A computer has a memory access time of 38 ns and does not use cache. Cache with an access time of 10 ns is added. The computer then runs 90% faster. Estimate the hit ratio.
What will be an ideal response?
Speedup = (Told ? Tnew)/Tnew,
Tnew = Told/(1 + speedup) = 40 ns/(1 + 90%) = 38/1.9 = 20 ns
Tave = hTcache + (1 ? h)Tmem or h = (Tmem ? Tave)/( Tmem ? Tcache) = (38 ?20)/(38 ? 10) = 18/28 = 0.64
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