GATE IT 2004 | Question: 67

Question

In a particular Unix OS, each data block is of size $1024$ bytes, each node has $10$ direct data block addresses and three additional addresses: one for single indirect block, one for double indirect block and one for triple indirect block. Also, each block can contain addresses for $128$ blocks. Which one of the following is approximately the maximum size of a file in the file system?

• A. $512$ MB
• B. $2$ GB
• C. $8$ GB
• D. $16$ GB

Comments

$Ans:B$

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1.Maximum file size

2.Total size of file system

3.maximum possible file size

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$\left(2^{7}\right)^3 \times 2^{10}=2^{31}=2GB$

Answer 1

Answer: (B)

Maximum file size $= 10 \times 1024$ Bytes $+ 1 \times 128 \times 1024$ Bytes $+ 1\times 128\times 128 \times 1024$ Bytes $+ 1 \times 128 \times 128 \times 128 \times 1024$ Bytes $=$ approx $2 \ GB$.

Comments

@Gupta731

U can do that one with the same way..

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Each block can contain 128 addresses
where in the other question size of a block is 128B and size of a add. is 8B so no. of addresses in one block are 128/8.

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we have a number of blocks = DB Size / DBA =128, in question they have mentioned that

Which one of the following is approximately the maximum size of a file in the file system

(128)^3 * 2^10

(2)^31

2GB

Answer 2

Size of File/Levels	Direct Address	Indirect Address-1	Indirect Address-2	Indirect Address-3
Level-0	10*1024 B	1	1	1
Level -1		1*128*1024 B	1*128*1024 B	1*128*1024 B
Level- 2			1*128*128*1024 B	1*128*128*1024 B
Level -3				1*128*128*128*1024 B

Add the Principle Diagonal Values = 2.01575 GB = Approx. 2.0 GB

Comments

why multiplied by 1 in indirect address (though it will not effect the result but just curious about it )

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it indicates file has 1 indirect address

Answer 3

maximum file size=(10+x+x²+x³)*Blocksize

x=no of block addresses in a block=2⁷

maximum file size=(10+x+x²+x³)*2¹⁰ B = (10+2⁷+2¹⁴+2²¹)*2¹⁰  = 2³¹B = 2GB

 

Answer 4

Maximum size of the File System = Summation of size of all the data blocks whose addresses belongs to the node.

Given:
Size of 1 data block = 1024 Bytes
No. of addresses which 1 data block can contain = 128 (just like 128 page table entry stored in a block)

Now, Maximum File Size can be calculated as:
10 direct addresses will point to 10 data blocks in the memory = 10*1024
1 single indirect entry will point to 1 block which again  points to 128 data block in the memory (just like multi level paging) = 128*1024
1 doubly indirect entry will point to 128*128 data block = 128*128*1024
1 triple indirect entry will point to 128*128*128 data block = 128*128*128*1024

Hence,
Max File Size = 10*1024 + 128*1024 + 128*128*1024 + 
                128*128*128*1024 Bytes
              = 2113674*1024 Bytes
              = 2.0157 GB ~ 2GB