Question 7769 – GATE 2019
April 23, 2024Operating-Systems
April 23, 2024File-System
| Question 3 |
A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is
| 3 KBytes | |
| 35 KBytes | |
| 280 KBytes | |
| dependent on the size of the disk |
So, one direct block addressing will point to 8 disk blocks = 8*128 B = 1 KB
Singly Indirect block addressing will point to 1 disk block which has 128/8 disc block addresses = (128/8)*128 B = 2 KB
Doubly indirect block addressing will point to 1 disk block which has 128/8 addresses to disk blocks which in turn has 128/8 addresses to disk blocks = 16*16*128 B = 32 KB
Maximum possible file size = 1 KB + 2 KB + 32 KB = 35 KB
So, one direct block addressing will point to 8 disk blocks = 8*128 B = 1 KB
Singly Indirect block addressing will point to 1 disk block which has 128/8 disc block addresses = (128/8)*128 B = 2 KB
Doubly indirect block addressing will point to 1 disk block which has 128/8 addresses to disk blocks which in turn has 128/8 addresses to disk blocks = 16*16*128 B = 32 KB
Maximum possible file size = 1 KB + 2 KB + 32 KB = 35 KB