Suppose a computer has a file system for a 128 GB disk, where each disk block…
Suppose a computer has a file system for a 128 GB disk, where each disk block is 8 KB. If the OS for this computer uses a FAT, what is the smallest amount of memory that could possibly be used for the FAT (assuming the entire FAT is in memory)?
Answer: A. 48MB — There has to be a FAT entry for each disk block. Since the disk is 128 GB (237 bytes) and a disk block is 213 bytes, the number of disk blocks (and thus the…
- A.
48MB
- B.
16MB
- C.
8MB
- D.
32MB
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Correct answer: A
There has to be a FAT entry for each disk block. Since the disk is 128 GB (237 bytes) and a disk block is 213 bytes, the number of disk blocks (and thus the number of FAT entries) is 237 / 213 = 224 . Since there are 224 entries, a block number (disk address) requires a minimum of 24 bits = 3 bytes. So, the minimum amount of space occupied by the FAT is the number of entries (224) times the 3 bytes per entry, 224 * 3 = 16 MB * 3 = 48 MB.