COMPUTING VALID INEQUALITIES FOR GENERAL INTEGER
PROGRAMS USING AN EXTENSION OF MAXIMAL DUAL
FEASIBLE FUNCTIONS TO NEGATIVE ARGUMENTS
J¨urgen Rietz
1
, Cl´audio Alves
1
, J. M. Val´erio de Carvalho
1
and Franc¸ois Clautiaux
2
1
Dept. Produc¸˜ao e Sistemas, Universidade do Minho, 4710-057 Braga, Portugal
2
Universit´e des Sciences et Technologies de Lille, LIFL UMR CNRS 8022, INRIA Bˆatiment INRIA
Parc de la Haute Borne, 59655 Villeneuve d’Ascq, France
K
eywords:
Integer programming, Dual feasible functions, Valid inequalities.
Abstract:
Dual feasible functions (DFFs) were used with much success to compute bounds for several combinatorial
optimization problems and to derive valid inequalities for some linear integer programs. A major limitation of
these functions is that their domain remains restricted to the set of positive arguments. To tackle more general
linear integer problems, the extension of DFFs to negative arguments is essential. In this paper, we show how
these functions can be generalized to this case. We explore the properties required for DFFs with negative
arguments to be maximal, we analyze additional properties of these DFFs, we prove that many classical
maximal DFFs cannot be extended in this way, and we present some non-trivial examples.
1 INTRODUCTION
Dual feasible functions (DFFs) were introduced in
(Johnson, 1973), and used since then to compute
bounds for different combinatorial optimization prob-
lems and valid inequalities for integer linear programs
(see for example (Nemhauser and Wolsey, 1998),
(Fekete and Schepers, 2001) and (Clautiaux et al.,
2010)). To ensure the quality of the bounds, one has to
resort to maximal DFFs. The criteria for a DFF to be
maximal were described first by Carlier and N´eron in
(Carlier and N´eron, 2007). Recently, in (Rietz et al.,
2011), some of the strongest maximal DFFs of the lit-
erature were analyzed with respect to their worst cases
in the computation of lower bounds.
In (Clautiaux et al., 2010), the authors showed that
DFFs could be used to compute valid inequalities for
integer programs. However, all the DFFs developed
until now apply exclusively to positive data. This
fact constitutes a clear restriction for their use in the
computation of valid inequalities for general integer
programs. The extension of DFFs to negative argu-
ments is not trivial. It raises different issues that are
addressed in this paper.
Example 1. The function f
FS,1
was defined in (Fekete
and Schepers, 2001) for 0 ≤ x ≤ 1 and k ∈ IN\ {0} as
f
FS,1
(x) =

x if (k+ 1)x ∈ ZZ
⌊(k+ 1)x⌋/k otherwise
,
but it cannot be extended as a maximal DFF to x ∈
IR with the same formula. Let 0 < ε <
1
k+1
. Then
f
FS,1
(
k+2
k+1
− ε) =
k+1
k
>
k+2
k+1
= f
FS,1
(
k+2
k+1
), and hence
f
FS,1
would not be monotonous.
The paper is organized as follows. The definition
and the characteristics of maximal DFFs with a do-
main that is the whole set of real numbers are intro-
duced in the next section. Additional properties of
these functions and some tools to construct maximal
DFFs follow in Section 3. Several non-trivial exam-
ples of general DFFs (with positive and negative ar-
guments) are presented in Section 4. In Section 5, we
show through an example how these functions apply
to general integer linear programs.
2 DEFINITIONS AND ESSENTIAL
PROPERTIES
The notion of (maximal/extremal) dual feasible
function can be extended to domain and range IR.
The defining conditions of these functions remain
nearly the same as for the DFFs restricted to positive
39
Rietz J., Alves C., M. Valério de Carvalho J. and Clautiaux F..
COMPUTING VALID INEQUALITIES FOR GENERAL INTEGER PROGRAMS USING AN EXTENSION OF MAXIMAL DUAL FEASIBLE FUNCTIONS
TO NEGATIVE ARGUMENTS.
DOI: 10.5220/0003751700390047
In Proceedings of the 1st International Conference on Operations Research and Enterprise Systems (ICORES-2012), pages 39-47
ISBN: 978-989-8425-97-3
Copyright
c
 2012 SCITEPRESS (Science and Technology Publications, Lda.)

