Agra and Constantino (2006). Finally we show (via an example)that when n = 3, the non-zero coefficients of the continuous variables can take different values.
1 Introduction
In this paper we study the set S = SLP ∩ (Rm × Zn) where
SLP ={
(x, y) ∈ Rm × Rn :m∑i=1
xi +n∑j=1
cjyj ≥ b, u ≥ x ≥ 0, y ≥ 0}
and u, c, b > 0. The case where n = 1, known as the single arc flow set, was first studied by Magnanti,Mirchandani, and Vachani [6]. They gave an explicit characterization of the convex hull of S via residualcapacity inequalities (see Section 1.1).
We study the properties of facet-defining inequalities for the case n ≥ 2 and characterize the convexhull of S when n = 2. More precisely, we first prove in Section 2 that in any facet-defining inequality, thenumber of distinct non-zero coefficients of the continuous variables is at most 2n − n. We then study thecase when n = 2 in detail, and show in Section 3 that all non-trivial facet-defining inequalities for conv(S)
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are of the form: ∑i∈I
xi + γ1y1 + γ2y2 ≥ β
where I ⊆ {1, . . . ,m} and x+ γ1y1 + γ2y2 ≥ β is facet-defining for the set
Q(b′, u′) = conv{
(x, y) ∈ R× Z2 : x+ c1y1 + c2y2 ≥ b′, u′ ≥ x ≥ 0, y ≥ 0},
where b′ = b−∑
i 6∈I ui and u′ =∑
i∈I ui.In other words, when n = 2, all facets of conv(S) can be obtained from three-variable relaxations.
Throughout, Q(b, u) will denote the special case of conv(S) when m = 1 and n = 2. In other words,Q(b, u) is the convex hull of nonnegative (x, y1, y2) ∈ R3, with y1, y2 integral and x ≤ u, satisfyingx+ c1y1 + c2y2 ≥ b.
We then analyze the facial structure of Q(b, u) in Section 4. We show that non-trivial facet-defininginequalities either have a zero coefficient for the x variable and therefore are facet-defining for
conv{
(x, y) ∈ R× Z2 : c1y1 + c2y2 ≥ b− u, y ≥ 0},
or they can be obtained from a relaxation in which the upper bound is dropped:
Q(b,∞) = conv{
(x, y) ∈ R× Z2 : x+ c1y1 + c2y2 ≥ b, x ≥ 0, y ≥ 0}.
For Q(b,∞) for any b > 0, Agra and Constantino [2] gave a polynomial-time algorithm to enumerateall facet-defining inequalities. Note that S has an exponential number of relaxations of type Q(b, u), onefor each I ⊆ {1, . . . ,m}, and therefore our result does not lead directly to a polynomial-time separationalgorithm for S. However, given a fixed objective function, we show that optimization over S can be carriedout by solving m three variable problems, and thus the separation problem is also polynomially solvableusing the ellipsoid algorithm.
Finally, in Section 5, we show that our results cannot be generalized to n ≥ 3. In particular, we presenta facet of a particular set with n = 3 such that the non-zero coefficients of the continuous variables in theassociated inequality take different values.
1.1 Related Literature
Magnanti, Mirchandani, and Vachani [6] studied the single arc design problem as a subproblem of thenetwork loading problem. In particular, they studied the set{
(x, y) ∈ Rm × Z :m∑i=1
xi ≤ cy, u ≥ x ≥ 0, y ≥ 0}.
The network loading problem is the problem of choosing arc capacities of minimum cost in a networkso as to enable flows of different quantities (ui) between m pairs of nodes, or equivalently to enable amulticommodity flow. On any arc, capacities must be chosen in integral multiples of a fixed constant (say
2
c). The single arc design problem is the subproblem which enforces the fact that the flow through an arc forany commodity must be bounded above by the corresponding demand value (ui), and the sum of flows is atmost the chosen capacity (cy) on the arc.
By replacing each variable xi by ui − xi, it is clear that the above set is equivalent to the set
{(x, y) ∈ Rm × Z :
m∑i=1
xi + cy ≥ b, u ≥ x ≥ 0, y ≥ 0},
where b =∑m
i=1 ui (this is a special case of the set S described earlier, when n = 1). Magnanti et. al.showed that the residual capacity inequalities give the convex hull of solutions of this set, and Atamturk andRajan [4] give a polynomial-time separation algorithm for these inequalities.
When n = 1, Theorem 3.11 reduces to the result of Magnanti et. al. More precisely, when n = 1, ourresult implies that every facet of S is of the form
∑i∈I xi+γy ≥ β where I ⊆ {1, . . . ,m} and x+γy ≥ β
is facet-defining for the two-variable set conv{(x, y) ∈ R×Z : x+cy ≥ b′, u′ ≥ x ≥ 0, y ≥ 0}where b′ =b −
∑i 6∈I ui and u′ =
∑i∈I ui. But the two-variable set has only one nontrivial facet-defining inequality,
which is given either by the basic mixed-integer inequality, namely x+ (b′ − b b′c cc)y ≥ (b′ − b b′c c)db′
c e orby y ≥ d(b′−u′)/ce when I = ∅ . Thus our results yield an alternative proof of the result of Magnanti et al.
Atamturk and Gunluk [1] describe the set studied by Magnanti et. al. as the splittable flow arc set whichthey studied as a subproblem of the multicommodity network design problem. In particular, they studied themore general version which is equivalent to S with n = 1 and arbitrary b, and used the results in Magnantiet. al. to prove that the residual capacity inequalities still give the convex hull in this case.
Later Magnanti, Mirchandani, Vachani [7] stated that their results for n = 1 extend directly to give theconvex hull for a special case of the single arc problem with two capacities (S with n = 2) in which all thedata ui and cj are integer, c1 = 1 and c2 > 1. Note that in this case the associated capacity variable y1 canbe treated as continuous. Yaman [10] studied an extension of this version in which the capacity variablesalso impose an integer lower bound on the sum of the flows.
Wolsey and Yaman [9] study the continuous knapsack set with divisible capacities, i.e., c1| · · · |cn, andshow that the coefficients of the continuous variables in any facet-defining inequality lie in {0, 1} for allvalues of m and n.
2 Coefficients of continuous variables in facet-defining inequalities
In this section we consider a non-trivial facet-defining inequality
m∑i=1
αixi +n∑j=1
γjyj ≥ β
of conv(S) and study the properties of the vector α. Let F = {(x, y) ∈ S : αx+ γy = β} denote the set ofpoints in S that satisfy this inequality as equality.
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2.1 Basic properties of conv(S)
We start off with some basic polyhedral properties of conv(S). Let ei ∈ Rm denote the unit vector with aone in the ith component and zeros elsewhere and e ∈ Rm denote the vector of all ones. We let ei stand forthe unit vector in Rn with a one in the ith component. We call the following inequalities that appear in thedescription of SLP , bound inequalities: xi ≥ 0 and xi ≤ ui for i = 1, . . . ,m; and yj ≥ 0 for j = 1, . . . , n.We call the inequality ex + cy ≥ b, the capacity inequality. We also refer to all of these inequalities astrivial inequalities and the associated facets as trivial facets.
Lemma 2.1. The following properties hold for S:
(a) conv(S) is full-dimensional and its recession cone is C = {(0, y) ∈ Rm × Rn : y ≥ 0}.
(b) The bound inequalities are facet-defining.
(c) Assume that∑m
i=1 ui > b. Then the capacity inequality is facet-defining if and only if b ≥ cj for allj = 1, . . . n.
(d) If αx+ γy ≥ β is a non-trivial facet-defining inequality, then α ≥ 0, γ > 0 and β > 0.
(e) All non-trivial facets of S are bounded.
