Efficient Management of Fragmented Replica in Data Grids

Efficient Management of Fragmented Replica in Data Grids

Chao-Chin Wu (National Changhua University of Education, Taiwan), Lien-Fu Lai (National Changhua University of Education, Taiwan) and Jia-Xian Lai (National Changhua University of Education, Taiwan)
Copyright: © 2012 |Pages: 8
DOI: 10.4018/jghpc.2012040105
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Rather than replicating a file completely, Chang et al. proposed a fragmented replication technique to cope with the problem that only partial content of the replica are required in a local application. Furthermore, they also proposed two server selection algorithms for replica retrieval. However, their algorithms do not always find an optimal solution. To address the problem, in this paper, the authors propose a replica selection algorithm to improve the fragmented replica retrieval efficiency in this paper. It is a heuristic considering not only the transmission time but also the number of available servers for each block. Simulation results show that the proposed algorithm can improve the retrieval efficiency up to 12%.
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Chang et al. introduced the concept of fragmented replicas and proposed algorithms for their retrieval (Chang & Chen, 2007). Though the transmission time can be improved by either the AR algorithm or the OOCA algorithm, their proposed algorithms do not always find an optimal solution. For instance, the OOCA will select the appropriate server for data blocks one by one. In each run, the algorithm will select the server having the minimum accumulative transfer time after serving the subsequent block.

We give an example with the following assumptions to explain why the OOCA algorithm cannot get the optimal solution. There are four servers, S1, S2, S3 and S4, in the system and the transferring times of one block by the servers are 6, 13, 8, and 17, respectively. The target file consists of four data blocks, B1, B2, B3 and B4. The possession relationship between servers and blocks is represented by a table as shown in Figure 1. The value of each element is either 1 or 0. If the server Si has the block Bj, the value of the corresponding element e(i, j) is 1.

Figure 1.

The relation between servers and blocks. Si has a replica of Bj if (Si, Bj) is 1.


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