Question

I want someone to explain how you reach these answers. Thank you!

For the memory system which has 4M words of main memory, a cache of 128K words where the

catch is organized as a 2 way set associative cache with a block size of 64 words, how many bits

are there in the TAG, SET, and WORD fields?

a. Main Memory: 4M words = 2 2 * 2 20

b. Cache: 128K words = 2 7 * 2 10

c. Way: 2 = 2 1

d. Block: 64 words = 2 6

e. WORD: 6

f. SET: 7

g. TAG: 22 – 17 – 1 = 6

Answer 1

Main memory 4 M words = 2^2 * 2^20 [1M = 2^20] so total 22 bits Cache 128 k words = 2^7 * 2^10 [1 k = 2^10] 1 block = 64 words = 2^6 2 way set auociative Total blocks = 128k/64 = 2^17/2^6 = 2^11 No of sets in cache = total blocks/set auociativity = 2^11/2 = 2^10 So set offset = 10 Block has 64 words = 2^6 so word field (offset) = 6 bits TAG: = main memory bits - [set offset + wordoffset] = 22 - 16 = 6 So TAG = 6 SET = 10