ISRO CSE 2014 | Question: 35

Question

Using the page table shown below, translate the physical address 25 to virtual address. The address length is 16 bits and page size is 2048 words while the size of the physical memory is four frames.

 

Page      Present (1-ln, 0-0ut)            Frame

0                   1                                        3

1                   1                                        2

2                   1                                        0

3                   0                                        –

 

• A. 25
• B. 6169
• C. 2073
• D. 4121

Comments

It's a beautiful question. Need very fundamental knowledge of how virtual to physical as well as physical to virtual mapping is done.

Answer 1

Virtual Address is given=16 bit

Physical Address= 4(physical frame) *2048=13 bit

No of pages= 2^16/2048= 2^5

Physical Address 25 =  0000000011001 (binary in 13 bit)

in which 2 bit are represent frame no and 11 bit are word i.e 00 00000011001

in given table 00 frame is mapped in page no 2

page no 2 is represented by 00010 (b/c total pages is 32)

Now convert 25 physical address(13 bit) to virtual address(16 bit) is

00010 00000011001 (16 bit)=4121

Answer 2

PA BITS=13 AS(4 FRAMES =>2BIT,,,,,2048 WORDS=>11BIT)

REPRESENT PHYSICAL ADDRESS 25 IN BINARY HAVING 13 BITHAT IS

00 00000011001 

FIRST 2 BITS ARE FRAME NUMBER=FRAME # 0.

NOW IT PAGE TABLE FRAME #  0 CORRESPONDS TO PAGE  # 2...

AND 2 IN BINARY =00010(TAKEN 5 BIT COZ LA IS 16 BIT:IN WHICH 11 BIT IS FOR WORD SO 5 BIT FOR PAGE)

THEREFORE THE BINARY REPRESENTATION :

00010  00000011001 =4096+25=4121

Comments

correct....v good...asu

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THEN WHATS IS PAS(PHYSICAL ADDRESS SPACE)   AND  LAS(LOGICAL ADDRESS)

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PARTICULARLY TAKING PAGE NO 2 YOU CAN TAKE ANY NUMBER

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y u calculated only for frame no. "zero"which corresponded for page2?

i mean u could hv seen frame no. "two"or "three" which corresponds to page no."1"or "0"respectively

Answer 3

Physical address: 

Frame No.(a)

Frame Offset (b)

Virtual Address:

Page No. (c)

Page Offset (b)

Page offset (b) = 11

 

Physical Address = 25 = 0000 0000 0001 1001

 

Virtual Address: ......000 0001 1001 //(last 11 bits same, as Frame offset = Page offset)

 

c = 2, because Page Number 2 corresponds to Frame Number 0. We know frame Number is 0 because in the Physical Address the "a" portion is 0.

 

So, Virtual address: 0010 000 0001 1001 //(First "c" then "b" = Virtual Address)

 

= 4121.

 

Option D

Answer 4

physical address=frame number *frame size+offset
25=0*2048+25
o frame is for page 3
so
virtual address space is
2*2048+25=4121