By Allan Borodin (auth.), Siu-Wing Cheng, Chung Keung Poon (eds.)
This e-book constitutes the refereed complaints of the second one foreign convention on Algorithmic elements in info and administration, AAIM 2006, held in Hong Kong, China in June 2006.
The 34 revised complete papers offered including abstracts of two invited talks have been rigorously reviewed and chosen from 263 submissions. The papers hide themes from parts resembling on-line scheduling, video game and finance, information buildings and algorithms, computational geometry, optimization, graph, and string.
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Extra resources for Algorithmic Aspects in Information and Management: Second International Conference, AAIM 2006, Hong Kong, China, June 20-22, 2006. Proceedings
If such a machine does not exist, discard job j (job j is thus lost). Clearly the Simple Algorithm satisﬁes immediate notiﬁcation. The following theorem can be easily shown with the similar analysis as in . For the sake of completeness we also give a proof. Theorem 2. The competitive ratio of algorithm LS is 2. Proof. Consider the following instance with 2m jobs. The ﬁrst m jobs are available at time zero, each of which has a deadline 52 . And the other m ones with a common deadline 32 come at time 12 .
We assume the goods of the requests is partible. If a given request has overmany goods in the sense of the current capacity of the server, it can be divided into several partitions. The server is allowed to load some partitions of them. And the rest of them can be considered as the goods of new requests. In this paper we assume that zi ≤ Z, ∀i ∈ N. It is also assumed that the sequence R = (r1 , r2 , · · · , rm ) of requests is given in order of non-decreasing release times, that is, 0 ≤ t1 ≤ t2 ≤ · · · ≤ tm .
Proof. (By induction) We ﬁrst prove that d1 ≥ d1 ; Otherwise d1 < d1 . According to Lemma 2 all the job with deadline at least d1 must be accepted by the algorithm. Let j∗ be the last job, with deadline at least d1 , which is started before Ci,min in σ. Let s∗ be the starting time of the job j∗ . If s∗ ≥ si , then all the jobs in job set Ai must have deadline at least dt+1 , it is a contradiction. So it must be processed before time si (see Figure 5). According to the algorithm, P (s∗ ) contains only job j∗ .