arguments. These conditions are stated in the sequel.
Definition 1. A function f : IR → IR is called a dual
feasible function (DFF), if for all n ∈ IN and all
x
1
,... ,x
n
∈ IR with
n
∑
i=1
x
i
≤ 1, we have
n
∑
i=1
f(x
i
) ≤ 1.
Definition 2. A DFF f : IR → IR is a maximal dual
feasible function (MDFF), if there is no other DFF
g : IR → IR with g(x) ≥ f(x) for all x ∈ IR.
Definition 3. A MDFF f : IR → IR is extremal, if any
MDFFs g,h : IR → IR with 2f(x) = g(x) + h(x) for all
x ∈ IR are necessarily identical to f.
Note that the identity function f
id
is clearly a DFF.
Any DFF f : IR → IR has the following properties:
•
n
∑
i=1
x
i
≤ 0 =⇒
n
∑
i=1
f(x
i
) ≤ 0, especially f(x) ≤ 0
for all x ≤ 0
• 0 < x ≤ 1 =⇒ f(x) ≤ 1/⌊1/x⌋
• if f(x
1
) > 0 for a certain x
1
∈ IR, then f(x) < 0 for
all x < 0.
In a general integer linear program with the deci-
sion variables x
1
,... ,x
n
∈ IN and a set of coefficients
a
1
,... , a
n
∈ IR, if the inequality
n
∑
i=1
a
i
x
i
≤ 1 is re-
quired, then the following inequality obtained by ap-
plying a DFF f : IR → IR
n
∑
i=1
x
i
× f(a
i
) ≤ 1
is valid because
n
∑
i=1
a
i
x
i
=
n
∑
i=1
x
i
∑
j=1
a
i
≤ 1,
and hence 1 ≥
n
∑
i=1
x
i
∑
j=1
f(a
i
).
In the following proposition, we show that
MDFFs with domain IR are different from those with
domain [0,1].
Proposition 1. For every c ∈ [0,1], the function f :
IR → IR with f(x) := cx for all x ∈ IR is a MDFF.
Proof. For this proof, we resort to the definitions 1
and 2. Let n ∈ IN\{0} and x
1
,... ,x
n
∈ IR with
n
∑
i=1
x
i
≤
1 be given. Then,
n
∑
i=1
f(x
i
) = c×
n
∑
i=1
x
i
≤ c, and hence
f is a DFF. Suppose that there is a DFF g : IR → IR
with g(x) ≥ cx, for all x ∈ IR, and g(x
0
) > cx
0
for a
certain x
0
∈ IR. Since g(−x
0
) ≥ f(−x
0
), it follows
that g(x
0
)+ g(−x
0
) > cx
0
− cx
0
= 0. That is a contra-
diction. Since f is not dominated by another DFF g,
the assertion follows.
The following theorem characterizes the MDFFs.
It is inspired on the theorem by Carlier and N´eron
(Carlier and N´eron, 2007), but here the domain and
range are IR and not only an interval.
Theorem 1. Let f : IR → IR be a given function.
(a) If f satisfies the following conditions, then f is a
MDFF:
1. f(0) = 0;
2. f is superadditive, i.e. for all x
1
,x
2
∈ IR, it holds
that
f(x
1
+ x
2
) ≥ f(x
1
) + f (x
2
); (1)
3. there is an ε > 0, such that f(x) ≥ 0 for all x ∈
(0,ε);
4. for all x ∈ IR, it holds that
f(x) + f(1− x) = 1. (2)
(b) If f is a MDFF, then the above properties (1.)–(3.)
hold for f , but not necessarily (4.);
(c) If f satisfies the above conditions (1.)–(3.), then f
is monotonously increasing.
Proof. The proof is made in the following order: first,
we prove (c), then (b), and finally (a) is proved.
(c) If f satisfies the first three conditions, then for
any x > 0 it follows that n := ⌊x/ε⌋ + 1 ∈ IN \ {0}
and 0 < x/n < ε. Hence, we have f(x/n) ≥ 0 and
f(x) ≥ n× f(x/n) ≥ 0. Therefore, the monotonicity
follows immediately from f(x
2
) ≥ f(x
1
)+ f(x
2
− x
1
)
for any x
1
,x
2
∈ IR with x
1
≤ x
2
. The remaining
proof is partially similar to Theorem 1 of (Carlier and
N´eron, 2007).
(b) Let f : IR → IR be a MDFF. We prove the prop-
erties (1.)–(3.). One has f(0) ≤ 0 due to the condition
for DFFs. On the other hand, f(x) < 0 for a certain
x ≥ 0 is impossible, because f is maximal and setting
f(x) to zero cannot violate the condition for DFFs.
Assume that f(x
1
+ x
2
) < f(x
1
) + f(x
2
) for certain
x
1
,x
2
∈ IR. Define a function g : IR → IR as
g(x) :=

f(x) if x 6= x
1
+ x
2
f(x
1
) + f (x
2
) otherwise
.
Since f is a MDFF, g must violate the defining condi-
tion for a DFF. Replacing g(x
1
+ x
2
) by g(x
1
) + g(x
2
)
and x
1
+ x
2
by two ones x
1
and x
2
leads to a violation
if x
1
,x
2
6= 0, because of the definition of g. That is a
contradiction.
(a) The converse direction is to prove that if f
satistfies the conditions (1.)–(4.), then f is a MDFF.
ICORES 2012 - 1st International Conference on Operations Research and Enterprise Systems
40

For any n ∈ IN and x
1
,... ,x
n
∈ IR with
n
∑
i=1
x
i
≤ 1,
the superadditivity condition (2.) yields
n
∑
i=1
f(x
i
) ≤
f(
n
∑
i=1
x
i
). Let x
0
:= 1 −
n
∑
i=1
x
i
≥ 0. Therefore, we
have f(x
0
) ≥ 0. Because of f(1 − x
0
) + f(x
0
) = 1,
it follows that f is a DFF. Let g : IR → IR be a DFF
with g(x) > f(x) for a certain x ∈ IR. Since g is a
DFF, one has g(1 − x) + g(x) ≤ 1. It follows that
g(1 − x) ≤ 1 − g(x) < 1 − f(x) = f(1 − x) due to
(2), hence g does not dominate f. Therefore, f is a
MDFF.
The third condition is necessary for the assertion
(a) as it can be shown through a counter example. The
following function f : IR → IR obeys only the 1st, 2nd
and 4th condition of the theorem and it is not a DFF
(see Figure 1):
f(x) :=









