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Integer Programming, Part 1
Rudi Pendavingh
Technische Universiteit Eindhoven
May 18, 2016
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 1 / 37
Linear Inequalities and Polyhedra Farkas’ Lemma / Linear Programming
Prerequisites for the course:
Theorem (Farkas’ Lemma)
The system Ax ≤ b is infeasible if and only if the system uA = 0, ub < 0, u ≥ 0 is feasible.
Let P := {x : Ax ≤ b} and D := {u : uA = c , u ≥ 0}.
Theorem (Linear Programming Duality)
If P and D are both nonempty, then
max{cx : x ∈ P} = min{ub : u ∈ D}.
Theorem (Complementary slackness)
Let x∗ ∈ P and u∗ ∈ D. Then x∗ and u∗ are both optimal if and only if
u∗i (aix∗ − bi ) = 0 for all i
where ai denotes the i-the row of matrix A.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 2 / 37
Linear Inequalities and Polyhedra Affine, Convex, and Conic Combinations
Let x1, . . . , xk ∈ Rn and λ1 . . . , λk ∈ R. Then
x = λ1x1 + · · ·+ λkx
k
is a linear combination of x1, . . . , xk .
x is an affine combination if∑
i λi = 1
x is a convex combination if∑
i λi = 1 and λi ≥ 0 for all i
x is a nonnegative combination if λi ≥ 0 for all i
Let S ⊆ Rn be any set of vectors.
The
linear hullaffine hull
convex hullcone
of S is the set of all
linearaffine
convexnonnegative
combinations of x1, . . . , xk ∈ S .
A set S is affine if S = aff(S), convex if S = conv (S), a cone if S = cone (S).
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 3 / 37
Linear Inequalities and Polyhedra Polyhedra and the Theorem of Minkowski-Weyl
Let C ⊆ Rn.
C is a polyhedral cone if C = {x : Ax ≤ 0} for some matrix A.
C is finitely generated if C = cone {r1, . . . , rq} for some vectors r1, . . . , rq
Theorem (Minkowski-Weyl for cones)
Let C ⊆ Rn. Then C is a finitely generated cone if and only if C is a polyhedral cone.
Example
Let C := {x ∈ R2 : x1 − x2 ≤ 0,−2x1 + x1 ≤ 0} be a polyhedral cone. ThenC = cone {(1, 1), (1, 2)}, so C is also a finitely generated cone.
Example
Let C := cone {(1, 5), (1, 1), (4, 1)} be a finitely generated cone. ThenC := {x ∈ R2 : x1 − 4x1 ≤ 0.− 5x1 + x2 ≤ 0}, so C is also a polyhedral cone.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 4 / 37
Linear Inequalities and Polyhedra Polyhedra and the Theorem of Minkowski-Weyl
P ⊆ Rn is a polyhedron if P = {x : Ax ≤ b} for some A, b.
Q ⊆ Rn is a polytope if Q = conv {v1, . . . , vp} for some vectors v1, . . . , vp
the Minkowski sum of A,B ⊆ Rn is A + B := {a + b : a ∈ A, b ∈ B}
Theorem (Minkowski-Weyl for polyhedra)
Let P ⊆ Rn. Then P is a polyhedron if and only if P = Q + C for some polytope Q and somefinitely generated cone C .
Proof.
