TestBook TestSeries

Question

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

Comments

https://gateoverflow.in/77917/testbook-os-test

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@pavan it is asking effective acccess time

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address translation time

= TLB HR * (TLb access time) + TLb MR * ( TLb access time + page table access)

= 0.99(20) + 0.01(20 + 100)

= 21ns

 

Actual data access time

= Page HR * (MM access time)

+ page MR * ( write back the dirty page + time to swap page into MM + MM access time  )

= 0.95 * (100) + 0.05 * ( 5000000 + 5000000 + 100 )

= 500100 ns

 

Total = 21 + 500100 = 500121 ns

Answer 1

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 C is the right answer.

Comments

in formula you said 2*page swap time but in solution didn’t multiply by 2 ?

why?