Question

An operating system uses shortest remaining time first scheduling algorithm for pre-emptive scheduling of processes. Consider the following set of processes with their arrival times and CPU burst times (in milliseconds):

Process	Arrival Time	Burst Time
P1	0	12
P2	2	4
P3	3	6
P4	8	5

The average waiting time (in milliseconds) of the process is _____

Answer 1

Gantt chart:

Process	P1	P2	P2	P3	P3	P4	P1
Time(AT to CT )	0 to 2	2 to 3	3 to 6	6 to 8	8 to 12	12 to 17	17 to 27

Calculating average waiting time:

Process	Arrival time (AT)	Burst Time (BT)	Completion time (CT)	Turnaround time (TAT)	Waiting Time (WT)
P1	0	12	27	27	15
P2	2	4	6	4	0
P3	3	6	12	9	3
P4	8	5	17	11	4

 

Average waiting time \(= \frac{{15\; + \;0\; +\; 3\; + \;4}}{4} = \frac{{22}}{4} = 5.5\)

Hence 5.5 is the correct answer.

Important Point:

TAT = CT – AT

WT = TAT – BT