UGC NET CSE | December 2025 | Part 2 | Question: 31

Question

Consider the following set of processes assumed to have arrived at time $0$ in order $P_{1}, P_{2}, P_{3}, P_{4}$ and $P_{5}$ with the length of CPU burst given in Milliseconds.

Process	Burst Time	Priority
$P_1$	$10$	$3$
$P_2$	$1$	$1$
$P_3$	$2$	$4$
$P_4$	$1$	$5$
$P_5$	$5$	$2$

What is the average waiting time (in milliseconds) using priority scheduling.

• A. $8$
• B. $8.2$
• C. $8.3$
• D. $8.4$

Answer 1

$\text{Turnaround time:}$ The interval from the time of submission of a process to the time of completion is the turnaround time. Turnaround time is the sum of the periods spent waiting to get into memory, waiting in the ready queue, executing on the CPU,and doing I/O.

.$\boxed{\text{Turnaround time = Completion time - Arrival time}}$

$\text{Waiting time:}$ Waiting time is the sum of the periods spent waiting in the ready queue.

$\boxed{\text{Waiting time = Turnaround time - Burst time}}$