Shortest Remaining Time First (Preemptive SJF) Scheduling Algorithm
In previous post, we have discussed Set 1 of SJF i.e. non-preemptive. In this post we will discuss the preemptive version of SJF known as Shortest Remaining Time First (SRTF).
In the Shortest Remaining Time First (SRTF) scheduling algorithm, the process with the smallest amount of time remaining until completion is selected to execute. Since the currently executing process is the one with the shortest amount of time remaining by definition, and since that time should only reduce as execution progresses, processes will always run until they complete or a new process is added that requires a smaller amount of time.
Examples to show working of Preemptive Shortest Job First CPU Scheduling Algorithm:
Example-1: Consider the following table of arrival time and burst time for five processes P1, P2, P3, P4 and P5.
Process | Burst Time | Arrival Time |
---|---|---|
P1 | 6 ms | 2 ms |
P2 | 2 ms | 5 ms |
P3 | 8 ms | 1 ms |
P4 | 3 ms | 0 ms |
P5 | 4 ms | 4 ms |
The Shortest Job First CPU Scheduling Algorithm will work on the basis of steps as mentioned below:
At time = 0,
- Process P4 arrives and starts executing
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
0-1ms | P4 | 0ms | 1ms | 3ms | 2ms |
At time= 1,
- Process P3 arrives.
- But, as P4 has a shorter burst time. It will continue execution.
- Thus, P3 will wait till P4 gets executed.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
1-2ms | P4 | 0ms | P3 | 1ms | 2ms | 1ms |
P3 | 1ms | 0ms | 8ms | 8ms |
At time =2,
- Process P1 arrives with burst time = 6
- As the burst time of P1 is more than that of P4
- Thus, P4 will continue its execution.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
2-3ms | P3, P1 | |||||
P3 | 1ms | 0ms | 8ms | 8ms | ||
P1 | 2ms | 0ms | 6ms | 6ms |
At time = 3,
- Process P4 will finish its execution.
- Then, the burst time of P3 and P1 is compared.
- Process P1 is executed because its burst time is less as compared to P3.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
3-4ms | P3 | 1ms |
P3 |
0ms | 8ms | 8ms |
P1 | 2ms | 1ms | 6ms | 5ms |
At time = 4,
- Process P5 arrives.
- Then the burst time of P3, P5, and P1 is compared.
- Process P5 gets executed first among them because its burst time is lowest, and process P1 is preempted.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
4-5ms | P3 | 1ms | P3, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
P5 | 4ms | 1ms | 4ms | 3ms |
At time = 5,
- Process P2 arrives.
- The burst time of all processes are compared,
- Process P2 gets executed as its burst time is lowest among all.
- Process P5 is preempted.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
5-6ms | P3 | 1ms | P3, P5, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
P5 | 4ms | 0ms | 3ms | 3ms | ||
P2 | 5ms | 1ms | 2ms | 1ms |
At time = 6,
- Process P2 will keep executing.
- It will execute till time = 8 as the burst time of P2 is 2ms
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
6-7ms | P3 | 1ms |
P3, P5, P1 |
0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
P5 | 4ms | 0ms | 3ms | 3ms | ||
At time=7,
- The process P2 finishes its execution.
- Then again the burst time of all remaining processes is compared.
- The Process P5 gets executed because its burst time is lesser than the others.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
7-10ms | P3 | 1ms | P3, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
At time = 10,
- The process P5 will finish its execution.
- Then the burst time of the remaining processes P1 and P3 is compared.
- Thus, process P1 is executed as its burst time is less than P3
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
10-15ms | P3 | 1ms | P3 | 0ms | 8ms | 8ms |
At time = 15,
- The process P1 finishes its execution and P3 is the only process left.
- P3 will start executing.
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
15-23ms |
At time = 23,
- Process P3 will finish its execution.
