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Disk Scheduling Techniques

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Question

In a particular Unix OS, each data block is of size 1024 bytes, each node has 10 direct data blocked addresses and three additional addresses; one for single indirect block, one for double indirect block and one for triple indirect block. Also, each block can contain addresses for 128 blocks. Which one of the follwoing is approximately the maximum size of a file in the system?

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Solution

Maximum file size = [direct DBA + Number of (Data block size/DBA)+Number of (Data block size/DBA) $^{2}$ +Number (Data block size/DBA) $^{3}$ ]*
Data block size.
Data Block size = 1024 byte
(Dist Block size/DBA) = Number of Disk Block
address Store inside One Block.
Maximum file size = [10 + 1* (128)+ 1 (128 128)+1(128128128)]*1024 Byte
= Approx 2 GB.

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