IBM Analytics
Jonas Schweiger
CPLEX Optimization, IBM Italy
with Pierre Bonami, Andrea Lodi, Andrea Tramontani
Max Clique Cuts for (Standard) Quadratic Programs
1
c
2015
IBM Corporation
IBM Analytics
Table of Contents
1 (Extremely) Short introduction to Quadratic Programming
2 Max Clique Cuts
Separation
Connection to RLT
2
c
2015
IBM Corporation
IBM Analytics
Standard Quadratic Programs (SQP)
x T Qx
X
s. t.
xi = 1
min
i
x ≥0
(
)
X
x ∈ Rd xi = 1, x ≥ 0 = Standard Simplex
i
3
c
2015
IBM Corporation
IBM Analytics
Standard Quadratic Programs (SQP)
x T Qx
X
s. t.
xi = 1
−→
min
min
s. t.
i
hQ, Y i
X
xi = 1
i
x ≥0
x ≥0
Y = xx T
(
)
X
x ∈ Rd xi = 1, x ≥ 0 = Standard Simplex
i
3
c
2015
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McCormick
Convex envelope of z = xy :
25
z ≥ xy + xy − xy
z ≥ xy + xy − xy
0
z ≤ xy + xy − xy
z ≤ xy + xy − xy
5
−25
−5
0
x
4
0
5 −5
y
c
2015
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Relaxation-Linearization-Technique (RLT)
Idea
5
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
c
2015
IBM Corporation
IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 1: Bound Constraints
Constraint 1: xi ≤ xi
Constraint 2: xj ≤ xj
5
c
2015
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IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 1: Bound Constraints
Constraint 1: xi ≤ xi
Constraint 2: xj ≤ xj
5
0 ≤ xi − xi xj − xj
c
2015
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IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 1: Bound Constraints
Constraint 1: xi ≤ xi
Constraint 2: xj ≤ xj
0 ≤ xi − xi xj − xj
= xi xj −xi xj − xi xj + xi xj
|{z}
Yij
5
c
2015
IBM Corporation
IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 1: Bound Constraints
Constraint 1: xi ≤ xi
Constraint 2: xj ≤ xj
0 ≤ xi − xi xj − xj
= xi xj −xi xj − xi xj + xi xj
|{z}
Yij
This is one of the 4 McCormick inequalities!
5
c
2015
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IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 2
Constraint:
P
Variable: xj
5
i
xi = 1
X
i
xi xj = xj
|{z}
Yij
c
2015
IBM Corporation
IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 2
Constraint:
P
i
xi = 1
Variable: xj
X
i
xi xj = xj
|{z}
Yij
What happens if some Yij has not been generated?
X
i∈Vj
5
Yij +
X
xi xj = xj
i6∈Vj
c
2015
IBM Corporation
IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 2
Constraint:
P
i
xi = 1
Variable: xj
X
i
xi xj = xj
|{z}
Yij
What happens if some Yij has not been generated?
X
i∈Vj
5
Yij +
X
xi xj ≥ xj
i6∈Vj
c
2015
IBM Corporation
IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 2
Constraint:
P
i
xi = 1
Variable: xj
X
i
xi xj = xj
|{z}
Yij
What happens if some Yij has not been generated?
X
Yij +
i∈Vj
5
xi xj ≥ xj
i6∈Vj
i∈Vj
X
X
Yij +
X
i6∈Vj
xi xj ≤ xj
c
2015
IBM Corporation
IBM Analytics
Relaxation-Linearization-Technique (RLT)
Idea
1
Multiply linear constraints/variables
2
Reformulate using linearization variables
Example 2
Constraint:
P
i
xi = 1
Variable: xj
X
i
xi xj = xj
|{z}
Yij
What happens if some Yij has not been generated?
X
X
i∈Vj
i6∈Vj
X
X
i∈Vj
5
Yij +
Yij +
i6∈Vj
xi xj ≥ xj
xi xj ≤ xj
Use local bounds in
Branch&Bound-tree!
c
2015
IBM Corporation
IBM Analytics
Numbers - SQPs of size 30
CPLEX
Solution status
Optimal
RLT
108
140
2497.28
398.92
590.60
34.32
Separation time in seconds
Mean
Max
0.00
0.00
0.00
0.00
Number of cuts
Mean
Max
0.00
0.00
23.59
30.00
219202.41
34971.71
12054.18
1925.67
957.11
2496.79
62.09
127.86
Runtime in seconds
Mean
Shifted Geomean (s = 10)
Nodes
Mean
Shifted Geomean (s = 100)
Root gap
Mean
Max
150 randomly generated instances of size 30
Timelimit 7200 seconds
6
c
2015
IBM Corporation
IBM Analytics
Table of Contents
1 (Extremely) Short introduction to Quadratic Programming
2 Max Clique Cuts
Separation
Connection to RLT
7
c
2015
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IBM Analytics
Maximum Clique as SQP
Theorem (Motzkin-Straus)
Let A be the adjacency matrix of a simple graph G and ω(G ) its clique
number (i.e. the size of a maximum clique in G ). Then
(
)
X
1
max x T Ax xi = 1, x ≥ 0 = 1 −
ω(G )
i
8
c
2015
IBM Corporation
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Maximum Clique as SQP
Theorem (Motzkin-Straus)
Let A be the adjacency matrix of a simple graph G and ω(G ) its clique
number (i.e. the size of a maximum clique in G ). Then
(
)
X
1
max x T Ax xi = 1, x ≥ 0 = 1 −
ω(G )
i
Observations
8
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2015
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Maximum Clique as SQP
Theorem (Motzkin-Straus)
Let A be the adjacency matrix of a simple graph G and ω(G ) its clique
number (i.e. the size of a maximum clique in G ). Then
(
)
X
1
