Inode Questions

Question

Consider an indexed file allocation using index nodes (inodes). An inode contains among other things, 14 direct indexes, one indirect index, two double indexes, and three triple indexes. If the system contains the 10485 files, size of disk is 1 PB and disk sector is 1 Kbytes.
 
1. What is the size of inode in this allocation scheme?_____Bytes

2. How much space of disk is required to store inode of all the files?____Bytes

3. How large of a file can be addressed by the double indirect block alone?____MB

4. What is the maximum file size possible in this allocation scheme?____GB

Answer 1

Block size = sector size = 1KB

Block pointer size = (40-10) = 30 bits

Now there is one inode for each file in the system.

1. Size of inode = (14+2+3)*(block pointer size) = 19*30bits = 570bits= 72 Bytes.

2. space of disk is required to store inode of all the files = 72*10485 = 754920 bytes.

3. Size of file Addressed by double indirect block pointer = (274*274*1KB) = 75.076 MB
   [ As the No of entries in each block = (8*1024/30) = 274 ]

4. Maximum file size possible = 14*1KB + 274*1KB + 2*274*274*1KB +3*274*274*274*1KB = almost 62 GB

Comments

Thank You!

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You have taken PB as 40 it must be 50
And also no. Of entries will be 20 not 19 bcoz 14+1+2+3 =20
Please correct if i m wrong.
Answer should be 20×40 bits = 800 bits or 100 bytes

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There are 20 Entries in total. why are you taking 19 ?mystylecse

 

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why [ As the No of entries in each block = (8*1024/30) = 274 ] in 3^rd point?

it is 1024/40 bits==25 bytes naa?

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@tvs6 Number of entries (Block Pointers that can be stored in 1 block) = total bits available in each block divided by size of each pointer, 

as size of each pointer is 30 bits, 

and size of each block is 1024*8 bits, 

so total number of pointers that can be stored in a block = floor(1024*8/30) = 273.

Answer 2

Block size = 1KB, and its offset is 10 bits.

address size for 1 PB = 50 bits.

Block Pointer Size = 50 - 10 = 40 bits = 5 Bytes.

1. Size of each Inode = (14+1+2+3) * 5 B = 100 B.

2. Space required to store all inodes:

   size each file have their own inode, so

   total space required for inodes = 10485 * 100 B = 1048500 B = 1MB (aprox.)

3. Max file size using double indirect pointer:

    number of address 1 block can have = 1 KB / 5B = 204.8 = 204 (only lower bound number of address can be stored).

    total size using double indirect pointer = 204 * 204 * 1 KB = 41616 B = 40.64 MB.

4. max file size possible = 14*1 KB + 1*204*1 KB + 2*204*204*1KB + 3*204*204*204*1KB = 24.37 GB (aprox)