2014/7
■
The continuous knapsack set
Sanjeeb Dash, Oktay Günlük
and Laurence A. Wolsey
DISCUSSION PAPER
Center for Operations Research
and Econometrics
Voie du Roman Pays, 34
B-1348 Louvain-la-Neuve
Belgium
http://www.uclouvain.be/core
CORE DISCUSSION PAPER
2014/7
The continuous knapsack set
Sanjeeb DASH 1, Oktay GÜNLÜNK 2
and Laurence A. WOLSEY 3
February 2014
Abstract
We study the convex hull of the continuous knapsack set which consists of a single inequality constraint
with n non-negative integer and m non-negative bounded continuous variables. When n = 1, this set is a
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 the
continuous variables is bounded by 2n − n. Our next result is to show that when n = 2, this upper bound
is actually 1. This implies that when n = 2, the coefficients of the continuous variables in any facetdefining inequality are either 0 or 1 after scaling, and that all the facets can be obtained from facets of
continuous knapsack sets with m = 1. The convex hull of the sets with n = 2 and m = 1 is then shown to
be given by facets of either two-variable pure-integer knapsack sets or continuous knapsack sets with n
= 2 and m = 1 in which the continuous variable is unbounded. The convex hull of these two sets 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.
Keywords: continuous knapsack set, single arc flow set, convex hull, facet coefficients.
JEL classification: 90C11, 90C26.
1
IBM Watson Research Center, New York, USA.
IBM Watson Research Center, New York, USA.
3
Université catholique de Louvain, CORE, B-1348 Louvain-la-Neuve, Belgium.
E-mail: [email protected]
2
Part of the research of the third author was carried out during a stay at the IBM Watson Research Center, Yorktown
Heights and a stay at ETH Zurich, supported by FIM - Institute for Mathematical Research, ETH Zurich. This text
presents research results of the Belgian Program on Interuniversity Poles of Attraction initiated by the Belgian State,
Prime Minister’s Office, Science Policy Programming, contract no. P7/36, Combinatorial Optimization:
Metaheuristics and Exact Methods. The scientific responsibility is assumed by the authors.
1 Introduction
In this paper we study the set S = SLP ∩ (Rm × Zn ) where
n
m
o
n
X
X
cj yj ≥ b, u ≥ x ≥ 0, y ≥ 0
xi +
SLP = (x, y) ∈ Rm × Rn :
j=1
i=1
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 residual
capacity inequalities (see Section 1.1).
We study the properties of facet-defining inequalities for the case n ≥ 2 and characterize the convex
hull of S when n = 2. More precisely, we first prove in Section 2 that in any facet-defining inequality, the
number of distinct non-zero coefficients of the continuous variables is at most 2n − n. We then study the
case when n = 2 in detail, and show in Section 3 that all non-trivial facet-defining inequalities for conv(S)
are of the form:
X
xi + γ1 y1 + γ2 y2 ≥ β
i∈I
where I ⊆ {1, . . . , m} and x + γ1 y1 + γ2 y2 ≥ β is facet-defining for the set
n
o
Q(b′ , u′ ) = conv (x, y) ∈ R × Z2 : x + c1 y1 + c2 y2 ≥ b′ , u′ ≥ x ≥ 0, y ≥ 0 ,
P
P
where b′ = b − i6∈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, satisfying
x + c1 y1 + c2 y2 ≥ b.
We then analyze the facial structure of Q(b, u) in Section 4. We show that non-trivial facet-defining
inequalities either have a zero coefficient for the x variable and therefore are facet-defining for
n
o
conv (x, y) ∈ R × Z2 : c1 y1 + c2 y2 ≥ b − u, y ≥ 0 ,
or they can be obtained from a relaxation in which the upper bound is dropped:
n
o
Q(b, ∞) = conv (x, y) ∈ R × Z2 : x + c1 y1 + c2 y2 ≥ b, x ≥ 0, y ≥ 0 .
For Q(b, ∞) for any b > 0, Agra and Constantino [2] gave a polynomial-time algorithm to enumerate
all facet-defining inequalities. Note that S has an exponential number of relaxations of type Q(b, u), one
for each I ⊆ {1, . . . , m}, and therefore our result does not lead directly to a polynomial-time separation
algorithm for S. However, given a fixed objective function, we show that optimization over S can be carried
out by solving m three variable problems, and thus the separation problem is also polynomially solvable
using the ellipsoid algorithm.
Finally, in Section 5, we show that our results cannot be generalized to n ≥ 3. In particular, we present
a facet of a particular set with n = 3 such that the non-zero coefficients of the continuous variables in the
associated inequality take different values.
2
1.1 Related Literature
Magnanti, Mirchandani, and Vachani [6] studied the single arc design problem as a subproblem of the
network loading problem. In particular, they studied the set
n
(x, y) ∈ Rm × Z :
m
X
i=1
o
xi ≤ cy, u ≥ x ≥ 0, y ≥ 0 .
The network loading problem is the problem of choosing arc capacities of minimum cost in a network
so as to enable flows of different quantities (ui ) between m pairs of nodes, or equivalently to enable a
multicommodity flow. On any arc, capacities must be chosen in integral multiples of a fixed constant (say
c). The single arc design problem is the subproblem which enforces the fact that the flow through an arc for
any commodity must be bounded above by the corresponding demand value (ui ), and the sum of flows is at
most 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
n
(x, y) ∈ Rm × Z :
m
X
i=1
o
xi + cy ≥ b, u ≥ x ≥ 0, y ≥ 0 ,
Pm
where b = 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 Atamtürk and
Rajan [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, our
P
result 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′ =
P
P
b − i6∈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′ − ⌊ bc ⌋c)y ≥ (b′ − ⌊ bc ⌋)⌈ bc ⌉ or
by y ≥ ⌈(b′ − u′ )/c⌉ when I = ∅ . Thus our results yield an alternative proof of the result of Magnanti et al.
Atamtürk and Günlük [1] describe the set studied by Magnanti et. al. as the splittable flow arc set which
they studied as a subproblem of the multicommodity network design problem. In particular, they studied the
more general version which is equivalent to S with n = 1 and arbitrary b, and used the results in Magnanti
et. 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 the
convex hull for a special case of the single arc problem with two capacities (S with n = 2) in which all the
data ui and cj are integer, c1 = 1 and c2 > 1. Note that in this case the associated capacity variable y1 can
be treated as continuous. Yaman [10] studied an extension of this version in which the capacity variables
also 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 , and
show that the coefficients of the continuous variables in any facet-defining inequality lie in {0, 1} for all
values of m and n.