⇒: If P = {x : Ax ≤ b}, consider CP := {(x , y) : Ax − by ≤ 0, x ∈ Rn, y ∈ R}.⇐: If P = Q + C = conv {v1, . . . , vp}+ cone {r1, . . . , rq}, consider
CP := cone
{(v1
1
), . . . ,
(vp
1
),
(r1
0
), . . . ,
(rq
0
)}In either case, apply Minkowski-Weyl for cones.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 5 / 37
Linear Inequalities and Polyhedra Polyhedra and the Theorem of Minkowski-Weyl
Example
Consider P = {x ∈ R2 : x1 ≥ 0, x2 ≥ 0, x1 + x2 ≥ 2, x1 + 2x2 ≥ 3}. Then P = Q + C , where
Q = conv
{(30
),
(11
),
(02
)}, C = cone
{(10
),
(01
)}
Example
Consider P = {x ∈ R3 : x1 ≥ 0, x2 ≥ 0, x1 + x2 ≥ 1}. Then P = Q + C , where
Q = conv
1
00
,
010
, C = cone
1
00
,
010
,
001
,
00−1
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 6 / 37
Linear Inequalities and Polyhedra Lineality Space and Recession Cone
Let P ⊆ Rn be a nonempty polyhedron.
the lineality space of P is lin(P) := {r ∈ Rn : x + λr ∈ P for all x ∈ P and λ ∈ R}the recession cone of P is rec(P) := {r ∈ Rn : x + λr ∈ P for all x ∈ P and λ ∈ R+}
Theorem
Suppose P = {x : Ax ≤ b} = conv {v1, . . . , vp}+ cone {r1, . . . , rq}. Then
lin(P) = {x : Ax = 0} and
rec(P) = {r : Ar ≤ 0} = cone {r1, . . . , rq}
Example
Consider again the polyhedron P = {x ∈ R3 : x1 ≥ 0, x2 ≥ 0, x1 + x2 ≥ 1}. Then
lin(P) = {x ∈ R3 : x1 = x2 = 0}
rec(P) = {r ∈ R3 : r1 ≥ 0, r2 ≥ 0} = cone
1
00
,
010
,
001
,
00−1
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 7 / 37
Linear Inequalities and Polyhedra Implicit Equalities, Affine Hull, and Dimension
The system Ax ≤ b consists of inequalities aix ≤ bi for i ∈ Maix ≤ bi is an implicit equality of Ax ≤ b if aix = bi for all x such that Ax ≤ bA=x ≤ b= denotes the subsystem of Ax ≤ b containing all implicit equalitiesfor any P ⊆ Rn, dim(P) := dim(aff(P))
Theorem
Let P = {x ∈ Rn : Ax ≤ b} be a nonempty polyhedron. Then
aff(P) = {x : A=x = b=} = {x : A=x ≤ b=}
Furthermore, dim(P) = n − rank(A=).
Example
The assignment polytope
P =
x ∈ Rn2:∑i
xij = 1 for all j ,∑j
xij = 1 for all i , xij ≥ 0 for all i , j
has dimension n2 − 2n + 1
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 8 / 37
Linear Inequalities and Polyhedra Faces
Let P ⊆ Rn.
An inequality cx ≤ δ is valid for P if cx ≤ δ for all x ∈ P
A face of P is a set F of the form
F = P ∩ {x ∈ Rn : cx = δ}
where cx ≤ δ is a valid inequality for P
Let P = {x : aix ≤ bi for i ∈ M}, and for each I ⊆ M put
FI := {x ∈ Rn : aix = bi for all i ∈ I , and aix ≤ bi for all i ∈ M \ I}
Theorem
For each I ⊆ M, FI is a face of P. Conversely, if F is a nonempty face of P, then F = FI forsome I ⊆ M.
It follows that the number of faces of polyhedron P is finite.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 9 / 37
Linear Inequalities and Polyhedra Minimal Representation and Facets
Let P = {x : aix ≤ bi for i ∈ M}.The inequality ajx ≤ bj is redundant if it is valid for
{x : aix ≤ bi for all i ∈ M \ {j}}
”aix ≤ bi for i ∈ M” is a minimal representation of P if aix ≤ bi is irredundant for i ∈ M
A face F of P is proper if F 6= ∅ and F 6= P
A facet of P is an proper face of P not strictly contained in another proper face of P
Theorem
For each facet F of P there is an inequality aix ≤ bi so that
F = P ∩ {x ∈ Rn : aix = bi}.
