Generating cutting planes for mixed-integer programs from multi-row
relaxations
Quentin Louveaux
University of Liège - Montefiore Institute
April 2011
Based on joint works with
Laurent Poirrier (Liège)
Kent Andersen (Aarhus)
Santanu Dey (Atlanta)
Robert Weismantel (Magdeburg)
Laurence Wolsey (Louvain-la-Neuve)
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
1 / 24
Outline
Introduction to two-row cuts
Some extensions
The split rank question
The separation question
Future work : theory and practice
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
2 / 24
The framework : Mixed-Integer Programming
We consider the set
XMIP = {x ∈ Rn |Ax = b
xj ∈ Z+ , for j ∈ J
xi ≥ 0, for i 6∈ J }
for which we want to find valid inequalities.
Definition
An inequality aT x ≤ a0 is valid for XMIP if aT x ≤ a0 holds for all x ∈ XMIP .
This task is however hard in general. We therefore need to consider relaxations of XMIP .
Definition
A set Y is a relaxation of XMIP if XMIP ⊆ Y .
Important Result
A valid inequality for Y is also valid for XMIP .
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
3 / 24
The framework : Mixed-Integer Programming
We consider the set
XMIP = {x ∈ Rn |Ax = b
xj ∈ Z+ , for j ∈ J
xi ≥ 0, for i 6∈ J }
for which we want to find valid inequalities.
Definition
An inequality aT x ≤ a0 is valid for XMIP if aT x ≤ a0 holds for all x ∈ XMIP .
This task is however hard in general. We therefore need to consider relaxations of XMIP .
Definition
A set Y is a relaxation of XMIP if XMIP ⊆ Y .
Important Result
A valid inequality for Y is also valid for XMIP .
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
3 / 24
The framework : Mixed-Integer Programming
We consider the set
XMIP = {x ∈ Rn |Ax = b
xj ∈ Z+ , for j ∈ J
xi ≥ 0, for i 6∈ J }
for which we want to find valid inequalities.
Definition
An inequality aT x ≤ a0 is valid for XMIP if aT x ≤ a0 holds for all x ∈ XMIP .
This task is however hard in general. We therefore need to consider relaxations of XMIP .
Definition
A set Y is a relaxation of XMIP if XMIP ⊆ Y .
Important Result
A valid inequality for Y is also valid for XMIP .
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
3 / 24
The simplest relaxation and cut generation : Chvátal-Gomory cuts
Based on a single-row relaxation in the context of pure integer problems.
Chvátal-Gomory cuts
P
Let XMIP ⊆ X = {x ∈ Zn+P| ni=1 ai xi ≤ b}.
n
The inequality
i=1 bai cxi ≤ bbc
is valid for conv(XMIP ).
We can derive a function that gives the coefficient of the cut in terms of the initial
coefficient of the inequality.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
4 / 24
The simplest relaxation and cut generation : Chvátal-Gomory cuts
Based on a single-row relaxation in the context of pure integer problems.
Chvátal-Gomory cuts
P
Let XMIP ⊆ X = {x ∈ Zn+P| ni=1 ai xi ≤ b}.
n
The inequality
i=1 bai cxi ≤ bbc
is valid for conv(XMIP ).
We can derive a function that gives the coefficient of the cut in terms of the initial
coefficient of the inequality.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
4 / 24
Basic Mixed-Integer Rounding
Consider the basic set
X = {(x, y ) ∈ Z × R+ |x ≤ b + y }.
Only one inequality is missing to describe conv(X ).
It is called the Mixed-Integer-Rounding Inequality (MIR)
x ≤ bbc +
y
,
1−f
where f = b − bbc is the fractional part of b.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
5 / 24
Basic Mixed-Integer Rounding
Consider the basic set
X = {(x, y ) ∈ Z × R+ |x ≤ b + y }.
MIR CUT
Only one inequality is missing to describe conv(X ).
It is called the Mixed-Integer-Rounding Inequality (MIR)
x ≤ bbc +
y
,
1−f
where f = b − bbc is the fractional part of b.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
5 / 24
Mixed-Integer-Rounding
Generalization of the MIR to two integer variables
Consider the basic set
X = {(x, y ) ∈ Z2 × R+ | a1 x1 + a2 x2 ≤ b + y }.
