Solution

The correct answer is option 4.

CONCEPT:

Shortest Job First (non-preemptive): Among the processes present in the ready queue at any instant, the one with the smallest burst time is selected and run to completion.

Round Robin (time quantum = 10): Processes are executed in FCFS order for a fixed time slice (quantum). If a process is not finished after its quantum, it is placed at the end of the ready queue.

Formulae (for reference):

TAT = CT − AT    and    WT = TAT − BT

Given:

P₁(BT=9, AT=0),  P₂(30,0),  P₃(4,0),  P₄(8,2),  P₅(11,6)

Non-preemptive SJF – Gantt Chart:

P3

P4

P1

P5

P2

0   4   12   21   32   62

Completion order: P₃, P₄, P₁, P₅, P₂ → Second last = P₅.

Round Robin (TQ = 10) – Gantt Chart:

P1(9)

P2(10/30)

P3(4)

P4(8)

P5(10/11)

P2(10/20)

P5(1/1)

P2(10/10)

0  9  19  23  31  41  51  52  62

Completion times: P₁=9, P₃=23, P₄=31, P₅=52, P₂=62 → Second last = P₅.

Therefore, SJF: P₅; RR: P₅ — correct option is 4.