循环罗宾调度算法

Question

所以再一次，用JAVA做一些调度算法，但问题是，我在互联网上找到的所有例子都没有回答我的问题，所以我没有别的地方可问，于是我写了这篇文章：

如果量子时间为3，并处理：

Name - ArrivalTime - BurstTime
P0 - 0 - 5
P1 - 6 - 5
P2 - 6 - 9
P3 - 8 - 2

因此，我没有找到任何例子来说明这一点，如果过程P1还没有到达，但是量子结束了呢？所以P0在3ms内执行，还剩2ms，量子已经结束，P1还没有到达。程序会等待P1，还是会完成P0，并且仍然有1ms的等待时间(直到P1到达)？

Answer 1

调度算法只调度等待运行的进程。

一项执行可以是：

T0 : Waiting Process = [P0] ; Executed Process = P0(1-2-3) 
T3 : Waiting Process = [P0] ; Executed Process = P0(4-5) => P0 finished
T5 : Waiting Process = [] ; Executed Process = Nothing
T6 : Waiting Process = [P1, P2] ; Executed Process = P1(1-2-3)
T9 : Waiting Process = [P2, P3, P1] ; Executed Process = P2(1-2-3)
T12 : Waiting Process = [P3, P1, P2] ; Executed Process = P3(1-2) => P3 finished
T14 : Waiting Process = [P1, P2] ; Executed Process = P1(4-5) => P1 finished
T16 : Waiting Process = [P2] ; Executed Process = P2(4-5-6)
T19 : Waiting Process = [P2] ; Executed Process = P2(7-8-9) => P2 finished

在T3上，只有P0在等待运行，因此它将在下一个时间段内执行。