Question

A virtual memory has a page size of 32 words. There are 16 pages and 8 blocks. The associative memory page table contains the following entries
Page Block
a) For the following CPU addresses find the virtual addresses: B5, 2A (hex)

b) For the following virtual addresses find the CPU addresses: 1A9, 016, 149 (hex)

Page Block
0 3
4 5
7 2
9 4
12 0
13 1
14 6
15 7

Answer 1

Step 1 ;-

Data given is :-

Page Size = 32Words

Total Pages is 16

And Total blocks are 8

So, that means there is 2 pages in 1 block and it is 2 way associative

Offset bit will be = log 32 =5bits

Block address bits = log 8 = 3 bits

Virtual address Size = 16 * 32 = 512 words

Virtual address bits = log 512 = 9bits

and Outer page table address bits will be = 9 - ( 5 + 3 ) = 1 bit

Step 2 :-(a)

(1) CPU address given in Hexadecimal is = B5

In binary will be = 10110101

Where

Block address( 3 bits )	Offset( 5 bits )
101	10101

In 5th block ( 101 ) there is 4th page present

So, Virtual address in binary 9bits is = 010010101 = 4th page in binary(4 bits ) + offset( 5 bits)

And In Hexadecimal will be = 095H

(2)

CPU address given in Hexadecimal is = 2A

In binary will be = 00101010

Where

Block address( 3 bits )	Offset( 5 bits )
001	01010

In 1st block ( 001 ) there is 13th page present

So, Virtual address in binary 9bits is = 110101010 = 13th page in binary(4 bits ) + offset( 5 bits)

And In Hexadecimal will be = 1AA H

Step 3 :-(b)

(1) Virtual Address given in hexadecimal is =1A9 H

In Binary form 9 bits of virtual address is = 110101001

Where

Page Address ( 4bits)	Offset(5 bits )
1101	01001

In 13th page (1101) , there is 1st Block present

So,CPU address will be in binary ( 8 bits ) is = 00101001 =1st Block address ( 3bits ) + Offset in( 5bits )

CPU address in Hexadecimal = 29 H

(2)

Virtual Address given in hexadecimal is =016 H

In Binary form 9 bits of virtual address is = 000010110

Where

Page Address ( 4bits)	Offset(5 bits )
0000	10110

In 0th page (0000) , there is 3rd Block present

So,CPU address will be in binary ( 8 bits ) is = 01110110 =1st Block address ( 3bits ) + Offset in( 5bits )

CPU address in Hexadecimal = 76 H

(3)

Virtual Address given in hexadecimal is =149 H

In Binary form 9 bits of virtual address is = 101001001

Where

Page Address ( 4bits)	Offset(5 bits )
1010	01001

10th page (1010) is not present in any block as we can see in the table

So,There is a Page Fault