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76 votes
76 votes

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?

  1. Least-recently-used
  2. First-in-first-out
  3. Last-in-first-out
  4. Most-recently-used
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Best answer
82 votes
82 votes

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) pages frames are replaced and then only most recently used ones are replaced.

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35 votes
35 votes

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.

17 votes
17 votes

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 20th 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


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.)

17 votes
17 votes

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.

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