Integer Programming
5.1 The Integer Hull of a Polyhedron
As linear programs, integer programs can be infeasible or unbounded. It is not easy to decide whether PI = ∅ for a polyhedron P. But if an integer program is feasible we can decide whether it is bounded by simply considering the LP relaxation.
Proposition 5.1. Let P = {x : Ax ≤ b}be some rational polyhedron whose integer hull is nonempty, and let c be some vector. Then max{cx : x ∈ P} is bounded if and only ifmax{cx :x∈ PI}is bounded.
Proof: Suppose max{cx : x ∈ P}is unbounded. Then Theorem 3.22 the dual LP min{yb : y A = c, y ≥ 0} is infeasible. Then by Corollary 3.21 there is a rational (and thus an integral) vector z withcz <0 and Az ≥0. Let y ∈ PI be some integral vector. Then y−kz∈ PI for allk∈N, and thus max{cx :x∈ PI}
is unbounded. The other direction is trivial. 2
Definition 5.2. Let Abe an integral matrix. Asubdeterminantof AisdetBfor some square submatrixB ofA(defined by arbitrary row and column indices). We write(A)for the maximum absolute value of the subdeterminants ofA.
Lemma 5.3. LetC = {x: Ax ≥0}be a polyhedral cone, Aan integral matrix.
ThenC is generated by a finite set of integral vectors, each having components with absolute value at most(A).
Proof: By Lemma 3.11,C is generated by some of the vectorsy1, . . . ,yt, such that for each i, yi is the solution to a system M y = b where M consists ofn
linearly independent rows of
A I
andb = ±ej for some unit vectorej. Set zi := |detM|yi. By Cramer’s rule, zi is integral with||zi||∞≤(A). Since this holds for eachi, the set{z1, . . . ,zt}has the required properties. 2
A similar lemma will be used in the next section:
Lemma 5.4. Each rational polyhedral coneC is generated by a finite set of in-tegral vectors {a1, . . . ,at} such that each integral vector inC is a nonnegative integral combination ofa1, . . . ,at. (Such a set is called aHilbert basisforC.) Proof: LetC be generated by the integral vectorsb1, . . . ,bk. Leta1, . . . ,at be all integral vectors in the polytope
{λ1b1+. . .+λkbk: 0≤λi ≤1 (i =1, . . . ,k)}
We show that{a1, . . . ,at}is a Hilbert basis forC. They indeed generateC, because b1, . . . ,bk occur among thea1, . . . ,at.
For any integral vectorx∈C there areµ1, . . . , µk≥0 with x = µ1b1+. . .+µkbk = µ1b1+. . .+ µkbk+
(µ1− µ1)b1+. . .+(µk− µk)bk, so x is a nonnegative integral combination ofa1, . . . ,at. 2
An important basic fact in integer programming is that optimum integral and fractional solutions are not too far away from each other:
Theorem 5.5. (Cook et al. [1986]) Let Abe an integralm×n-matrix andb∈ Rm,c∈Rn arbitrary vectors. Let P:= {x: Ax ≤b}and suppose thatPI = ∅. (a) Supposey is an optimum solution ofmax{cx : x ∈ P}. Then there exists an
optimum integral solutionzofmax{cx:x∈ PI}with||z−y||∞≤n(A). (b) Suppose y is a feasible integral solution ofmax{cx : x ∈ PI}, but not an
optimal one. Then there exists a feasible integral solutionz∈ PI withcz>cy and||z−y||∞≤n(A).
Proof: The proof is almost the same for both parts. Let first y ∈ P arbitrary.
Let z∗ be an optimum integral solution of max{cx : x ∈ PI}. We split Ax ≤ b into two subsystems A1x≤b1,A2x≤b2 such thatA1z∗≥ A1yandA2z∗ <A2y.
Thenz∗−ybelongs to the polyhedral cone C:= {x: A1x≥0,A2x≤0}. C is generated by some vectors xi (i = 1, . . . ,s). By Lemma 5.3, we may assume that xi is integral and||xi||∞≤(A)for alli.
Since z∗−y ∈ C, there are nonnegative numbersλ1, . . . , λs with z∗ −y = s
i=1λixi. We may assume that at mostn of theλi are nonzero.
Forµ=(µ1, . . . , µs)with 0≤µi ≤λi (i =1, . . . ,s) we define zµ:=z∗−
s i=1
µixi =y+ s
i=1
(λi−µi)xi
94 5. Integer Programming
and observe thatzµ∈ P: the first representation ofzµimplies A1zµ≤ A1z∗ ≤b1; the second one implies A2zµ≤ A2y≤b2.
Case 1: There is somei∈ {1, . . . ,s}withλi ≥1 andcxi >0. Letz:=y+xi. We have cz > cy, showing that this case cannot occur in case (a). In case (b), when y is integral, z is an integral solution of Ax ≤ b such that cz > cy and
||z−y||∞= ||xi||∞≤(A).
