[PPT] - Strong Convex Nonlinear Relaxations of the Pooling Problem One PowerPoint Presentation

SLIDE 1

Strong Convex Nonlinear Relaxations of the Pooling Problem

One Relaxation to Rule Them All? Claudia D’Ambrosio

LIX, ´ Ecole Polytechnique

Jeff Linderoth Jim Luedtke

Univ. of Wisconsin-Madison

Jonas Schweiger

IBM CPLEX

20th Combinatorial Optimization Workshop Aussois, France, January, 2016

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 1 / 24

SLIDE 2

Warning!

We will go an excited and dangerous quest—much like building a human pyramid

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 2 / 24

SLIDE 3

Warning!

We will go an excited and dangerous quest—much like building a human pyramid Such quests are frought with peril, and it will turn out to be much more difficult than we thought

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 2 / 24

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The Pooling Problem

Inputs I Pools L Outputs J λki αkj Nodes N = I ∪ L ∪ J Arcs A (i, j) ∈ (I × L) ∪ (L × J) ∪ (I × J)

n which materials flow

Material attributes: K Arc capacities: uij Node capacities: Ci, i ∈ N Attribute requirements αkj, k ∈ K, j ∈ J Pooling Problem Maximize linear function of flow through network, obeying capacities, and meeting quality attribute requirements

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 3 / 24

SLIDE 5

Attribute Blending: Bilinear

Inputs have associated attribute concentrations λki, k ∈ K, i ∈ I The concentration of an attribute in pool is the weighted average of the concentrations of its inputs—This results in bilinear constraints.

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 4 / 24

SLIDE 6

Attribute Blending: Bilinear

Inputs have associated attribute concentrations λki, k ∈ K, i ∈ I The concentration of an attribute in pool is the weighted average of the concentrations of its inputs—This results in bilinear constraints. Variables xij: Flow on (i, j) ∈ A qiℓ: Proportion of flow to pool ℓ ∈ L from input i ∈ I. (qiℓ =

xiℓ

j∈J xℓj )

Note also that

i∈I qiℓ = 1

∀ℓ ∈ L

wilj = qilxlj (flow from i through pool ℓ to output j) xil =

j∈J

wiℓj xℓj =

i∈I

wiℓj

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 4 / 24

SLIDE 7

Attribute Blending: Bilinear

Inputs have associated attribute concentrations λki, k ∈ K, i ∈ I The concentration of an attribute in pool is the weighted average of the concentrations of its inputs—This results in bilinear constraints. Variables xij: Flow on (i, j) ∈ A qiℓ: Proportion of flow to pool ℓ ∈ L from input i ∈ I. (qiℓ =

xiℓ

j∈J xℓj )

Note also that

i∈I qiℓ = 1

∀ℓ ∈ L

wilj = qilxlj (flow from i through pool ℓ to output j) xil =

j∈J

wiℓj xℓj =

i∈I

wiℓj Start Strong! We use the PQ-formulation (Sahinidis and Tawarmalani (2005) as our starting point

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 4 / 24

SLIDE 8

Linearizing Attribute Restriction

The concentration requirement constraints can be written as

i∈I

(λki − αkj)xik +

j∈J
i∈I

(λki − αkj)qijxjk ≤ 0 ∀k ∈ K, j ∈ J

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 5 / 24

SLIDE 9

Linearizing Attribute Restriction

The concentration requirement constraints can be written as

i∈I

(λki − αkj)xik +

j∈J
i∈I

(λki − αkj)qijxjk ≤ 0 ∀k ∈ K, j ∈ J Since wiℓj = qiℓxℓj, the attribute constraints have the (linear) form

i∈I

(λki − αkj)xij +

i∈I
ℓ∈L

(λki − αkj)wiℓj ≤ 0 ∀k ∈ K, j ∈ J

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 5 / 24

SLIDE 10

Linearizing Attribute Restriction

The concentration requirement constraints can be written as

i∈I

(λki − αkj)xik +

j∈J
i∈I

(λki − αkj)qijxjk ≤ 0 ∀k ∈ K, j ∈ J Since wiℓj = qiℓxℓj, the attribute constraints have the (linear) form

i∈I

(λki − αkj)xij +

i∈I
ℓ∈L

(λki − αkj)wiℓj ≤ 0 ∀k ∈ K, j ∈ J The feasible region of the pooling problem is P ∩ B where P is a polyhedron, and B = {(x, q, w) | wiℓj = qiℓxℓj ∀i ∈ I, ℓ ∈ L, j ∈ J} The only nonlinearities/nonconvexities in the problem are in B

