ME-OS

Question

i got 44 as answer. givne answer is 48
they took PTE to be 3 Bytes
correct answer is?

Answer 1

Given logical address  =  48 bits and the addressing is byte addresable type..

So logical address space  = 2⁴⁸ B

Given page size  = 16 kB  =  2¹⁴ B

So no of pages  =   2^48-14   = 2³⁴

So no of entries in page table  =  2³⁴

Also no of frames  =  16 GB / 16 kB  =  2²⁰

So no of frame bits  = log₂ 2²⁰  =  20 bits

So no of bits in 1 page table entry  = 20 + 2(for protection as mentioned in question)

                                                   =  22 bits

So page table size  =  2³⁴ * 22

                             =  22 * 16 Gb 

                             =  22 * 2 GB

                             =  44 GB

Hence size of page table = 44 GB

                             

Comments

"each address defers to byte in memory"  what does it means ??

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It means memory is Byte addressable.

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Here is "contradiction" btw 44GB OR 48GB

https://gateoverflow.in/112333/pte-paging

Answer 2

44 gb ia correct 

Comments

"each address defers to byte in memory"  what does it means ??