in Operating System
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2 votes
2 votes

In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is

  1. 100 ns
  2. 68.0 ns
  3. 68.5 ns
  4. 78.5 ns
in Operating System
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1 comment

@MiNiPanda

 
paging is used then why page table is not considered ?
 
is best answer on page hit , only one memory access is used where actually it takes two memory access, one for page table and one for actual page
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3 Answers

8 votes
8 votes
Best answer
Answer is -  c)  68.5 ns

Main memory and secondary memory are hierarchical. But as per convention page fault service time includes primary memory access time. Here we are not given page fault service time but secondary memory access time. So,
Effective Memory Access Time = $0.35 \times 10 + 0.65 \times (100+10) = 3.5 + 65+6.5 = 75.$

This is not among the choices. So, lets try excluding primary memory access time for secondary memory access:

Effective Memory Access Time = $0.35 \times 10 + 0.65 \times 100 = 3.5 + 65 = 68.5.$
edited by

2 Comments

Should we use hit and try method in GATE!!!! πŸ˜‚
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 This is not blind hit and trial. First hierarchical access is assumed. Then, simultaneous access is assumed.

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2 votes
2 votes
Answer must be d) as not only the time to do address translation of page table but also the time to access a page table from  memory is required

4 Comments

Yes, you should know both ways and take from options. Had no option been given, 75 is the best pick here. But if "page fault service time" was given instead of "secondary memory access time", 68.5 is the answer.
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just becoz option is not available we can't neglect the main memory access time e.g while accessing secondary memory

i don't think in GATE such ambiguity will appear..if yes..then can we can select 68.5 as the answer???
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It has happened in GATE also - not sure if they gave Mark to All though. For any ambiguity type questions, if choice is given you are supposed to try all options and take the matching one.
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0 votes
0 votes
Answer must be D

1 comment

Time required to access a page in memory = Look in page table + look in memory

Had it been given that time to access main memory =10ns then we would have done 0.35(10+10) and for page miss part we have to try out all options. First by treating it like .65(100+10) then if options don’t match then considering 100ns as page fault service time we take .65(100)

So answer should be either 75 or 68.5 as described in best answer.

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Answer:

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