We define
f = b − bbc, f1 = a1 − ba1 c, f2 = a2 − ba2 c
and assume f1 ≤ f ≤ f2 .
The inequality
„
«
f2 − f
y
ba1 cx1 + ba2 c +
x2 ≤ bbc +
1−f
1−f
is valid for conv(X ) and is called Mixed-Integer-Rounding.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
6 / 24
Mixed-Integer-Rounding
Generalization of the MIR to two integer variables
Consider the basic set
X = {(x, y ) ∈ Z2 × R+ | a1 x1 + a2 x2 ≤ b + y }.
We define
f = b − bbc, f1 = a1 − ba1 c, f2 = a2 − ba2 c
and assume f1 ≤ f ≤ f2 .
The inequality
«
„
y
f2 − f
ba1 cx1 + ba2 c +
x2 ≤ bbc +
1−f
1−f
is valid for conv(X ) and is called Mixed-Integer-Rounding.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
6 / 24
Mixed-Integer-Rounding
Generalization of the MIR to two integer variables
Consider the basic set
X = {(x, y ) ∈ Z2 × R+ | a1 x1 + a2 x2 ≤ b + y }.
We define
f = b − bbc, f1 = a1 − ba1 c, f2 = a2 − ba2 c
and assume f1 ≤ f ≤ f2 .
The inequality
«
„
y
f2 − f
ba1 cx1 + ba2 c +
x2 ≤ bbc +
1−f
1−f
is valid for conv(X ) and is called Mixed-Integer-Rounding.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
6 / 24
Comparison of the Chvatal-Gomory and the MIR
If we plot the functions that give the coefficients of the valid inequality in terms of the
coefficients of the initial inequality, we obtain
Chvátal cut
MIR cut
The MIR inequality is always stronger than the Chvátal-Gomory cut.
Even if we only have integer variables !
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
7 / 24
Comparison of the Chvatal-Gomory and the MIR
If we plot the functions that give the coefficients of the valid inequality in terms of the
coefficients of the initial inequality, we obtain
Chvátal cut
MIR cut
f
The MIR inequality is always stronger than the Chvátal-Gomory cut.
Even if we only have integer variables !
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
7 / 24
The geometry of MIR cut
Basic model
x = f + s1 − s2
x ∈ Z, s1 , s2 ∈ R+
Example :
11
+ s1 − s2
4
x ∈ Z, s1 , s2 ∈ R+ .
(1)
x=
Valid inequality :
4
s2 ≥ 1
3
By substituting (1), we can see that it is equivalent to
4s1 +
x ≤ 2 + 4s1 .
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
8 / 24
The geometry of MIR cut
Basic model
x = f + s1 − s2
x ∈ Z, s1 , s2 ∈ R+
f
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
2
3
4
5
x
Example :
11
+ s1 − s2
4
x ∈ Z, s1 , s2 ∈ R+ .
(1)
x=
Valid inequality :
4
s2 ≥ 1
3
By substituting (1), we can see that it is equivalent to
4s1 +
x ≤ 2 + 4s1 .
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
8 / 24
The geometry of MIR cut
Basic model
x = f + s1 − s2
x ∈ Z, s1 , s2 ∈ R+
1
0
0
1
1
s2
f
1
0
0
1
1
0
0
1
2
3
s1
1
0
0
1
1
0
0
1
4
5
x
Example :
11
+ s1 − s2
4
x ∈ Z, s1 , s2 ∈ R+ .
(1)
x=
Valid inequality :
4
s2 ≥ 1
3
By substituting (1), we can see that it is equivalent to
4s1 +
x ≤ 2 + 4s1 .
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
8 / 24
Why using such a model ?
We consider the LP simplex optimal tableau.
Example : One row of the LP optimal tableau after the second step is
x1 −
1
9
10
s1 + s3 =
.
4
4
4
Relaxing nonnegativity of x1 and integrality of s1 , s3 , we obtain naturally the model and
can read off the cut
9
1
s1 + s3 ≥ 1
2
2
that always cuts off the LP point.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
9 / 24
Why using such a model ?
We consider the LP simplex optimal tableau.
Example : One row of the LP optimal tableau after the second step is
x1 −
1
9
10
s1 + s3 =
.
