Shortest Remaining Time First (SRTF) With predicted Time

CPU scheduling algorithms are essential components of operating systems that determine the order in which processes are executed on a computer's central processing unit (CPU). These algorithms aim to optimize CPU utilization, reduce waiting times, and enhance system performance by efficiently managing the execution of tasks in a multi-tasking environment. Various algorithms, such as First-Come, First-Served (FCFS), Shortest Job Next (SJN), Round Robin (RR), and Priority Scheduling, are employed to achieve these objectives, each with its own set of advantages and limitations. In this article, we study about Shortest Remaining Time First CPU Scheduling algorithm.

What is the SRTF Algorithm?

Shortest Remaining Time First (SRTF) is the preemptive version of the Shortest Job Next (SJN) algorithm, where the processor is allocated to the job closest to completion. It is also known as Shortest Job First with dynamic priority. It is a CPU scheduling algorithm of the Operating System.

In SRTF with predicted time, each process is assigned a priority based on its expected remaining burst time. The process with the highest priority (i.e., the shortest predicted remaining burst time) is selected to execute next. The predicted burst time is updated dynamically as the process runs, and the priority is recalculated based on the edited expected burst time.

Advantages of SRTF

The advantage of SRTF with predicted time is that:

• It can reduce the number of preemptions and context switches, as processes with shorter predicted burst times will be given higher priority.
• The accuracy of the predicted burst times can affect the performance of the algorithm, as inaccurate predictions can lead to poor scheduling decisions.

Disadvantages of SRTF

• Shortest Remaining Time First (SRTF) is a CPU scheduling algorithm in which the process with the smallest remaining burst time is scheduled next. 
• This algorithm can be impractical in certain scenarios, such as when the burst times of the processes are not known in advance or when there are frequent arrivals and departures of processes.

Pseudo Code

1. Initialize the ready queue with all the processes.
2. While the ready queue is not empty:
   a. Find the process with the smallest predicted remaining burst time from the input.
   b. Execute the process for a unit of time.
   c. Update the predicted remaining burst time of the processes based on it's past execution history.
   d. If the process has completed its execution, remove it from the ready queue.
   e. If a new process arrives, add it to the ready queue.

C Program

#include <stdio.h>
#include <stdbool.h>

#define MAX_PROCESSES 100

//declaring the stucture
struct process {
    int pid;
    int arrival_time;
    int burst_time;
    int remaining_time;
};

int main() {
    int n;
    struct process processes[MAX_PROCESSES];
    bool completed[MAX_PROCESSES] = {false};
    int current_time = 0;
    int shortest_time = 0;
    int shortest_index = 0;

    printf("Enter the number of processes: ");
    scanf("%d", &n);

    for (int i = 0; i < n; i++) {
        printf("Enter arrival time and burst time for process %d: ", i+1);
        scanf("%d %d", &processes[i].arrival_time, &processes[i].burst_time);
        processes[i].pid = i+1;
        processes[i].remaining_time = processes[i].burst_time;
    }

    printf("\n");

    while (true) {
        bool completed_all = true;
        for (int i = 0; i < n; i++) {
            if (!completed[i]) {
              int all_completed = false;
                if (processes[i].arrival_time <= current_time &&
                    processes[i].remaining_time < processes[shortest_index].remaining_time) {
                    shortest_index = i;
                }
            }
        }
        if (completed_all) {
            break;
        }
        if (shortest_index != -1) {
            processes[shortest_index].remaining_time--;
            if (processes[shortest_index].remaining_time == 0) {
                completed[shortest_index] = true;
            }
        }
        current_time++;
        shortest_index = -1;
    }

    printf("Process\tArrival Time\tBurst Time\tWaiting Time\tTurnAround Time\n");

    int total_waiting_time = 0;
    int total_turnaround_time = 0;
    for (int i = 0; i < n; i++) {
        int waiting_time = 0;
        int turnaround_time = 0;
        waiting_time = current_time - processes[i].burst_time - processes[i].arrival_time;
        turnaround_time = current_time - processes[i].arrival_time;
        total_waiting_time += waiting_time;
        total_turnaround_time += turnaround_time;
        printf("%d\t%d\t\t%d\t\t%d\t\t%d\n", processes[i].pid, processes[i].arrival_time, processes[i].burst_time, waiting_time, turnaround_time);
    }

    float avg_waiting_time = (float) total_waiting_time / n;
    float avg_turnaround_time = (float) total_turnaround_time / n;

    printf("The Average Waiting Time: %.2f\n", avg_waiting_time);
    printf("The Average Turnaround Time: %.2f\n", avg_turnaround_time);

