Question

How many total bits are required for a 1-way set associative cache with 16 KB of data and 8-word blocks, assuming a 32-bit address?

Answer 1

Given that 1-way set associative cache with 16 Kb of data and 8-word blocks, and 32 but address.

assuming each word is 1 byte, that implies 8-words means 8 bytes of data, therefore each block size is 8 bytes.

To refer 8 bytes of information we need 3 bits (i.e, 2³ = 8).

Therefore word offset = 3 bits.

cache size is 16 Kilo bytes,

The number of blocks in the cache = cache size / block size = 16 Kb / 8 bytes = (16*1024 bytes) / 8 bytes = 2*1024 blocks = 2048 blocks, and the cache is 1-way set associative which means we have to refer each block as an independent. Which means we have 2048 sets.

To refer these 2048 sets we need 11 bits (i.e, 2¹¹ = 2048)

Therefore set index = 11 bits.

Therefore total number of bits required to refer the cache is = set index + word offset = 11 bits + 3 bits = 14 bits.

Note:

tag bits = total memory width - (set index + word offset) = 32 - (11+3) = 32 -14 = 18 bits.