GATE - 2011 | OS | Consider the following table of arrival time and burst time for three processes
Posted on 13-02-2022
GATE - 2011 [Operating System]
Question:
Consider the following table of arrival time and burst time for three processes P0, P1 and P2.
The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?
|
A |
5.0 ms |
|
B |
4.33 ms |
|
C |
6.33 ms |
|
D |
7.33 ms |
Solution:
Option (A) is Correct.
CT = Completion time
TAT = Turn Around Time
WT = Waiting Time
TAT = CT - AT
WT = TAT - BT
Gantt Chart using pre-emptive shortest job first scheduling algorithm,
Avg. WT = 4+0+11/3 = 5ns