Effective memory access time

Question

A demand paging uses a TLB and single level page table stored in main memory. The memory access time is 5 micro sec and the page fault service time is 25 ms. If 70% of access are in TLB and of the remaining, 20% is not present in main memory. The effective memory access time in ms is ..........................

Comments

Source of the question?

Answer 1

Answer : 1.505 ms

EMAT = TLB hit * (Mem access time) + TLB Miss [ 0.2 * ( Page fault Service time) + 0.8 *(Page table access + Mem access time) ]

Substitute values..

TLB Hit --> 70% so TLB miss --> 30%

Page Table access = Mem access time = 5 micro sec

Page fault service time = 25 ms

$EMAT = 0.7 \times 5 + 0.3 \times [0.2 \times 25000 + 0.8 \times (5+5)] \\=3.5 + 0.3 \times [5008] = 3.5+ 1502.4\\=1505.9 \mu s$

Answer 2

Page Fault doesnot occur=80%

TLB hit =70%

TLB access time=0

MM access time=5 micro sec.

Effective access time=.8(.7X5+.3X(5+5))=5.2microsec.

Page Fault occur=20%

It will include 25000 microsec.

So, Effective access time= 0.2(.7X5+.3X(5+5)+25000)=5001.3micro sec.

So, total Effective access time =5001.3+5.2=5006.5 micro sec.

Comments

@Arjun Sir , yes here so huge page fault service time

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@srestha, here question says "of the remaining, 20% is not present in main memory".. here of the remaining is 30% i.e TLB miss ..so 20% of that 30% is not in memory.. so how you assumed overall 20% page faults ??

I have just begin memory questions solving.. so kindly tell where i am wrong ?

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@gate yes, you are correct.

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thanks sir