Question

Design Principles Canvas Ā 8 pts Question 35 Five batch jobs, A, B, C, D, and E, arrive at a computer system at time 0, 1, 2, 3, 4, respectively. And they have service time of 7.1. 3,5, and 4, respectively. The new jobs arrive a little ahead of the expiration of time slices. For each of the following scheduling algorithms, determine the turnaround time for each process for all jobs. Ignore process switching overhead. • Round robin with a time quantum of 2 units. The new processes arrive before the time quantum expires. • Feedback: preemptive, multilevel feedback, when a process has time quantum expires, OS will feed it back into a lower- priority queue, with time quantum extended to 2 • SRT: Shortest remaining time, a preemptive algorithm (preemption points are whenever a new process arrives) • HRRN: Highest response ratio next: non-preemptive Secondary Scheduling Policy: FCFS, if the first policy cannot identify the next process, FCFS will be used Determine the turnaround time T, E D C | В A RR, q=2 FCFS TO SRT TI HRRNT MacBook Pro Question 36 ya $ esc # 3 4 2 R. Е. W

Answer 1

Question - 35 process А BT 7 - Flo-oom I 2 3 G D E 4 Round Robin - Tq=2 Sum = 42 Turn around time Sum= 62 Average waiting time = 8.40 Average Jurnaround time 12.40 waiting time FOES :- waiting Time Sum= 32 Turn around time Sum=52 Average waiting time = 6,40 Average Turnaround time = 10.40 SRTF:- Turnaround time Sum = 40 Average TAT = 8 Average WT = 4. 40 YEARS OF EXCELLENCE classmate

2 HORN :- (non -premptive) Turnaround time Sum= 51 waiting time Sum= 31 Average Average TAT= 10:20 WT - 6,20 classmate

Arr_time[1] : 0 Arr_time[2] : 1 Arr_time[3] : 2 Arr_time[4] : 3 Arr_time[5] : 4 Burst_time[1] : 7 Burst_time[2] : 1 Burst_time[3] : 3 Burst_time[4] : 5 Burst_time[5] : 4 Enter the time quantum : 2 ROUND ROBIN BT СТ АТ 1 2 3 1 2 3 4 1 3 5 4 TAT 19 3 12 20 18 WT 19 2 10 17 14 4. Sum of Completion time : 72.00 Sum of Waiting time: 42.00 Sum of Turnaround time: 62.00 Average Completion time : 14.40 Average Waiting time : 8.40 Average Turnaround time : 12.40 Process returned 0 (0x0) execution time : 15.345 s Press any key to continue. Activate Windo

Enter the no.of processes : 5 Arr_time[1] : 0 Arr_time[2] : 1 Arr_time[3] : 2 Arr_time[4] : 3 Arr_time [5] : 4 Burst_time[1] : 7 Burst_time[2] : 1 Burst_time[3] : 3 Burst_time[4] : 5 Burst_time[5] : 4 FCFS AT WT BT 7 1 PNO P1 P2 P3 P4 P5 CT 7 8 11 16 20 TAT 7 7 9 13 16 2 3 4 3 5 4 6 6 8 12 Sum of turnaround time : 52.00 Sum of waiting time: 32.00 Average Waiting time : 6.40 Average Turnaround time : 10.40 execution time : 50.059 s Process returned o (0x0) Press any key to continue. Activate Windows Go to Settings to activate Wil

Enter the number of Processes: 5 Arr_time[1] : 0 Arr_time[2] : 1 Arr_time [3] : 2 Arr_time[4] : 3 Arr_time[5] : 4 Burst_time[1]: 7 Burst_time[2]: 1 Burst_time[3]: 3 Burst_time[4]: 5 Burst_time[5]: 4 SRTE AT BT TAT 20 WT 13 0 1 pid 1 2 3 4 5 1 2 3 4 1 3 5 4 CT 20 2 5 14. 9 3 11 5 6 1 Average waiting time = 4.000000 Average Turnaround time = 8.000000 Process returned 34 (0x22) execution time : 13.134 s Press any key to continue. Activate Windows Go to Settings to activate

Enter the no.of processes : 5 Arr_time[1] : 0 Arr_time[2] : 1 Arr_time[3] : 2 Arr_time[4] : 3 Arr_time[5] : 4 Burst_time[1] : 7 Burst_time [2] : 1 Burst_time[3] : 3 Burst_time[4] : 5 Burst_time [5] : 4 HRRN AT TAT WT BT 7 1 PNO P1 P2 P3 P4 P5 1 2 4 3 СТ 7 8 11 15 20 3 4 5 9 11 17 6 6 7 12 Sum of turnaround time : 51.00 Sum of waiting time: 31.00 Average Waiting time : 6.20 Average Turnaround time : 10.20 Process returned O (Oxo) execution time : 23.053 s Press any key to continue. Activate Windows Go to Settings to activate