Question

Which bits of the address would be used in the tag, index and offset in a two-way set associative cache with 1-word blocks and a total capacity of 512 words

Answer 1

Size of block = 1 word = 8 bytes

number of blocks = size of cache / size of word

= 512 word /1 word = 512 blocks

In a two-way set associative cache, 2 blocks will form 1 set. So number of index = 512/2=256=2^8

Size of offset = 3 bits

size of index = 8 bits

Size of tag = 64-11=53 bits