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 Aussois 2019 | Christopher Hojny: Strong IP Formulations | 1

Example I – The Good � i =1 2 n − i x i ≤ 2 n − 3 � x ∈ { 0, 1 } n : � n X = Pro: Con: ◮ small IP formulation ◮ 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] n − 1 � x ∈ [0, 1] n : � � x i ≤ n − 2 i =1 ◮ facet description is numerically stable Aussois 2019 | Christopher Hojny: Strong IP Formulations | 2

Example II – The Bad � � x ∈ { 0, 1 } 2 n : � n i =1 2 n − i ( x i + x n + i ) ≤ 2 n − 1 X = Question: Can we use facet description of conv( X )? ◮ consists of Θ (3 n ) inequalities [Kaibel, Loos 2011] ◮ separable in linear time [Loos 2010] ◮ contains badly scaled inequalities, e.g., n − 1 � 2 n − 1 − i ( x i + x n + i ) ≤ 2 n − 1 x n + x 2 n + i =1 ◮ facet description is numerically instable Aussois 2019 | Christopher Hojny: Strong IP Formulations | 3

Questions Every X ⊆ { 0, 1 } n admits IP formulation with { 0, ± 1 } -inequalities, e.g., via infeasibility cuts n � � x ∈ { 0, 1 } n \ X , � (1 − ¯ x i ) x i + ¯ x i (1 − x i ) ≥ 1 ∀ ¯ i =1 but 1. Are such IP formulations strong? 2. If not, what is the minimum size of coefficients in strong IP formulations? Aussois 2019 | Christopher Hojny: Strong IP Formulations | 4

Outline Measure of Size of Coefficients Measure of Strength of IP Formulations The Lower Bound Applications Aussois 2019 | Christopher Hojny: Strong IP Formulations | 5

Measure of Size of Coefficients Inequality a ⊤ x ≤ β is numerically more stable the smaller the ratio � | a i | � ρ ( a ) := max | a j | : a j � = 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 Aussois 2019 | Christopher Hojny: Strong IP Formulations | 6

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 ∈ Z n \ X ◮ strongest formulation uses facets of conv( X ) cutting off all points in R n \ 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 λ Z n we have A ¯ x ∈ 1 of X if for every ¯ x ≤ b iff ¯ x ∈ conv( X ). Aussois 2019 | Christopher Hojny: Strong IP Formulations | 7

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 )) ∩ Z n . ⇔ � Find bounds on coefficients in general IP formulations. Aussois 2019 | Christopher Hojny: Strong IP Formulations | 8

Lower Bound on Coefficients Theorem [H. 2018] Let P ⊆ R n be a full-dimensional integral polytope. Let x ∈ Z n \ P be not cut off by any valid box constraint. ◮ ¯ ◮ ¯ Ax ≤ ¯ b consist of all facet inequalities for P cutting off ¯ x . s := ¯ x − ¯ ◮ ¯ A ¯ b . If some technical assumptions hold, every valid inequality c ⊤ x ≤ δ cutting off ¯ x fulfills | c j | ≥ min k {| ¯ | c i | A ki | − ¯ s k } s k } . max k {| ¯ A kj | + ¯ Aussois 2019 | Christopher Hojny: Strong IP Formulations | 9

Proof Idea Consider P = conv { x ∈ R 2 : 2 x 1 + 7 x 2 ≤ 14, x 1 ≥ 0, x 2 ≥ 0 } . Question: Is there an IP formulation with smaller ρ -value? Idea: x ∈ Z 2 \ P ◮ select ¯ ◮ derive bounds on coefficients in any valid inequality cutting of ¯ x Aussois 2019 | Christopher Hojny: Strong IP Formulations | 10

Proof Idea Consider P = conv { x ∈ R 2 : 2 x 1 + 7 x 2 ≤ 14, x 1 ≥ 0, x 2 ≥ 0 } . Question: Is there an IP formulation with smaller ρ -value? Idea: x ∈ Z 2 \ P ◮ select ¯ ◮ 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. Aussois 2019 | Christopher Hojny: Strong IP Formulations | 10

Example Consider P = conv { x ∈ R 2 : 2 x 1 + 7 x 2 ≤ 14, x 1 ≥ 0, x 2 ≥ 0 } and select ¯ � 4 � x = . 1 Aussois 2019 | Christopher Hojny: Strong IP Formulations | 11

