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Question: Consider a file with an i-width of 32 bits and an i-height of 16. The block size is 4 kB. Assume that the file system uses one reference counter for hard links in the inode and stores file size and file type as the last two entries in the inode. Assume that the file system uses the single indirect method in addition to listing the blocks directly in the inode to store larger files with a single entry reserved for the single indirect method in the inode. What is the maximal file size the file system supports? The only options are 4148 KB, 4096 KB and 4144 KB
To calculate the maximal file size the file system supports, we need to consider the following factors: 1. The size of the inode (i-width * i-height): 32 bits * 16 = 512 bits = 64 bytes 2. The size of the reference counter, file size and file type: 4 * 3 = 12 bytes 3. Remaining space in the inode for direct block addresses: 64 bytes - 12 bytes = 52 bytes 4. Number of direct block addresses in bytes: 52 bytes - 1 entry (reserved for single indirect method) = 48 bytes 5. Number of direct block addresses: 48 bytes / 4 bytes (1 address) = 12 addresses 6. Size of file covered by the direct block addresses: 12 addresses * 4 kB = 48 kB 7. Number of entries that can fit in a single indirect block: 4 kB / 4 bytes (1 address) = 1024 8. Size of file covered by the single indirect block: 1024 entries * 4 kB = 4096 kB Now, considering both direct and single indirect blocks: Maximal file size = 48 kB (direct blocks) + 4096 kB (single indirect block) = 4144 kB Therefore, the maximal file size the file system supports is 4144 KB.
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