cutting plane separators in scip · cut selection strategy. efﬁcacy, i.e., distance of the...

Cutting Plane Separators in SCIP

Kati WolterZuse Institute Berlin

DFG Research Center MATHEONMathematics for key technologies

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General Cutting Plane Method

MIP

min{cT x : x ∈ X MIP}, X MIP := {x ∈ Zn × Rm : Ax ≤ b}

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MIP


Observation

. If the data are rational, then

I conv(X MIP) is a rational polyhedron

I we can formulate the MIP as

min{cT x : x ∈ conv(X MIP)}︸︷︷︸LP

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MIP


Problem (in general)

. Complete linear description of conv(X MIP)?

. Number of constraints needed to describeconv(X MIP) is extremely large

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MIP


Idea

. Construct a polyhedron Q with

I conv(X MIP) ⊆ Q ⊆ X LP

I min{cT x : x ∈ conv(X MIP)}=min{cT x : x ∈ Q}

Start with X LP andadd inequalities which are valid for X MIP

(but violated by the current LP solution)

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Main Solving Loop

Domain Propagation Solve LP

Pricing

Separation

Domain Propagation

LP Solving

Constraint Enforcement

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Two Classes of Cuts

General cuts

. Complemented mixed integerrounding cuts

. Gomory mixed integer cuts

. Strong Chvátal-Gomory cuts

. Implied bound cuts

Problem specific cuts

. 0-1 knapsack problem

. 0-1 single node flow problem

. Stable set problem

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Two Classes of Cuts

General cuts









I want to solve general MIPs!Why do I care about cutting planes for special problems?

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Two Classes of Cuts

General cuts









Which plugins of SCIP generate cuts?

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Constraint Handlers and Separators

Constraint Handlers

. Check a given solution for feasibility w.r.t. all constraints of their type

. Provide linear relaxations of their constraints (in advance or on the fly)

. Provide additional problem specific cuts

Separators

. Provide general cuts

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Constraint Handler or Separator?

Type of cuts?

Separator

GMI

Can your cons be expressed by a“small” nr. of existing cons types?

Can you represent and processyour cons in a more efficient way? Constraint handler

LOP

Separator

0-1 SNFP

Constraint handler

0-1 KP

General cuts Problem specific cuts

Yes No

No Yes

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Some Remarks

. By default, only globally valid cuts are generated.

I FREQ = 0 and SEPAFREQ = 0

. Cuts are generated in rounds.

I MAXROUNDSROOT

. Cuts are first selected in a separation storage.

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Cut Selection Strategy

. Efficacy, i.e., distance of the hyperplane to the LP sol er

. Orthogonality with respect to the other cuts or

. Parallelism with respect to the objective function pr

⇒ Select cuts with largest value ofer + wo ∗ or + wp ∗ pr

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Consequence

. Cut as deep as possible into the current LP polyhedron

. Select cuts that are pairwise almost orthogonal

. Prefer cuts that are close to being parallel to the objective function

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Weights of the criteria can be adjusted

. ORTHOFAC = 1.0

. OBJPARALFAC = 0.0001

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Other useful parameters

. MINEFFICACYROOT = 0.01

. MAXCUTSROOT = 2000

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Outline

1 Complemented Mixed Integer Rounding Cuts

2 Gomory Mixed Integer and Strong Chvátal-Gomory Cuts

3 Implied Bound Cuts

4 Cuts for the 0-1 Knapsack Problem

5 Cuts for the 0-1 Single Node Flow Problem

6 Cuts for the Stable Set Problem

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Outline







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MIR Inequality

Elementary mixed integer set

X := {(x , s) ∈ Z× R : x ≤ a0 + s (I)s ≥ 0 (II)}

x ≤ ba0c+s

1− fa0︸︷︷︸MIR inequality

(fa0 := a0 − ba0c)

s

x

a0

X

conv(X )

(I)

(II)

(III)

(IV )

MIR

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MIR Inequality


X := {(x , s) ∈ Z× R : x ≤ a0 + s (I)s ≥ 0 (II)}

Inequalities (I) and (II) do not suffice todescribe conv(X ).

x ≤ ba0c+s


(fa0 := a0 − ba0c)

s

x

a0

X

conv(X )

(I)

(II)

(III)

(IV )

MIR

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MIR Inequality


X := {(x , s) ∈ Z× R : x ≤ a0 + s (I)s ≥ 0 (II)}

Disjunctive argument

If an inequality is valid for X 1 and X 2 itis also valid for X 1 ∪ X 2.

x ≤ ba0c+s


(fa0 := a0 − ba0c)

s

x

a0

X

conv(X )

(I)

