Question
Download Solution PDFConsider the following set of processes and the length of CPU burst time given in milliseconds:
|
Process |
CPU Burst time (ms) |
|
P1 |
5 |
|
P2 |
7 |
|
P3 |
6 |
|
P4 |
4 |
Assume that processes being scheduled with Round-Robin Scheduling Algorithm with Time Quantum 4ms. Then the waiting time for P4 is _________ ms.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFScheduling Algorithm: Round-Robin
Time Quantum: 4 milliseconds
Gantt chart:
|
P1 |
P2 |
P3 |
P4 |
P1 |
P2 |
P3 |
0 4 8 12 16 17 20 22
Process Table:
|
Process |
Arrival Time (AT) |
Burst Time (BT) |
Completion Time (CT) |
Turnaround Time (TAT) |
Waiting Time (WT) |
|
P1 |
0 |
5 |
17 |
17 |
12 |
|
P2 |
0 |
7 |
20 |
20 |
13 |
|
P3 |
0 |
6 |
22 |
22 |
16 |
|
P4 |
0 |
4 |
16 |
16 |
12 |
Therefore, the waiting time for P4 is 12 milliseconds.
Formula used:
TAT = CT – AT
WT = TAT – BT
Important Points:
Since arrival time is not given of processes is not given consider all the process arrive at time = 0 and by convention, the process with lowest id associated with it executes first, that is, P1 followed by P2 and so on.
Last updated on Jul 7, 2025
-> The UGC NET Answer Key 2025 June was released on the official website ugcnet.nta.ac.in on 06th July 2025.
-> The UGC NET June 2025 exam will be conducted from 25th to 29th June 2025.
-> The UGC-NET exam takes place for 85 subjects, to determine the eligibility for 'Junior Research Fellowship’ and ‘Assistant Professor’ posts, as well as for PhD. admissions.
-> The exam is conducted bi-annually - in June and December cycles.
-> The exam comprises two papers - Paper I and Paper II. Paper I consists of 50 questions and Paper II consists of 100 questions.
-> The candidates who are preparing for the exam can check the UGC NET Previous Year Papers and UGC NET Test Series to boost their preparations.