the value function of a mixed-integer linear program with a …ted/files/talks/value_opt08.pdf ·...

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IntroductionStructure of The Value FunctionConstructing the Value Function

Current Work

The Value Function of a Mixed Integer LinearPrograms with a Single Constraint

MENAL GUZELSOYTED RALPHSISE Department

COR@L LabLehigh University

[email protected]

OPT 2008, Atlanta, March 14, 2008

Thanks: Work supported in part by the National Science Foundation

Ted Ralphs, Menal Guzelsoy The Value function of a Mixed-Integer Linear Program with a Single Constraint

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Outline

1 Introduction

2 Structure of The Value FunctionDefinitionsLinear ApproximationProperties

3 Constructing the Value FunctionMaximal Subadditive ExtensionExtending the Value FunctionProcedures

4 Current Work


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Motivation

The goal of this work is to study the structure of the valuefunction of a general MILP.Eventually, we hope this will lead to methods for approximationuseful for

sensitivity analysiswarm startingother methods that requires dual information

Computing the value function (or even an approximation) isdifficult even in a small neighborhood

Our approach is to begin by considering the value functions ofvarious single-row relaxations.


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Previous Work

Johnson [1973,1974,1979], Jeroslow [1978] : the theory ofsubadditive duality for integer linear programs.Pure Integer Programs:

Jeroslow [1982] : Gomory functions - maximum of finitely manysubadditive functions.Lasserre [2004] : generating functions, two-sided Z transformation.Loera et al. [2004] : generating functions, global test set.

Mixed Integer Programs:Jeroslow [1982] : minimum of finitely many Gomory functions.Blair [1995] : Jeroslow formula - consisting of a Gomory functionand a correction term.


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DefinitionsLinear ApproximationProperties

Definitions

We consider the primal problem

minx∈S

cx, (P)

c ∈ Rn, S = {x ∈ Zr+ × Rn−r

+ | a′x = b} with a ∈ Qn, b ∈ R.

The value function of (P) is

z(d) = minx∈S(d)

cx,

where for a given d ∈ R, S(d) = {x ∈ Zr+ × Rn−r

+ | a′x = d}.

Assumptions: Let I = {1, . . . , r}, C = {r + 1, . . . , n}, N = I ∪ C.

z(0) = 0 =⇒ z : R → R ∪ {+∞},N+ = {i ∈ N | ai > 0} 6= ∅ and N− = {i ∈ N | ai < 0} 6= ∅,r < n, that is, |C| ≥ 1 =⇒ z : R → R.


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Example

min 12 x1 + 1

2 x3 + 2x4 + x5 + 34 x6

s.t x1 −32 x2 + 3

2 x3 + x4 − x5 + 13 x6 = b and

x1, x2, x3 ∈ Z+, x4, x5, x6 ∈ R+.

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

Fz


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Lower Bound (LP Relaxation)

The value function of the LP relaxation yields a lower bound.

In this case, it has a convenient closed form:

FL(d) = max{ud | ζ ≤ u ≤ η, u ∈ R} =

ηd if d > 0,0 if d = 0,ζd if d < 0.

where

η = min{ci

ai| i ∈ N+} and ζ = max{

ci

ai| i ∈ N−}.

FL ≤ z.


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Upper Bound (Continuous Relaxation)

To get an upper bound, we consider only the continuousvariables:

FU(d) = min{∑

i∈C

cixi |∑

i∈C

aixi = d, xi ≥ 0 ∀i ∈ C} =

ηCd if d > 00 if d = 0ζCd if d < 0

where

ηC = min{ci

ai| i ∈ C+ = {i ∈ C | ai > 0}} and

ζC = max{ci

ai| i ∈ C− = {i ∈ C | ai < 0}}

By convention:C+ ≡ ∅ → ηC = ∞.C− ≡ ∅ → ζC = −∞.

