Page Faults in LFU | Implementation - GeeksforGeeks

02 Jun, 2021
Overview :
In this, it is using the concept of paging for memory management, a page replacement algorithm is needed to decide which page needs to be replaced when the new page comes in. Whenever a new page is referred to and is not present in memory, the page fault occurs and the Operating System replaces one of the existing pages with a newly needed page. LFU is one such page replacement policy in which the least frequently used pages are replaced. If the frequency of pages is the same, then the page that has arrived first is replaced first.
Example :
Given a sequence of pages in an array of pages[] of length N and memory capacity C, find the number of page faults using the Least Frequently Used (LFU) Algorithm.
Example-1 :
Input : N = 7, C = 3
pages = { 1, 2, 3, 4, 2, 1, 5 }
Output :
Page Faults = 6
Page Hits = 1
Explanation :
Capacity is 3, thus, we can store maximum 3 pages at a time.
Page 1 is required, since it is not present,  
it is a page fault : page fault = 1
Page 2 is required, since it is not present,  
it is a page fault : page fault = 1 + 1 = 2
Page 3 is required, since it is not present,  
it is a page fault : page fault = 2 + 1 = 3
Page 4 is required, since it is not present,  
it replaces LFU page 1 : page fault = 3 + 1 = 4
Page 2 is required, since it is present,  
it is not a page fault : page fault = 4 + 0 = 4
Page 1 is required, since it is not present,  
it replaces LFU page 3 : page fault = 4 + 1 = 5
Page 5 is required, since it is not present,  
it replaces LFU page 4 : page fault = 5 + 1 = 6
Example-2 :
Input : N = 9, C = 4
pages = { 5, 0, 1, 3, 2, 4, 1, 0, 5 }
Output :
Page Faults = 8
Page Hits = 1
Explanation :
Capacity is 4, thus, we can store maximum 4 pages at a time.
Page 5 is required, since it is not present,  
it is a page fault : page fault = 1
Page 0 is required, since it is not present,  
it is a page fault : page fault = 1 + 1 = 2
Page 1 is required, since it is not present,  
it is a page fault : page fault = 2 + 1 = 3
Page 3 is required, since it is not present,  
it is a page fault : page fault = 3 + 1 = 4
Page 2 is required, since it is not present,  
it replaces LFU page 5 : page fault = 4 + 1 = 5
Page 4 is required, since it is not present,  
it replaces LFU page 0 : page fault = 5 + 1 = 6
Page 1 is required, since it is present,  
it is not a page fault : page fault = 6 + 0 = 6
Page 0 is required, since it is not present,  
it replaces LRU page 3 : page fault = 6 + 1 = 7
Page 5 is required, since it is not present,  
it replaces LFU page 2 : page fault = 7 + 1 = 8
Algorithm :
Step-1 : Initialize count as 0.
Step-2 : Create a vector / array of size equal to memory capacity.
            Create a map to store frequency of pages 
Step-3 : Traverse elements of pages[]
Step-4 : In each traversal:
       if(element is present in memory):
            remove the element and push the element at the end  
            increase its frequency
       else:
            if(memory is full) 
                 remove the first element and decrease frequency of 1st element
            Increment count  
            push the element at the end and increase its frequency
      Compare frequency with other pages starting from the 2nd last page                  
      Sort the pages based on their frequency and time at which they arrive
      if frequency is same, then, the page arriving first must be placed first       
Following is the implementation of the above algorithm.
C++
// C++ program to illustrate// page faults in LFU  #include <bits/stdc++.h>using namespace std;  /* Counts no. of page faults */int pageFaults(int n, int c, int pages[]){    // Initialise count to 0    int count = 0;      // To store elements in memory of size c    vector<int> v;    // To store frequency of pages    unordered_map<int, int> mp;      int i;    for (i = 0; i <= n - 1; i++) {          // Find if element is present in memory or not        auto it = find(v.begin(), v.end(), pages[i]);          // If element is not present        if (it == v.end()) {              // If memory is full            if (v.size() == c) {                  // Decrease the frequency                mp[v[0]]--;                  // Remove the first element as                // It is least frequently used                v.erase(v.begin());            }              // Add the element at the end of memory            v.push_back(pages[i]);            // Increase its frequency            mp[pages[i]]++;              // Increment the count            count++;        }        else {              // If element is present            // Remove the element            // And add it at the end            // Increase its frequency            mp[pages[i]]++;            v.erase(it);            v.push_back(pages[i]);        }          // Compare frequency with other pages        // starting from the 2nd last page                         int k = v.size() - 2;          // Sort the pages based on their frequency         // And time at which they arrive        // if frequency is same        // then, the page arriving first must be placed first        while (mp[v[k]] > mp[v[k + 1]] && k > -1) {            swap(v[k + 1], v[k]);            k--;        }    }      // Return total page faults    return count;}  /* Driver program to test pageFaults function*/int main(){      int pages[] = { 1, 2, 3, 4, 2, 1, 5 };    int n = 7, c = 3;      cout << "Page Faults = " << pageFaults(n, c, pages)         << endl;    cout << "Page Hits = " << n - pageFaults(n, c, pages);    return 0;}  // This code is contributed by rajsanghavi9.
Page Faults = 6
Page Hits = 1
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