Applied Test Series:

Question

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Suppose that you wish to design a virtual memory system with the following characteristics:
- The size of a page table entry is 4 bytes.
- Each page table must fit into a single physical frame.
- The system must be able to support virtual address spaces as large as $2^{38}$ bytes ( 256 GB )

Suppose that you decide to use a multi-level paging scheme with no more than two levels of page tables. The minimum page size that your system must have is $\qquad$ KB

Comments

@ramakrushna Same question discuss here...

https://gateoverflow.in/365347/Applied-test-series

Ans is 16 KB.

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Hii @Yaman Sahu As we know the outer page table contains base address of each pages which are presents in inner level page table if required multiple pages .So if inner level page table has only 1 page which means outer page table has only 1 entry corresponding to that page.Is this understanding is correct ??

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@lalitver10 Yes

Answer 1

Hence, psge size is 16KB

Comments

👍🏻👍🏻👍🏻

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Thanks for the detailed explanation.

Answer 2

Lets say ,page size = 2^x byte.

so no of bits required for offset =x.

now page table entry = 4 B

now ,max no of entry one page can store is 2^x/4 =2^(x-2).

so each inner and outer level page table fit completely in one page frame . so we can write,

x-2

x-2

x

outer level                                          inner level                                      offset,

now it sums up to virtual address space .2^38.

(x-2)+(x-2)+x=38

x=14.

so page size =16 KB

Comments

Its also very nice way to approach. thanks Kabir5454

Answer 3

Hope this helps