现有3个同时到达的作业J1、J2和J3,它们的执行时间分别是T1、T2和T3,且T1image.pngT2image.pngT3。系统按单道方式运行且采用短作业优先算法,则平均周转时间是______。
- A.T1+T2+T3
- B.(T1+T2+T3)/3
- C.(3T1+2T2+T3)/3
- D.(T1+2T2+3T3)/3
正确答案及解析
正确答案
解析
短作业优先(shortest Job First, SJF)调度算法用于进程调度时称为短进程优先调度算法,该调度算法既可以用于作业调度也可以用于进程调度。在作业调度中,短作业优先调度算法每次从后备作业队列中选择估计运行时间最短的一个或几个作业,将它们调入内存,分配必要的资源,创建进程并放入就绪队列。由于3个作业同时到达,所以作业执行的次序依次是J1、J2和J3,故周转时间分别为:J1: T1;J2: T1+T2;J3: T1+T2+T3;则平均周转时间:(T1+(T1+T2)+(T1+T2+T3))/3=(3T1+2T2+T3)/3
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