speedup

Question

suppose that execution time for a program is proportional to instruction fetch time ,assume that fetching an instruction from the cache takes 1 time unit ,but fetching it from main memory takes 10 time units.Also,assume that a requested instruction is found in the cache with probability 0.96.Finally,assume that if an instruction is not found in the cache it must first be fetched from main memory into the cache and then etched from the cache to be exexuted.Compute the speedup ratio?

Answer 1

"I think Speedup ratio is asked in terms of the system with or without cache"

Without cache, each instruction fetch requires 10 time units.
With cache on average each instruction fetch requires = 0.96*1 + .04(11) = .96 + .44 = 1.4 time units

Speedup Ratio = 10/1.4 = 7.142 or 71.42% 
System will be 71.42% faster with cache than without cache!