FU ≥ z


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Example (cont’d)

We have η = 12 , ζ = 0, ηC = 2 and ζC = −1. Consequently,

FL(d) =

{

12 d if d ≥ 00 if d < 0

and FU(d) =

{

2d if d ≥ 0−d if d < 0

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

ηC

η

ζ

ζC


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Observations

{η = ηC}⇐⇒{z(d) = FU(d) = FL(d) ∀d ∈ R+}

{ζ = ζC}⇐⇒{z(d) = FU(d) = FL(d) ∀d ∈ R−}

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

ηC

η

ζ

ζC


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Observations

Let

d+U = sup{d ≥ 0 | z(d) = FU(d)}

d−

U = inf{d ≤ 0 | z(d) = FU(d)},

d+L = inf{d > 0 | z(d) = FL(d)}

d−

L = sup{d < 0 | z(d) = FL(d)}.

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

d−Ld−U d+U

d+L


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Observations

z(d) = FU(d) ∀d ∈ (d−

U , d+U )

d+L ≥ d+

U if d+L > 0 and d−

L ≤ d−

U if d−

L < 0if b ∈ {d ∈ R | z(d) = FL(d)}, then, z(kb) = kFL(b), k ∈ Z+

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

2d−Ld−L

d−U d+Ud+L 2d+L


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Observations

Notice the relation between FU and the linear segments of z: {ηC, ζC}

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

2d−Ld−L

d−U d+Ud+L 2d+L


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Redundant Variables

Let T ⊆ C be such that

t+ ∈ T if and only if ηC < ∞ and ηC =ct+

at+and similarly,

t− ∈ T if and only if ζC > −∞ and ζC =ct−

at−.

and define

ν(d) = min cIxI + cTxT

s.t. aIxI + aTxT = dxI ∈ ZI

+, xT ∈ RT+

Then

ν(d) = z(d) for all d ∈ R.

The variables in C\T are redundant.

z can be represented with at most 2 continuous variables.


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Jeroslow Formula

Let M ∈ Z+ be such that for any t ∈ T, Maj

at∈ Z for all j ∈ I.

Then there is a Gomory function g such that

z(d) = mint∈T

{g(⌊d⌋t) +ct

at(d − ⌊d⌋t)} ∀d ∈ R

where ⌊d⌋t = atM

⌊

Mdat

⌋

.

The Gomory function above is the value function of a relatedPILP:

g(q) = min cIxI + 1M cTxT + z(ϕ)v

s.t aIxI + 1M aTxT + ϕv = q

xI ∈ ZI+, xT ∈ ZT

+, v ∈ Z+

for all q ∈ R, where ϕ = − 1M

∑

t∈T at.


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Piecewise-Linearity of the Value Function

For t ∈ T, setting

ωt(d) = g(⌊d⌋t) +ct

at(d − ⌊d⌋t) ∀d ∈ R,

we can writez(d) = min

t∈Tωt(d) ∀d ∈ R

For t ∈ T,ωt is piecewise linear with finitely many linear segments on anyclosed interval andeach of those linear segments has a slope of ηC if t = t+ or ζC ift = t−.

Thus, z is also piecewise-linear with finitely many linearsegments on any closed interval.

Furthermore, each of those linear segments has a slope of ηC orζC.


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Structure of Linear Pieces

Theorem

If the value function z is linear on an interval U ⊂ R, then there existsa y ∈ ZI

+ such that y is the integral part of an optimal solution for anyd ∈ U. Consequently, for some t ∈ T, z can be written as

z(d) = cI y +ct

at(d − aI y) ∀d ∈ U.

Furthermore, for any d ∈ U, we have d − aI y ≥ 0 if t = t+ andd − aI y ≤ 0 if t = t−.


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Example (cont’d)

T = {4, 5} and hence, x6 is redundant. ηC = 2 and ζC = −1,

0 d1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

z

U1 = [0, 1/2], U2 = [1/2, 1], U3 = [1, 7/6], U4 = [7/6, 3/2], . . .y1 = (0 0 0), y2 = (1 0 0), y3 = (1 0 0), y4 = (0 0 1), . . .

z(d) =

2d if d ∈ U1

−d + 3/2 if d ∈ U2

2d − 3/2 if d ∈ U3

−d + 2 if d ∈ U4

. . .Ted Ralphs, Menal Guzelsoy The Value function of a Mixed-Integer Linear Program with a Single Constraint

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Continuity

ωt+ is continuous from the right and ωt− is continuous from theleft.