3x− 2 if x < 0
−x if 0 ≤ x < 1/2
1/2 if x = 1/2
2− x if 1/2 < x ≤ 1
3x otherwise
(3)
The first condition is obviously fulfilled. The fourth
is also checked easily. If x < 0, then 1 − x > 1 and
f(x) + f(1− x) = 3x − 2+ 3(1− x) = 1. If 0 ≤ x <
1/2, then f(x) + f(1− x) = − x + 2− (1− x) = 1. To
check the superadditivity, assume that x
1
≤ x
2
. If x
2
<
0 or x
1
> 1, then the proof is trivial. If x
2
> 1 and
0 ≤ x
1
+ x
2
< 1/2, then f(x
1
+ x
2
) − f(x
1
) − f(x
2
) =
−(x
1
+x
2
)−(3x
1
−2)−3x
2
= 2−4(x
1
+x
2
) > 0. The
other cases are left to the reader.
-
6
x
y
1
1
2
3
4
p
p
p
`
`
`
`
f
Figure 1: The need for monotonicity.
The following proposition simplifies the proof of
a given real function to be a MDFF by Theorem 1.
Proposition 2. If the function f : IR → IR satisfies (2)
for all x ≤ 1/2, then (2) holds for all x ∈ IR. If addi-
tionally the inequality (1) holds for all x
1
,x
2
∈ IR with
(x
1
+ x
2
≤ 2/3 and) x
1
≤ x
2
≤
1−x
1
2
, then the inequal-
ity (1) is true for all x
1
,x
2
∈ IR.
Proof. If x > 1/2, then z := 1 − x < 1/2, and hence
f(z) + f(1− z) = 1 due to (2). That implies f(x) +
f(1− x) = 1. This symmetry will be assumed for the
entire remaining proof.
The condition x
1
≤ x
2
≤
1−x
1
2
implies x
1
+ x
2
≤
2/3 and x
1
≤ 1/3, because x
1
≤
1−x
1
2
leads to 3x
1
≤ 1
and therefore x
1
+ x
2
≤
1+x
1
2
≤
1+1/3
2
=
2
3
. Obvi-
ously, the inequality (1) is valid if and only if it is
true after swapping x
1
against x
2
. Therefore, x
1
≤ x
2
can be enforced without loss of generality. Now we
prove that the inequality (1) holds for all x
1
,x
2
∈ IR,
if it is true for all x
1
,x
2
∈ IR with x
1
+ x
2
≤ 2/3. If
x
1
+ x
2
> 2/3, then x
2
> 1/3 due to x
1
≤ x
2
. Hence
1− x
2
< 2/3 and f(x
1
) + f(1− x
2
− x
1
) ≤ f(1 − x
2
)
according to the inequality (1). The symmetry (2)
yields f(x
1
) + 1 − f(x
1
+ x
2
) ≤ 1− f(x
2
), and hence
f(x
1
) + f(x
2
) ≤ f(x
1
+ x
2
), as needed. Therefore,
x
1
+ x
2
≤ 2/3 can be assumed in the rest of the
proof, and hence x
1
≤
1
3
≤
1−x
1
2
. If 2x
2
> 1 − x
1
,
then let x
3
:= 1 − x
1
− x
2
<
1−x
1
2
. Due to the previ-
ous parts of the proof and the prerequisites, the su-
peradditivity rule (1) can be used, implying f(x
1
) +
f(x
3
) ≤ f(x
1
+ x
3
). The symmetry rule (2) yields
f(x
1
)+ 1− f(1− x
3
) ≤ 1− f(1− x
1
− x
3
), and hence
f(x
1
)+ f(1− x
1
− x
3
) = f(x
1
)+ f(x
2
) ≤ f(1− x
3
) =
f(x
1
+ x
2
).
3 ADDITIONAL PROPERTIES OF
MDFFS
In this section, some further tools to (dis)prove that a
given function is a MDFF are provided.
Proposition 3. If f : IR → IR is a MDFF, then for
all ¯x ∈ IR the limits lim
x↑ ¯x
f(x) and lim
x↓ ¯x
f(x) exist and
lim
x↑0
f(x) = inf
¯x∈IR
{lim
x↑ ¯x
f(x) − lim
x↓ ¯x
f(x)}.
Proof. f is monotonously increasing and defined for
all real arguments. To verify the existence of the
left and right limits at a certain ¯x ∈ IR, choose any
sequences (x
n
) and (y
n
) of real numbers with x
0
<
··· < x
n
< ·· · < ¯x < · ·· < y
n
< ·· · < y
0
, and lim
n→∞
x
n
=
lim
n→∞
y
n
= ¯x. Then f(x
0
) ≤ ··· ≤ f(x
n
) ≤ · · · ≤ f( ¯x) ≤
··· ≤ f(y
n
) ≤ ··· ≤ f(y
0
). Any monotonous and
bounded sequence converges. Therefore, the claimed
limits exist.
The superadditivity rule (1) implies f(−2) ≤
f( ¯x − 1) − f ( ¯x + 1) ≤ lim
x↑ ¯x
f(x) − lim
x↓ ¯x
f(x) for every
¯x ∈ IR, and hence a := inf
¯x∈IR
{lim
x↑ ¯x
f(x) − lim
x↓ ¯x
f(x)} is fi-
nite. Because of f(x) ≥ 0 for all x ≥ 0, it follows that
lim
x↑0
f(x) ≥ a. For any ε < 0, there is an ¯x ∈ IR with
lim
x↓ ¯x
f(x) − lim
x↑ ¯x
f(x) > ε − a. The superadditivity (1)
implies f(ε) ≤ f ( ¯x +
ε
2
) − f ( ¯x−
ε
2
). The monotonic-
ity of f implies f( ¯x +
ε
2
) ≤ lim
x↑ ¯x
f(x) and f ( ¯x −
ε
2
) ≥
lim
x↓ ¯x
f(x), and hence ε − a < f( ¯x −
ε
2
) − f( ¯x +
ε
2
) ≤
COMPUTING VALID INEQUALITIES FOR GENERAL INTEGER PROGRAMS USING AN EXTENSION OF
MAXIMAL DUAL FEASIBLE FUNCTIONS TO NEGATIVE ARGUMENTS
41

− f(ε), i.e. f(ε) < a − ε. Since this holds for all
ε < 0, it follows that lim
x↑0
f(x) ≤ a. Together with
lim
x↑0
f(x) ≥ a, the assertion follows.
The next proposition shows in contrast to the case
of domain and range [0,1] that the set of differentiable
MDFFs is much stronger restricted.
Proposition 4. If the function f : IR → IR possesses
the properties (1.) and (2.) of Theorem 1, and if it is
continuously differentiable at the points 0 and a ∈ IR,
then f
′
(a) = f
′
(0).
Proof. Let h ∈ IR, h > 0. The superadditivity of f
yields f(a+ h) ≥ f(a) + f(h) and f(a) ≥ f(a+ h) +
f(−h), and hence f(h) ≤ f(a+h) − f(a) ≤ − f(−h).
Since f(0) = 0 and h > 0, this can be rewritten
as
f(h)− f (0)
h
≤
f(a+h)− f(a)
h
≤
f(0)− f (−h)
h
=
f(−h)− f (0)
−h
.
Using the limit h ↓ 0 yields due to the assumed contin-
uous differentiabilityof f the inequality chain f
′
(0) ≤
f
′
(a) ≤ f
′
(0), and hence f
′
(a) = f
′
(0).
Corollary 1. Any continuously differentiable MDFF
f : IR → IR has the form f(x) = cx with c ∈ [0,1].
Proof. The derivative is constant, and hence f(x) =
cx+d with certain constants c,d ∈ IR. Since f(0) = 0,
it follows that f(x) = cx. Definition 1 yields f(1) ≤ 1,
hence c ≤ 1. Since f(x) ≤ 0 for x < 0, one has c ≥
0.
Proposition 5. Let f : IR → IR be a superadditive
function. If there is an a ∈ IR \ {0} with f(a) +
f(−a) = 0, then the function g : IR → IR with g(x) :=
f(x) − x× f(a)/a (for all x ∈ IR) is periodic with pe-
riod a.
Proof. The superadditivity of f implies f(x + a) ≥
f(x) + f(a) and f(x) ≥ f(x + a) + f(−a) for all
x ∈ IR. Hence, f(x + a) ≤ f(x) − f(−a) = f(x) +
f(a) because of f(a) = − f(−a), and finally f(x +
a) = f(x) + f(a) for all x ∈ IR. That yields g(x +
a) − g(x) = f(x+ a) − f(x) − (x+ a) × f(a)/a+ x×
f(a)/a = f(a) − a × f(a)/a = 0.
An example of this kind of MDFFs is the Burdett and
Johnson function f
BJ,1
(see Proposition 12).
If a function is given, which satisfies most of the
demands, but is not symmetric, then sometimes a
MDFF can be constructed from it like it was done
in the Theorem 1 of (Clautiaux et al., 2010), but not
generally.
Proposition 6. Let f : IR → IR be a function with
f(1) > 0 and satisfying the conditions (1.)–(3.) of our
Theorem 1. Define the function g : IR → IR as
g(x) :=



