File-System
Question 1 |
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 |
Question 1 Explanation:
It’s given disk block is of size 128B.
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
Question 2 |
The data blocks of a very large file in the Unix file system are allocated using
contiguous allocation | |
linked allocation | |
indexed allocation | |
an extension of indexed allocation |
Question 2 Explanation:
An Unix file system can utilizes an extension of indexed allocation. Because it uses direct blocks, single indirect blocks, double indirect blocks and triple indirect blocks.
Question 3 |
In the index allocation scheme of blocks to a file, the maximum possible size of the file depends on
the size of the blocks, and the size of the address of the blocks. | |
the number of blocks used for the index, and the size of the blocks. | |
the size of the blocks, the number of blocks used for the index, and the size of the address of the blocks. | |
None of the above |
Question 3 Explanation:
No. of addressable blocks using one Index block (A) = Size of block/Size of block address.
No. of block addresses available for addressing one file(B) = No. of Maximum blocks we can use for the Index * No. of addressable blocks using one Index block(A)
Size of File = B * Size of Block
No. of block addresses available for addressing one file(B) = No. of Maximum blocks we can use for the Index * No. of addressable blocks using one Index block(A)
Size of File = B * Size of Block
There are 3 questions to complete.