Conversely, if the inequality aix ≤ bi is irredundant and not an implied equation, thenF = P ∩ {x ∈ Rn : aix = bi} is a facet of P.
So in a minimal representation of P, each inequality is an implied equation or defines a facet.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 10 / 37
Linear Inequalities and Polyhedra Minimal Faces
Let P be a nonempty polyhedron.
a set F is a minimal face of P if F is a face of P which does not contain a proper face
Theorem
Let P = {x ∈ Rn : Ax ≤ b}. The a nonempty face F of P is a minimal face if and only ifF = {x : A′x = b′} for some subsystem A′x ≤ b′ of Ax ≤ b such that rank(A′) = rank(A).
a set F is a vertex of P if F is a face of P of dimension 0, i.e. F = {x}P is pointed if P has a vertex
Theorem
Let P = {x ∈ Rn : Ax ≤ b}. If P is pointed, then equivalent:
{x} is a vertex
x satisfies n linearly independent inequalities from Ax ≤ b with equality
there are no x ′, x ′′ ∈ P \ {x} so that x is a convex combination of x ′, x ′′.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 11 / 37
Linear Inequalities and Polyhedra Decomposition Theorem for Polyhedra
Let P be any nonempty polyhedron. Put
t := dim(lin(P))
let F1, . . . ,Fp be the minimal faces of P; let v i ∈ F1
let R1, . . . ,Rq be the (t + 1)-dimensional faces of rec(P); let r i ∈ Ri \ lin(P)
Theorem
P = conv {v1, . . . , vp} + cone {r1, . . . , rp} + lin(P)
Proof.
The special case where t = 0 reduces to Minkowski-Weyl.If t > 0, let a1, . . . , at be a basis of lin(P). Then P = Q + lin(P) where
Q := {x ∈ P : aix = 0 for i = 1, . . . t}
We have dim(lin(Q)) = 0, and hence Q = conv {v1, . . . , vp} + cone {r1, . . . , rp}.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 12 / 37
Homework
Sections 3.1, 3.2, 3.3, 3.14 are prerequisite knowledge for the course. Review thesesections if necessary, and verify that you are able to make exercises 3.1, 3.2, 3.30.
Read sections 3.4 — 3.12.
Make exercises 3.7, 3.10, 3.16, 3.33.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 13 / 37
Perfect Formulations Properties of Integral Polyhedra
A polyhedron P ⊆ Rn is integral if P = conv (P ∩ Zn).
If P is integral, then max{cx : x ∈ P ∩ Zn} = max{cx : x ∈ P} is an ordinary linear program.
We want to write combinatorial optimization problems as optimization over integral polyhedra.
We next develop some tools for proving that a polyhedron P is integral.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 14 / 37
Perfect Formulations Properties of Integral Polyhedra
Theorem
P ⊆ Rn a rational polyhedron. Then equivalent:
1 P = conv (P ∩ Zn)
2 F ∩ Zn 6= ∅ for every minimal face F of P
3 max{cx : x ∈ P} attained by integral vector x ∈ P ∩Zn for each c ∈ Rn, if attained at all
4 max{cx : x ∈ P} ∈ Z ∪ {∞} for each c ∈ Zn
Proof.
Some of the implications are straightforward...
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 15 / 37
Perfect Formulations Properties of Integral Polyhedra
Theorem
P ⊆ Rn a rational polyhedron. Then equivalent:
1 P = conv (P ∩ Zn)
2 F ∩ Zn 6= ∅ for every minimal face F of P
3 max{cx : x ∈ P} attained by integral vector x ∈ P ∩Zn for each c ∈ Rn, if attained at all
4 max{cx : x ∈ P} ∈ Z ∪ {∞} for each c ∈ Zn
Proof.