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 5 / 24

SLIDE 11

The “IP” approach

We wish to optimize a linear function over conv(P ∩ B), so we can first try to create the relaxation P ∩ conv(B)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 6 / 24

SLIDE 12

The “IP” approach

We wish to optimize a linear function over conv(P ∩ B), so we can first try to create the relaxation P ∩ conv(B) But by the result of Sauron G¨ unl¨ uk et al., the standard “McCormick relaxation” describes the convex hull of this non-convex set B wiℓj ≤ Cjqil wiℓj ≤ xℓj wiℓj ≥ Cjqil + xℓj − C

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 6 / 24

SLIDE 13

The “IP” approach

We wish to optimize a linear function over conv(P ∩ B), so we can first try to create the relaxation P ∩ conv(B) But by the result of Sauron G¨ unl¨ uk et al., the standard “McCormick relaxation” describes the convex hull of this non-convex set B wiℓj ≤ Cjqil wiℓj ≤ xℓj wiℓj ≥ Cjqil + xℓj − C Upshot If we want to “beat” the PQ formulation (which is actually quite hard to do), we will need to look at more of the problem

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 6 / 24

SLIDE 14

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 7 / 24

SLIDE 15

Our Quest—Seek Simple Sets Extract a simple but nontrivial set and attempt to convexify

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 7 / 24

SLIDE 16

Our Quest—Seek Simple Sets Extract a simple but nontrivial set and attempt to convexify Intuition for Pooling Will need to include more than just the nonconvexity wiℓj = qiℓxℓj Attribute constraints on outputs are important Idea: Focus on a single output and attribute

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 7 / 24

SLIDE 17

Make It Easier. Focus on Single Output, Single Attribute.

Inputs I Pools L Output J Fix output j and attribute k.

(Drop these indices)

Relevant constraints

i∈I

(λi − α)

γi

xi+

i∈I
ℓ∈L

(λi − α)

γi

wiℓ ≤ 0

i∈I∪L

xi ≤ C

i∈I

qiℓ = 1, ∀ℓ ∈ L wiℓ = qiℓxℓ, ∀i ∈ I, ℓ ∈ L

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 8 / 24

SLIDE 18

Make It Easier. Focus on Single Output, Single Attribute.

Inputs I Pools L Output J Fix output j and attribute k.

(Drop these indices)

Relevant constraints

i∈I

(λi − α)

γi

xi+

i∈I
ℓ∈L

(λi − α)

γi

wiℓ ≤ 0

i∈I∪L

xi ≤ C

i∈I

qiℓ = 1, ∀ℓ ∈ L wiℓ = qiℓxℓ, ∀i ∈ I, ℓ ∈ L Still Too Hard! Try to focus on a single pool ℓ ∈ L and consider rest as a single “by-pass”?

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 8 / 24

SLIDE 19

Make it Easier-er. Focus on Single Pool.

Inputs I Pool ℓ Output j Separate pool ℓ ∈ L from the rest Gray squiggles are now aggregated into variables y and z

i∈I

γixi +

i∈I
t∈L\{ℓ}

γiwit

y

+

i∈I

γiwiℓ ≤ 0

i∈I∪L\{ℓ}

xi

z

+xℓ ≤ C, z ≥ 0, βz ≤ y ≤ βz β, β are the best and worst inputs that can each

utput

β = min

i∈I γi

β := max

i∈I γi

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 9 / 24

SLIDE 20

Single Output, Single Pool Set

Inputs I Pool ℓ Output j wi x z Now also drop the index ℓ for the pool, letting X be the set of (x, y, z, q, w) ∈ R3+2|I| that satisfy y +

