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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:

1. How is the logical-to-physical address mapping accomplished in this system? (For the indexed allocation, assume that a file is always less than 512 blocks long.)

2. 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?

#### Solution Preview

Let X is the starting address of the file

a. Contiguous:- Now, since files are stored in contiguous memory block , the file system only stores the starting address of the file

i. To reach to X, We first need to divide it by 512.
Y=X/512 -> Y+X will give us the starting of the block.
Z=X%512 -> Z will be the displacement in that block.

ii. Since, files are stored in contiguous block we will need to make just 1 read to go to block 4 from block 10.(We know that it is 6 blocks away from this block , so we can directly ...

#### Solution Summary

For each of the three allocation strategies (contiguous, linked, and indexed), the solution answers various questions.

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