KVS PGT DEC-2013 PART-A
September 25, 2024GATE 2019
September 25, 2024ISRO CS 2011
| Question 4 |
The following table shows the processes in the ready queue and time required for each process for completing its job.
If round-robin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
If round-robin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
| 27 ms | |
| 26.2 ms | |
| 27.5 ms | |
| 27.2 ms |
Question 4 Explanation:
→In the round robin algorithm, time slices (also known as time quanta) are assigned to each process in equal portions and in circular order, handling all processes without priority (also known as cyclic executive).
→Given scheduling algorithm is round robin with time quantum value is 5ms.
→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
→Given scheduling algorithm is round robin with time quantum value is 5ms.
→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
Correct Answer: B
Question 4 Explanation:
→In the round robin algorithm, time slices (also known as time quanta) are assigned to each process in equal portions and in circular order, handling all processes without priority (also known as cyclic executive).
→Given scheduling algorithm is round robin with time quantum value is 5ms.
→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
→Given scheduling algorithm is round robin with time quantum value is 5ms.
→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2