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9 votes
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Suppose:
TLB lookup time = 20 ns
TLB hit ratio = 80%
memory access time = 75 ns
swap page time = 500,000 ns
75% of pages are dirty
OS uses a 3 level page table

What is the effective access time (EAT) if we assume the page fault rate is 15% ?
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Best answer
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13 votes

EAT=No Page Fault(TLB hit+TLB miss)+ Page Fault (TLB hit+TLB miss+ Page swap time)

=0.85(0.8(20+75)+0.2(20+3$\times$75+75))+0.15(0.8(20+75)+0.2(20+3$\times$75+75)+500000)

=119+75021=75140 ns

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4 votes
4 votes

we can also solve like this -

Answer:

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