i∈I

γiwi ≤ 0, z + x ≤ C,

i∈I

qi = 1, wi = qix, ∀i ∈ I βz ≤ y ≤ βz x, w, z, q ≥ 0

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 10 / 24

SLIDE 21

Single Output, Single Pool Set

Inputs I Pool ℓ Output j wi x z Now also drop the index ℓ for the pool, letting X be the set of (x, y, z, q, w) ∈ R3+2|I| that satisfy y +

i∈I

γiwi ≤ 0, z + x ≤ C,

i∈I

qi = 1, wi = qix, ∀i ∈ I βz ≤ y ≤ βz x, w, z, q ≥ 0 Argh! X is still too complicated One last simplification trick

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 10 / 24

SLIDE 22

More Tricks

“BILBO BAGGINS! Do not take me for some conjuror of cheap tricks!”

—Gandalf

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 11 / 24

SLIDE 23

More Tricks

“BILBO BAGGINS! Do not take me for some conjuror of cheap tricks!”

—Gandalf

Trying to define a tractable relaxation beyond the standard McCormick relaxation is (in my mind) one of the most clever and important contributions of this work.

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 11 / 24

SLIDE 24

A Slight Extended Formulation

Define Variables u =

i∈I γiwi : y + i∈I γiwi ≤ 0

⇒ y + u ≤ 0 t =

i∈I γiqi : u = i∈I γiwi = i∈I γiqi

i ′∈I wi ′ = tx

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 12 / 24

SLIDE 25

A Slight Extended Formulation

Define Variables u =

i∈I γiwi : y + i∈I γiwi ≤ 0

⇒ y + u ≤ 0 t =

i∈I γiqi : u = i∈I γiwi = i∈I γiqi

i ′∈I wi ′ = tx

WLOG, assume output capacity is C = 1. The set Y := {(u, x, t, y, z, w, q) ∈ R5+2|I| | Inequalities below are satisfied} is an extended formulation of X: u = tx y + u ≤ 0 z + x ≤ 1 y ≥ βz y ≤ βz x, z ≥ 0 x =

i∈I

wi t =

i∈I

γiqi 1 =

i∈I

qi w, q ≥ 0 wi = qix, ∀i ∈ I

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 12 / 24

SLIDE 26

The Final Trick

Let Y′ ⊇ Y be the relaxation of the extended formulation obtained by relaxing wi = qix ∀i ∈ I. Then Proju,x,t,y,z(Y′) = T, where T is the 5-variable set The set T ⊂ R5 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ βz y ≤ βz x, z ≥ 0 t ∈ [γ, γ] Inequalities valid for T can be “lifted” to be valid for X by the simple formulae: x =

i∈I

wi t =

i∈I

γiqi u =

i∈I

γiwi

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 13 / 24

SLIDE 27

The Final Trick

Let Y′ ⊇ Y be the relaxation of the extended formulation obtained by relaxing wi = qix ∀i ∈ I. Then Proju,x,t,y,z(Y′) = T, where T is the 5-variable set The set T ⊂ R5 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ βz y ≤ βz x, z ≥ 0 t ∈ [γ, γ] Inequalities valid for T can be “lifted” to be valid for X by the simple formulae: x =

i∈I

wi t =

i∈I

γiqi u =

i∈I

γiwi Case #1 For now, assume “standard case”: β < 0 < β and γ < 0 < γ

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 13 / 24

SLIDE 28

The Fellowship of the Pool

Our “magic” for finding inequalities for nonlinear sets

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 14 / 24

SLIDE 29

The Fellowship of the Pool

Our “magic” for finding inequalities for nonlinear sets Teamwork!

1

Jonas: Characterize extreme points of conv(T). (Most) extreme points are solution to (1-D) parameterized system of equations

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 14 / 24

SLIDE 30

The Fellowship of the Pool

Our “magic” for finding inequalities for nonlinear sets Teamwork!

