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Mice21 [21]
1 year ago
9

onsider a file system on a disk that has both logical and physical block sizes of 512 bytes. assume that the information about e

ach file is already in memory. for each of the three allocation strategies (contiguous, linked, and indexed), answer the following question: if we are currently at logical block 10 (the last block accessed was block 10) and want to access logical block 4, how many physical blocks must be read from the disk?
Computers and Technology
1 answer:
anygoal [31]1 year ago
4 0

For three disk allocation strategies (contiguous, linked, and indexed),

in this case, the  physical blocks must be read from the disk are 1,4 and 2 respectively.

In the disk, every block has a sequential number. The majority of disks today do logical I/O in hardware. Physical I/O is seldom ever required of operating systems today.

File access uses virtual I/O. Similar to disk logical I/O is disk virtual I/O. The distinction is that with virtual I/O, as opposed to physical I/O, the blocks that make up a file are

You could want to transfer from virtual block 10 to virtual block 4 in a contiguous allocation. Since the file is contiguous and you are at logical block 10 + Z, block four can be accessed by reading logical block 4 + Z. Without any intermediaries, that can be done directly.

Each block in the linked structure has 511 bytes of data and one byte that serves as an offset to the following block.

Read the first block, determine the offset to the second block, read the second block, determine the offset to the fourth block, and then read the fourth block in order to gain access to the fourth block.

If blocks are numbered from 1 to 10, then then is the "4" correct. The remainder of the solutions, however, make the assumption that block numbering begins at 0, using division to obtain offsets.

Assume there is a contiguous index for the indexed way of allocating files. Once more, it is assumed that each offset in the index is one byte in size.

Let's say that Z serves as the index's origin. Then, the byte at block Z + 3 DIV 512 and the offset at 3 MOD 512 (3 = 4 - 1) make up the entry for block 4 (numbered 1 to 4).

You must read the index and the data block in order to locate the block (2 reads).

To learn more about disk allocation strategies click here:

brainly.com/question/15083707

#SPJ4

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See attachment for explanation.

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