2 votes 2 votes Suppose that a cache is 20 times faster than main memory and cache memory can be used 80% of the time. The speed-up factor that can be achieved by using the cache is _________. CO and Architecture co-and-architecture cache-memory + – prathams asked Dec 10, 2015 • retagged Nov 13, 2017 by Arjun prathams 1.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 7 votes 7 votes let tm be access time of memory so tc=tm/20 time without cache =tm with cache =0.8*tm /20 +0.2*tm =4.8/20 tm speed up= time without cache/time with cache =tm/(4.8tm/20) =20/4.8 =4.16 Pooja Palod answered Dec 10, 2015 • selected Dec 10, 2015 by prathams Pooja Palod comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Time taken by Main Memory 20X " " " Cache Memory X For access the data time taken without cache 20X " " with cache 0.8*X +0.2*20X =0.8X+4X=4.8X Speedup=20X /4.8X =4.16 srestha answered Dec 10, 2015 srestha comment Share Follow See 1 comment See all 1 1 comment reply vjabhishek commented Dec 31, 2017 reply Follow Share thanx fr d explanation 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Speed up = 1 / ( (1-% of the time cache is used)+(% of the time cache is used/speed up using cache)) Soumyashree answered Dec 11, 2015 Soumyashree comment Share Follow See all 0 reply Please log in or register to add a comment.