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TLB lookup time (th) = 20 ns

TLB hit ratio (ht) = 99%

Memory access time (ts) = 100 ns

Page fault rate (hp) =0.05%

Swap page time (in or out) (tp) = 5000,000 ns

What is the effective access time (EAT) if we assume that all pages currently in main memory are dirty?

  • 5118.91 ns

  • 51.21 ns

  • 5120.9 ns

  • None of the above

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1 Answer

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EAT = TLB Hit  Rate* [TLB access time + MM access time] + TLB Miss Rate * [ No Page Fault Rate * (TLB access time + page table access + MM access time) + Page Fault Rate * (2 * Swap Page Time)]

=.99 * (20 +100) + .01 * [.95 * (20 + 100 + 100) + .05 * (5000000)]

=5120.89 ns

Hence Option is the right answer.