GATE CSE 2014 | Set 2 | Question: 33

Question

A computer has twenty physical page frames which contain pages numbered $101$ through $120$. Now a program accesses the pages numbered $\text{1, 2, ..., 100}$ in that order, and repeats the access sequence THRICE. Which one of the following page replacement policies experiences the same number of page faults as the optimal page replacement policy for this program?

• A. Least-recently-used
• B. First-in-first-out
• C. Last-in-first-out
• D. Most-recently-used

Comments

What the hell should i refer that he took 1 to 20 instead of 100 to 120.....

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acc to my observations

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Optimal - Removes 20 — used farthest in future

MRU     - Removes 20 — most recently used 

LRU      - Removes 1 — least recently used

FIFO     - Removes 1 — first loaded

LIFO     - Removes 20 — last page loaded into memory.

But LIFO ≠ MRU 
LIFO: Remove the page that was loaded into memory most recently.
MRU: Remove the page that was accessed most recently.

 

Answer 1

It will be (D) i.e Most-recently-used.

To be clear "repeats the access sequence THRICE" means totally the sequence of page numbers are accessed $4$ times though this is not important for the answer here. 

If we go optimal page replacement algorithm it replaces the page which will be least used in near future.  

Now we have frame size $20$ and reference string is

$$1, 2, \dots, 100,1, 2, \dots, 100,1, 2, \dots, 100, 1, 2, \dots, 100$$

First $20$ accesses will cause page faults - the initial pages are no longer used and hence optimal page replacement replaces them first. Now, for page $21$, according  to reference string page 1 will be used again after $100$ and similarly $2$ will be used after $1$ so, on and so the least likely to be used page in future is page $20$. So, for $21^{\text{st}}$ reference page $20$ will be replaced and then for $22^{\text{nd}}$ page reference, page $21$ will be replaced and so on which is MOST RECENTLY USED page replacement policy.  

PS: Even for Most Recently Used page replacement at first all empty (invalid) page frames are replaced and then only most recently used ones are replaced.

Comments

Everyone just forgot the fact that the 20 pages were previously loaded with pages 101 to 120!

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@JAINchiNMay yes exactlly, in MRU, other than 101 all rest remain there forever, as only 1 frame remain most recently used all the time, so
its not behaving like Optimal poolicy where all 101....120 are replaced first, so how is htat correct ans

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@JAINchiNMay yes exactlly, in MRU, other than 101 all rest remain there forever, as only 1 frame remain most recently used all the time, so
its not behaving like Optimal poolicy where all 101....120 are replaced first, so how is that correct ans

Answer 2

Correct Answer is MRU. Let's see how

we can convert this problem to 3 pages and 5 page accesses which is accessed 3 times (you can access any no. of times above 2 it does not matter) because ratio of page fault will be same.

Note- Whenever there is repeated sequence like 12345 12345 then optimal convert itself into MRU & whenever there is reverse sequence like12345 54321 then optimal convert itself into LRU.

Comments

I tried with 4 frames and 5 page access, and got the same result. I think it will come to same conclusion because here it does not really matter how many frames of page accesses you use but the rotation and the order of the same rotation is something that really matters, that make us jump to this results irrespective of change in number m,n.

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@Nitesh Singh 2  Nicely Explained. 

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Time saver

Answer 3

The optimal page replacement algorithm swaps out the page whose next use will occur farthest in the future. In the given question, the computer has 20 page frames and initially page frames are filled with pages numbered from 101 to 120. Then program accesses the pages numbered 1, 2, …, 100 in that order, and repeats the access sequence THRICE. The first 20 accesses to pages from 1 to 20 would definitely cause page fault. When 21st is accessed, there is another page fault. The page swapped out would be 20 because 20 is going to be accessed farthest in future. When 22nd is accessed, 21st is going to go out as it is going to be the farthest in future. The above optimal page replacement algorithm actually works as most recently used in this case. As a side note, the first 100 would cause 100 page faults, next 100 would cause 81 page faults (1 to 19 would never be removed), the last 100 would also cause 81 page faults.

Comments

Aware1111 After the first round the sequence we will be having is 1,2,3,4.............19,100. 100 will be placed in the last frame as 20 will be replaced by 21, then 21 with 22 and so on till 100. Now when the second sequence starts till 1-19 there will be a hit now 20 will now be replaced with 19 as 19 is the will be not used for the longest time, similarly 20 by 21 till 99. Now when in the second sequence we come across 100 it will be a hit. Hence in the second sequence, we get 20 hits

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Mistake  in qestiion.
101 to 120 already present in process framees
If we apply MRU :
101,102,103....120 ->insert 1 : 1,102,103,104...120. Now most recently used is 1 so insert 2 replaces 1 : 2,102,103...120 This keeps repeating. we dont get a single hit here. SO 300 page faults. whereas optimal gives 260 pagefaults. .
 

