Chapter 9 Homework Solutions 9.1, 9.4, 9.14, 9.16 9.1 Consider the following set of processes: Process Name A B C D E Arrival Time 0 1 3 9 12 Processing Time 3 5 2 5 5 A B 0 1 A B 0 1 C 2 3 D 4 5 6 7 8 C 2 3 9 E 10 11 D 4 5 6 7 8 9 12 13 14 15 16 17 18 19 13 14 15 16 17 18 19 E 10 11 12 Process Name A B C D E Arrival Time 0 1 3 9 12 Processing Time 3 5 2 5 5 Average 9.1 Consider the following set of processes: Process Name A B C D E Arrival Time 0 1 3 9 12 Processing Time 3 5 2 5 5 A B 0 1 A B 0 1 C 2 3 D 4 5 6 7 8 C 2 3 9 E 10 11 D 4 5 6 7 8 9 12 13 14 15 16 17 18 19 13 14 15 16 17 18 19 E 10 11 12 9.4 Favors I/O-bound processes: If a process uses too much processor time, it will be moved to a lower-priority queue. This leaves I/O-bound processes in the higher-priority queues. 9.14 Favors I/O-bound processes: If a process uses too much processor time, it will be moved to a lower-priority queue. This leaves I/O-bound processes in the higher-priority queues. 9.16 Five batch jobs, A through E, arrive at a computer center at essentially the same time. They have estimated running times of 15, 9, 3, 6 and 12 minutes, respectively. Their (externally defined) priorities are 6, 3, 7, 9 and 4, respectively, with lower value corresponding to higher priorities. For each of the following scheduling algorithms, determine the turnaround time for each process and the average turnaround time for all five jobs. Ignore process switching overhead. Explain how you arrived at your answers. In the last three cases, assume that only one job at a time runs until it finishes and that all jobs are processor bound. (a) round robin with a time quantum of 1 minute (b) priority scheduling (c) FCFS (run in order 15, 9, 3, 6 and 12) (d) Shortest job next. (a) round robin with a time quantum of 1 minute Sequence with which processes will get 1 min of processor time: The turnaround time for each process: A = 45 min, B = 35 min, C = 13 min, D = 26 min, E = 42 min The average turnaround time is (45+35+13+26+42) / 5 = 32.2 min (b) Priority scheduling The average turnaround time is: (9+21+36+39+45) / 5 = 30 min (c) FCFS (run in order 15, 9, 3, 6 and 12) The average turnaround time is: (15+24+27+33+45) / 5 = 28.8 min (d) Shortest job next. The average turnaround time is: (3+9+18+30+45) / 5 = 21 min