Analysis of a Compound bin Packing Algorithm
نویسندگان
چکیده
while w(d) = 4=15 and we conclude that, in any event, w(B 3) 1. Case 3. Suppose jB 3 j = 4. B 3 cannot contain an X 1 item, since 1=2 + 3(1=5) > 1. Neither can it contain two X 2 items, since 2(1=3) + 2(1=5) > 1. Similarly, it cannot contain four X 3 items, since each has size greater than 1/4. However, if it contains three items of type X 3 and one of type X 4 , then w(B 3) 3(4=15) + 1=5 = 1. Thus B 3 must contain exactly one X 2 item. If the other three items have weight less than or equal to 1/5, w(B 3) 2=5 + 3(1=5) = 1. If there were two X 3 items, s(B 3) > 1=3 + 2(1=4) + 1=5 > 1. Thus B 3 must contain exactly one X 3 item. Let B 3 = fb; c; d; eg, with b of type X 2 , and c of type X 3. If w(b) < 2=5, then w(B 3) 3=10 + 4=15 + 2(1=5) < 1. Thus w(b) = 2=5 and w(c) = 4=15. This means c must be available when b is packed. If b is the largest item in some bin B of the B2F packing, then B would contain two X 2 items and another item since s(b) + s(c) + s(d) + s(e) 1 implies that 2s(b) + s(c) < 1. This cannot happen, however, so it must be that B = fx; bg where s(x) > 1 0 2s(c), since b was not replaced by two smaller items. Because s(c) < 1 0 1=3 0 2=5 = 4=15, we know s(x) > 7=15. Thus B is the third exceptional bin (s(b) < 1 0 1=4 0 2=5 = 7=20) and again w(B 3) 1. Now, to complete our proof of Theorem A, we note that w(B) = 4=5 for all but at most four B2F bins (the three exceptional bins and the last bin), so that P x2L w(x) (4/5) (B2F(L) 0 4). At the same time, w(B 3) 1 for all B 3 in the optimal packing ensures P x2L w(x) OPT(L). Combining these two inequalities yields B2F(L) (5/4)OPT(L) + 4, as desired. 2 33 Case 2. Suppose jB 3 j = 3. The largest item in B 3 must have a weight exceeding 1/3, and so must …
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ورودعنوان ژورنال:
- SIAM J. Discrete Math.
دوره 4 شماره
صفحات -
تاریخ انتشار 1991