i.4 polyhedral theory 1. integer programming 2011 2 objective of study: want to know how to...

I.4 Polyhedral Theory

1

Integer Programming 2011 2

Objective of Study: want to know how to describe the convex hull of the so-lution set to the IP problem S = {xZ+

n : Ax b} using linear inequalities (at least approximately). Which inequalities are necessary? How can we identify the inequalities necessary (or strong) to describe the convex hull?

Def 1.1: Given S Rn, x Rn is a convex combination of points of S if there exists {xi}i=1

t S such that x = i=1t ixi, i=1

t i = 1, R+t.

Convex hull of S:

set of all convex combination of points in S (inside description)

Intersection of all convex sets containing S (outside description)

linearly dependent: {x1, … , xk} some xi can be described as a linear combina-tion of the remaining xj‘s.

linearly independent: x1, … , xk Rn linearly independent i=1k ixi = 0 im-

plies i = 0 i. (i.e. not linearly dependent)

Subspace;

The set closed under addition and scalar multiplication (or arbitrary linear combination) of elements in the set.

Given a set A Rn,

Inside description: (subspace generated by A Rn)

Outside description: linear hull of A Rn, (intersection of all subspaces con-taining A)

Prop: x1, … , xk Rn linearly independent and x0 = ixi,

Then (1) all I’s unique and (2) {x1, … , xk}{x0}\{xj} linearly independent j 0.

Def: Rank of a matrix A: mn : maximum number of linearly independent rows of A (= maximum number of linearly independent columns of A)

Basis of A Rn : linearly independent subset of A which generates all of A.

(minimal generating set in A, maximal independent set in A)

Rank of A Rn : basis size Integer Programming 2011 3

Basis equicardinality property Prop 1.2: The following statements are equivalent:

a) {xRn: Ax = b} b) rank(A) = rank(A, b)

ixi is an affine combination if i = 1.

Affine space: the set closed under affine combination, i.e. x1, … , xk L, i=1k i

= 1 ixi L.

Note: affine space L = S + {a}, for some a L, S: subspace. Constrained form: subspace S = {x: Ax = 0}, affine space L = {x: Ax b}

Affine span, affine hull of A Rn. Def: x1, … , xk Rn affinely dependent if some xi can be expressed as an

affine combination of remaining xj, j i. Otherwise affinely independent.

Def 1.4: x1, … , xk Rn affinely independent if the unique solution of i=1k ixi

= 0, i=1k i = 0 is i = 0 for i = 1, … , k.


Prop 1.3: The following statements are equivalent:a. x1, … , xk Rn are affinely independent.

b. x2 – x1, … , xk – x1 are linearly independent.

c. (x1, -1), … , (xk, -1) Rn+1 are linearly independent.

Def: 1. Affine rank of A Rn is the maximum number of affinely independent points in

A.

2. dim(L) = dim(S), L is affine space and L = S + {a}

Maximum number of affinely independent points in Rn is n+1. (n linearly in-dependent points + 0 vector)


Prop 1.4: If {xRn: Ax = b} , maximum number of affinely independent so-lutions of Ax = b is n+1-rank(A). (Compare with rank(A) + nullity(A) = n)

Solution of Ax = b is translation of solution of Ax = 0.

Solution set of Ax = 0 is null space (orthogonal subspace) of rows of A whose dimension is n – rank(A) affine rank is (n+1) – rank(A).

Def: p Rn, H subspace, then the projection of p on H is q H such that p-q H. S Rn, the projection of S on H is denoted by projH(S) = {q: q is projec-

tion of p on H for some p S}.


2.Definitions of Polyhedra and Dimension

Def: Polyhedron is the set of points that satisfy a finite number of linear in-equalities, i.e. P = {xRn : Ax b}. (outside description)

Bounded polyhedron is polytope (convex hull of finitely many points)

T Rn is a convex set if x1, x2 T implies that x1 + (1- )x2 T for all 0 1.

Cone C Rn : x C x C, R+1 (we only consider convex, polyhedral

cones) Prop: Polyhedron is a convex set. Prop: P = {xRn : Ax 0} is a cone.

Def: A polyhedron P is of dimension k, denoted dim(P) = k, if the maximum number of affinely independent points in P is k+1. (dimension of the small-est affine space containing P.)

Def: A polyhedron P Rn is full-dimensional if dim(P) = n.


Notation:

M = {1, … , m}, m: number of constraints

M= = {iM: aix = bi, xP}

M = {iM: aix < bi, for some xP} = M\M=

(A=, b=), (A, b) are corresponding rows of (A, b)

P = {xRn: A=x = b=, Ax b} Note that if i M , then (ai, bi) cannot be written as a linear combination of

the rows of (A=, b=).

Def: inner point x: aix < bi, for all i M

Def: interior point x: aix < bi, for all i M

Prop 2.3: Every nonempty polyhedron P has an inner point.

Pf) If M = , every point of P is inner. For each i M, there exists xiP such that aixi < bi. Let x* = (1/|M|)iM xi P,

then akx* < bk, for all k M, hence inner point. Integer Programming 2011 8

Prop 2.4: P Rn (P ), then dim(P) + rank(A=, b=) = n. (see Prop 1.4)

Pf) Suppose rank(A=) = rank(A=, b=) = n-k, 0 k n.

dim{x: A=x = 0} = k k linearly independent points, say y1, … , yk.

