operating system

Question

Answer 1

Given Page size (PS) = 16KB

Physical address (PA) = 32 bit so physical address space (PAS) = 2³² Bytes

Page table size (PTS) = 22M

No. frames= $\frac{PAS}{PS}$

No.of frames= 2³² /2¹⁴​​​​​ = 2¹⁸ frames, hence no of bits used for frames i.e f= 18 bits.

Entry size (e) = f + 1(valid bit) + 1(modified bit) + 2(modified bit)= 22 bit = 3B(approx)

Page table size PTS = 22MB = 2^{​​​​​​}2 * 2²⁰ bytes

PTS = 22 * 2²⁰ = No. of pages * entry size

No. of pages =(22 * 2²⁰)/3 = 2²³ pages (approx) 

Offset = log_{​​​​2}(PS) = 14 bits 

Logical address = 23+ 14 = 37 bits

Hence 37 bits is the correct answer

Comments

Hi,

The page size is given to be 16 KB and that comes out to be 2^14 right ?

While calculating the no of frames, the frame bits will come out to be 2^32 / 2^14 = 2^18 i.e. 18 bits.

So in the final calculation the page bits in virtual address will be 20 and offset bits were 14, adding up to 34 bits.

Hence the answer should be 34.

Correct me if I am wrong.

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Answer will be 37 only

Yes no of bits for frame is 18 bit, now we calculate the entry size as 18+1+1+2=22B=3B(approx)

No. of bits for pages comes out to be 23 bits

VA=23+14=37 bit

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Thanks for correction.

Looks fine now

Answer 2

Given Page size = 16KB

Physical address size = 32 bits (2^32 Bytes)

Page table size PTS => 2​​​​​​2 * 2^20 bytes(  22MB)

No.of frames= 2^32 /2^14​​​​​ = 2^18 frames, hence no of bits used for frames i.e f= 18 bits.

Entry size (e) = f + 1(valid bit) + 1(modified bit) + 2(modified bit)= 22 bit

let the virtual address space = x + 14 ( 2^x is number of page entries and 2^14 is page size )

size of page table = 2^x * 22 = 22MB

from here x = 20 

so virtual address space = 20 + 14 (34) ans