1

Jonas: Characterize extreme points of conv(T). (Most) extreme points are solution to (1-D) parameterized system of equations

2

Claudia: Use characterization to create an approximating (extreme-point description) polyhedron. Run PORTA to get an inequality description

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 14 / 24

SLIDE 31

The Fellowship of the Pool

Our “magic” for finding inequalities for nonlinear sets Teamwork!

1

Jonas: Characterize extreme points of conv(T). (Most) extreme points are solution to (1-D) parameterized system of equations

2

Claudia: Use characterization to create an approximating (extreme-point description) polyhedron. Run PORTA to get an inequality description

3

Jim: Identify parameterized set of inequalities in PORTA output, and prove validity of conjectured inequality

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 14 / 24

SLIDE 32

The Fellowship of the Pool

Our “magic” for finding inequalities for nonlinear sets Teamwork!

1

Jonas: Characterize extreme points of conv(T). (Most) extreme points are solution to (1-D) parameterized system of equations

2

Claudia: Use characterization to create an approximating (extreme-point description) polyhedron. Run PORTA to get an inequality description

3

Jim: Identify parameterized set of inequalities in PORTA output, and prove validity of conjectured inequality

4

Jeff: Confuse Things. Attempt Jokes for Talks.

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 14 / 24

SLIDE 33

Extreme Point Characterization

Extreme points of conv(T) are given by a short list of “simple” extreme points, and the following, where we set t = ^ t

1

For ^ t ∈ [γ, 0]:

t x z y u ^ t

β β−^ t −^ t β−^ t −^ tβ β−^ t ^ tβ β−^ t

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 15 / 24

SLIDE 34

Extreme Point Characterization

Extreme points of conv(T) are given by a short list of “simple” extreme points, and the following, where we set t = ^ t

1

For ^ t ∈ [γ, 0]:

t x z y u ^ t

β β−^ t −^ t β−^ t −^ tβ β−^ t ^ tβ β−^ t

2

For ^ t ∈ [0, γ]:

t x z y u ^ t

−β ^ t−β ^ t ^ t−β β^ t ^ t−β −β^ t ^ t−β

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 15 / 24

SLIDE 35

Finite Approximation

Consider polyhedron defined by finite subset of these extreme points Use Porta (or other tool) to find inequality description

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 16 / 24

SLIDE 36

Finite Approximation

Consider polyhedron defined by finite subset of these extreme points Use Porta (or other tool) to find inequality description Look for parameterized set(s)

f inequalities

Example Porta Output

INEQUALITIES SECTION ( 1) - 23524x1+ 5600x2 + 1599x4- 8681x5 <= -11200 ( 2) - 22004x1+ 5600x2 + 1579x4- 8301x5 <= -11200 ( 3) - 20044x1+ 5600x2 + 1509x4- 7811x5 <= -11200 ( 4) - 18764x1+ 5600x2 + 1429x4- 7491x5 <= -11200 ( 5) - 18044x1+ 5600x2 + 1369x4- 7311x5 <= -11200 ( 6) - 17812x1+ 5600x2 + 1347x4- 7253x5 <= -11200 ( 7) - 16924x1+ 5600x2 + 1249x4- 7031x5 <= -11200 ( 8) - 15524x1+ 5600x2 + 1039x4- 6681x5 <= -11200 ( 9) - 11452x1+ 2800x2 + 797x4- 4263x5 <= -5600 (10) - 10292x1+ 2800x2 + 767x4- 3973x5 <= -5600 (11) - 10156x1+ 2800x2 + 761x4- 3939x5 <= -5600 (12) - 8356x1+ 2800x2 + 611x4- 3489x5 <= -5600 (13) - 8252x1+ 2800x2 + 597x4- 3463x5 <= -5600 (14) - 7492x1+ 2800x2 + 467x4- 3273x5 <= -5600 (15) - 5428x1+ 1400x2 + 393x4- 2057x5 <= -2800 (16) - 5356x1+ 1400x2 + 391x4- 2039x5 <= -2800 (17) - 4396x1+ 1400x2 + 331x4- 1799x5 <= -2800 (18) - 4460x1+ 1120x2 + 317x4- 1675x5 <= -2240 (19) - 4228x1+ 1120x2 + 311x4- 1617x5 <= -2240 (20) - 4172x1+ 1120x2 + 309x4- 1603x5 <= -2240 (21) - 3956x1+ 1120x2 + 299x4- 1549x5 <= -2240 (22) - 3428x1+ 1120x2 + 255x4- 1417x5 <= -2240 . . . D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 16 / 24