(II)

(III)

(IV )

MIR

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MIR Inequality


X := {(x , s) ∈ Z× R : x ≤ a0 + s (I)s ≥ 0 (II)}

HereX 1 := X ∩ {(x , s) : x ≥ da0e (III)}X 2 := X ∩ {(x , s) : x ≤ ba0c (IV )}

x ≤ ba0c+s


(fa0 := a0 − ba0c)

s

x

a0

Xconv(X )

(I)

(II)

(III)

(IV )

MIR

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MIR Inequality


X := {(x , s) ∈ Z× R : x ≤ a0 + s (I)s ≥ 0 (II)}


Inequality valid for X 1 and X 2

x ≤ ba0c+s

1− fa0︸︷︷︸(I)+fa0 (III) and (II)+(1−fa0 )(IV )

(fa0 := a0 − ba0c)s

x

a0

Xconv(X )

(I)

(II)

(III)

(IV )

MIR

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MIR Inequality


X := {(x , s) ∈ Z× R : x ≤ a0 + s (I)s ≥ 0 (II)}


Inequality valid for X 1 ∪ X 2 = X

x ≤ ba0c+s


(fa0 := a0 − ba0c)s

x

a0

X

conv(X )

(I)

(II)

(III)

(IV )

MIR

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C-MIR InequalityMixed knapsack set

X MK := {(x , s) ∈ Zn+ × R+ :

∑j∈N

ajxj ≤ a0 + s,

xj ≤ bj for all j ∈ N}

. N = {1, . . . , n}

. a0 and aj are rational numbers for all j ∈ N

. bj are nonnegative rational numbers for all j ∈ N

d

Fα(d)

−1 1 2 3α

1

2

3

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C-MIR Inequality

Mixed knapsack set

X MK := {(x , s) ∈ Zn+ × R+ :

∑j∈N

ajxj ≤ a0 + s,


MIR function

Fα : R → Rd 7→ Fα(d) := bdc+ (fd−α)+

1−α

. 0 ≤ α < 1

. fd := d − bdc

. d+ := max{d , 0} d

Fα(d)

−1 1 2 3α

1

2

3

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C-MIR Inequality

Mixed knapsack set

X MK := {(x , s) ∈ Zn+ × R+ :

∑j∈N

ajxj ≤ a0 + s,


MIR inequality∑j∈N

Ffa0(aj)xj ≤ ba0c+

s1− fa0

C-MIR inequality

. Divide cons by δ ∈ Q+\{0}

. Complement int vars (xj = bj − x̄j )

. MIR inequality

d

Fα(d)

−1 1 2 3α

1

2

3

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Example

∑j∈N

ajxj ≤ a0 + s ∑j∈N


s1− fa0

1x1 + 4x2 ≤ 112 + s

Bounds: x1, x2 ≤ 2

1x1 + 4x2 ≤ 5 + 2s

1x1 + 1x2 ≤ 2 + 2s

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Example

∑j∈N



s1− fa0

1x1 + 4x2 ≤ 112 + s


1x1 + 4x2 ≤ 5 + 2s

1x1 + 1x2 ≤ 2 + 2s

For δ = 1

. f 112

= 112 −

⌊ 112

⌋= 1

2

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Example

∑j∈N



s1− fa0

1x1 + 4x2 ≤ 112 + s


1x1 + 4x2 ≤ 5 + 2s

1x1 + 1x2 ≤ 2 + 2s

For δ = 1

. f 112

= 112 −

⌊ 112

⌋= 1

2

d

F 12(d)

−1 1 212

1

2

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Example

∑j∈N



s1− fa0

1x1 + 4x2 ≤ 112 + s


1x1 + 4x2 ≤ 5 + 2s

1x1 + 1x2 ≤ 2 + 2s

For δ = 4, x1 = 2 − x̄1

. − 14 x̄1 + x2 ≤ 7

8 + 14 s

. f 78

= 78 −

⌊ 78

⌋= 7

8

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Example

∑j∈N



s1− fa0

1x1 + 4x2 ≤ 112 + s


1x1 + 4x2 ≤ 5 + 2s

1x1 + 1x2 ≤ 2 + 2s

For δ = 4, x1 = 2 − x̄1

. − 14 x̄1 + x2 ≤ 7

8 + 14 s

. f 78

= 78 −

⌊ 78

⌋= 7

8 d

F 78(d)

−1 1 278

1

2

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Example

∑j∈N



s1− fa0

1x1 + 4x2 ≤ 112 + s


1x1 + 4x2 ≤ 5 + 2s

−1x̄1 + 1x2 ≤ 0 + 2s

For δ = 4, x1 = 2 − x̄1

. − 14 x̄1 + x2 ≤ 7

8 + 14 s

. f 78

= 78 −

⌊ 78

⌋= 7

8 d

F 78(d)

