Better Algorithms for Online Bin Stretching

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 the 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 1.5 for any number of bins. We also show a specialized algorithm for three bins with a stretching factor of 11/8 = 1.375.

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

“…The medium items of size in (3,4] are packed in their own bins (four or five per bin). Similarly, large items of size in (6,9] are packed in pairs in their own bins. Finally, the huge items of size larger than 9 are handled similarly as in the previous papers: If possible, they are packed with the regular items, otherwise each in their own bin.…”

“…Second, as the item is not put in any existing regular bin R, we have s(R) + s(T ) > 6 and this also holds later when more items are packed into any of these bins. A new regular bin R with s(R) ≤ 4 can be created only from a tiny bin; note that a bin created from an empty bin by a regular item is either tiny or has size in (4,6]. If another regular bin with size at most 4 already exists, then both the size of the tiny bin and the size of the new item are larger than 2 and thus the new regular bin has size more than 4.…”

“…If B ∈ E, either B has at most two (new) non-large items and s(B ) ≤ 6+6 = 12, or it has at least three items and s(B ) > 5+5+5 = 15; both options are impossible for B ∈ C . If B ∈ R, it has old items of total size in (3,6]. Either B has a single new item and s(B ) ≤ 6 + 6 = 12, or it has at least two new items and s(B ) > 3 + 5 + 5 = 13; both options are impossible for B ∈ C .…”

Online bin stretching with three bins

“…The only remaining case is when s(A) < 15 throughout the algorithm and several items with size in the interval I · · = (22 − s(A), 11 − 1 2 c) arrive. These items are packed into C. Note that I ⊆ (7,11) and that the lower bound of I may decrease during the course of the algorithm.…”

“…We start the proof of s(B ←j ) + s(r) > 6.8 by restating (3), (7), and (8) Before summing up the inequalities, we multiply the first one by 8, the second by 2 and the third by 2. In total, we have: Finally, using the bound s(j) < 4 and noting that (82 − 48)/5 = 6.8, we get s(B ←j ) + s(r) > 6.8.…”