Discussion
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 turn around time of these processes will be
Answer: B
Round robin with a time slice of 4 ms gives the order A (0-4), B (4-8), C (8-12), D (12-16), A (16-20), C (20-24), D (24-25) and C (25-26). B finishes at 8, A at 20, D at 25 and C at 26.
\[ \text{turnaround} = \text{finish} - \text{arrival} \]
\[ A: 20,\; B: 7,\; C: 24,\; D: 22 \]
\[ \frac{20 + 7 + 24 + 22}{4} = 18.25 \text{ ms} \]
This is option (B). Options (A), (C) and (D) come from a different order of the queue or from using waiting time in place of turnaround time.
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