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did you got the answer ?? i have same doubt!

 

Answer 4

The optimal algorithm gives 340 page faults(assuming 3 sequences from 1-100): 

We have 20 frames containing 20 pages:  101, 102, 103, 104.....120. Since these pages are never referred again in future, we will replace them all. So, we now have 1, 2, 3, .... 17, 18, 19, 20 in the frames. 20 would not be used for the longest duration, so we remove 20 and load 21. Repeating this till we reach 100.

Now, we have 1, 2, 3,... 17, 18 , 19, 100. In second repetition we get 19 hits (for 1-19). Now 100 is needed in this repetition but 19 in next.

 (1, 2, 3, 4, .... 19, 20,.... 99, 100  | 1, 2, 3,... 19, 20, 100)

So removing 19 and loading 20^th page. At this instance main memory has

1, 2, 3, .... 20, ... 99, 100

20 is the best candidate to remove because it'd be needed later than the rest (like 1 , 2, 3...) in next iteration. So removing 20 and loading 21. Again to accommodate 22, 21 is best candidate to be removed. Repeating this process, we reach 100, which is already present in memory. So total hits become 20. Before the next repetition, main memory has

1, 2, 3, ...18, 99, 100

At the beginning, there would be total 18 hits. Since we need 99 and 100 in this iteration, 18 shall be removed, followed by 19, 20.... till 98. So at the end, again we have 20 hits. In the third iteration too, 17 shall be replaced, 98, 99, 100 already found in main memory, summing up the total hits to 20. So total page faults with optimal algorithm are 100+(100-20)+(100-20)+(100-20)=340

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This clearly is MRU or most recently used where the latest referred page gets removed. For eg, in first iteration, since 19 was referred last, we removed it. But the frames were not empty at the beginning. Only optimal algorithm looks at the future demands and decides the best course, all the other algorithms make decisions based on previous entries. Hence neither MRU nor LIFO would eliminate all the 20 entries 101, 102, .... 120.

In fact MRU and LIFO in such circumstances would yield same results because they would keep replacing the last entry and perform worst. Thus to answer, we must assume the frames as empty, in which case MRU would better LIFO (because LIFO would remove the last entered page such as 100, thus reducing hits by 1.)

Comments

great answer.

FIFO=LRU=400 page faults

MRU=optimal=340 pf

LIFO=343 pf

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If there are 300 pages given(1 to 100, 1 to 100, 1 to 100)..then how page faults occur more than 340 ?

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There are 3 ADDITIONAL repetitions, therefore in total there will 400 pages in the sequence

Answer 5

The page frame access sequence is

1,2,3.......19,20,21.....99,100,
1,2,3.......19,20,21.....99,100,
1,2,3.......19,20,21.....99,100,
1,2,3.......19,20,21.....99,100,

Using OPTIMAL replacement method from first 1....100 , first 20 page fault will take place as there are no common pages ,,from 21 to 100 again page fault will take place (these page fault will be equivalent to using MRU replacement method ). after these replacement the page frames will contain [1.....19,100] no pages .Considering the next sequence,no page fault from 1 to 19 and at the 20 page no ,19 will get replaced from the page frame and in the further sequence till 99 there will be page fault(these will be equivalent to using MRU ) and the resulting pages in frames will be 1...18 ,99,100 . going through the further sequence we will get the total page fault as 260 ......... replacing the pages using optimal method is equivalent to MRU replacement method thus answer (D)

Comments

arjun sir i am getting the same no of page faults for LIFO also. LIFO will also behave as MRU after the frames 1-20 are present in the frames. please clear this huge doubt.

And as you are saying- " Pages 101-120 are not referenced by the current process. So, they are replaced first even with MRU policy." Isnt it the case for LIFO too????

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problem with LIFO is :

initially we had 101 ,102,....120 when we load frame no1 then last in was 120 and 1 replaced in 120 ,then when frame 2 come last used was frame 1 again frame 1 replaced, subsequently in the same page frame(i.e 10th frame) we have to replace all the 300 next pages. so number of miss will be 300. but in optimal we have 260 missed. as explained above by arjun sir.

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Sir, how can MRU policy replace the pages that aren't in it's sequence? MRU derives which pages to remove only from it's own sequence.

Answer 6

Optimal PRA(Page Replacement Algorithm) behaves like MRU PRA  when the sequence is repeated only in case, where the frames are empty.

Here, the frames are already initialized with 101 to 120. So, none of the given options are correct.