Testbook Test Series: Operating System - Virtual Memory

Question

Comments

page table is divide into pages?

I suppose it should be "each page table has $2^{12}$ entries"

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yes, without the no. of entry in a page we cant say anything .

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I think, they mean that 1st level page table is divided into 2^12 pages considering given Page size

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yes..page table is divided into 2^12 pages wat..////.is wrong..paging is applied on page table...

Answer 1

The logical adress =p1+p2+d.

Here p1=12bits p2=12bits and d=12 bits. Therefore logical adress size=36 bits

Comments

Answer is given as 37

Answer 2

logical address

no. of pages P

page size

now no. of pages are P of logical address is further used for 1st level table...as

P

P1( no. of pages of page table)12bit

P2(page size of page tableIi.e 4KW)12 bit

  no . of frames = p.a.s/ frame size
frame size =  p.a.s/ no. of frames 
frame size = 128MW/ 2¹⁴ = 2¹³ W
 now frame size same as page size...
page size = 2¹³ W

so P = P1 +P2

AND  page size of logical address is same as frame size

logical addess

P1+P2= 12+12= 24 BIT

PAGE SIZE = FRAME SIZE==>13 BIT

LOGICAL ADDRESS = 37 BIT

Comments

phulwani...why paging..evolved just read the book... else  we would have continued with contiguous memory allocation..

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@Tauhin Gangwar

so in above explaination by @Tauhin...we are assuming page size of first level is equal to page table size of second table that is why 12+12 ??

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Even if we consider this logic as true then inner level pages will require 2B for addressing each frame bcoz 2^14 frames are there. And size of page table is 2^12 then 2^12/2 gives 2^11 entries in inner page table.

Then answer should be 12+11+13 which gives 36 bit LA

Answer 3

Please see if this approach is right or wrong.

Let Logical Address Space be x words.

Now, Page Size =  Frame Size = $\frac{Physical Address Space}{No. of Frames}$ = $\frac{2^{27}}{2^{14}}$ = 2¹³ words.

Number of Pages = $\frac{x}{2^{13}}$

Page Table Entry = Number of bits to address each frame = 14bits = 1.75B

Number of pages in inner Page Table = $\frac{x * 1.75}{2^{13} * 2^{12}}$ (Each page of a Page Table is 4KW)

Now, $\frac{x * 1.75}{2^{13} * 2^{12}}$ = 2¹² (Page Table is divided into 2¹² Pages)

Solving, we get, x = 1.14 * 2³⁶ $\approx$ 2¹ * ³⁶ = 2³⁷ = 37bits.

Please check this!!

Answer 4

PAS = $2^{27}$W

with $2^{14}$ Frames

$\Rightarrow $ size of frame$ = $ size of Page $ =2^{13}$W

LAS =$13+12+12=37$ bits

Comments

How did you take 12+12 plz explain?

Answer 5

Logical address- p1+p2+d

We have d =13 bits and p2=12bits p1=12bits
Frame size=page size= PAS/no of frames = 128Mw/2^14=2^13
 

So LA= 37bits.

P1 is no of pages in first page table.

P2 is offset of each page of first page table.

d= offset value of page