−1 1 278

1

2

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Example

∑j∈N



s1− fa0

1x1 + 4x2 ≤ 112 + s


1x1 + 4x2 ≤ 5 + 2s

1x1 + 1x2 ≤ 2 + 2s

For δ = 4, x1 = 2 − x̄1

. − 14 x̄1 + x2 ≤ 7

8 + 14 s

. f 78

= 78 −

⌊ 78

⌋= 7

8 d

F 78(d)

−1 1 278

1

2

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Outline of the Separation Algorithm

Mixed integer constraints (MIP)

⇓ Aggregation heuristic

Aggregated mixed integer constraint

⇓ Bound substitution heuristic

Mixed knapsack constraint (relaxation of MIP)

⇓

Cut generation heuristic

. Choose δ ∈ Q+\{0}

. Choose U ⊆ N

Violated c-MIR inequality

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Outline of the Separation AlgorithmMixed integer constraints (MIP)





⇓


. Choose δ ∈ Q+\{0}

. Choose U ⊆ N


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Efficiency of the Separation Algorithm

Aggregation heuristic

. Prefer constraints with

I large LP solution value of the dual variableI small densityI small slack

and which

I have not been involved in an aggregation before

Reducing the separation time

. Limit the number of starting constraints

I Use the same criterion as in the aggregation heuristicI MAXFAILS = 150

. In addition

I MAXCUTS = 100 and MAXROUNDS = 50

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Outline







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GMI and Strong CG cuts

Gomory mixed integer (GMI) inequalities

. Equivalent to MIR inequalities

. Pure integer case: dominate CG inequalities

Strong Chvátal-Gomory (CG) inequalities

. No dominance relation to MIR inequalities

. Pure integer case: dominate CG inequalities

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GMI – Sepa Algo






⇓


. Choose δ ∈ Q+\{0} and U ⊆ N


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GMI – Sepa Algo


⇓


. Weights of A−1B

Row of the simplex tableau – for integer var with fractional LP val

⇓

Bound substitution heuristic

. Nearly turned off


⇓


. δ = 1 and U is nearly empty


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Strong CG – Sepa AlgoMixed integer constraints (MIP)

⇓


. Weights of A−1B

Row of the simplex tableau – no real vars with negative coeffs

⇓


. Nearly turned off


⇓


. δ = 1 and U is nearly empty

. Apply strong CG function

Violated strong CG inequality

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Outline







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Implied Bound Cuts

Extended in preprocessing and probing:

Implication graph

Represents logical implications of the form

x = v → y ≤ b

x = v → y ≥ b.

. x is a binary variable and v ∈ {0, 1}

. y ∈ [l , u] is an arbitrary variable

Used to separate implied bound cuts of the form

. y ≤ cx + d (variable upper bounds)

. y ≥ cx + d (variable lower bounds).

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Implied Bound Cuts

Logical implication Implied bound cut

x = 0 → y ≤ b y ≤ (u − b)x + b =

{b : x = 0u : x = 1

}

x = 0 → y ≥ b y ≥ (l − b)x + b =

{b : x = 0l : x = 1

}

x = 1 → y ≤ b y ≤ (b − u)x + u =

{u : x = 0b : x = 1

}

x = 1 → y ≥ b y ≥ (b − l)x + l =

{l : x = 0b : x = 1

}

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Outline







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Cuts for the 0-1 Knapsack Problem0-1 knapsack polytope

conv(X BK ), X BK := {x ∈ {0, 1}n :∑j∈N

ajxj ≤ a0}

. N = {1, . . . , n}

. a0 and aj are integers for all j ∈ N

. 0 ≤ aj ≤ a0 for all j ∈ N

. 5x1 + 6x2 + 2x3 + 2x4 ≤ 8

. Minimal cover: C = {2, 3, 4}

. x2 + x3 + x4 ≤ 2defines a facet ofconv(X BK ∩ {x ∈ {0, 1}4 : x1 = 0})

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Cuts for the 0-1 Knapsack Problem

0-1 knapsack polytope

conv(X BK ), X BK := {x ∈ {0, 1}n :∑j∈N

ajxj ≤ a0}

Minimal cover:

C ⊆ N

with

.∑

j∈C aj > a0

.∑

j∈C\{i} aj ≤ a0 ∀i ∈ C

. 5x1 + 6x2 + 2x3 + 2x4 ≤ 8



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conv(X BK ), X BK := {x ∈ {0, 1}n :∑j∈N

ajxj ≤ a0}

Minimal cover inequality:∑j∈C

xj ≤ |C| − 1

defines a facet of conv(X BK∩{x ∈ {0, 1}n : xj = 0 ∀j ∈ N\C})

