GATE CSE 2023 | Question: 48

Question

Consider a computer system with $57$-bit virtual addressing using multi-level tree-structured page tables with $\mathrm{L}$ levels for virtual to physical address translation. The page size is $4 \mathrm{~KB}(1 \mathrm{~KB}=1024 \mathrm{~B})$ and a page table entry at any of the levels occupies $8$ bytes.

The value of $\mathrm{L}$ is ______________.

Comments

I don’t think this question is accurate. I remember that they asked how many block accesses are required to get desired data with this paging system, which is 5 block accesses to convert virtual address to physical address and 1 block access to retrieve the data from desired location. 5 + 1 = 6.

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@jomboy

The question was asking for $L$, which was defined as the number of levels in the Multi-level paging.

Detailed Solution is here:

Multi-Level Paging, Number of Levels

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The Level of Paging is 5

Answer 1

Given,

Virtual Address Space $=57$ bits

Page size  $=4$ KB =$12$ bits

Page number= $(57-12)=45$ bits

Page table entry= $8$ bytes.

Each page can contain =$\frac{4 KB}{8B}=\frac{2^{12}}{2^{3}}=2^9$ page table entries. So, we need $9$ bits to index the page table .

So, Number of levels = $\lceil \frac{45}{9} \rceil=5$ .

So ,Value of L is $5$ .

 

 

Answer 2

L value is 5 .