GATE CSE 2012 | Question: 41

Question

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

Comments

Both are the same approach. Look carefully.

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$(2^4)^2*2^7B$

=> $2^{15}B$

=> 32KB approx.

https://gateoverflow.in/1045/gate2004-49?show=330294#a330294

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                   Pictorial explanation                               

Answer 1

Direct block addressing will point to $8$ disk blocks $= 8 \times 128 \ B = 1 \ KB$

Singly Indirect block addressing will point to $1$ disk block which has $128/8$ disc block addresses $= (128/8) \times 128 \ B = 2 \ KB$

Doubly indirect block addressing will point to $1$ disk block which has $128/8$ addresses to disk blocks which in turn has $128/8$ addresses to disk blocks  $= 16 \times 16 \times 128 \ B= 32 \ KB$

Total $= 35 \ KB$

Answer is (B).

Comments

@Arjun sir, could you please clarify whether the ans should be 35KB, as given in the Best Answer or 32 KB as mentioned in one of the comments?

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What does 300 GB means here ? Does it means we have a disk of size 300GB but we only can use only 35KB of those 300GB ?

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@KanuSwavik in the question it is telling disk size is 300GB and we know that a disk may contain more than one file and each file have I-Node. In this way each file may contain some limit to store data in it . Hence to store file it needs only file descriptor not whole disk space needed..hence here disk size given is unnecessary to the given question (may be it is given for generate confussion to the student in the examination hall to spent more time in this easy question) thankyouu soo much……. keep learning…..

Answer 2

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 block 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

Comments

Nice explanation .....

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this should be taken as the best answer.

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Helpful! Good explanation.