    return 0;
}

Output

C++ Code

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

struct process {

    int pid, arrival_time, burst_time, remaining_time, predicted_time;
};

bool cmp(process a, process b) {
    if (a.arrival_time == b.arrival_time) {
        if (a.predicted_time == b.predicted_time) {
            return a.pid < b.pid;
        }
        return a.predicted_time < b.predicted_time;
    }
    return a.arrival_time < b.arrival_time;
}

void SRTF(vector<process>& processes) {
    int n = processes.size();
    int current_time = 0, completed = 0;
    vector<int> waiting_time(n, 0);
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;

    while (completed < n) {
        for (int i = 0; i < n; i++) {
            if (processes[i].arrival_time <= current_time && processes[i].remaining_time > 0) {
                pq.push(make_pair(processes[i].predicted_time, processes[i].pid));
            }
        }
        if (pq.empty()) {
            current_time++;
            continue;
        }
        int pid = pq.top().second;
        pq.pop();
        processes[pid].remaining_time--;
        current_time++;
        if (processes[pid].remaining_time == 0) {
            completed++;
            waiting_time[pid] = current_time - processes[pid].arrival_time - processes[pid].burst_time;
        }
        else {
            processes[pid].predicted_time = processes[pid].remaining_time + current_time;
        }
    }

    cout << "PID\tArrival Time\tBurst Time\tWaiting Time\n";
    int total_waiting_time = 0;
    for (int i = 0; i < n; i++) {
        cout << processes[i].pid << "\t" << processes[i].arrival_time << "\t\t" << processes[i].burst_time << "\t\t" << waiting_time[i] << "\n";
        total_waiting_time += waiting_time[i];
    }
    cout << "Average waiting time = " << (float)total_waiting_time / n << "\n";
}

int main() {
    int n;
    cout << "Enter the number of processes: ";
    cin >> n;
    vector<process> processes(n);
    for (int i = 0; i < n; i++) {
        processes[i].pid = i;
        cout << "Enter the arrival time and burst time of process " << i << ": ";
        cin >> processes[i].arrival_time >> processes[i].burst_time;
        processes[i].remaining_time = processes[i].burst_time;
        processes[i].predicted_time = processes[i].burst_time;
    }
    sort(processes.begin(), processes.end(), cmp);
    SRTF(processes);
    return 0;
}

Output

Java Code

import java.util.*;

public class SRTF {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter the number of processes: ");
        int n = sc.nextInt();
        int[] processId = new int[n];
        int[] arrivalTime = new int[n];
        int[] burstTime = new int[n];
        int[] remainingTime = new int[n];
        int[] predictedTime = new int[n];
        int[] completionTime = new int[n];
        int[] turnaroundTime = new int[n];
        int[] waitingTime = new int[n];
        boolean[] isCompleted = new boolean[n];
        int currentTime = 0;
        int completed = 0;

        for (int i = 0; i < n; i++) {
            System.out.println("Enter details for process " + (i + 1));
            System.out.print("Process ID: ");
            processId[i] = sc.nextInt();
            System.out.print("Arrival time: ");
            arrivalTime[i] = sc.nextInt();
            System.out.print("Burst time: ");
            burstTime[i] = sc.nextInt();
            remainingTime[i] = burstTime[i];
            System.out.print("Predicted time: ");
            predictedTime[i] = sc.nextInt();
        }

        while (completed!=n) {
            int shortest=-1;
            for (int i=0; i<n;i++) {
                if (!isCompleted[i] && arrivalTime[i] <= currentTime) {
                    if (shortest == -1 || remainingTime[i] < remainingTime[shortest]) {
                        shortest=i;
                    }
                }
            }

            if (shortest == -1) {
                currentTime++;
                continue;
            }

            remainingTime[shortest]--;

            if (remainingTime[shortest] == 0) {
                completionTime[shortest] = currentTime + 1;
                turnaroundTime[shortest] = completionTime[shortest] - arrivalTime[shortest];
                waitingTime[shortest] = turnaroundTime[shortest] - burstTime[shortest];
                isCompleted[shortest] = true;
                completed++;
            }

            currentTime++;
        }

        System.out.println("\nProcess\tArrival Time\tBurst Time\tPredicted Time\tCompletion Time\tTurnaround Time\tWaiting Time");
        for (int i = 0; i < n; i++) {
            System.out.println(processId[i] + "\t\t" + arrivalTime[i] + "\t\t" + burstTime[i] + "\t\t" + predictedTime[i] + "\t\t\t" + completionTime[i] + "\t\t\t" + turnaroundTime[i] + "\t\t\t" + waitingTime[i]);
        }

        double totalTurnaroundTime = 0;
        double totalWaitingTime = 0;
        for (int i = 0; i < n; i++) {
            totalTurnaroundTime += turnaroundTime[i];
            totalWaitingTime += waitingTime[i];
        }
        System.out.println("\nAverage Turnaround Time: " + (totalTurnaroundTime / n));
        System.out.println("Average Waiting Time: " + (totalWaitingTime / n));
    }
}