3
2 Coefficients of continuous variables in facet-defining inequalities
In this section we consider a non-trivial facet-defining inequality
m
X
αi xi +
i=1
n
X
γj yj ≥ β
j=1
of conv(S) and study the properties of the vector α. Let F = {(x, y) ∈ S : αx + γy = β} denote the set of
points in S that satisfy this inequality as equality.
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 a
one in the ith component and zeros elsewhere and e ∈ Rm denote the vector of all ones. We let ēi stand for
the unit vector in Rn with a one in the ith component. We call the following inequalities that appear in the
description 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 as
trivial 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.
P
(c) Assume that m
i=1 ui > b. Then the capacity inequality is facet-defining if and only if b ≥ cj for all
j = 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, ⌈b/c1 ⌉ē1 ), pi = w + (ui ei , 0) for
i = 1, . . . , m, and qj = w + (0, ēj ) for j = 1, . . . , n. These points are affinely independent as the points
pi − w and qj − w are linearly independent. The recession cone of conv(S) is C as c > 0 and the continuous
variables are bounded.
(b) For each i = 1, . . . , m, there are m + n points with xi = 0 among the (n + m + 1) affinely
independent points given above. Furthermore by adding (ui ei , 0) to each of these n + m points one again
obtains affinely independent points. Therefore the bound constraints on the continuous variables are facetdefining. 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 the
coordinate 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, note
that in this case the inequality cannot be tight if yj ≥ 1. Now assume that b ≥ cj for all j = 1, . . . n.
4
P
b−cj
Let µ = m
i=1 ui . The following points are affinely independent: qj = ( µ u, ēj ) ∈ S for j = 1, . . . , n.
Now consider the point p0 = ( b−ǫ
µ u, 0) 6∈ S for some ǫ > 0. If ǫ is small enough, the following points
pi = p0 + (ǫei , 0) ∈ S for i = 1, . . . , m. These points are affinely independent from the points constructed
above 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 this
inequality. For any i = 1, . . . , m, there is a point (x, y) ∈ F with xi < ui , and for some small ǫ > 0, we
have (x, y) + (ǫei , 0) ∈ S . This implies that α ≥ 0. In addition, as the recession cone of conv(S) is C, we
have γ ≥ 0. As the inequality is not implied by the non-negativity constraints, β > 0. Finally, if γi = 0 for
some i in {1, . . . , n}, then αx + γy ≥ β is violated by (0, ⌈b/ci ⌉ēi ) ∈ S. Therefore γ > 0.
(e) As x ≤ u for all (x, y) ∈ S and γ > 0 for all non-trivial facet-defining inequalities, the claim
holds.
Note that above we have only given some necessary conditions for the capacity inequality to be facetdefining. 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 be
obtained from a lower-dimensional relaxation of S. Remember that F denotes the set of points in S that
satisfy this inequality as equality. We next present two observations that lead to the main result of this
section. First we consider the case when some of the entries of α are zero.
Lemma 2.2. If αm = 0, then
Pm−1
i=1
αi xi +
Pn
j=1 γj yj
≥ β is facet-defining for the set
n
m−1
o
n
X
X
cj yj ≥ b − um , y ≥ 0, u′ ≥ x ≥ 0
xi +
S ′ = conv (x, y) ∈ Rm−1 × Zn :
j=1
i=1
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 for
P
Pn
′
conv(S) ∩ Xm , then m−1
i=1 αi xi +
j=1 γj yj ≥ β is valid for S . We next argue that the inequality is
facet-defining for S ′ .
Let pk = (xk , y k ) be a collection of m + n affinely independent points in F , which exist as conv(F )
is a facet of conv(S). Let p̂k = (x̂k , y k ) ∈ Rm−1 × Zn be obtained from pk by deleting the last entry of
xk for all k = 1, . . . , m + n. Also let α̂ ∈ Rm−1 be obtained from α by deleting the last entry. Notice
Pm k
Pm k
P
k
k
′
that m−1
i=1 xi − um and therefore p̂ ∈ S for all k = 1, . . . , m + n.
i=1 x̂i =
i=1 xi − xm ≥
Furthermore, α̂x̂k = αxk , and consequently α̂x̂k + γy k = β for all k = 1, . . . , m + n. As the affine rank of
{p̂1 , . . . , p̂m+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.
5
Pn
Corollary 2.3. If α = 0, then γy ≥ β is facet-defining for S ′ = conv{y ∈ Zn :
j=1 cj yj ≥ b −
Pm
Pm
i=1 ui , y ≥ 0}. In addition, if
i=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.
P
Pn
Lemma 2.4. If αm−1 = αm , then m−1
i=1 αi xi +
j=1 γj yj ≥ β is facet-defining for the set
n
m−1
o
n
X
X
ci yi ≥ b, y ≥ 0, u′ ≥ x ≥ 0
xi +
S ′′ = conv (x, y) ∈ Rm−1 × Zn :
j=1
i=1
where
u′i
= ui for i = 1, . . . , m − 2 and
u′m−1
= um−1 + um .
Pn−1
P
Proof. The proof is very similar to the proof of Lemma 2.2. First we observe that i=1
αi xi + nj=1 γj yj ≥
β is valid for S ′′ provided that αx + γy ≥ β is valid for S. Then, we modify the points pi defined in
Lemma 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, α̂1 , . . . , α̂t } for all i = 1, . . . , m where α̂1 , . . . , α̂t are distinct positive
P
P
numbers. Then ti=1 α̂i x̂i + nj=1 γj yj ≥ β is facet-defining for
n
t
o
n
X
X
cj yj ≥ b − u0 , y ≥ 0, û ≥ x̂ ≥ 0
x̂i +
Ŝ = conv (x̂, y) ∈ Rt × Zn :
j=1
i=1
where ûj =
P
k:αk =α̂j
uk for j = 1, . . . , t and u0 =
P
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 form
Ŝ above, obtained by combining continuous variables, it is not always possible to lift them in the obvious
way 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 are
distinct (α may have a single component). Remember that F denotes the set of points in S that satisfy this
inequality 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 , ŷ), (x2 , ŷ) ∈ F then x1 = x2 .