Proof. (a) Consider the following n + m + 1 points in S: w = (0, db/c1ee1), pi = w + (uiei, 0) fori = 1, . . . ,m, and qj = w + (0, ej) for j = 1, . . . , n. These points are affinely independent as the pointspi−w and qj−w are linearly independent. The recession cone of conv(S) is C as c > 0 and the continuousvariables are bounded.
(b) For each i = 1, . . . ,m, there are m + n points with xi = 0 among the (n + m + 1) affinelyindependent points given above. Furthermore by adding (uiei, 0) to each of these n + m points one againobtains affinely independent points. Therefore the bound constraints on the continuous variables are facet-defining. Similarly for j = 2, . . . , n, there are n + m points among the n + m + 1 points with yj = 0
and consequently the non-negativity constraints for these constraints are facet-defining. As the choice of thecoordinate corresponding to y1 in the construction of w is arbitrary, y1 ≥ 0 is facet-defining as well.
(c) To see that the capacity inequality cannot be facet-defining if b < cj for some j = 1, . . . n, notethat in this case the inequality cannot be tight if yj ≥ 1. Now assume that b ≥ cj for all j = 1, . . . n.Let µ =
∑mi=1 ui. The following points are affinely independent: qj = (
b−cjµ u, ej) ∈ S for j = 1, . . . , n.
Now consider the point p0 = ( b−εµ u, 0) 6∈ S for some ε > 0. If ε is small enough, the following pointspi = p0 + (εei, 0) ∈ S for i = 1, . . . ,m. These points are affinely independent from the points constructedabove and therefore the capacity inequality is facet-defining.
(d) Let F = {(x, y) ∈ S : αx + γy = β} be the set of integral points in the facet defined by thisinequality. For any i = 1, . . . ,m, there is a point (x, y) ∈ F with xi < ui, and for some small ε > 0, wehave (x, y) + (εei, 0) ∈ S . This implies that α ≥ 0. In addition, as the recession cone of conv(S) is C, we
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have γ ≥ 0. As the inequality is not implied by the non-negativity constraints, β > 0. Finally, if γi = 0 forsome i in {1, . . . , n}, then αx+ γy ≥ β is violated by (0, db/cieei) ∈ S. Therefore γ > 0.
(e) As x ≤ u for all (x, y) ∈ S and γ > 0 for all non-trivial facet-defining inequalities, the claimholds.
Note that above we have only given some necessary conditions for the capacity inequality to be facet-defining. Characterizing sufficiency conditions is somewhat involved.
2.2 Facets of conv(S) obtainable from relaxations
We next study the conditions under which a non-trivial facet-defining inequality αx + γy ≥ β can beobtained from a lower-dimensional relaxation of S. Remember that F denotes the set of points in S thatsatisfy this inequality as equality. We next present two observations that lead to the main result of thissection. First we consider the case when some of the entries of α are zero.
Lemma 2.2. If αm = 0, then∑m−1
i=1 αixi +∑n
j=1 γjyj ≥ β is facet-defining for the set
S′ = conv{
(x, y) ∈ Rm−1 × Zn :m−1∑i=1
xi +n∑j=1
cjyj ≥ b− um, y ≥ 0, u′ ≥ x ≥ 0}
where u′i = ui for i = 1, . . . ,m− 1 .
Proof. First notice that S′ is obtained by deleting xm from the set of points conv(S) ∩ Xm where Xm =
{(x, y) ∈ Rn+m : xm = um}. Therefore, if αx + γy ≥ β is valid for conv(S), and consequently forconv(S) ∩ Xm, then
∑m−1i=1 αixi +
∑nj=1 γjyj ≥ β is valid for S′. We next argue that the inequality is
facet-defining for S′.Let pk = (xk, yk) be a collection of m + n affinely independent points in F , which exist as conv(F )
is a facet of conv(S). Let pk = (xk, yk) ∈ Rm−1 × Zn be obtained from pk by deleting the last entry ofxk for all k = 1, . . . ,m + n. Also let α ∈ Rm−1 be obtained from α by deleting the last entry. Noticethat
∑m−1i=1 xki =
∑mi=1 x
ki − xm ≥
∑mi=1 x
ki − um and therefore pk ∈ S′ for all k = 1, . . . ,m + n.
Furthermore, αxk = αxk, and consequently αxk + γyk = β for all k = 1, . . . ,m+n. As the affine rank of{p1, . . . , pm+n} is one less than the affine rank of {p1, . . . , pm+n}, we conclude that the claim is true.
Applying this observation repeatedly, we make the following observation when α = 0.
Corollary 2.3. If α = 0, then γy ≥ β is facet-defining for S′ = conv{y ∈ Zn :∑n
j=1 cjyj ≥ b −∑mi=1 ui, y ≥ 0}. In addition, if
∑mi=1 ui ≥ b, then the facet is one of the non-negativity facets associated
with y.
We next consider the case when some of the entries of the coefficient vector α are the same.
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Lemma 2.4. If αm−1 = αm, then∑m−1
i=1 αixi +∑n
j=1 γjyj ≥ β is facet-defining for the set
S′′ = conv{
(x, y) ∈ Rm−1 × Zn :m−1∑i=1
xi +n∑j=1
ciyi ≥ b, y ≥ 0, u′ ≥ x ≥ 0}
where u′i = ui for i = 1, . . . ,m− 2 and u′m−1 = um−1 + um .
Proof. The proof is very similar to the proof of Lemma 2.2. First we observe that∑n−1
i=1 αixi+∑n
j=1 γjyj ≥β is valid for S′′ provided that αx + γy ≥ β is valid for S. Then, we modify the points pi defined inLemma 2.2 by combining the last two entries of the continuous variables. The resulting (lower dimensional)points are in S′′ and have the desired affine rank to conclude the proof.
Theorem 2.5. Assume that αi ∈ {0, α1, . . . , αt} for all i = 1, . . . ,m where α1, . . . , αt are distinct positivenumbers. Then
∑ti=1 αixi +
∑nj=1 γjyj ≥ β is facet-defining for
S = conv{
(x, y) ∈ Rt × Zn :t∑i=1
xi +n∑j=1
cjyj ≥ b− u0, y ≥ 0, u ≥ x ≥ 0}
where uj =∑
k:αk=αjuk for j = 1, . . . , t and u0 =
∑k:αk=0 uk.
Proof. Applying Lemma 2.2 and Lemma 2.4 repeatedly proves the claim.
We note that the reverse is not true in the sense that given a facet of a lower dimensional set of the formS above, obtained by combining continuous variables, it is not always possible to lift them in the obviousway to obtain facets of the original set S.
2.3 Bounding the number of distinct coefficients in facet-defining inequalities
We next consider a facet-defining inequality αx + γy ≥ β such that α > 0 and all of the entries of α aredistinct (α may have a single component). Remember that F denotes the set of points in S that satisfy thisinequality as equality. We start off with a technical observation that we use later.
Lemma 2.6. Assume that α > 0 has all distinct coefficients. If (x1, y), (x2, y) ∈ F then x1 = x2.
Proof. Clearly αx1 + γy = αx2 + γy = β and therefore αx1 = αx2. Assume x1 6= x2. If α ∈ R, thenthe result trivially follows. Otherwise there must exist two indices i and j such that x1
i 6= x2i and x1
j 6= x2j
and therefore x = 12x
1 + 12x
2 has ui > xi > 0 and uj > xj > 0. Note that (x, y) ∈ F ⊆ S. Now assumeαi > αj and notice that for some small ε > 0 a new point x′ obtained by reducing xi by ε and increasingxj by ε gives (x′, y) ∈ S. This point (x′, y), however, violates the facet-defining inequality as αx′ < αx, acontradiction.