(4)⇒ (2): We may assume P = {x : Ax ≤ b} for some A, b with integer entries
Consider a minimal face F . F = {x : AF x = bF} for a subsystem AF x ≤ bF of Ax ≤ b
Suppose F ∩ Zn = ∅; then uA ∈ Zn and ub 6∈ Z for some u. W.l.o.g. u ≥ 0
With c := uA, z := ub, we have cx = uAx ≤ ub = z for all x ∈ P, with equality if x ∈ F .
Then max{cx : x ∈ P} = z 6∈ Z
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 16 / 37
Perfect Formulations Properties of Integral Polyhedra
Theorem
P ⊆ Rn a rational polyhedron. Then equivalent:
1 P = conv (P ∩ Zn)
2 F ∩ Zn 6= ∅ for every minimal face F of P
3 max{cx : x ∈ P} attained by integral vector x ∈ P ∩Zn for each c ∈ Rn, if attained at all
4 max{cx : x ∈ P} ∈ Z ∪ {∞} for each c ∈ Zn
Proof.
(2)⇒ (1): We may again assume P = {x : Ax ≤ b} for some A, b with integer entries
Let F 1, . . . ,F p be the minimal faces of P, and pick any v i ∈ F i ∩ Zn for each i .
Then P = conv {v1, . . . , vp}+ rec(P)
rec(P) = {r : Ar ≤ 0} = cone {r1, . . . , rq} for some rational r i , since A is rational.
Scaling the r j , we may assume r j ∈ Zn for all j .
So P = conv {v1, . . . , vp}+ cone {r1, . . . , rq} with v i , r j ∈ Zn, hence P = conv (P ∩ Zn)
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 17 / 37
Perfect Formulations Total Unimodularity
We next consider the question:
Which matrices A are such that {x : Ax ≤ b, x ≥ 0} is integral for all b ∈ Zn?
A matrix A is totally unimodular (TU) if det(B) ∈ {−1, 0, 1} for every square submatrix B of A
Lemma
If B is an integer matrix such that det(B) ∈ {−1, 1}, then B−1 is an integer matrix.
Theorem (Hoffman and Kruskal)
A is TU ⇔ {x : Ax ≤ b, x ≥ 0} is integral for each b ∈ Zn.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 18 / 37
Perfect Formulations Total Unimodularity
If A is a matrix, then an equitable bicoloring of A is a partiton of the columns into ‘red’ and‘blue’ columns so that ∑
j red
Aj
−∑
j blue
Aj
∈ {−1, 0, 1}m
Here Aj is the j-th column of A.
Theorem
A TU ⇐⇒ every column submatrix admits an equitable bicoloring.
Proof.=⇒:Let B be a column submatrix of A. Since A is TU, so is B. The polyhedron
P := {x : b1
2B1c ≤ Bx ≤ d1
2B1e, 0 ≤ x ≤ 1}
contains an x ∈ Zn. Let the i-th column be red if xi = 0, blue otherwise.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 19 / 37
Perfect Formulations Total Unimodularity
If A is a matrix, then an equitable bicoloring of A is a partiton of the columns into ‘red’ and‘blue’ columns so that ∑
j red
Aj
−∑
j blue
Aj
∈ {0, 1}mHere Aj is the j-th column of A.
Theorem
A TU ⇐⇒ every column submatrix admits an equitable bicoloring.
Proof.⇐=:Suppose not. Let B be a smallest submatrix of A with δ := det(B) 6∈ {−1, 0, 1}. Then δB−1
is a 0,±1-matrix. Let d be the first column of δB−1. Then Bd = δe1. If δ is even, then δ = 0.If δ is odd, then δ = 1.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 20 / 37
Perfect Formulations Total Unimodularity
If A is a matrix, then an equitable bicoloring of A is a partiton of the columns into ‘red’ and‘blue’ columns so that ∑
j red
Aj
−∑
j blue
Aj
∈ {0, 1}mHere Aj is the j-th column of A.