Proof. Clearly αx1 + γ ŷ = αx2 + γ ŷ = β and therefore αx1 = αx2 . Assume x1 6= x2 . If α ∈ R, then
the result trivially follows. Otherwise there must exist two indices i and j such that x1i 6= x2i and x1j 6= x2j
and therefore x̂ = 12 x1 + 12 x2 has ui > x̂i > 0 and uj > x̂j > 0. Note that (x̂, ŷ) ∈ F ⊆ S. Now assume
αi > αj and notice that for some small ǫ > 0 a new point x′ obtained by reducing x̂i by ǫ and increasing
x̂j by ǫ gives (x′ , ŷ) ∈ S. This point (x′ , ŷ), however, violates the facet-defining inequality as αx′ < αx̂, a
contradiction.
6
Lemma 2.7. Assume that α > 0 has all distinct coefficients. Then F contains a subset of m + n affinely
independent points {(xi , y i ) : i = 1, . . . , m + n} such that conv(y 1 , . . . , y m+n ) is (i) full-dimensional, (ii)
has m + n integral vertices and contains no other integer points.
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 , y i ) : i = 1, . . . , m + n} stand for the one with minimum
number of integer points in conv(Ly ), where Ly = {y 1 , . . . , y m+n }.
(i) If conv(Ly ) is not full-dimensional, then there exists 0 6= γ ′ ∈ Rn , β ′ ∈ R such that γ ′ y i = β ′ for
i = 1, . . . , m + n. But this means that points in F satisfy the equation γ ′ y = β ′ , which contradicts the fact
that αx + γy = β is uniquely defined up to multiplication by a scalar and α > 0.
P
i
(ii) Suppose conv(Ly ) contains an integer point ȳ which is not a vertex. Then ȳ = m+n
i=1 µi y for some
Pm+n
Pm+n
µi ≥ 0 with i=1 µi = 1. Let x̄ = i=1 µi xi . Then (x̄, ȳ) is contained in F . Assume ȳ ∈ Ly . In
this case, let y k = ȳ for some k ≥ 0. Then we can assume µk = 0. Furthermore, Lemma 2.6 implies that
x̄ = xk , and therefore (xk , y k ) is a convex combination of other points in L, which contradicts the affine
independence of points in L.
Therefore we can assume all points in Ly are vertices of conv(Ly ), and ȳ 6∈ Ly . As the points in L are
affinely independent, for some index j ∈ {1, . . . , m + n} with µj > 0, the set L′ = L ∪ {(x̄, ȳ)}\{(xj , y j )}
is affinely independent. L′ also has the property that the convex hull of L′y is strictly contained in the convex
hull of Ly and has fewer integral points (it does not contain y j ). 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 continuous 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 distinct
non-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 proof
of Lemma 2.7. Suppose Ly contains more than 2n integer points. Then it contains at least two distinct
points, say y k and y l , with the same odd/even parity (that is, for all i: yik is odd if and only if yil is odd).
Consequently, ŷ = (y k + y l )/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 the
proof.
Corollary 2.9. If n = 2, then all facet-defining inequalities for conv(S) can be obtained from relaxations
of the form Ŝ 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 in
facet-defining inequalities of disjunctive relaxations of S. More precisely, we consider a disjunctive cut
7
αx + γy ≥ β for conv(S) that can be derived using the |K|-term disjunction D = ∪k∈K D k where
D k = (x, y) ∈ Rm × Rn : Ak y ≥ dk
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 (D k ∩ S LP ))
that is distinct from the bound constraints on the x variables. Note that αi ≥ 0 for all i (as the related facet
must contain a point with xi < ui and if αi < 0, this point can be perturbed to obtain a new point in Q
violating the inequality). Clearly some of the sets D k ∩ S LP can be empty. Without loss of generality,
assume that D k ∩ S LP 6= ∅ for k ∈ K̄ = {1, . . . , |K̄|} and D k ∩ S LP = ∅ for k > |K̄|.
As the inequality αx + γy ≥ β is valid for
D k ∩ S LP = (x, y) ∈ Rm × Rn : ex + cy ≥ b, Ak y ≥ dk , y ≥ 0, u ≥ x ≥ 0
for k ≤ |K̄|, there exist nonnegative multipliers θ k (associated with ex + cy ≥ b), η k (associated with
Ak y ≥ dk ), and λk (associated with −x ≥ −u) that yield the valid inequality
(θ k e − λk )x + (θ k c + η k Ak )y ≥ θ k b + η k dk − λk u
for D k ∩ S LP where α ≥ (θ k e − λk ), γ ≥ (η k Ak + θ k c) and β ≤ (θ k b + η k dk − λk u). Furthermore, as
αx + γy ≥ β is facet-defining for Q by assumption, we have (here ej is a unit vector in Rn with a one in
the j component)
αi = max{θ k − λki }, γj = max{θ k cj + η k Ak ej }, and β = min{θ k b + η k dk − λk u).
k∈K̄
k∈K̄
k∈K̄
Note that if θ̂ = mink∈K̄ {θ k } > 0, then decreasing all entries of the vector θ by θ̂ yields a stronger valid
inequality for Q. This is not possible as αx + γy ≥ β is facet-defining for Q. Consequently, we conclude
that mink∈K̄ {θ k } = 0. Without loss of generality, we assume that θ |K̄| ≥ θ |K̄|−1 ≥ . . . ≥ θ 1 = 0.
Using the multipliers θ and η (but not λ) it is easy to see that for all k ∈ K̄ the inequality θ k ex+γy ≥ β k ,
where β k = θ k b + η k dk , is valid for D k ∩ S LP . Consequently, for all k ∈ K̄ we can define the following
relaxation of the set D k ∩ S LP :
W k = (x, y) ∈ Rm × Rn : θ k ex + γy ≥ β k , u ≥ x ≥ 0 ⊇ D k ∩ S LP .
Notice that αx + γy ≥ β is valid for each W k and consequently, it is valid for W = conv(∪k∈K̄ W 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 associated
disjunctive relaxation, the vector α has at most 2(t − 1) distinct non-zero coefficients.
