LP Relaxations of Mixed Integer Programs - Eaton.math.rpi.edu

LP Relaxations of Mixed Integer Programs

John E. Mitchell

Department of Mathematical SciencesRPI, Troy, NY 12180 USA

February 2015

Mitchell LP Relaxations 1 / 29

LP Relaxations

LP relaxations

We want to solve an integer program by solving a sequence of linearprograms. We can relax the integrality constraints to give an LPrelaxation. Let

S := {x ∈ Zn : Ax ≥ b, x ≥ 0}= Zn ∩ P

where P := {x ∈ Rn : Ax ≥ b, x ≥ 0}.

Since we restrict x ≥ 0, the polyhedron P has extreme points if it isnonempty.


LP Relaxations

Convex hull of S

TheoremThe convex hull of S is a polyhedron.

The proof is in Nemhauser and Wolsey.

It is important to note that the extreme rays of conv(S) are the extremerays of P.


LP Relaxations

Equivalent LP formulation

Our integer program of interest is

min{cT x : x ∈ S} (IP).

We relate this to the linear program

min{cT x : x ∈ conv(S)} (CIP).

We don’t know the constraints of the linear program (CIP) explicitly.


LP Relaxations

(IP) and (CIP)

r r r r r rr r r r r

r r r

r r

S

�� cT x = z∗


LP Relaxations

(IP) and (CIP)


r r r

r r

S

�� @

@@

��HH

HHH

Hconv(S)

�� cT x = z∗


LP Relaxations

Value of (IP) and (CIP)

TheoremFor any c ∈ Rn, we have:

1 The objective value of (IP) is unbounded below if and only if theobjective value of (CIP) is unbounded below.

2 If (CIP) has a bounded optimal value then it has an optimalsolution that is an optimal solution to (IP). This optimal solution isan extreme point of conv(S).

3 If x0 is an optimal solution to (IP) then it is an optimal solutionto (CIP).

The proof is in the text. It relies on the fact that the minimizer of aconcave function over a convex set is at an extreme point.


LP Relaxations

Possible cases for an integer linear program

Corollary(IP) is either unbounded or infeasible or has an optimal solution.


LP Relaxations

Set up LP relaxation

So in principle, could solve (IP) by solving the linear program (CIP).But we don’t know a polyhedral description of conv(S).

Look at the LP relaxation

min {cT x : Ax ≥ b, x ≥ 0} = min {cT x : x ∈ P}. (LP)


LP Relaxations

Compare (IP) to LP relaxation

We can relate (LP) to (IP) in the following theorem. The proof exploitsthe fact that S ⊆ P.

TheoremLet zLP and z IP be the optimal values of (LP) and (IP) respectively.

1 If P = ∅ then S = ∅.2 If zLP is finite then either S = ∅ or z IP is finite. If S 6= ∅ then

z IP ≥ zLP .3 If xLP is optimal for (LP) and if xLP ∈ Zn then xLP is optimal

for (IP).


LP Relaxations

Cutting plane algorithm illustration

HHHHHH

HH

��

PPPPP

��

��

PPP

��

Solve linear program relaxation

{x ∈ Rn+ : Ax ≥ b}


r r r

r r�� @

@@

��HH

HHH

Hconv(S) tSoln to LP relaxation


LP Relaxations


HHHHHH

HH

��

PPPPP

��

��

PPP

��


{x ∈ Rn+ : Ax ≥ b}


r r r

r r�� @

@@

��HH

HHH

Hconv(S) tSoln to LP relaxation

��

��

��

Cutting planeaT

1 x = b1


LP Relaxations


HHHHHH

HH

��

PPPPP

��

��

PPP

��


{x ∈ Rn+ : Ax ≥ b, aT

1 x ≥ b1}


r r r

r r�� @

@@

��HH

HHH

Hconv(S)

��

��

tSoln to LP relaxation


LP Relaxations


HHHHHH

HH

��

PPPPP

��

��

PPP

��


{x ∈ Rn+ : Ax ≥ b, aT

1 x ≥ b1}


r r r

r r�� @

@@

��HH

HHH

Hconv(S)