f(x)
f(1)
if x <
1
2
1/2 if x = 1/2
1− g(1− x) otherwise
.
If g(x) + g(y) ≤ g(x + y) holds for all x < 0 and y ∈
[
1
2
,
1−x
2
], then g is a MDFF, but not generally.
Proof. The function g satisfies obviously the condi-
tions (1.), (3.) and (4.) of Theorem 1 by construction.
We show the superadditivity of g under the additional
constraint. According to Proposition 2, choose any
x,y ∈ IR with x ≤ y ≤
1−x
2
. Five cases have to be dis-
tinguished:
1. x,y,x + y < 1/2: the superadditivity of f yields
g(x) + g(y) ≤ g(x+ y), because f(1) > 0;
2. x,y < 1/2= x+y: the superadditivity of f implies
f(1) ≥ 2 × f(1/2) and f(x) + f(y) ≤ f(1/2) ≤
1
2
× f(1), and hence g(x) + g(y) ≤
1
2
= g(x+ y);
3. x,y < 1/2 < x + y: the superadditivity of f leads
to f(x) + f(y) + f(1 − x − y) ≤ f(1), and hence
g(x) + g(y) + g(1 − x − y) ≤ 1 and by symmetry
g(x) + g(y) ≤ g(x+ y);
4. x = 0: this case is trivial because of g(0) = 0;
5. x < 0,
1
2
≤ y ≤
1−x
2
: this case is explicitly given in
the prerequisites.
There are no other cases, because x > 0 leads to
1−x
2
<
1
2
. Therefore, g is superadditive due to Proposition 2.
A counter-example to the superadditivity of g
without the additional constraint arises from f(x) :=
⌊3x⌋ for all x ∈ IR. This function satisfies all the
conditions (1.)–(3.) of Theorem 1, but the resulting
function g is not superadditive. We get g(−1/3) =
−1/3, g(7/12) = 1 − g(5/12) = 1 − ⌊
5
4
⌋/3 = 2/3
and g(1/6) = g(1/4) = 0, and hence g(−1/3) +
g(1/2) = 1/6 > g(1/6) and g(−1/3) + g(7/12) =
1/3 > g(1/4).
Define frac(x) := x − ⌊x⌋ as an abbreviation for
the non-integer part of any real expression x. The fol-
lowing proposition helps in proving superadditivity.
Proposition 7. Let f : IR → IR and g : [0,1) → IR be
superadditive functions such that for all x,y,z ∈ IR
with 0 < y ≤ z < 1 and y+ z ≥ 1 it holds that
f(x+ 1) − f(x) ≥ g(y) + g(z) − g(y+ z− 1). (4)
Then the function h : IR → IR defined by h(x) :=
f(⌊x⌋) + g(frac(x)) is superadditive.
ICORES 2012 - 1st International Conference on Operations Research and Enterprise Systems
42

Proof. Choose any x,y ∈ IR. To verify h(x +
y) ≥ h(x) + h(y), the non-integer parts of x,y need
to be considered. If frac(x) + frac(y) < 1, then
frac(x+y) = frac(x)+frac(y) and ⌊x+ y⌋ = ⌊x⌋+⌊y⌋,
and hence h(x + y) − h(x) − h(y) = f(⌊x⌋ + ⌊y⌋) −
f(⌊x⌋) − f (⌊y⌋) + g(frac(x) + frac(y)) − g(frac(x)) −
g(frac(y)) ≥ 0, because of the superadditivityof f and
g.
The other case is frac(x) + frac(y) ≥ 1 leading
to frac(x + y) = frac(x) + frac(y) − 1 and ⌊x + y⌋ =
⌊x⌋ + ⌊y⌋ + 1. The prerequisite (4) brings f(⌊x⌋ +
⌊y⌋ + 1) ≥ f(⌊x⌋ + ⌊y⌋) + g(frac(x)) + g(frac(y)) −
g(frac(x) + frac(y) − 1), and hence h(x + y) −
h(x) − h(y) = f(⌊x⌋ + ⌊y⌋ + 1) − f(⌊x⌋) − f(⌊y⌋) +
g(frac(x) + frac(y) − 1) − g(frac(x)) − g(frac(y)) ≥
f(⌊x⌋ + ⌊y⌋) − f(⌊x⌋) − f(⌊y⌋) ≥ 0.
Proposition 7 becomes more useful in conjunction
with the following proposition about composed
functions.
Proposition 8. The composition f(g(·)) of superad-
ditive functions f, g : IR → IR is superadditive, if the
inner function g is additive or if the outer function f
is monotonously increasing, but not generally.
Proof. If g is additive, then g(x + y) = g(x) + g(y)
for all x,y ∈ IR. The superadditivity of f implies in
this case f(g(x + y)) = f(g(x) + g(y)) ≥ f(g(x)) +
f(g(y)). If f is monotonously increasing, then
one gets g(x + y) ≥ g(x) + g(y) and f(g(x + y)) ≥
f(g(x) + g(y)) ≥ f(g(x)) + f(g(y)). If f is not
monotonously increasing and g is not additive, then
the composition needs not to be superadditive, as the
following counter-example shows. Let f be the func-
tion (3) and g be the Burdett and Johnson function
f
BJ,1
(see Proposition 12) with parameter C = 9/2.
One gets g(1/9) = 0 and g(2/9) = 1/4, and hence
f(g(1/9)) = f(0) = 0 and f(g(2/9)) = f(1/4) =
−1/4 < 2× f(g(1/9)).
4 EXAMPLES
In this section, we present and analyze non-trivial
examples of MDFF whose domain and range is the
set of real numbers IR.
Proposition 9. Let a,b ∈ IR with 0 ≤ a ≤ 1 and a ≤
b. The following function f : IR → IR satisfies all the
conditions (1.)–(4.) of Theorem 1, and hence it is a
MDFF:
f(x) =