Output

Enter the number of processes: 3
Enter details for process 1
Process ID: 1
Arrival time: 0
Burst time: 3
Predicted time: 2
Enter details for process 2
Process ID: 3
Arrival time: 4
Burst time: 9
Predicted time: 6
Enter details for process 3
Process ID: 2
Arrival time: 4
Burst time: 10
Predicted time: 7

Process Arrival Time    Burst Time      Predicted Time  Completion Time Turnaround Time Waiting Time
1       0       3       2       3       3       0
3       4       9       6       13      9       0
2       4       10      7       23      19      9

Average Turnaround Time: 10.333333333333334
Process ID: 4
Arrival time: 5
Burst time: 12
Predicted time: 9

Process Arrival Time    Burst Time      Predicted Time  Completion Time Turnaround Time Waiting Time
1       0       5               4               5               5               0
2       3       9               6               14              11              2
3       4       10              7               24              20              10
4       5       12              9               36              31              19

Average Turnaround Time: 16.75
Average Waiting Time: 7.75

JavaScript Code

class Process {
    constructor(id, arrivalTime, burstTime, predictedTime) {
      this.id = id;
      this.arrivalTime = arrivalTime;
      this.burstTime = burstTime;
      this.predictedTime = predictedTime;
      this.remainingTime = burstTime;
    }
  }
  
  function SRTF(processes) {
    let currentTime = 0;
    let completedProcesses = [];
    let readyQueue = [];
    let runningProcess = null;
  
    while (completedProcesses.length < processes.length) {
      // Check if there are any processes that have arrived
      for (let i = 0; i < processes.length; i++) {
        if (processes[i].arrivalTime <= currentTime && !completedProcesses.includes(processes[i])) {
          readyQueue.push(processes[i]);
        }
      }
  
      // Sort readyQueue by remaining time
      readyQueue.sort((a, b) => {
        if (a.remainingTime === b.remainingTime) {
          return a.predictedTime - b.predictedTime;
        }
        return a.remainingTime - b.remainingTime;
      });
  
      // If runningProcess has completed, add it to completedProcesses and set runningProcess to null
      if (runningProcess && runningProcess.remainingTime === 0) {
        completedProcesses.push(runningProcess);
        runningProcess = null;
      }
  
      // If there are processes in the readyQueue, start the one with the shortest remaining time
      if (readyQueue.length > 0) {
        const shortestProcess = readyQueue.shift();
        if (runningProcess !== shortestProcess) {
          runningProcess = shortestProcess;
        }
      }
  
      // Decrement remaining time of runningProcess and increment currentTime
      if (runningProcess) {
        runningProcess.remainingTime--;
        currentTime++;
      } else {
        currentTime++;
      }
    }
  
    return completedProcesses;
  }
  
  // Example usage
  const processes = [
    new Process(1, 0, 6, 4),
    new Process(2, 2, 4, 2),
    new Process(3, 4, 2, 1),
    new Process(4, 5, 3, 3)
  ];
  const completedProcesses = SRTF(processes);
  console.log(completedProcesses);

Output

[
  Process {
    id: 3,
    arrivalTime: 4,
    burstTime: 2,
    predictedTime: 1,
    remainingTime: -1
  },
  Process {
    id: 2,
    arrivalTime: 2,
    burstTime: 4,
    predictedTime: 2,
    remainingTime: -4
  },
  Process {
    id: 4,
    arrivalTime: 5,
    burstTime: 3,
    predictedTime: 3,
    remainingTime: -9
  },
  Process {
    id: 1,
    arrivalTime: 0,
    burstTime: 6,
    predictedTime: 4,
    remainingTime: -1
  }
]

Conclusion

In conclusion, SRTF is a CPU scheduling algorithm that prioritizes tasks based on their remaining execution time. It is an extension of the SJF algorithm and can reduce the average waiting time for all tasks. However, it requires knowledge of the remaining execution time of each task, which may not always be available, and the predicted execution times can be used as a solution for the program.

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Article Tags :

• Operating Systems
• Operating Systems-CPU Scheduling