Online bin stretching with three bins

Abstract

Abstract. Online Bin Stretching is a semi-online variant of bin packing in which the algorithm has to use the same number of bins as an optimal packing, but is allowed to slightly overpack the bins. The goal is to minimize the amount of overpacking, i.e., the maximum size packed into any bin. We give an algorithm for Online Bin Stretching with a stretching factor of 11/8 = 1.375 for three bins. Additionally, we present a lower bound of 45/33 = 1.36 for Online Bin Stretching on three bins and a lower bound of 19/14 for four and five bins that were discovered using a computer search.

A two-phase algorithm for bin stretching with stretching factor 1.5

“…Indeed, (7) together with (6) yields w(A) + w(A) ≥ −5 and summing this with all the other bounds, namely w(F) > 5 from Lemma 2.13 and w(B i ) ≥ 0 for whole blocks B i ∈ D from Lemma 2.15, leads to w(L) > 0, which is a contradiction. Now we prove (7). The items n and n from A together with the first two special bins in A have size more than 36.…”

“…The paper of Gabay et al [10] showed a lower bound of 19/14 ≈ 1.357 for m = 3 using a computer search. Extending their methods, the authors of this paper were able to reach a lower bound of 15/11 = 1.36 for m = 3 in [7] as well as a bound of 19/14 for m = 4 and m = 5. The preprint [10] was updated in 2015 [11] to include a lower bound of 19/14 for m = 4 bins.…”

“…Summing this with s(H) ≥ 12|H| from Lemma 2.8(iii) now yields s(L) > 12|L|, a contradiction. Thus (7) always leads to a contradiction. We now distinguish subcases depending on |A| and in each case we either prove (8) or obtain a contradiction directly.…”

“…If i is medium and there is a medium-item bin where i fits, pack it there. (7) If i is large and there is a large-item bin where i fits, pack it there. (8) If i is huge:…”

“…For m = 2, it is easy to achieve a stretching factor 4/3, which is optimal. Thus the first interesting case is m = 3; the currently best algorithm given by the authors of this paper [7] has a stretching factor of 11/8 = 1.375.…”