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Longest Remaining Time First (LRTF) CPU Scheduling Program

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  • Difficulty Level : Hard
  • Last Updated : 21 Jun, 2022

Prerequisite – CPU Scheduling | Longest Remaining Time First (LRTF) algorithm 
We have given some process with arrival time and Burst Time and we have to find the completion time (CT), Turn Around Time(TAT), Average Turn Around Time (Avg TAT), Waiting Time(WT), Average Waiting Time (AWT) for the given processes.

Example: Consider the following table of arrival time and burst time for four processes P1, P2, P3 and P4. 

Process   Arrival time   Burst Time
P1            1 ms          2 ms
P2            2 ms          4 ms
P3            3 ms          6 ms
p4            4 ms          8 ms  

Gantt chart will be as following below,

Since, completion time (CT) can be directly determined by Gantt chart, and

Turn Around Time (TAT)
= (Completion Time) - (Arrival Time)

Also, Waiting Time (WT)
= (Turn Around Time) - (Burst Time) 

Therefore,

Output: 

Total Turn Around Time = 68 ms
So, Average Turn Around Time = 68/4 = 17.00 ms

And, Total Waiting Time = 48 ms
So, Average Waiting Time = 12.00 ms 

Algorithm – 

  • Step-1: Create a structure of process containing all necessary fields like AT (Arrival Time), BT(Burst Time), CT(Completion Time), TAT(Turn Around Time), WT(Waiting Time).
  • Step-2: Sort according to the AT;
  • Step-3: Find the process having Largest Burst Time and execute for each single unit. Increase the total time by 1 and reduce the Burst Time of that process with 1.
  • Step-4: When any process have 0 BT left, then update the CT(Completion Time of that process CT will be Total Time at that time).
  • Step-2: After calculating the CT for each process, find TAT and WT.
(TAT = CT - AT) 
(WT  = TAT - BT) 

Implementation of Algorithm –  

C++




#include <bits/stdc++.h>
 
using namespace std;
 
// creating a structure of a process
struct process {
    int processno;
    int AT;
    int BT;
 
    // for backup purpose to print in last
    int BTbackup;
    int WT;
    int TAT;
    int CT;
};
 
// creating a structure of 4 processes
struct process p[4];
 
// variable to find the total time
int totaltime = 0;
int prefinaltotal = 0;
 
// comparator function for sort()
bool compare(process p1, process p2)
{
    // compare the Arrival time of two processes
    return p1.AT < p2.AT;
}
 
// finding the largest Arrival Time among all the available
// process at that time
int findlargest(int at)
{
    int max = 0, i;
    for (i = 0; i < 4; i++) {
        if (p[i].AT <= at) {
            if (p[i].BT > p[max].BT)
                max = i;
        }
    }
 
    // returning the index of the process having the largest BT
    return max;
}
 
// function to find the completion time of each process
int findCT()
{
 
    int index;
    int flag = 0;
    int i = p[0].AT;
    while (1) {
        if (i <= 4) {
            index = findlargest(i);
        }
 
        else
            index = findlargest(4);
        cout << "Process executing at time " << totaltime
             << " is: P" << index + 1 << "\t";
 
        p[index].BT -= 1;
        totaltime += 1;
        i++;
 
        if (p[index].BT == 0) {
            p[index].CT = totaltime;
            cout << " Process P" << p[index].processno
                 << " is completed at " << totaltime;
        }
        cout << endl;
 
        // loop termination condition
        if (totaltime == prefinaltotal)
            break;
    }
}
 
int main()
{
 
    int i;
 
    // initializing the process number
    for (i = 0; i < 4; i++) {
        p[i].processno = i + 1;
    }
 
    // cout<<"arrival time of 4 processes : ";
    for (i = 0; i < 4; i++) // taking AT
    {
        p[i].AT = i + 1;
    }
 
    // cout<<" Burst time of 4 processes : ";
    for (i = 0; i < 4; i++) {
 
        // assigning {2, 4, 6, 8} as Burst Time to the processes
        // backup for displaying the output in last
        // calculating total required time for terminating
        // the function().
        p[i].BT = 2 * (i + 1);
        p[i].BTbackup = p[i].BT;
        prefinaltotal += p[i].BT;
    }
 
    // displaying the process before executing
    cout << "PNo\tAT\tBT\n";
 
    for (i = 0; i < 4; i++) {
        cout << p[i].processno << "\t";
        cout << p[i].AT << "\t";
        cout << p[i].BT << "\t";
        cout << endl;
    }
    cout << endl;
 
    // sorting process according to Arrival Time
    sort(p, p + 4, compare);
 
    // calculating initial time when execution starts
    totaltime += p[0].AT;
 
