1 votes 1 votes CO and Architecture co-and-architecture cache-memory ace-test-series associative-memory + – Na462 asked Jan 21, 2019 edited Mar 3, 2019 by I_am_winner Na462 687 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Na462 commented Jan 21, 2019 reply Follow Share I am getting 19 bits as the answer anybody verify 0 votes 0 votes Kunal Kadian commented Jan 21, 2019 reply Follow Share Yes it should be 16+3 = 19. Assuming memory is byte addressable 1 votes 1 votes manisha11 commented Aug 6, 2019 reply Follow Share @Kunal Kadian 3 came from?? 0 votes 0 votes Manas Mishra commented Aug 6, 2019 reply Follow Share @manisha11 tag bits will obviously be needed and for block we will take as byte addressable and so 64 bits = 64/8 = 8 bytes = 2^3 , so min value of x = 16+3 = 19 0 votes 0 votes manisha11 commented Aug 6, 2019 reply Follow Share oh yes 64/8 thanks 0 votes 0 votes Himanshu Kumar Gupta commented Sep 8, 2020 reply Follow Share I am also getting 19 as answer Any one please Verifiy…. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes To get minimun value of X , you should consider that it is block addressable where block size should be 64 bits. So minimun number of bits will be only for tag bits. So x=16. Rishabh Agrawal answered Jan 21, 2019 Rishabh Agrawal comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments kman30 commented Jan 21, 2019 reply Follow Share @Rishabh Agrawal does this mean here one word is of 64bits ? 0 votes 0 votes Rishabh Agrawal commented Jan 21, 2019 reply Follow Share Actually somewhere it is Block addressable while somewhere it is word addressable.Just different terminology. 0 votes 0 votes Rishabh Agrawal commented Jan 21, 2019 reply Follow Share @kman30 Yes 0 votes 0 votes Please log in or register to add a comment.