Lemma 2.7. Assume that α > 0 has all distinct coefficients. Then F contains a subset of m + n affinelyindependent points {(xi, yi) : i = 1, . . . ,m + n} such that conv(y1, . . . , ym+n) is (i) full-dimensional, (ii)has m+ n integral vertices and contains no other integer points.
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Proof. As conv(F ) ⊆ Rm+n has dimension m + n − 1, F contains m + n affinely independent points.Among all such sets of points, let L = {(xi, yi) : i = 1, . . . ,m + n} stand for the one with minimumnumber of integer points in conv(Ly), where Ly = {y1, . . . , ym+n}.
(i) If conv(Ly) is not full-dimensional, then there exists 0 6= γ′ ∈ Rn, β′ ∈ R such that γ′yi = β′ fori = 1, . . . ,m+ n. But this means that points in F satisfy the equation γ′y = β′, which contradicts the factthat αx+ γy = β is uniquely defined up to multiplication by a scalar and α > 0.
(ii) Suppose conv(Ly) contains an integer point y which is not a vertex. Then y =∑m+n
i=1 µiyi for some
µi ≥ 0 with∑m+n
i=1 µi = 1. Let x =∑m+n
i=1 µixi. Then (x, y) is contained in F . Assume y ∈ Ly. In
this case, let yk = y for some k ≥ 0. Then we can assume µk = 0. Furthermore, Lemma 2.6 implies thatx = xk, and therefore (xk, yk) is a convex combination of other points in L, which contradicts the affineindependence of points in L.
Therefore we can assume all points in Ly are vertices of conv(Ly), and y 6∈ Ly. As the points in L areaffinely independent, for some index j ∈ {1, . . . ,m+n} with µj > 0, the set L′ = L∪{(x, y)}\{(xj , yj)}is affinely independent. L′ also has the property that the convex hull of L′y is strictly contained in the convexhull of Ly and has fewer integral points (it does not contain yj). This contradicts the definition of L.
We next bound the number of distinct values of the entries of α when S has an arbitrary number of con-tinuous variables. By Theorem 2.5, one needs to consider the case when α has distinct non-zero coefficients.
Theorem 2.8. If αx+ γy ≥ β is a facet-defining inequality for conv(S) then α has at most 2n − n distinctnon-zero entries.
Proof. As the claim holds for trivial facet-defining inequalities, we only consider non-trivial inequalities.First assume that α > 0 and has all distinct coefficients and let Ly ⊆ Rn be defined as in the proofof Lemma 2.7. Suppose Ly contains more than 2n integer points. Then it contains at least two distinctpoints, say yk and yl, with the same odd/even parity (that is, for all i: yki is odd if and only if yli is odd).Consequently, y = (yk + yl)/2 ∈ conv(Ly) ∩ Zn which contradicts Lemma 2.7. Therefore when α > 0
and has all distinct coefficients m + n ≤ 2n. Combining Theorem 2.5 with this observation completes theproof.
Corollary 2.9. If n = 2, then all facet-defining inequalities for conv(S) can be obtained from relaxationsof the form S presented in Theorem 2.5 that have 2 continuous variables.
2.4 Bounding the number of distinct coefficients in disjunctive cuts
We next derive an upper bound on the number of distinct positive coefficients of continuous variables infacet-defining inequalities of disjunctive relaxations of S. More precisely, we consider a disjunctive cutαx+ γy ≥ β for conv(S) that can be derived using the |K|-term disjunction D = ∪k∈KDk where
Dk ={
(x, y) ∈ Rm × Rn : Aky ≥ dk}
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for k ∈ K. We assume that Rm × Zn ⊆ D and therefore αx + γy ≥ β is a valid inequality for conv(S).Furthermore, we assume that αx+ γy ≥ β defines a facet of the disjunctive relaxation of conv(S)
Q = conv(∪k∈K(Dk ∩ SLP ))
that is distinct from the bound constraints on the x variables. Note that αi ≥ 0 for all i (as the related facetmust contain a point with xi < ui and if αi < 0, this point can be perturbed to obtain a new point in Qviolating the inequality). Clearly some of the sets Dk ∩ SLP can be empty. Without loss of generality,assume that Dk ∩ SLP 6= ∅ for k ∈ K = {1, . . . , |K|} and Dk ∩ SLP = ∅ for k > |K|.
As the inequality αx+ γy ≥ β is valid for
Dk ∩ SLP ={
(x, y) ∈ Rm × Rn : ex+ cy ≥ b, Aky ≥ dk, y ≥ 0, u ≥ x ≥ 0}
for k ≤ |K|, there exist nonnegative multipliers θk (associated with ex + cy ≥ b), ηk (associated withAky ≥ dk), and λk (associated with −x ≥ −u) that yield the valid inequality
(θke− λk)x+ (θkc+ ηkAk)y ≥ θkb+ ηkdk − λku
for Dk ∩ SLP where α ≥ (θke − λk), γ ≥ (ηkAk + θkc) and β ≤ (θkb + ηkdk − λku). Furthermore, asαx + γy ≥ β is facet-defining for Q by assumption, we have (here ej is a unit vector in Rn with a one inthe j component)
αi = maxk∈K{θk − λki }, γj = max
k∈K{θkcj + ηkAkej}, and β = min
k∈K{θkb+ ηkdk − λku).
Note that if θ = mink∈K{θk} > 0, then decreasing all entries of the vector θ by θ yields a stronger validinequality for Q. This is not possible as αx + γy ≥ β is facet-defining for Q. Consequently, we concludethat mink∈K{θk} = 0. Without loss of generality, we assume that θ|K| ≥ θ|K|−1 ≥ . . . ≥ θ1 = 0.
Using the multipliers θ and η (but not λ) it is easy to see that for all k ∈ K the inequality θkex+γy ≥ βk,where βk = θkb + ηkdk, is valid for Dk ∩ SLP . Consequently, for all k ∈ K we can define the followingrelaxation of the set Dk ∩ SLP :
W k ={
(x, y) ∈ Rm × Rn : θkex+ γy ≥ βk, u ≥ x ≥ 0}⊇ Dk ∩ SLP .
Notice that αx + γy ≥ β is valid for each W k and consequently, it is valid for W = conv(∪k∈KW k).Furthermore, as W is a relaxation of Q, the inequality αx+ γy ≥ β is facet-defining for W .
Theorem 2.10. Given a t-term disjunction and a facet-defining inequality αx+ γy ≥ β for the associateddisjunctive relaxation, the vector α has at most 2(t− 1) distinct non-zero coefficients.
Proof. Using the notation introduced in the the preceding discussion, αx + γy ≥ β is valid for W k forall k ∈ K, and consequently there exists a non-negative vector λk ∈ Rm for each k ∈ K such thatαi ≥ θk − λki and β ≤ βk − λku. As αx + γy ≥ β is facet-defining for the associated disjunctive
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relaxation, αi = maxk∈K{θk − λki } and β = mink∈K{βk − λku}. Clearly, λki ≥ θk − αi and λki ≥ 0
implying λki ≥ (θk − αi)+ for all i = 1, . . . ,m and k ∈ K. Without loss of generality, we also assume thatλki = (θk − αi)+ for all i = 1, . . . ,m and k ∈ K.
We next argue that if αi, αj 6∈ {θ1, . . . , θ|K|}, then max{αi, αj} > θl > min{αi, αj} for some l. Asαi = maxk∈K{θk − λki } ≤ θ|K| and θ1 = 0 we have θ1 ≤ αi ≤ θ|K| for all i. Assume that there existsdistinct v, w together with an index k1 such that θk1+1 > αv, αw > θk1 . Clearly, λkv = λkw = 0 whenk ≤ k1 and λkv , λ
kw > 0, otherwise. Let ε > 0 be sufficiently small and define:
α(ε) =
αi i 6∈ {u, v}
αv − ε/uv i = v
αw + ε/uw i = w
α(−ε) =
αi i 6∈ {u, v}
αv + ε/uv i = v
αw − ε/uw i = w
and similarly,
λki (ε) =
λki i 6∈ {u, v} and k ∈ K
λki = 0 i ∈ {u, v} and k ≤ k1
λkv + ε/uv i = v and k ≥ k1 + 1
λkw − ε/uw i = w and k ≥ k1 + 1
λki (−ε) =
λki i 6∈ {u, v} and k ∈ K
λki = 0 i ∈ {u, v} and k ≤ k1
λkv − ε/uv i = v and k ≥ k1 + 1
λkw + ε/uw i = w and k ≥ k1 + 1.
