paging os

Question

In a two level paging environment a page table is divided into 2^12 pages each of size 4 KW. The memory is word addressable, The physical address space is 128 MW. Memory is divided into 2^14 frames. Length of the logical address is(in bits) __________

Comments

37 bits?

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yes explain once!

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@focus _GATE , I am having the same doubt as you. 

I cant understand the framing of the statement 

a page table is divided into 2^12 pages each of size 4 KW

Can you explain with diagram of levels of paging so that it is easy to understand which level is for 2^12 and which level is for 4KW ?

Answer 1

Check now...

Comments

anirudh :

Means  Page table 2: 2^nd page table contain 2¹² pages in page table. which take 12 bits .

            Page table 1: Number of bits required to represent page size of page table or  word number of page of page table . which is 4KW = 2¹² Words. = 12 bits

suppose if we do three time paging than the first time paging we done we will get  3rd  level  page table, second time paging we do we get 2^nd level page table and third time paging done we get 1^st level page table .!

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yes......

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@Anusha Motamarri , I am having the same problem understanding the framing of the statement

Could you please tell now, If you have got the right perspective?