max x T Ax xi = 1, x ≥ 0 = 1 −
ω(G )
i
Observations
Identification of variables xi with nodes in the graph G = (V , E )
X
X
x T Ax =
2xi xj =
2Yij
(i,j)∈E
8
(i,j)∈E
c
2015
IBM Corporation
IBM Analytics
Maximum Clique as SQP
Theorem (Motzkin-Straus)
Let A be the adjacency matrix of a simple graph G and ω(G ) its clique
number (i.e. the size of a maximum clique in G ). Then
(
)
X
1
max x T Ax xi = 1, x ≥ 0 = 1 −
ω(G )
i
Proposition
If A is the adjacency matrix of some graph G and x in the standard
simplex the following inequality holds:
x T Ax ≤ 1 −
8
1
ω(G )
c
2015
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Max Clique Cuts
Proposition
Every Max Clique cut
hY , Ai ≤ 1 −
1
ω(G )
is valid for the set
(
)
T
(x, Y ) | Y = xx ,
X
xi = 1, x ≥ 0
i
9
c
2015
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Max Clique Cuts
Proposition
Every Max Clique cut
hY , Ai ≤ 1 −
1
ω(G )
is valid for the set
(
)
T
(x, Y ) | Y = xx ,
X
xi = 1, x ≥ 0
i
Can this be of any use?
9
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2015
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Max Clique Cuts
Proposition
Every Max Clique cut
hY , Ai ≤ 1 −
1
ω(G )
is valid for the set
(
)
T
(x, Y ) | Y = xx ,
X
xi = 1, x ≥ 0
i
Can this be of any use? Sure!
In the solution process Y only approximates xx T !
9
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2015
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Separation Problem
Problem
Let (x ∗ , Y ∗ ) be the solution of a suitable relaxation of an SQP. The
task is to find a graph G with adjacency matrix A such that
hA, Y ∗ i > 1 −
1
.
ω(G )
Bilevel Problem
First Level: Compute graph
Second Level: Computs its clique number
10
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2015
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Separation is MIP for fixed clique number k
Separate point (x ∗ , Y ∗ ):
1
k
X
|S| (|S| − 1)
s. t.
Aij ≤
− 1 for all S ⊆ V , |S| = k + 1
2
max
hA, Y ∗ i − 1 +
i,j∈S
i<j
11
Aij = Aji
for all i, j ∈ V
Aii = 0
for all i ∈ V
Aij ∈ {0, 1}
for all i, j ∈ V
c
2015
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Max Clique Cuts for Complete Bipartite Graphs
M
M̄
Clique size 2
12
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2015
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Max Clique Cuts for Complete Bipartite Graphs
M
M̄
Clique size 2
Clean structure:
XX
1
2 xi xj ≤ 1 −
2
|{z}
i∈M i6∈M
12
Yij
c
2015
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Separation: Complete Bipartite Graphs
Given: Relaxation solution (x ∗ , Y ∗ )
Find: Partition M such that
XX
i∈M i6∈M
13
2Yij∗ >
1
2
c
2015
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Separation: Complete Bipartite Graphs
Given: Relaxation solution (x ∗ , Y ∗ )
Find: Partition M such that
XX
2Yij∗ >
i∈M i6∈M
1
2
Solve Integer Quadratic Program:
max
XX
2Yij∗ zi (1 − zj ) −
i∈V j∈V
1
2
s. t. z ∈ {0, 1}V
Value > 0: Found violated cut: M = { i | zi = 1}
Value ≤ 0: Y ∗ fulfills all clique cuts for bipartite graphs
13
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2015
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Numbers - SQPs of size 30
CPLEX
RLT
108
140
140
2497.28
398.92
590.60
34.32
548.51
29.02
Separation time in seconds
Mean
Max
0.00
0.00
0.00
0.00
1.23
22.30
Number of cuts
Mean
Max
0.00
0.00
23.59
30.00
74.00
288.00
219202.41
34971.71
12054.18
1925.67
4978.40
1156.95
957.11
2496.79
62.09
127.86
23.76
93.28
Solution status
Optimal
Runtime in seconds
Mean
Shifted Geomean (s = 10)
Nodes
Mean
Shifted Geomean (s = 100)
Root gap
Mean
Max
Bipartite
150 randomly generated instances of size 30
Timelimit 7200 seconds
14
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2015
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Connecting RLT and Max Clique Cuts
Fix set M and do some algebra:
X
xi
=
1
i∈V
15
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2015
IBM Corporation
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Connecting RLT and Max Clique Cuts
Fix set M and do some algebra:
RLT equalities
X
xi xj =
xj
i∈V
15
c
2015
IBM Corporation
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Connecting RLT and Max Clique Cuts
Fix set M and do some algebra:
Sum up RLT equalities for M
XX
j∈M i∈V
15
xi xj =
X
xj
j∈M
c
2015
IBM Corporation
IBM Analytics
Connecting RLT and Max Clique Cuts
Fix set M and do some algebra:
Sum up RLT equalities for M
XX
xi xj =
j∈M i∈V
Reformulate
XX
j∈M i6∈M
15
X
xj
j∈M
xi xj =
X
j∈M
xj −
XX
xi xj
j∈M i∈M
c
2015
IBM Corporation
IBM Analytics
Connecting RLT and Max Clique Cuts
Fix set M and do some algebra:
Sum up RLT equalities for M
XX
xi xj =
j∈M i∈V
Reformulate
XX
X
xj
j∈M
xi xj =
X
j∈M i6∈M
j∈M
XX
X
xj −
XX
j∈M i∈M
2

