GATE CSE 2004 | Question: 46

Question

Consider the following set of processes, with the arrival times and the CPU-burst times gives in milliseconds.$$\small \begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time}\\\hline \text{P1} & 0 & 5\\ \text{P2} & 1 & 3\\ \text{P3} & 2 & 3 \\ \text{P4} & 4 & 1\\\hline  \end{array}$$What is the average turnaround time for these processes with the preemptive shortest remaining processing time first (SRPT) algorithm?

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

Comments

Here after the process p2 why p3 is not scheduled....and why p4 got scheduled

Answer 1

$$\small\begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Waiting Time = } & \textbf{Turnaround Time =} \\ & \textbf{(Turnaround Time - Burst time)} & \textbf{(Completion Time - Arrival Time )}\\\hline \text{P1} & 7 & 12 \\\hline \text{P3} & 0 & 3\\\hline \text{P2} & 3 & 6 \\\hline \text{P4} & 0 & 1\\\hline  \end{array}$$
Average turnaround time $=\frac{12+3+6+1}{4}=\frac{22}{4}=5.5$

Correct Answer: $A$

Comments

can someone just edit and put all TATs in bracket i.e (12+3+6+1)/4 .Thankyou👍

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@Lakshman Patel RJIT please look into what @PSDesai09 has mentioned.

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Fixed now 👍

Answer 2

Ans: Average Turnaround Time is : 5.5

Comments

AFTER GRANT CHART  WE FOUND TAT OF ALL GIVEN PROCESSE   

TAT OF P1=12       ,TAT OF P2=3        TAT OF P3= 6     TAT OF P4=1

SO AVG TAT =TATAL SUM OF TAT /TOTAL NO OF PROCESS

                          22/4=5.5

Answer 3

AFTER GRANT CHART  WE FOUND TAT OF ALL GIVEN PROCESSE   

TAT OF P1=12       ,TAT OF P2=3        TAT OF P3= 6     TAT OF P4=1

SO AVG TAT =TATAL SUM OF TAT /TOTAL NO OF PROCESS

                          22/4=5.5