Question 9390 – Heap-Tree
December 13, 2023UGC NET CS 2016 July- paper-2
December 13, 2023Process-Scheduling
| Question 16 |
Consider the following processes, with the arrival time and the length of the CPU burst given in milliseconds. The scheduling algorithm used is preemptive shortest remaining-time first.
The average turn around time of these processes is _________ milliseconds.
| 8.25 | |
| 8.26 | |
| 8.27 | |
| 8.28 |
Question 16 Explanation:
Here the scheduling algorithm used is preemptive shortest remaining-time first.
To answer the question we need to design the gantt chart:
In this algorithm, the processes will be scheduled on the CPU which will be having least remaining burst time.
Turnaround Time (TAT) = Completion Time (CT) – Arrival Time (AT)
TAT for P1 = 20 – 0 = 20,
TAT for P2 = 10 – 3 = 7,
TAT for P3 = 8 – 7 = 1,
TAT for P4 = 13 – 8 = 5.
Total TAT = 20 + 7 + 1 + 5 = 33 / 4 = 8.25 (Avg. TAT)
To answer the question we need to design the gantt chart:
In this algorithm, the processes will be scheduled on the CPU which will be having least remaining burst time.
Turnaround Time (TAT) = Completion Time (CT) – Arrival Time (AT)
TAT for P1 = 20 – 0 = 20,
TAT for P2 = 10 – 3 = 7,
TAT for P3 = 8 – 7 = 1,
TAT for P4 = 13 – 8 = 5.
Total TAT = 20 + 7 + 1 + 5 = 33 / 4 = 8.25 (Avg. TAT)
Correct Answer: A
Question 16 Explanation:
Here the scheduling algorithm used is preemptive shortest remaining-time first.
To answer the question we need to design the gantt chart:
In this algorithm, the processes will be scheduled on the CPU which will be having least remaining burst time.
Turnaround Time (TAT) = Completion Time (CT) – Arrival Time (AT)
TAT for P1 = 20 – 0 = 20,
TAT for P2 = 10 – 3 = 7,
TAT for P3 = 8 – 7 = 1,
TAT for P4 = 13 – 8 = 5.
Total TAT = 20 + 7 + 1 + 5 = 33 / 4 = 8.25 (Avg. TAT)
To answer the question we need to design the gantt chart:
In this algorithm, the processes will be scheduled on the CPU which will be having least remaining burst time.
Turnaround Time (TAT) = Completion Time (CT) – Arrival Time (AT)
TAT for P1 = 20 – 0 = 20,
TAT for P2 = 10 – 3 = 7,
TAT for P3 = 8 – 7 = 1,
TAT for P4 = 13 – 8 = 5.
Total TAT = 20 + 7 + 1 + 5 = 33 / 4 = 8.25 (Avg. TAT)
1 Comment
Answer is wrong….check it and update it…