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Page tables are stored in memory , which has access time of 100 ns. The TLB holding 8 page table entries, has an access time of 10 ns. Using execution of process , it is found that 85 % of the time, a required page table entry exist in TLB and only 2 % of the total references causes page fault. Page replacement time is 2 ms . Calculate the effective memory access time , assuming page memory access requires 2 memory accesses and TLB requires one memory access.

A) 38120 ns B) 40000 ns C) 40120 ns D) None

2 Answers

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Page table access time=$100 ns$

TLB has 8 PTE and access time of a PTE is $10 ns$, that means TLB access time $10 ns$

Page Table are in TLB i.e. TLB  hit time$=85$%

Page Fault is for $=2$%

TLB miss time $=13$%

For TLB hit =$1$ memory access

For TLB miss=$2$ memory access

So, Effective Memory Access Time$=0.85\left (100+10 \right )+0.13\left ( 100+100 \right )+\frac{2}{100}\times 2000000$

$=40119.50ns\simeq 40120ns.$

Ans $C)$

https://www.inf.ed.ac.uk/teaching/courses/os/slides/10-paging16.pdf

Similar ques from book https://gateoverflow.in/211302/operating-system-by-harris

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D

0.85(10+110)+0.15(13/15(10+100+100)+2/15(10+100+2000000+100)=40125

85% time there will be a TLB hit so it will take (10+100) ns.

15% of the time two things can happen:

  • X% of the time page fault will occur that is it will take (10+100+2000000+100) ns.
  • (1-X)% of the time no page fault so it will take (10+100+100) ns.

Page fault occurs 2% of the total memory accesses so X% of 15 should be equal to 2.

i.e. 15 * X/100 = 2        X=200/15.

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