8
Proof. Using the notation introduced in the the preceding discussion, αx + γy ≥ β is valid for W k for
all k ∈ K̄, and consequently there exists a non-negative vector λk ∈ Rm for each k ∈ K̄ such that
αi ≥ θ k − λki and β ≤ βk − λk u. As αx + γy ≥ β is facet-defining for the associated disjunctive
relaxation, αi = maxk∈K̄ {θ k − λki } and β = mink∈K̄ {βk − λk u}. Clearly, λki ≥ θ k − αi and λki ≥ 0
implying λki ≥ (θ k − αi )+ for all i = 1, . . . , m and k ∈ K̄. Without loss of generality, we also assume that
λki = (θ k − αi )+ for all i = 1, . . . , m and k ∈ K̄.
We next argue that if αi , αj 6∈ {θ 1 , . . . , θ |K̄| }, then max{αi , αj } > θ l > min{αi , αj } for some l. As
αi = maxk∈K̄ {θ k − λki } ≤ θ |K̄| and θ 1 = 0 we have θ 1 ≤ αi ≤ θ |K| for all i. Assume that there exists
distinct v, w together with an index k1 such that θ k1 +1 > αv , αw > θ k1 . Clearly, λkv = λkw = 0 when
k ≤ k1 and λkv , λkw > 0, otherwise. Let ǫ > 0 be sufficiently small and define:
α(ǫ) =
and similarly,



α

 i
αv − ǫ/uv



αw + ǫ/uw



λki




 λk = 0
i
k
λi (ǫ) =
k


λv + ǫ/uv




λk − ǫ/uw
w
i 6∈ {u, v}
α(−ǫ) =
i=v
i=w



α

 i
αv + ǫ/uv



αw − ǫ/uw



λki




 λk = 0
i
k
λi (−ǫ) =
k


λv − ǫ/uv




λk + ǫ/uw
w
i 6∈ {u, v} and k ∈ K̄
i ∈ {u, v} and k ≤ k1
i = v and k ≥ k1 + 1
i = w and k ≥ k1 + 1
i 6∈ {u, v}
i=v
i=w
i 6∈ {u, v} and k ∈ K̄
i ∈ {u, v} and k ≤ k1
i = v and k ≥ k1 + 1
i = w and k ≥ k1 + 1.
Notice that for all k, we have λk (ǫ)u = λk (−ǫ)u = λk u. Consequently, β = mink∈K̄ {βk − λk (ǫ)u} =
mink∈K̄ {βk − λk (−ǫ)u}. Furthermore, as θ k1+1 > αv , αw > θ k1 by assumption and ǫ > 0 is small enough,
we also have αi (ǫ) = maxk∈K̄ {θ k −λki (ǫ)} and αi (−ǫ) = maxk∈K̄ {θ k −λki (−ǫ)} for all i. Therefore, both
αǫ x + γy ≥ β and α−ǫ x + γy ≥ β are valid inequalities for conv(∪k∈K̄ W k ). But in this case αx + γy ≥ β
cannot be facet-defining as α = (αǫ + α−ǫ )/2. We can therefore conclude that if αu , αw 6∈ {θ 1 , . . . , θ |K|},
then max{αu , αw } > θ l > min{αu , αw } for some l. Consequently, there can only be at most one αi that
lies between two consecutive θ entries. As θ 1 = 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 disjunction
with at most 2n -terms [3]. Consequently, Theorem 2.10 implies that if αx + γy ≥ β is facet-defining for
conv(S) then α has at most 2(2n − 1) distinct non-zero entries. This bound is weaker than that given by
Theorem 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.
9
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 inequality
for conv(S) with n = 2 (i.e., y ∈ R2 ), then all nonzero components of α are equal. The proof is by
contradiction; we assume Theorem 3.11 is not true and consider a minimal counterexample to Theorem 3.11
with n = 2, i.e., a set S with n = 2 and a facet-defining inequality αx + γy ≥ β such that S has as few
variables as possible. We can assume that α > 0 and its coefficients are all distinct. The reason for this
assumption is the following. Let S and αx + γy ≥ β form a minimal counterexample – i.e., it is facetdefining for conv(S) and the nonzero coefficients of α are not all equal, and m + n is as small as possible. If
a component of α is zero or a pair of components of α are nonzero but equal, we can apply either Lemma 2.2
or Lemma 2.4 and obtain a facet-defining inequality of a set S ′ with fewer variables than S, but with the
facet-defining inequality having the same set of nonzero α values as before. This would contradict the
assumption of minimality of S. Therefore, in a minimal counterexample, α > 0 and its coefficients are all
distinct. In this case, Corollary 2.9 implies that m ≤ 2.
If m = 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) for
arbitrary 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 that
αx + γy ≥ β is a nontrivial facet-defining inequality for conv(S) and F is the set of points in S lying on
the 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 with
αi < αj , we have xi = ui . Furthermore, if 0 < xi < ui for i = 1, . . . , m, then α has all its coefficients
equal.
Proof. If there is some index i 6= j with αi < αj such that xi < ui , then letting ǫ = min{ui − xi , xj }, we
see that (x, y) + ǫei − ǫej ∈ 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 pair
of 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.
10
Proof. Let (x, y) ∈ F with x 6= 0. Suppose ex + cy = b + ǫ for some ǫ > 0. By definition, xi > 0 for some
i ∈ {1, . . . , m}; then (x − min{xi , ǫ}ei , y) ∈ S but violates αx + γy ≥ β (as α > 0), a contradiction to
the fact that this inequality is valid for conv(S). If the second part of the Lemma is not true, then each point
in F satisfies ex + cy = b by the first part of the Lemma. As conv(S) is full-dimensional, this means that
αx + γy ≥ β is a scalar multiple of ex + cy ≥ b, which contradicts the nontriviality of αx + γy ≥ β.
Lemma 3.4. Assume α > 0 has all distinct coefficients. Let (x̂, ŷ), (x̄, ȳ) ∈ F . If αx̂ ≤ αx̄, then x̂ ≤ x̄.
Therefore αx̂ = αx̄ if and only if x̂ = x̄.
Proof. Let the conditions of the Lemma be true, but assume αx̂ ≤ αbarx and x̂j > x̄j for some index
j ∈ {1, . . . , m}. The fact that x̂j > 0 implies (by Lemma 3.2) that x̂i = ui ≥ x̄i for all i 6= j with αi < αj .
