CPU scheduler will decide which process should be given the CPU for its execution. For this it uses different algorithm to choose among the process. One among that algorithm is FCFS algorithm. In this algorithm the process which arrives first is given the CPU after finishing its request only it will allow the CPU to execute another process. Write a c program to implement the first come first serve CPU scheduling algorithm Sample output: Enter number of processes:3 Enter the burst times: Gantt chartPi P2 P3 Process Burst Time Waiting Time Turnaround Time 8 P1 12 P3 0 13 Average Waiting Time is 7.000000 Average Turn around Time is 14.333333 22

CPU scheduler will decide which process should be given the CPU for its execution. For this it uses different algorithm to choose among the process. One among that algorithm is FCFS algorithm. In this algorithm the process which arrives first is given the CPU after finishing its request only it will allow the CPU to execute another process. Write a c program to implement the first come first serve CPU scheduling algorithm Sample output: Enter number of processes:3 Enter the burst times: Gantt chartPi P2 P3 Process Burst Time Waiting Time Turnaround Time 8 P1 12 P3 0 13 Average Waiting Time is 7.000000 Average Turn around Time is 14.333333 22

Name: CPU scheduler will decide which processshould be given the CPU for it
Uploaded: 2024-03-20T06:51:23.000Z
Channel: Bartleby
Description: CPU scheduler will decide which processshould be given the CPU for itsexecution. For this it uses differentalgorithm to choose among the process.One among that algorithm is FCFSalgorithm. In this algorithm the processwhich arrives first is given the

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

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();}
In this project, you'll implement one of the following CPU scheduling algorithms by writing a CPU simulator. Shortest Job First (SJF)(Non-preemptive)The shortest job first algorithm is a priority based scheduling algorithm that associates with each process the length of the process's next CPU burst. When CPU is available, it is assigned to the process that has the smallest next CPU burst. Shortest Remaining Time Next (SRTN)(Preemptive)The shortest remaining process next scheduling algorithm is the preemptive Shortest Job First algorithm. With this scheduling algorithm, the process in the ready queue with the shortest execution time is chosen to execute. If a "new process" arrives in the ready queue with a CPU service time less than the remaining time of the current process, preempt. First Come First Serve (FCFS) (recommended)The first come first serve algorithm is simplest CPU-scheduling algorithm. In this algorithm, the process at the head of the queue is allowed to execute until it…
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();}
Given the list of processes, their CPU burst times, arrival times and priorities implement SJF,Priority and Round Robin scheduling algorithms on the processes with preemption. For each ofthe scheduling policies, compute and print the completion Time(CT), Turnaround Time(TAT),and Waiting Time(WT) for each process using C Programming.Waiting time: Processes need to wait in the process queue before execution starts and inexecution while they get preempted. Turnaround time: Time elapsed by each process to get completely served. (Difference betweensubmission time and completion time).
In executing Jobs A and B using the priority scheduling algorithm, what will happen ifboth jobs have:a. unequal priorities but equal CPU cycles?b.equal priorities but different CPU cvcles?
In executing Jobs A and B using the priority scheduling algorithm, what will happen ifboth jobs have:a. unequal priorities but equal CPU cycles?b. equal priorities but different CPU cvcles?
The ready queue is subdivided into many queues by the multilevel scheduling algorithm, and each of these queues has its own scheduling technique or policy. Choose one: Is it accurate or erroneous? The time a process spends waiting should be minimised by the CPU scheduler. Choose one: OFalse O True
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];…
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…
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();}
When this option is available, the I/O queue should be prioritised above the process execution queue. In what ways may this be beneficial? Who knows what will take place if there is a disruption in the I/O. It is unclear at this time if this will have an effect on the maximum burst rate of the CPU. I'm not really clear on the point you're trying to make here.
Let the processes P1, P2, P3, and P4 be given. They arrive in the system at the same time in this order. The processes have the following service times (in time units): Process P1 P2 P3 P4 Service time (CPU burst) 53 17 68 24 For the scheduling method "Round Robin" (RR), specify the average execution time (average turnaround time) and the number of context switches. The time quantum q is set to 20 time units. You must show your calculations. You will not have to draw Gantt charts. Note: For RR, processes execute in order of their ID (i.e., P1-P2-P3-P4).