��

��

tSoln to LP relaxation

Cutting planeaT

2 x = b2


LP Relaxations


HHHHHH

HH

��

PPPPP

��

��

PPP

��


{y ∈ Rn+ : Ax ≥ b, aT

1 x ≥ b1, aT2 x ≥ b2}


r r r

r r�� @

@@

��HH

HHH

Hconv(S) �

��

d Solution to LP relaxation


LP Relaxations


HHHHHH

HH

��

PPPPP

��

��

PPP

��


{x ∈ Rn+ : Ax ≥ b, aT

1 x ≥ b1, aT2 x ≥ b2}


r r r

r r�� @

@@

��HH

HHH

Hconv(S) �

��

d Solution to LP relaxation

Solution is integral,so optimal to IP


LP Relaxations

A cutting plane approach

An integer program has many LP relaxations.

The aim of a cutting plane approach is to add linear constraints to theoriginal LP relaxation in order to tighten up the LP relaxation.

These linear constraints are satisfied by every point in S but they mightnot be satisfied by every point in P.

Adding these constraints gives an improved LP relaxation of theinteger program.

We would like to improve the relaxation so that it resembles conv(S) insome sense, at least in the neighborhood of an optimal solution to (IP).


Describing Polyhedra by Facets

Polyhedral description of conv(S)

We would like to get a polyhedral description of conv(S).

We’d like as compact a description as possible.

We first set up some notation.

Given a matrix A ∈ Rm×n and a vector b ∈ Rm, we construct thepolyhedron

Q := {x ∈ Rn : Ax ≥ b}.

Note: We have changed notation from Section 1. In particular, anynonnegativity constraints are included in the system Ax ≥ b.



Example 0

Full-dimensional example

Q :=

x ∈ R2 :

x1 + x2 ≤ 6x2 ≤ 6

x1 − x2 ≤ 22x1 + x2 ≤ 10

x1 ≥ 0x2 ≥ 0



Example 1

First example with implicit equality in R2

Q :=

x ∈ R2 :

x1 + x2 ≤ 3x1 + x2 ≥ 3

3x1 + x2 ≥ 3x1 ≥ 0

x2 ≥ 0



Example 2

Second example with implicit equality in R2

Q :=

x ∈ R2 :

2x1 + x2 ≤ 4x1 + x2 ≥ 4x1 ≥ 0

x2 ≥ 0



Implicit equality constraints

We construct three sets of indices:

M := {1, . . . ,m}

M= := {i ∈ M : ATi x = bi ∀x ∈ Q}

M≥ := M \M=

= {i ∈ M : ATi x > bi for some x ∈ Q}.

We also use the notation A=, A≥, b= and b≥ in an analogous way.



Example 0


Q :=

x ∈ R2 :

x1 + x2 ≤ 6 M≥

x2 ≤ 6 M≥

x1 − x2 ≤ 2 M≥

2x1 + x2 ≤ 10 M≥

x1 ≥ 0 M≥

x2 ≥ 0 M≥



Example 1


Q :=

x ∈ R2 :

x1 + x2 ≤ 3 M=

x1 + x2 ≥ 3 M=

3x1 + x2 ≥ 3 M≥

x1 ≥ 0 M≥

x2 ≥ 0 M≥



Example 2


Q :=

x ∈ R2 :

2x1 + x2 ≤ 4 M=

x1 + x2 ≥ 4 M=

x1 ≥ 0 M=

x2 ≥ 0 M≥



Valid inequalities

DefinitionThe inequality πT x ≥ π0 is a valid inequality for Q if it is satisfied by allpoints in Q. We also represent the inequality using the pair (π, π0).

DefinitionLet πT x ≥ π0 be a valid inequality for Q. Let H = {x ∈ Rn : πT x = π0}so F := Q ∩ H is a face of Q. Then (π, π0) represents F . If F 6= ∅ then(π, π0) supports Q.

We extend the notation M=, M≥ to the face Fusing the notation M=

F , M≥F .