The fact that x̄j < uj implies (by Lemma 3.2) that x̄i = 0 ≤ x̂i for all i 6= j with αi > αj . Then x̂i ≥ x̄i
for i = 1, . . . , m (as all coefficients of α are distinct), and x̂j > x̄j which implies that αx̂ > αx̄ (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-defining
inequality αx + γy ≥ β with α2 > α1 > 0. By Theorem 2.1 we have γ > 0 and β > 0. We define F to be
the set of integral points in S satisfying αx + γy = β. Let L = {(xi , y i ) : i = 1, . . . , 4} ⊂ R2 × R2 be a
set of affinely independent points in F which has the properties in Lemma 2.7. We refer to these points as
p1 , . . . , p4 . As β > 0, these points are also linearly independent. As before, let Ly = {y 1 , . . . , y 4 }, and let
Q = conv(Ly ).
Lemma 3.5. Q is a parallelogram.
Proof. Lemma 2.7 implies that Q ⊆ R2 is full-dimensional, has four vertices (namely y 1 , . . . , y 4 ), and
contains no other integer points. In R2 , such a set can only be a parallelogram. To see this, let the vertices
of the quadrilateral Q be y 1 , . . . , y 4 in clockwise order, with the interior angles (in degrees) between the
edges defining these vertices equal to θ1 , . . . , θ4 . If Q is not a parallelogram, we can assume, without loss
of generality, that θ1 + θ2 > 180. Furthermore, we can assume that either θ4 + θ1 ≥ 180 or θ3 + θ2 ≥ 180.
In the first case, y 4 + y 2 − y 1 is contained in Q (and distinct from y 3 ), and in the second case y 3 + y 1 − y 2
is contained in Q (and distinct from y 4 ), a contradiction. Therefore Q is a parallelogram.
We next assume that the points in L are sorted by nondecreasing values of γy i , i.e., γy 1 ≤ γy 2 ≤ γy 3 ≤
γy 4 , and therefore by nonincreasing values of αxi . Lemma 3.4 implies that xi ≥ xj for j > i. If y i = y k ,
then Lemma 2.6 implies that xi = xk which contradicts the distinctness of pi and pk , so we can assume all
y i s are distinct.
Lemma 3.6. If γy 1 = γy 2 and γy 3 = γy 4 , then the points pi (i = 1, . . . , 4) are linearly dependent.
11
Proof. If the conditions of the Lemma are satisfied, then Lemma 3.4 implies that x1 = x2 and x3 = x4 .
Next observe that the y i s are all distinct. As 0 6= γ ∈ R2 , and 0 6= y 2 − y 1 and 0 6= y 4 − y 3 are orthogonal
to γ, we infer that y 2 − y 1 is a scalar multiple of y 4 − y 3 (from the fact that these vectors lie in R2 ), and
therefore y 2 − y 1 − µ(y 4 − y 3 ) = 0 for some nonzero scalar µ. As x2 − x1 = 0 and x4 − x3 = 0, it follows
that p2 − p1 − µ(p4 − p3 ) = 0.
As the points in L are linearly independent, we conclude that the conditions of Lemma 3.6 cannot hold.
Using the fact that γy i is non-decreasing and either γy 1 < γy 2 or γy 3 < γy 4 , leads to the following
observation.
Corollary 3.7. For the points in L we either have γy 1 < γy 2 or γy 3 < γy 4 and therefore γy 1 < γy 4 .
Lemma 3.8. The points y 1 and y 4 form opposite corners of the parallelogram Q.
Proof. By Corollary 3.7 we have γy 1 < γy 4 . Assume the result is not true and that y 1 and y 4 define adjacent
corners of Q. Then the other adjacent corner of Q to y 1 is defined by either y 3 (in which case y 4 − y 1 =
y 2 − y 3 ) or by y 2 (in which case y 4 − y 1 = y 3 − y 2 ). In the first case we have 0 < γ(y 4 − y 1 ) = γ(y 2 − y 3 )
which contradicts the fact that γy 2 ≤ γy 3 . Therefore y 4 − y 1 = y 3 − y 2 and 0 < γ(y 4 − y 1 ) = γ(y 3 − y 2 ).
This combined with γy 1 ≤ γy 2 ≤ γy 3 ≤ γy 4 implies that γy 1 = γy 2 and γy 3 = γy 4 . Lemma 3.6 then
implies 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 cy 4 > b.
Proof. For i = 1, 2, there is a point (x̄, ȳ) ∈ L with x̄i < ui and a point (x′ , y ′ ) with x′i > 0. This follows
directly 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 + cy 4 = cy 4 = b, then exk + cy k > b for some k < 4 (again, because we are dealing with
a nontrivial facet-defining inequality). But Lemma 3.3 implies that xk = 0 = x4 ⇒ γy k = γy 4 = β.
Therefore, if we switch the points (xk , y k ) and (x4 , y 4 ), L is still sorted by nondecreasing values of γy i and
cy 4 > 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. If
α > 0, then α1 = α2 .
Proof. Suppose the coefficients of α are distinct and assume that 0 < α1 < α2 . We know that y 1 and y 4
form nonadjacent corners of Q, and y 2 and y 3 form the remaining corners of Q. Therefore
y4 − y3 = y2 − y1 .
12
(1)
From the equations αxi + γy i = β for i = 1, . . . , 4, we conclude that
α(x3 − x4 ) + γ(y 3 − y 4 ) = 0 = α(x1 − x2 ) + γ(y 1 − y 2 ).
Using equation (1) we can conclude that
α(x3 − x4 ) = α(x1 − x2 ).
(2)
Furthermore, given that 0 < γ(y 2 − y 1 ) = γ(y 4 − y 3 ), from Corollary 3.7, we have αx3 > αx4 and
αx1 > αx2 . Therefore x1 , x2 , x3 6= 0 as x4 = 0 by Lemma 3.9. Using Lemma 3.3 we have
exi + cy i = b for i = 1, . . . , 3,
and consequently, (x4 , y4 ) cannot lie on the capacity constraint, implying
cy 4 = ex4 + cy 4 > b.
From this we can conclude that
e(x3 − x4 ) + c(y 3 − y 4 ) < 0 = e(x1 − x2 ) + c(y 1 − y 2 )
Again using equation (1) we infer that
e(x3 − x4 ) < e(x1 − x2 ).