(1+ b)x if x ≤ 0
(1− a)x if 0 ≤ x ≤ 1/4
(1+ a)x−
a
2
if
1
4
≤ x ≤
3
4
(1− a)x+ a if
3
4
≤ x ≤ 1
(1+ b)x− b for x ≥ 1
Proof. The function f is piecewise linear and contin-
uous. In particular, we have f(0) = 0, f(
1
4
) =
1−a
4
,
f(
3
4
) =
3+a
4
and f(1) = 1. The conditions (1.), (3.)
and (4.) can be checked easily. Only the superadditiv-
ity condition (2.) needs a large case distinction. For
this purpose, choose any x,y ∈ IR with x ≤ y (with-
out loss of generality) and x + y ≤ 2/3 according to
Proposition 2. Define
d(x,y) := f(x+ y) − f (x) − f(y).
We have to prove that d(x,y) ≥ 0. This function is
also piecewise linear and continuous.
1. If x + y ≤ 0, then x ≤ 0 due to x ≤ y, and hence
d(x,y) = (1+ b)(x + y − x) − f(y) = (1 + b)y −
f(y). Except for
3
4
< y < 1, the desired inequality
f(y) ≤ (1+ b)y is obvious. Since d is piecewise
linear and continuous, we get d(x,y) ≥ 0 for the
excluded case too.
2. If
1
4
≤ x+ y ≤
3
4
, then x ≤
3
8
, y ≥
1
8
, and the fol-
lowing subcases arise:
(a) x ≤ 0 and
1
4
≤ y ≤
3
4
yields d(x,y) = (1 +
a) × (x + y) −
a
2
− (1 + b)x − (1 + a)y +
a
2
=
(a− b)x ≥ 0.
(b) y ≥ 1 yields x < 0 and d(x,y) = (1+ a) × (x+
y)−
a
2
−(1+b)x−(1+b)y+b = (a−b)×(x+
y) + b−
a
2
≥
3a−3b
4
+ b −
a
2
=
a+b
4
≥ 0.
(c) x ≤ 0∧
3
4
< y < 1 needs not to be analyzed, be-
cause d is continuous and piecewise linear.
(d) 0 < x ≤ y ≤
1
4
gives d(x,y) = (1 + a) × (x +
y) −
a
2
− (1− a) × (x+ y) = 2a × (x+ y) −
a
2
≥
2a
4
−
a
2
= 0.
(e) 0 < x ≤
1
4
< y <
3
4
brings d(x,y) = (1 + a) ×
(x + y) −
a
2
− (1 − a)x − (1 + a)y +
a
2
= (1 +
a)x− (1− a)x= 2ax ≥ 0.
(f)
1
4
< x ≤ y ≤
3
8
leads to d(x,y) = (1+ a) × (x+
y) −
a
2
− (1+ a) × (x+ y) + a=
a
2
≥ 0.
COMPUTING VALID INEQUALITIES FOR GENERAL INTEGER PROGRAMS USING AN EXTENSION OF
MAXIMAL DUAL FEASIBLE FUNCTIONS TO NEGATIVE ARGUMENTS
43

3. The remaining cases 0 < x+y<
1
4
or
3
4
< x+y < 1
need no further analysis, because setting z := x+y
yields d(x,y) = f(z) − f(x) − f(z − x), and this
expression is a continuous and piecewise linear
function in both arguments x and z.
The next function has a simple structure similar
to the function f
CCM,1
by Carlier, Clautiaux and
Moukrim (see e.g. (Clautiaux et al., 2010)), but it
is still different. Moreover, we will see that f
CCM,1
cannot be generalized to be a MDFF with domain IR.
Proposition 10. Let b ∈ IR, b ≥ 1. The following func-
tion f : IR → IR is a MDFF:
f(x) =



b× ⌊2x⌋ if x < 1/2
1/2 if x = 1/2
1− b× ⌊2− 2x⌋ if x > 1/2
(5)
Proof. f satisfies obviously the conditions (1.), (3.)
and (4.) of Theorem 1. Only the superadditivity (2.)
needs a more careful check. Choose for this purpose
any x,y ∈ IR with x ≤ y ≤
1−x
2
according to Proposi-
tion 2, and hence x ≤ 1/3 and therefore f(x) ≤ 0. A
case distinction follows.
1. If y < 1/2 and x + y < 1/2, then f(x) + f(y) =
b× (⌊2x⌋ + ⌊2y⌋) ≤ b× ⌊2x+ 2y⌋ = f(x+ y).
2. If y < 1/2 ≤ x + y, then f(x) ≤ 0, f(y) ≤ 0 and
f(x+ y) ≥ 1/2 > f(x) + f(y).
3. If y = 1/2 and x+y < 1/2, then f(x+ y)− f(x) =
b× (⌊2x+ 2y⌋ − ⌊2x⌋) = b× (⌊2x+ 1⌋ − ⌊2x⌋) =
b× 1 ≥ 1 > f(y).
4. If y = 1/2 ≤ x+ y, then f(x+ y) ≥ 1/2 = f(y) ≥
f(x) + f(y).
5. If y > 1/2 > x+ y, then f(x+ y) − f(x) − f(y) =
b×(⌊2x+2y⌋−⌊2x⌋+⌊2−2y⌋)−1≥ b×(⌊2x+
2y+ 1− 2y⌋ − ⌊2x⌋) − 1= b − 1 ≥ 0.
6. If x < 0, then x + y ≤ x +
1−x
2
=
1+x
2
<
1
2
, and
hence the case x+ y = 1/2 < y is impossible.
7. If y > 1/2 and x+ y > 1/2, then f(x+ y)− f(x) −
f(y) = 1− b× ⌊2− 2x− 2y⌋ − b× ⌊2x⌋ − 1+ b×
⌊2−2y⌋ ≥ b×(⌊2−2y⌋−⌊2−2x−2y+2x⌋ ) ≥ 0.
A generalization of this function runs into difficulties,
as the following proposition shows.
Proposition 11. Let a, b,c,d ∈ IR and the MDFF f :
IR → IR be defined as follows:
f(x) =