- The overall execution of the processes will be as shown below:
Time Instance | Process | Arrival Time | Waiting Table | Execution Time | Initial Burst Time | Remaining Burst Time |
---|---|---|---|---|---|---|
0-1ms | P4 | 0ms | 1ms | 3ms | 2ms | |
1-2ms | P4 | 0ms | P3 | 1ms | 2ms | 1ms |
P3 | 1ms | 0ms | 8ms | 8ms | ||
2-3ms | P3, P1 | |||||
P3 | 1ms | 0ms | 8ms | 8ms | ||
P1 | 2ms | 0ms | 6ms | 6ms | ||
3-4ms | P3 | 1ms |
P3 |
0ms | 8ms | 8ms |
P1 | 2ms | 1ms | 6ms | 5ms | ||
4-5ms | P3 | 1ms | P3, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
P5 | 4ms | 1ms | 4ms | 3ms | ||
5-6ms | P3 | 1ms | P3, P5, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
P5 | 4ms | 0ms | 3ms | 3ms | ||
P2 | 5ms | 1ms | 2ms | 1ms | ||
6-7ms | P3 | 1ms | P3, P5, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
P5 | 4ms | 0ms | 3ms | 3ms | ||
7-10ms | P3 | 1ms | P3, P1 | 0ms | 8ms | 8ms |
P1 | 2ms | 0ms | 5ms | 5ms | ||
10-15ms | P3 | 1ms | P3 | 0ms | 8ms | 8ms |
15-23ms |
Gantt chart for above execution:

Gantt chart for SRTF
Now, lets calculate average waiting time and turn around time:
As we know,
- Turn Around time = Completion time – arrival time
- Waiting Time = Turn around time – burst time
Process | Completion Time | Turn Around Time | Waiting Time |
---|---|---|---|
P1 | 15 | 15-2 = 13 | 13-6 = 7 |
P2 | 7 | 7-5 = 2 | 2-2 = 0 |
P3 | 23 | 23-1 = 22 | 22-8 = 14 |
P4 | 3 | 3-0 = 3 | 3-3 = 0 |
P5 | 10 | 10-4 = 6 | 6-4 = 2 |
Now,
- Average Turn around time = (13 + 2 + 22 + 3 + 6)/5 = 9.2
- Average waiting time = (7 + 0 + 14 + 0 + 2)/5 = 23/5 = 4.6
Some of the key characteristics of SRTF
- Preemptive: SRTF is a preemptive algorithm, which means that the currently running process can be interrupted if a new process arrives with a shorter burst time. This helps in ensuring that the processes with the shortest burst times are executed first.
- Dynamic: SRTF is a dynamic algorithm, which means that it can adapt to changes in the arrival time and burst time of processes. It constantly re-evaluates the remaining burst time of each process and schedules the process with the shortest remaining time.
- Low waiting time: SRTF is known for its low waiting time. By selecting the process with the shortest remaining burst time, it ensures that the processes with the shortest burst times are executed first, which reduces the average waiting time of processes.
- SRTF has a higher complexity than other scheduling algorithms like FCFS (First Come First Serve) and RR (Round Robin), because it requires frequent context switches and preemptions.
Implementation of SRTF Algorithm:
Approach:
- Traverse until all process gets completely executed.
- Find process with minimum remaining time at every single time lap.
- Reduce its time by 1.
- Check if its remaining time becomes 0
- Increment the counter of process completion.
- Completion time of current process = current_time + 1;
- Calculate waiting time for each completed process.
- wt[i]= Completion time – arrival_time-burst_time
- Increment time lap by one.
- Find turnaround time (waiting_time + burst_time).
