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

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answer is B

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

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as given in the question.

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and three additional addresses: one for single indirect block, one for double indirect block and one for triple indirect block

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

Comments

best answer

Answer 5

Its 2GB

Comments

pls somebody make it understand to mee pls pls explain
a big thanks in advance

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They are asking size of the file. That means all we need to calculate is number of blocks*size of block to know the size of file. since files are being stored as blocks. 

 To calculate total number of blocks 

Block size =1024 bytes

1)Directly given blocks   10*1024 bytes

2)single indirect block    1*128*1024 bytes  (since 1 block is pointing to a block which has 128 address of 128 blocks each of size 1024. since it is given each block will have address of 128 other blocks

3)double, triple indirects with same logics 

totally adding them gives total block sizes

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now clear..!! :)

Answer 6

Block Size = 2^10B

No. of block pointer in one block = 2^7

Total no. of block pointers = 10+2^7+2^7*2^7+2^7*2^7*2^7

                            = 10+2^7+2^14+2^21

Max file Size = (10+2^7+2^14+2^21)*1KB

              = 2113674KB

              = 2.01575660706 MB (option B)