ωt+ and ωt− are both lower-semicontinuous.

Theorem

If z is discontinuous at a right-hand-side b ∈ R, then there existsa y ∈ ZI

+ such that b − aI y = 0.

z is lower-semicontinuous.

ηC < ∞ if and only if z is continuous from the right.

ζC > −∞ if and only if z is continuous from the left.

Both ηC and ζC are finite if and only if z is continuous everywhere.


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Example

min x1 − 3/4x2 + 3/4x3

s.t 5/4x1 − x2 + 1/2x3 = b, x1, x2 ∈ Z+, x3 ∈ R+.

ηc = 3/2, ζC = −∞.

-1

-2

12

1

32

2

52

−

12−

32 -1−

52 -2-3−

72 11

2 2 352

32

72

−

12

−

32

−

52

For each discontinuous point di, we have di − (5/4yi1 − yi

2) = 0 and eachlinear segment has the slope of ηC = 3/2.


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Maximal Subadditive ExtensionExtending the Value FunctionProcedures

Maximal Subadditive Extension

Let f : [0, h]→R, h > 0 be subadditive and f (0) = 0.

The maximal subadditive extension of f from [0, h] to R+ is

fS(d) =

f (d) if d ∈ [0, h]

infC∈C(d)

∑

ρ∈C

f (ρ) if d > h ,

C(d) is the set of all finite collections {ρ1, ..., ρR} such thatρi ∈ [0, h], i = 1, ..., R and

PRi=1 ρi = d.

Each collection {ρ1, ..., ρR} is called an h-partition of d.We can also extend a subadditive function f : [h, 0]→R, h < 0 to R

−

similarly.

fS is subadditive and if g is any other subadditive extension of ffrom [0, h] to R+, then g ≤ fS (maximality).


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What if we use z as the seed function?

We can change the “inf” to “min”:

Lemma

Let the function f : [0, h]→R be defined by f (d) = z(d) ∀d ∈ [0, h].Then,

fS(d) =

z(d) if d ∈ [0, h]

minC∈C(d)

∑

ρ∈C

z(ρ) if d > h .

For any h > 0, z(d) ≤ fS(d) ∀d ∈ R+.

Observe that for d ∈ R+, fS(d)→z(d) while h→∞.

Is there an h < ∞ such that fS(d) = z(d) ∀d ∈ R+?


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We can get the value function by extending it from a specificneighborhood.

Theorem

Let dr = max{ai | i ∈ N} and dl = min{ai | i ∈ N} and let the functionsfr and fl be the maximal subadditive extensions of z from the intervals[0, dr] and [dl, 0] to R+ and R−, respectively. Let

F(d) =

{

fr(d) d ∈ R+

fl(d) d ∈ R−

then, z = F.

Outline of the Proof.z ≤ F : By construction.z ≥ F : Using MILP duality, F is dual feasible.

In other words, the value function is completely encoded by thebreakpoints in [dl, dr] and 2 slopes.


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Constructing The Value Function

Two questions

Can we obtain the list of breakpoints efficiently?Can we obtain z(d) for some d 6∈ [dl, dr] from the resultingencoding?

We address the second question first.

Consider evaluating

z(d) = minC∈C(d)

∑

ρ∈C

z(ρ) for d 6∈ [dl, dr].

Can we limit |C|, C ∈ C(d)? Yes!

Can we limit |C(d)|? Yes!


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Theorem

Let d > dr and let kd ≥ 2 be the integer such that d ∈ ( kd2 dr,

kd+12 dr].

Then

z(d) = min{kd

∑

i=1

z(ρi) |kd

∑

i=1

ρi = d, ρi ∈ [0, dr], i = 1, ..., kd}.

Therefore, |C| ≤ kd for any C ∈ C(d).

How about |C(d)|?


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Lower Break Points

Set Ψ be the lower break points of z in [0, dr].

Theorem

For any d ∈ R+\[0, dr] there is an optimal dr-partition C ∈ C(d) suchthat |C\Ψ| ≤ 1.

