409 views
0 votes
0 votes
Consider an indexed file allocation using index nodes (inodes). An inode contains among other things, 7 indexes, one indirect index, one double index, and one triple index. If the disk sector is 512 bytes, which is the closest option for maximum file size in this allocation scheme?

1 Answer

1 votes
1 votes
We must first determine the size of an index.

For a 2-byte index, we can have 65536 disk blocks, i.e. 512 * 65536 = 32MB. But a triple index structure can express more disk blocks.

Therefore, we go to a 4-byte indexing scheme (3-byte indexing scheme are not attractive and are not sufficient).

A 4-byte indexing scheme therefore gives a maximum file size of

7*512 + 128*512 + 128*128*512 + 128*128*128*512 = 108219952 or about 1Gbytes

Related questions

0 votes
0 votes
1 answer
3
lalitver10 asked Jan 22, 2022
648 views
Question→ Given a system with 3 processes where each process requires at least 2 resources to complete their execution, then the largest number of resources which will ...