23 votes 23 votes A computer system has a $4 \ K$ word cache organized in 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: $15, 40$ $6, 4$ $7, 2$ $4, 6$ CO and Architecture gate1995 co-and-architecture cache-memory normal + – Kathleen asked Oct 8, 2014 • recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 10.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Sanandan commented Sep 15, 2020 reply Follow Share option D) 4,6 is the answer. 0 votes 0 votes Vichitra_kanya commented Jan 15, 2021 reply Follow Share NO. of bits required are- block size= 64 = $2^{6}$ (6 bits) cache size = 4k = $2^{12}$ number of lines= $\frac{cache size}{block size}$ = $\frac{4K}{64}$ = 64 = $2^{6}$ number of set = $\frac{number of lines}{block per set}$ = $\frac{64}{4}$ = 16 = $2^{4}$ (4 bits) 4 6 Tag SET NO BLOCK OFFSET Ans: D. 4,6 4 votes 4 votes Please log in or register to add a comment.
Best answer 34 votes 34 votes Number of sets $=\dfrac{4K}{(64\times 4)}=16$ So, we need $4$-bits to identify a set $\Rightarrow$ SET $= 4$ bits. $64$ words per block mean WORD is $6$-bits. So, the answer is an option (D). Arjun answered Oct 29, 2014 • edited Apr 25, 2021 by Lakshman Bhaiya Arjun comment Share Follow See all 4 Comments See all 4 4 Comments reply set2018 commented Aug 22, 2017 reply Follow Share determine the physical address size ? 0 votes 0 votes joshi_nitish commented Aug 22, 2017 reply Follow Share can not be determined with given information. 9 votes 9 votes praveenb commented Dec 2, 2018 reply Follow Share hi, can i find out size of main memory with the available information? 0 votes 0 votes soram commented Sep 10, 2020 reply Follow Share No, the given data is not enough. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes ……………………………………………………. Mohitdas answered Nov 11, 2021 Mohitdas comment Share Follow See all 0 reply Please log in or register to add a comment.