Case 2: For alli∈ {1, . . . ,s},λi ≥1 impliescxi ≤0. Let z :=zλ = z∗−
s i=1
λixi.
z is an integral vector of P withcz≥cz∗ and
||z−y||∞ ≤ s
i=1
(λi− λi)||xi||∞ ≤ n(A).
Hence in both (a) and (b) this vectorz does the job. 2 As a corollary we can bound the size of optimum solutions of integer pro-gramming problems:
Corollary 5.6. If P= {x∈Qn :Ax ≤b}is a rational polyhedron andmax{cx: x ∈ PI}has an optimum solution, then it also has an optimum integral solution x withsize(x)≤13n(size(A)+size(b)).
Proof: By Proposition 5.1 and Theorem 4.4, max{cx :x ∈ P}has an optimum solution y with size(y) ≤ 4n(size(A)+size(b)). By Theorem 5.5(a) there is an optimum solution x of max{cx : x ∈ PI} with ||x −y||∞ ≤ n(A). By Propositions 4.1 and 4.3 we have
size(x) ≤ 2 size(y)+2nsize(n(A))
≤ 8n(size(A)+size(b))+2nlogn+4nsize(A)
≤ 13n(size(A)+size(b)). 2
Theorem 5.5(b) implies the following: given any feasible solution of an integer program, optimality of a vector x can be checked simply by testing x+y for a finite set of vectors y that depend on the matrix A only. Such a finite test set (whose existence has been proved first by Graver [1975]) enables us to prove a fundamental theorem on integer programming:
Theorem 5.7. (Wolsey [1981], Cook et al. [1986]) For each integral m×n -matrix Athere exists an integral matrix M whose entries have absolute value at mostn2n(A)n, such that for each vectorb∈Qmthere exists a vectord with
{x: Ax ≤b}I = {x:M x ≤d}.
Proof: We may assume A=0. LetC be the cone generated by the rows of A.
Let
L := {z∈Zn :||z||∞≤n(A)}.
For each K ⊆L, consider the cone
CK := C∩ {y:zy≤0 for all z∈K}.
By the proof of Theorem 3.24 and Lemma 5.3, CK = {y : U y ≤ 0}for some matrixU (whose rows are generators of{x: Ax ≤0}and elements of K) whose entries have absolute value at mostn(A). Hence, again by Lemma 5.3, there is a finite setG(K)of integral vectors generatingCK, each having components with absolute value at most(U)≤n!(n(A))n ≤n2n(A)n.
LetM be the matrix with rows
K⊆LG(K). SinceC∅=C, we may assume that the rows of Aare also rows of M.
Now letbbe some fixed vector. If Ax ≤b has no solution, we can complete b to a vectord arbitrarily and have{x :M x≤d} ⊆ {x :Ax ≤b} = ∅.
If Ax ≤b contains a solution, but no integral solution, we setb:=b−A1l, where Aarises from Aby taking the absolute value of each entry. Then Ax ≤b has no solution, since any such solution yields an integral solution of Ax ≤b by rounding. Again, we completeb tod arbitrarily.
Now we may assume that Ax ≤ b has an integral solution. For y ∈ C we define
δy := max{yx : Ax ≤b,x integral} (this maximum is bounded if y∈C). It suffices to show that
{x: Ax ≤b}I =
x:yx ≤δy for each y∈
K⊆L
G(K)
. (5.1) Here “⊆” is trivial. To show the converse, letcbe any vector for which
max{cx : Ax ≤b, x integral}
is bounded, and letx∗be a vector attaining this maximum. We show thatcx ≤cx∗ for all x satisfying the inequalities on the right-hand side of (5.1).
By Proposition 5.1 the LP max{cx : Ax ≤b}is bounded, so by Theorem 3.22 the dual LP min{yb:y A=c, y≥0}is feasible. Hencec∈C.
Let K¯ := {z ∈ L : A(x∗+z)≤ b}. By definition cz ≤ 0 for all z ∈ ¯K, so c∈CK¯. Thus there are nonnegative numbersλy (y∈G(K¯))such that
c =
y∈G(K¯)
λyy.
Next we claim that x∗ is an optimum solution for max{yx :Ax ≤b, xintegral}
for each y ∈ G(K¯): the contrary assumption would, by Theorem 5.5(b), yield a vector z∈ ¯K withyz >0, which is impossible since y∈CK¯. We conclude that
96 5. Integer Programming
Thus the inequalitycx ≤cx∗ is a nonnegative linear combination of the inequal-ities yx ≤δy for y∈G(K¯). Hence (5.1) is proved. 2
See Lasserre [2004] for a similar result.