了解最短的作业优先算法(非抢先) [英] Understanding the Shortest Job First Algorithm (Non-preemptive)

问题描述

最短的作业优先算法如下图所示:

The shortest job first algorithm is shown in the following image:

如果首先是最短的作业/接下来是最短的流程，则顺序不应为: P1→P5→P3→P4→P2?因为这是最低到最高服务时间的顺序.
为什么进程2排第二?

If it is shortest job first/shortest process next, shouldn't the order be: P1 → P5 → P3 → P4 → P2 ? Since that's the order of lowest to highest service times.
Why does process 2 come second?

我知道我们是否使用突发时间，这就是顺序，但是我不知道服务时间和突发时间之间有什么区别.

I know if we use burst times instead, that would be the order, but I have no idea what the differences between service time and burst times are.

任何帮助解释该图形的人将不胜感激.

Any help would be much appreciated explaining that graphic.

推荐答案

问题中的图像遵循正确的顺序，即:

The image in the question follows the correct order which is:

P1→P2→P5→P3→P4

P1 → P2 → P5 → P3 → P4

说明:
P1到达time = 0，因此将首先执行它.此过程的服务时间为3.因此，此过程在time=3处完成.

Explanation:
P1 is arrived at time = 0 , so it will be executed first. Service Time of this process is 3. So this process is completed at time=3.

在time=3，只有一个到达的进程是P2.所有其他进程将在稍后到达.因此，此过程现在已执行.此过程的服务时间为6，因此该过程在time=3+6=9处完成.

At time=3, there is only one process that is arrived which is P2. All other processes arrive later. So this process is now executed. Service time of this process is 6, so this process is completed at time=3+6=9.

现在在time=9，有三个过程分别是P3，P4和P5(分别到达time= 4, 6 and 8).由于P5的服务时间是2，与P3和P4的服务时间相比是最短的，因此现在执行P5并在time=9+2=11处完成.

Now at time=9, there are three processes which are P3, P4 and P5 (which arrived at time= 4, 6 and 8 respectively). Since the service time of P5 is 2 which is minimum as compared to that of P3 and P4, so P5 is now executed and it gets completed at time=9+2=11.

在time=11，我们有两个进程，分别是P3和P4(分别到达time= 4 and 6).由于P3的服务时间是4，比P4的服务时间短，因此P4现在被执行，并在time= 11+4=15

At time=11, we have two processes which are P3 and P4 (which are arrived at time= 4 and 6 respectively). Since the service time of P3 is 4 which is less as compared to that of P4, so P4 is executed now and it gets completed at time= 11+4=15

在time=15，我们只有一个进程，即P4.这样就可以执行了.由于此过程的服务时间为5，因此它在time = 15+5 = 20

At time=15, we have only one process which is P4. So it is executed now. Since service time of this process is 5, so it gets completed at time = 15+5 = 20

这篇关于了解最短的作业优先算法(非抢先)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！