算法设计:最长剩余时间优先(LRTF)CPU调度程序

2021年3月12日12:56:58 发表评论 802 次浏览

本文概述

先决条件–

CPU调度|最长剩余时间优先(LRTF)算法

我们给出了到达时间和爆发时间的一些过程, 我们必须找到完成时间(CT), 周转时间(TAT), 平均周转时间(Avg TAT), 等待时间(WT), 平均等待时间(AWT) )。

例子:考虑下表中四个进程P1, P2, P3和P4的到达时间和突发时间。

Process   Arrival time   Burst Time
P1            1 ms          2 ms
P2            2 ms          4 ms
P3            3 ms          6 ms
p4            4 ms          8 ms

甘特图将如下所示,

最长剩余时间优先(LRTF)CPU调度程序1

由于可以通过甘特图直接确定完成时间(CT), 并且

Turn Around Time (TAT)
= (Complition Time) - (Arival Time)

Also, Waiting Time (WT)
= (Turn Around Time) - (Burst Time)

因此,

输出如下:

Total Turn Around Time = 68 ms
So, Average Turn Around Time = 68/4 = 17.00 ms

And, Total Waiting Time = 48 ms
So, Average Waiting Time = 12.00 ms

算法–

  • 第1步:创建一个包含所有必填字段的过程结构, 例如AT(到达时间), BT(突发时间), CT(完成时间), TAT(转身时间), WT(等待时间)。
  • 第2步:根据AT排序;
  • 步骤3:查找具有最大突发时间的过程, 并为每个单元执行。将总时间增加1, 并将该过程的突发时间减少1。
  • 步骤4:当任何进程的BT剩余为0时, 则更新CT(该进程的完成时间CT将为该时间的总时间)。
  • 第2步:在为每个过程计算CT之后, 找到TAT和WT。
    (TAT = CT - AT) 
    (WT  = TAT - BT) 

算法的实现–

C ++

#include <bits/stdc++.h>
  
using namespace std;
  
// creating a structure of a process
struct process {
     int processno;
     int AT;
     int BT;
  
     // for backup purpose to print in last
     int BTbackup;
     int WT;
     int TAT;
     int CT;
};
  
// creating a structe of 4 processes
struct process p[4];
  
// variable to find the total time
int totaltime = 0;
int prefinaltotal = 0;
  
// comparator function for sort()
bool compare(process p1, process p2)
{
     // compare the Arrival time of two processes
     return p1.AT < p2.AT;
}
  
// finding the largest Arrival Time among all the available
// process at that time
int findlargest( int at)
{
     int max = 0, i;
     for (i = 0; i < 4; i++) {
         if (p[i].AT <= at) {
             if (p[i].BT > p[max].BT)
                 max = i;
         }
     }
  
     // returning the index of the process having the largest BT
     return max;
}
  
// function to find the completion time of each process
int findCT()
{
  
     int index;
     int flag = 0;
     int i = p[0].AT;
     while (1) {
         if (i <= 4) {
             index = findlargest(i);
         }
  
         else
             index = findlargest(4);
         cout << "Process executing at time " << totaltime
              << " is: P" << index + 1 << "\t" ;
  
         p[index].BT -= 1;
         totaltime += 1;
         i++;
  
         if (p[index].BT == 0) {
             p[index].CT = totaltime;
             cout << " Process P" << p[index].processno 
                  << " is completed at " << totaltime;
         }
         cout << endl;
  
         // loop termination condition
         if (totaltime == prefinaltotal)
             break ;
     }
}
  
int main()
{
  
     int i;
  
     // initializing the process number
     for (i = 0; i < 4; i++) {
         p[i].processno = i + 1;
     }
  
     // cout<<"arrival time of 4 processes : ";
     for (i = 0; i < 4; i++) // taking AT
     {
         p[i].AT = i + 1;
     }
  
     // cout<<" Burst time of 4 processes : ";
     for (i = 0; i < 4; i++) {
  
         // assigning {2, 4, 6, 8} as Burst Time to the processes
         // backup for displaying the output in last
         // calculating total required time for terminating 
         // the function().
         p[i].BT = 2 * (i + 1);
         p[i].BTbackup = p[i].BT;
         prefinaltotal += p[i].BT;
     }
  
     // displaying the process before executing
     cout << "PNo\tAT\tBT\n" ;
  
     for (i = 0; i < 4; i++) {
         cout << p[i].processno << "\t" ;
         cout << p[i].AT << "\t" ;
         cout << p[i].BT << "\t" ;
         cout << endl;
     }
     cout << endl;
  
     // soritng process according to Arrival Time
     sort(p, p + 4, compare);
  
     // calculating initial time when execution starts
     totaltime += p[0].AT;
  
     // calculating to terminate loop
     prefinaltotal += p[0].AT;
     findCT();
     int totalWT = 0;
     int totalTAT = 0;
     for (i = 0; i < 4; i++) {
         // since, TAT = CT - AT
         p[i].TAT = p[i].CT - p[i].AT;
         p[i].WT = p[i].TAT - p[i].BTbackup;
  
         // finding total waiting time
         totalWT += p[i].WT;
  
         // finding total turn around time
         totalTAT += p[i].TAT;
     }
  
     cout << "After execution of all processes ... \n" ;
  
