循环调度算法

Question

再次，在 JAVA 中做一些调度算法，但问题是我在互联网上找到的所有示例都没有回答我的问题，所以我无处可问，我发了这个帖子：

如果我有 3 的量子时间和进程：

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

所以我没有找到任何例子来说明如果进程 P1 没有到达，但是量子结束了怎么办？所以 P0 在 3 毫秒内执行，还剩下 2 毫秒，时间片已经结束，P1 还没有到达。程序会等待 P1 还是会在完成 P0 后仍然等待 1 毫秒（直到 P1 到达）？

Answer 1

调度算法只调度等待运行的进程。

一个执行可能是：

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，所以会在下一个时间段执行。