Question

Consider a file system on a disk that has both logical and physical block sizes of 512 bytes. Assume that the information about each file is already in memory. For each of the three allocation strategies (contiguous, linked, and indexed), answer these questions:

If we are currently at logical block 5 (the last block accessed was blocked 4) and want to access logical block 15, how many physical blocks must be read from the disk?

Answer 1

If we are currently at logical block 5 (the last block accessed was block 4) and want to access logical block 15, the number of physical blocks must be read from the disk are given as follows:

Contiguous

Divide the given logical address by 512. E and F represents the quotient and the remainder respectively.

Assume Y to be the starting address of the file.

Add E to Y to get the physical block number.

The displacement into the block is represented by F.

So, if the logical block is 5, the number of blocks as per the contiguous method, the readable blocks are 1.

Linked

Divide the given logical address by 511. E and F represents the quotient and the remainder respectively.

So, for chasing the linked list, the number of blocks obtained are A+1.

In this the number of displacements would be B+1.

So, if the logical block is 5, the number of blocks as per the linked method, the readable blocks are 15.

Indexed

Divide the given logical address by 512. E and F represents the quotient and the remainder respectively.

To get the indexed block into the memory, the address of the physical block is there in the index block which is at the location of A.

The displacement here is B.

So, if the logical block is 5, the number of blocks as per the indexed method, the readable blocks are 7.5