Question

How many disk accesses are needed to bring byte 4090 of a file into memory when the file is stored using double indirect indexed allocation? Assume that only the file's FCB is in memory, block pointers require 32 bits, and that blocks hold 1024 bytes each.

Answer 1

A double indirect block contains pointers to single indirect blocks. A single indirect block contains pointers to data blocks.

Each block holds 1024 bytes. Block pointers are 32 bits.

Block pointers per block = (1024 * 8) / 32 = 256

It means one single indirect block will contain pointers for 256 blocks.

One double indirect block will contain pointers for 256 single indirect blocks.

Dat block 29 Indihect s12e 519 248 Double Scannedwith canhe

It means first entry in double indirect index will point to first 256 data blocks of file.

1st data block of file - byte 0 to 1023

2nd data block of file - byte 1024 to 2047

3rd data block of file - byte 2048 to 3071

4th data block of file - byte 3072 to 4095 (This block will contain byte 4090. This block needs to be accessed.)

1st access will be for double indirect index block.

2nd access will be for single indirect index block. Now we already know that we have to access 4th data block of file. So we will directly access that block, whose pointer is stored in single indirect index block.

3rd access will be for the data block containing data byte 4090.

Hence total 3 disk access are needed.