0 votes 0 votes Consider a direct mapped cache of size 16 KB and block size is 4 words. The word length is 16 bits. Find the number of bits needed for cache indexing if CPU generates 32 bit address CO and Architecture memory-management + – Deepika Bagaria asked Jan 30, 2016 Deepika Bagaria 801 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes block size=4*2=8 bytes no of blocks =16kB/8=2048so for cache indexing 11 bits are required 3 bit for offset (assuming byte addressing; 2 for word addressing) and 32-14=18 for tag (19 if word addressing is used) Pooja Palod answered Jan 30, 2016 • selected Feb 29, 2016 by Arjun Pooja Palod comment Share Follow See all 3 Comments See all 3 3 Comments reply papesh commented Feb 29, 2016 reply Follow Share it should be word offset which needed only 2 bits...please confirm? 0 votes 0 votes Arjun commented Feb 29, 2016 reply Follow Share 2 in case of word addressing. But, in both cases index bits remain 11- we dont need to find offset/tag bits to calculate this. 0 votes 0 votes papesh commented Feb 29, 2016 reply Follow Share here which type of addressing we need to consider. i know if not given anything then we have to use byte addressing. but here word size is given...so byte addressing or word addressing ? i learn from some source that index is (block offset + word offset)... means i'm confused. 0 votes 0 votes Please log in or register to add a comment.