    // calculating to terminate loop
    prefinaltotal += p[0].AT;
    findCT();
    int totalWT = 0;
    int totalTAT = 0;
    for (i = 0; i < 4; i++) {
        // since, TAT = CT - AT
        p[i].TAT = p[i].CT - p[i].AT;
        p[i].WT = p[i].TAT - p[i].BTbackup;
 
        // finding total waiting time
        totalWT += p[i].WT;
 
        // finding total turn around time
        totalTAT += p[i].TAT;
    }
 
    cout << "After execution of all processes ... \n";
 
    // after all process executes
    cout << "PNo\tAT\tBT\tCT\tTAT\tWT\n";
 
    for (i = 0; i < 4; i++) {
        cout << p[i].processno << "\t";
        cout << p[i].AT << "\t";
        cout << p[i].BTbackup << "\t";
        cout << p[i].CT << "\t";
        cout << p[i].TAT << "\t";
        cout << p[i].WT << "\t";
        cout << endl;
    }
 
    cout << endl;
    cout << "Total TAT = " << totalTAT << endl;
    cout << "Average TAT = " << totalTAT / 4.0 << endl;
    cout << "Total WT = " << totalWT << endl;
    cout << "Average WT = " << totalWT / 4.0 << endl;
    return 0;
}

Java




// Java Program to implement
// longest remaining time first
import java.util.*;
 
class GFG
{
 
  // creating a class of a process
  static class process {
    int processno;
    int AT;
    int BT;
 
    // for backup purpose to print in last
    int BTbackup;
    int WT;
    int TAT;
    int CT;
  }
 
  static process[] p = new process[4];
 
  // variable to find the total time
  static int totaltime = 0;
  static int prefinaltotal = 0;
 
  // finding the largest Arrival Time among all the
  // available process at that time
  static int findlargest(int at)
  {
    int max = 0, i;
    for (i = 0; i < 4; i++) {
      if (p[i].AT <= at) {
        if (p[i].BT > p[max].BT)
          max = i;
      }
    }
 
    // returning the index of the process having the
    // largest BT
    return max;
  }
 
  // function to find the completion time of each process
  static void findCT()
  {
 
    int index;
    int flag = 0;
    int i = p[0].AT;
    while (true) {
      if (i <= 4) {
        index = findlargest(i);
      }
 
      else
        index = findlargest(4);
      System.out.print("Process executing at time "
                       + totaltime + " is: P"
                       + (index + 1) + "\t");
 
      p[index].BT -= 1;
      totaltime += 1;
      i++;
 
      if (p[index].BT == 0) {
        p[index].CT = totaltime;
        System.out.println(
          " Process P" + p[index].processno
          + " is completed at " + totaltime);
      }
      System.out.println();
 
      // loop termination condition
      if (totaltime == prefinaltotal)
        break;
    }
  }
 
  public static void main(String[] args)
  {
    int i;
 
    // initializing the process number
    for (i = 0; i < 4; i++) {
      p[i] = new process();
      p[i].processno = i + 1;
    }
 
    for (i = 0; i < 4; i++) // taking AT
    {
      p[i].AT = i + 1;
    }
 
    for (i = 0; i < 4; i++) {
 
      // assigning {2, 4, 6, 8} as Burst Time to the
      // processes backup for displaying the output in
      // last calculating total required time for
      // terminating the function().
      p[i].BT = 2 * (i + 1);
      p[i].BTbackup = p[i].BT;
      prefinaltotal += p[i].BT;
    }
 
    // displaying the process before executing
    System.out.print("PNo\tAT\tBT\n");
 
    for (i = 0; i < 4; i++) {
      System.out.print(p[i].processno + "\t");
      System.out.print(p[i].AT + "\t");
      System.out.println(p[i].BT + "\t");
    }
    System.out.println();
    Arrays.sort(p, (process p1, process p2) -> {
      return p1.AT - p2.AT;
    });
 
    // calculating initial time when execution starts
    totaltime += p[0].AT;
 
    // calculating to terminate loop
    prefinaltotal += p[0].AT;
    findCT();
    int totalWT = 0;
    int totalTAT = 0;
    for (i = 0; i < 4; i++) {
      // since, TAT = CT - AT
      p[i].TAT = p[i].CT - p[i].AT;
      p[i].WT = p[i].TAT - p[i].BTbackup;
 
      // finding total waiting time
      totalWT += p[i].WT;
 
      // finding total turn around time
      totalTAT += p[i].TAT;
    }
 
    System.out.print(
      "After execution of all processes ... \n");
 
    // after all process executes
    System.out.print("PNo\tAT\tBT\tCT\tTAT\tWT\n");
 
    for (i = 0; i < 4; i++) {
      System.out.print(p[i].processno + "\t");
      System.out.print(p[i].AT + "\t");
      System.out.print(p[i].BTbackup + "\t");
      System.out.print(p[i].CT + "\t");
      System.out.print(p[i].TAT + "\t");
      System.out.println(p[i].WT + "\t");
    }
 
