GATE IT 2005 | Question: 60

Question

We wish to schedule three processes $P1$, $P2$ and $P3$ on a uniprocessor system. The priorities, CPU time requirements and arrival times of the processes are as shown below.
$$\begin{array}{|c|c|c|c|} \hline \textbf{Process} & \textbf{Priority} & \textbf{CPU time} & \textbf{Arrival time}\\
&  & \textbf{required} & \textbf{(hh:mm:ss)}
\\\hline \text{P1} & \text{10 (highest)} & 20\text{ sec} &  00:00:05 \\\hline \text{P2} & 9 & 10 \text{ sec}&  00:00:03 \\\hline \text{P3} & \text{8 (lowest)} & 15 \text{ sec}&  00:00:00 \\\hline  \end{array}$$

We have a choice of preemptive or non-preemptive scheduling. In preemptive scheduling, a late-arriving higher priority process can preempt a currently running process with lower priority. In non-preemptive scheduling, a late-arriving higher priority process must wait for the currently executing process to complete before it can be scheduled on the processor.
What are the turnaround times (time from arrival till completion) of $P2$ using preemptive and non-preemptive scheduling respectively?

• A. $30$ sec, $30$ sec
• B. $30$ sec, $10$ sec
• C. $42$ sec, $42$ sec
• D. $30$ sec, $42$ sec

Answer 1

Answer will be (D).

TAT = Completion Time - Arrival Time.

The Gannt Chart for Non Preemptive scheduling will be $(0)P3,(15)P1,(35)P2(45).$

From above this can be inferred easily that completion time for $P2$ is $45$, for $P1$ is $35$ and $P3$ is $15.$

Gantt Chart for Preemptive-  $(0)P3,(1)P3,(2)P3,(3)P2,(4)P2,(5)P1,(25)P2,(33)P3(45)$.

Similarly take completion time from above for individual processes and subtract it from the Arrival time to get TAT.

Comments

The table can be modified before solving as 

	Priority	CPU TIME	AT
P1	10	20	5
P2	9	10	3
P3	8	15	0

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How is p1 getting executed before p2 as its arrival time is 5?

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because when p3 finished, it was at sec 15. both p1 and p2 were waiting and p1 has higher priority than p2 to be executed before p2

Answer 2

BEST ANSWER

Answer 3

Explanation