MADE EASY Basic Full Length 3

Question

Consider a page size of 1 KB and that each page table entry takes 4 B. If every page table fits into a single page and 3 level paging are used then size of virtual memory is ________(in bits).

Comments

@Ashwin,

i have written same as yours :)

---

Thank you guys for your comments.

I was just confirming the ans  as no of bits required to represent virtual memory or size of logical address.

and as usual mistakes in test series given ans is 34 hence this questions must be wrong.

---

@nitish

Ya, I also thought to calculate it in bits, you are right they should ask size of virtual address in bits then it would be 34 bits.

Answer 1

Here Page size =1 KB =1024B =2^10 B =2^13 bits

Page table entry size  = 4 B = 32bits =2^5 bits

No of page table entries in outer page table that fits in one page = 1024 B / 4B = 2^8 entries

So this entries will point to one page in 2nd page table

Total no of entries in 2nd page table = 2^8 * 2^ 8 = 2^16

Total no of entries in 3rd page table = 2^8 * 2^ 8 *2^8 = 2^24

these entries will point to no of pages in virtual memory

there fore virtual memory size = 2^24 pages * 1024 B (page size)  = 2^34 B = 2^37 bits =137438953472 bits