GO Classes 2023 | IIITH Mock Test 1 | Question: 26

Answer 1

The process chart looks flow looks like:

(0 to 2s) -→ Running process: P1; Job remaining: 5s (gets preempted by incoming process P2 since execution time of P2 is lesser than the remaining time of P1)

(2 to 4s) -→ Running process: P2; Job remaining: 2s (gets preempted by incoming process P3 since P3 has least execution time among P1, P2 and P3 having remaining executions times equal to 5, 2 and 1 respectively)

(4 to 5s) -→ Running process: P3 (completes execution. Processes available in queue are P1, P2 and P4 having remaining time equal to 5, 2 and 4s. Process P2 gets picked.)

(5 to 7s) --→ Running process: P2 (completes execution. Processes available in queue are P1 and P4 having remaining time equal to 5 and 4s. Process P4 gets picked.)

(7 to 11s) --→ Running process: P4 (completes execution. Processes available in queue is P1 and gets picked.)

( 11 to 16s) --→ Running process: P4 (completes execution)

 

Waiting time = (Finish time – Arrival time) – Process duration

Wait time (P1) = (16 – 0) – 7 = 9

Wait time (P2) = (7 – 2) – 4 = 1

Wait time (P2) = (5 – 4) – 1 = 0

Wait time (P4) = (11 – 5) – 4 = 2

Average wait time = (9 + 1+ 0 + 2) / 4 = 12 / 4 = 3

Hence, answer: (B)

Answer 2

AT = Arrival Time, BT = Burst Time, CT = Completion Time, TAT = Turnaround Time, WT = Waiting Time

Formula: TAT = CT – AT and WT = TAT – BT

Process	AT	BT	CT	TAT	WT
P1	0	7	16	16	9
P2	2	4	7	5	1
P3	4	1	5	1	0
P4	5	4	11	6	2

 

Gantt chart: 

$\therefore$ Average(WT) = (9+1+0+2)/4 = 3

Answer: B