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Consider a system with 2-levels of paging and a TLB with hit rate of 95% and TLB access time of 1ns. Find the effective memory access time if there’s a data cache whose hit rate is 85% and cache access time is 1ns, and main memory access time is 100ns.

100ns.

  1.   27ns
  2.  ​​​​​​​ 25ns
  3.  ​​​​​​​ 30ns
  4.  ​​​​​​​ 20ns
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method "1"

EMAT=TLB HIT( Cache Hit( TLB A.T + Cache A.T) + Cache Miss( TLB A.T + Cache A.T + M/M A.T)
            +TLB Miss( Cache Hit( TLB A.T + 2*M/M A.T + Cache A.T) + Cache Miss(TLB A.T + 2*M/M A.T + Cache A.T + M/MAT 

 =.95[.85*(1+1)+.15(1+1+100)] + .05[.85(1+200+1) +.15(1+200+1+100)

= 26.99 =27

Method "2"

EMAT=.95[1+.85(1)+.15(1+100)]  + .05[1+200+.85(1)+.15(1+100) ]

=27

similar example :-

https://gateoverflow.in/100831/operating-system-cache?show=100831#q100831

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