Question 1:
find the average Turn Around Time and Waiting Time of following processes using Non-Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 8 |
| P2 | 1 | 4 |
| P3 | 2 | 9 |
| P4 | 3 | 5 |
| P5 | 4 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 8 | 8 | 8 - 0 = 8 | 8 - 8 = 0 |
| P5 | 2 | 10 | 10 - 4 = 6 | 6 - 2 = 4 |
| P2 | 4 | 14 | 14 - 1 = 13 | 13 - 4 = 9 |
| P4 | 5 | 19 | 19 - 3 = 16 | 16 - 5 = 11 |
| P3 | 9 | 28 | 28 - 2 = 26 | 26 - 9 = 17 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (8 + 6 + 13 + 16 + 26) / 5 = 13.8
Average WT = (0 + 4 + 9 + 11 + 17) / 5 = 8.2
Question 2:
find the average Turn Around Time and Waiting Time of following processes using Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 7 |
| P2 | 2 | 4 |
| P3 | 4 | 1 |
| P4 | 5 | 4 |
| P5 | 7 | 3 |
| P6 | 8 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 7 | 21 | 21 - 0 = 21 | 21 - 7 = 14 |
| P2 | 4 | 7 | 7 - 2 = 5 | 5 - 4 = 1 |
| P3 | 1 | 5 | 5 - 4 = 1 | 1 - 1 = 0 |
| P4 | 4 | 16 | 16 - 5 = 11 | 11 - 4 = 7 |
| P5 | 3 | 10 | 10 - 7 = 3 | 3 - 3 = 0 |
| P6 | 2 | 12 | 12 - 8 = 4 | 4 - 2 = 2 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (21 + 5 + 1 + 11 + 3 + 14) / 6 = 9.16
Average WT = (14 + 1 + 0 + 7 + 0 + 2) / 6 = 4
Question 3:
find the average Turn Around Time and Waiting Time of following processes using Non-Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 5 |
| P2 | 1 | 3 |
| P3 | 2 | 8 |
| P4 | 3 | 6 |
| P5 | 4 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 5 | 5 | 5 - 0 = 5 | 5 - 5 = 0 |
| P5 | 2 | 7 | 7 - 4 = 3 | 3 - 2 = 1 |
| P2 | 3 | 10 | 10 - 1 = 9 | 9 - 3 = 6 |
| P4 | 6 | 16 | 16 - 3 = 13 | 13 - 6 = 7 |
| P3 | 8 | 24 | 24 - 2 = 22 | 22 - 8 = 14 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (5 + 3 + 9 + 13 + 22) / 5 = 10.4
Average WT = (0 + 1 + 6 + 7 + 14) / 5 = 5.6
Question 4:
find the average Turn Around Time and Waiting Time of following processes using Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 6 |
| P2 | 1 | 4 |
| P3 | 2 | 9 |
| P4 | 3 | 5 |
| P5 | 4 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 6 | 12 | 12 - 0 = 12 | 12 - 6 = 6 |
| P2 | 4 | 6 | 6 - 1 = 5 | 5 - 4 = 1 |
| P3 | 9 | 26 | 26 - 2 = 24 | 24 - 9 = 15 |
| P4 | 5 | 17 | 17 - 3 = 14 | 14 - 5 = 9 |
| P5 | 2 | 7 | 7 - 4 = 3 | 3 - 2 = 1 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (12 + 5 + 24 + 14 + 3) / 5 = 11.6
Average WT = (6 + 1 + 15 + 9 + 1) / 5 = 6.4
Question 5:
find the average Turn Around Time and Waiting Time of following processes using Non-Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 3 |
| P2 | 1 | 6 |
| P3 | 3 | 8 |
| P4 | 5 | 5 |
| P5 | 6 | 4 |
| P6 | 7 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 3 | 3 | 3 - 0 = 3 | 3 - 3 = 0 |
| P2 | 6 | 9 | 9 - 1 = 8 | 8 - 6 = 2 |
| P3 | 8 | 28 | 28 - 3 = 25 | 25 - 8 = 17 |
| P4 | 5 | 20 | 20 - 5 = 15 | 15 - 5 = 10 |
| P5 | 4 | 15 | 15 - 6 = 9 | 9 - 4 = 5 |
| P6 | 2 | 11 | 11 - 7 = 4 | 4 - 2 = 2 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (3 + 8 + 25 + 15 + 9 + 4) / 6 = 10.6
Average WT = (0 + 2 + 17 + 10 + 5 + 2) / 6 = 6
Question 6:
find the average Turn Around Time and Waiting Time of following processes using Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 8 |
| P2 | 1 | 4 |
