Computer Arch.

Question

Given a 32 bit processor with 16 MB main memory, 32 KB 4 way set associative on chip cache and block size of 16 words, The number of tag bits in memory address format are : ?

Here we will take the length of instruction as 32 bits or 24 ? Why ?

Answer 1

32 bit processor means vitual address 32 bits

main memeory or physical address is 2^24 that is 24 bits

32 KB 4 way set associative on chip cache and block size of 16 words

block size of 16 words = 2^4   (here i have a doubt ,in this qs word size is not given,should i take standard word size 4 B..then block size will be 64 B,then 2^6)

cache size = 2^15

no of blocks  = 2^9

no of sets = 2^9 /4 = 2^7.

so tag bits = 24 - 6- 7 =11 (considering word size as 4 B...dont know i am right or wrong)