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Plz explain its concept ?? A file system with 300 GByte d

Plz explain its concept ??

A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1
indirect block address and 1 doubly indirect block address. The size of each disk block is 128
Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in
this file system is

(A) 3 KBytes    (B) 35 KBytes
(C) 280 KBytes (D) dependent on the size of the disk

 

2Comments
Arvind Rawat @arvind.rawat
16 Dec 2014 02:56 pm

Total number of possible addresses stored in a disk block = 128/8 = 16

Maximum number of addressable bytes due to direct address block = 8*128
Maximum number of addressable bytes due to 1 single indirect address block = 16*128
Maximum number of addressable bytes due to 1 double indirect address block = 16*16*128

The maximum possible file size = 8*128 + 16*128 + 16*16*128 = 35KB

jayendra @jayendra
16 Dec 2014 02:57 pm

Here, 8 direct disk block address:

means, you have 8 addresses which are pointing to 8 disk blocks. So, with this 8 direct DBA you are getting = 8*128 bytes (no of disk blocks * size of disk block)

1 indirect disk block address:

means, you have a disk block, in which only disk bock addresses are there and each of the DBA is pointing to disk block. therefore you need to find how many DBA you can store in 1 disk block. which are,

no of address in 1 block = (Disk block size) / Disk block address in bytes

here, it is 128/8 = 16 address in 1 block and each is pointing to one disk block. therefore with this 1 indirect block address you are getting  16*128 Bytes

doubly indirect block address:

means, you have a disk block in which disk block addresses are there(same like singly indirect). but now these blocks have again only address of next level disk blocks. which are same as you got in singly indirect. therefore with doubly indirect 16*16*128 bytes.

total:

therefore maximum space = maximum file size = direct + singly + doubly

= (8*128) + (16*128) + (16*16*128)

= 1024B + 2048B + 32768B

=35840 Bytes

=35 Kbytes