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C++ Program for Shortest Job First (SJF) scheduling(preemptive)
Given process, the burst time of a process respectively and a quantum limit; the task is to find and print the waiting time, turnaround time and their respective average time using Shortest Job First Scheduling preemptive method.
What is the shortest job first scheduling?
Shortest job first scheduling is the job or process scheduling algorithm that follows the nonpreemptive scheduling discipline. In this, scheduler selects the process from the waiting queue with the least completion time and allocate the CPU to that job or process. Shortest Job First is more desirable than FIFO algorithm because SJF is more optimal as it reduces average wait time which will increase the throughput.
SJF algorithm can be preemptive as well as non-preemptive. Preemptive scheduling is also known as shortest-remaining-time-first scheduling. In Preemptive approach, the new process arises when there is already executing process. If the burst of newly arriving process is lesser than the burst time of executing process than scheduler will preempt the execution of the process with lesser burst time.
What is the Turnaround time, waiting time and completion time?
- Completion Time is the time required by the process to complete its execution
Turnaround Time is the time interval between the submission of a process and its completion.
Turnaround Time = completion of a process – submission of a process
Waiting Time is the difference between turnaround time and burst time
Waiting Time = turnaround time – burst time
Example
We are given with the processes P1, P2, P3, P4 and P5 having their corresponding burst time given below
| Process | Burst Time | Arrival Time |
|---|---|---|
| P1 | 4 | 0 |
| P2 | 2 | 1 |
| P3 | 8 | 2 |
| P4 | 1 | 3 |
| P5 | 9 | 4 |
Since the arrival time of P1 is 0 it will be the first one to get executed till the arrival of another process. When at 1 the process P2 enters and the burst time of P2 is less than the burst time of P1 therefore scheduler will dispatch the CPU with the process P2 and so on.
Average waiting time is calculated on the basis of gantt chart. P1 have to wait for (0+4)4, P2 have to wait for 1, P3 have to wait for 7, P4 have to wait for 3 and P5 have to wait for 15. So, their average waiting time will be −
Algorithm
Start
Step 1-> Declare a struct Process
Declare pid, bt, art
Step 2-> In function findTurnAroundTime(Process proc[], int n, int wt[], int tat[])
Loop For i = 0 and i < n and i++
Set tat[i] = proc[i].bt + wt[i]
Step 3-> In function findWaitingTime(Process proc[], int n, int wt[])
Declare rt[n]
Loop For i = 0 and i < n and i++
Set rt[i] = proc[i].bt
Set complete = 0, t = 0, minm = INT_MAX
Set shortest = 0, finish_time
Set bool check = false
Loop While (complete != n)
Loop For j = 0 and j < n and j++
If (proc[j].art <= t) && (rt[j] < minm) && rt[j] > 0 then,
Set minm = rt[j]
Set shortest = j
Set check = true
If check == false then,
Increment t by 1
Continue
Decrement the value of rt[shortest] by 1
Set minm = rt[shortest]
If minm == 0 then,
Set minm = INT_MAX
If rt[shortest] == 0 then,
Increment complete by 1
Set check = false
Set finish_time = t + 1
Set wt[shortest] = finish_time - proc[shortest].bt -proc[shortest].art
If wt[shortest] < 0
Set wt[shortest] = 0
Increment t by 1
Step 4-> In function findavgTime(Process proc[], int n)
Declare and set wt[n], tat[n], total_wt = 0, total_tat = 0
Call findWaitingTime(proc, n, wt)
Call findTurnAroundTime(proc, n, wt, tat)
Loop For i = 0 and i < n and i++
Set total_wt = total_wt + wt[i]
Set total_tat = total_tat + tat[i]
Print proc[i].pid, proc[i].bt, wt[i], tat[i]
Print Average waiting time i.e., total_wt / n
Print Average turn around time i.e., total_tat / n
Step 5-> In function int main()
Declare and set Process proc[] = { { 1, 5, 1 }, { 2, 3, 1 }, { 3, 6, 2 }, { 4, 5, 3 } }
Set n = sizeof(proc) / sizeof(proc[0])
Call findavgTime(proc, n)
StopExample
#include <bits/stdc++.h>
using namespace std;
//structure for every process
struct Process {
int pid; // Process ID
int bt; // Burst Time
int art; // Arrival Time
};
void findTurnAroundTime(Process proc[], int n, int wt[], int tat[]) {
for (int i = 0; i < n; i++)
tat[i] = proc[i].bt + wt[i];
}
//waiting time of all process
void findWaitingTime(Process proc[], int n, int wt[]) {
int rt[n];
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;
while (complete != n) {
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;
}
// decrementing the remaining time
rt[shortest]--;
minm = rt[shortest];
if (minm == 0)
minm = INT_MAX;
// If a process gets completely
// executed
if (rt[shortest] == 0) {
complete++;
check = false;
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 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);
cout << "Processes " << " Burst time " << " Waiting time " << " Turn around time\n";
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;
}
// main function
int main() {
Process proc[] = { { 1, 5, 1 }, { 2, 3, 1 }, { 3, 6, 2 }, { 4, 5, 3 } };
int n = sizeof(proc) / sizeof(proc[0]);
findavgTime(proc, n);
return 0;
}Output
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