Question

Consider the following set of processes, with the length of the CPU burst given in milliseconds:

Process	Burst Time	Priority
P1	5	4
P2	3	1
P3	1	2
P4	7	2
P5	4	3

The processes are assumed to have arrived in the order P1 , P2 , P3 , P4 , P5 , all at time 0. 

a. Draw four Gantt charts that illustrate the execution of these processes using the following scheduling algorithms: FCFS, SJF, nonpreemptive priority (a larger priority number implies a higher priority), and RR (quantum = 2).

 b. What is the turnaround time of each process for each of the scheduling algorithms in part a?

 c. What is the waiting time of each process for each of these scheduling algorithms? 

d. Which of the algorithms results in the minimum average waiting time (over all processes)?

> please try to give the answer as soon as possible with in 2 hrs. thanks in advance.

Answer 1

FCFS Rorace Gantt chart Pi P2 5 P3 Pч | Ps ។ 16 20 Waiting Turnaround time time Process Arrival Burst time P, Completion Time 5 time 0 5 8 3 8 8 g 1 Le تم م م 0 0 0 0 16 ។ 16 it I 20 16 4 PS 20 Turn around time- completion Arrival time Time Turn around time waiting time Average tomaround time = Buerst time 5+8+9+16+20 5 - 11.6 - 58 5 time = o+5+8+9+16 Average waiting - 7.6 - 33 38 5

S JF Gantt chant Į Ps ! P2 PSIP, IP 4 8 13 20 waiting Porocess Arrival Bwist tme completion Torin around time time time time 8. P, 13 5 13 P2 o 1 4 니 2) - w マク २५ 7 20 20 20 i – i o 13 8 J འཤ Average turn around time = = 13+ ultit 2018 5 2 에 동 = g. 2 Averge 8 titot 13th waiting time 3 5 = 26 26 s = 5.2

Non-preemptive polarity Chantt chant Ps P3 Ipy Pi o P2 17 )0 9 20 5 Turnaround / waiting completion Porocess seronity Arrival time Burst time time time time 5 o 2) 5 5 Pi 0 17 가 20 P2 20 3 jo 10 1 2 ) it or own 7 2 17 10 17 2) 5 9 3 g Ps 5+20 tot 17tg 1) time Averge turin caround = اي up 5 12.2 time 11 Averge walting ot 17t9t lots 5 41 5 8-2

Round Robin (quantom=2) Queve: - D. / / / / 4 / 54 / X PARA Gantt chart P, P2 | P₂ Py PSIP, 2 4 5 7 9 P2) Pyl Pol P, Pylly) 12 14 16 17 19 20 Process Arrival time Burst time 5 completion Turnaround time time 17 1 1카 Waiting P time 72 32 3 12 12 or 23 ) 5 N in a d 4 니 Py 7 20 20 13 Ps O 4 16 16 12 Averge turn around time 2 17+12+5420416 5 70 70 5 = = 14 Average = = 12tg + 2+ 3+ 12 Waiting time - = - d) 쁘 SJ have time. minimum average waiting