Test by Bikram | Operating Systems | Test 2 | Question: 17

Question

The address sequence generated by tracing a particular program, executing in a pure demand paging system with $100$ records per page, with $1$ free main memory frame is recorded as follows: $560,\  540, \ 430, \ 390, \ 350, \ 450, \ 102, \ 198, \ 754, \ 785$. The page fault rate is _______

Comments

560 -> PF -> 560-659 in memory

540 -> PF -> 540-639 in memory

430 -> PF -> 430-529 in memory

390 -> PF -> 390-489 in memory

350 -> PF -> 350-449 in memory

450 -> PF -> 450-549 in memory

102 -> PF -> 102-201 in memory

198 -> Hit

754 -> PF -> 754-853 in memory

785 -> Hit

Second approach:
Frame: 560, 430, 390, 450, 102, 754
PF: 6

Which one to follow??

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@Ashwani Kumar 2

The records(data) of a process already exist in pages. If we need to access a record, we'll have to fetch that whole page — and not carve out a page of our own taking the record number as the location for starting address.

Using this knowledge, required page sequence will be: 6,6,5,4,4,5,2,2,8,8.

So, Page Faults = 6.

And page fault rate = 6/10.

Answer 1

Take like 0,1,2,3,4,5,6,7,8,9
according to question it contain 100 pages, so page no. 0 contain 0-99, page no 2 have 100-199 and so on.

Now just check require page and  don't count repeated page twice.

Hence Page fault rate = 6 /10 =0.6

Comments

560 - 1 page fault. Records 560-659 in memory

540 - 2 page faults. Records 540-639 in memory

430- 3 page faults. Records 430-529 in memory

390- 4 page faults. Records 390-489 in memory

350 - 5 page faults. Records 350-449 in memory

450 - 6 page faults. Records 450-549 in memory

102 - 7 page faults. Records 102-201 in memory

198 - 7 page faults. Records 102-201 in memory

754 - 8 page faults. Records 754-853 in memory

785 - 8 page faults. Records 754-853 in memory

 8 page faults.---Rate 0.8.

Where went wrong??

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explain in detail plz.

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see this https://gateoverflow.in/2619/gate1995_2-7

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SIr what is wrong in my answer (mentioned above)!!

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yes @swap_it

sorry for late reply , i was thinking actually what went wrong in your answer .

in your answer

350 - 5 page faults. Records 350-449 in memory

450 - 6 page faults. Records 450-549 in memory

thiese 2 are redundant i think.

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@SWAP_IT

THE PROBLEM IS NOT IN YOUR ANSWER ...BUT IN UR AND OUR LOGIC

WE CANNOT SAY THAT OUR ANSWER IS RIGHT OR YOUR (BCZ WE HERE DOES NOT INFORMATIONS THAT EACH PAGE IS FIXED ONE ( 0-99) LIKE OR FROM X - X+99)

SO TAKING APPROACH ACCORDING FIRST UR ANSWER IS WRONG OUR RIGHT ND TAKING SECOND APPROACH ..U R RIGHT WE ARE WRONG...

BUT HERE THEY DID NOT GIVE ABOUT SECOND APPROACH SO WE GENERALLY TAKE FIRST ONE AS DEFAULT.

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sir, 

350 - 5 page faults. Records 350-449 in memory

450 - 6 page faults. Records 450-549 in memory

why are they redundant ? 

the page frame loaded in the memory before the above two is  Records 390-489 , so the next record 350 will create a page fault.

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ok!! got it.

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Sir if we have a page fault of 670 say then we would pick 600-700 or 670-769? Which one to pick and what is the logic behind it?

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@ saxena0612 

It is mention in question " system with 100 records per page " 

so if we have a page fault of 670 say then we would pick 670-769 .

for more clarity see this https://gateoverflow.in/2619/gate1995_2-7

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@Bikram Sir In the question link you mentioned above ,@Arjun Sir picked up the logic that a page consist of serial 100 records.So if there is a page fault of 430 a page consist of records 400-499 has been picked up from memory. Is it intuitive? Or we have to rely on fact that a page has a serial order so when there is a page fault occur of any mid value say X34 we ultimately get the complete record of X00-X99 if there are 100 records per page?

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@saxena

yes ,

we have to rely on fact that a page has a serial order .

So when there is a page fault occur of any mid value say X34 we ultimately get the complete record of X00-X99  ... yes correct assumption.

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how you took total 10 pages..??

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@shubham shukla 5 

see the question please , it gave total 10 pages ..