Notice that for all k, we have λk(ε)u = λk(−ε)u = λku. Consequently, β = mink∈K{βk − λk(ε)u} =
mink∈K{βk−λk(−ε)u}. Furthermore, as θk1+1 > αv, αw > θk1 by assumption and ε > 0 is small enough,we also have αi(ε) = maxk∈K{θk−λki (ε)} and αi(−ε) = maxk∈K{θk−λki (−ε)} for all i. Therefore, bothαεx+ γy ≥ β and α−εx+ γy ≥ β are valid inequalities for conv(∪k∈KW k). But in this case αx+ γy ≥ βcannot be facet-defining as α = (αε + α−ε)/2. We can therefore conclude that if αu, αw 6∈ {θ1, . . . , θ|K|},then max{αu, αw} > θl > min{αu, αw} for some l. Consequently, there can only be at most one αi thatlies between two consecutive θ entries. As θ1 = 0, we conclude that the vector α has at most 2(t − 1)
distinct non-zero coefficients.
Note that any facet-defining inequality for S can be generated as a disjunctive cut from a disjunctionwith at most 2n-terms [3]. Consequently, Theorem 2.10 implies that if αx + γy ≥ β is facet-defining forconv(S) then α has at most 2(2n − 1) distinct non-zero entries. This bound is weaker than that given byTheorem 2.8, however Theorem 2.10 still leads to useful observations:
Corollary 2.11. If D is a split disjunction, then α has at most 2 distinct coefficients.
3 Characterizing facet-defining inequalities when n = 2
In this section we show Theorem 3.11, namely that if αx+ γy ≥ β is a nontrivial facet-defining inequalityfor conv(S) with n = 2 (i.e., y ∈ R2), then all nonzero components of α are equal. The proof is by
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contradiction; we assume Theorem 3.11 is not true and consider a minimal counterexample to Theorem 3.11with n = 2, i.e., a set S with n = 2 and a facet-defining inequality αx + γy ≥ β such that S has as fewvariables as possible. We can assume that α > 0 and its coefficients are all distinct. The reason for thisassumption is the following. Let S and αx + γy ≥ β form a minimal counterexample – i.e., it is facet-defining for conv(S) and the nonzero coefficients of α are not all equal, and m+n is as small as possible. Ifa component of α is zero or a pair of components of α are nonzero but equal, we can apply either Lemma 2.2or Lemma 2.4 and obtain a facet-defining inequality of a set S′ with fewer variables than S, but with thefacet-defining inequality having the same set of nonzero α values as before. This would contradict theassumption of minimality of S. Therefore, in a minimal counterexample, α > 0 and its coefficients are alldistinct. In this case, Corollary 2.9 implies that m ≤ 2.
Ifm = 1 and α has a single component, there is nothing to prove. So we make the following assumption.
Assumption 3.1. Suppose n = 2. If S and αx+ γy ≥ β form a minimal counterexample to Theorem 3.11,then m = 2, and 0 < α1 < α2.
We start off by proving some properties of integral points contained in nontrivial facets of conv(S) forarbitrary n, and then focus on the case n = 2 and m = 2.
3.1 Properties of integral points on facets of conv(S)
We next study properties of integral points on facets of conv(S) for general n. Throughout we assume thatαx + γy ≥ β is a nontrivial facet-defining inequality for conv(S) and F is the set of points in S lying onthe corresponding facet.
Lemma 3.2. Let (x, y) ∈ F and let xj > 0 for some j ∈ {1, . . . ,m}. Then for every index i 6= j withαi < αj , we have xi = ui. Furthermore, if 0 < xi < ui for i = 1, . . . ,m, then α has all its coefficientsequal.
Proof. If there is some index i 6= j with αi < αj such that xi < ui, then letting ε = min{ui − xi, xj}, wesee that (x, y) + εei − εej ∈ S but violates αx+ γy ≥ b.
For the second part of the Lemma, assume 0 < xi < ui for i = 1, . . . ,m. If αk 6= αl for any pairof indices k, l ∈ {1, . . . ,m}, then either αk < αl or αk > αl, and the first part of the Lemma implies,respectively, that xk = uk or xl = ul, a contradiction.
Lemma 3.3. Assume α > 0 has all distinct coefficients. If (x, y) ∈ F with x 6= 0, then ex + cy = b.Therefore F contains a point (x, y) with x = 0.
Proof. Let (x, y) ∈ F with x 6= 0. Suppose ex+ cy = b+ ε for some ε > 0. By definition, xi > 0 for somei ∈ {1, . . . ,m}; then (x −min{xi, ε}ei, y) ∈ S but violates αx + γy ≥ β (as α > 0), a contradiction tothe fact that this inequality is valid for conv(S). If the second part of the Lemma is not true, then each pointin F satisfies ex + cy = b by the first part of the Lemma. As conv(S) is full-dimensional, this means thatαx+ γy ≥ β is a scalar multiple of ex+ cy ≥ b, which contradicts the nontriviality of αx+ γy ≥ β.
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Lemma 3.4. Assume α > 0 has all distinct coefficients. Let (x, y), (x, y) ∈ F . If αx ≤ αx, then x ≤ x.Therefore αx = αx if and only if x = x.
Proof. Let the conditions of the Lemma be true, but assume αx ≤ αbarx and xj > xj for some indexj ∈ {1, . . . ,m}. The fact that xj > 0 implies (by Lemma 3.2) that xi = ui ≥ xi for all i 6= j with αi < αj .The fact that xj < uj implies (by Lemma 3.2) that xi = 0 ≤ xi for all i 6= j with αi > αj . Then xi ≥ xi
for i = 1, . . . ,m (as all coefficients of α are distinct), and xj > xj which implies that αx > αx (as α > 0),a contradiction. The second part of the Lemma follows trivially from the first.
Note that Lemma 3.4 implies Lemma 2.6.
3.2 Properties of S when n = 2,m = 2.
We now focus on the case n = 2 and m = 2 and assume that there exists a nontrivial facet-defininginequality αx+ γy ≥ β with α2 > α1 > 0. By Theorem 2.1 we have γ > 0 and β > 0. We define F to bethe set of integral points in S satisfying αx + γy = β. Let L = {(xi, yi) : i = 1, . . . , 4} ⊂ R2 × R2 be aset of affinely independent points in F which has the properties in Lemma 2.7. We refer to these points asp1, . . . , p4. As β > 0, these points are also linearly independent. As before, let Ly = {y1, . . . , y4}, and letQ = conv(Ly).
Lemma 3.5. Q is a parallelogram.
Proof. Lemma 2.7 implies that Q ⊆ R2 is full-dimensional, has four vertices (namely y1, . . . , y4), andcontains no other integer points. In R2, such a set can only be a parallelogram. To see this, let the verticesof the quadrilateral Q be y1, . . . , y4 in clockwise order, with the interior angles (in degrees) between theedges defining these vertices equal to θ1, . . . , θ4. If Q is not a parallelogram, we can assume, without lossof generality, that θ1 + θ2 > 180. Furthermore, we can assume that either θ4 + θ1 ≥ 180 or θ3 + θ2 ≥ 180.In the first case, y4 + y2 − y1 is contained in Q (and distinct from y3), and in the second case y3 + y1 − y2
is contained in Q (and distinct from y4), a contradiction. Therefore Q is a parallelogram.
