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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

asked in Operating System by Veteran (14.6k points) | 332 views
what if page fault occured in case of TLB hit also. Should we not consider that in this?

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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.

 

answered by Active (2.1k points)

@Viral Kapoor 

how 13/15 comes in your equation? 

can u explain how you got those accesses time and what all are your considerations?


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