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4 Answers

Best answer
15 votes
15 votes

Here Logical address (LA)=32bit

and  Physical address (PA)=32bit

Page size=4KB

Page table size =Number of entries in page table *page table entry size

                        =$\frac{2^{32}}{2^{12}}\times$(page table contain frame number+additional bit(valid/invalid,protection bit etc.))

                       =$2^{20}\times 20\ bit$

                       =4 MB(apx).

So C is correct ans .

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2 votes
2 votes

Page table for 32-bit address space with 4K byte pages has 232 / 212 = 220 entries. If each entry is 32 bits, need 4M bytes of memory to store page table.Correct answer is 4M bytes.Page table for 32-bit address space with 4K byte pages has 232 / 212 = 220 entries.

If each entry is 32 bits, need 4M bytes of memory to store page table.

1 votes
1 votes
the answer is 4M bytes.

let p  bits represent page number and d represent offset.

page size=2^d=4k=4*2^10; 2^d=2^12 therefore d=12;

p+d=32; p=20;

size of page table=max number of pages*size of each entry=2^p*4=2^20*4=4M
–3 votes
–3 votes

Option D, is correct answer.

In a Paging Concept, by default we take single level Paging,

IN A SINGLE LEVEL PAGING

Page Size = Frame Size = Page Table Size

because, page table will also load in main memory into one of the frame.

so, here

Page Size = Page Table Size = 4 K bytes

Answer:

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