Theorem
A TU ⇐⇒ every column submatrix admits an equitable bicoloring.
The signed incidence matrix of a directed graph D = (V ,A) is the V × A matrix AD withentries ava ∈ {−1, 0, 1} such that
ava =
−1 if v is the tail of a1 if v is the head of a0 otherwise
Corollary
The incidence matrix of any directed graph D = (V ,A) is totally unimodular.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 21 / 37
Homework
Read 4.1, 4.2, and first two pages of 4.3. Try to fully understand all calls to othertheorems in the proof of Thm 4.1.
Make exercises 4.3, 4.4, 4.5, 4.6.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 22 / 37
Perfect Formulations Total Dual Integrality
A rational system Ax ≤ b is totally dual integral (TDI) if
min{yb : yA = c, y ≥ 0}
either has an integral optimal solution or is infeasible for each c ∈ Zn.
Theorem
If Ax ≤ b is TDI and b integral, then {x : Ax ≤ b} is integral.
Proof.
If Ax ≤ b is TDI and b integral, then min{yb : yA = c , y ≥ 0} ∈ Z ∪ {∞} for each c ∈ Zn.Hence max{cx : Ax ≤ b} ∈ Z ∪ {∞} for each c ∈ Zn, so that {x : Ax ≤ b} is integral.
We next show:
Theorem
If P is rational, then there exists a TDI system Ax ≤ b so that P = {x : Ax ≤ b}.
Theorem
If P is (rational and) integral, then there exists a TDI system Ax ≤ b so thatP = {x : Ax ≤ b} and b integral.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 23 / 37
Perfect Formulations Total Dual Integrality
Theorem
If P is rational, then there exists a TDI system Ax ≤ b so that P = {x : Ax ≤ b}.
Proof.
if P = ∅, then P = {x : 0x ≤ −1}P is rational, so we may assume P = {x : Mx ≤ d} for integer M, d
Put C := {c ∈ Zn : c = λM, 0 ≤ λ ≤ 1}, δc = max{cx : x ∈ P} for all c ∈ C
let Ax ≤ b be the system of all inequalities cx ≤ δc for c ∈ C
P ⊆ {x : Ax ≤ b} ⊆ {x : Mx ≤ d} = P, so we have equality throughout
to show that Ax ≤ b is TDI, consider a c ∈ Zn so that max{cx : Ax ≤ b} is finite
max{cx : Ax ≤ b} = max{cx : Mx ≤ d} = min{yd : yM = c, y ≥ 0} has an optimalsolution y∗; put λ = y∗ − by∗c, c ′ := λM, c ′′ := by∗cM;
min{yd : yM = c , y ≥ 0} = min{yd : yM = c ′, y ≥ 0}+ min{yd : yM = c”, y ≥ 0}, soλ, by∗c are optimal solutions for the latter two problems
min{yd : yM = c ′, y ≥ 0} = max{c ′x : x ∈ P} = δc ′ , and c ′x ≤ δc ′ is a row of Ax ≤ b
min{yb : yA = c , y ≥ 0} has an integer optimal solution obtained from by∗c and c ′
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 24 / 37
Perfect Formulations Total Dual Integrality
Theorem
If P is integral, then there exists a TDI system Ax ≤ b so that P = {x : Ax ≤ b} and bintegral.
Proof.
If P is integer, then in the previous proof δc := max{cx : x ∈ P} ∈ Z for each c ∈ C ⊆ Zn
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 25 / 37
Perfect Formulations Fundamental Theorem of Integer Programming
Theorem
Let A,G be rational matrices, b a rational vector. Put
P := {(x , y) : Ax + Gy ≤ b}, S := {(x , y) : Ax + Gy ≤ b, x ∈ Zn}.
Then:
1 ∃A′,G ′, b′ rational so that conv (S) = {(x , y) : A′x + G ′y ≤ b′}.2 if S 6= ∅, then the recession cones of P and conv (S) coincide.
