Question

Consider the following four processes with the arrival time and length of CPU burst given in milliseconds :

Process	Arrival Time	Burst Time
P1	0	8
P2	1	4
P3	2	9
P4	3	5

 

The average waiting time for preemptive SJF scheduling algorithm is

1. 6.5 ms
2. 7.5 ms 
3. 6.75 ms
4. 7.75 ms

Answer 1

Correct Answer - Option 1 : 6.5 ms

The correct answer is option 1.

CONCEPT:

Shortest Job First:

In the SJF scheduling algorithm, the process with the lowest burst time, among the list of available processes in the ready queue, is going to be scheduled next. Preemptive SJF scheduling is the shortest remaining time first

Formula:

TAT=CT-AT

WT=TAT-BT

Gantt Chart:

P1

P2

P4

P1

P3

0                      1                        5                      10                      17                    26             

Process	Arrival Time(AT)	Burst Time(BT)	Completion Time(CT)	TAT=CT-AT	WT=TAT-BT
P1	0	8	17	17	9
P2	1	4	5	4	0
P3	2	9	26	24	15
P4	3	5	10	7	2

 

Average waiting time= \(\frac{(9+0+15+2) }{4}\)=6.5

∴ Hence the correct answer is 6.5.