In practice, a computer spends a fraction of the time performing internal operations as well as memory accesses. Consequently, the effective speedup ratio is reduced, because cache memory has no effect on internal operations. The average time required to execute an instruction can be written as.

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tave Finternal teye+Fmemory[hte + (1 -h)(te + ta)] teye where Finternal F memory fraction of time spent doing internal operations (0 to 1) = fraction of time spent doing memory accesses operations (0 to 1) Teye clock cycle time t = cache access time expressed in clock cycles ta delay penalty paid when not accessing cache memory expressed in clock cycles For the following systems, calculate the average cycle time a. Finternal = 20%, teye = 20 ns, te = 1, ta = 3, h = 0.95 b. Finternal = 50%, teye 20 ns, t= 1, ta = 3, h = 0.9 =