UGC NET CSE | October 2020 | Part 2 | Question: 17

Question

Suppose you have a Linux file system where the block size is $2K$ bytes, a disk address is $32$ bits, and an $i-$node contains the disk addresses of the first $12$ direct blocks of file, a single indirect block and a double indirect block. Approximately, what is the largest file that can be represented by an $i-$node?

• A. $513$ Kbytes
• B. $513$ MBytes
• C. $537$ Mbytes
• D. $537$ KBytes

Answer 1

total size of file system= {no of direct dba + (db size/dba)+((db size/dba)^2+…..}*db size

Answer 2

• Size of the disk block = 2048 Bytes

• disk address = 32 bits = 4 Bytes

• Number of addresses per block = 2048/4 = 512 = 2⁹

• Maximum file size = 12 Direct Blocks + 1 indirect block + 1 double indirect block

• = 12 + 2⁹ + 2⁹* 2⁹

• = 12 + 2⁹ + 2¹⁸

• The size of each block = 211 Bytes

• Maximum size of the file = ( 12 + 2⁹ + 2¹⁸ ) × 2¹¹ Bytes

• ∴ Maximum size of the file = 513.0234 ≈ 513 MBytes