Program to implement Shortest Remaining Time First:
C++
// C++ program to implement Shortest Remaining Time First // Shortest Remaining Time First (SRTF) #include <bits/stdc++.h> using namespace std; struct Process { int pid; // Process ID int bt; // Burst Time int art; // Arrival Time }; // Function to find the waiting time for all // processes void findWaitingTime(Process proc[], int n, int wt[]) { int rt[n]; // Copy the burst time into rt[] for ( int i = 0; i < n; i++) rt[i] = proc[i].bt; int complete = 0, t = 0, minm = INT_MAX; int shortest = 0, finish_time; bool check = false ; // Process until all processes gets // completed while (complete != n) { // Find process with minimum // remaining time among the // processes that arrives till the // current time` for ( int j = 0; j < n; j++) { if ((proc[j].art <= t) && (rt[j] < minm) && rt[j] > 0) { minm = rt[j]; shortest = j; check = true ; } } if (check == false ) { t++; continue ; } // Reduce remaining time by one rt[shortest]--; // Update minimum minm = rt[shortest]; if (minm == 0) minm = INT_MAX; // If a process gets completely // executed if (rt[shortest] == 0) { // Increment complete complete++; check = false ; // Find finish time of current // process finish_time = t + 1; // Calculate waiting time wt[shortest] = finish_time - proc[shortest].bt - proc[shortest].art; if (wt[shortest] < 0) wt[shortest] = 0; } // Increment time t++; } } // 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 << " P\t\t" << "BT\t\t" << "WT\t\t" << "TAT\t\t\n" ; // Calculate total waiting time and // total turnaround 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\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; } // Driver code int main() { Process proc[] = { { 1, 6, 2 }, { 2, 2, 5 }, { 3, 8, 1 }, { 4, 3, 0}, {5, 4, 4} }; int n = sizeof (proc) / sizeof (proc[0]); findavgTime(proc, n); return 0; } |
Java
// Java program to implement Shortest Remaining Time First // Shortest Remaining Time First (SRTF) class Process { int pid; // Process ID int bt; // Burst Time int art; // Arrival Time public Process( int pid, int bt, int art) { this .pid = pid; this .bt = bt; this .art = art; } } public class GFG { // Method to find the waiting time for all // processes static void findWaitingTime(Process proc[], int n, int wt[]) { int rt[] = new int [n]; // Copy the burst time into rt[] for ( int i = 0 ; i < n; i++) rt[i] = proc[i].bt; int complete = 0 , t = 0 , minm = Integer.MAX_VALUE; int shortest = 0 , finish_time; boolean check = false ; // Process until all processes gets // completed while (complete != n) { // Find process with minimum // remaining time among the // processes that arrives till the // current time` for ( int j = 0 ; j < n; j++) { if ((proc[j].art <= t) && (rt[j] < minm) && rt[j] > 0 ) { minm = rt[j]; shortest = j; check = true ; } } if (check == false ) { t++; continue ; } // Reduce remaining time by one rt[shortest]--; // Update minimum minm = rt[shortest]; if (minm == 0 ) minm = Integer.MAX_VALUE; // If a process gets completely // executed if (rt[shortest] == 0 ) { // Increment complete complete++; check = false ; // Find finish time of current // process finish_time = t + 1 ; // Calculate waiting time wt[shortest] = finish_time - proc[shortest].bt - proc[shortest].art; if (wt[shortest] < 0 ) wt[shortest] = 0 ; } // Increment time t++; } } // Method to calculate turn around time static 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]; } // Method to calculate average time static void findavgTime(Process proc[], int n) { int wt[] = new int [n], tat[] = new int [n]; int 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 System.out.println( "Processes " + " Burst time " + " Waiting time " + " Turn around time" ); // Calculate total waiting time and // total turnaround time for ( int i = 0 ; i < n; i++) { total_wt = total_wt + wt[i]; total_tat = total_tat + tat[i]; System.out.println( " " + proc[i].pid + "\t\t" + proc[i].bt + "\t\t " + wt[i] + "\t\t" + tat[i]); } System.out.println( "Average waiting time = " + ( float )total_wt / ( float )n); System.out.println( "Average turn around time = " + ( float )total_tat / ( float )n); } // Driver Method public static void main(String[] args) { Process proc[] = { new Process( 1 , 6 , 1 ), new Process( 2 , 8 , 1 ), new Process( 3 , 7 , 2 ), new Process( 4 , 3 , 3 )}; findavgTime(proc, proc.length); } } |