Reformulate
j∈M i6∈M
15
xi xj =
j∈M
xi xj
xj − 
X
xj 
j∈M
c
2015
IBM Corporation
IBM Analytics
Connecting RLT and Max Clique Cuts
Fix set M and do some algebra:
Sum up RLT equalities for M
XX
xi xj =
j∈M i∈V
Reformulate
XX
X
xj
j∈M
xi xj =
X
xj −
j∈M i6∈M
j∈M
XX
X
XX
j∈M i∈M
2

Reformulate
j∈M i6∈M
xi xj =
xj − 
j∈M
With g (z) = z − z 2
j∈M i6∈M
15
xi xj = g 
X
xj 
j∈M

XX
xi xj

X
xj 
j∈M
c
2015
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Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Any tangent on g will give an upper bound

XX
j∈M i6∈M
16
xi xj = g 

X
xj 
j∈M
c
2015
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Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Constant Upper Bound

XX
j∈M i6∈M
16
xi xj = g 

X
xj 
j∈M
c
2015
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Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Constant Upper Bound ⇒ Max Clique Cut

XX
j∈M i6∈M
16
xi xj = g 

X
j∈M
1
1
xj  ≤ =
4
2
1
1−
2
c
2015
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Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Only one variable and tangent at 0

X X
j∈M i6∈M
16
xi xj = g 

X
xj 
j∈M
c
2015
IBM Corporation
IBM Analytics
Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Only one variable and tangent at 0 ⇒ RLT inequality