Proof.
P = conv {v1, . . . , v t}+ cone {r1, . . . , rq} for some rational v i , integral r j .
Consider T := {∑
i λivi +∑
j µj rj :∑λi = 1, λ ≥ 0, 0 ≤ µ ≤ 1}
Put TI := {(x , y) ∈ T : x ∈ Zn}, RI := {∑
j µj rj : µ ∈ Zq
+}S = TI + RI ; hence conv (S) = conv (TI ) + conv (RI )
conv (TI ) is a rational polyhedron,
rec(conv (S)) = conv (RI ) = cone {r1, . . . , rq} = rec(P) is a rational cone
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 26 / 37
Homework
Read sections 4.6, 4.8.
Make exercises 4.9, 4.11, 4.18, 4.20.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 27 / 37
Perfect Formulations Matchings in Graphs
A matching in an undirected graph G = (V ,E ) is a set of pairwise disjoint edges M ⊆ E .
IP (Maximum cardinality matching)
max∑
e∈E xe∑e∈δ(v) xe ≤ 1 v ∈ V
x ∈ {0, 1}E
LetP(G ) := conv {x ∈ {0, 1}E :
∑e∈δ(v)
xe ≤ 1 for all v ∈ V }
and put
Q(G ) := {x ∈ RE :∑
e∈δ(v)
xe ≤ 1 for all v ∈ V , x ≥ 0}
Is Q(G ) integral for all undirected graphs G = (V ,E ), i.e. is P(G ) = Q(G )?
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 28 / 37
Perfect Formulations Matchings in Graphs
The incidence matrix of an undirected graph G = (V ,E ) is the V × E matrix AG with entriesave ∈ {0, 1} such that
ave =
{1 if v is incident with e0 otherwise
Clearly
Q(G ) = {x ∈ RE :∑
e∈δ(v)
xe ≤ 1 for all v ∈ V , x ≥ 0} = {x ∈ RE : AGx ≤ 1, x ≥ 0}
Lemma
The incidence matrix of G is totally unimodular if and only if G is bipartite.
Theorem
If G is bipartite, then P(G ) is integral.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 29 / 37
Perfect Formulations Matchings in Graphs
P(G ) := conv {x ∈ {0, 1}E :∑
e∈δ(v)
xe ≤ 1 for all v ∈ V }
Lemma
If G = (V ,E ) then for any U ⊆ V with |U| odd the inequality∑e∈E [U]
xe ≤|U| − 1
2
is valid for P(G ).
Is
P(G ) = {x ∈ RE :∑
e∈δ(v)
xe ≤ 1 for all v ∈ V ,∑
e∈E [U]
xe ≤|U| − 1
2for all odd U ⊆ V , x ≥ 0}
??
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 30 / 37
Perfect Formulations Matchings in Graphs
Theorem
Let G = (V ,E ) be a graph. The polyhedron
{x ∈ RE :∑
e∈δ(v)
xe = 1 for all v ∈ V ,∑
e∈δ(U)
xe ≥ 1 for all odd U ⊆ V , x ≥ 0}
is integral.
Proof.
Suppose not.
let G = (V ,E ) be a counterexample to the theorem with |V |+ |E | as small as possible.
let x be a fractional vertex of the polyhedron; there is some odd U so that∑
e∈δ(U) xe = 1
x is a convex combination of integer vectors in the polyhedron, contradiction
Applying the theorem to a ’doubled’ version of G , it then follows that
P(G ) = {x ∈ RE :∑
e∈δ(v)
xe ≤ 1 for all v ∈ V ,∑
e∈E [U]
xe ≤|U| − 1
2for all odd U ⊆ V , x ≥ 0}
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 31 / 37
Perfect Formulations Spanning Trees
A spanning tree of G = (V ,E ) is a set T ⊆ E so that (V ,T ) is acyclic and connected.