. 5x1 + 6x2 + 2x3 + 2x4 ≤ 8



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conv(X BK ), X BK := {x ∈ {0, 1}n :∑j∈N

ajxj ≤ a0}

Sequential up-lifting:

. Used to strengthen minimalcover inequalities

. 5x1 + 6x2 + 2x3 + 2x4 ≤ 8



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Sequential Up-lifting

. X BK = {x ∈ {0, 1}4 : 5x1 + 6x2 + 2x3 + 2x4 ≤ 8},

. C = {2, 3, 4}∑j∈C

xj ≤ 2 valid for X 0 := X BK ∩ {x ∈ {0, 1}4 : x1 = 0}

(I)∑j∈C

xj + α1x1 ≤ 2 valid for X 1 := X BK

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. X BK = {x ∈ {0, 1}4 : 5x1 + 6x2 + 2x3 + 2x4 ≤ 8},

. C = {2, 3, 4}∑j∈C


(I)∑j∈C


Case 1: Inequality (I) is valid for X 1 ∩ {x ∈ {0, 1}4 : x1 = 0}

⇔∑j∈C

xj + α1 · 0 ≤ 2 is valid for X 0

⇔ α1 ∈ [−∞,∞]

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. X BK = {x ∈ {0, 1}4 : 5x1 + 6x2 + 2x3 + 2x4 ≤ 8},

. C = {2, 3, 4}∑j∈C


(I)∑j∈C


Case 2: Inequality (I) is valid for X 1 ∩ {x ∈ {0, 1}4 : x1 = 1}

⇔∑j∈C

xj + α1 · 1 ≤ 2 is valid for all x with 6x2 + 2x3 + 2x4 ≤ 8− 5

⇔ max{∑j∈C

xj : 6x2 + 2x3 + 2x4 ≤ 8− 5, x ∈ {0, 1}4 }+ α1 · 1 ≤ 2

⇔ α1 ≤ 2− 1 = 1

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. X BK = {x ∈ {0, 1}4 : 5x1 + 6x2 + 2x3 + 2x4 ≤ 8},

. C = {2, 3, 4}∑j∈C


(I)∑j∈C


Result: Inequality (I) is valid for X 1 for α1 ≤ 1

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. (j1, . . . , jt) lifting sequence of the variables in N\C

. X i := X BK ∩ {x ∈ {0, 1}n : xji+1 = . . . = xjt = 0}

∑j∈C

xj ≤ |C| − 1 valid for X 0

∑j∈C

xj + αj1xj1 ≤ |C| − 1 valid for X 1

...∑j∈C

xj +t∑

k=1

αjk xjk ≤ |C| − 1 valid for X t = X BK

. Different lifting sequences may lead to different inequalities!

. Use sequential up- and down-lifting!

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Sequential Up- and Down-lifting

TheoremIf C ⊆ N is a minimal cover for X BK and (C1, C2) is any partition of C withC1 6= ∅, then inequality ∑

j∈C1

xj ≤ |C1| − 1

defines a facet of

conv( X BK ∩ {x ∈ {0, 1}n : xj = 0 for all j ∈ N\C,xj = 1 for all j ∈ C2} ).

. Up-lifting: variables in N\C

. Down-lifting: variables in C2

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Outline of the Separation Algorithm

Step 1 (Initial cover)

. Determine an initial cover C for X BK

Step 2 (Minimal cover and partition)

. Make the initial cover minimal by removing vars from C

. Find a partition (C1, C2) of C with C1 6= ∅

Step 3 (Lifting)

. Determine a lifting sequence of the variables in N\C1

. Lift the inequality∑

j∈C1xj ≤ |C1| − 1 using sequential up- and down-lifting

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Outline







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Cuts for the 0-1 Single Node Flow Problem0-1 single node flow set

X SNF := {(x , y) ∈ {0, 1}n × Rn+ :

∑j∈N1

yj −∑j∈N2

yj ≤ b,

yj ≤ ujxj for all j ∈ N}

. N = {1, . . . , n}

. (N1, N2) is a partition of N

. b is a rational number

. uj are nonnegative rational numbers for all j ∈ N

y1 ≤ 3x1

y2 ≤ 3x2

y3≤ 3x3

y4 ≤1x4

y5 ≤ 1x5

2

(C1, C2) = ({1, 2}, {4, 5})

6− 2 = 2 + 2

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Cuts for the 0-1 Single Node Flow Problem

