Question

C++

==================================

Define a problem utilizing Queues and Priority Queues. Write the design, code and display output.

Include source code and output. If no output explain the reason why and what you are going to do make sure it does not happen again aka learning from your mistakes.

Problem:
Design:
Code:
Output:

Answer 1

Problem on Queue:

PROBLEM: LRU Cache Implementation

DESIGN: Queue is implemented using a doubly linked list. The maximum size of the queue will be equal to the total number of frames available(cache size). The most recently used pages will be near front end and the least recently pages will be near the rear end.

A Hash with page number as key and address of the corresponding queue node as value.

CODE:

#include <bits/stdc++.h>
using namespace std;

class LRUCache {
    // store keys of cache
    list<int> dq;

    // store references of key in cache
    unordered_map<int, list<int>::iterator> ma;
    int csize; // maximum capacity of cache

public:
    LRUCache(int);
    void refer(int);
    void display();
};

// Declare the size
LRUCache::LRUCache(int n)
{
    csize = n;
}

// Refers key x with in the LRU cache
void LRUCache::refer(int x)
{
    // not present in cache
    if (ma.find(x) == ma.end()) {
        // cache is full
        if (dq.size() == csize) {
            // delete least recently used element
            int last = dq.back();

            // Pops the last elmeent
            dq.pop_back();

            // Erase the last
            ma.erase(last);
        }
    }

    // present in cache
    else
        dq.erase(ma[x]);

    // update reference
    dq.push_front(x);
    ma[x] = dq.begin();
}

// Function to display contents of cache
void LRUCache::display()
{

    // Iterate in the deque and print
    // all the elements in it
    for (auto it = dq.begin(); it != dq.end();
         it++)
        cout << (*it) << " ";

    cout << endl;
}

// Driver Code
int main()
{
    LRUCache ca(4);

    ca.refer(1);
    ca.refer(2);
    ca.refer(3);
    ca.refer(1);
    ca.refer(4);
    ca.refer(5);
    ca.display();

    return 0;
}

OUTPUT: 5 4 1 3

Problem on priority queues

PROBLEM: Priority CPU Scheduling

DESIGN: Priority scheduling is one of the most common scheduling algorithms in batch systems. Each process is assigned a priority. Process with the highest priority is to be executed first and so on.
Processes with the same priority are executed on first come first served basis. Priority can be decided based on memory requirements, time requirements or any other resource requirement.

implementation:

1- First input the processes with their burst time
   and priority.
2- Sort the processes, burst time and priority
   according to the priority.
3- Now simply apply FCFS algorithm.

CODE:

#include<bits/stdc++.h>
using namespace std;

struct Process
{
    int pid; // Process ID
    int bt;   // CPU Burst time required
    int priority; // Priority of this process
};

// Function to sort the Process acc. to priority
bool comparison(Process a, Process b)
{
    return (a.priority > b.priority);
}

// Function to find the waiting time for all
// processes
void findWaitingTime(Process proc[], int n,
                     int wt[])
{
    // waiting time for first process is 0
    wt[0] = 0;

    // calculating waiting time
    for (int i = 1; i < n ; i++ )
        wt[i] = proc[i-1].bt + wt[i-1] ;
}

// Function to calculate turn around time
void findTurnAroundTime( Process proc[], int n,
                         int wt[], int tat[])
{
    // calculating turnaround time by adding
    // bt[i] + wt[i]
    for (int i = 0; i < n ; i++)
        tat[i] = proc[i].bt + wt[i];
}

//Function to calculate average time
void findavgTime(Process proc[], int n)
{
    int wt[n], tat[n], total_wt = 0, total_tat = 0;

    //Function to find waiting time of all processes
    findWaitingTime(proc, n, wt);

    //Function to find turn around time for all processes
    findTurnAroundTime(proc, n, wt, tat);

    //Display processes along with all details
    cout << "\nProcesses "<< " Burst time "
         << " Waiting time " << " Turn around time\n";

    // Calculate total waiting time and total turn
    // around time
    for (int i=0; i<n; i++)
    {
        total_wt = total_wt + wt[i];
        total_tat = total_tat + tat[i];
        cout << "   " << proc[i].pid << "\t\t"
             << proc[i].bt << "\t    " << wt[i]
             << "\t\t " << tat[i] <<endl;
    }

    cout << "\nAverage waiting time = "
         << (float)total_wt / (float)n;
    cout << "\nAverage turn around time = "
         << (float)total_tat / (float)n;
}

void priorityScheduling(Process proc[], int n)
{
    // Sort processes by priority
    sort(proc, proc + n, comparison);

    cout<< "Order in which processes gets executed \n";
    for (int i = 0 ; i < n; i++)
        cout << proc[i].pid <<" " ;

    findavgTime(proc, n);
}

// Driver code
int main()
{
    Process proc[] = {{1, 10, 2}, {2, 5, 0}, {3, 8, 1}};
    int n = sizeof proc / sizeof proc[0];
    priorityScheduling(proc, n);
    return 0;
}

OUTPUT:

Order in which processes gets executed
1 3 2
Processes Burst time Waiting time Turn around time
1        10     0         10
3        8     10         18
2        5     18         23

Average waiting time = 9.33333
Average turn around time = 17