We next assume that the points in L are sorted by nondecreasing values of γyi, i.e., γy1 ≤ γy2 ≤ γy3 ≤γy4, and therefore by nonincreasing values of αxi. Lemma 3.4 implies that xi ≥ xj for j > i. If yi = yk,then Lemma 2.6 implies that xi = xk which contradicts the distinctness of pi and pk, so we can assume allyis are distinct.
Lemma 3.6. If γy1 = γy2 and γy3 = γy4, then the points pi(i = 1, . . . , 4) are linearly dependent.
Proof. If the conditions of the Lemma are satisfied, then Lemma 3.4 implies that x1 = x2 and x3 = x4.Next observe that the yis are all distinct. As 0 6= γ ∈ R2, and 0 6= y2 − y1 and 0 6= y4 − y3 are orthogonalto γ, we infer that y2 − y1 is a scalar multiple of y4 − y3 (from the fact that these vectors lie in R2), andtherefore y2− y1−µ(y4− y3) = 0 for some nonzero scalar µ. As x2− x1 = 0 and x4− x3 = 0, it followsthat p2 − p1 − µ(p4 − p3) = 0.
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As the points in L are linearly independent, we conclude that the conditions of Lemma 3.6 cannot hold.Using the fact that γyi is non-decreasing and either γy1 < γy2 or γy3 < γy4, leads to the followingobservation.
Corollary 3.7. For the points in L we either have γy1 < γy2 or γy3 < γy4 and therefore γy1 < γy4.
Lemma 3.8. The points y1 and y4 form opposite corners of the parallelogram Q.
Proof. By Corollary 3.7 we have γy1 < γy4. Assume the result is not true and that y1 and y4 define adjacentcorners of Q. Then the other adjacent corner of Q to y1 is defined by either y3 (in which case y4 − y1 =
y2− y3) or by y2 (in which case y4− y1 = y3− y2). In the first case we have 0 < γ(y4− y1) = γ(y2− y3)
which contradicts the fact that γy2 ≤ γy3. Therefore y4− y1 = y3− y2 and 0 < γ(y4− y1) = γ(y3− y2).This combined with γy1 ≤ γy2 ≤ γy3 ≤ γy4 implies that γy1 = γy2 and γy3 = γy4. Lemma 3.6 thenimplies that p1, . . . , p4 are linearly dependent, a contradiction.
Lemma 3.9. The point x1 > 0 with x11 = u1 and x4 = 0. Further, one can assume that cy4 > b.
Proof. For i = 1, 2, there is a point (x, y) ∈ L with xi < ui and a point (x′, y′) with x′i > 0. This followsdirectly from the fact that conv(S) is full-dimensional and αx + γy ≥ β is not equal to xi ≤ ui or xi ≥ 0
for i = 1, 2. As x1 ≥ · · · ≥ x4 by Lemma 3.4, we have x12 > 0, and from Lemma 3.2, the first part of
the result follows. If x4 6= 0, then Lemma 3.4 implies that x1, . . . , x4 6= 0 which contradicts Lemma 3.3.Therefore x4 must be 0.
If ex4 + cy4 = cy4 = b, then exk + cyk > b for some k < 4 (again, because we are dealing witha nontrivial facet-defining inequality). But Lemma 3.3 implies that xk = 0 = x4 ⇒ γyk = γy4 = β.Therefore, if we switch the points (xk, yk) and (x4, y4), L is still sorted by nondecreasing values of γyi andcy4 > b and x4 = 0.
Combining these observations, we next show that a nontrivial facet-defining inequality αx + γy ≥ β
with α2 > α1 > 0 cannot exist.
Theorem 3.10. Let m = 2, n = 2 and consider a nontrivial facet-defining inequality αx + γy ≥ β. Ifα > 0, then α1 = α2.
Proof. Suppose the coefficients of α are distinct and assume that 0 < α1 < α2. We know that y1 and y4
form nonadjacent corners of Q, and y2 and y3 form the remaining corners of Q. Therefore
y4 − y3 = y2 − y1. (1)
From the equations αxi + γyi = β for i = 1, . . . , 4, we conclude that
α(x3 − x4) + γ(y3 − y4) = 0 = α(x1 − x2) + γ(y1 − y2).
Using equation (1) we can conclude that
α(x3 − x4) = α(x1 − x2). (2)
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Furthermore, given that 0 < γ(y2 − y1) = γ(y4 − y3), from Corollary 3.7, we have αx3 > αx4 andαx1 > αx2. Therefore x1, x2, x3 6= 0 as x4 = 0 by Lemma 3.9. Using Lemma 3.3 we have
exi + cyi = b for i = 1, . . . , 3,
and consequently, (x4, y4) cannot lie on the capacity constraint, implying
cy4 = ex4 + cy4 > b.
From this we can conclude that
e(x3 − x4) + c(y3 − y4) < 0 = e(x1 − x2) + c(y1 − y2)
Again using equation (1) we infer that
e(x3 − x4) < e(x1 − x2). (3)
We now have two cases.Case 1: x2
1 = u1. Scale α such that α1 < 1 and α2 = 1. Recall that x11 = u1 and x1
2 > 0 (byLemma 3.9). As x1 − x2 6= 0, we conclude that x1 − x2 is nonzero only in the second coordinate. Asx3 6= 0 = x4, x3
1 > 0 by Lemma 3.2, and therefore x3 − x4 is definitely nonzero in the first coordinate.Therefore α(x1 − x2) = e(x1 − x2) and α(x3 − x4) < e(x3 − x4). But this combined with (3) and (2)leads to a contradiction.
Case 2: x21 < u1. Scale α such that α1 = 1 and α2 > 1. Lemma 3.2 implies that x2
2 = 0. As x12 > 0,
α(x1 − x2) > e(x1 − x2). Furthermore x3 ≤ x2 which implies that x31 < u1 and therefore x3
2 = 0 byLemma 3.2. Therefore α(x3 − x4) = e(x3 − x4). Combining this fact and α(x1 − x2) > e(x1 − x2) with(3) and (2) leads to a contradiction.
Therefore, we have shown that α1 and α2 must be the same.
Combining Theorems 2.5 and 3.10 leads to the following result:
Theorem 3.11. When n = 2, all non-trivial facet-defining inequalities for conv(S) are of the form:∑i∈I
xi + γ1y1 + γ2y2 ≥ β
where I ⊆ {1, . . . ,m} and x+ γ1yj + γ2y2 ≥ β is facet-defining for the set
Q(b′, u′) = conv{
(x, y) ∈ R× Z2 : x+ c1y1 + c2y2 ≥ b′, u′ ≥ x ≥ 0, y ≥ 0}.
where b′ = b−∑
i 6∈I ui and u′ =∑
i∈I ui.
Therefore, when n = 2, all nontrivial facet-defining inequalities of conv(S) are obtainable from facetsof 3-variable mixed-integer sets of the form Q(b, u), we next study sets of this form (or equivalently, the setS when m = 1 and n = 2).
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4 The structure of Q(b, u)
Given fixed positive integers c1, c2, for any positive integers b, u (u can also be infinity) recall that the set
Q(b, u) = conv({(x, y) ∈ R1+ × Z2
+ : x+ c1y1 + c2y2 ≥ b, x ≤ u}).
The main result we will prove in this section is that
Q(b, u) = Q(b,∞) ∩ {(x, y) ∈ R3 : x ≤ u} ∩ (R× P≥(b− u))
whereP≥(b− u) = conv({y ∈ Z2
+ : c1y1 + c2y2 ≥ b− u}).
