Question 12404 – TIFR PHD CS & SS 2014
December 25, 2023Question 14291 – Computer-Organization
December 25, 2023Cache
| Question 13 |
| 2 |
Cache size = 2KB
Block size = 64 Bytes = 2^6 bytes. So the OFFSET is 6 bits long.
The TAG field is 22 bits. And the physical address is 32 bits.
Since it is given as a set associative cache, the physical address is divided as (TAG, SET, OFFSET).
From the given information we can find the SET number = 32 – 22 – 6 = 4 bits.
Using the 4 SET bits, the number of sets possible = 2^4 = 16.
Number of blocks in the cache = Cache size / Block size = 2KB / 64 = 2^11/2^6 = 2^5.
Number of sets in the cache = Number of blocks in the cache / Number of blocks in each set
Number of blocks in each set is also called the associativity.
16 = 2^5/associativity
Associativity = 2^5/16 = 2
Cache size = 2KB
Block size = 64 Bytes = 2^6 bytes. So the OFFSET is 6 bits long.
The TAG field is 22 bits. And the physical address is 32 bits.
Since it is given as a set associative cache, the physical address is divided as (TAG, SET, OFFSET).
From the given information we can find the SET number = 32 – 22 – 6 = 4 bits.
Using the 4 SET bits, the number of sets possible = 2^4 = 16.
Number of blocks in the cache = Cache size / Block size = 2KB / 64 = 2^11/2^6 = 2^5.
Number of sets in the cache = Number of blocks in the cache / Number of blocks in each set
Number of blocks in each set is also called the associativity.
16 = 2^5/associativity
Associativity = 2^5/16 = 2