Consider the following set of processes, with the length of the CPU-burst and arrival time given in milliseconds:

Process ID	Arrival Time	Bust Time	Turn around Time
P1	0	1	1
P2	2	2	2
P3	2	2	4
P4	5	3	4
P5	7	4	6
P6	10	6	9

What are the average turn around time when round-robin (time quantam – 3 unit) and first come first serve (FCFS) CPU Scheduling algorithm?

(A) Round-Robin – 3.6, FCFS – 3.5
(B) Round-Robin – 4, FCFS – 4
(C) Round-Robin – 4.3, FCFS – 3.5
(D) Round-Robin – 4.3, FCFS – 4.3


Answer: (D)

Explanation: Since arrival time of every process are 0, so every processes available initially.
When the burst time of all jobs is less than time quantum and arrival time is also very early(No waiting in ready queue) then Round robin exactly behave like FCFS scheduling.
According to round-robin processes are scheduled in CPU accordingly:

Turn around time = Completion time – arrival time.

Process ID	Arrival Time	Bust Time	Turn around Time
P1	0	1	1
P2	2	2	2
P3	2	2	4
P4	5	3	4
P5	7	4	6
P6	10	6	9

So, option (D) is correct.


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