Let x* be an inner point (existence guaranteed)Þ x* + yi P for small > 0 and x*, x* + y1, … , x* + yk, affinely indepen-

dent (since (x* + yi) – x* linearly independent)Þ dim(P) k dim(P) + rank(A=, b=) nNow suppose dim(P) = k and x0, x1, … , xk are affinely indep. points of P.

xi – x0 are linearly independent and A=(xi – x0) = 0

nullity(A=) k rank(A=) = rank(A=, b=) n-k

dim(P) + rank(A=, b=) n

Cor: P is full-dimensional if and only if P has an interior point.


3. Describing Polyhedra by Facets

Def: x 0 [or (, 0)] is called a valid inequality for P if it is satisfied by all

x P.

(, 0) valid if and only if max{x: x P} 0

Def: (, 0) valid. F = {x P: x = 0} is called a face of P and (, 0) repre-

sents F. F is said to be proper if F and F P.

F max{x: x P} = 0

If F , we say that (, 0) supports P.

Prop: F P nonempty face of P c Rn such that cx is maximized over P precisely on F.


Prop 3.1: P = {xRn : Ax b} with equality set M= M, F is a nonempty face of P. Then, F is a polyhedron and F = {xRn : aix = bi, iMF

=, aix bi, iMF},

where MF= M=, MF

= M \ M=. The number of distinct faces of P is finite.

Pf) Let F be the set of optimal solutions to 0 = max{ x: Ax b}. Let u* be

an optimal solution to min{ ub: uA = , u 0}, I* = { i: ui* > 0}.

Let F* = {xRn : aix = bi, iI*, aix bi, iM \ I* }

Show F = F*

1) Suppose x F*, then x = u*Ax = iI* ui*aix = iI* ui

*bi = 0 (u*A = from u*

dual feasible) x F, hence F* F.

2) Suppose x P \ F*, then akx bk for some k I*.

Þ x = iI* ui*aix < iI* ui

*bi = 0 x F, hence F F*.

(showed x F* x F, i.e. x F x F* )

From 1), 2), F = F*, F polyhedron. Since F P, the equality set (AF, bF

) of

F must have the required property.

Finally, since M is finite, possible equality set MF= is finite, so the number of

distinct faces is finite. Integer Programming 2011 11

Def: F is a facet if dim(F) = dim(P) – 1.

Prop 3.2: If F is a facet of P, akx bk, k M representing F.

Pf) dim(F) = dim(P) - 1 rank(AF=, bF

=) = rank(A=, b=) + 1. Prop 3.3: For each facet F of P, one of the inequalities representing F is nec-

essary in the description of P.

Pf) Let PF be the polyhedron obtained by dropping inequalities representing

F. Show PF \ P .

Let x* be an inner point of F, and arx br be an inequality representing F.

ar linearly independent of rows of A= does not exist x such that xA= = ar

By thm of alternatives, y such that A=y = 0, ary > 0 (thm of alternatives for subspaces)

x* inner point of F aix* < bi, i M \{inequalities representing F}

Now ai(x* + y) = aix* + aiy = bi, i M=

ai(x* + y) = aix* + aiy < bi, i M \{inequalities representing F}

ar(x* + y) = arx* + ary > br

x* + y PF \ P for small > 0.

(For pf of thm of alt, may consider (P) min 0x, xA = ar, (D) max ary, Ay = 0)Integer Programming 2011 12

Prop 3.4: Every inequality representing a face of P of dimension less than dim(P) – 1 is irrelevant to the description of P.

When two inequalities (1, 01) and (2, 0

2) are equivalent in the description of

P?

{x: A=x = b=, x 0} = {x: A=x = b=, ( + A=)x 0 + b=} >0, R|M=|

Hence equivalent if (2, 02) = (1, 0

1) + (A=, b=) for some >0, R|M=|

Thm 3.5: a. P full-dimensional P has a unique representation (to within positive scalar

multiplication) by a finite set of inequalities.

b. If dim(P) = n-k, k>0, then

P = {xRn : aix = bi, i = 1, … , k, aix bi, i = k+1, … , k+t}.

For i = 1, … , k, (ai, bi) are a maximal set of linearly independent rows of (A=, b=)

For i = k+1, … , k+t, (ai, bi) is any inequality from the equivalence class of in-

equalities representing Fi.


Thm 3.6: F = {x P: x = 0} proper face of P. Then the following two state-

ments are equivalent:1) F is a facet of P.

2) If x = 0 x F, then (, 0) = (A=, 0 + b=) for some R1 and R|M=|

Pf) 2) 1) : Let L = {(, 0)Rn+1: (, 0) satisfies (, 0)=( A=, 0 + b=)}

(generated set)

L’ = {(, 0)Rn+1: x = 0, x F} (constrained set)

L L’ since x A=x = 0 + b= x F.

By hypothesis of 2), L’ L L = L’ (L, L’ subspaces)

Suppose dim(P) = n –k rank(A=, b=) = k dim(L) = k+1

(F proper face (, 0) linearly independent of (A=, b=))

Suppose x1, … , xr maximal affinely independent points in F.

(continued)


Þ D = rank r

Consider (, 0)DT = 0

Maximum number of affinely independent solution is (n+1) + 1 – rank(D) = n+2-r dim(L’) = n+1-r = k+1 r = n-k F is a facet.

1) 2):

As above, L L’. Show L = L’

Suppose dim(P) = (n-k), F facet n-k affinely independent points in F

Similarly dim(L’) = k+1, dim(L) = k+1 and L L’

L = L’


i.4 polyhedral theory 1. integer programming 2011 2 objective of study: want to know how to...

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