Question

A cache memory unit with capacity of N words and block size of B words is to be designed. If it is designed as a direct mapped cache, the length of the TAG field is 10 bits. If the cache unit is now designed as 16-way set-associative cache, the length of the TAG field is _______ bits.

Answer 1

Formula:

Cache size = N words

No. of bits to address cache size = log₂N

Block size = B words

Block offset = log₂B

No. of blocks in the cache = N/B

No. of bits to represent blocks = log₂ (\(\frac{N}{B}\))

No. of sets in cache = (\(\frac{N}{B}\)) / 16

Bits to represent sets = log₂ (\(\frac{N}{B}\)) / 16)

For direct mapper cache,

Tag (10)

Cache line (log2 (\(\frac{N}{B}\)))

Block offset (log2B)

 

For set associative cache,

Tag (Let, X)

Set No. log2 (\(\frac{N}{B}\)) / 16)

Block offset (log2B)

 

Here, direct mapped cache is compared with set associative cache, but in both the cases block offset is same so neglect it.

Calculation:

10 + log₂ (\(\frac{N}{B}\)) = x + log₂ (\(\frac{N}{B}\)) / 16)

10 + log₂ (\(\frac{N}{B}\)) = x + log₂ (\(\frac{N}{B}\)) – log₂ 16

10 = x – 4

X = 14

Tag bits for 16-bit set associative cache are 14.