UGC NET CSE | January 2025 | Part 2 | Question: 96

Question

Consider the following set of processes with the arrival time and length of CPU Burst time given in milli-seconds ($\mathrm{ms}$):

$\begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time} \\ \hline P_1 & 0 & 5 \\ \hline P_2 & 2 & 3 \\ \hline P_3 & 2 & 2 \\ \hline P_4 & 5 & 3 \\ \hline P_5 & 6 & 1 \\ \hline \end{array}$

What is the average turnaround time for these Processes with First Come First Serve (FCFS) scheduling Algorithm?

• A. $5.3$ $\mathrm{ms}$
• B. $6.4$ $\mathrm{ms}$
• C. $7.0$ $\mathrm{ms}$
• D. $8.2$ $\mathrm{ms}$

Answer 1

P1

P2

P3

P4

P5

       5      8     10   13     14

completion time for 

p1-5

p2-8

p3-10

p4-13

p5-14

Formula for TurnAroundTime=Completion Time- Arrival Time

TurnAroudTime for P1,P2,P3,P4,P5 are 5, 6, 8, 8, 8 respctively

average turnaround time=(5+6+8+8+8)/5=35/5=7

Therefore the answer is 7

OPTION C