Question

Assume a cache with 2048 blocks, a 4-word block size, and a

32-bit address. For each of the following configurations, find the total number of bits for each cache

block and the total numbers of bits for the entire cache.

a. Direct-mapped

b. Two-way set associative

c. Four-way set associative

d. Fully-associative

Answer 1

a. In case of direct mapping, different blocks of main memory maps uniquely to one of the all blocks of cache memory with i^th block of main memory maps to i%N block of cache where N is number of blocks in cache.

Since there are 2048 blocks, so total number of bits for each block = log₂ 2048 = 11 bits

And then since there are 4 words in a block, so number of bits to access a word = log₂ 4= 2 bits

So total number of bits for entire cache =, 11+2=13 bits

b. In Two-way set associative memory, 2 consecutive address block (with even and odd block pair like (0,1),(2,3)..) combined as single set. So 2048 blocks will make 2048/2=1024 sets.

And so the set in which i^th block of cache is mapped to i%N set where N is number of sets.

Since there are 1024 sets, so number of bits to identify a set= log₂ 1024 = 10 bits

And then 2 bits required due to 4 words in a block.

So total bits for each cache =10+2= 12 bits

c. Four-way set associative cache will have total 2048/4=512 sets.

So number of bits to identify a set = log₂ 512 = 9 bits

Total bits for each cache = 9+2=11 bits

d. In fully associative memory entire block is treated as a single set so that any available cache can be assigned with any main memory block. So 0 bit is needed to identify any set and 2 bit is needed to identify the word out of 4 words in a block.

So total bits for cache =0+2= 2 bits

Please comment for any clarification.