Round-Robin Scheduling Algorithm in OS - c

Question

I have been trying to understand how the round robin concept and how the algorithm works. I have tried running this code in ubuntu, and i'm unable to get the answer that i wanted.
So based on the Round Robin Scheduling Algorithm; Let's say there are 3 processes. Where Processor 1 - Burst Time is 24, Processor 2 - Burst time is 3 and Processor 3 - Burst time is 3. and the Time Quantum is 3.
Based on this information, the waiting time for P1 is 6, P2 is 4 and P3 is 7. So the Turn Around Time is P1 is 30, P2 is 7 and P3 is 10.
Average Turn Around time is 15.6667 and The average waiting time is 5.667
Based on the code below, if i run it, it would return me; for waiting time - P1 is 6, P2 is 3 and P3 is 6, Turn around time P1 is 30, P2 is 6, P3 is 9.
And the Average Turn Around time is 15.0000 and The average waiting time is 5.000
I'm unable to figure out the error. Can any one help me with this problem and provide an explanation to error and solution?
#include<stdio.h>
#include<curses.h>
int main()
{
int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max;
float awt=0,att=0,temp=0;
clear();
printf("Enter the no of processes -- ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("\nEnter Burst Time for process %d -- ", i+1);
scanf("%d",&bu[i]);
ct[i]=bu[i];
}
printf("\nEnter the size of time slice -- ");
scanf("%d",&t);
max=bu[0];
for(i=1;i<n;i++)
if(max<bu[i])
max=bu[i];
for(j=0;j<(max/t)+1;j++)
for(i=0;i<n;i++)
if(bu[i]!=0)
if(bu[i]<=t)
{
tat[i]=temp+bu[i];
temp=temp+bu[i];
bu[i]=0;
}
else
{
bu[i]=bu[i]-t;
temp=temp+t;
}
for(i=0;i<n;i++)
{
wa[i]=tat[i]-ct[i];
att+=tat[i];
awt+=wa[i];
}
printf("\n\tPROCESS\t BURST TIME \t WAITING TIME\tTURNAROUND TIME\n");
for(i=0;i<n;i++)
{
printf("\t%d \t %d \t\t %d \t\t %d \n",i+1,ct[i],wa[i],tat[i]);
}
printf("\nThe Average Turnaround time is -- %f",att/n);
printf("\nThe Average Waiting time is -- %f ",awt/n);
getch();
}

Answer 1

The code is returning the right answer.
All the processes arrived at time 0.
P1 - 0-3 21 units remaining
P2 - 3-6 0 units remaining
P3 - 6-9 0 units remaining
P1 - 9-30 0 units remaining
P1 waited 6 units, P2 waited 3 units and P3 waited 6 units.
Note that Waiting time is the amount of time a process is waiting without being executed after given to the scheduler. Turnaround time is the total time a process took to complete its execution (Waiting time + Burst time).
Avg waiting time: (6+3+6) / 3 = 15
Average Turnaround time = (30+6+9) / 3 = 15