| P3 | 2 | 9 |
| P4 | 3 | 5 |
| P5 | 5 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 8 | 19 | 19 - 0 = 19 | 19 - 8 = 11 |
| P2 | 4 | 5 | 5 - 1 = 4 | 4 - 4 = 0 |
| P3 | 9 | 28 | 28 - 2 = 26 | 26 - 9 = 17 |
| P4 | 5 | 12 | 12 - 3 = 9 | 9 - 5 = 4 |
| P5 | 2 | 7 | 7 - 5 = 2 | 2 - 2 = 0 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (19+4+26+9+2) / 5 = 12
Average WT = (11+0+17+4+0) / 5 = 6.4
Question 7:
find the average Turn Around Time and Waiting Time of following processes using Non-Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 2 |
| P2 | 1 | 3 |
| P3 | 2 | 4 |
| P4 | 3 | 6 |
| P5 | 4 | 1 |
| P6 | 5 | 5 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 2 | 2 | 2 - 0 = 2 | 2 - 2 = 0 |
| P2 | 3 | 5 | 5 - 1 = 4 | 4 - 3 = 1 |
| P3 | 4 | 10 | 10 - 2 = 8 | 8 - 4 = 4 |
| P4 | 6 | 21 | 21 - 3 = 18 | 18 - 6 = 12 |
| P5 | 1 | 6 | 6 - 4 = 2 | 2 - 1 = 1 |
| P6 | 5 | 15 | 15 - 5 = 10 | 10 - 5 = 5 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (2+4+8+18+2+10) / 6 = 7.33
Average WT = (0 + 1 + 4 +12 + 1+5) / 6 = 3.83
Question 8:
find the average Turn Around Time and Waiting Time of following processes using Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 5 |
| P2 | 1 | 2 |
| P3 | 3 | 6 |
| P4 | 4 | 3 |
| P5 | 5 | 4 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 5 | 7 | 7 - 0 = 7 | 7 - 5 = 2 |
| P2 | 2 | 3 | 3 - 1 = 2 | 2 - 2 = 0 |
| P3 | 6 | 20 | 20 - 3 = 17 | 17 - 6 = 11 |
| P4 | 3 | 10 | 10 - 4 = 6 | 6 - 3 = 3 |
| P5 | 4 | 14 | 14 - 5 = 9 | 9 - 4 = 5 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (7 + 2 + 17 +6 + 9) / 5 = 8.2
Average WT = (2 + 0 + 11 + 3 + 5) / 5 = 4.2
Question 9:
find the average Turn Around Time and Waiting Time of following processes using Non-Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 8 |
| P2 | 1 | 3 |
| P3 | 2 | 7 |
| P4 | 3 | 4 |
| P5 | 4 | 2 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 8 | 8 | 8 - 0 = 8 | 8 - 8 = 0 |
| P2 | 3 | 13 | 13 - 1 = 12 | 12 - 3 = 9 |
| P3 | 7 | 24 | 24 - 2 = 22 | 22 - 7 = 15 |
| P4 | 4 | 17 | 17 - 3 = 14 | 14 - 4 = 10 |
| P5 | 2 | 10 | 10 - 4 = 6 | 6 - 2 = 4 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (8 + 6 + 12 + 14 + 22) / 5 = 12.4
Average WT = (0 + 4 + 9 + 10 + 15) / 5 = 7.6
Question 10:
find the average Turn Around Time and Waiting Time of following processes using Preemptive SJF(SHORTEST JOB FIRST) process scheduling algorithm?
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 5 |
| P2 | 1 | 7 |
| P3 | 2 | 4 |
| P4 | 3 | 1 |
| P5 | 4 | 3 |
Solution :
Formulas:
- Turn Around Time (TAT) = Completion Time (CT) - Arrival Time (AT)
- Waiting Time (WT) = Turn Around Time (TAT) - Burst Time (BT)
Gantt Chart
| Process | Burst Time | Completion Time | Turn Around Time (TAT) | Waiting Time (WT) |
|---|---|---|---|---|
| P1 | 5 | 6 | 6 - 0 = 6 | 6 - 5 = 1 |
| P2 | 7 | 20 | 20 - 1 = 19 | 19 - 7 = 12 |
| P3 | 4 | 13 | 13 - 2 = 11 | 11 - 4 = 7 |
| P4 | 1 | 4 | 4 - 3 = 1 | 1 - 1 = 0 |
| P5 | 3 | 9 | 9 - 4 = 5 | 5 - 3 = 2 |
Formula for average TAT and WT:
- Average Turn Around Time (Average TAT) = Σ(TAT) / Number of Processes
- Average Waiting Time (Average WT) = Σ(WT) / Number of Processes
Average TAT = (6 + 19 + 11 + 1 + 5) / 5 = 8.4
Average WT = (1 + 12 + 7 + 0 + 2) / 5 = 4.4