In other words, we will show that every nontrivial facet-defining inequality for Q(b, u) either defines afacet of Q(b,∞) (i.e., u plays no role) or the coefficient of the x variable is zero and the inequality (whentreated as an inequality on the variables y1 and y2) defines a facet of P≥(b− u). The latter set correspondsto the convex hull of integer points in Q(b, u) with x = u.
As discussed before, Agra and Constantino gave a polynomial-time algorithm to enumerate the facetsof P≥(b) for any b (and therefore for P≥(b − u)), and then extended their algorithm to obtain all facets ofQ(b,∞). Our result implies that one can thus use their algorithm to get all nontrivial facets of Q(b, u).
To analyze the facets of Q(b, u), we study lower-dimensional sets, defined in the space of the integervariables only. Accordingly, in addition to P≥(b− u), we also define
P≤(b) = conv({y ∈ Z2+ : c1y1 + c2y2 ≤ b}),
P (b− u, b) = conv({y ∈ Z2+ : b− u ≤ c1y1 + c2y2 ≤ b}).
We next study the integral points of Q(b, u) that lie on the capacity constraint. The projection of thesepoints on the y-coordinates gives the set P (b − u, b), the convex hull of all integer points between twoparallel hyperplanes.
4.1 The Convex Hull of P (b− u, b)
Let c ∈ R2+ with c > 0, and consider real numbers b, u > 0 with b− u ≥ 0.
Theorem 4.1. P (b− u, b) = P≤(b) ∩ P≥(b− u).
Proof. For ease of notation, we let Q0 = P (b− u, b), Q1 = P≤(b) and Q2 = P≥(b− u). Furthermore, let
P0 = {y ∈ R2+ : b− u ≤ cy ≤ b},
P1 = {y ∈ R2+ : cy ≤ b},
P2 = {y ∈ R2+ : cy ≥ b− u}.
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Therefore Qi = conv(Pi∩Z2) for i = 0, 1, 2. Note that P0 = P1∩P2 and therefore Q0 ⊆ Q1∩Q2. Q2 is afull-dimensional anti-blocking polyhedron with facets defined by inequalities of the form y1 ≥ 0 or y2 ≥ 0
or cy ≥ γ for c > 0 and γ > 0. Similarly Q1 is a blocking polyhedron; if full-dimensional, its facets aredefined by inequalities of the form y1 ≥ 0 or y2 ≥ 0 or cy ≤ γ for c ≥ 0 and γ > 0.
Assume that Q1 ∩ Q2 6⊆ Q0. Then Q1 ∩ Q2 is not an integral polyhedron; otherwise Q1 ∩ Q2 wouldbe the convex hull of integer points in P1 ∩ P2, and thus would be contained in Q0. We can assume Q1 isfull-dimensional, for if it were not, then Q1 ⊆ {y ∈ R2 : y1 = 0} ∪ {y ∈ R2 : y2 = 0}, and in that case,Q1 ∩Q2 is easily seen to be an integral polyhedron.
Let v be a nonintegral vertex of Q1 ∩ Q2. As Q1, Q2 ⊂ R2, we can assume v = f1 ∩ f2, wheref1 = conv(p1, p2) is a facet of Q1 and f2 = conv(q1, q2) is a facet of Q2. Let c1y ≤ γ1 be the inequalitydefining f1, and let c2y ≥ γ2 be the inequality defining f2; these inequalities are unique (up to multiplicationby a scalar) as Q1, Q2 are full-dimensional.
As v is nonintegral it cannot equal any of the endpoints of f1 or f2 and therefore must lie in the relativeinterior of each facet. Furthermore, one of the end points of f1, say p2, must strictly satisfy c2y ≥ γ2. Thenp1 strictly violates c2y ≥ γ2. This implies that p1 ∈ R2
+ \ P2 as c2y ≥ γ2 is valid for all integral points inP2. As f2 is entirely contained in P2 and intersects f1 = conv(p1, p2), this means that p2 must be containedin P2; thus p2 ∈ P1∩P2. Similarly, we can assume q1 strictly violates c1y ≤ γ1 and therefore q1 ∈ R2
+ \P1.Furthermore, q2 strictly satisfies c1y ≤ γ1 and also belongs to P1, otherwise f2 would not intersect f1 whichis contained in P1. Therefore, q2 ∈ P1 ∩ P2. In other words, we have
c1q2 < γ1, c1q1 > γ1, (4)
c2p2 > γ2, c2p1 < γ2, (5)
cp1 < b− u, b− u ≤ cp2 ≤ b, (6)
cq1 > b, b− u ≤ cq2 ≤ b. (7)
It is clear that the lines c1y = γ1 and c2y = γ2 have different slopes as the points p1, p2, q1, q2 and v arenot collinear. Then c1
2/c11 6= c2
2/c21 (here c1
1 stands for the first component of c1, c12 for the second component,
etc.). Note that c2 > 0 and so c22/c
21 is a positive number. As for c1, c1
1 may equal zero, in which case wewill think of c1
2/c11 as the ‘number’∞ and greater than any positive number. We can assume, without loss
of generality, that c12/c
11 > c2
2/c21; if c1
2/c11 < c2
2/c21, we can switch the coefficients of c and construct an
instance with the desired relationship of slopes of the lines c1y = γ1 and c2y = γ2. See Figure 1 for adepiction of p1, p2, q1, q2 and v.
Let δ = p2 − q2 ∈ Z2. As c1q2 < γ1 and c1p2 = γ1, we have c1δ = c1(p2 − q2) > 0. Similarly,as c2p2 > γ2 and c2q2 = γ2, we have c2δ > 0. Note that as c1 ≥ 0, c1δ > 0 implies that at least onecomponent of δ is positive. Also, as cp2 ≤ b and cq2 ≥ b− u, we have
cδ ≤ u. (8)
Case 1: δ1 ≥ 0. Clearly p1 + δ ∈ Z2. Then c1(p1 + δ) > γ1 as c1δ > 0. Further, cp1 < b − u and(8) together imply that c(p1 + δ) < b. Finally, as the second component of p1 is greater than the second
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y1
y2
p1
p2
q1
q2
vc1y ≤ γ1
c2y ≥ γ2
δ
cy ≥ b− u cy ≤ b
Figure 1: Facets f1 and f2 and their intersection v
component of q2 (because of the relationship between the slopes of the lines c1y = γ1 and c2y = γ2) andq2 + δ = p2 ∈ R2
+, we have p1 + δ ∈ R2+. In other words, p1 + δ is an integral point in P1 which violates
c1y ≤ γ1 a contradiction to the fact that this inequality is valid for all integral points in P1.Case 2: δ1 < 0, δ2 > 0. First q1 − δ ∈ Z2. Next, c2(q1 − δ) < γ2 as c2δ > 0. Further, cq1 > b and (8)
imply that c(q1 − δ) > b− u. Finally, as the second component of q1 is greater than the second componentof p2 and p2 − δ = q2 ∈ R2
+, we have q1 − δ ∈ R2+. In other words, q1 − δ is an integral point in P2 which
violates c2y ≥ γ2. This contradicts the fact that c2y ≥ γ2 is valid for all integral points in P2.
We note that a closely related result appears in an unpublished manuscript of Basu, Bonami, Confortiand Cornuejols where the authors study integer programs with two variables where one of the variables hasboth an upper and a lower bound.
4.2 The Convex Hull of Q(b, u)
We need the following easy lemma before we prove the main result of this section in Theorem 4.3.
Lemma 4.2. Let y1, y2, y3 ∈ Z2 be three distinct points such that conv(y1, y2, y3) contains no other integerpoint and let y2, y3 lie on the line γy = β where the coefficients of γ ∈ Z2 are coprime integers. Thenβ − 1 ≤ γy1 ≤ β + 1.
