Question

A 4-way set associative cache has 64 blocks of 16 words. How many bits are there in the ""tag"" field of the 15-bit address for this architecture

Answer 1

3 bits

Step-by-step explanation:

Given a 4- way set associative cache that has 64 blocks of 16 words.

Therefore, the number of sets cache has:

`(64)/(4) = 16`

Now,

Cache data size is 16kB

The number of cache blocks can be calculated as:

`16kB/16 = 1024 bytes/16 byte* 16 = 256 cache blocks`

Now,

Total sets =
`(cache blocks)/(associative sets)`

Total sets =
`(256)/(4) = 64`

Now,

`2^(n) = 64`

n = 6

For 15 bit address for the architecture, the bits in tag field is given by:

15 - (6 + 6) = 3 bits

Thus the tag field will have 3 bits