(3)
We now have two cases.
Case 1: x21 = u1 . Scale α such that α1 < 1 and α2 = 1. Recall that x11 = u1 and x12 > 0 (by
Lemma 3.9). As x1 − x2 6= 0, we conclude that x1 − x2 is nonzero only in the second coordinate. As
x3 6= 0 = x4 , x31 > 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 x22 = 0. As x12 > 0,
α(x1 − x2 ) > e(x1 − x2 ). Furthermore x3 ≤ x2 which implies that x31 < u1 and therefore x32 = 0 by
Lemma 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:
X
xi + γ1 y1 + γ2 y2 ≥ β
i∈I
where I ⊆ {1, . . . , m} and x + γ1 yj + γ2 y2 ≥ β is facet-defining for the set
n
o
Q(b′ , u′ ) = conv (x, y) ∈ R × Z2 : x + c1 y1 + c2 y2 ≥ b′ , u′ ≥ x ≥ 0, y ≥ 0 .
P
P
where b′ = b − i6∈I ui and u′ = i∈I ui .
13
Therefore, when n = 2, all nontrivial facet-defining inequalities of conv(S) are obtainable from facets
of 3-variable mixed-integer sets of the form Q(b, u), we next study sets of this form (or equivalently, the set
S when m = 1 and n = 2).
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 + c1 y1 + c2 y2 ≥ 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))
where
P≥ (b − u) = conv({y ∈ Z2+ : c1 y1 + c2 y2 ≥ b − u}).
In other words, we will show that every nontrivial facet-defining inequality for Q(b, u) either defines a
facet of Q(b, ∞) (i.e., u plays no role) or the coefficient of the x variable is zero and the inequality (when
treated as an inequality on the variables y1 and y2 ) defines a facet of P≥ (b − u). The latter set corresponds
to 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 facets
of P≥ (b) for any b (and therefore for P≥ (b − u)), and then extended their algorithm to obtain all facets of
Q(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 integer
variables only. Accordingly, in addition to P≥ (b − u), we also define
P≤ (b) = conv({y ∈ Z2+ : c1 y1 + c2 y2 ≤ b}),
P (b − u, b) = conv({y ∈ Z2+ : b − u ≤ c1 y1 + c2 y2 ≤ b}).
We next study the integral points of Q(b, u) that lie on the capacity constraint. The projection of these
points on the y-coordinates gives the set P (b − u, b), the convex hull of all integer points between two
parallel 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).
14
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}.
Therefore Qi = conv(Pi ∩ Z2 ) for i = 0, 1, 2. Note that P0 = P1 ∩ P2 and therefore Q0 ⊆ Q1 ∩ Q2 . Q2 is a
full-dimensional anti-blocking polyhedron with facets defined by inequalities of the form y1 ≥ 0 or y2 ≥ 0
or ĉy ≥ γ̂ for ĉ > 0 and γ̂ > 0. Similarly Q1 is a blocking polyhedron; if full-dimensional, its facets are
defined by inequalities of the form y1 ≥ 0 or y2 ≥ 0 or ĉy ≤ γ̂ for ĉ ≥ 0 and γ̂ > 0.
Assume that Q1 ∩ Q2 6⊆ Q0 . Then Q1 ∩ Q2 is not an integral polyhedron; otherwise Q1 ∩ Q2 would
be the convex hull of integer points in P1 ∩ P2 , and thus would be contained in Q0 . We can assume Q1 is
full-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 , where
f1 = conv(p1 , p2 ) is a facet of Q1 and f2 = conv(q1 , q2 ) is a facet of Q2 . Let c1 y ≤ γ 1 be the inequality
defining f1 , and let c2 y ≥ γ 2 be the inequality defining f2 ; these inequalities are unique (up to multiplication
by 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 relative
interior of each facet. Furthermore, one of the end points of f1 , say p2 , must strictly satisfy c2 y ≥ γ 2 . Then
p1 strictly violates c2 y ≥ γ 2 . This implies that p1 ∈ R2+ \ P2 as c2 y ≥ γ 2 is valid for all integral points in
P2 . As f2 is entirely contained in P2 and intersects f1 = conv(p1 , p2 ), this means that p2 must be contained
in P2 ; thus p2 ∈ P1 ∩ P2 . Similarly, we can assume q1 strictly violates c1 y ≤ γ 1 and therefore q1 ∈ R2+ \ P1 .
Furthermore, q2 strictly satisfies c1 y ≤ γ 1 and also belongs to P1 , otherwise f2 would not intersect f1 which
is contained in P1 . Therefore, q2 ∈ P1 ∩ P2 . In other words, we have
c1 q2 < γ 1 ,
c1 q1 > γ 1 ,
(4)
c2 p2 > γ 2 ,
c2 p1 < γ 2 ,
(5)
cp1 < b − u,
b − u ≤ cp2 ≤ b,
(6)
cq1 > b,
b − u ≤ cq2 ≤ b.
(7)
It is clear that the lines c1 y = γ 1 and c2 y = γ 2 have different slopes as the points p1 , p2 , q1 , q2 and v are
not collinear. Then c12 /c11 6= c22 /c21 (here c11 stands for the first component of c1 , c12 for the second component,
etc.). Note that c2 > 0 and so c22 /c21 is a positive number. As for c1 , c11 may equal zero, in which case we
will think of c12 /c11 as the ‘number’ ∞ and greater than any positive number. We can assume, without loss
of generality, that c12 /c11 > c22 /c21 ; if c12 /c11 < c22 /c21 , we can switch the coefficients of c and construct an
instance with the desired relationship of slopes of the lines c1 y = γ 1 and c2 y = γ 2 . See Figure 1 for a
depiction of p1 , p2 , q1 , q2 and v.