Impicit equalities for faces

PropositionIf F is a nonempty face of Q then F is a polyhedron and

F = {x ∈ Rn : aTi x = bi ∀i ∈ M=

F , aTi x ≥ bi ∀i ∈ M≥

F },

where M=F ⊇ M= and M≥

F ⊆ M≥.



Facets

Proposition

If F is a facet of Q then there exists some inequality aTk x ≥ bk with

k ∈ M≥ that represents F .

Proof.Now dim(F ) = dim(P)-1, so rank(A=

F ,b=F ) = rank(A=,b=) + 1.

Therefore, we only need one inequality in M≥ that is in M=F , and this

inequality represents the facet.



One inequality per facet

PropositionFor each facet F of Q, one of the inequalities representing F isnecessary in the description of Q.

Proposition

Every inequality aTi x ≥ bi for i ∈ M≥ that represents a face of Q of

dimension less than dim(Q)− 1 is irrelevant to the description of Q.

Thus, later we will emphasize finding facets of the convex hull of S.



Minimal representation of polyhedron

With these propositions, we can now state two theorems showing howa polyhedron can be minimally represented using linear equalities andinequalities.

Theorem (full dimensional case)A full dimensional polyhedron Q has a unique (to within scalarmultiplication) minimal representation by a finite set of linearinequalities. In particular, for each facet Fi of Q with i = 1, . . . , t , thereis an inequality aT

i x ≥ bi (unique to within scalar multiplication)representing Fi and

Q = {x ∈ Rn : aTi x ≥ bi , i = 1, . . . , t}.



Example 0


Q :=

x ∈ R2 :

x1 + x2 ≤ 6 M≥

x2 ≤ 6 M≥

x1 − x2 ≤ 2 M≥

2x1 + x2 ≤ 10 M≥

x1 ≥ 0 M≥

x2 ≥ 0 M≥



Minimal representation of polyhedron

Theorem (not full dimensional)If dim(Q) = n − k with k > 0 then let t be the number of facets Fi of Q,and we have

Q = {x ∈ Rn : aTi x = bi for i = 1, . . . , k ,

aTi x ≥ bi for i = k + 1, . . . , k + t}.

The set {(ai ,bi) : i = 1, . . . , k} is a maximal set of linearlyindependent rows of (A=,b=).For i = k + 1, . . . , k + t , (ai ,bi) is any inequality from the equivalenceclass of inequalities representing the facet Fi of Q.



Example 1


Q :=

x ∈ R2 :

x1 + x2 ≤ 3 M=

x1 + x2 ≥ 3 M=

3x1 + x2 ≥ 3 M≥

x1 ≥ 0 M≥

x2 ≥ 0 M≥



Example 1


Q :=

x ∈ R2 :

x1 + x2 = 3 M=

x1 + x2 ≥ 3 M=

3x1 + x2 ≥ 3 M≥

x1 ≥ 0 M≥

x2 ≥ 0 M≥



Example 1


Q :=

x ∈ R2 :

x1 + x2 ≤ 3 M=

x1 + x2 = 3 M=

3x1 + x2 ≥ 3 M≥

x1 ≥ 0 M≥

x2 ≥ 0 M≥



Example 2


Q :=

x ∈ R2 :

2x1 + x2 ≤ 4 M=

x1 + x2 ≥ 4 M=

x1 ≥ 0 M=

x2 ≥ 0 M≥



Example 2


Q :=

x ∈ R2 :

2x1 + x2 ≤ 4 M=

x1 + x2 = 4 M=

x1 = 0 M=

x2 ≥ 0 M≥



Example 2


Q :=

x ∈ R2 :

2x1 + x2 = 4 M=

x1 + x2 ≥ 4 M=

x1 = 0 M=

x2 ≥ 0 M≥



Example 2


Q :=

x ∈ R2 :

2x1 + x2 = 4 M=

x1 + x2 = 4 M=

x1 ≥ 0 M=

x2 ≥ 0 M≥



Summary

The aim in a cutting plane approach is to try to develop a minimaldescription of the polyhedron conv(S), at least in the neighborhoodof an optimal point to (IP).

This pair of theorems tells us that it is enough to determine all thefacets of conv(S), as well as a description of the affine hull of conv(S).


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