UGC NET CSE | November 2017 | Part 3 | Question: 49

Question

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

$\begin{array}{ccc} \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time} \\ P_1 & 0 & 8 \\ P_2 & 1 & 4 \\ P_3 & 2 & 9 \\ P_4 & 3 & 5 \end{array}$

The average waiting time for preemptive SJF scheduling algorithm is ____________

• A. $6.5$ ms
• B. $7.5$ ms
• C. $6.75$ ms
• D. $7.75$ ms

Answer 1

GANTT CHART

t=0      t=1               t=5                 t=10                    t=17           t=26  

P1

P2

P4

P1

P3

 

$Process$

$AT$

$BT$

TAT= CT-AT

WT=TAT-BT

P1	0	8	17 – 0 =17	9
P2	1	4	5 – 1=4	0
P3	2	9	26 – 2=24	15
P4	3	5	10 – 3 =7	2

 

AVG WT=  (9+0+15+2)/4=  6.5 ms