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UGC NET CSE | June 2013 | Part 3 | Question: 58

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Consider the following processes with time slice of 4 milliseconds (I/O requests are ignored):

Process A B C D
Arrival time 0 1 2 3
CPU Cycle 8 4 9 5

The average turnaround time of these processes will be

  1. 19.25 milliseconds
  2. 18.25 milliseconds
  3. 19.5 milliseconds
  4. 18.5 milliseconds
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Answer : 18.25 milliseconds

Process AT BT CT TAT
A 0 8 20 20
B 1 4 8 7
C 2 9 26 24
D 3 5 25 22
A B C D A C D C

0            4             8              12           16              20         24             25          26

Avg TAT = 20 + 7 + 24 + 22 / 4 = 18.25 

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Answer B

The system uses Round robin algorithm of time quantum = 4ms

| o A |4 B |8 C |12 D | 16 A |20 C|24| D |25 C |26

average turn around time = (Finished time- start time) of each process/ no of processes.

=(20-0)+(8-1)+(26-2)+(25-3) /4 = 18.25 ms

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