Unix Inode file system

Question

Consider a file system that uses UNIX like inodes to keep track of the sectors allocated to files. Assume that disk blocks are 1 KB in size, disk block addresses are 32 bits and the inode has space for 8 direct blocks, 1 singly indirect blocks, 1 doubly indirect block and 1 triply indirect block. What is the largest disk drive that could be fully utilized by this system?

(A) 2³² bytes
(B) 2³⁴ bytes
(C) 2¹⁰ bytes
(D) 2⁴² byte
Is there asked about total size of file system?
whats difference between them?

Comments

max file size=2^34

max disk drive  size=2^42

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pooja please give explanation.

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How to calculate max disk drive size???

Answer 1

The direct blocks store addresses of disk blocks. The size of each disk block is 1 KB = 2¹⁰ bytes.
There are 8 direct blocks, which can address: 8 × 1 KB = 2³ × 2¹⁰ bytes = 2¹³ bytes.
Size of each disk block is 1 KB and size of each disk block address is 32 bit. So, each indirect block can addresses: 2¹⁰ bytes / 4 bytes = 2⁸ blocks.
Thus,
1 single indirect block addresses 2⁸ × 1 KB = 2⁸ × 2¹⁰ bytes = 2¹⁸ bytes
1 double indirect block addresses 2⁸ × 2⁸ × 2¹⁰ bytes = 2²⁶ bytes
1 triple indirect block addresses 2⁸ × 2⁸ × 2⁸ × 2¹⁰ bytes = 2³⁴ bytes
Thus, the maximum file size is: 2¹³ + 2¹⁸ + 2²⁶ + 2³⁴ bytes ≈ 2³⁴ bytes ≈ 16 GB.

AND

Each block must be addressable in order to be used. The block addresses are 32 bits long. Thus, we can address 2³² blocks. Each block is 1 KB = 2¹⁰ bytes.
Thus, we can address 2³² × 2¹⁰ bytes = 2⁴² bytes.

Answer 2

Disk Block Address = 32-bit = 4B

Disk Block Size = 1KB 

#of Disk Block Pointer that can fit in one block pointer = 1024/4 = 256

Direct pointer maximum file size = 8 * 1KB = 8KB

Due to Single Indirect Pointer maximum file size = 256 * 1KB = 256KB

Due to Double Indirect Pointer maximum file size = 256 * 256 * 1KB

Due to Triple Indirect Pointer maximum file size = 256 * 256 * 256 * 1KB

so, total file size (approx) = 2^34 Bytes.

Comments

For total file size why you have not used: -

[no of direct disk block pointer + (DBS/DBA) + (DBS/DBA)^2 + (DBS/DBA)^3)*DBS

where DBS = Disk Block size.

DBA = Disk Block address size?

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I do agree it would be more standard to use the formula , for people who are new to this and are wondering how to utilize the formula to calculate the max file size.  

*********Calculation for 32 bit address size , 10 direct block & 1 indirect ***********

no of direct disk block pointer + (DBS/DBA) + (DBS/DBA)^2 + (DBS/DBA)^3)*DBS

DBS = Disk Block size= 4,096 bytes = 4KB

DBA = Disk Block address size = 32 bits = 4 byte

DBS/DBA = 4,096/4 = 1024

10 direct address entries= 10

one primary indirect block= 1024

max file size = (1024+10)*4KB= 4136 KB

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*********Calculation for 24 bit address size, 10 direct block & 1 indirect ***********

(This is an unstandard address size, but some professors use it to test their students )  

Max file size = no of direct disk block pointer + (DBS/DBA) + (DBS/DBA)^2 + (DBS/DBA)^3)*DBS

DBS = Disk Block size= 1024 bytes = 1KB

DBA = Disk Block address size = 24 bits = 3 byte

DBS/DBA = 1024/3 = 341.3333 = 341

10 direct address entries= 10

one primary indirect block= 341

max file size = (341+10)*1KB= 352 KB = 360448 bytes