PTE- paging

Question

consider a paging system with 48bit virtual address space.Each address defers to a byte in memory.suppose the size of page is 16KB and the main memory size is 16GB.The minimum size of page table with each entry need 2 protection bits is _____ (in GB)

now what should be the aswer? i got PTE as 22bits. should i round it to 3bytes and make answer as 48GB or shuld i keep it as it is and write the answer as 44GB?

Comments

"Each address defers to a byte in memory".

So, I will go with 48 GB ..( PTE = 3 byte).

One more discussion link- 

https://www.facebook.com/groups/core.cs/permalink/1411897255509190/?comment_id=1413428322022750&notif_t=group_comment_mention&notif_id=1485571013068291

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@vijay yes i agree with you that it shuld be 48Bytes. wanted to confirm. :) thank you

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Yup! i think that means memory is byte adressable! so we should go for 3 byte

Answer 1

Already resolved in the above comments:

Comments

@Debashish / @Anusha whenever the term "Byte addressable memory" comes should we round off? What if there is no mention of byte addressing. We usually by default assume byte addressable and solve. Should we round off in that case too?

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Same doubt.......here

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@Arjun Sir Could you please confirm this ? Should we byte align or leave as it is?

Answer 2

no of entries in page table :2^34
no of frames are 2^20
so PTE size will be 20 +2= 22 bits k
now further we have page table size as = 2^34 * 22 bits convert it to BYTE we have (2^34 * 22)/8
= 44GB

Comments

how do we store only some part of a byte? 22/8= 2.75 bytes. (2bytes+3/4th of another byte)
are u going to utilize remaining 1/4th part of 3rd byte?