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KornShell(ksh)调度算法(SRT)

发布时间:2020-12-16 01:15:31 所属栏目:安全 来源:网络整理
导读:我已经获得了一个从txt文件中读取模拟进程的任务,如下所示. ID: 35; Arrival_Time: 0; Total_Exec_Time: 4;ID: 65; Arrival_Time: 2; Total_Exec_Time: 6;ID: 10; Arrival_Time: 3; Total_Exec_Time: 3;ID: 124; Arrival_Time: 5; Total_Exec_Time: 5;ID: 18
我已经获得了一个从txt文件中读取模拟进程的任务,如下所示.
ID: 35; Arrival_Time: 0; Total_Exec_Time: 4;
ID: 65; Arrival_Time: 2; Total_Exec_Time: 6;
ID: 10; Arrival_Time: 3; Total_Exec_Time: 3;
ID: 124; Arrival_Time: 5; Total_Exec_Time: 5;
ID: 182; Arrival_Time: 6; Total_Exec_Time: 2;

我必须从(先到先得,最短剩余时间,循环q = 2)的选择中完成两个算法.我需要根据我选择的两个算法打印出当时运行的当前时间和进程.我已经成功完成了FCFS.我的下一个方法是关于SRT,除了我对算法背后的逻辑有一些严重的问题.

我目前正在尝试一种迭代方法(在下面发布),它在一定程度上起作用(直到当前时间9),但我觉得这可能只是一个幸运的巧合.

有没有人对这个算法有任何建议,或者其他两个.我已经在这项任务上工作了好几天,并决定把我的骄傲吸收并堆叠起来.

注意:这是我第一次使用shell脚本,所以我的代码可能有点乱.我仍在努力了解KornShell(ksh).

file="/path/to/file.txt"
  IFS=': ;'
  i=0
  while read -r f1 f2 f3 f4 f5 f6  
    do 
      integer id[i]="$f2" #id array
      integer at[i]="$f4" #arrival time array
      integer et[i]="$f6" #exec time array
      integer rt[i]=0 #run time so far
      integer current[i]=i

      ((i++))
    done <"$file"

  integer curr_index=0
  integer currTime=0
  let totalProcesses=${#at[@]}
  let totalProcesses=totalProcesses-1
  let totalRunTime=0
  for x in ${et[@]}; do
    let totalRunTime+=$x
  done 

  scheduleTask () { 
    currTime=$1
    for y in ${current[@]}; do
      if (( rt[$y] < et[$y] )); then
        #if the program is not finished,keep going
        if (( at[$y] < $currTime )); then
          #if the program is in que,keep going
          let diff=et[$y]-rt[$y]#not currently using
          let currDiff=et[$curr_index]-rt[$curr_index] #not currently using         
          if (( et[$y] <= et[$curr_index] )); then #is this broken?
            curr_index=$y
          fi
        fi
      else
        echo "${id[$y]} RAN ${rt[$y]} out of ${et[$y]} seconds"

        unset current[$y]
      fi
    done
  }

  for (( i = 0; i < $totalRunTime; i++ )); do
    echo "================================="
    scheduleTask $i 
    ((rt[$curr_index]++))
    print "ttcurrent time: $i"
    print "tttcurrent process: ${id[$curr_index]}"
    echo "================================="
  done

SRT的正确输出应该如下所示.

=================================
        current time: 0
            current process: 35
=================================
=================================
        current time: 1
            current process: 35
=================================
=================================
        current time: 2
            current process: 35
=================================
=================================
        current time: 3
            current process: 35
=================================
=================================
        current time: 4
            current process: 10
=================================
=================================
        current time: 5
            current process: 10
=================================
=================================
        current time: 6
            current process: 10
=================================
=================================
        current time: 7
            current process: 182
=================================
=================================
        current time: 8
            current process: 182
=================================
=================================
        current time: 9
            current process: 124
=================================
=================================
        current time: 10
            current process: 124
=================================
=================================
        current time: 11
            current process: 124
=================================
=================================
        current time: 12
            current process: 124
=================================
=================================
        current time: 13
            current process: 124
=================================
=================================
        current time: 14
            current process: 65
=================================
=================================
        current time: 15
            current process: 65
=================================
=================================
        current time: 16
            current process: 65
=================================
=================================
        current time: 17
            current process: 65
=================================
=================================
        current time: 18
            current process: 65
=================================
=================================
        current time: 19
            current process: 65
=================================
我仍然相对较新的堆栈溢出,并且对于家庭作业的想法和意见是天真的.我正在讨论删除这个问题,但在阅读这篇文章后( https://meta.stackexchange.com/questions/10811/how-to-ask-and-answer-homework-questions),我认为我的问题符合指南,因此值得跟上.

我想出了最短剩余时间算法.我很感激没有人回答这个问题,我自己搞定算法(在TA的帮助下)是值得的.因此,我提供的答案将具有基本的伪逻辑,而没有实际的代码.

shortest = the first process read from the input(assuming it has already arrived)
while there are still processes to be run
     process = next process (out of processes that have not completed yet)
     if (process arrival time <= currentTime) #process arrived 
           if (process execution time < shortest execution time)
                 shortest = process

注意:这与我从TA(编写作业)收到的帮助大致相同,这就是我觉得发布这个答案的原因.

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