# Program for FCFS Scheduling | Set 2 (Processes with different arrival times)

We have already discussed FCFS Scheduling of processes with same arrival time. In this post, scenario when processes have different arrival times are discussed. Given n processes with their burst times and arrival times, the task is to find average waiting time and average turn around time using FCFS scheduling algorithm.
FIFO simply queues processes in the order they arrive in the ready queue. Here, the process that comes first will be executed first and next process will start only after the previous gets fully executed.

1. Completion Time: Time at which process completes its execution.
2. Turn Around Time: Time Difference between completion time and arrival time. Turn Around Time = Completion Time – Arrival Time
3. Waiting Time(W.T): Time Difference between turn around time and burst time.
Waiting Time = Turn Around Time – Burst Time

```Process     Wait Time : Service Time - Arrival Time
P0 	                   0 - 0   = 0
P1 	                   5 - 1   = 4
P2 	                   8 - 2   = 6
P3 	                   16 - 3  = 13

Average Wait Time: (0 + 4 + 6 + 13) / 4 = 5.75
```

Service Time : Service time means amount of time after which a process can start execution. It is summation of burst time of previous processes (Processes that came before)

Changes in code as compare to code of FCFS with same arrival time:
To find waiting time: Time taken by all processes before the current process to be started (i.e. burst time of all previous processes) – arrival time of current process
wait_time[i] = (bt[0] + bt[1] +…… bt[i-1] ) – arrival_time[i]

Implementation:

```1- Input the processes along with their burst time(bt)
and arrival time(at)
2- Find waiting time for all other processes i.e. for
a given process  i:
wt[i] = (bt[0] + bt[1] +...... bt[i-1]) - at[i]
3- Now find turn around time
= waiting_time + burst_time for all processes
4- Average waiting time =
total_waiting_time / no_of_processes
5- Average turn around time =
total_turn_around_time / no_of_processes
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

```// C++ program for implementation of FCFS
// scheduling with different arrival time
#include<iostream>
using namespace std;

// Function to find the waiting time for all
// processes
void findWaitingTime(int processes[], int n, int bt[],
int wt[], int at[])
{
int service_time[n];
service_time[0] = 0;
wt[0] = 0;

// calculating waiting time
for (int i = 1; i < n ; i++)
{
// Add burst time of previous processes
service_time[i] = service_time[i-1] + bt[i-1];

// Find waiting time for current process =
// sum - at[i]
wt[i] = service_time[i] - at[i];

// If waiting time for a process is in negative
// before CPU becomes idle so its waiting time is 0
if (wt[i] < 0)
wt[i] = 0;
}
}

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

// Function to calculate average waiting and turn-around
// times.
void findavgTime(int processes[], int n, int bt[], int at[])
{
int wt[n], tat[n];

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

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

// Display processes along with all details
cout << "Processes " << " Burst Time " << " Arrival Time "
<< " Waiting Time " << " Turn-Around Time "
<< " Completion Time \n";
int total_wt = 0, total_tat = 0;
for (int i = 0 ; i < n ; i++)
{
total_wt = total_wt + wt[i];
total_tat = total_tat + tat[i];
int compl_time = tat[i] + at[i];
cout << " " << i+1 << "\t\t" << bt[i] << "\t\t"
<< at[i] << "\t\t" << wt[i] << "\t\t "
<< tat[i]  <<  "\t\t " << compl_time << endl;
}

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

// Driver code
int main()
{
// Process id's
int processes[] = {1, 2, 3};
int n = sizeof processes / sizeof processes[0];

// Burst time of all processes
int burst_time[] = {5, 9, 6};

// Arrival time of all processes
int arrival_time[] = {3, 3, 6};

findavgTime(processes, n, burst_time, arrival_time);

return 0;
}
```

Output:

```Processes  Burst Time  Arrival Time  Waiting Time  Turn-Around Time  Completion Time
1		5		3		0		 5		 8
2		9		3		2		 11		 14
3		6		6		8		 14		 20
Average waiting time = 3.33333
Average turn around time = 10
```

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.5 Average Difficulty : 2.5/5.0
Based on 17 vote(s)