4
4
4
Relaxing nonnegativity of x1 and integrality of s1 , s3 , we obtain naturally the model and
can read off the cut
1
9
s1 + s3 ≥ 1
2
2
that always cuts off the LP point.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
9 / 24
Why using such a model ?
We consider the LP simplex optimal tableau.
Example : One row of the LP optimal tableau after the second step is
x1 −
1
9
10
s1 + s3 =
.
4
4
4
Relaxing nonnegativity of x1 and integrality of s1 , s3 , we obtain naturally the model and
can read off the cut
1
9
s1 + s3 ≥ 1
2
2
that always cuts off the LP point.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
9 / 24
Rounding : from 1D to 2D (or more)
How can we round simultaneously two variables without aggregating ?
y
3
2
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
Quentin Louveaux (University of Liège)
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
2
3
4
5
Multi-row relaxations
x
April 2011
10 / 24
Rounding : from 1D to 2D (or more)
How can we round simultaneously two variables without aggregating ?
y
3
2
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
Quentin Louveaux (University of Liège)
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
2
3
4
5
Multi-row relaxations
x
April 2011
10 / 24
Rounding : from 1D to 2D (or more)
How can we round simultaneously two variables without aggregating ?
y
3
2
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
Quentin Louveaux (University of Liège)
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
2
3
4
5
Multi-row relaxations
x
April 2011
10 / 24
Rounding : from 1D to 2D (or more)
How can we round simultaneously two variables without aggregating ?
y
3
2
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
Quentin Louveaux (University of Liège)
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
2
3
4
5
Multi-row relaxations
x
April 2011
10 / 24
Fundamental problem with two constraints
The simplex tableau
x1 − ā11 s1 −· · ·− ā1n sn = b̄1
..
.
xm −ām1 s1 −· · ·−āmn sn = b̄m
- Select two rows
- Relax the integrality requirements of the non-basic variables
- Relax the nonnegativity requirements of the basic variables
The model
„
x1
x2
«
„
=
f1
f2
«
+
«
n „
X
r1j
sj
r2j
j=1
x1 , x2 ∈ Z, sj ∈ R+
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
11 / 24
Fundamental problem with two constraints
The simplex tableau
x1 − ā11 s1 −· · ·− ā1n sn = b̄1
..
.
xm −ām1 s1 −· · ·−āmn sn = b̄m
- Select two rows
- Relax the integrality requirements of the non-basic variables
- Relax the nonnegativity requirements of the basic variables
The model
„
x1
x2
«
„
=
f1
f2
«
+
«
n „
X
r1j
sj
r2j
j=1
x1 , x2 ∈ Z, sj ∈ R+
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
11 / 24
Extension of the MIR model to two constraints
The geometry
„
x1
x2
«
„
=
1/4
1/2
«
„
+
2
1
«
„
s1 +
1
1
«
„
s2 +
2s1 + 2s2 + 4s3 + s4 +
r
−3
2
«
„
s3 +
0
−1
«
„
s4 +
1
−2
«
s5
12
s5 ≥ 1
7
3
x
2
r 2
r
1
f
x
1
r 4
r5
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
12 / 24
Extension of the MIR model to two constraints
The geometry
„
x1
x2
«
„
=
1/4
1/2
«
„
+
2
1
«
„
s1 +
1
1
«
„
s2 +
2s1 + 2s2 + 4s3 + s4 +
r
−3
2
«
„
s3 +
0
−1
«
„
s4 +
1
−2
«
s5
12
s5 ≥ 1
7
3
x
2
r 2
r
1
f
x
1
r 4
r5
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
12 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
5
Classification of the facets in 3 categories :
disection cuts, lifted 2 variable-cuts, MIR cuts
The facets from the first 2 categories are never MIR cuts
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
5
Classification of the facets in 3 categories :
disection cuts, lifted 2 variable-cuts, MIR cuts
The facets from the first 2 categories are never MIR cuts
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
5
Classification of the facets in 3 categories :
disection cuts, lifted 2 variable-cuts, MIR cuts
The facets from the first 2 categories are never MIR cuts
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
5
Classification of the facets in 3 categories :
disection cuts, lifted 2 variable-cuts, MIR cuts
The facets from the first 2 categories are never MIR cuts
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
r
3
x
2
r 2
r
1
f
x
1
r 4
r5
4
Every facet is completely determined by exactly 3 or 4 rays
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
5
Classification of the facets in 3 categories :
disection cuts, lifted 2 variable-cuts, MIR cuts
The facets from the first 2 categories are never MIR cuts
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
r3
x
2
r 2
r
1
f
x
1
r 4
r5
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
Fundamental problem with two constraints
Main results
1
Every facet can be written
n
X
αj sj ≥ 1,
j=1
with αj ≥ 0.
