In this section, we will review the main results of the chapter, give some examples, explanations, and some proofs that show their connection.
The main purpose of this section is an analysis of the reformulation methods. This is done in Theorem3, which proves an upper bound on the number of B&B nodes, when branching on the last variable in the reformulations.
Theorems 4 and 5 show that an integral vector p, which is “near parallel” to acan be extracted from the transformation matrices of the reformulations. The notion of near parallelness that we use is stronger than just requiringsin(a, p) to be small. The relationship of the two parallelness concepts is clarified in Proposition2.
Theorem6proves an upper bound oniwidth(p,(KP)), wherepis an integral vector. A novelty of the bound is that it does not depend onβ1 andβ2, only on their difference. We show through examples that this bound is quite useful whenp is a near parallel vector found according to Theorems4and5.
In the end, a transference result between branching directions in the original, and reformulated problems completes the proof of Theorem3.
Theorem 3. Supposekak ≥ 2(n/2+1)n. Then
(1) iwidth(en, (KP-R)) ≤ ⌊f(a)(2kvk+(β2−β1))⌋+ 1.
(2) iwidth(en−1, (KP-N)) ≤ ⌊2g(a)kvk⌋+ 1.
Givenaandpintegral vectors, we will need the notion of their near parallelness. The obvious thing would be to require that|sin(a, p)|is small. Instead, we will write a decomposition
a=λp+r, withλ∈Q, r ∈Qn, r⊥p, (DECOMP)
and ask for krk /λ to be small. The following proposition clarifies the connection of the two near parallelness concepts and shows two useful consequences of the latter one.
Proposition 2. Suppose thata, p ∈ Zn,andrandλare defined to satisfy (DECOMP). Assume w.l.o.g.
λ >0.Then
(2) For anyM there existsa, p withkak≥M such that the inequality in (1) is strict.
(3) Denote bypi andai theith component ofp anda.Ifkrk/λ <1, andpi 6= 0, then the signs of
pi andai agree. Also, ifkrk/λ <1/2,then⌊ai/λ⌉=pi.
Proof Statement (1) follows from
sin(a, p) = krk/kak ≤ krk/kλpk ≤ krk/λ, (4.2.2)
where in the last inequality we used the integrality ofp. To see (2), consider the family ofa andp vectors
a = m2+ 1, m2 , p = m+ 1, m (4.2.3)
withman integer. Lettingλ andr be defined as in the statement of the proposition, a straightforward computation (or experimentation) shows that asm→ ∞
sin(a, p) → 0, krk/λ → 1/√2. Statement (3) is straightforward from
ai/λ = pi+ri/λ. (4.2.4)
The next two theorems show how the near parallel vectors can be found from the transformation matrices of the reformulations.
Theorem 4. Supposekak ≥ 2(n/2+1)n. LetU be a unimodular matrix such that the columns of a I U
are LLL-reduced andpthe last row ofU−1. Definer andλto satisfy (DECOMP), and assume w.l.o.g. λ >0. Then (1) kpk(1+krk2)1/2 ≤kakf(a); (2) λ≥1/f(a); (3) krk/λ≤2f(a).
Theorem 5. Supposekak ≥ 2(n/2+1)n. LetV be a matrix whose columns are an LLL-reduced basis ofN(a),ban integral column vector withab= 1, andpthe(n−1)strow of(V, b)−1. Definerandλ
to satisfy (DECOMP), and assume w.l.o.g. λ >0.
Thenr 6= 0, and (1) kpkkrk≤kakg(a); (2) krk/λ≤2g(a).
It is important to note thatp is integral, butλandr may not be. Also, the measure of parallelness toa, i.e., the upper bound onkrk/λ is quite similar for thepvectors found in Theorems4and5, but their length can be quite different. Whenkakis large, thepvector in Theorem4 is guaranteed to be much shorter thana by λ ≥1/f(a). On the other hand, thepvector from Theorem 5may be much
longer thana:the upper bound onkpkkrk does not guarantee any bound onkpk, sincer can be fractional.
The following example illustrates this:
Example 5. Consider the vector
a =
3488, 451, 1231, 6415, 2191
We computedp1, r1, λ1 according to Theorem4: p1 = 62, 8, 22, 114, 39 , r1 = 0.2582, 0.9688, −6.5858, 2.0554, −2.9021 , λ1 = 56.2539, kr1k/λ1 = 0.1342. (4.2.6)
We also computedp2, r2, λ2 according to Theorem5; notekp2k>kak: p2 = 12204, 1578, 4307, 22445, 7666 r2 = −0.0165, −0.0071, 0.0194, 0.0105, −0.0140 λ2 = 0.2858 kr2k/λ2 = 0.1110. (4.2.7)
Theorem6below gives an upper bound on the number of B&B nodes when branching on a hyper- plane in (KP).
Theorem 6. Suppose thata=λp+r, withp≥0.Then
iwidth(p,(KP)) ≤ krkkvk λ + β2−β1 λ + 1. (4.2.8)
This bound is quite strong for near parallel vectors computed from Theorems4and5. For instance, leta, p1, r1, λ1 be as in Example5. Ifβ1 =β2 in a knapsack problem with weight vectoraand each xi is bounded between 0and 3, then Theorem 6implies that the integer width is at most one. At the other extreme, it also implies that the integer width is at most one, if eachxi is bounded between0and 1, andβ2−β1 ≤39. However, this bound does not seem as useful, whenpis a “simple” vector, say a unit vector. Note that the assumption thatp≥0is only to simplify the proofs.
We now complete the proof of Theorem3, based on a simple transference result between branching directions, taken from [31].
Let us denote byQ, QR, andQN the feasible sets of the LP relaxations of (KP), of (KP-R), and of (KP-N), respectively.
First, letU andpbe the transformation matrix, and the near parallel vector from Theorem4. It was shown in [31] thatiwidth(p, Q) = iwidth(pU, QR). ButpU =±en, so
iwidth(p, Q) = iwidth(en, QR). (4.2.9)
On the other hand,
iwidth(p, Q) ≤ krkkvk λ + β2−β1 λ + 1 ≤ ⌊f(a)(2kvk+(β2−β1))⌋+ 1 (4.2.10)
with the first inequality coming from Theorem 6 and the second from using the bounds on1/λ and krk/λ from Theorem4. Combining (4.2.9) and (4.2.10) yields (1) in Theorem3.
Now letV andpbe the transformation matrix, and the near parallel vector from Theorem5. It was shown in [31] thatiwidth(p, Q) = iwidth(pV, QN).ButpV =±en−1, so
iwidth(en−1, QN) = iwidth(p, Q). (4.2.11)
On the other hand,
iwidth(p, Q) ≤ krkkvk λ + 1 ≤ ⌊g(a)(2kvk)⌋+ 1. (4.2.12)
with the first inequality coming from Theorem6and the second from using the bound onkrk /λ in Theorem5. Combining (4.2.11) and (4.2.12) yields (2) in Theorem3.