In particular, we only need to consider the collection

Λ(d) ≡ {H ∪ {µ} | H ∈ C(d − µ) ∩ Ψkd−1,∑

ρ∈H

ρ + µ = d,

µ ∈ [0, dr]}

In other words,

z(d) = minC∈Λ(d)

∑

ρ∈C

z(ρ) ∀d ∈ R+\[0, dr]

Observe that the set Λ(d) is finite.


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Example (cont’d)

For the interval[

0, 32

]

, we have Ψ = {0, 1, 32}. For b = 25

8 , C = { 18 , 3

2 , 32}

is an optimal dr-partition with |C\Ψ| = 1.

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

z(d)U-bpL-bp


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Combinatorial Approach

We can formulate the problem of evaluating z(d) ford ∈ R+\[0, dr] as a Constrained Shortest Path Problem.

Among the paths (feasible partitions) of size kb with exactly kb − 1edges (members of each partition) of each path chosen from Ψ,we need to find the minimum-cost path with a total length of d.


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Recursive Construction

Set Ψ(p) to the set of the lower break points of z in the interval(0, p] p ∈ R+.

1 Let p := dr.2 For any d ∈

`

p, p + p2

˜

, let

z(d) = min{z(ρ1) + z(ρ2) | ρ1 + ρ2 = d, ρ1 ∈ Ψ(p), ρ2 ∈ (0, p]}

Let p := p + p2 and repeat this step.


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We can also do the following:

z(d) = minj

gj(d) ∀d ∈(

p, p +p2

]

where, for each dj ∈ Ψ(p), the functions gj :[

0, p + p2

]

→R ∪ {∞} aredefined as

gj(d) =

z(d) if d ≤ dj,z(dj) + z(d − dj) if dj < d ≤ p + dj,∞ otherwise.

Because of subadditivity, we can then write

z(d) = minj

gj(d) ∀d ∈(

0, p +p2

]

.


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Example (cont’d)

Extending the value function of (2) from[

0, 32

]

to[

0, 94

]

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)


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Example (cont’d)


0, 32

]

to[

0, 94

]

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

g1g2


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Example (cont’d)


0, 32

]

to[

0, 94

]

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)


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Example (cont’d)


0, 94

]

to[

0, 278

]

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)


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Example (cont’d)


0, 94

]

to[

0, 278

]

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)

g2g1

g4

g3


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Example (cont’d)


0, 94

]

to[

0, 278

]

0d

1-1-2-3 3 42-4−

32 −

12−

52−

72

52

32

12

12

32

52

72

1

2

3

FL(d)FU(d)

z(d)


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Finding the Breakpoints

Note thatz overlaps with FU = ηCd in a right-neighborhood of origin and withFU = ζCd in a left-neighborhood of origin.The slope of each linear segment is either ηC or ζC.Furthermore, if both ηC and ζC are finite, then the slopes of linearsegments alternate between ηC and ζC (continuity).For d1, d2 ∈ [0, dr] (or [dl, 0]), if z(d1) and z(d2) are on the line withthe slope of ηC (or ζC), then z is linear over [d1, d2] with the sameslope (subadditivity).

With these observations, we can formulate a finite algorithm toevaluate z in [dl, dr].


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Example (cont’d)

ηC —— ζC ——

0 1 2

32

12

12

32

1

2

Figure: Evaluating z in [0, 32 ]


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Example (cont’d)

ηC —— ζC ——

0 1 2

32

12

12

32

2

1

Figure: Evaluating z in [0, 32 ]


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Current Work

Computational experiments

Extending our results to bounded case.

Extending results to multiple rows.

Approximating the value function of a general MILP using thevalue functions of single constraint relaxations.

Applying results to bilevel programming.


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Jeroslow Formula - General MILP

Let the set E consist of the index sets of dual feasible bases ofthe linear program

min{1M

cCxC :1M

ACxC = b, x ≥ 0}

where M ∈ Z+ such that for any E ∈ E , MA−1E aj ∈ Zm for all j ∈ I.

Theorem (Jeroslow Formular)

There is a g ∈ G m such that

z(d) = minE∈E

g(⌊d⌋E) + vE(d − ⌊d⌋E) ∀d ∈ Rm with S(d) 6= ∅,

where for E ∈ E , ⌊d⌋E = AE⌊A−1E d⌋ and vE is the corresponding basic

feasible solution.


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