i will try to Answer …...
In Set Associative Memory
| Tag Bit |
Number Of Set
|
Page Size |
We Know Cache Size Is
Number Of Set X Number Of Page In Each Set X Size Of Each Page = 256KB
Number Of Set X 16 X Size Of Each Page = 256KB
Number Of Set X Size Of Each Page = 2^18 / 2^4
Number Of Set X Size Of Each Page = 2^14
Number Of Bits ( Number Of Set X Size Of Each Page) = 14
Total Size of Physical Address = 36 bits
Then Tag Size = 36 – 14
= 22 bit
