cop4600-fa07-hw3Solutions

# cop4600-fa07-hw3Solutions - Homework # 3 (Solution) 1)...

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Homework # 3 (Solution) 1) Since i-node has 4 direct block addresses of 1 KB each and 1 indirect block address. File size due to direct block address= 4*1KB= 4 KB Size of indirect block address=1KB Each address takes 4 bytes. Hence number of block addresses in the indirect block address = 1024 bytes/ 4 bytes = 256 256 block address can store 256*1KB = 256 KB file size. Maximum file size= 256 KB + 4 KB = 260 KB 2) Assume 1 represents used block in bitmap, blank represents unused or assume a 0 in it. Initial bitmap 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 16 1 1 1 1 1 1 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 1 1 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 1 1 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 1 1 st file creation =10 blocks To minimize disk arm motion, the block should be allocated in such a way that they are as close to each other as possible. So in this case, we have 10 contiguous blocks available from 36 to 45 , so we will choose that. 1

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## This homework help was uploaded on 04/18/2008 for the course COP 3530 taught by Professor Davis during the Fall '08 term at University of Florida.

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cop4600-fa07-hw3Solutions - Homework # 3 (Solution) 1)...

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