Proof. As γ, y1, y2 are integral, so is β. We can also assume, without loss of generality, that γy1 ≤ β
(by multiplying γ, β by -1 if necessary). Also, there is nothing to prove if γy1 = β, so we assume γy1
is an integer less than β. As γ has coprime coefficients, there is a 2 × 2 unimodular matrix U such that
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γ = γU = (1, 0). Consider the points yi = U−1yi − U−1y3 for i = 1, . . . , 3. Then conv(y1, y2, y3)
contains no integer points other than y1, y2, y3. Furthermore,
γUU−1yj = γyj ⇒ γyj = γyj − γy3 = 0 for j = 2, 3 and γy1 < 0.
Therefore y3 = (0, 0), y2 lies on the line y1 = 0 and y11 < 0. As conv(y2, y3) contains no integer points
other than y2, y3, it follows that y2 is either (0, 1) or (0,−1). In either case, if y11 ≤ −2 then conv(y1, y2, y3)
has an integer point besides y1, y2, y3. Therefore y11 = γy1 = −1 and γy1 = β − 1.
Theorem 4.3.
Q(b, u) = Q(b,∞) ∩ {(x, y) ∈ R× R2 : x ≤ u} ∩ (R× P≥(b− u)). (9)
Proof. Q(b, u) is a subset of each of the three sets on the right-hand side of (9), and therefore Q(b, u) is asubset of their intersection. To prove the reverse inclusion, we next show that each facet-defining inequalityof Q(b, u) is a valid inequality for one of the three right-hand-side sets.
Any trivial facet-defining inequality for Q(b, u) is either valid for Q(b,∞) or for {(x, y) ∈ R × R2 :
x ≤ u}. Therefore let αx + γy ≥ β define a nontrivial facet F of Q(b, u). Scale the inequality so that γis integral and the components of γ are coprime. As Q(b, u) is a special case of the set conv(S) studied inLemma 2.1, we can conclude thatQ(b, u) is full-dimensional and α ≥ 0, γ > 0 and β > 0. Let α = 0. ThenLemma 2.2 implies that γy ≥ β is facet-defining for P≥(b − u) (this is exactly the set S′ in Lemma 2.2).Therefore 0x + γy ≥ β is facet-defining for R × P≥(b − u). We henceforth assume that α > 0. We willshow that under this condition αx+ γy ≥ β defines a facet of Q(b,∞).
Let L = {(x1, y1), (x2, y2), (x3, y3)} be a subset of three affinely independent integral points on Fsuch that conv(y1, y2, y3) is full-dimensional and contains no other integer points. These points exist byLemma 2.7. Without loss of generality, assume that x1 ≤ x2 ≤ x3. If xi > 0 for i = 1, 2, 3 then (xi, yi)
lies on the capacity inequality, contradicting the nontriviality of F , therefore x3 > 0. If all three points in Lsatisfy xi = 0, then F is defined by x ≥ 0 contradicting the nontriviality of F , therefore x1 = 0. We nextconsider two cases.
Case 1: Let x2 = 0. Recall that x3 = 0 and u ≥ x1 > 0. Therefore γy1 < γy2 = γy3 = β. As γis integral (by scaling) and so is y2, β is an integer. By Lemma 2.7, conv(y1, y2, y3) is a full-dimensionalpolyhedron in R2 containing no integer points other than y1, y2, y3. Lemma 4.2 implies that γy1 = β − 1.
We will now show that αx+ γy ≥ β is valid and facet-defining for Q(b,∞). Clearly, if this inequalityis valid, then it is facet-defining as the inequality is tight for the points (xi, yi) for i = 1, 2, 3 which arecontained in Q(b,∞). Suppose αx+γy ≥ β is not valid for Q(b,∞). Then there is a point in Q(b,∞) thatviolates this inequality and its x-coordinate is strictly larger than u. Therefore, for some α > α, αx+γy ≥ βis facet-defining for Q(b,∞); this cannot be facet-defining for Q(b, u) (as it would be implied by the validinequalities αx + γy ≥ β and x ≥ 0 for Q(b, u)). There must be an integral point (x, y) with x > 0 inQ(b,∞) satisfying αx+ γy = β; let (x, y) be chosen so that x is as small as possible. For any such point,x > u, otherwise αx+ γy ≥ β would be facet-defining for Q(b, u).
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Let the convex hull of {y, y2, y3} be the triangle H , and let F stand for the set of integral points on thefacet of Q(b,∞) defined by αx + γy ≥ β. Suppose H contains some integral point y′ different from thethree vertices of H . Then it equals λ1y + λ2y
2 + λ3y3 where λ1 + λ2 + λ3 = 1 and 0 ≤ λ1, λ2, λ3 < 1. If
λ1 = 0, then y′ is contained in the convex hull of y2, y3 which contradicts the assumption that this convexhull contains no other integer points besides y2, y3. Therefore 0 < λ1 < 1. Let x′ = (β − γy′)/α. Then(x′, y′) satisfies αx + γy = β and (x′, y′) = λ1(x, y) + λ2(x2, y2) + λ3(x3, y3). In other words, (x′, y′)
is a point in F with u < x′ < x which contradicts our assumption on the minimality of x. Thereforeconv(y, y2, y3) contains no other integer points besides y, y2, y3, and Lemma 4.2 implies that γy = β − 1
and αx = 1. But then 1 = αx > αx1 = 1 as x1 ≤ u < x and α < α. Thus we obtain a contradiction.Case 2: Let x2 > 0. Recall that x1 = 0 and x3 > 0. Let Fc denote the face of Q(b, u) defined by the
capacity inequality. As before, we can argue that (x2, y2) and (x3, y3) lie on Fc. Now consider the set ofpoints in Fc satisfying αx+γy ≥ β. As points on Fc satisfy x = b− cy, substituting for x in αx+γy ≥ β,we get
α(b− cy) + γy ≥ β
and therefore(γ − αc)y ≥ β − αb
as a valid inequality for {(x, y) ∈ Fc : αx+ γy ≥ β}. Let γ′ = γ − αc and β′ = β − αb. Then
α(x+ cy ≥ b) + (0x+ γ′y ≥ β′) = (αx+ γy ≥ β), (10)
where multiplying an inequality by a nonnegative number and adding two inequalities has the usual meaning.As αx + γy ≥ β is valid for all points in Q(b, u), all integral points (x, y) ∈ Fc satisfy γ′y ≥ β′;
furthermore for such points we have x + cy = b with 0 ≤ x ≤ u ⇒ b − u ≤ cy ≤ b. Also, as theinequality αx + γy ≥ β is tight for the points (x2, y2), (x3, y3) which also lie in Fc, we have γ′y2 = β′
and γ′y3 = β′. In other words, γ′y ≥ β′ is both valid for P (b − u, b) and facet-defining. By our previousresults, P (b − u, b) = P≥(b − u) ∩ P≤(b). Therefore γ′y ≥ β′ defines a facet of either P≥(b − u) or ofP≤(b).
Case 2a: Let γ′y ≥ β′ define a facet of P≥(b − u). In other words, γ′y ≥ β′ is valid for all integraly ≥ 0 with cy ≥ b − u. But any integral (x, y) ∈ Q(b, u) satisfies y ≥ 0 and cy ≥ b − u. Therefore0x + γ′y ≥ β′ is a valid inequality for Q(b, u). But then by (10), αx + γy ≥ β is the sum of two distinctvalid inequalities for Q(b, u) and cannot define a facet of Q(b, u), a contradiction.
