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GO Classes DPP | OS |PAGING

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2 2 votes

Consider a system with a $32$-bit logical address space and a page size of $4 ~\mathrm{KB}$. The system uses a two-level paging scheme where the first level (Outer Page Table) index is $10 ~\mathrm{bits}$. If each page table entry (PTE) at both levels is $4 ~\mathrm{bytes}$, what is the size of the inner page table $($in $\mathrm{KB})$?

  1. $1 ~\mathrm{KB}$
     
  2. $8 ~\mathrm{KB}$
     
  3. $16 ~\mathrm{KB}$
     
  4. $4 ~\mathrm{KB}$

2 Answers

1 1 vote
Since the logical address is $32$ bits and the page size is $4 \mathrm{~KB}\left(2^{12}\right)$, the page offset is $12$ bits. With $10$ bits for the outer index, the inner index must be $32-12-10=10$ bits, leading to $2^{10}$ entries of $4$ bytes each.

Correct option : D
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