X
xi xj = g 

xj  ≤ xj
i6∈M
16
c
2015
IBM Corporation
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Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Local Bounds on
P
j∈M

XX
j∈M i6∈M
16
xi xj = g 
xj

X
xj 
j∈M
c
2015
IBM Corporation
IBM Analytics
Looking at g (z) = z − z 2
g (z)
0.25
0
0
0.2
0.4
0.6
0.8
1
z
Local Bounds on
P
j∈M

XX
j∈M i6∈M
16
xi xj = g 
xj ⇒ secant approximation

X
j∈M

xj  ≥ g̃ 

X
xj 
Use local bounds in
Branch&Bound-tree!
j∈M
c
2015
IBM Corporation
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Separation of tangents
Separate point (x ∗ , Y ∗ ):
Find set M such that
XX
j∈M i6∈M
Tangent at α =
17
P
i∈M
!
Yij∗ > g
X
xi∗
i∈M
xi∗ separates the point
c
2015
IBM Corporation
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Separation of tangents
Separate point (x ∗ , Y ∗ ):
Find set M such that
XX
!
Yij∗ > g
j∈M i6∈M
Tangent at α =
max
XX
P
i∈M
X
xi∗
i∈M
xi∗ separates the point
Yij∗ zi (1 − zj ) − (α − α2 )
i∈V j∈V
s. t. α =
X
xi∗ zi
i∈V
z ∈ {0, 1}
Value > 0: Tangent at α is violated
Value ≤ 0: No violated tangent exists
17
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2015
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Separation of tangents
Separate point (x ∗ , Y ∗ ):
Find set M such that
XX
!
Yij∗ > g
j∈M i6∈M
Tangent at α =
max
XX
P
i∈M
X
xi∗
i∈M
xi∗ separates the point
Yij∗ zi (1 − zj ) − (α − α2 )
0.6
i∈V j∈V
s. t. α =
X
xi∗ zi
0.4
i∈V
z ∈ {0, 1}
0.2
Value > 0: Tangent at α is violated
Value ≤ 0: No violated tangent exists
17
0
0
0.2 0.4 0.6 0.8
1
c
2015
IBM Corporation
IBM Analytics
Numbers - SQPs of size 30
CPLEX
RLT
Bipartite
108
140
140
142
2497.28
398.92
590.60
34.32
548.51
29.02
470.30
35.16
Separation time in seconds
Mean
Max
0.00
0.00
0.00
0.00
1.23
22.30
320.59
7144.10
Number of cuts
Mean
Max
0.00
0.00
23.59
30.00
74.00
288.00
664.18
4380.00
219202.41
34971.71
12054.18
1925.67
4978.40
1156.95
1792.66
543.90
957.11
2496.79
62.09
127.86
23.76
93.28
14.70
93.28
Solution status
Optimal
Runtime in seconds
Mean
Shifted Geomean (s = 10)
Nodes
Mean
Shifted Geomean (s = 100)
Root gap
Mean
Max
Tangents
150 randomly generated instances of size 30
Timelimit 7200 seconds
18
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2015
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Numbers - SQPs of size 30
CPLEX
RLT
Bipartite
Tangents
TanGap
108
140
140
142
144
2497.28
398.92
590.60
34.32
548.51
29.02
470.30
35.16
403.75
30.99
Separation time in seconds
Mean
Max
0.00
0.00
0.00
0.00
1.23
22.30
320.59
7144.10
103.44
3218.00
Number of cuts
Mean
Max
0.00
0.00
23.59
30.00
74.00
288.00
664.18
4380.00
489.99
3037.00
219202.41
34971.71
12054.18
1925.67
4978.40
1156.95
1792.66
543.90
1789.36
542.01
957.11
2496.79
62.09
127.86
23.76
93.28
14.70
93.28
14.72
93.28
Solution status
Optimal
Runtime in seconds
Mean
Shifted Geomean (s = 10)
Nodes
Mean
Shifted Geomean (s = 100)
Root gap
Mean
Max
150 randomly generated instances of size 30
Timelimit 7200 seconds
19
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2015
IBM Corporation
IBM Analytics
Numbers - SQPs of size 50
CPLEX
RLT
Bipartite
Tangents
TanGap
51
124
128
140
140
15973.21
6765.49
4323.70
314.59
3814.38
248.47
2233.79
286.07
1850.86
206.38
Separation time in seconds
Mean
Max
0.00
0.00
0.00
0.00
12.46
353.10
1885.63
24156.00
1465.41
23491.70
Number of cuts
Mean
Max
0.00
0.00
39.84
50.00
168.30
1534.00
1778.76
8690.00
1159.55
3552.00
286841.90
102931.16
13882.70
3393.86
5931.25
1694.21
4228.04
746.12
4110.92
710.99
1543.18
3997.98
57.33
101.71
22.93
93.73
14.21
93.73
14.27
93.73
Solution status
Optimal
Runtime in seconds
Mean
Shifted Geomean (s = 10)
Nodes
Mean
Shifted Geomean (s = 100)
Root gap
Mean
Max
150 randomly generated instances of size 50
Timelimit 21600 seconds
20
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2015