15
y2
2
2
q1 c y ≥ γ
p1
c1 y ≤ γ 1
v
p2
q2 δ
cy ≥ b − u
cy ≤ b
y1
Figure 1: Facets f1 and f2 and their intersection v
Let δ = p2 − q2 ∈ Z2 . As c1 q2 < γ 1 and c1 p2 = γ 1 , we have c1 δ = c1 (p2 − q2 ) > 0. Similarly,
as c2 p2 > γ2 and c2 q2 = γ 2 , we have c2 δ > 0. Note that as c1 ≥ 0, c1 δ > 0 implies that at least one
component 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) > γ 1 as c1 δ > 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
component of q2 (because of the relationship between the slopes of the lines c1 y = γ 1 and c2 y = γ 2 ) and
q2 + δ = p2 ∈ R2+ , we have p1 + δ ∈ R2+ . In other words, p1 + δ is an integral point in P1 which violates
c1 y ≤ γ 1 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) < γ 2 as c2 δ > 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 component
of p2 and p2 − δ = q2 ∈ R2+ , we have q1 − δ ∈ R2+ . In other words, q1 − δ is an integral point in P2 which
violates c2 y ≥ γ 2 . This contradicts the fact that c2 y ≥ γ 2 is valid for all integral points in P2 .
We note that a closely related result appears in an unpublished manuscript of Basu, Bonami, Conforti
and Cornuéjols where the authors study integer programs with two variables where one of the variables has
both 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.
16
Lemma 4.2. Let y 1 , y 2 , y 3 ∈ Z2 be three distinct points such that conv(y 1 , y 2 , y 3 ) contains no other integer
point and let y 2 , y 3 lie on the line γy = β where the coefficients of γ ∈ Z2 are coprime integers. Then
β − 1 ≤ γy 1 ≤ β + 1.
Proof. As γ, y 1 , y 2 are integral, so is β. We can also assume, without loss of generality, that γy 1 ≤ β
(by multiplying γ, β by -1 if necessary). Also, there is nothing to prove if γy 1 = β, so we assume γy 1
is an integer less than β. As γ has coprime coefficients, there is a 2 × 2 unimodular matrix U such that
γ̄ = γU = (1, 0). Consider the points ȳ i = U −1 y i − U −1 y 3 for i = 1, . . . , 3. Then conv(ȳ 1 , ȳ 2 , ȳ 3 )
contains no integer points other than ȳ 1 , ȳ 2 , ȳ 3 . Furthermore,
γU U −1 y j = γy j ⇒ γ̄ ȳ j = γy j − γy 3 = 0 for j = 2, 3 and γ̄ ȳ 1 < 0.
Therefore ȳ 3 = (0, 0), ȳ 2 lies on the line y1 = 0 and ȳ11 < 0. As conv(ȳ 2 , ȳ 3 ) contains no integer points
other than ȳ 2 , ȳ 3 , it follows that ȳ 2 is either (0, 1) or (0, −1). In either case, if ȳ11 ≤ −2 then conv(ȳ 1 , ȳ 2 , ȳ 3 )
has an integer point besides ȳ 1 , ȳ 2 , ȳ 3 . Therefore ȳ11 = γ̄ ȳ 1 = −1 and γy 1 = 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 a
subset of their intersection. To prove the reverse inclusion, we next show that each facet-defining inequality
of 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 γ
is integral and the components of γ are coprime. As Q(b, u) is a special case of the set conv(S) studied in
Lemma 2.1, we can conclude that Q(b, u) is full-dimensional and α ≥ 0, γ > 0 and β > 0. Let α = 0. Then
Lemma 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 will
show that under this condition αx + γy ≥ β defines a facet of Q(b, ∞).
Let L = {(x1 , y 1 ), (x2 , y 2 ), (x3 , y 3 )} be a subset of three affinely independent integral points on F
such that conv(y 1 , y 2 , y 3 ) is full-dimensional and contains no other integer points. These points exist by
Lemma 2.7. Without loss of generality, assume that x1 ≤ x2 ≤ x3 . If xi > 0 for i = 1, 2, 3 then (xi , y i )
lies on the capacity inequality, contradicting the nontriviality of F , therefore x3 > 0. If all three points in L
satisfy xi = 0, then F is defined by x ≥ 0 contradicting the nontriviality of F , therefore x1 = 0. We next
consider two cases.
Case 1: Let x2 = 0. Recall that x3 = 0 and u ≥ x1 > 0. Therefore γy 1 < γy 2 = γy 3 = β. As γ
is integral (by scaling) and so is y 2 , β is an integer. By Lemma 2.7, conv(y 1 , y 2 , y 3 ) is a full-dimensional
polyhedron in R2 containing no integer points other than y 1 , y 2 , y 3 . Lemma 4.2 implies that γy 1 = β − 1.
We will now show that αx + γy ≥ β is valid and facet-defining for Q(b, ∞). Clearly, if this inequality
is valid, then it is facet-defining as the inequality is tight for the points (xi , y i ) for i = 1, 2, 3 which are
17
contained in Q(b, ∞). Suppose αx + γy ≥ β is not valid for Q(b, ∞). Then there is a point in Q(b, ∞) that
violates this inequality and its x-coordinate is strictly larger than u. Therefore, for some α̂ > α, α̂x+γy ≥ β
is facet-defining for Q(b, ∞); this cannot be facet-defining for Q(b, u) (as it would be implied by the valid
inequalities αx + γy ≥ β and x ≥ 0 for Q(b, u)). There must be an integral point (x̂, ŷ) with x̂ > 0 in
Q(b, ∞) satisfying α̂x̂ + γ ŷ = β; let (x̂, ŷ) be chosen so that x̂ is as small as possible. For any such point,
x̂ > u, otherwise α̂x + γy ≥ β would be facet-defining for Q(b, u).