2
If αj = 0 for some j, then the facet is an MIR cut
3
Every facet is tangent to exactly 3 or 4 important integer points
4
Every facet is completely determined by exactly 3 or 4 rays
5
Classification of the facets in 3 categories :
disection cuts, lifted 2 variable-cuts, MIR cuts
The facets from the first 2 categories are never MIR cuts
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
13 / 24
The geometry
The projection picture
2s1 + 2s2 + 4s3 + s4 +
We project the n + 2-dim space onto
the x-space
r
12
s5 ≥ 1
7
3
x
2
The facet is represented by a polygon
Lα
There is no integer point in the
interior of Lα
r
1
f
The coefficients are a ratio of
distances on the figure
α1 α3
Quentin Louveaux (University of Liège)
r 2
x
1
r 4
r5
Multi-row relaxations
April 2011
14 / 24
The geometry
The projection picture
2s1 + 2s2 + 4s3 + s4 +
We project the n + 2-dim space onto
the x-space
r
12
s5 ≥ 1
7
3
x
2
The facet is represented by a polygon
Lα
There is no integer point in the
interior of Lα
r
1
f
The coefficients are a ratio of
distances on the figure
α1 α3
Quentin Louveaux (University of Liège)
r 2
x
1
r 4
r5
Multi-row relaxations
April 2011
14 / 24
The geometry
The projection picture
2s1 + 2s2 + 4s3 + s4 +
We project the n + 2-dim space onto
the x-space
r
12
s5 ≥ 1
7
3
x
2
The facet is represented by a polygon
Lα
1111111
0000000
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
r 2
There is no integer point in the
interior of Lα
f
The coefficients are a ratio of
distances on the figure
α1 α3
x
1
r 4
r
Quentin Louveaux (University of Liège)
r1
Multi-row relaxations
5
April 2011
14 / 24
The geometry
The projection picture
2s1 + 2s2 + 4s3 + s4 +
We project the n + 2-dim space onto
the x-space
r
12
s5 ≥ 1
7
3
x
2
The facet is represented by a polygon
Lα
There is no integer point in the
interior of Lα
r
1
f
The coefficients are a ratio of
distances on the figure
α1 α3
Quentin Louveaux (University of Liège)
r 2
x
1
r 4
r5
Multi-row relaxations
April 2011
14 / 24
The geometry
The projection picture
2s1 + 2s2 + 4s3 + s4 +
We project the n + 2-dim space onto
the x-space
r
12
s5 ≥ 1
7
3
x
2
The facet is represented by a polygon
Lα
There is no integer point in the
interior of Lα
r 2
f
The coefficients are a ratio of
distances on the figure
α1 α3
x
1
r 4
r
Quentin Louveaux (University of Liège)
r1
Multi-row relaxations
5
April 2011
14 / 24
The geometry
The projection picture
2s1 + 2s2 + 4s3 + s4 +
We project the n + 2-dim space onto
the x-space
r
12
s5 ≥ 1
7
3
x
2
The facet is represented by a polygon
Lα
There is no integer point in the
interior of Lα
r 2
f
The coefficients are a ratio of
distances on the figure
α1 α3
x
1
r 4
r
Quentin Louveaux (University of Liège)
r1
Multi-row relaxations
5
April 2011
14 / 24
Improvements of the model
„
x1
x2
«
„
=
f1
f2
«
«
n „
X
r1j
sj
+
r2j
j=1
x1 , x2 ∈ Z, sj ∈ R+
Taking care of the integrality of the nonbasic variables sj .