0-1 single node flow set

X SNF := {(x , y) ∈ {0, 1}n × Rn+ :

∑j∈N1

yj −∑j∈N2

yj ≤ b,


. External demand b

. Inflow arcs N1

. Outflow arcs N2

. Capacities

I uj , if j is open (xj = 1)I 0, if j is closed (xj = 0)

. Flow conservation constraint

I inflow − outflow ≤ demand

y1 ≤ 3x1

y2 ≤ 3x2

y3≤ 3x3

y4 ≤1x4

y5 ≤ 1x5

2

(C1, C2) = ({1, 2}, {4, 5})

6− 2 = 2 + 2

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X SNF := {(x , y) ∈ {0, 1}n × Rn+ :

∑j∈N1

yj −∑j∈N2

yj ≤ b,


Flow cover:

(C1, C2)

with

. C1 ⊆ N1 and C2 ⊆ N2

.∑j∈C1

uj −∑j∈C2

uj = b + λ

. λ > 0

y1 ≤ 3x1

y2 ≤ 3x2

y3≤ 3x3

y4 ≤1x4

y5 ≤ 1x5

2

(C1, C2) = ({1, 2}, {4, 5})

6− 2 = 2 + 2

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X SNF := {(x , y) ∈ {0, 1}n × Rn+ :

∑j∈N1

yj −∑j∈N2

yj ≤ b,


Flow cover inequalities:

. SGFCI dominated by

I LSGFCII part. c-MIR inequ. for part.

mixed knapsack relaxation

. EGFCI dominated by

I LFCII part. c-MIR inequ. for part.

mixed knapsack relaxation

y1 ≤ 3x1

y2 ≤ 3x2

y3≤ 3x3

y4 ≤1x4

y5 ≤ 1x5

2

(C1, C2) = ({1, 2}, {4, 5})

6− 2 = 2 + 2

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⇓


. Choose δ ∈ Q+\{0} and U ⊆ N


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⇓ Transformation

0-1 single node flow constraint

⇓


. C1, L1 ⊆ N1\C1: variable ub

. C2, L2 ⊆ N2\C2: variable ub


⇓


. δ > λ (→ dom. SGFCI and EGFCI)

. U = C1 ∪ C2


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Efficiency of the Separation Algorithm


. Extend the candidate set for the value of δ(based on coefficients in 0-1 single node flow constraint)

Extension

. Separate c-MIR inequalities based on flow packs in addition

Reducing the separation time

. Use a similar strategy as in the separator for the c-MIR cut

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Outline







33 / 36

Cuts for the Stable Set Problem“Conflict graph”

G = (V , E)

. V contains a node for each binary variable xi and its complement x̄i

. (xi , xj) ∈ E ⇔ In any feas sol, xi and xj cannot be 1 at the same time

x1

x1

x̄1

x̄1

x2

x2x2

x̄2

x3

x̄3

x̄3

x4

x4

x̄4

x1 + x2 + x̄3 ≤ 1

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Cuts for the Stable Set Problem

“Conflict graph”

G = (V , E)

x1

x1

x̄1

x̄1

x2

x2x2

x̄2

x3

x̄3

x̄3

x4

x4

x̄4

x1 + x2 + x̄3 ≤ 1

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G = (V , E)

Stable set:S ⊆ V

with

. u, v ∈ S ⇒ (u, v) /∈ E

Relaxation of the MIP:

. Each feas sol corresponds toa stable set in the conflict graph

⇒ Stable set polytope isa relaxation of the feas region

x1

x1 x̄1

x̄1

x2

x2

x2

x̄2

x3

x̄3

x̄3

x4

x4 x̄4

x1 + x2 + x̄3 ≤ 1

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G = (V , E)

Clique:C ⊆ V

with

. u, v ∈ S ⇒ (u, v) ∈ E

Clique inequality:∑j∈C

xj ≤ 1

x1

x1 x̄1

x̄1

x2x2

x2

x̄2

x3

x̄3

x̄3

x4

x4

x̄4

x1 + x2 + x̄3 ≤ 1

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G = (V , E)

Separation algo:

. Branch-and-bound algo for themaximum weighted clique problem

x1

x1 x̄1

x̄1

x2x2

x2

x̄2

x3

x̄3

x̄3

x4

x4

x̄4

x1 + x2 + x̄3 ≤ 1

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Two Classes of Cuts

General cuts









35 / 36

Cutting Plane Separators in SCIP

Kati WolterZuse Institute Berlin

DFG Research Center MATHEONMathematics for key technologies

36 / 36

cutting plane separators in scip · cut selection strategy. efﬁcacy, i.e., distance of the...

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