0
1.6kviews
Consider the following set of processes, with the length of the CPU burst given in milliseconds

Draw Gantt chart that illustrate the execution of these processes using Round-Robin algorithm with time quantum of 5 milliseconds.

enter image description here

2 Answers
2
75views

Gantt Chart

Gantt Chart

Waiting Period

enter image description here

Average Waiting Period:- =(15+20+25+29)/4
= (89)/4
= 22.25

2
35views

Round Robin Scheduling

Turn Around Time = Process Completion Time – Process Arrival Time

Waiting time = Turn Around time – Burst time

Round Robin

Process Arrival Time Burst Time Completion Time Turn Around Time Waiting Time
P1 0 10 25 25 – 0 = 25 25 – 10 = 15
P2 1 9 29 29 – 1 = 28 28 – 9 = 19
P3 2 12 37 37 – 2 = 35 35 – 12 = 23
P4 3 6 35 35 – 3 = 32 32 – 6 = 26

Average Turn Around Time = (25 + 28 +35 + 32)/4 = 30 ms

Average Waiting time = (15 + 19 + 23 + 26)/4 = 20.75 ms

Please log in to add an answer.