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GO Classes DPP | OS | INDEXED ALLOCATION

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2 2 votes

A file system uses INDEXED ALLOCATION with a single-level index block. The disk block size is $\mathbf{4}$ KB, and each disk block address requires $\mathbf{4}$ bytes. What is the maximum possible size of a file in this system?

  1. $2 ~\text{MB}$
     
  2. $4 ~\text{MB}$
     
  3. $8 ~\text{MB}$
     
  4. $16 ~\text{MB}$

1 Answer

2 2 votes
  • Disk Block Size $=4 \mathrm{~KB}=2^{12}$ bytes.
     
  • Disk Block Address (Pointer) Size $=4$ bytes $=2^2$ bytes.
     
  • Number of pointers per Index Block $=\frac{\text { Block Size }}{\text { Address Size }}=\frac{4096}{4}=1024$ pointers.
     

In Single-Level Indexed Allocation, one index block contains pointers to data blocks.

  • Total Data Blocks $=1024$.
     
  • Max File Size $=$ Number of pointers × Block Size.
     
  • Max File Size $=1024 \times 4 \mathrm{~KB}=4096 \mathrm{~KB}=4 \mathrm{MB}$.
     

Correct Option: (B)

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