3.1k views

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                                        –

1. 25
2. 6169
3. 2073
4. 4121
| 3.1k views

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)

00010 00000011001 (16 bit)=4121

by (171 points)
selected

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

by Boss (10.9k points)
0
correct....v good...asu
0
0
PARTICULARLY TAKING PAGE NO 2 YOU CAN TAKE ANY NUMBER
0
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

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