SLIDE 37

Finite Approximation

Consider polyhedron defined by finite subset of these extreme points Use Porta (or other tool) to find inequality description Look for parameterized set(s)

f inequalities

Some “Jim Magic” happens... Example Porta Output

INEQUALITIES SECTION ( 1) - 23524x1+ 5600x2 + 1599x4- 8681x5 <= -11200 ( 2) - 22004x1+ 5600x2 + 1579x4- 8301x5 <= -11200 ( 3) - 20044x1+ 5600x2 + 1509x4- 7811x5 <= -11200 ( 4) - 18764x1+ 5600x2 + 1429x4- 7491x5 <= -11200 ( 5) - 18044x1+ 5600x2 + 1369x4- 7311x5 <= -11200 ( 6) - 17812x1+ 5600x2 + 1347x4- 7253x5 <= -11200 ( 7) - 16924x1+ 5600x2 + 1249x4- 7031x5 <= -11200 ( 8) - 15524x1+ 5600x2 + 1039x4- 6681x5 <= -11200 ( 9) - 11452x1+ 2800x2 + 797x4- 4263x5 <= -5600 (10) - 10292x1+ 2800x2 + 767x4- 3973x5 <= -5600 (11) - 10156x1+ 2800x2 + 761x4- 3939x5 <= -5600 (12) - 8356x1+ 2800x2 + 611x4- 3489x5 <= -5600 (13) - 8252x1+ 2800x2 + 597x4- 3463x5 <= -5600 (14) - 7492x1+ 2800x2 + 467x4- 3273x5 <= -5600 (15) - 5428x1+ 1400x2 + 393x4- 2057x5 <= -2800 (16) - 5356x1+ 1400x2 + 391x4- 2039x5 <= -2800 (17) - 4396x1+ 1400x2 + 331x4- 1799x5 <= -2800 (18) - 4460x1+ 1120x2 + 317x4- 1675x5 <= -2240 (19) - 4228x1+ 1120x2 + 311x4- 1617x5 <= -2240 (20) - 4172x1+ 1120x2 + 309x4- 1603x5 <= -2240 (21) - 3956x1+ 1120x2 + 299x4- 1549x5 <= -2240 (22) - 3428x1+ 1120x2 + 255x4- 1417x5 <= -2240 . . .

“If you sit on the door-step long enough, I daresay you will think of something.”

Bilbo

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 16 / 24

SLIDE 38

Wizard!

A set T ⊂ R5 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ −3z y ≤ 3z x, z ≥ 0 t ∈ [−2, 1]

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 17 / 24

SLIDE 39

Wizard!

A set T ⊂ R5 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ −3z y ≤ 3z x, z ≥ 0 t ∈ [−2, 1] Magic! The following 2 nonlinear inequalities are valid for T (u + 3x)(u + 2x) ≤ 3x(t + 2) 3(x − u) − 2(u + 2x) + 3y + 2(u + 2x)2 y + u + 2x ≤ 3(1 − t)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 17 / 24

SLIDE 40

LOTR Theorem 1: If β < 0, then the following inequality is valid for T: (u − βx)(u − γx) ≤ −βx(t − γ)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 18 / 24

SLIDE 41

LOTR Theorem 1: If β < 0, then the following inequality is valid for T: (u − βx)(u − γx) ≤ −βx(t − γ) Proof is “easy” Tight for all of the “small ^ t” extreme points—by-pass is needed to make up for excess concentration from the pool Inequality is second-order cone representable

