考虑以下一组要安排在单个 CPU 系统上执行的作业。
Job Arrival Time Size (msec) Priority
J1 0 10 2 (Silver)
J2 2 8 1 (Gold)
J3 3 3 3 (Bronze)
J4 10 4 2 (Silver)
J5 12 1 3 (Bronze)
J6 15 4 1 (Gold) 以下哪一项是为这组作业提供最短等待时间的调度策略的正确顺序?
(一)非抢占式SJF
(C)非抢占式 SJF < 抢占式优先级调度 < FCFS < 非抢占式优先级调度
(D)非抢占式 SJF < 抢占式优先级调度 < 非抢占式优先级调度 < FCFS
答案： (C)
说明： 1. FCFS 调度甘特图： _%7C_Question_50_0.jpg)
Job Arrival Time Size (msec) Priority Turn around time Waiting time
J1 0 10 2 (Silver) 0 0
J2 2 8 1 (Gold) 16 8
J3 3 3 3 (Bronze) 18 15
J4 10 4 2 (Silver) 15 11
J5 12 1 3 (Bronze) 14 13
J6 15 4 1 (Gold) 15 11平均等待时间
= (0 + 8 + 15 + 11 + 13 + 11) / 6 = 9.67 2. 非抢占式 SJF 调度的甘特图： _%7C_Question_50_1.jpg)
Job Arrival Time Size (msec) Priority Turn around time Waiting time
J1 0 10 2 (Silver) 0 0
J2 2 8 1 (Gold) 28 20
J3 3 3 3 (Bronze) 10 7
J4 10 4 2 (Silver) 8 4
J5 12 1 3 (Bronze) 2 1
J6 15 4 1 (Gold) 7 3平均等待时间
= (0 + 20 + 7 + 4 + 1 + 3) / 6 = 5.83 3. 非抢占式优先级调度甘特图： _%7C_Question_50_2.jpg)
Job Arrival Time Size (msec) Priority Turn around time Waiting time
J1 0 10 2 (Silver) 0 0
J2 2 8 1 (Gold) 16 8
J3 3 3 3 (Bronze) 26 23
J4 10 4 2 (Silver) 16 12
J5 12 1 3 (Bronze) 18 17
J6 15 4 1 (Gold) 15 3平均等待时间
= (0 + 8 + 23 + 12 + 17 + 3) / 6 = 10.5 4. 抢占式优先级调度甘特图： _%7C_Question_50_3.jpg)
Job Arrival Time Size (msec) Priority Turn around time Waiting time
J1 0 10 2 (Silver) 0 12
J2 2 8 1 (Gold) 8 0
J3 3 3 3 (Bronze) 26 3
J4 10 4 2 (Silver) 16 12
J5 12 1 3 (Bronze) 18 17
J6 15 4 1 (Gold) 4 0平均等待时间
= (12 + 0 + 3 + 12 + 17 + 0) / 6 = 7.33 因此，非抢占式 SJF < 抢占式优先级调度 < FCFS < 非抢占式优先级调度。
所以，选项(C)是正确的。
这个问题的测验