Practice Type | COA | Memory Mapping |

Question

Tag field size is ‘x’ bits in a 2 way block set associative mapped cache memory. The tag field size (in bits) when the system is upgraded to 8 way block set associative mapped cache (while keeping the memory capacity and block size unchanged) is

Options:

• X + 8

• X + 9

• X + 3

• X + 2

Answer 1

 

 

Let say cache block size is = 2^b byte --> bit  required = b bits
Number of block in the cache = 2^m --> bits required = m bits 
Associativity = 2^1 
Hence  in set no  field bits requered = log (2^m/2^1) = m-1

TAG (x bits)

SET NO(m-1)

BLOCK OFFSET(b bits)

Now current associativity = 8 =2^3

Hence  in set no  field bits requered = log (2^m/2^3) = m-3
 

TAG (ybits)

SET NO (m-3)

BLOCK OFFSET(b)

 

 

Now we can equate , x + (m - 1) = y+ (m-3)

                                     y = x + m - 1 - m + 3

                                    y = x + 2  ANS --> (D)