A system has 32 bit Virtual adress and 8KB page size.main memory is 128 MB..the size of page table, when page table ,when pt entry has 1 valid bit 1 modified bit and frame bits

1 month ago

Solution:

By default the address translation of logical (virtual address) into physical address is done by page table.But the question is :

How it works ?? Well for that we need to consider the page table entry which by default contains frame number

A physical memory (main memory) page is also known as frame. Hence number of frames in main memory will decide how many frame bits we require for each page table. In addition to this we can have other bits in each page table entry like valid bit to determine whether the referred page is in valid address space or not, then dirty bits , protection bits etc are also made part of page table entry sometimes..

But unless mentioned they are not included in page table entry. Only frame no is considered..

We know

Page table  size = No of logical pages * Page table entry size(Frame bits + Valid  bits + Modified bits)

So no of pages  =   Virtual address space / Page size

=   232 / 213

=   219

No of frame bits  =  log2(No of frames)

=  log2(227 / 213)

=  14 bits

In addition one valid and one modified bit

So page table entry size  =  16 bits  =  2 B

So page table size   =  No of pages * Page table entry size

=  220 B

=  1 MB

Hence size of page table required = 1 MB

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Oct 20, 2021

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