Paging question

Question

A computer system implements a 36 bit virtual address. Page size is 4KB and size of physical memory is 30 bits. The approximate size of page table in the system is ___ MB ?

Is it 36MB or 48MB ?

Comments

If you perform theoretically, then it is possible. But when you see conceptual wise, then it is not possible, as processor fetch a complete word, not a fraction of word.

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In the question Page size indicating that memory is byte addressable so if you want to store one bit only still you have to take one Byte block..

https://en.wikipedia.org/wiki/Byte_addressing

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Why it is not possible conceptual vise?

Answer 1

There is a issue of byte - allignment or word allignment associated the given question which is : 

If  a word (page table entry in this context) of a memory segment which is a page table is not a multiple of 8 in case the memory is byte addressable , then the word size is rounded to next higher byte (if it is byte addressable) for proper access of page table entries . Else what will happen is same memory address will be shared between two page table entries which is not correct. Hence there is a need of byte allignment .

Hence here page table size  =  Number of page table entries *  Page table entry size

                                         =  (Logical address space / Page size)  * Page table entry size

Page table entry size          =  log₂ (Number of frames)

                                         =  log₂ (Physical address space / Page size)

                                         =  log₂ (2³⁰ / 2¹²)

                                         =  18 bits

But it is not a multiple of  8 , hence rounded to next higher multiple of 8 which is  =  24 bits  =  3 B

Hence page table size        =   (2³⁶ / 2¹²) * 3 B

                                        =   2²⁴  *  3 B

           Page table size      =   48 MB

Comments

Hello Habib , shivam , shubhanshu , i know 2.5B is not practically possible but my point was , which answer is acceptable , i have seen at many places they provided answer like 36MB way...

theoretically 36MB is right i consider. in numeric section , which one is answer?

Answer 2

Given,

LA= 36 bits, MM= 30 bits, Page size= 4kB.

Now, number of bytes in a page= Page size/ Byte size = 4kB/1B = 4k(Since,byte addressable).

Thus number of bits required to represent an offset = log(4k) = log(2^2 × 2^10) = 12 bits.

Hence, number of bits for Frame = 30-12=18 bits ~ 3B( 1B=8 bits)

Number of bits for Page = 36-12 =14 bits.

Thus,number of entries in the page table = 2^24 = 2^20 × 2^4 = 16M entries.

Therefore, size of the page table = 16M × 3B = 48MB.

Answer 3

 page table size = page number* ptes(page table entry size)