GATE CSE 2019 | Question: 42

Question

The index node (inode) of a Unix -like file system has $12$ direct, one single-indirect and one double-indirect pointers. The disk block size is $4$ kB, and the disk block address is $32$-bits long. The maximum possible file size is (rounded off to $1$ decimal place) ____ GB

Comments

In the question the disk block size is explicitly mentioned as 4 kB, not 4KB, and 

kB means 1000 Bytes whereas KB means 1024 Bytes 

it's depend upon what is the base you are taking is it 2 or 10, but as per me, we have to take 1024 but not 1000.

but hope GATE key have range of values ! 

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$(2^{10})^2*2^{12}B$

=> $2^{32}B$

=> $4$ GB

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

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Answer 1

Given $12$ direct, $1$ single indirect, $1$ double indirect pointers

Size of Disk block $= 4kB$

Disk Block Address $= 32 \text{ bit } = 4B$

Number of addresses= Size of disk block/address size $= \frac{4kB}{4B} = 2^{10}$

Maximum possible file size$ = 12*4kB\  + \ 2^{10} * 4kB \ + \ 2^{10} * 2^{10} * 4kB$

$= 4.00395 \  GB \simeq 4 \ GB$

Hence $4GB$ is the correct answer

Comments

should we consider only the most significant term or the addition of all terms, because in some palces they only consider the most significant term...

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anchitjindal07 Please refer “The Design of the UNIX Operating System” by Maurice J. Bach

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here we can consider most significant term because if we add the terms then they can change only maximum of 4-5 digits places  and our significant term has 7 digits and they want answer upto 1 decimal place . so , we can consider :)

Answer 2

Disk block size $= 4$ KB $= 2^{12}$ Bytes

Disk block address $= 32$ bits $= 4$ Bytes $= 2^2$ Bytes

No of addresses in a block $= 2^{12}/ 2^2 = 2^{10}$

Maximum file size $= (20 + 2^{10} + (2^{10})^2)* 2^{12} = X$

$= X / 2^{30}$ GB

$= 4.0$ GB

Comments

Maximum file size =(20+210+(210)2)∗212=X

Here why 20 is added, why not 12 added?

Answer 3

1GB = $2^{30}B$

DB = Disk Block

DBA length or size = 32 bits

= 4 B

#DBAs that can be stored per DB = DB size/ DBA size

= $\frac{2^{12}}{2^2}$

= $2^{10}$

TOTAL file size

= $(\#DirectPtr + \sum_i(\#Ptr_i * \#DBAs)^{i} )  * DB size$

Where i = indirectionLevel (single = 1, double = 2, etc)

Each direct pointer points to a single DB and for indirect pointer it depends on the indirectionLevel (single, double, triple or etc) to point to that many DBs. Thus we get max file size by multiplying total DB pointers with DBsize.

= $(12 + \sum (1*2^{10})^1 \ + (1*2^{10})^2 )*2^{12}\ B$

=$4. 004\ GB$ or

= 4.0(ANS)

Comments

Can anyone elaborate how the calculation is done at the end ?

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2^32  = 4GB approx.

Answer 4

Simple calculation –  

= 4GB+ 4MB + 48KB

To convert everything into GB we have to multiply by 1/1GB which is (1 / 1024*1024*1024) B

So, 4MB = 4 * 1024 * 1024 B and now into 4 *1024 * 1024 / 1024 * 1024 * 1024

4MB = (4 / 1024) GB = 0.00390625 GB

48 KB = 48 * 1024 / 1024 * 1024 * 1024 = 48 / 1024 * 1024 = 0.0000457763671875 GB

Maximum possible file size = 4GB + 0.00390625 GB +

0.0000457763671875 GB

= 4.00395 GB ≃4 GB

Answer 5

inode: Inode(index node) is a data structure that contains all metadata(including data block(s) pointer) of a file.

  Block Size = 4KB

  Block address = 4B

  # of block pointers in a block = 2^12/2^2 = 2^10

  Total data block pointers  = 12+2^10+2^10*2^10

  Maximum possible file size = (12+2^10+2^20)*2^12

                             = 4299210752/2^30

                             = 4.00395202637 GB

Answer 6

Given disk block size=4KB = 2^12 B

disk block address= 32 bits = 4 B

Now number of entries in one block (require when we use indirect pointers) = 2^12 B/4B = 2^10 entries

So there are 2^10 entries in one block

wrt Single Indirect pointer, file size = 12*4KB

wrt Indirect pointer, file size = 2^10 * 4KB

wrt Double indrect pointer, file size = 2^10*2^10*4KB

So Maximum file size is the summation of all of these:

=(12+2^10+2^10*2^10)*4KB

=1048588*2^10 B

=4 GB approx(after rounding off to 1 decimal place)