ostep homework(8)

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第8章 调度：多级反馈队列 homework详解

实验环境

笔者采用的是Ubuntu22.04的虚拟机。
本次作业需要用到cpu-sched-mlfq文件，按照readme提示操作即可。

实验内容

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$ python3 mlfq.py -n 2 -j 2 -m 10 -M 0 -c  
……  
Job List:  
  Job  0: startTime   0 - runTime   8 - ioFreq   0  
  Job  1: startTime   0 - runTime   4 - ioFreq   0  


Execution Trace:  

[ time 0 ] JOB BEGINS by JOB 0  
[ time 0 ] JOB BEGINS by JOB 1  
[ time 0 ] Run JOB 0 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 7 (of 8) ]  
……  
[ time 7 ] Run JOB 0 at PRIORITY 1 [ TICKS 2 ALLOT 1 TIME 0 (of 8) ]  
[ time 8 ] FINISHED JOB 0  
[ time 8 ] Run JOB 1 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 3 (of 4) ]  
……  
[ time 11 ] Run JOB 1 at PRIORITY 1 [ TICKS 6 ALLOT 1 TIME 0 (of 4) ]  
[ time 12 ] FINISHED JOB 1  

Final statistics:  
  Job  0: startTime   0 - response   0 - turnaround   8  
  Job  1: startTime   0 - response   8 - turnaround  12  

  Avg  1: startTime n/a - response 4.00 - turnaround 10.00

给出限制条件：队列数为2，工作数为2，工作最大长度为10，I/O最大长度为0。

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实例1

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$ python3 mlfq.py -l 0,200,0 -q 10 -n 3 -c  
……  

Job List:  
  Job  0: startTime   0 - runTime 200 - ioFreq   0  


Execution Trace:  

[ time 0 ] JOB BEGINS by JOB 0  
[ time 0 ] Run JOB 0 at PRIORITY 2 [ TICKS 9 ALLOT 1 TIME 199 (of 200) ]  
……  
[ time 9 ] Run JOB 0 at PRIORITY 2 [ TICKS 0 ALLOT 1 TIME 190 (of 200) ]  
[ time 10 ] Run JOB 0 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 189 (of 200) ]  
……  
[ time 19 ] Run JOB 0 at PRIORITY 1 [ TICKS 0 ALLOT 1 TIME 180 (of 200) ]  
[ time 20 ] Run JOB 0 at PRIORITY 0 [ TICKS 9 ALLOT 1 TIME 179 (of 200) ]  
……  
[ time 199 ] Run JOB 0 at PRIORITY 0 [ TICKS 0 ALLOT 1 TIME 0 (of 200) ]  
[ time 200 ] FINISHED JOB 0  

Final statistics:  
  Job  0: startTime   0 - response   0 - turnaround 200  

  Avg  0: startTime n/a - response 0.00 - turnaround 200.00

单个长工作，其运行时间为200ms，3个队列，时间片长度为10ms。

实例2

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$ python3 mlfq.py -l 0,180,0:100,20,0 -q 10 -n 3 -c  
……  

Job List:  
  Job  0: startTime   0 - runTime 180 - ioFreq   0  
  Job  1: startTime 100 - runTime  20 - ioFreq   0  


Execution Trace:  

[ time 0 ] JOB BEGINS by JOB 0  
[ time 0 ] Run JOB 0 at PRIORITY 2 [ TICKS 9 ALLOT 1 TIME 179 (of 180) ]  
……  
[ time 9 ] Run JOB 0 at PRIORITY 2 [ TICKS 0 ALLOT 1 TIME 170 (of 180) ]  
[ time 10 ] Run JOB 0 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 169 (of 180) ]  
……  
[ time 19 ] Run JOB 0 at PRIORITY 1 [ TICKS 0 ALLOT 1 TIME 160 (of 180) ]  
[ time 20 ] Run JOB 0 at PRIORITY 0 [ TICKS 9 ALLOT 1 TIME 159 (of 180) ]  
……  
[ time 99 ] Run JOB 0 at PRIORITY 0 [ TICKS 0 ALLOT 1 TIME 80 (of 180) ]  
[ time 100 ] JOB BEGINS by JOB 1  
[ time 100 ] Run JOB 1 at PRIORITY 2 [ TICKS 9 ALLOT 1 TIME 19 (of 20) ]  
……  
[ time 109 ] Run JOB 1 at PRIORITY 2 [ TICKS 0 ALLOT 1 TIME 10 (of 20) ]  
[ time 110 ] Run JOB 1 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 9 (of 20) ]  
……  
[ time 119 ] Run JOB 1 at PRIORITY 1 [ TICKS 0 ALLOT 1 TIME 0 (of 20) ]  
[ time 120 ] FINISHED JOB 1  
[ time 120 ] Run JOB 0 at PRIORITY 0 [ TICKS 9 ALLOT 1 TIME 79 (of 180) ]  
……  
[ time 199 ] Run JOB 0 at PRIORITY 0 [ TICKS 0 ALLOT 1 TIME 0 (of 180) ]  
[ time 200 ] FINISHED JOB 0  

Final statistics:  
  Job  0: startTime   0 - response   0 - turnaround 200  
  Job  1: startTime 100 - response   0 - turnaround  20  

