Extreme functions with an arbitrary number of slopes Amitabh Basu - PowerPoint PPT Presentation

Extreme functions with an arbitrary number of slopes Amitabh Basu ∗ Michele Conforti † Marco Di Summa † Joseph Paat ∗ ∗ Johns Hopkins University † Dipartimento di Matematica, Universit‘a degli Studi di Padova, Italy. Aussois 2016 J.Paat JHU

1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1 The Problem Let b ∈ (0 , 1) , k ∈ N with k ≥ 2 . Find π : [0 , 1] → R + so that (i) π (0) = π (1) = 0 , (ii) (Subadditivity) π ( r 1 + r 2 ) ≤ π ( r 1 ) + π ( r 2 ) , for all r 1 , r 2 ∈ [0 , 1] , (iii) (Symmetry) π ( r ) + π ( b − r ) = 1 , for all r ∈ [0 , 1] , (iv) (Extreme) If π 1 , π 2 satisfy (i)-(iii) and π = π 1 + π 2 then π = π 1 = π 2 . 2 (v) (k-slopes) Piecewise linear, continuous and has k different slopes. J.Paat JHU

The Problem Let b ∈ (0 , 1) , k ∈ N with k ≥ 2 . Find π : [0 , 1] → R + so that (i) π (0) = π (1) = 0 , (ii) (Subadditivity) π ( r 1 + r 2 ) ≤ π ( r 1 ) + π ( r 2 ) , for all r 1 , r 2 ∈ [0 , 1] , (iii) (Symmetry) π ( r ) + π ( b − r ) = 1 , for all r ∈ [0 , 1] , (iv) (Extreme) If π 1 , π 2 satisfy (i)-(iii) and π = π 1 + π 2 then π = π 1 = π 2 . 2 (v) (k-slopes) Piecewise linear, continuous and has k different slopes. 1 Ex. k = 2 , b = 1 / 2 0.8 0.6 Why do we care? 0.4 0.2 0.2 0.4 0.6 0.8 1 J.Paat JHU