Python3
# Python3 program to implement Shortest Remaining Time First # Shortest Remaining Time First (SRTF) # Function to find the waiting time # for all processes def findWaitingTime(processes, n, wt): rt = [ 0 ] * n # Copy the burst time into rt[] for i in range (n): rt[i] = processes[i][ 1 ] complete = 0 t = 0 minm = 999999999 short = 0 check = False # Process until all processes gets # completed while (complete ! = n): # Find process with minimum remaining # time among the processes that # arrives till the current time` for j in range (n): if ((processes[j][ 2 ] < = t) and (rt[j] < minm) and rt[j] > 0 ): minm = rt[j] short = j check = True if (check = = False ): t + = 1 continue # Reduce remaining time by one rt[short] - = 1 # Update minimum minm = rt[short] if (minm = = 0 ): minm = 999999999 # If a process gets completely # executed if (rt[short] = = 0 ): # Increment complete complete + = 1 check = False # Find finish time of current # process fint = t + 1 # Calculate waiting time wt[short] = (fint - proc[short][ 1 ] - proc[short][ 2 ]) if (wt[short] < 0 ): wt[short] = 0 # Increment time t + = 1 # Function to calculate turn around time def findTurnAroundTime(processes, n, wt, tat): # Calculating turnaround time for i in range (n): tat[i] = processes[i][ 1 ] + wt[i] # Function to calculate average waiting # and turn-around times. def findavgTime(processes, n): wt = [ 0 ] * n tat = [ 0 ] * n # Function to find waiting time # of all processes findWaitingTime(processes, n, wt) # Function to find turn around time # for all processes findTurnAroundTime(processes, n, wt, tat) # Display processes along with all details print ( "Processes Burst Time Waiting" , "Time Turn-Around Time" ) total_wt = 0 total_tat = 0 for i in range (n): total_wt = total_wt + wt[i] total_tat = total_tat + tat[i] print ( " " , processes[i][ 0 ], "\t\t" , processes[i][ 1 ], "\t\t" , wt[i], "\t\t" , tat[i]) print ( "\nAverage waiting time = %.5f " % (total_wt / n) ) print ( "Average turn around time = " , total_tat / n) # Driver code if __name__ = = "__main__" : # Process id's proc = [[ 1 , 6 , 1 ], [ 2 , 8 , 1 ], [ 3 , 7 , 2 ], [ 4 , 3 , 3 ]] n = 4 findavgTime(proc, n) # This code is contributed # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to implement Shortest Remaining Time First // Shortest Remaining Time First (SRTF) using System; public class Process { public int pid; // Process ID public int bt; // Burst Time public int art; // Arrival Time public Process( int pid, int bt, int art) { this .pid = pid; this .bt = bt; this .art = art; } } public class GFG { // Method to find the waiting // time for all processes static void findWaitingTime(Process []proc, int n, int []wt) { int []rt = new int [n]; // Copy the burst time into rt[] for ( int i = 0; i < n; i++) rt[i] = proc[i].bt; int complete = 0, t = 0, minm = int .MaxValue; int shortest = 0, finish_time; bool check = false ; // Process until all processes gets // completed while (complete != n) { // Find process with minimum // remaining time among the // processes that arrives till the // current time` for ( int j = 0; j < n; j++) { if ((proc[j].art <= t) && (rt[j] < minm) && rt[j] > 0) { minm = rt[j]; shortest = j; check = true ; } } if (check == false ) { t++; continue ; } // Reduce remaining time by one rt[shortest]--; // Update minimum minm = rt[shortest]; if (minm == 0) minm = int .MaxValue; // If a process gets completely // executed if (rt[shortest] == 0) { // Increment complete complete++; check = false ; // Find finish time of current // process finish_time = t + 1; // Calculate waiting time wt[shortest] = finish_time - proc[shortest].bt - proc[shortest].art; if (wt[shortest] < 0) wt[shortest] = 0; } // Increment time t++; } } // Method to calculate turn around time static 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]; } // Method to calculate average time