IP (Maximum weight spanning tree)
max∑
e∈E wexe∑e∈E [S] xe ≤ |S | − 1 S ⊆ V , S 6= ∅∑e∈E xe = |V | − 1
x ∈ {0, 1}E
P(G ) := {x ∈ RE :∑
e∈E [S]
xe ≤ |S | − 1 for all nonempty S ⊆ V ,∑e∈E
xe = |V | − 1, x ≥ 0}
Theorem
P(G ) is integral.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 32 / 37
Perfect Formulations Submodular Polyhedra
Let N be a finite set. A function f : 2N → R is submodular if
f (S) + f (T ) ≥ f (S ∩ T ) + f (S ∪ T )
Example
Let G = (V ,E ) be an undirected graph and let f : 2V → R be defined by f (S) := |δ(S)|.Then f is submodular.
Let G = (V ,E ) be an undirected graph and let f : 2E → R be defined by
f (S) := |V | −# of components of (V ,S)
Then f is submodular.
The submodular polyhedron is
P(f ) := {x ∈ RN :∑j∈S
xj ≤ f (S) for all S ⊆ N}
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 33 / 37
Perfect Formulations Submodular Polyhedra
The submodular polyhedron is
P(f ) := {x ∈ RN :∑j∈S
xj ≤ f (S) for all S ⊆ N}
Theorem
Suppose f : 2N → Z is submodular and f (∅) = 0. Then P(f ) is integral, and the system∑j∈S xj ≤ f (S) for all S ⊆ N is TDI.
Proof.
Let c ∈ Zn. Consider max{cx :∑
j∈S xj ≤ f (S) for all S ⊆ N}. W.l.o.g. c1 ≥ c2 ≥ . . . ≥ cn.A feasible solution x is obtained by setting xj := f (Sj)− f (Sj−1), where Sj := {1, . . . , j}.The dual is
min{∑S⊆N
f (S)yS :∑j∈S
yS = cj for all j ∈ N, y ≥ 0}
We construct an integral optimal solution y for this problem so that cx =∑
S⊆N f (S)ySproving optimality of both integral solutions.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 34 / 37
Perfect Formulations
Theorem (Nash-Williams’ orientation theorem)
A 2k-connected undirected graph G can be oriented so as to become a k-connected directedgraph.
Proof.
Let D = (V ,A) arise from G by arbitrarily orienting the edges.
if there is an x ∈ ZA so that
0 ≤ x ≤ 1, x [δ−(U)]− x [δ+(U)] ≤ |δ−(U)| − k for all U ⊆ V (∗)
then we are done by reversing the arcs a so that xa = 1.
We consider the polyhedron P := {x ∈ RA : (∗)}. Then x = 121 ∈ P, so P 6= ∅
We will prove TDI-ness of (∗). Then PI = P 6= ∅, hence P ∩ ZA 6= ∅.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 35 / 37
Perfect Formulations
Proof.
We prove TDI-ness of
0 ≤ x ≤ 1, x [δ−(U)]− x [δ+(U)] ≤ |δ−(U)| − k for all U ⊆ V (∗)
Consider the LP dual to maximizing cx over (∗).
Let z be an optimal dual solution so that∑
U zU |U| · |V \ U| minimal.
Then U := {U | zU > 0} is cross-free, i.e. if T ,U ∈ U , then
U ⊆ T , T ⊆ U, U ∩ T = ∅ or U ∪ T = V
The restricted system
0 ≤ x ≤ 1, x [δ−(U)]− x [δ+(U)] ≤ |δ−(U)| − k for all U ∈ U
is totally unimodular.
So the dual optimum is determined by this TU system, and hence is integral.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 36 / 37
Homework
Read sections 4.3, 4.4, 4.5, 4.7.
Make exercises 4.9, 4.12, 4.14, 4.18.
Rudi Pendavingh (TU/e) Integer Programming, Part 1 May 18, 2016 37 / 37
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