Case 2b: Let γ′y ≥ β′ define a facet of P≤(b). If αx+ γy ≥ β does not define a facet of Q(b,∞) thenthere exists an integral point (x, y) ∈ Q(b,∞) \ Q(b, u) with αx + γy < β and x > u and x + cy ≥ b.Therefore x ≥ max{u, b−cy}. If u ≥ b−cy, then setting x to u, we get a point which violates αx+γy ≥ βbut belongs to Q(b, u), a contradiction. So we can assume that setting x to b − cy, we get an integral point(x, y) ∈ Q(b,∞) \ Q(b, u) which violates αx + γy ≥ β and lies on the capacity constraint. But theny ∈ P≤(b) and therefore γ′y ≥ β′. This, combined with x + cy = b and (10) implies that αx + γy ≥ β, acontradiction.
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5 Final Remarks
We have shown that
conv({
(x, y) ∈ Rm+ × Z2+ :
m∑i=1
xi +2∑j=1
cjyj ≥ b, x ≤ u}
)
=⋂T⊆M
{(x, y) ∈ Rm+ × Z2
+ : (∑i∈T
xi, y) ∈ PT}⋂{
(x, y) ∈ Rm+ × R2+ : x ≤ u
},
where M = {1, . . . ,m} and
PT = conv((xT , y) ∈ R1+ × Z2
+ : xT + cy ≥ b− u(M \ T )).
For a fixed set T ⊆ M, it is possible to use the results of Agra and Constantino [2] to enumerate allfacet-defining inequalities for PT in polynomial-time. As there is an exponential number of choices for theset T , this does not lead directly to a polynomial time separation algorithm for S.
For general m ≥ 1, the optimization problem min{px + qy : (x, y) ∈ S} can be solved by solvingat most m three-variable integer programs. Assume that the variables are indexed in such a way that p1 ≤· · · ≤ pm and consider an optimal solution (x, y). x must be an optimal solution of the linear programmin{px :
∑mi=1 xi ≥ b − cy, 0 ≤ x ≤ u}. An optimal extreme point solution can be constructed greedily
and has at most one variable strictly between its bounds. It follows that there is an (alternative)-optimalsolution (x, y) with xi = ui for i < k, xi = 0 for i > k, and 0 ≤ xk ≤ uk for some value of k ∈ {1, . . . ,m}.But then (x, y) is also an optimal solution to
∑i:i<k
piui + min{pkxk + qy : xk +
2∑j=1
cjyj ≥ b−∑i:i<k
ui, 0 ≤ xk ≤ uk, y ∈ Z2+}.
Thus it suffices to solve the m three variable problems obtained as k varies from 1 to m. Each of theseproblems can be solved in polynomial-time using the results of Agra and Constantino [2].
It follows that the separation problem for conv(S) can be solved in polynomial time using the ellipsoidalgorithm. However it would be preferable to have a more practical algorithm: a natural conjecture is thatit suffices to order the variables such that x
∗1u1≥ · · · ≥ x∗m
umand then separate over the m + 1 sets PTi where
Ti = {i, i + 1, . . . ,m} and i = 1, . . . ,m + 1. This provides a polynomial algorithm in the case of n = 1,see Atamturk and Rajan [4] and in a mixing set variant, see Di Summa and Wolsey [5].
With three or more integer variables, the sets
PnT = conv((xT , y) ∈ R1+ × Zn+ : xT + cy ≥ b− u(M \ T ))
still lead to valid relaxations for conv(S), but the main results of Section 4 do not generalize. For anarbitrary number of integer variables, similar results to those of Section 4 hold when the coefficients aredivisible (Wolsey and Yaman [9]); in particular similar relaxations give a complete description of conv(S).
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For the remainder of this section, we assume that P≥(b), P (b − u, b), P≤(b) and Q(b, u) are defined ina similar fashion to the definitions in Section 4 when y ∈ R3. For example, for given coefficients c1, c2, c3,P≥(b) = conv{y ∈ Z3 : c1y1 + c2y2 + c3y3 ≥ b}.
Remark 5.1. Theorem 4.1 does not generalize to the case n = 3.
Consider the set
P (94, 97) = conv{y ∈ Z3+ : 94 ≤ 5y1 + 13y2 + 22y3 ≤ 97}.
The point (83 ,
23 ,
103 ) lies in P≤(97)∩P≥(94) as it can be written as 1/3(1, 0, 4)+1/3(1, 2, 3)+1/3(6, 0, 3) ∈
P≤(97) and as 1/3(2, 0, 4)+2/3(3, 1, 3) ∈ P≥(94). However it is cut off by the valid inequality (also facet-defining) y3 ≤ 3 of P (94, 97). To see that all solutions of P (94, 97) satisfy y3 ≤ 3, first note that y3 ≤ 4
is a valid inequality: y3 ≤ 97/22 follows from the nonnegativity of y and we can round down the right-hand-side of this inequality as y3 is integral. Observe that if (y1, y2, 4) is a non-negative integral point inP (94, 97), then 5y1 + 13y2 ∈ [6, 9] which is not possible.
Remark 5.2. Theorem 4.3 does not generalize in the case n = 3.
Consider the set
Q(97, 3) = conv{(x, y) ∈ R1+ × Z3
+ : x+ 5y1 + 13y2 + 22y3 ≥ 97, x ≤ 3}.
The point (53 ,
83 ,
23 ,
103 ) = 1/3(4, 1, 0, 4)+1/3(0, 1, 2, 3)+1/3(1, 6, 0, 3) ∈ Q(97,∞). From Remark 5.1, it
lies in R1+×P≥(94) and clearly 0 ≤ x ≤ 3. However it is cut off by the valid inequality (also facet-defining)
x + 5y1 + 13y2 + 21y3 ≥ 94 of Q(97, 3). To see validity, note that this inequality is valid when y3 ≤ 3
(simply add −y3 ≥ −3 to the capacity inequality). There are no points in Q(97, 3) with y3 ≥ 5. Finally, forpoints (x, y) in Q(97, 3) with y3 = 4, x + 5y1 + 13y2 ≥ 9. Clearly x + 5y1 + 13y2 ≥ 9 and 0 ≤ x ≤ 3
together imply that x + 5y1 + 13y2 ≥ 10 as there is no solution to x + 5y1 = 9 with y1 a non-negativeinteger and x ∈ [0, 3]. But then x+ 5y1 + 13y2 + 21y3 ≥ 94.
Proposition 5.3. The convex hull of the following set has a facet-defining inequality with distinct nonzerocoefficients for the continuous variables:
S = {(x, y) ∈ R2+ × Z3
+ : x1 + x2 + 5y1 + 13y2 + 22y3 ≥ 97, x1, x2 ≤ 3}.
Proof. We will show thatx1 + 2x2 + 5y1 + 13y2 + 21y3 ≥ 94 (11)
is facet-defining for S. As before, it is easy to see that (11) is valid for points in S with y3 ≤ 3. There areno points in S with y3 ≥ 5. To see the validity when y3 = 4, consider a point (x, y) ∈ S with y3 = 4.Then x1 + x2 + 5y1 + 13y2 ≥ 9. Therefore either x1 + x2 + 5y1 + 13y2 = 9 (and y2 = 0 and x2 ≥ 1 as0 ≤ x1 ≤ 3 and y1 is integral) or x1 + x2 + 5y1 + 13y2 ≥ 10. In either case, x1 + 2x2 + 5y1 + 13y2 ≥ 10
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and the validity of (11) follows when y3 = 4. The linearly independent points (0, 0, 2, 0, 4), (0, 0, 1, 2, 3),(1, 0, 6, 0, 3), (3, 1, 1, 0, 4), (3, 0, 3, 1, 3) show that it defines a facet.
Previously, the smallest instance with a nontrivial facet with distinct coefficients for the x variables ofwhich we are aware had n = 8 integer variables [8].
Some questions remain open. Are there some more general conditions under which Theorem 4.1 holds?Theorem 4.1 is a necessary condition for Theorem 4.3 to hold. Are there cases in which it is also sufficient?
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