    System.out.println();
    System.out.println("Total TAT = " + totalTAT);
    System.out.println("Average TAT = "
                       + (totalTAT / 4.0));
    System.out.println("Total WT = " + totalWT);
    System.out.println("Average WT = " + totalWT / 4.0);
  }
}
 
// This code is contributed by Karandeep Singh

Python3




# Python3 program to implement
# Longest Remaining Time First
 
# creating a structure of 4 processes
p = []
for i in range(4):
    p.append([0, 0, 0, 0, 0, 0, 0])
 
# variable to find the total time
totaltime = 0
prefinaltotal = 0
 
# finding the largest Arrival Time
# among all the available process
# at that time
def findlargest(at):
    max = 0
    for i in range(4):
        if (p[i][1] <= at):
            if (p[i][2] > p[max][2]) :
                max = i
     
    # returning the index of the
    # process having the largest BT
    return max
 
# function to find the completion
# time of each process
def findCT(totaltime):
    index = 0
    flag = 0
    i = p[0][1]
    while (1):
        if (i <= 4):
            index = findlargest(i)
        else:
            index = findlargest(4)
        print("Process execute at time ",
                    totaltime, end = " ")
        print(" is: P", index + 1,
                        sep = "", end = " ")
        p[index][2] -= 1
        totaltime += 1
        i += 1
        if (p[index][2] == 0):
                p[index][6] = totaltime
                print("Process P", p[index][0],
                           sep = "", end = " ")
                print(" is completed at ",
                     totaltime, end = " ")
        print()
         
        # loop termination condition
        if (totaltime == prefinaltotal):
            break
 
# Driver code
if __name__ =="__main__":
     
    # initializing the process number
    for i in range(4):
        p[i][0] = i + 1
 
    for i in range(4): # taking AT
        p[i][1] = i + 1
 
    for i in range(4):
 
        # assigning 2, 4, 6, 8 as Burst Time
        # to the processes backup for displaying
        # the output in last calculating total
        # required time for terminating the function().
        p[i][2] = 2 * (i + 1)
        p[i][3] = p[i][2]
        prefinaltotal += p[i][2]
 
    # displaying the process before executing
    print("PNo\tAT\tBT")
 
    for i in range(4):
        print(p[i][0], "\t",
              p[i][1], "\t", p[i][2])
    print()
     
    # sorting process according to Arrival Time
    p = sorted(p, key = lambda p:p[1])
 
    # calculating initial time when
    # execution starts
    totaltime += p[0][1]
 
    # calculating to terminate loop
    prefinaltotal += p[0][1]
    findCT(totaltime)
    totalWT = 0
    totalTAT = 0
    for i in range(4):
         
        # since, TAT = CT - AT
        p[i][5] = p[i][6]- p[i][1]
        p[i][4] = p[i][5] - p[i][3]
 
        # finding total waiting time
        totalWT += p[i][4]
 
        # finding total turn around time
        totalTAT += p[i][5]
 
    print("\nAfter execution of all processes ... ")
 
    # after all process executes
    print("PNo\tAT\tBT\tCT\tTAT\tWT" )
 
    for i in range(4):
        print(p[i][0], "\t", p[i][1], "\t",
              p[i][3], "\t", end = " ")
        print(p[i][6], "\t",
              p[i][5], "\t", p[i][4])
    print()
    print("Total TAT = ", totalTAT)
    print("Average TAT = ", totalTAT / 4.0)
    print("Total WT = ", totalWT)
    print("Average WT = ", totalWT / 4.0)
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

Output: 

PNo    AT    BT
1    1    2    
2    2    4    
3    3    6    
4    4    8    

Process executing at time 1 is: P1    
Process executing at time 2 is: P2    
Process executing at time 3 is: P3    
Process executing at time 4 is: P4    
Process executing at time 5 is: P4    
Process executing at time 6 is: P4    
Process executing at time 7 is: P3    
Process executing at time 8 is: P4    
Process executing at time 9 is: P3    
Process executing at time 10 is: P4    
Process executing at time 11 is: P2    
Process executing at time 12 is: P3    
Process executing at time 13 is: P4    
Process executing at time 14 is: P2    
Process executing at time 15 is: P3    
Process executing at time 16 is: P4    
Process executing at time 17 is: P1     Process P1 is completed at 18
Process executing at time 18 is: P2     Process P2 is completed at 19
Process executing at time 19 is: P3     Process P3 is completed at 20
Process executing at time 20 is: P4     Process P4 is completed at 21
After execution of all processes ... 
PNo    AT    BT    CT    TAT    WT
1    1    2    18    17    15    
2    2    4    19    17    13    
3    3    6    20    17    11    
4    4    8    21    17    9    

Total TAT = 68
Average TAT = 17
Total WT = 48
Average WT = 12 

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