a+ b× ⌊cx+ d⌋ if x < 1/2
1/2 if x = 1/2
1− f(1− x) otherwise
.
Then f is necessarily the function (5) with b ≥ 1.
Proof. The superadditivity implies f(x/2) ≥ x/2 for
all x ∈ IN, because f(1/2) = 1/2. Therefore, we have
bc 6= 0. Hence, the function f possesses gaps of at
least the size |b|. According to Proposition 3, it fol-
lows that lim
x↑0
f(x) ≤ −|b| < 0 = f(0). Therefore, we
have c > 0 and d ∈ ZZ, causing ⌊cx + d⌋ = ⌊cx⌋ + d.
If d 6= 0, then replace a by a + bd and after that d
by zero. That does not change f. Therefore, we
may assume d = 0 for the rest of the proof. We get
0 = f(0) = a + b × ⌊c × 0⌋ = a. The assumption
0 < c < 2 leads to f(
1
2
−
1
c
) − f(
−1
c
) = b× ⌊
c
2
− 1⌋ −
b× ⌊−1⌋ = b× (−1) − b× (−1) = 0 in contradiction
to
1
2
= f (
1
2
) ≤ f(
1
2
−
1
c
) − f(
−1
c
) according to the su-
peradditivity rule. Therefore, we have that c ≥ 2.
Suppose now that c > 2. Let ε :=
1−frac(c)
2c
. Then
f(1+ ε−
1
c
) = 1− f(
1
c
−ε) = 1, because 0 < ε <
1
c
<
1
2
. Since f(
1−⌊c⌋
c
) = b× (1− ⌊c⌋), the superadditivity
rule implies 1+b×(1−⌊c⌋) ≤ f(1+ε−
1
c
+
1−⌊c⌋
c
) =
f(1 + ε −
⌊c⌋
c
) = f (
frac(c)
c
+ ε) = f(
1+frac(c)
2c
) = 0,
and hence b ≥
1
⌊c⌋−1
. On the other hand,
1
c
<
1
2
and
f(
1
c
) = b yield due to ⌊ c⌋ ×
1
c
≤ 1 and the superaddi-
tivity the contradiction 1 ≥ f(⌊c⌋×
1
c
) ≥ ⌊c⌋× f (
1
c
) =
⌊c⌋×b≥ ⌊c⌋×
1
⌊c⌋−1
> 1. Hence, c > 2 is impossible,
such that finally c = 2 follows.
Due to the superadditivity rule, f(−
1
2
) = −b,
f(
1
4
) = 0 and f(
3
4
) = 1 yield f(−
1
2
) + f(
3
4
) ≤ f(
1
4
),
and hence b ≥ 1.
If more general functions are allowed like g :
IR → IR with parameters b, c, d ∈ IR and a function
f : [0,1) → IR according to
g(x) :=



b× ⌊cx+ d⌋ + f(frac(cx+ d)),x < 1/2,
1/2, x = 1/2,
1− g(1− x), otherwise,
then, there are more possibilities. Since ⌊cx + d⌋ =
⌊cx + frac(d)⌋ + ⌊d⌋ and frac(cx + d) = frac(cx +
frac(d)), we can assume 0 ≤ d < 1. Otherwise, the
additional constant b × ⌊d⌋ shall become a part of f,
such that d can be replaced by frac(d), leading to the
same function g. Some necessary conditions for g to
be a MDFF are the following:
• f(d) = 0, because of 0 ≤ d < 1 and g(0) = 0;
• bc > 0, because g must be monotonously increas-
ing and g(0) = 0 and g(x) < 0 for x < 0. If we had
bc = 0 then g would be constant or periodic for
x < 1/2, and bc < 0 would yield lim
x→−∞
g(x) = +∞;
ICORES 2012 - 1st International Conference on Operations Research and Enterprise Systems
44

