Question 1:

Given the following processes with their arrival times and burst times, calculate the turnaround time and waiting time for each process using FCFS scheduling.

Answer :

FCFS Scheduling executes processes in the order they arrive.

1. Completion Times:
   • P1: Completes at 0 + 5 = 5
   • P2: Completes at 5 + 3 = 8
   • P3: Completes at 8 + 8 = 16
   • P4: Completes at 16 + 6 = 22
2. Turnaround Times (TAT):
   • P1: TAT = 5 – 0 = 5
   • P2: TAT = 8 – 1 = 7
   • P3: TAT = 16 – 2 = 14
   • P4: TAT = 22 – 3 = 19
3. Waiting Times (WT):
   • P1: WT = 0
   • P2: WT = 5 – 1 = 4
   • P3: WT = 8 – 2 = 6
   • P4: WT = 16 – 3 = 13

Average Turnaround Time = (5 + 7 + 14 + 19) / 4 = 11.25
Average Waiting Time = (0 + 4 + 6 + 13) / 4 = 5.75

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Question 2:

Consider the following processes that arrive at different times. Calculate the completion time, turnaround time, and waiting time for each process using FCFS scheduling.

Answer 2:

FCFS Scheduling executes processes in the order they arrive.

1. Order of Execution: P4 -> P3 -> P1 -> P2
2. Completion Times:
   • P4: Completes at 0 + 3 = 3
   • P3: Completes at 3 + 8 = 11
   • P1: Completes at 11 + 6 = 17
   • P2: Completes at 17 + 2 = 19
3. Turnaround Times (TAT):
   • P4: TAT = 3 – 0 = 3
   • P3: TAT = 11 – 1 = 10
   • P1: TAT = 17 – 2 = 15
   • P2: TAT = 19 – 5 = 14
4. Waiting Times (WT):
   • P4: WT = 0
   • P3: WT = 3 – 1 = 2
   • P1: WT = 11 – 2 = 9
   • P2: WT = 17 – 5 = 12

Average Turnaround Time = (3 + 10 + 15 + 14) / 4 = 10.5
Average Waiting Time = (0 + 2 + 9 + 12) / 4 = 5.75

Gantt Chart

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Question 3:

Given the following processes with their arrival times and burst times, determine the Gantt chart, and calculate the average turnaround time and average waiting time using FCFS scheduling.

Answer 3:

FCFS Scheduling executes processes in the order they arrive.

1. Completion Times:
   • P1: Completes at 0 + 4 = 4
   • P2: Completes at 4 + 5 = 9
   • P3: Completes at 9 + 3 = 12
   • P4: Completes at 12 + 2 = 14
   • P5: Completes at 14 + 4 = 18
2. Turnaround Times (TAT):
   • P1: TAT = 4 – 0 = 4
   • P2: TAT = 9 – 1 = 8
   • P3: TAT = 12 – 2 = 10
   • P4: TAT = 14 – 3 = 11
   • P5: TAT = 18 – 4 = 14
3. Waiting Times (WT):
   • P1: WT = 0
   • P2: WT = 4 – 1 = 3
   • P3: WT = 9 – 2 = 7
   • P4: WT = 12 – 3 = 9
   • P5: WT = 14 – 4 = 10

Average Turnaround Time = (4 + 8 + 10 + 11 + 14) / 5 = 47 / 5 = 9.4
Average Waiting Time = (0 + 3 + 7 + 9 + 10) / 5 = 29 / 5 = 5.8

Gantt Chart

Tutorial on First Come First Served Process Scheduling