retagged by
5,785 views
0 votes
0 votes
A block-set-associative cache consists of a total of 64 blocks, divided into 4-block sets.
The main memory contains 4096 blocks, each consisting of 32 words. Assuming a 32-bit
byte-addressable address space, how many bits are there in each of the Tag, Set, and word
fields?
retagged by

1 Answer

0 votes
0 votes
Assume 1 word = 1 Byte ( as nothing is mention )

so Block size = 32 word = 32 Byte =  5 bits

so Block offset  = 5 bits

 No of blocks  in cache memory  = 64  and cache is 4 ways set associative

so No of sets = 64block / 4 ways = 16 sets

So Set number bits = 4 bit

Main memory size = No of block * block size =  4096 * 32 Bytes = 128 KB = 17 bits

so Tag bits =   17 –  ( set bits + Block offset bits )= 17- 9  =  8 bits

Related questions

0 votes
0 votes
0 answers
2