Question

Consider the following set of processes, with the arrival times and the CPU-burst times given in milliseconds

Process	Arrival time	Burst time
P1 P2 P3 P4	0 1 2 4	5 3 3 1

What is the average turnaround time for these processes with the preemptive Shortest Remaining Processing Time first (SRPT) algorithm?

• A. 6.25
• B. 5.50
• C. 6.00
• D. 5.75

Answer 1

The correct option is B 5.50

Process Arrival Time Burst Time

$P_1$ 0 5 $P_2$ 1 3 $P_3$ 2 3 $P_4$ 4 1

The Gantt chart for SRPT CPU scheduling algorithm is

$P_1$

$P_2$

$P_2$

$P_2$

$P_4$

$P_3$

$P_3$

$P_3$

$P_1$

$P_1$

$P_1$

$P_1$

0 1 2 3 4 5 6 7 8 9 10 11 12

Process $P_1$ arrive at time 0, at time unit 1 the remaining time of $P_1$ is 4 and remaining time of $P_2$ is 3 so $P_1$ preempted. At the end of time unit 2 the remaining time of $P_2$ is 2 and $P_3$ is 3 so $P_2$ is not preempted At time unit 4 the remain time of $P_2$ is 0 and $P_4$ is scheduled. The same process repeated upto time unit 12.
The turnaround time is the interval between submission and completion of a given process. Calculate turnarround time for each process as

$P_1=12-0=12$
$P_2=4-1=3$
$P_3=6,P_4=1$

So average turnaround time
$=\frac{12+3+6+1}{4}=\frac{22}{4}=5.50$