多级反馈队列调度是如何工作的？

Question

我一直在阅读 Ritchie“操作系统：合并 UNIX 和 Windows”第 4.3 章关于调度的内容。而且我很难理解这种调度算法是如何工作的，因为我的测试有点与我的导师相矛盾。

这是一个示例问题： 2 多级反馈队列，1 CPU

-Time quantum is 5ms for both queues
-At time t=0 process P1(8ms) arrives
-At time t=2 process P2(7ms) arrives
-At time t=6 process P3(10ms) arrives

进程的最终状态是什么，队列，并且 CPU 在时间 t=7?

所以根据我的理解，这应该是结果。

 T0 => Q1(P1(8ms) Running), Q()
 T1 => Q1(), Q2(P1(3ms) Ready) *preemptively moved to Q2*
 T2 => Q1(P2(7ms) Running), Q2(P1(3ms) Ready)
 T3 => Q1(), Q2(P2(2ms) Ready, P1(3ms) Ready) *preemptively moved to Q2*
 T4 => Q1(), Q2(P2(2ms) Ready, P1(3ms) Running)
 T5 => Q1(), Q2(P2(2ms) Ready) *P1(3ms) executed for 3ms and terminated*
 T6 => Q1(P3(10ms), Ready), Q2(P2(2ms) Running)
 T7 => Q1(P3(10ms), Running), Q2() *P2(2ms) executed for 2ms and terminated*

然而这个问题的答案如下：

时间 t=7:

-P1 has been executed for 5ms and it is in the READY state of the 2nd queue
-P2 has been executed for 2ms and it is in the RUNNING state
-P3 has arrived and it is in the READY state of the 1st queue

我很困惑，网上的视频一点帮助都没有。

Answer 1

我认为您误解了 time=7 的意思。您的计算表明您将其视为："What will the state of the system be after 7 intervals of time?"

他们在这里问的是："What will the state of the system be 7 milliseconds after it started running?"

这个问题的答案是：

-P1 has been executed for 5ms and it is in the READY state of the 2nd queue
-P2 has been executed for 2ms and it is in the RUNNING state
-P3 has arrived and it is in the READY state of the 1st queue

解释：P1在时间=0（即系统启动后0毫秒）到达，因此进入就绪队列Q1并立即在运行ning上启动处理器。 5 毫秒后，它被抢占并移动到 Q2，等待运行剩余的 3 毫秒。此时（系统启动后5ms）P2已经到达，正在Q1等待。所以，它开始运行ning。在 6 ms P3 到达 Q1。因此，系统启动后 7 毫秒：P1 在 Q2 等待，还剩 3 毫秒。 P2 还剩运行 5 毫秒。而P3在Q1等待

多级反馈队列调度是如何工作的？

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scheduling