slide 53

slide 55

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2. (5 pt) The following is the map of Storage Blocks Allocation for three files (ref: Slide-53) | 1:free 2:file1 3:file1 4:file2 5:file2 6:file3 7:file3 8:free 9:file1 10:free | 11:file2 | 12:file1 | 13:file2 | 14:file3 | 15:file3 | 16:file2 Show the Storage Allocation Table (ref: Slide-55) Unit Pointer Unit Pointer UnitPointer Pointer 5 9 Unit 13 14 6 10 11 12 15 16 Shortest Job First (SJF) Scheduling Arrival Bourst Time Process ID 1 0. 10 2 0 3 0. The scheduling order for SJF is P3, P2, P1 The Gantt Chart for the schedule is: P3 P2 o 2 5 - Waiting time for P = 5; P, = 2; P, = 0 - Average waiting time: (5+ 2 + 0)/3 = 2.33 This scheduling policy is much better than FCFS. More Example of SJF Process Arrival Burst Time 3 SJF scheduling chart . . . , 9 Average waiting time = [0 + (6-5) + (9-4) + (16-2)]/ 4 = 5.0 16 First-Come, First-Served (FCFS) Scheduling Process ID Arrival Burst Time 10 0 0 Suppose that the processes arrive in the order: P, , P2, P3 The Gantt Chart for the schedule is: P2 P3 P1 10 13 15 Waiting time for P, = 0; P, = 10; P, = 13 - Average waiting time: (0 + 10 + 13)/3 = 7.67 Example of RR with Time Quantum = 4 Process ID Arrival 0 0 0 Burst Time 10 3 2 2 3 The Gantt chart is: | Pi | P2|P3 | P1 | Pi 0 4 7 9 13 15 Average Waiting Time P, (0+5+0) + P2 (4) + P3 (7) => 16 + 3 = 5.33 - Better than FCFS, but not as good as SJF More Example of SJF Process Arrival Burst Time O 3 4 0 SJF scheduling chart PAP, PP, 6 16 24 Average waiting time = (3 + 16 + 9 + 0) / 4 = 7 2. (5 pt) The following is the map of Storage Blocks Allocation for three files (ref: Slide-53) | 1:free 2:file1 3:file1 4:file2 5:file2 6:file3 7:file3 8:free 9:file1 10:free | 11:file2 | 12:file1 | 13:file2 | 14:file3 | 15:file3 | 16:file2 Show the Storage Allocation Table (ref: Slide-55) Unit Pointer Unit Pointer UnitPointer Pointer 5 9 Unit 13 14 6 10 11 12 15 16 Shortest Job First (SJF) Scheduling Arrival Bourst Time Process ID 1 0. 10 2 0 3 0. The scheduling order for SJF is P3, P2, P1 The Gantt Chart for the schedule is: P3 P2 o 2 5 - Waiting time for P = 5; P, = 2; P, = 0 - Average waiting time: (5+ 2 + 0)/3 = 2.33 This scheduling policy is much better than FCFS. More Example of SJF Process Arrival Burst Time 3 SJF scheduling chart . . . , 9 Average waiting time = [0 + (6-5) + (9-4) + (16-2)]/ 4 = 5.0 16 First-Come, First-Served (FCFS) Scheduling Process ID Arrival Burst Time 10 0 0 Suppose that the processes arrive in the order: P, , P2, P3 The Gantt Chart for the schedule is: P2 P3 P1 10 13 15 Waiting time for P, = 0; P, = 10; P, = 13 - Average waiting time: (0 + 10 + 13)/3 = 7.67 Example of RR with Time Quantum = 4 Process ID Arrival 0 0 0 Burst Time 10 3 2 2 3 The Gantt chart is: | Pi | P2|P3 | P1 | Pi 0 4 7 9 13 15 Average Waiting Time P, (0+5+0) + P2 (4) + P3 (7) => 16 + 3 = 5.33 - Better than FCFS, but not as good as SJF More Example of SJF Process Arrival Burst Time O 3 4 0 SJF scheduling chart PAP, PP, 6 16 24 Average waiting time = (3 + 16 + 9 + 0) / 4 = 7