  Avg  1: startTime n/a - response 0.00 - turnaround 110.00

长工作运行期间进入了一个短工作，由于短工作优先级高，故优先执行短工作。书上写长工作从最低优先队列开始运行，这里是从最高优先队列运行的，不过不影响结果。
实例3

1
$ python3 mlfq.py  -l 0,40,2:0,200,0 -q 10 -n 3 -i 5 -c

第8章 调度：多级反馈队列 homework详解

实验环境

笔者采用的是Ubuntu22.04的虚拟机。
本次作业需要用到cpu-sched-mlfq文件，按照readme提示操作即可。

实验内容

1

$ python3 mlfq.py -n 2 -j 2 -m 10 -M 0 -c
……
Job List:
  Job  0: startTime   0 - runTime   8 - ioFreq   0
  Job  1: startTime   0 - runTime   4 - ioFreq   0


Execution Trace:

[ time 0 ] JOB BEGINS by JOB 0
[ time 0 ] JOB BEGINS by JOB 1
[ time 0 ] Run JOB 0 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 7 (of 8) ]
……
[ time 7 ] Run JOB 0 at PRIORITY 1 [ TICKS 2 ALLOT 1 TIME 0 (of 8) ]
[ time 8 ] FINISHED JOB 0
[ time 8 ] Run JOB 1 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 3 (of 4) ]
……
[ time 11 ] Run JOB 1 at PRIORITY 1 [ TICKS 6 ALLOT 1 TIME 0 (of 4) ]
[ time 12 ] FINISHED JOB 1

Final statistics:
  Job  0: startTime   0 - response   0 - turnaround   8
  Job  1: startTime   0 - response   8 - turnaround  12

  Avg  1: startTime n/a - response 4.00 - turnaround 10.00

给出限制条件：队列数为2，工作数为2，工作最大长度为10，I/O最大长度为0。

2

实例1

$ python3 mlfq.py -l 0,200,0 -q 10 -n 3 -c
……

Job List:
  Job  0: startTime   0 - runTime 200 - ioFreq   0


Execution Trace:

[ time 0 ] JOB BEGINS by JOB 0
[ time 0 ] Run JOB 0 at PRIORITY 2 [ TICKS 9 ALLOT 1 TIME 199 (of 200) ]
……
[ time 9 ] Run JOB 0 at PRIORITY 2 [ TICKS 0 ALLOT 1 TIME 190 (of 200) ]
[ time 10 ] Run JOB 0 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 189 (of 200) ]
……
[ time 19 ] Run JOB 0 at PRIORITY 1 [ TICKS 0 ALLOT 1 TIME 180 (of 200) ]
[ time 20 ] Run JOB 0 at PRIORITY 0 [ TICKS 9 ALLOT 1 TIME 179 (of 200) ]
……
[ time 199 ] Run JOB 0 at PRIORITY 0 [ TICKS 0 ALLOT 1 TIME 0 (of 200) ]
[ time 200 ] FINISHED JOB 0

Final statistics:
  Job  0: startTime   0 - response   0 - turnaround 200

  Avg  0: startTime n/a - response 0.00 - turnaround 200.00

单个长工作，其运行时间为200ms，3个队列，时间片长度为10ms。

实例2

$ python3 mlfq.py -l 0,180,0:100,20,0 -q 10 -n 3 -c
……

Job List:
  Job  0: startTime   0 - runTime 180 - ioFreq   0
  Job  1: startTime 100 - runTime  20 - ioFreq   0


Execution Trace:

[ time 0 ] JOB BEGINS by JOB 0
[ time 0 ] Run JOB 0 at PRIORITY 2 [ TICKS 9 ALLOT 1 TIME 179 (of 180) ]
……
[ time 9 ] Run JOB 0 at PRIORITY 2 [ TICKS 0 ALLOT 1 TIME 170 (of 180) ]
[ time 10 ] Run JOB 0 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 169 (of 180) ]
……
[ time 19 ] Run JOB 0 at PRIORITY 1 [ TICKS 0 ALLOT 1 TIME 160 (of 180) ]
[ time 20 ] Run JOB 0 at PRIORITY 0 [ TICKS 9 ALLOT 1 TIME 159 (of 180) ]
……
[ time 99 ] Run JOB 0 at PRIORITY 0 [ TICKS 0 ALLOT 1 TIME 80 (of 180) ]
[ time 100 ] JOB BEGINS by JOB 1
[ time 100 ] Run JOB 1 at PRIORITY 2 [ TICKS 9 ALLOT 1 TIME 19 (of 20) ]
……
[ time 109 ] Run JOB 1 at PRIORITY 2 [ TICKS 0 ALLOT 1 TIME 10 (of 20) ]
[ time 110 ] Run JOB 1 at PRIORITY 1 [ TICKS 9 ALLOT 1 TIME 9 (of 20) ]
……
[ time 119 ] Run JOB 1 at PRIORITY 1 [ TICKS 0 ALLOT 1 TIME 0 (of 20) ]
[ time 120 ] FINISHED JOB 1
[ time 120 ] Run JOB 0 at PRIORITY 0 [ TICKS 9 ALLOT 1 TIME 79 (of 180) ]
……
[ time 199 ] Run JOB 0 at PRIORITY 0 [ TICKS 0 ALLOT 1 TIME 0 (of 180) ]
[ time 200 ] FINISHED JOB 0

Final statistics:
  Job  0: startTime   0 - response   0 - turnaround 200
  Job  1: startTime 100 - response   0 - turnaround  20

  Avg  1: startTime n/a - response 0.00 - turnaround 110.00

长工作运行期间进入了一个短工作，由于短工作优先级高，故优先执行短工作。书上写长工作从最低优先队列开始运行，这里是从最高优先队列运行的，不过不影响结果。
实例3

$ python3 mlfq.py  -l 0,40,2:0,200,0 -q 10 -n 3 -i 5 -c

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