in Operating System
416 views
0 votes
0 votes

original question https://gateoverflow.in/3600/gate2006-it-56

Please see the best answer of Arjun Sir.

My doubt is for paging method - Given is 2 level paging.

For P1 we need 195 pages and which can be accommodated in one PT and same for P2 and P3 and P4 have 364 pages/enteries and thus it should need 2 PT.

But in question every process needs one more PT, why?

as P1,P2,P3 needs - 2 PT and P4 needs 3PT

in Operating System
by
416 views

4 Comments

See

maximum size of a segment is 256 KB

and each page size is 1KB

So, it can accommodate upto 256 pages

Now, P1 has 195 pages

So, it should accommodate in 1st level

but as it requires $195\times 4$ bits

So, we need to go for 2nd level Page Table

right?
1
1

chk this line

overhead of each process would be P + page_count * 4 bytes for paging

https://cs.stackexchange.com/questions/6867/processes-and-segmentation 

1
1
Thank you, Ma'am!
0
0

Please log in or register to answer this question.

Related questions