24. [NEW] Worked Example: FCFS / SJF / SRTF / Round Robin

This is the single most common practical follow-up after "explain the scheduling algorithms" — a numeric Gantt-chart problem. Practice this pattern.

Given processes (arrival time, burst time):

Process Arrival Burst
P1	0	5
P2	1	3
P3	2	8
P4	3	6

FCFS (run in arrival order):

Gantt: | P1(0-5) | P2(5-8) | P3(8-16) | P4(16-22) |

Waiting time = start time − arrival time. Turnaround = completion − arrival.

Process Completion Turnaround Waiting
P1	5	5	0
P2	8	7	4
P3	16	14	6
P4	22	19	13

Average waiting = (0+4+6+13)/4 = 5.75. Average turnaround = (5+7+14+19)/4 = 11.25.

SJF, non-preemptive (pick shortest available burst each time the CPU is free):

Gantt: | P1(0-5) | P2(5-8) | P4(8-14) | P3(14-22) |

(At t=5, ready set is {P2(3), P3(8), P4 not yet arrived at t=3? P4 arrives at 3, so ready = {P2(3), P3(8), P4(6)} → pick P2.) At t=8, ready = {P3(8), P4(6)} → pick P4. Then P3 last.

Average waiting = ((0)+(5-1)+(14-3)+(8-2))/4 = (0+4+11+6)/4 = 5.25. Lower than FCFS — shorter jobs finishing earlier drags the average down, illustrating why SJF minimizes average waiting time.

SRTF (preemptive SJF): at every new arrival, compare remaining burst of the running process against the new arrival's burst, and switch to whichever is shorter. This usually beats non-preemptive SJF on average waiting time but adds context-switch overhead and can still starve long jobs.

Round Robin (quantum = 4):

Gantt: | P1(0-4) | P2(4-7) | P3(7-11) | P4(11-15) | P1(15-16) | P3(16-20) | P4(20-22) |

RR gives every process the CPU roughly on a rotating basis, so response time is far better than FCFS/SJF even though total completion may not be minimal.

Takeaway for interviews: be ready to (1) build the Gantt chart, (2) compute waiting/turnaround per process, (3) average them, and (4) explain the trade-off in one line — e.g., "SJF minimizes average waiting time but can starve long jobs; Round Robin trades throughput for fairness and responsiveness."

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