The Problem Let b ∈ (0 , 1) , k ∈ N with k ≥ 2 . Find π : [0 , 1] → R + so that (i) π (0) = π (1) = 0 , (ii) (Subadditivity) π ( r 1 + r 2 ) ≤ π ( r 1 ) + π ( r 2 ) , for all r 1 , r 2 ∈ [0 , 1] , (iii) (Symmetry) π ( r ) + π ( b − r ) = 1 , for all r ∈ [0 , 1] , (iv) (Extreme) If π 1 , π 2 satisfy (i)-(iii) and π = π 1 + π 2 then π = π 1 = π 2 . 2 (v) (k-slopes) Piecewise linear, continuous and has k different slopes. 1 Ex. k = 2 , b = 1 / 2 0.8 0.6 Why do we care? 0.4 0.2 0.2 0.4 0.6 0.8 1 J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ R n × m . Consider the feasible region Let A B be a basis from col( A ) Ax = b A B x B + A N x N = b x ∈ Z m x ≥ 0, x B ∈ Z n , x N ∈ Z m − n x ≥ 0, − A − 1 B A N x N = x B − A − 1 − A − 1 B A N x N = x B − A − 1 B b B b x B , x N ≥ 0, x N ≥ 0, x B ∈ Z n , x N ∈ Z m − n x B ∈ Z n , x N ∈ Z m − n � � · The convex hull of Step 4 is B A N x ∈ Z n − A − 1 x ∈ Z m − n : − A − 1 B A N x N ∈ Z n − A − 1 − A − 1 B b B b + Gomory’s corner polyhedron x N ≥ 0, b b �∈ Z n (i.e. x N = 0) then we · If A − 1 x N ∈ Z m − n want to separate it from the solutions J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ R n × m . Consider the feasible region Let A B be a basis from col( A ) Ax = b A B x B + A N x N = b x ∈ Z m x ≥ 0, x B ∈ Z n , x N ∈ Z m − n x ≥ 0, − A − 1 B A N x N = x B − A − 1 − A − 1 B A N x N = x B − A − 1 B b B b x B , x N ≥ 0, x N ≥ 0, x B ∈ Z n , x N ∈ Z m − n x B ∈ Z n , x N ∈ Z m − n � � · The convex hull of Step 4 is B A N x ∈ Z n − A − 1 x ∈ Z m − n : − A − 1 B A N x N ∈ Z n − A − 1 − A − 1 B b B b + Gomory’s corner polyhedron x N ≥ 0, b b �∈ Z n (i.e. x N = 0) then we · If A − 1 x N ∈ Z m − n want to separate it from the solutions J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ R n × m . Consider the feasible region Let A B be a basis from col( A ) Ax = b A B x B + A N x N = b x ∈ Z m x ≥ 0, x B ∈ Z n , x N ∈ Z m − n x ≥ 0, Rearrange − A − 1 B A N x N = x B − A − 1 − A − 1 B A N x N = x B − A − 1 B b B b x B , x N ≥ 0, x N ≥ 0, x B ∈ Z n , x N ∈ Z m − n x B ∈ Z n , x N ∈ Z m − n � � · The convex hull of Step 4 is B A N x ∈ Z n − A − 1 x ∈ Z m − n : − A − 1 B A N x N ∈ Z n − A − 1 − A − 1 B b B b + Gomory’s corner polyhedron x N ≥ 0, b b �∈ Z n (i.e. x N = 0) then we · If A − 1 x N ∈ Z m − n want to separate it from the solutions J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ R n × m . Consider the feasible region Let A B be a basis from col( A ) Ax = b A B x B + A N x N = b x ∈ Z m x ≥ 0, x B ∈ Z n , x N ∈ Z m − n x ≥ 0, Rearrange Drop nonnegativity − A − 1 B A N x N = x B − A − 1 − A − 1 B A N x N = x B − A − 1 B b B b on x B x B , x N ≥ 0, x N ≥ 0, x B ∈ Z n , x N ∈ Z m − n x B ∈ Z n , x N ∈ Z m − n � � · The convex hull of Step 4 is B A N x ∈ Z n − A − 1 x ∈ Z m − n : − A − 1 B A N x N ∈ Z n − A − 1 − A − 1 B b B b + Gomory’s corner polyhedron x N ≥ 0, b b �∈ Z n (i.e. x N = 0) then we · If A − 1 x N ∈ Z m − n want to separate it from the solutions J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ R n × m . Consider the feasible region Let A B be a basis from col( A ) Ax = b A B x B + A N x N = b x ∈ Z m x ≥ 0, x B ∈ Z n , x N ∈ Z m − n x ≥ 0, Rearrange Drop nonnegativity − A − 1 B A N x N = x B − A − 1 − A − 1 B A N x N = x B − A − 1 B b B b on x B x B , x N ≥ 0, x N ≥ 0, x B ∈ Z n , x N ∈ Z m − n x B ∈ Z n , x N ∈ Z m − n ∼ = � � · The convex hull of Step 4 is B A N x ∈ Z n − A − 1 x ∈ Z m − n : − A − 1 B A N x N ∈ Z n − A − 1 B A N x N ∈ Z n − A − 1 − A − 1 − A − 1 B b B b B b + Gomory’s corner polyhedron x N ≥ 0, x N ≥ 0, b b �∈ Z n (i.e. x N = 0) then we · If A − 1 x N ∈ Z m − n x N ∈ Z m − n want to separate it from the solutions J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ R n × m . Consider the feasible region Let A B be a basis from col( A ) Ax = b A B x B + A N x N = b x ∈ Z m x ≥ 0, x B ∈ Z n , x N ∈ Z m − n x ≥ 0, Rearrange Drop nonnegativity − A − 1 B A N x N = x B − A − 1 − A − 1 B A N x N = x B − A − 1 B b B b on x B x B , x N ≥ 0, x N ≥ 0, x B ∈ Z n , x N ∈ Z m − n x B ∈ Z n , x N ∈ Z m − n ∼ = · The convex hull of Step 4 is B A N x N ∈ Z n − A − 1 − A − 1 B b Gomory’s corner polyhedron x N ≥ 0, b b �∈ Z n (i.e. x N = 0) then we · If A − 1 x N ∈ Z m − n want to separate it from the solutions J.Paat JHU