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Heuristic Separation
Remember: Every graph will do!
21
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2015
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IBM Analytics
Heuristic Separation
Remember: Every graph will do!
Support graph of Y ∗
E = (i, j) | Yij∗ > 0, i 6= j
Compute clique number with graph lib
21
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2015
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Heuristic Separation
Remember: Every graph will do!
Support graph of Y ∗
E = (i, j) | Yij∗ > 0, i 6= j
Compute clique number with graph lib
Lifting
Adding edges is allowed as long as the clique number does not increase.
→ Stronger Cuts
21
c
2015
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Heuristic Separation
Remember: Every graph will do!
Support graph of Y ∗
E = (i, j) | Yij∗ > 0, i 6= j
Compute clique number with graph lib
Lifting
Adding edges is allowed as long as the clique number does not increase.
→ Stronger Cuts
Clique reduction
Removing one edge per clique reduces the clique number by 1.
→ More cuts
21
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2015
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Generalization
Theorem
(
)
T
Y | Y = xx ,
X
xi = 1, x ≥ 0
i
22
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2015
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Generalization
Theorem
(
)
T
Y | Y = xx ,
X
αi xi ≤ β, x ≥ 0
i
22
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2015
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Generalization
Theorem
For every graph G = (V , E ) and α ≥ 0, the cut
X
1
2αi αj xi xj ≤ β 2 1 −
ω(G )
|{z}
(i,j)∈E
Yij
is valid for
(
)
T
Y | Y = xx ,
X
αi xi ≤ β, x ≥ 0
i
22
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2015
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Numbers on Quadratic Knapsack
min x T Qx
s. t. ax ≤ b
x ∈ [0, 1]
Solved in Root
Root Gap
Mean
Shifted Geomean (s = 0.5)
Max
23
CPLEX
RLT
Bipartite
6 / 150
25 / 150
39 / 150
5.65
1.99
66.94
0.29
0.22
2.00
0.11
0.10
0.89
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
24
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
Max Clique Cuts can considerably improve relaxation
24
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
Max Clique Cuts can considerably improve relaxation
Max Clique Cuts can reduce number of nodes and running time
24
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
Max Clique Cuts can considerably improve relaxation
Max Clique Cuts can reduce number of nodes and running time
Max Clique Cuts can be separated (fast)
24
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
Max Clique Cuts can considerably improve relaxation
Max Clique Cuts can reduce number of nodes and running time
Max Clique Cuts can be separated (fast)
Max Clique Cuts have connections to RLT methodology
24
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
Max Clique Cuts can considerably improve relaxation
Max Clique Cuts can reduce number of nodes and running time
Max Clique Cuts can be separated (fast)
Max Clique Cuts have connections to RLT methodology
Max Clique Cuts ROCK!
24
c
2015
IBM Corporation
IBM Analytics
Conclusions
Max Clique Cuts
Max Clique Cuts can considerably improve relaxation
Max Clique Cuts can reduce number of nodes and running time
Max Clique Cuts can be separated (fast)
Max Clique Cuts have connections to RLT methodology
Max Clique Cuts ROCK!
Outlook
Recognize structure in general quadratic problems
24
c
2015
IBM Corporation
IBM Analytics
Jonas Schweiger
CPLEX Optimization, IBM Italy
with Pierre Bonami, Andrea Lodi, Andrea Tramontani
Max Clique Cuts for (Standard) Quadratic Programs
25
c
2015
IBM Corporation
IBM Analytics
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