A page-replacement algorithm should minimize the number of page faults. We can a

Question

A page-replacement algorithm should minimize the number of page faults. We can achieve this minimization by distributing heavily used pages evenly over all of memory, rather than having them compete for a small number of page frames. We can associate with each page frame a counter of the number of pages associated with that frame. Then, to replace a page, we can search for the page frame with the smallest counter. a. Define a page-replacement algorithm using this basic idea. Specifically address these problems i. What the initial value of the counters is ii. When counters are increased iii. When counters are decreased iv. How the page to be replaced is selected b. How many page faults occur for your algorithm for the following reference string, for four page frames? 1, 2, 3, 4, 5, 3, 4, 1, 6, 7, 8, 7, 8, 9, 7, 8, 9, 5, 4, 5, 4, 2 c. What is the minimum number of page faults for an optimal page replacement strategy for the reference string in part b with four page frames?

Explanation / Answer

Solution:

Note: a is solved completely as per Chegg guidelines, please repost others.

Before proceeding let me tell you I am talking about Least Recently Used algorithm(LRU) which is a page replacement algorithm for replacing the page in the cache.

i)

Here, two counters are used.

basically, they are header and in the beginning, they both are pointing to the same address, which is at frame 0 of the cache memory.

ii)

The data at which LRU is pointing will be replaced with the new data in that block and MRU will start to point at that location.

iii)

If the block which is getting accessed is already in the set then MRU will start pointing at that address or data.

iv)

The page which is least recently used in the memory in the past gets replaced irrespective of its use in the future.

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

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