Round Robin CPU Scheduling PROCESS AT BT CT TAT WT P1 4 P2 1 5 P3 2 2 P4 3 1 P5 4 P6 6 3 Avg TAT= | Avg WT= Time Quantum = 2 %3D
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- Modiflow y the beprogram given to include response time program;FCFS CPU SCHEDULING ALGORITHM #include<stdio.h>#include<conio.h>main(){int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg;clrscr();printf("\nEnter the number of processes -- "); scanf("%d", &n);for(i=0;i<n;i++){printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]);}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0];for(i=1;i<n;i++){wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");for(i=0;i<n;i++){printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]);}printf("\nAverage Waiting Time -- %f", wtavg/n);printf("\nAverage Turnaround Time -- %f", tatavg/n); getch();}Consider Context Switch time of 2 secs and modify below program accordingly. FCFS CPU SCHEDULING ALGORITHM: #include<stdio.h> #include<conio.h> main() { int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg; clrscr(); printf("\nEnter the number of processes -- "); scanf("%d", &n); for(i=0;i<n;i++) { printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]); } wt[0] = wtavg = 0; tat[0] = tatavg = bt[0]; for(i=1;i<n;i++) { wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i]; } printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n"); for(i=0;i<n;i++) { printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i],…Consider Context Switch time of 2 secs and modify below program accordingly Program: FCFS CPU SCHEDULING ALGORITHM:#include<stdio.h>#include<conio.h>main(){int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg;clrscr();printf("\nEnter the number of processes -- "); scanf("%d", &n); for(i=0;i<n;i++){printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]);}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0]; for(i=1;i<n;i++){wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");for(i=0;i<n;i++){printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]);}printf("\nAverage Waiting Time -- %f", wtavg/n);printf("\nAverage Turnaround Time -- %f", tatavg/n); getch();}
- Computer Science A multiprogrammed system with a single CPU has a very large number of CPU boundprocesses running on it that all have the same priority. Could you think of a CPU schedulingalgorithm that would combine the lowest possible response times with the highest possiblethroughput and no starvationUSE ROUND-ROBIN CPU SCHEDULING WITH TIME QUANTUM OF 7 millisecondsA comparative analysis is conducted on various operating systems, including Linux, Windows, Mac, and Android, with a focus on their respective multithreading methodologies. The CPU Scheduling Algorithm can be characterized in a comparable manner?
- Consider Context Switch time of 2 secs and modify below program accordingly Program: SJF CPU SCHEDULING ALGORITHM: #include<stdio.h>#include<conio.h>using namespace std; int main(){int p[20], bt[20], wt[20], tat[20], i, k, n, temp; float wtavg, tatavg;printf("\nEnter the number of processes--"); scanf("%d", &n);for(i=0;i<n;i++){p[i]=i;printf("Enter Burst Time for Process %d--", i); scanf("%d", &bt[i]);}for(i=0;i<n;i++)for(k=i+1;k<n;k++)if(bt[i]>bt[k]){temp=bt[i]; bt[i]=bt[k];bt[k]=temp;temp=p[i];p[i]=p[k];p[k]=temp;}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0]; for(i=1;i<n;i++){wt[i] =wt[i-1]+bt[i-1];tat[i] =tat[i-1]+bt[i];wtavg = wtavg + wt[i];tatavg = tatavg + tat[i];}printf("\n\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n"); for(i=0;i<n;i++)printf("\n\t P%d \t\t %d \t\t %d \t\t %d", p[i], bt[i], wt[i], tat[i]);printf("\nAverage Waiting Time--%f", wtavg/n);printf("\nAverage Turnaround Time--%f", tatavg/n); getch();}NOTE: "Exekveringstid(ms)" means execution time or burst time in milliseconds Assume that a system with a CPU at one time has the following processes queued in the CPU's READY queue: Calculate the waiting time for the seven processes: 1) If FCFS is being used 2) If LRU is being used 3) If RR with time quantum of 8msImagine that you have implemented an algorithm in a program which has 40% parallelizable code. Now consider the following two cases assuming that scheduling overhead is ignored: -The program is executed on a single processor. -The program is executed on a parallel system with 4 processors. Compute the speed-up gained in the case where multiple processors are used.
- Consider Context Switch time of 2 secs and modify below program accordingly. SJF CPU SCHEDULING ALGORITHM #include<stdio.h> #include<conio.h> using namespace std; int main() { int p[20], bt[20], wt[20], tat[20], i, k, n, temp; float wtavg, tatavg; printf("\nEnter the number of processes--"); scanf("%d", &n); for(i=0;i<n;i++) { p[i]=i; printf("Enter Burst Time for Process %d--", i); scanf("%d", &bt[i]); } for(i=0;i<n;i++) for(k=i+1;k<n;k++) if(bt[i]>bt[k]) { temp=bt[i]; bt[i]=bt[k]; bt[k]=temp; temp=p[i]; p[i]=p[k];…Under what circumstances does Round Robin Scheduling provide the lowest CPU usage rate?In a timesharing OS we have the following cpu timeline for two tasks X and Y. The timeslice is 1s.Both tasks are available in the system at the same time t=9:00:00.000 and order of arrival is the obviousX followed by Y. (The decimals reflect milliseconds if they showup in an indicated time reference.) There are no other processes (tasks) in the system other than X,Y. 1234567890 XYXYX--YXY The time line 1 indicates that at t=9:00:00s task X starts its execution and when t=9:00:01s is reached task Y takes over. The time line 1 indicates the 'first second' and time line 0 indicates the 'tenth second' above.Task Y completes its execution at t=9:00:10s, the completion of the tenth second since X started its execution. Task X has completed its execution earlier. (a) What is the total number of context switches starting from prior to t=9:00:00s (e.g. t=8:59:59.999) through thecompletion of $Y$? answer is 16s (b) What is the turnaround time for Y? answer is 10s (c) What is…
