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Assuming n objects in the system, let S be the total amount of disk space for these objects and their replicas. Let Rj be the optimal number of instances for object j, Sj to denote the size of object j, and Fj to represent the access frequency (%) of object j. The problem is to determine Rj for each object j (1 Յ n j Յ n) while satisfying ͚j ϭ Rj и Sj Յ S. There exist several algorithms to solve this problem (54). A simple one known as the Hamilton method computes the number of instances per object j based on its frequency (see (53).

This is eliminated in the following manner. The algorithm is first performed virtually—that is, in main memory, as a compaction algorithm on the free lists. Once completed, the entire sequence of operations that have been performed determine the ultimate destination of each of the modified sections. The scheduler constructs a list of these sections. This list is in- 735 serted into a queue of house keeping I/Os. Associated with each element of the queue is an estimated amount of time required to perform the task.

In this study, we focus on the scenario where replacement is required; that is, ABSENT(Z) Ͼ free_disk_space. We define latency time observed by a request referencing Z (ᐉ(Z)) as the amount of time elapsed from the arrival of the request to the onset of the display. It is a function of DISK(Z) and BTertiary. If DISK(Z) ϭ size(SZ,1), then the maximum value for ᐉ(Z) is the worst reposition time of the tertiary storage device. ) If Ͼ size(SZ,1), then (ᐉ(Z) ϭ 0 (due to assumed optimization). , DISK(Z) Ͻ size(SZ,1)), the system determines the starting address of the nondisk resident portion of Z (missing), and ᐉ(Z) is defined as the total sum of (1) the repositioning of tertiary to the physical location corresponding to missing, and (2) the materialization time of the remainder of the first slice, DISK(Z) size(block) × (size(SZ,1 ) − DISK (Z)) BTertiary The average (expected value of) latency as a function of requests can be defined as µ= heat(x) ∗ (x) (18) heat(x) ∗ ( (x) − µ)2 (19) x The variance is σ2 = x By deleting a portion of an object, we may increase its latency time resulting in a higher Ȑ and ␴2.

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37.Multimedia by John G. Webster (Editor)


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