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Suppose that cache  access time is 7 ns, main memory  access time is 50 ns and disk space  access time is 1200 ns. If the hit rate of cache is 60% and of main memory hierarchy is 70 percent, the average memory access time is _____________ (in ns).

 

Answer is 171 ns.

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Given:
cache  access time is 7 ns, m

ain memory  access time is 50 ns

disk space  access time is 1200 ns.

hit rate of cache is 60%

main memory hierarchy is 70 percent,

Asked :
the average memory access time is 
Sol:

Avg Time:

Hit in Cache * Access(cache)+ MissCache * Hit Main* ( Access Main+Access Cache)+MissCache* MissMain (Accesscache+AccessMain+AccessDisk)
=>0.6*7+0.4*0.7*(50+7)+0.4*0.3(7+50+1200)

=>4.2+0.28*57+0.12*1257
=>4.2+15.96+150.84

=>171

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Taverage = Access time of cache + Miss rate of cache*  Access time of Main memory + miss rate of cache * miss rate of main memory* access time of Disk

= 7 +  0.4 * 50 + 0.4* 0.3* 1200

= 7 + 20 + 144

= 171 ns

:)

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