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 18 / 24

SLIDE 42

LOTR Theorem 1: If β < 0, then the following inequality is valid for T: (u − βx)(u − γx) ≤ −βx(t − γ) Proof is “easy” Tight for all of the “small ^ t” extreme points—by-pass is needed to make up for excess concentration from the pool Inequality is second-order cone representable LOTR Theorem 2 If β > 0 and γ < 0, then the following inequality is valid for T if y ≥ 0: β(γx − u) + γ(u − γx) + (γ − γ)y + (−γ)(u − γx)2 y + u − γx ≤ β(γ − t)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 18 / 24

SLIDE 43

LOTR Theorem 1: If β < 0, then the following inequality is valid for T: (u − βx)(u − γx) ≤ −βx(t − γ) Proof is “easy” Tight for all of the “small ^ t” extreme points—by-pass is needed to make up for excess concentration from the pool Inequality is second-order cone representable LOTR Theorem 2 If β > 0 and γ < 0, then the following inequality is valid for T if y ≥ 0: β(γx − u) + γ(u − γx) + (γ − γ)y + (−γ)(u − γx)2 y + u − γx ≤ β(γ − t) Proof is more complicated Tight for the “large ^ t” extreme points: pool needs to make up for excess concentration from by-pass

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 18 / 24

SLIDE 44

LOTR Inequality 2

Inequality is only valid if y > 0 Let f(x, y, u) :=    y ≤ 0 (γ − γ)y +

(−γ)(u−γx)2 y+u−γx

y > 0. f(x, y, u) is convex, and the following inequality is valid for T: f(x, y, u) ≤ β(γ − t) − β(γx − u) − γ(u − γx) := ℓ(t, x, u) Reduces to inequality from last slide when y > 0, and also valid when y ≤ 0 Can approximate with linear inequalities. Just separate when necessary by just using tangent inequalities Jeff-jecture: The set {(x, y, u, E) ∈ R4 | f(x, y, u) ≤ E} is not basic semi-algebraic ⇒ Approximating with tangent cutting planes is the best we’re going to be able to do! (in the 5D-space)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 19 / 24

SLIDE 45

A Conjecture

“I don’t know, and I would rather not guess.”

Frodo

The set T ⊂ R5 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ βz y ≤ βz x, z ≥ 0 t ∈ [γ, γ] conv(T) (u − βx)(u − γx) ≤ −βx(t − γ) f(x, y, u) ≤ ℓ(t, x, u) Linear Inequalities from T u ≥ γx u ≥ t + γ(1 − x) u ≤ γx u ≤ t − γ(1 − x)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 20 / 24

SLIDE 46

Other cases

Previous results assumed always good and bad inputs, and also good and bad bypass: β < 0 < β and γ < 0 < γ Important Fact: Not always true in sparse networks (for a particular attribute/product/pool combination)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 21 / 24

SLIDE 47

Other cases

Previous results assumed always good and bad inputs, and also good and bad bypass: β < 0 < β and γ < 0 < γ Important Fact: Not always true in sparse networks (for a particular attribute/product/pool combination) Case #2: All inputs (to pool) “good” (γ ≤ 0) LOTR nonlinear inequality 2 + Following (new) linear inequality −γ(β − γ)x − βt + (γ − γ)y + (β − γ)u ≤ −βγ

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 21 / 24

SLIDE 48

Other cases

Previous results assumed always good and bad inputs, and also good and bad bypass: β < 0 < β and γ < 0 < γ Important Fact: Not always true in sparse networks (for a particular attribute/product/pool combination) Case #2: All inputs (to pool) “good” (γ ≤ 0) LOTR nonlinear inequality 2 + Following (new) linear inequality −γ(β − γ)x − βt + (γ − γ)y + (β − γ)u ≤ −βγ Case #3: All inputs to pool “bad” (γ ≥ 0) LOTR nonlinear inequality 1 + Following (new) linear inequality (γ − β)(γx − u) ≤ −β(γ − t)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 21 / 24