• f must be monotonous, namely non-increasing if
b < 0 and non-decreasing if b > 0, because of the
needed monotonicity of g;
• f must be bounded, namely | f(0) − lim
x↑1
f(x)| ≤
|b|, otherwise g would not be monotonous;
• if d = 0, then f must be superadditive in the inter-
val [0,min{1,
c
2
}) if c > 0 and in the entire domain
[0,1) if c < 0, because of the needed superadditiv-
ity of g.
An example of this type of functions g is presented in
the following proposition.
Proposition 12. Let C ∈ IR, C ≥ 1. The function
f
BJ,1
: IR → IR due to Burdett and Johnson (Bur-
dett and Johnson, 1977) with f
BJ,1
(x) = (⌊Cx⌋ +
max{0,
frac(Cx)−frac(C)
1−frac(C)
})/⌊C⌋ is a MDFF.
Proof. The conditions (1.) and (3.) of Theorem 1 are
obviously satisfied. To prove the other ones, choose
any x, y ∈ IR. We have C − Cx = ⌊C⌋ + frac(C) −
⌊Cx⌋ − frac(Cx) and
• either frac(Cx) > frac(C), and hence
⌊C⌋ × ( f
BJ,1
(x) + f
BJ,1
(1− x))
= ⌊Cx⌋ + ⌊C−Cx⌋
+max{0,
frac(Cx) − frac(C)
1− frac(C)
}
+max{0,
frac(C−Cx) − frac(C)
1− frac(C)
}
= ⌊C⌋ − 1+
frac(Cx) − frac(C)
1− frac(C)
+
frac(C) + 1− frac(Cx) − frac(C)
1− frac(C)
= ⌊C⌋ − 1+
1− frac(C)
1− frac(C)
= ⌊C⌋,
• or frac(Cx) ≤ frac(C) and therefore ⌊C⌋ ×
( f
BJ,1
(x) + f
BJ,1
(1− x)) = ⌊Cx⌋ + ⌊C−Cx⌋ + 0+
0 = ⌊Cx⌋ + ⌊C⌋ − ⌊Cx⌋ = ⌊C⌋,
such that the symmetry (2) is verified. The su-
peradditivity f
BJ,1
(x + y) ≥ f
BJ,1
(x) + f
BJ,1
(y) is ob-
viously valid for frac(Cx) ≤ frac(C) or frac(Cy) ≤
frac(C). Therefore, assume frac(Cx) > frac(C)
and frac(Cy) > frac(C). We have Cx + Cy =
⌊Cx⌋ + ⌊Cy⌋ + frac(Cx) + frac(Cy) and d := ⌊C⌋ ×
( f
BJ,1
(x + y) − f
BJ,1
(x) − f
BJ,1
(y)) = ⌊frac(Cx) +
frac(Cy)⌋ + max{0,
frac(frac(Cx)+frac(Cy))−frac(C)
1−frac(C)
} −
frac(Cx)+frac(Cy)−2frac(C)
1−frac(C)
. Three cases arise:
1. frac(Cx) + frac(Cy) < 1 yields
d =
frac(Cx)+frac(Cy)−frac(C)
1−frac(C)
−
frac(Cx)+frac(Cy)−2frac(C)
1−frac(C)
=
frac(C)
1−frac(C)
> 0;
2. 1 ≤ frac(Cx) + frac(Cy) ≤ 1 + frac(C)
brings d = 1 −
frac(Cx)+frac(Cy)−2frac(C)
1−frac(C)
=
1+frac(C)−frac(Cx)−frac(Cy)
1−frac(C)
≥ 0;
3. frac(Cx) + frac(Cy) > 1 + frac(C) gives
d = 1 +
frac(Cx)+frac(Cy)−1−frac(C)
1−frac(C)
−
frac(Cx)+frac(Cy)−2frac(C)
1−frac(C)
= 1+
frac(C)−1
1−frac(C)
= 0.
In all cases the superadditivity is also valid, such that
all conditions of Theorem 1 are satisfied.
The survey (Clautiaux et al., 2010) already stated
that f
BJ,1
, restricted to the domain [0,1], is a MDFF,
but without a complete proof. This function f
BJ,1
can be seen as a use of Propositions 7 and 8. The
next function f
LL,1
due to Letchford and Lodi is
built similarly, cf. (Clautiaux et al., 2010). On the
contrary, the improved function f
LL,2
of (Clautiaux
et al., 2010) cannot be extended to a MDFF with
domain IR.
Proposition 13. Let C ∈ IR \ IN, C > 1 and k ∈ IN,
k ≥ ⌈
1
frac(C)
⌉. The following function f
LL,1
: IR → IR
satisfies the conditions (1.)–(3.) of Theorem 1, but is
not symmetric and cannot be improved by Proposition
6 in spite of f
LL,1
(1) = 1:
f
LL,1
(x) :=
⌊Cx⌋ + max{0,
⌈(k− 1) ×
frac(Cx)−frac(C)
1−frac(C)
⌉
k
}
⌊C⌋
Proof. One gets f
LL,1
(0) = (0 + max{0,⌈(k − 1) ×
−frac(C)
1−frac(C)
⌉/k})/⌊C⌋ = 0 and f
LL,1
(1) = (⌊C⌋ +
max{0,0})/⌊C⌋ = 1. Moreover, for all x > 0 it
holds obviously that f
LL,1
(x) ≥ 0, because ⌊Cx⌋ ≥ 0
and ⌊C⌋ > 0. To prove the superadditivity, Propo-
sitions 7 and 8 are used. We set f(x) := x and
g(x) := max{0,⌈(k − 1) ×
x−frac(C)
1−frac(C)
⌉}/k in Proposi-
tion 7. We verify that for all x, y,z ∈ IR with 0 ≤ y ≤
z < 1 and y + z > 1 the inequality f(x + 1) − f(x) ≥
g(y) + g(z) − g(y + z − 1) holds. That is equivalent
to k ≥ max{0,⌈(k− 1) ×
y−frac(C)
1−frac(C)
⌉} + max{0,⌈(k −
1)×
z−frac(C)
1−frac(C)
⌉}−max{0,⌈(k−1)×
y+z−1−frac(C)
1−frac(C)
⌉}.
Since z < 1, it follows that ⌈(k − 1) ×
z−frac(C)
1−frac(C)
⌉ ≤
COMPUTING VALID INEQUALITIES FOR GENERAL INTEGER PROGRAMS USING AN EXTENSION OF
MAXIMAL DUAL FEASIBLE FUNCTIONS TO NEGATIVE ARGUMENTS
45