Leads to several papers on lifting
[Dey, Wolsey 2008], [Basu, Conforti, Cornuéjols, Zambelli 2009]
Results similar to what was obtained by Gomory, Johnson in 1970’s for the group
problem (since it is a similar model)
Taking care of bounds on the basic variables xi
The whole theory carries over except that we consider S-free sets instead of
lattice-point-free sets.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
15 / 24
Improvements of the model
„
x1
x2
«
„
=
f1
f2
«
«
n „
X
r1j
sj
+
r2j
j=1
x1 , x2 ∈ Z, sj ∈ R+
Taking care of the integrality of the nonbasic variables sj .
Leads to several papers on lifting
[Dey, Wolsey 2008], [Basu, Conforti, Cornuéjols, Zambelli 2009]
Results similar to what was obtained by Gomory, Johnson in 1970’s for the group
problem (since it is a similar model)
Taking care of bounds on the basic variables xi
The whole theory carries over except that we consider S-free sets instead of
lattice-point-free sets.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
15 / 24
S-free sets
11111111111
00000000000
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
2
00000000000
11111111111
00000000000
11111111111
{0,1}
−free
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
set
Negative coefficient for that ray
We again have triangles and quadrilaterals but with stronger coefficients.
Question : can we characterize these objects in a better way ?
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
16 / 24
S-free sets
11111111111
00000000000
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
2
00000000000
11111111111
00000000000
11111111111
{0,1}
−free
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
set
Negative coefficient for that ray
We again have triangles and quadrilaterals but with stronger coefficients.
Question : can we characterize these objects in a better way ?
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
16 / 24
S-free sets
11111111111
00000000000
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
2
00000000000
11111111111
00000000000
11111111111
{0,1}
−free
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
set
Negative coefficient for that ray
We again have triangles and quadrilaterals but with stronger coefficients.
Question : can we characterize these objects in a better way ?
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
16 / 24
S-free sets
11111111111
00000000000
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
2
00000000000
11111111111
00000000000
11111111111
{0,1}
−free
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
set
Negative coefficient for that ray
We again have triangles and quadrilaterals but with stronger coefficients.
Question : can we characterize these objects in a better way ?
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
16 / 24
Another improvement of the model
„
x1
x2
«
„
=
f1
f2
«
+
«
n „
X
r1j
sj
j
r2
j=1
x1 , x2 ∈ Z, sj ∈ R+
Can we consider bounds on the nonbasic variables sj ?
Known bounds or bounds induced by the best possible pivot ?
The facets can be similarly studied but we can now have pentagons in the case of exactly
one bound [Andersen, L., Weismantel 2010].
The minimal subsystem describing the facet might be more complicated.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
17 / 24
Another improvement of the model
„
x1
x2
«
„
=
f1
f2
«
+
«
n „
X
r1j
sj
j
r2
j=1
x1 , x2 ∈ Z, sj ∈ R+
Can we consider bounds on the nonbasic variables sj ?
Known bounds or bounds induced by the best possible pivot ?
The facets can be similarly studied but we can now have pentagons in the case of exactly
one bound [Andersen, L., Weismantel 2010].
The minimal subsystem describing the facet might be more complicated.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
17 / 24
A problem with a bound on a nonbasic variable
Idea : being able to cut off an edge instead cutting just a point.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
18 / 24
The split rank question
1
0
0
1
11
00
00
11
1
0
0
1
1
0
0
1
11
00
00
11
1
0
0
1
Every cut that was presented here is not a split cut i.e. cannot be obtained by any
one-row cut generators.
The following cut cannot be achieved with split cuts only
[Cook, Kannan, Schrijver 1990]
All the other cuts from the two-row plain model can be achieved in a finite number of
rounds of split cuts. [Dey, L. 2010]
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
19 / 24
The split rank question
1
0
0
1
11
00
00
11
1
0
0
1
1
0
0
1
11
00
00
11
1
0
0
1
Every cut that was presented here is not a split cut i.e. cannot be obtained by any
one-row cut generators.
The following cut cannot be achieved with split cuts only
[Cook, Kannan, Schrijver 1990]
All the other cuts from the two-row plain model can be achieved in a finite number of
rounds of split cuts. [Dey, L. 2010]
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
19 / 24
The separation question
We have a model PI :=
{ (x, s) ∈ R2 × Rn :
x
x
sj
=
∈
≥
f +
Z2
0
P
j
r j sj
}
and the general form of a cut
α1 s1 + . . . + αn sn ≥ 1,
How to compute a valid α ?
