3 votes 3 votes A computer system has a $8\; K$ word cache organized in block-set-associative manner with $8$ blocks per set and $32$ words per block. The number of bits in the SET and WORD fields of the main memory address format is: $6, 4$ $4, 5$ $5, 5$ $8, 5$ CO and Architecture go2025-coa-1 easy cache-memory + – gatecse asked Aug 3, 2020 gatecse 167 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 5 votes 5 votes Word Offset = $\log_2{32}=5.$ Number of sets $ = \dfrac{2^{13}}{2^3\times 2^5} = 32.$ So, set offset $ = \log_2 32 = 5.$ gatecse answered Aug 3, 2020 • selected Jul 31, 2021 by Arjun gatecse comment Share Follow See all 0 reply Please log in or register to add a comment.