     // after all process executes
     cout << "PNo\tAT\tBT\tCT\tTAT\tWT\n" ;
  
     for (i = 0; i < 4; i++) {
         cout << p[i].processno << "\t" ;
         cout << p[i].AT << "\t" ;
         cout << p[i].BTbackup << "\t" ;
         cout << p[i].CT << "\t" ;
         cout << p[i].TAT << "\t" ;
         cout << p[i].WT << "\t" ;
         cout << endl;
     }
  
     cout << endl;
     cout << "Total TAT = " << totalTAT << endl;
     cout << "Average TAT = " << totalTAT / 4.0 << endl;
     cout << "Total WT = " << totalWT << endl;
     cout << "Average WT = " << totalWT / 4.0 << endl;
     return 0;
}

Python3

# Python3 program to implement 
# Longest Remaining Time First 
  
# creating a structure of 4 processes
p = []
for i in range ( 4 ):
     p.append([ 0 , 0 , 0 , 0 , 0 , 0 , 0 ])
  
# variable to find the total time 
totaltime = 0
prefinaltotal = 0
  
# finding the largest Arrival Time 
# among all the available process 
# at that time 
def findlargest(at):
     max = 0
     for i in range ( 4 ):
         if (p[i][ 1 ] < = at):
             if (p[i][ 2 ] > p[ max ][ 2 ]) :
                 max = i 
      
     # returning the index of the 
     # process having the largest BT 
     return max
  
# function to find the completion
# time of each process 
def findCT(totaltime): 
     index = 0
     flag = 0
     i = p[ 0 ][ 1 ] 
     while ( 1 ): 
         if (i < = 4 ):
             index = findlargest(i) 
         else :
             index = findlargest( 4 ) 
         print ( "Process execute at time " , totaltime, end = " " )
         print ( " is: P" , index + 1 , sep = " ", end = " ")
         p[index][ 2 ] - = 1
         totaltime + = 1
         i + = 1
         if (p[index][ 2 ] = = 0 ): 
                 p[index][ 6 ] = totaltime 
                 print ( "Process P" , p[index][ 0 ], sep = " ", end = " ")
                 print ( " is completed at " , totaltime, end = " " )
         print ()
          
         # loop termination condition 
         if (totaltime = = prefinaltotal): 
             break
  
# Driver code 
if __name__ = = "__main__" :
      
     # initializing the process number 
     for i in range ( 4 ): 
         p[i][ 0 ] = i + 1
  
     for i in range ( 4 ): # taking AT
         p[i][ 1 ] = i + 1
  
     for i in range ( 4 ): 
  
         # assigning 2, 4, 6, 8 as Burst Time 
         # to the processes backup for displaying
         # the output in last calculating total
         # required time for terminating the function(). 
         p[i][ 2 ] = 2 * (i + 1 ) 
         p[i][ 3 ] = p[i][ 2 ] 
         prefinaltotal + = p[i][ 2 ] 
  
     # displaying the process before executing 
     print ( "PNo\tAT\tBT" )
  
     for i in range ( 4 ): 
         print (p[i][ 0 ], "\t" , p[i][ 1 ], "\t" , p[i][ 2 ])
     print ()
      
     # soritng process according to Arrival Time 
     p = sorted (p, key = lambda p:p[ 1 ]) 
  
     # calculating initial time when
     # execution starts 
     totaltime + = p[ 0 ][ 1 ]
  
     # calculating to terminate loop 
     prefinaltotal + = p[ 0 ][ 1 ] 
     findCT(totaltime) 
     totalWT = 0
     totalTAT = 0
     for i in range ( 4 ):
          
         # since, TAT = CT - AT 
         p[i][ 5 ] = p[i][ 6 ] - p[i][ 1 ] 
         p[i][ 4 ] = p[i][ 5 ] - p[i][ 3 ] 
  
         # finding total waiting time 
         totalWT + = p[i][ 4 ] 
  
         # finding total turn around time 
         totalTAT + = p[i][ 5 ] 
  
     print ( "\nAfter execution of all processes ... " )
  
     # after all process executes 
     print ( "PNo\tAT\tBT\tCT\tTAT\tWT" )
  
     for i in range ( 4 ): 
         print (p[i][ 0 ], "\t" , p[i][ 1 ], "\t" , p[i][ 3 ], "\t" , end = " " )
         print (p[i][ 6 ], "\t" , p[i][ 5 ], "\t" , p[i][ 4 ])
     print ()
     print ( "Total TAT = " , totalTAT)
     print ( "Average TAT = " , totalTAT / 4.0 )
     print ( "Total WT = " , totalWT)
     print ( "Average WT = " , totalWT / 4.0 )
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

输出如下:

PNo    AT    BT
1    1    2    
2    2    4    
3    3    6    
4    4    8    

Process executing at time 1 is: P1    
Process executing at time 2 is: P2    
Process executing at time 3 is: P3    
Process executing at time 4 is: P4    
Process executing at time 5 is: P4    
Process executing at time 6 is: P4    
Process executing at time 7 is: P3    
Process executing at time 8 is: P4    
Process executing at time 9 is: P3    
Process executing at time 10 is: P4    
Process executing at time 11 is: P2    
Process executing at time 12 is: P3    
Process executing at time 13 is: P4    
Process executing at time 14 is: P2    
Process executing at time 15 is: P3    
Process executing at time 16 is: P4    
Process executing at time 17 is: P1     Process P1 is completed at 18
Process executing at time 18 is: P2     Process P2 is completed at 19
Process executing at time 19 is: P3     Process P3 is completed at 20
Process executing at time 20 is: P4     Process P4 is completed at 21
After execution of all processes ... 
PNo    AT    BT    CT    TAT    WT
1    1    2    18    17    15    
2    2    4    19    17    13    
3    3    6    20    17    11    
4    4    8    21    17    9    

Total TAT = 68
Average TAT = 17
Total WT = 48
Average WT = 12

木子山

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