Question

Q.2] Answer the following questions

Process	Burst Time	Priority
P1	3	1
P2	8	3
P3	2	4
P4	4	5
P5	5	1

1. (21 points) Consider the set of processes shown in the table above, with the length of the CPU-burst time given in milliseconds. The processes are assumed to have arrived in the order P5, P4, P3, P2 , and P1, all approximately at time 0. Draw three Gantt charts illustrating the execution of these processes using SJF, Round-Robin (quantum size = 3), and a non-preemptive priority (a smaller priority number implies a higher priority) scheduling. Calculate average waiting time of each scheduling. Calculate average turnaround time of each scheduling.

2. (1 points) What is convoy effect?

Answer 1

2) WT - TAT-BT Noless TOTCT-AT AI RT (1 3 5 22-3=14 22-0-22 22 8. O ܕܐ 2-2-0 2-0-2 Г. 2 O 2 9-4-5 P. 9-0 = 9 9 9 J IS 14-5 =9. 14 14-0 =14 ال Gantt charti PS P3 AN Pu Pi 22. 2 5 AT= Arrival time ; BT = Burst time; CT: Completion Time TAT= Tourn Around Time. 5 +22+ 2+ atly 52 Ang. TATO 522 10.4 ms. 5 2+14+ 0 +5+9 А Aug WT - ulu = 6.my 5 CS Scanned with CamScanner

1:) Round Robin ' ( Time Quantam =3 wT = TAT-BT. Process TAT = (T-AT Т AT BT CT 11 14 14 P. 14 22 22 22 8 6 8 8 8 P AM 2 13 Pu 17 니 17 Il 5 16 16 Ps o gantt chart? Ps Pu PP pl Ps / Pu 1 P₂ P2 3 6 8 ll 14 16 17 20 22. 1 =15.ung. Tag = 14 +22+8 + 19+16 77 5 16 5 Aug 5 55 11+14+ 6+13+11 WT = ! - limy. . Aug 5 5 CS Scanned with CamScanner

(it) non- preemptive Priority WT CT TAT Process Rienity BT 5 8 8 1 Pi 3 8 16 16 8 3 P₂ 16 18 18 2 4 wo 18 22 22 5 Pu 4 5 O 1 a 5 s Ps gantt chartin is all lp ) Pu 22 5 8 16 18 69 13.8 my 8 + 16+ 18 t 22+5 11 1) TATE Avg S 5+ 8 + 16+ 1870 니다 5 5 9.4 my. = Aug WT= 5 Some process b) Convoy effect have to wait for long in ready queue for bigger process to Convery effect It is observed time is relase CPU. this known as in FCFS. CS Scanned with CamScanner
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