So for R ∈ R n × k and b ∈ R n we have the feasible region � + : Rx ∈ Z n + b � x ∈ Z k . J.Paat JHU

So for R ∈ R n × k and b ∈ R n we have the feasible region � + : Rx ∈ Z n + b � x ∈ Z k . If R = ( r 1 , r 2 , . . . , r k ) then R n R= . . . r 1 r 2 . . . r k . . . x = . . . ← 0 [ x ( r 1 ) x ( r 2 ) . . . x ( r k )] 0 → . . . J.Paat JHU

So for R ∈ R n × k and b ∈ R n we have the feasible region � + : Rx ∈ Z n + b � x ∈ Z k . If R = ( r 1 , r 2 , . . . , r k ) then R n R= . . . r 1 r 2 . . . r k . . . x = . . . ← 0 [ x ( r 1 ) x ( r 2 ) . . . x ( r k )] 0 → . . . J.Paat JHU

So for R ∈ R n × k and b ∈ R n we have the feasible region � + : Rx ∈ Z n + b � x ∈ Z k . If R = ( r 1 , r 2 , . . . , r k ) then R n R= . . . r 1 r 2 . . . r k . . . x = . . . ← 0 [ x ( r 1 ) x ( r 2 ) . . . x ( r k )] 0 → . . . Gomory and Johnson introduced the n-row infinite group relaxation � � x : R n → Z + : rx r ∈ Z n + b , x has finite support � R b ( R n , Z n ) := r ∈ R n J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R , we want to separate x = 0 from + : Rx ∈ Z n + b �� x ∈ Z k �� conv . A valid inequality looks like k � γ i x i ≥ 1 , i =1 where γ i ≥ 0 for each i ∈ [ k ]. J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R , we want to separate x = 0 from + : Rx ∈ Z n + b �� x ∈ Z k �� conv . A valid inequality looks like k � γ i x i ≥ 1 , i =1 where γ i ≥ 0 for each i ∈ [ k ]. We can write γ i =: π ( r i ). J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R , we want to separate x = 0 from + : Rx ∈ Z n + b �� x ∈ Z k �� conv . A valid inequality looks like k � γ i x i ≥ 1 , i =1 where γ i ≥ 0 for each i ∈ [ k ]. We can write γ i =: π ( r i ). A cut-generating function π : R n → R + satisfies � π ( r i ) x i ≥ 1 , r ∈ R n for every x ∈ R b ( R n , Z n ). J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R , we want to separate x = 0 from + : Rx ∈ Z n + b �� x ∈ Z k �� conv . A valid inequality looks like k � γ i x i ≥ 1 , i =1 where γ i ≥ 0 for each i ∈ [ k ]. We can write γ i =: π ( r i ). A cut-generating function π : R n → R + satisfies � π ( r i ) x i ≥ 1 , r ∈ R n for every x ∈ R b ( R n , Z n ). Andersen, Averkov, Balas, Basu, Borozan, Campelo, Conforti, Cornu´ ejols, Daniilidis, Dash, Dey, Gomory, G¨ unl¨ uk, Hildebrand, Hong, Johnson, K¨ oppe, Letchford, Li, Lodi,Miller, Molinaro, Richard, Wolsey, Yıldız, Zambelli, Zhou... J.Paat JHU

A famous cut-generating function — For b ∈ (0 , 1), the Gomory function is  1 b r , 0 ≤ r < b   1 GMI b ( r ) = 1 − b (1 − r ) , b ≤ r < 1  π ( r − j ) , r ∈ [ j , j + 1) , j ∈ Z \ { 0 } .  1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1 GMI 1 / 2 J.Paat JHU