static void findavgTime(Process []proc, int n) { int []wt = new int [n]; int []tat = new int [n]; int 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 Console.WriteLine( "Processes " + " Burst time " + " Waiting time " + " Turn around time" ); // Calculate total waiting time and // total turnaround time for ( int i = 0; i < n; i++) { total_wt = total_wt + wt[i]; total_tat = total_tat + tat[i]; Console.WriteLine( " " + proc[i].pid + "\t\t" + proc[i].bt + "\t\t " + wt[i] + "\t\t" + tat[i]); } Console.WriteLine( "Average waiting time = " + ( float )total_wt / ( float )n); Console.WriteLine( "Average turn around time = " + ( float )total_tat / ( float )n); } // Driver Method public static void Main(String[] args) { Process []proc = { new Process(1, 6, 1), new Process(2, 8, 1), new Process(3, 7, 2), new Process(4, 3, 3)}; findavgTime(proc, proc.Length); } } // This code has been contributed by 29AjayKumar |
Javascript
<script> // Javascript program to implement // Shortest Remaining Time First // Shortest Remaining Time First (SRTF) class Process { constructor(pid,bt,art) { this .pid = pid; // Process ID this .bt = bt; // Burst Time this .art = art; // Arrival Time } } // Method to find the waiting time for all // processes function findWaitingTime( proc,n,wt) { let rt = new Array(n); // Copy the burst time into rt[] for (let i = 0; i < n; i++) rt[i] = proc[i].bt; let complete = 0, t = 0, minm = Number.MAX_VALUE; let shortest = 0, finish_time; let check = false ; // Process until all processes gets // completed while (complete != n) { // Find process with minimum // remaining time among the // processes that arrives till the // current time` for (let j = 0; j < n; j++) { if ((proc[j].art <= t) && (rt[j] < minm) && rt[j] > 0) { minm = rt[j]; shortest = j; check = true ; } } if (check == false ) { t++; continue ; } // Reduce remaining time by one rt[shortest]--; // Update minimum minm = rt[shortest]; if (minm == 0) minm = Number.MAX_VALUE; // If a process gets completely // executed if (rt[shortest] == 0) { // Increment complete complete++; check = false ; // Find finish time of current // process finish_time = t + 1; // Calculate waiting time wt[shortest] = finish_time - proc[shortest].bt - proc[shortest].art; if (wt[shortest] < 0) wt[shortest] = 0; } // Increment time t++; } } // Method to calculate turn around time function findTurnAroundTime(proc,n,wt,tat) { // calculating turnaround time by adding // bt[i] + wt[i] for (let i = 0; i < n; i++) tat[i] = proc[i].bt + wt[i]; } // Method to calculate average time function findavgTime(proc,n) { let wt = new Array(n), tat = new Array(n); let 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 document.write( "Processes " + " Burst time " + " Waiting time " + " Turn around time<br>" ); // Calculate total waiting time and // total turnaround time for (let i = 0; i < n; i++) { total_wt = total_wt + wt[i]; total_tat = total_tat + tat[i]; document.write( " " + proc[i].pid + " " + proc[i].bt + " " + wt[i] + " " + tat[i]+ "<br>" ); } document.write( "Average waiting time = " + total_wt / n+ "<br>" ); document.write( "Average turn around time = " + total_tat / n+ "<br>" ); } // Driver Method let proc=[ new Process(1, 6, 1), new Process(2, 8, 1), new Process(3, 7, 2), new Process(4, 3, 3)]; findavgTime(proc, proc.length); // This code is contributed by rag2127 </script> |
P BT WT TAT 1 6 7 13 2 2 0 2 3 8 14 22 4 3 0 3 5 4 2 6 Average waiting time = 4.6 Average turn around time = 9.2
Time Complexity: O(N)
Auxiliary Space: O(N)
Advantages:
- Short processes are handled very quickly.
- The system also requires very little overhead since it only makes a decision when a process completes or a new process is added.
- When a new process is added the algorithm only needs to compare the currently executing process with the new process, ignoring all other processes currently waiting to execute.
Disadvantages:
- Like shortest job first, it has the potential for process starvation.
- Long processes may be held off indefinitely if short processes are continually added.
Source:Wiki
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