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Operating System: UNIX Inode

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Q1> Consider the Unix file node which maintains 12 direct disk block addresses, 1 single indirect, 1 double indirect, and 1 triple indirect disk block addresses. The disk block address requires 32 bits and DB size is 1 KB then calculate

1. total file size.

2. Max file size possible.

3. Max disk size.

Q2> Find the location of data byte no. 10000000 Byte in the file, location comprises of

<In which direct/ indirect pointers branch the address is>
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Given , disk block address requires 32 bits = 4 Bytes  and DB size is 1 KB = 210 Bytes .

If , The Disk Block Address = DBA  bits

The Disk Block Size = DBSize  Kbits

then maximum file size possible 

= { n0 + n1 ( DBSize / DBA ) + n1 ( DBSize / DBA )2 + n3 ( DBSize / DBA)3  }  * DB size

where n0 = direct disk block addresses,  

n1=  single indirect disk block addresses, 

n2 = doubly indirect DBAs 

 n3 = triple indirect DBAs.

Given , n0 = 12 , n1 = 1 , n2 = 1 , n3 = 1

Now,  DBSize / DBA = (210 / 4) = ( 210 / 22 ) =  28 

Hence , Max file size possible = 12 + { (1 * 256 ) + (1 * 2562  )+ (1 * 2563 ) } * 1024 B

=  ( 12 + 256 + 65536 + 16, 777 , 216  ) * 1024 Bytes

= 16, 843, 020 * 1024 Bytes

= 17, 247, 252, 480 Bytes

Reference :

https://gateoverflow.in/20557/elaborate-unix-inode

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= 210  * 232 

= 242  Bytes

Max disk size = 

(  DBSize / DBA)  = (210 / 4) = ( 210 / 22 ) =  28 

Max disk size = 2563 * 1024 = 17,179, 869, 184 = 234 Bytes

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