1 votes 1 votes $16$kB cache with line size $64$B uses $4$ – way set associative mapping. Main memory is $8$ MB and byte addressable. The size of extra space needed for storing tag information in bytes is _________ CO and Architecture tbb-coa-2 numerical-answers co-and-architecture cache-memory + – Bikram asked May 27, 2017 • retagged Sep 20, 2020 by ajaysoni1924 Bikram 360 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes Tag field =11bits Set field =6 bits Offset=6 bits Number of blocks = 4*sets = 4 *2?6 =256 Space for tag field = 256*11=2816 bits = 352 bytes Bikram answered May 27, 2017 • selected Nov 3, 2017 by Rishabh Gupta 2 Bikram comment Share Follow See 1 comment See all 1 1 comment reply rahuljai commented Dec 12, 2018 reply Follow Share but sir the memory is byte aligned and in some questions, we first divide tag bits by 8 to get size in Bytes. then we multiply by #lines. but here directly multiplied by #lines and divided by 8 . Please explain. :) 0 votes 0 votes Please log in or register to add a comment.