How to choose among valid α ?
The separation problem
Given (x̂, ŝ) ∈ R2 × Rn+ , either
prove that (x̂, ŝ) ∈ conv(PI ), or
P
P
find a valid inequality ni=1 αi si ≥ 1 for PI that minimizes ni=1 αi ŝi (maximize the
violation)
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
20 / 24
The separation question
We have a model PI :=
{ (x, s) ∈ R2 × Rn :
x
x
sj
=
∈
≥
f +
Z2
0
P
j
r j sj
}
and the general form of a cut
α1 s1 + . . . + αn sn ≥ 1,
How to compute a valid α ?
How to choose among valid α ?
The separation problem
Given (x̂, ŝ) ∈ R2 × Rn+ , either
prove that (x̂, ŝ) ∈ conv(PI ), or
P
P
find a valid inequality ni=1 αi si ≥ 1 for PI that minimizes ni=1 αi ŝi (maximize the
violation)
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
20 / 24
The cut generating LP for the 2-row model
r
3
x
2
r 2
r1
f
x
1
r 4
r5
What are extreme points of conv(PI ) ?
x = f + RS
They correspond to points (x, s) ∈ Z2 ×Rk+
such that support(s) ≤ 2.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
21 / 24
The cut generating LP for the 2-row model
r
3
x
2
r2
r1
f
x
1
r 4
r5
„
1
1
«
„
=
1
4
1
2
«
+
Quentin Louveaux (University of Liège)
1 1 1 2
r + r
4
4
Multi-row relaxations
April 2011
21 / 24
The cut generating LP for the 2-row model
r
3
x
2
r2
r1
f
x
1
r 4
r5
„
1
1
«
„
=
1
4
1
2
«
+
The polar
1 1 1 2
r + r
4
4
1
1
α1 + α2 ≥ 1
4
4
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
21 / 24
The cut generating LP for the 2-row model
r
3
x
2
r2
r1
f
x
1
r 4
r5
The polar
„
1
1
«
„
=
1
4
1
2
«
2
1 5
+ r2 +
r
3
12
Quentin Louveaux (University of Liège)
1
1
α1 + α2 ≥ 1
4
4
2
1
α2 +
α5 ≥ 1
3
12
Multi-row relaxations
April 2011
21 / 24
Complexity of writing the polar
For each cone, compute the integer hull.
For each integer point in each integer hull, compute the representation in the given
cone and generate one inequality for the polar
Quadratic complexity in the number of rays for the number of cones
Polynomial number of integer vertices in each cone (but may be large if the numbers
involved are large)
For a complete enumeration of the integer hulls, the rays must be available in
rationals
The complexity is still too large for a cut generating LP.
Theorem [L., Poirrier 2011]
The complexity of writing the polar can be reduced to linear in the number of rays.
We can avoid the computation of integer hulls.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
22 / 24
Complexity of writing the polar
For each cone, compute the integer hull.
For each integer point in each integer hull, compute the representation in the given
cone and generate one inequality for the polar
Quadratic complexity in the number of rays for the number of cones
Polynomial number of integer vertices in each cone (but may be large if the numbers
involved are large)
For a complete enumeration of the integer hulls, the rays must be available in
rationals
The complexity is still too large for a cut generating LP.
Theorem [L., Poirrier 2011]
The complexity of writing the polar can be reduced to linear in the number of rays.
We can avoid the computation of integer hulls.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
22 / 24
Complexity of writing the polar
For each cone, compute the integer hull.
For each integer point in each integer hull, compute the representation in the given
cone and generate one inequality for the polar
Quadratic complexity in the number of rays for the number of cones
Polynomial number of integer vertices in each cone (but may be large if the numbers
involved are large)
For a complete enumeration of the integer hulls, the rays must be available in
rationals
The complexity is still too large for a cut generating LP.
Theorem [L., Poirrier 2011]
The complexity of writing the polar can be reduced to linear in the number of rays.
We can avoid the computation of integer hulls.
