Strong IP Formulations Need Large Coefficients Christopher Hojny - - PowerPoint PPT Presentation

Strong IP Formulations Need Large Coefficients

Christopher Hojny Technische Universität Darmstadt Department of Mathematics Aussois Combinatorial Optimization Workshop 2019

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Example I – The Good

X =

• x ∈ {0, 1}n : n

i=1 2n−i xi ≤ 2n − 3

• Pro:

◮ small IP formulation

Con:

◮ badly scaled coefficients

numerical instabilities

Question: Is there a better IP formulation?

◮ tightest IP formulation consists of facet inequalities of conv(X) ◮ complete linear description [Laurent, Sassano 1992]

• x ∈ [0, 1]n :

n−1

• i=1

xi ≤ n − 2

• ◮ facet description is numerically stable

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Example II – The Bad

X =

• x ∈ {0, 1}2n : n

i=1 2n−i(xi + xn+i) ≤ 2n − 1

• Question: Can we use facet description of conv(X)?

◮ consists of Θ(3n) inequalities [Kaibel, Loos 2011] ◮ separable in linear time [Loos 2010] ◮ contains badly scaled inequalities, e.g.,

xn + x2n +

n−1

• i=1

2n−1−i(xi + xn+i) ≤ 2n−1

◮ facet description is numerically instable

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Questions

Every X ⊆ {0, 1}n admits IP formulation with {0, ±1}-inequalities, e.g., via infeasibility cuts

n

• i=1
• (1 − ¯

xi) xi + ¯ xi (1 − xi)

• ≥ 1

∀¯

x ∈ {0, 1}n \ X, but

• 1. Are such IP formulations strong?
• 2. If not, what is the minimum size of coefficients in strong IP formulations?

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Outline

Measure of Size of Coefficients Measure of Strength of IP Formulations The Lower Bound Applications

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Measure of Size of Coefficients

Inequality a⊤x ≤ β is numerically more stable the smaller the ratio

ρ(a) := max |ai| |aj| : aj = 0, i = j

• .

Definition The ρ-value of an IP formulation Ax ≤ b is max{ρ(a) : a is row of A}.

the smaller the ρ-value the higher the numerical stability

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Strong IP Formulations

Question: How to measure strength of IP formulation Ax ≤ b?

◮ IP formulation of X ⊆ {0, 1}n has to cut off points x ∈ Zn \ X ◮ strongest formulation uses facets of conv(X) cutting off all points

in Rn \ conv(X) Idea: Refine IP formulations by not only cutting of infeasible binary points but also points in a refinement of the integer lattice. Definition Let λ ∈ Z>0 and X ⊆ {0, 1}n. Then Ax ≤ b is called 1

λ-relaxation

• f X if for every ¯

x ∈ 1

λ Zn we have A¯

x ≤ b iff ¯ x ∈ conv(X).

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Measuring Strength of IP Formulations

Measure of Strength If Ax ≤ b is a 1

λ-relaxation of X, then it is the stronger the larger λ.

In particular, 1

λ-relaxations might have small coefficients, while facets of conv(X)

have large coefficients. But: How to estimate the size of coefficients in 1

λ-relaxations?

Observation Ax ≤ b, x ∈ [0, 1]n is 1

λ-relaxation of X

⇔

Ax ≤ λ b, x ∈ [0, λ]n is IP formulation of (λ conv(X)) ∩ Zn.

Find bounds on coefficients in general IP formulations.

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Lower Bound on Coefficients

Theorem [H. 2018] Let P ⊆ Rn be a full-dimensional integral polytope. Let

◮ ¯

x ∈ Zn \ P be not cut off by any valid box constraint.

◮ ¯

Ax ≤ ¯ b consist of all facet inequalities for P cutting off ¯ x.

◮ ¯

s := ¯ A¯ x − ¯ b. If some technical assumptions hold, every valid inequality c⊤x ≤ δ cutting off ¯ x fulfills

|ci| |cj| ≥ mink{| ¯

Aki| − ¯ sk} maxk{| ¯ Akj| + ¯ sk}.

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Proof Idea

Consider P = conv{x ∈ R2 : 2 x1 + 7 x2 ≤ 14, x1 ≥ 0, x2 ≥ 0}. Question: Is there an IP formulation with smaller ρ-value? Idea:

◮ select ¯

x ∈ Z2 \ P

◮ derive bounds on coefficients in any

valid inequality cutting of ¯ x

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Proof Idea

Consider P = conv{x ∈ R2 : 2 x1 + 7 x2 ≤ 14, x1 ≥ 0, x2 ≥ 0}. Question: Is there an IP formulation with smaller ρ-value? Idea:

◮ select ¯

x ∈ Z2 \ P

◮ derive bounds on coefficients in any

valid inequality cutting of ¯ x Lemma If P is a full-dimensional polytope, every valid inequality for P is a conic combination of facet defining inequalities.

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Example

Consider P = conv{x ∈ R2 : 2 x1 + 7 x2 ≤ 14, x1 ≥ 0, x2 ≥ 0} and select ¯ x =

4

1

• .

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Example

Consider P = conv{x ∈ R2 : 2 x1 + 7 x2 ≤ 14, x1 ≥ 0, x2 ≥ 0} and select ¯ x =

4

1

• .

