Question

The arrival time, priority, and durations of the CPU and I/O bursts for each of three processes $P_1$, $P_2$ and $P_3$ are given in the table below. Each process has a CPU burst followed by ab I/O burst followed by another CPU burst. Assume that each prcess has its own I/O resource.

Process	Arrival time	Priority	Burst durations CPU, I/O CPU
$P_1$	0	2	1,5,3
$P_2$	2	3 (lowest)	3,3,1
$P_3$	3	1 (highest)	2,3,1

The multi-programmed operating system uses preemptive priority scheduling. What are the finish times of the processes $P_1$ $P_2$ and $P_3$ ?

• A. 12, 17, 11
• B. 11, 16, 10
• C. 11, 15, 9
• D. 10, 15, 9

Answer 1

The correct option is D 10, 15, 9

At t = 0 only P1 is present so execute it for 1 unit (remaining CPU time of P1 = 0, now it will perform I/O).
At t = 1 no process is available for CPU so CPU will remain idle and P1 will perform I/O
(remaining I/O time of P1 = 4)
At t = 2 P2 arrives and is available for CPU so execute it for 1 unit and simultaneously P1 will do I/O (remaining CPU time of P2=2 and remaining I/O time of P1 = 3).

At t = 3 P3 also arrives and it is having highest priority among all processes available for CPU so we can execute P3 for its complete CPU burst (i.e. 2 unit) as all processes has arrived and simultaneously P1 will do I/O (remaining CPU time of P3 = 0, now P3 will perform I/O & remaining I/O time of P1 = 1).

At t = 5 only P2 is available for CPU so it will execute for 1 unit and P1 and P3 will perform I/O (remaining CPU time of p2=1, remaining I/O time of P3 = 2, remaining I/O time of P1 = 0, now P1 will perform CPU)

At t = 6 P2 and P1 are available for CPU but P1 will be selected because it is having highest priority then P2 so execute P1 for 1 unit (remaining CPU time of P1 = 2 & remaining I/O time of P3 is 1)

At t = 7 again P1 will be executed for 1 unit (remaining CPU time of P1 = 1 and remianing I/O time of P3 = 0, now P3 will perform CPU).

At t = 8 now P1, P2 & P3 are available for CPU so P3 will be selected based on highest priority and will be executed for 1 unit (remaining CPU time of P3 = 0, so P3 completed at t = 9).

At t = 9 P1 will be executed on CPU for 1 unit (remaining CPU time of P1 = 0, so P1 completed at t = 10).

At t = 10 only P2 is available for CPU so it will execute its remaining CPU burst (i.e. 1 unit) (remaining CPU time of P2 = 0, now P2 will perform I/O.)

At t = 11 P2 will perform I/O for 3 units and CPU will remain IDLE.

At t = 14 P2 will perform CPU again and P2 is completed at t = 15

P1	IDLE	P2	P3	P2	P1	P1	P3	P1	P2	IDLE	P2
0	1	2	3	5	6	7	8	9	10	11	14 15