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

Answer:

3 bits

Explanation:

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

Therefore, the number of sets cache has:

$\frac{64}{4} = 16$

Now,

Cache data size is 16kB

The number of cache blocks can be calculated as:

$16kB/16 = 1024 bytes/16 byte\times 16 = 256 cache blocks$

Now,

Total sets = $\frac{cache blocks}{associative sets}$

Total sets = $\frac{256}{4} = 64$

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