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Simple PIRATE. become disk resident) are required to be deleted from the disk, independent of the size of the victims. The following proof formalizes this statement and proves the optimality of simple PIRATE in minimizing the latency time of the system. Lemma 1: To minimize the average latency time of the system, during pass 2, PIRATE must delete those blocks corresponding to the object with the lowest heat, independent of the object sizes. Proof: Without loss of generality, assume F ϭ ͕X, Y͖, size(block) ϭ 1 and BTertiary ϭ 1.

In the first pass, it deletes from those objects (say i) whose disk resident portion is greater than the size of their first slice (Si,1). By doing so, it can ensure a zero latency time for requests that reference these objects in the future (by employing the pipelining mechanism). Note that PIRATE deletes objects at a granularity of a block. Moreover, it frees up only sufficient space to accommodate the pending request and no more than that. ) If the disk space made available by the first pass is insufficient, then simple PIRATE enters its second pass.

We use t to represent an atomic task and ␪ for a composite task. Similarly, T represents a set of atomic tasks while ⌰ is a set of composite tasks. A composite task, itself, is a set of atomic tasks; for example, ␪ ϭ ͕t1, t2, . , tn͕. Each atomic task has the same parameters as defined earlier, except for the release time r(t). Instead, each atomic tasks has a lag time denoted by §(t). Without loss of generality, we assume for a composite task ␪, §(t1) Յ §(t2) Յ и и и Յ §(tn). Subsequently, we denote the first atomic task in the set ␪ as car(␪); that is, car(␪) ϭ t1.

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37.Multimedia by John G. Webster (Editor)


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