SLIDE 49

Convex Hull: Case #3

The set T ⊂ R5, γ ≥ 0 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ βz y ≤ βz x, z ≥ 0 t ∈ [γ, γ] LOTR Convex Hull Theorem: R = conv(T) y + u ≤ 0 z + x ≤ 1 y − βz ≥ 0 γx ≤ u ≤ γx (γ − β)(γx − u) ≤ −β(γ − t) (u − βx)(u − γx) ≤ −βx(t − γ)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 22 / 24

SLIDE 50

Convex Hull: Case #3

The set T ⊂ R5, γ ≥ 0 u = xt y + u ≤ 0 z + x ≤ 1 y ≥ βz y ≤ βz x, z ≥ 0 t ∈ [γ, γ] LOTR Convex Hull Theorem: R = conv(T) y + u ≤ 0 z + x ≤ 1 y − βz ≥ 0 γx ≤ u ≤ γx (γ − β)(γx − u) ≤ −β(γ − t) (u − βx)(u − γx) ≤ −βx(t − γ) By validity and convexity of inequalities in R, we have: R = conv(R) ⊇ conv(T) ⊇ T In a long, arduous, 5 page proof, we (read Jim) show that ext(R) ⊆ T Which completes the proof that R = conv(T), since ext(R) ⊆ T ⇒ conv(ext(R)) ⊆ conv(T) ⇒ conv(R) ⊆ conv(T)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 22 / 24

SLIDE 51

Computational Results

“No One Believes Me”

J.R.R. Tolkein

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 23 / 24

SLIDE 52

Computational Results

“No One Believes Me”

J.R.R. Tolkein

McCormick relaxation of PQ-Formulation (LP) versus LP+LOTR inequalities, Optimal Value Instances from the literature (exclude those for which LP yields optimal value) These are small instances

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 23 / 24

SLIDE 53

Computational Results

“No One Believes Me”

J.R.R. Tolkein

McCormick relaxation of PQ-Formulation (LP) versus LP+LOTR inequalities, Optimal Value Instances from the literature (exclude those for which LP yields optimal value) These are small instances Preliminary results Instance zPQ zLOTR z∗ adhya1

766.31
697.01
549.8

adhya2

570.79
568.32
549.8

adhya3

571.33
570.60
561.0

adhya4

961.20
923.66
877.6

bental4

550.00
450.00
450.0

haverly1

500.00
400.00
400.0

haverly2

1000.00
600.00
600.0

haverly3

800.00
791.67
750.0

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 23 / 24

SLIDE 54

Computational Results

“No One Believes Me”

J.R.R. Tolkein

McCormick relaxation of PQ-Formulation (LP) versus LP+LOTR inequalities, Optimal Value Instances from the literature (exclude those for which LP yields optimal value) These are small instances Preliminary results Instance zPQ zLOTR z∗ adhya1

766.31
697.01
549.8

adhya2

570.79
568.32
549.8

adhya3

571.33
570.60
561.0

adhya4

961.20
923.66
877.6

bental4

550.00
450.00
450.0

haverly1

500.00
400.00
400.0

haverly2

1000.00
600.00
600.0

haverly3

800.00
791.67
750.0

These results are honestly not that great, but there is more to be done!

“Despair is only for those who see the end beyond all doubt. We do not.”

Gandalf

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 23 / 24

SLIDE 55

Conclusions

“All’s well that ends better”

The Gaffer

Even with five variables and one simple nonlinear inequality! Convex hull is not polyhedral. We conjecture that there is not a basic semi-algebraic description of the convex hull in the original space of variables A key contribution has been to find the “right” set to study

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 24 / 24

SLIDE 56

Conclusions

“All’s well that ends better”

The Gaffer

Even with five variables and one simple nonlinear inequality! Convex hull is not polyhedral. We conjecture that there is not a basic semi-algebraic description of the convex hull in the original space of variables A key contribution has been to find the “right” set to study Continuing Quest Computation! Prove remaining convex hull results Better approximation of by-pass (e.g., upper bounds on different inputs)

D’Ambrosio, Linderoth, Luedtke, Schweiger The Fellowship of the Pool Aussois 2016 24 / 24