GATE CSE 2025 | Set 1 | Question: 4

Question

Consider a demand paging memory management system with $32$-bit logical address, $20$-bit physical address, and page size of $2048$ bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

• A. $2^{21}$
• B. $2^{20}$
• C. $2^{22}$
• D. $2^{24}$

Comments

Reprsentation is very good !! well done

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Verify this PTE is equal to No of pages (process)  Always 

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@ankit2024 only for the single level paging

 

Answer 1

Answer- A.

Comments

A,,,https://correctbrain.com/buy/

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GATE 25 Set 1 and Set 2 are incomparable to each other. Many differences are there it seems.

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Quite easy set 2 as well. The multi level page table q had soe anomaly because outer PT had more bits than inner but procedurally it was same. And the hit miss q was easy as hell barely 2 mins question

Answer 2

 

Answer 3

The logical address size is given as 32 bits, which means the total logical address space is $2^{32}$ bytes since the memory is byte-addressable.

The page size is 2048 bytes, which can be written in powers of 2 as:

$$\text{Page Size} = 2048 \text{ bytes} = 2^{11} \text{ bytes}$$

The number of bits required for the page offset is 11 bits.

The number of bits required for the page number can be found by subtracting the page offset bits from the total logical address bits:

$$\text{Page Number bits} = 32 - 11 = 21 \text{ bits}$$

Alternatively, you can directly find the total number of pages by dividing the total logical address space by the page size:

$$\text{Total Pages} = \frac{2^{32}}{2^{11}} = 2^{21}$$

Since each page needs exactly one dedicated entry in the page table, the maximum number of entries in the page table is equal to the total number of pages, which is $2^{21}$.

Correct Option: A