k − 1, and hence the desired inequality is obvi-
ously fulfilled, if y ≤ frac(C). Therefore, as-
sume y > frac(C), such that the inequality becomes
k ≥ ⌈(k − 1) ×
y−frac(C)
1−frac(C)
⌉ + ⌈(k − 1) ×
z−frac(C)
1−frac(C)
⌉ −
max{0,⌈(k − 1) ×
y+z−1−frac(C)
1−frac(C)
⌉}. Since for all x ∈
IR it holds that x ≤ ⌈x⌉ < x + 1, one gets ⌈ (k − 1) ×
y−frac(C)
1−frac(C)
⌉ + ⌈(k − 1) ×
z−frac(C)
1−frac(C)
⌉ − max{0, ⌈ (k −
1) ×
y+z−1−frac(C)
1−frac(C)
⌉} < 2+ (k − 1) ×
y+z−2×frac(C)
1−frac(C)
−
(k − 1) ×
y+z−1−frac(C)
1−frac(C)
= 2 +
k−1
1−frac(C)
× (−2 ×
frac(C) + 1 + frac(C)) = k + 1. The left part of this
inequality is integer, implying that it is not above k.
Therefore, the chosen functions f and g satisfy the
prerequisite (4) of Proposition 7, such that the func-
tion x 7→ ⌊x⌋ + max{0,⌈(k − 1) ×
frac(x)−frac(C)
1−frac(C)
⌉/k}
is superadditive. This function can be composed with
the linear function x 7→ Cx according to Proposition 8.
Finally, dividing the entire expression by ⌊C⌋ ≥ 1 has
no influence on the superadditivity. Therefore, f
LL,1
is superadditive.
It remains to show that the additional constraint
of Proposition 6 is violated for some feasible pa-
rameter choices (and hence f
LL,1
is not symmet-
ric). Choose any C ∈ (1,2) and any enough large
odd k ∈ IN. Let x :=
−1
C
and y :=
1
2
. That
yields f
LL,1
(x) = ⌊−1⌋/⌊C⌋ = −1 and f
LL,1
(x +
y) = f
LL,1
(
C−2
2C
) = ⌊
C
2
− 1⌋ + max{0,⌈(k − 1) ×
frac(C/2−1)−frac(C)
1−frac(C)
⌉/k} = −1 + max{0,⌈(k − 1) ×
C/2−C+1
1−C+1
⌉/k} = −1+ max{0, ⌈
k−1
2
⌉/k} = −1+
k−1
2k
,
because k is odd, and finally f
LL,1
(x+ y) =
−1
2
−
1
2k
<
−1+
1
2
.
5 USING MDFFS TO COMPUTE
VALID INEQUALITIES
In this section, we demonstrate how the MDFFs can
be used to generate valid inequalities to solve integer
linear programs, and we illustrate their use through a
simple example. Let be given an instance
max c
⊤
x
s.t. Ax ≤ b
x ∈ IN
n
,
where A ∈ IR
m×n
, b ∈ IR
m
and c ∈ IR
n
whith
m,n ∈ IN \ {0}. We can take every non-negative
linear combination of the constraints to generate
another valid inequality, i.e. choose any u ∈ IR
m
+
\ {o}
to obtain one inequality u
⊤
Ax ≤ b
⊤
u. If f : IR → IR
is any monotonously increasing superadditive func-
tion, e.g. a MDFF, then the monotonicity of f
yields f(b
⊤
u) ≥ f(u
⊤
Ax), and this is not below
n
∑
j=1
f(
m
∑
i=1
a
ij
u
i
) × x
j
due to the superadditivity of f
and the condition x ∈ IN
n
.
Example 2. Consider the following integer linear
program (with negative coefficients):
max z := 10x
1
− 3x
2
s.t. 7x
1
− 2x
2
≤ 9
x
1
,x
2
∈ IN
The solution x:= (3, 6)
⊤
is feasible and yields z = 12.
In the sequel, we show that this solution is optimal by
showing that the following inequality
10x
1
− 3x
2
≤ 12
is in fact a valid inequality. Furthermore, we show
that this inequality can be derived using a MDFF with
domain and range IR.
Choose any u > 0 and a MDFF f to get the valid
inequality f (7u) × x
1
+ f(−2u) × x
2
≤ f (9u). We try
several functions f, namely according to
• Proposition 9: to get the desired inequality, we set
u := 1/9, yielding
3
4
≤ 7u ≤ 1, and hence
((1−a) ×
7
9
+a)× x
1
−
2
9
×(1+ b)× x
2
≤ 1 (6)
with 0 ≤ a ≤ 1 and b ≥ a. The inequality (6) be-
comes the sharpest for the smallest possible b, i.e.
for b = a. That yields
(
7
9
+
2
9
a) × x
1
− (
2
9
+
2
9
a) × x
2
≤ 1.
Choosing a := 1/14 yields
50
63
x
1
−
15
63
x
2
≤ 1 or
equivalently 10x
1
− 3x
2
≤
63
5
. Since the left hand
side is integer, it follows that z = 12 is optimal.
• Proposition 10: here we have to distinguish sev-
eral cases with respect to u > 0, but this function
fails to give the needed strong valid inequality.
For instance
3
14
< u ≤
2
9
yields the valid inequality
2x
1
− x
2
≤ 2, but it remains too weak.
• Proposition 12: We may use any C ≥ 1 and u >
0, yielding (⌊7Cu⌋+ max{0;
frac(7Cu)−frac(C)
1−frac(C)
})×
x
1
+ (⌊−2Cu⌋ + max{0;
frac(−2Cu)−frac(C)
1−frac(C)
}) ×
x
2
≤ ⌊9Cu⌋ + max{0;
frac(9Cu)−frac(C)
1−frac(C)
}, for in-
stance C := 10/7 and u := 1, yielding 10x
1
−
3x
2
≤ 12
3
4
. The choice C :=
13
7
and u :=
10
13
leads
directly to 10x
1
− 3x
2
≤ 12, as desired.
ICORES 2012 - 1st International Conference on Operations Research and Enterprise Systems
46

Note that the valid inequality 10x
1
− 3x
2
≤ 12 is
a Chv
´
atal-Gomory-inequality (cf. (Nemhauser and
Wolsey, 1998)), and it can be obtained by using u :=
10/7 and rounding down to the next integer.
6 CONCLUSIONS
In this paper, we generalized the notion of (maximal)
dual feasible functions to functions of which the do-
main comprises the entire set of real numbers. This
extension is important to allow the use of DFFs for de-
riving valid inequalities for any general integer linear
program. This generalization is also non-trivial. In-
deed, the well-knownsymmetry condition, which was
necessary for a DFF with domain [0,1] to be maximal,
does not hold for all MDFFs with domain IR. Further-
more, the influence of the conditions that characterize
these functions becomes more restrictive, i.e. many
well known classical MDFFs cannot be generalized to
domain IR. On the contrary, besides the MDFF f
BJ,1
,
some other non-trivial MDFFs were defined. Some
examples were proposed and discussed in this paper.
Finally, we illustrated through the use of an example
how valid inequalities could be derived using these
new MDFFs.
ACKNOWLEDGEMENTS
This work was partially supported by the Portuguese
Science and TechnologyFoundation through the post-
doctoral grant SFRH/BPD/45157/2008 for J¨urgen Ri-
etz and by the Algoritmi Research Center of the Uni-
versity of Minho for Cl´audio Alves and Jos´e Val´erio
de Carvalho.
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COMPUTING VALID INEQUALITIES FOR GENERAL INTEGER PROGRAMS USING AN EXTENSION OF
MAXIMAL DUAL FEASIBLE FUNCTIONS TO NEGATIVE ARGUMENTS
47