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
22 / 24
Computational results
10teams
air03
air04
air05
arki001
bell3a
bell5
blend2
cap6000
danoint
dcmulti
egout
fast0507
fiber
fixnet6
flugpl
gen
gesa2
gesa2 o
gesa3
gesa3 o
gt2
harp2
khb05250
l152lav
lseu
mas74
mas76
misc03
misc06
misc07
CG
Time (s)
21.371
0.045
769.765
558.691
0.525
0.072
0.242
0.076
2.212
0.832
0.374
0.127
453.502
0.225
0.340
0.061
0.544
0.245
0.424
0.449
0.612
0.110
0.311
0.114
10.855
0.042
0.086
0.054
0.057
0.065
0.129
CGLP
Time (s)
21.080
0.041
763.789
554.118
0.476
0.068
0.229
0.071
2.152
0.786
0.349
0.120
451.284
0.210
0.319
0.057
0.510
0.229
0.397
0.420
0.572
0.104
0.275
0.106
10.616
0.039
0.080
0.050
0.054
0.061
0.121
Quentin Louveaux (University of Liège)
CG
383
38
2438
2044
256
103
212
71
11
222
520
191
569
334
385
58
424
349
534
555
769
108
232
139
631
64
72
50
69
93
150
CG
iter
397
43
2443
2056
496
122
407
82
15
224
539
220
569
364
547
101
757
418
712
758
1030
191
306
173
654
70
72
50
83
109
162
Models
Rounds
5
5
5
5
5
5
5
5
5
5
5
5
2
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
6
17
18
8
16
19
14
7
11
18
19
5
14
17
16
20
13
15
15
18
12
11
17
18
10
11
8
7
14
9
Multi-row relaxations
Valid
cuts
103
38
709
757
57
77
122
25
3
21
378
161
245
296
317
40
296
157
240
370
548
92
37
101
240
45
7
7
52
77
120
Active
cuts
9
8
22
13
42
29
31
7
1
5
92
80
9
108
236
13
47
80
87
87
117
33
14
57
15
22
3
4
8
30
13
%gc
0.00
100.00
8.04
4.92
10.16
68.20
23.78
25.28
20.67
0.17
61.06
80.82
1.64
18.83
62.40
13.95
61.19
36.79
41.50
55.29
60.88
86.82
4.18
95.26
12.94
32.03
2.54
2.35
10.22
76.86
0.31
cplex
cuts
3
3
5
5
46
11
12
5
0
3
48
29
0
30
111
6
16
77
66
37
39
16
9
21
2
6
2
2
3
12
13
%gc
100.00
100.00
9.73
4.74
36.21
58.56
18.19
20.40
0.00
0.33
67.82
95.99
0.00
90.11
69.46
15.57
80.26
81.43
78.74
53.63
60.58
99.30
27.08
79.35
16.93
67.09
5.04
4.65
17.59
61.58
0.72
April 2011
23 / 24
Future work
Computational
Find the right choice of pairs of rows
Find the right basis
Go beyond the basis model
Theoretical
What type of cutting planes are needed to be able to generate all facets of the
integer hull of some given problem ?
Mixed-{0, 1} problems : only split cuts are enough
Mixed-{0, 1, 2} problems ? : we do not know. . .
Which lattice-free objects are needed in order to generate facets of the integer hull
of a given problem ?
Example : For a plain two-row problem, we only need splits and the
Cook-Kannan-Schrijver triangle.
In general, is it enough to consider maximal lattice-point-free polyhedra that have
integer vertices ?
Are some closures polyhedra ?
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
24 / 24
Future work
Computational
Find the right choice of pairs of rows
Find the right basis
Go beyond the basis model
Theoretical
What type of cutting planes are needed to be able to generate all facets of the
integer hull of some given problem ?
Mixed-{0, 1} problems : only split cuts are enough
Mixed-{0, 1, 2} problems ? : we do not know. . .
Which lattice-free objects are needed in order to generate facets of the integer hull
of a given problem ?
Example : For a plain two-row problem, we only need splits and the
Cook-Kannan-Schrijver triangle.
In general, is it enough to consider maximal lattice-point-free polyhedra that have
integer vertices ?
Are some closures polyhedra ?
Quentin Louveaux (University of Liège)
Multi-row relaxations
April 2011
24 / 24