Let the convex hull of {ŷ, y 2 , y 3 } be the triangle H, and let F̂ stand for the set of integral points on the
facet of Q(b, ∞) defined by α̂x + γy ≥ β. Suppose H contains some integral point y ′ different from the
three vertices of H. Then it equals λ1 ŷ + λ2 y 2 + λ3 y 3 where λ1 + λ2 + λ3 = 1 and 0 ≤ λ1 , λ2 , λ3 < 1. If
λ1 = 0, then y ′ is contained in the convex hull of y 2 , y 3 which contradicts the assumption that this convex
hull contains no other integer points besides y 2 , y 3 . Therefore 0 < λ1 < 1. Let x′ = (β − γy ′ )/α̂. Then
(x′ , y ′ ) satisfies α̂x + γy = β and (x′ , y ′ ) = λ1 (x̂, ŷ) + λ2 (x2 , y 2 ) + λ3 (x3 , y 3 ). In other words, (x′ , y ′ )
is a point in F̂ with u < x′ < x̂ which contradicts our assumption on the minimality of x̂. Therefore
conv(ŷ, y 2 , y 3 ) contains no other integer points besides ŷ, y 2 , y 3 , and Lemma 4.2 implies that γ ŷ = β − 1
and α̂x̂ = 1. But then 1 = α̂x̂ > αx1 = 1 as x1 ≤ u < x̂ and α < α̂. 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 , y 2 ) and (x3 , y 3 ) lie on Fc . Now consider the set of
points 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 the
inequality αx + γy ≥ β is tight for the points (x2 , y 2 ), (x3 , y 3 ) which also lie in Fc , we have γ ′ y 2 = β ′
and γ ′ y 3 = β ′ . In other words, γ ′ y ≥ β ′ is both valid for P (b − u, b) and facet-defining. By our previous
results, P (b − u, b) = P≥ (b − u) ∩ P≤ (b). Therefore γ ′ y ≥ β ′ defines a facet of either P≥ (b − u) or of
P≤ (b).
Case 2a: Let γ ′ y ≥ β ′ define a facet of P≥ (b − u). In other words, γ ′ y ≥ β ′ is valid for all integral
y ≥ 0 with cy ≥ b − u. But any integral (x, y) ∈ Q(b, u) satisfies y ≥ 0 and cy ≥ b − u. Therefore
0x + γ ′ y ≥ β ′ is a valid inequality for Q(b, u). But then by (10), αx + γy ≥ β is the sum of two distinct
valid 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, ∞) then
there exists an integral point (x̂, ŷ) ∈ Q(b, ∞) \ Q(b, u) with αx̂ + γ ŷ < β and x̂ > u and x̂ + cŷ ≥ b.
18
Therefore x̂ ≥ max{u, b−cŷ}. If u ≥ b−cŷ, then setting x̂ to u, we get a point which violates αx+γy ≥ β
but belongs to Q(b, u), a contradiction. So we can assume that setting x̂ to b − cŷ, we get an integral point
(x̂, ŷ) ∈ Q(b, ∞) \ Q(b, u) which violates αx + γy ≥ β and lies on the capacity constraint. But then
ŷ ∈ P≤ (b) and therefore γ ′ ŷ ≥ β ′ . This, combined with x̂ + cŷ = b and (10) implies that αx̂ + γ ŷ ≥ β, a
contradiction.
5 Final Remarks
We have shown that
2
m
o
n
X
X
2
c
y
≥
b,
x
≤
u
)
x
+
conv( (x, y) ∈ Rm
×
Z
:
j j
i
+
+
i=1
=
j=1
o\n
o
\ n
X
m
2
2
(x,
y)
∈
R
×
R
:
x
≤
u
,
(x, y) ∈ Rm
×
Z
:
(
x
,
y)
∈
P
i
T
+
+
+
+
T ⊆M
i∈T
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 all
facet-defining inequalities for PT in polynomial-time. As there is an exponential number of choices for the
set 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 solving
at most m three-variable integer programs. Assume that the variables are indexed in such a way that p1 ≤
· · · ≤ pm and consider an optimal solution (x̄, ȳ). x̄ must be an optimal solution of the linear program
P
min{px : m
i=1 xi ≥ b − cȳ, 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)-optimal
solution (x̃, ȳ) with x̃i = ui for i < k, x̃i = 0 for i > k, and 0 ≤ x̃k ≤ uk for some value of k ∈ {1, . . . , m}.
But then (x̃, ȳ) is also an optimal solution to
X
i:i<k
pi ui + min{pk xk + qy : xk +
2
X
cj yj ≥ b −
j=1
X
ui , 0 ≤ xk ≤ uk , y ∈ Z2+ }.
i:i<k
Thus it suffices to solve the m three variable problems obtained as k varies from 1 to m. Each of these
problems 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 ellipsoid
algorithm. However it would be preferable to have a more practical algorithm: a natural conjecture is that
∗
x∗
it suffices to order the variables such that u11 ≥ · · · ≥ uxm
and then separate over the m + 1 sets PTi where
m
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].
19
With three or more integer variables, the sets
PTn = 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 an
arbitrary number of integer variables, similar results to those of Section 4 hold when the coefficients are
divisible (Wolsey and Yaman [9]); in particular similar relaxations give a complete description of conv(S).
For the remainder of this section, we assume that P≥ (b), P (b − u, b), P≤ (b) and Q(b, u) are defined in
a similar fashion to the definitions in Section 4 when y ∈ R3 . For example, for given coefficients c1 , c2 , c3 ,
P≥ (b) = conv{y ∈ Z3 : c1 y1 + c2 y2 + c3 y3 ≥ 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 , 10
3 ) 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 facetdefining) 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 righthand-side of this inequality as y3 is integral. Observe that if (ȳ1 , ȳ2 , 4) is a non-negative integral point in
P (94, 97), then 5ȳ1 + 13ȳ2 ∈ [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 , 38 , 23 , 10
3 ) = 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
1
lies in R+ ×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, for
points (x̄, ȳ) in Q(97, 3) with ȳ3 = 4, x̄ + 5ȳ1 + 13ȳ2 ≥ 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-negative
integer and x ∈ [0, 3]. But then x̄ + 5ȳ1 + 13ȳ2 + 21ȳ3 ≥ 94.
Proposition 5.3. The convex hull of the following set has a facet-defining inequality with distinct nonzero
coefficients for the continuous variables:
S = {(x, y) ∈ R2+ × Z3+ : x1 + x2 + 5y1 + 13y2 + 22y3 ≥ 97, x1 , x2 ≤ 3}.
20
Proof. We will show that
x1 + 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 are
no points in S with y3 ≥ 5. To see the validity when y3 = 4, consider a point (x̄, ȳ) ∈ S with ȳ3 = 4.
Then x̄1 + x̄2 + 5ȳ1 + 13ȳ2 ≥ 9. Therefore either x̄1 + x̄2 + 5ȳ1 + 13ȳ2 = 9 (and y2 = 0 and x2 ≥ 1 as
0 ≤ x1 ≤ 3 and y1 is integral) or x̄1 + x̄2 + 5ȳ1 + 13ȳ2 ≥ 10. In either case, x̄1 + 2x̄2 + 5ȳ1 + 13ȳ2 ≥ 10
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 of
which 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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