Question

Given a 32-bit address, calculate the following values for a two-way set associative for:

Cache size: 32KB

Block size: 64B

i) The number of bits in the block offset field.

ii) The number of index bits.

iii) The number of sets in the cache.

iv) The number of tag bits.

Answer 1

Given that cache size s 32 Kb, and Block size is 64 Bytes, and two-way set associative.

assuming system is byte addressable.

The set associative address format is

Tag

Set index

Block offset

i) The number of bits in the block offset field:

Block size is 64 bytes, therefore to represent each byte we need 6 bits (2⁶ = 64 bytes). Therefore block offset is 6 bits.

ii) The number of index bits:

Total number of blocks = cache size / Block size = 32 Kb / 64 Bytes = (32 * 1024 bytes) / 64 Bytes = 512 blocks.

And the cache is two way set associative, therefore number of set = 512 blocks / 2 sets = 256 sets.

Therefore to represent these 256 sets we need 8 bits (2⁸ = 256 sets). Therefore set index bits are 8 bits.

iii) The number of sets in the cache:

From part(ii) it is clear that number of sets are 256.

iv) The number of tag bits:

Tag bits = total memory width - (block offset + set index bits ) = 32 - (8 bits + 6 bits) = 32 bits - 14 bits = 18 bits.

Therefore tag is 18 bits.