Example Consider P = conv { x ∈ R 2 : 2 x 1 + 7 x 2 ≤ 14, x 1 ≥ 0, x 2 ≥ 0 } and select ¯ � 4 � x = . 1 ◮ up to scaling, every valid inequality has form c ⊤ x ≤ δ : ⇔ (2 − α ) x 1 + (7 − β ) x 2 ≤ 14 ◮ inserting ¯ x yields (2 − α )¯ x 1 + (7 − β )¯ x 2 > 14 ⇔ 4 α + β < 1 ◮ together with α ≥ 0 and β ≥ 0 ρ ( c ) = 7 − β 2 − α > 6 + 4 α 2 − α ≥ 6 2 = 3 Aussois 2019 | Christopher Hojny: Strong IP Formulations | 11

Generalization to General Integer Polytopes Notation: ◮ B smallest box containing P with lower bounds ℓ and upper bounds u ◮ ¯ x ∈ ( B \ P ) ∩ Z n ◮ ¯ 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. Aussois 2019 | Christopher Hojny: Strong IP Formulations | 12

Condition I to Bound | c i | | c j | Sign Restrictions: ◮ For all rows k , k ′ of ¯ A , we have sgn( ¯ A kt ) = sgn( ¯ A k ′ t ) � = 0 ∀ t ∈ { i , j } . ◮ For every row k of A not corresponding to a box constraint, we have sgn( A kt ) ∈ { 0, sgn( ¯ A 1 t ) } ∀ t ∈ { i , j } . Ax ≤ ¯ ¯ s := ¯ x − ¯ x ∈ [ ℓ , u ] \ P Ax ≤ b remaining facets ¯ b facets violated by ¯ x ¯ A ¯ b Aussois 2019 | Christopher Hojny: Strong IP Formulations | 13

Condition II to Bound | c i | | c j | Box Restrictions: ◮ ¯ x i > ℓ i . ◮ ¯ x j < u j . ◮ For every inequality a ⊤ x ≤ β in Ax ≤ b not corresponding to a box constraint, we have n � x t + sgn( ¯ a t ¯ A 1 j ) e j ≤ β . t =1 Ax ≤ ¯ ¯ s := ¯ x − ¯ x ∈ [ ℓ , u ] \ P ¯ b facets violated by ¯ x Ax ≤ b remaining facets ¯ A ¯ b Aussois 2019 | Christopher Hojny: Strong IP Formulations | 14

Condition III to Bound | c i | | c j | Excess Restrictions: ◮ | ¯ s k for every row k of ¯ A ki | ≥ ¯ A . ◮ ¯ x j − ℓ j ≥ max k { ¯ s k } . x ∈ [ ℓ , u ] \ P Ax ≤ ¯ ¯ Ax ≤ b remaining facets s := ¯ x − ¯ ¯ b facets violated by ¯ x ¯ A ¯ b Aussois 2019 | Christopher Hojny: Strong IP Formulations | 15

Lower Bound on Coefficients Theorem [H. 2018] Let P ⊆ R n be a full-dimensional integral polytope. Let x ∈ Z n \ P be not cut off by any valid box constraint. ◮ ¯ ◮ ¯ Ax ≤ ¯ b consist of all facet inequalities for P cutting off ¯ x . s := ¯ x − ¯ ◮ ¯ A ¯ b . If Conditions I–III hold, every valid inequality c ⊤ x ≤ δ cutting off ¯ x fulfills | c j | ≥ min k {| ¯ | c i | A ki | − ¯ s k } s k } . max k {| ¯ A kj | + ¯ Aussois 2019 | Christopher Hojny: Strong IP Formulations | 16

Applications: IP formulations Using this theorem, one can show: ◮ Every IP formulation of n � 2 i x i ≤ 2 n � � x ∈ Z n +1 : + i =0 has ρ -value at least 2 n − 1 . 2 ◮ Consequently, there exist X ⊆ Z n that need exponentially large coefficients in any IP formulation. Aussois 2019 | Christopher Hojny: Strong IP Formulations | 17

Applications: 1 λ -relaxation Recall Ax ≤ b is 1 Ax ≤ λ b is IP formulation of ( λ conv( X )) ∩ Z n . λ -relaxation of X ⇔ ◮ The theorem implies every 1 3 -relaxation of n n � x ∈ { 0, 1 } 3 n +2 : x 3 n +1 + x 3 n +2 + � � 2 3( n − i )+4 x 3( i − 1)+1 + 2 3( n − i )+3) x 3( i − 1)+2 i =1 i =1 n n (2 3( n − i )+4 − 2 3( n − i )+2 ) x 3 i ≤ 1 + (2 3( n − i )+4 − 2 3( n − i )+2 ) � � � + i =1 i =1 n / 3 − 2 − 1 has ρ -value at least 3 . 2 ◮ result can be generalized to exponentially large class of knapsacks Aussois 2019 | Christopher Hojny: Strong IP Formulations | 18

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 Aussois 2019 | Christopher Hojny: Strong IP Formulations | 19