1 votes 1 votes A computer system has 4 K word cache organized in a block-set associative manner, with 4 blocks per set, 64 words per block. The number of bits in the SET and WORD fields of the main memory address format is 15, 4 6, 4 7, 2 4, 6 Operating System go-os-1 operating-system cache-memory co-and-architecture + – Bikram asked Sep 3, 2016 retagged Nov 13, 2017 by Arjun Bikram 314 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes 64 words per set - so $\lg 64 = 6$ bits for WORD. No. of sets = cache size/(block size * no. of blocks per set) = 4K/(64 * 4) = 16. So, no. of set bits = $\lg 16 = 4.$ Answer is - d)4, 6 go_editor answered Sep 3, 2016 selected Sep 13, 2016 by Arjun go_editor comment Share Follow See all 0 reply Please log in or register to add a comment.