◮ up to scaling, every valid inequality has form

c⊤x ≤ δ :⇔ (2 − α)x1 + (7 − β)x2 ≤ 14

◮ inserting ¯

x yields (2 − α)¯ x1 + (7 − β)¯ x2 > 14

⇔

4α + β < 1

◮ together with α ≥ 0 and β ≥ 0

ρ(c) = 7 − β

2 − α > 6 + 4α 2 − α ≥ 6 2 = 3

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Generalization to General Integer Polytopes

Notation:

◮ B smallest box containing P with

lower bounds ℓ and upper bounds u

◮ ¯

x ∈ (B \ P) ∩ Zn

◮ ¯

Ax ≤ ¯ b facets violated by ¯ x

◮ Ax ≤ b remaining facets

Idea: To find lower bounds on the ρ-value, use the previous lemma to find upper and lower bounds on coefficients in inequalities cutting of given ¯ x. Problem: Bounds depend on conic multipliers.

Introduce further conditions to get rid of multipliers.

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Condition I to Bound |ci|

|cj|

Sign Restrictions:

◮ For all rows k, k′ of ¯

A, we have sgn( ¯ Akt) = sgn( ¯ Ak′t) = 0

∀t ∈ {i, j}.

◮ For every row k of A not corresponding to a box constraint, we have

sgn(Akt) ∈ {0, sgn( ¯ A1t)}

∀t ∈ {i, j}.

¯ x ∈ [ℓ, u] \ P ¯ Ax ≤ ¯ b facets violated by ¯ x Ax ≤ b remaining facets ¯ s := ¯ A¯ x − ¯ b

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Condition II to Bound |ci|

|cj|

Box Restrictions:

◮ ¯

xi > ℓi.

◮ ¯

xj < uj.

◮ For every inequality a⊤x ≤ β in Ax ≤ b not corresponding to a box constraint,

we have

n

• t=1

at ¯ xt + sgn( ¯ A1j) ej ≤ β.

¯ x ∈ [ℓ, u] \ P ¯ Ax ≤ ¯ b facets violated by ¯ x Ax ≤ b remaining facets ¯ s := ¯ A¯ x − ¯ b

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Condition III to Bound |ci|

|cj|

Excess Restrictions:

◮ | ¯

Aki| ≥ ¯ sk for every row k of ¯ A.

◮ ¯

xj − ℓj ≥ maxk{¯ sk}.

¯ x ∈ [ℓ, u] \ P ¯ Ax ≤ ¯ b facets violated by ¯ x Ax ≤ b remaining facets ¯ s := ¯ A¯ x − ¯ b

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Lower Bound on Coefficients

Theorem [H. 2018] Let P ⊆ Rn be a full-dimensional integral polytope. Let

◮ ¯

x ∈ Zn \ P be not cut off by any valid box constraint.

◮ ¯

Ax ≤ ¯ b consist of all facet inequalities for P cutting off ¯ x.

◮ ¯

s := ¯ A¯ x − ¯ b. If Conditions I–III hold, every valid inequality c⊤x ≤ δ cutting off ¯ x fulfills

|ci| |cj| ≥ mink{| ¯

Aki| − ¯ sk} maxk{| ¯ Akj| + ¯ sk}.

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Applications: IP formulations

Using this theorem, one can show:

◮ Every IP formulation of

• x ∈ Zn+1

+

:

n

• i=0

2ixi ≤ 2n has ρ-value at least 2n−1

2

.

◮ Consequently, there exist X ⊆ Zn that need exponentially large coefficients in

any IP formulation.

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Applications: 1

λ-relaxation

Recall Ax ≤ b is 1

λ-relaxation of X

⇔

Ax ≤ λ b is IP formulation of (λ conv(X)) ∩ Zn.

◮ The theorem implies every 1 3-relaxation of

• x ∈ {0, 1}3n+2 : x3n+1 + x3n+2 +

n

• i=1

23(n−i)+4x3(i−1)+1 +

n

• i=1

23(n−i)+3)x3(i−1)+2 +

n

• i=1

(23(n−i)+4 − 23(n−i)+2)x3i ≤ 1 +

n

• i=1

(23(n−i)+4 − 23(n−i)+2)

• has ρ-value at least 3

n/3−2−1

2

.

◮ result can be generalized to exponentially large class of knapsacks

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Consequences

Corollary There exist sets X ⊆ {0, 1}n that need exponentially large coefficients in every 1

λ-relaxation if λ ≥ 3.

Remark: result can be generalized to case λ ≥ 2 Consequences:

◮ every X ⊆ {0, 1}n admits 1-relaxation with {0, ±1}-coefficients. ◮ 1 2-relaxations may need large coefficients ◮ strong IP formulations need exponentially large coefficients in general

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Consequences

Corollary There exist sets X ⊆ {0, 1}n that need exponentially large coefficients in every 1

λ-relaxation if λ ≥ 3.

Remark: result can be generalized to case λ ≥ 2 Consequences:

◮ every X ⊆ {0, 1}n admits 1-relaxation with {0, ±1}-coefficients. ◮ 1 2-relaxations may need large coefficients ◮ strong IP formulations need exponentially large coefficients in general

Thank You For Your Attention

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