Answer

Note: Question states “four processes” but lists five (P1–P5). Solution is given for all five as per table.

Time quantum $q = 3\,ms$. Arrival Times (AT) and Burst Times (BT) as given.

Process	AT	BT
P1	0	12
P2	0	10
P3	1	4
P4	4	10
P5	2	12

Assumptions

• Arrival order for same AT: P1 before P2.
• New arrivals during a quantum are enqueued before the preempted process is appended to the ready queue (standard RR).
• Context switch overhead = 0.

Step-by-Step Execution

Time	Running	Ready Queue after slice	Remarks
0–3	P1 (12→9)	[P2, P3, P5, P1]	P3 (1), P5 (2) arrived
3–6	P2 (10→7)	[P3, P5, P1, P4, P2]	P4 (4) arrived
6–9	P3 (4→1)	[P5, P1, P4, P2, P3]
9–12	P5 (12→9)	[P1, P4, P2, P3, P5]
12–15	P1 (9→6)	[P4, P2, P3, P5, P1]
15–18	P4 (10→7)	[P2, P3, P5, P1, P4]
18–21	P2 (7→4)	[P3, P5, P1, P4, P2]
21–22	P3 (1→0) completes at 22	[P5, P1, P4, P2]	Quantum truncated to 1
22–25	P5 (9→6)	[P1, P4, P2, P5]
25–28	P1 (6→3)	[P4, P2, P5, P1]
28–31	P4 (7→4)	[P2, P5, P1, P4]
31–34	P2 (4→1)	[P5, P1, P4, P2]
34–37	P5 (6→3)	[P1, P4, P2, P5]
37–40	P1 (3→0) completes at 40	[P4, P2, P5]
40–43	P4 (4→1)	[P2, P5, P4]
43–44	P2 (1→0) completes at 44	[P5, P4]
44–47	P5 (3→0) completes at 47	[P4]
47–48	P4 (1→0) completes at 48	[]

Gantt Chart

0   3   6   9  12  15  18  21 22  25  28  31  34  37  40  43 44  47 48
|P1 |P2 |P3 |P5 |P1 |P4 |P2 |P3|P5 |P1 |P4 |P2 |P5 |P1 |P4 |P2|P5 |P4|

Mermaid:

Calculation Table

PID	AT	BT	CT	TAT = CT−AT	WT = TAT−BT
P1	0	12	40	40	28
P2	0	10	44	44	34
P3	1	4	22	21	17
P4	4	10	48	44	34
P5	2	12	47	45	33
Sum		48		194	146

Final Answer

• Average Turn-Around Time $= \dfrac{194}{5} = \mathbf{38.8\,ms}$
• Average Waiting Time $= \dfrac{146}{5} = \mathbf{29.2\,ms}$

If strictly only four processes are considered (excluding P5), CTs are P1=28, P2=32, P3=10, P4=36 → Avg TAT = 26.5 ms, Avg WT = 18.5 ms (quantum 3).