Longest Remaining Time First (LRTF) CPU Scheduling Program
We have given some processes with arrival time and Burst Time and we have to find the completion time (CT), Turn Around Time(TAT), Average Turn Around Time (Avg TAT), Waiting Time(WT), Average Waiting Time (AWT) for the given processes.
Prerequisite: CPU Scheduling | Longest Remaining Time First (LRTF) algorithm
LRTF is a preemptive scheduling algorithm. Its tie-breaker is FCFS and if FCFS does not breaks the tie then, we use process id as the tie-breaker.
Example: Consider the following table of arrival time and burst time for four processes P1, P2, P3, and P4.
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
Gantt chart will be as following below,
Since completion time (CT) can be directly determined by Gantt chart, and
Turn Around Time (TAT) = (Completion Time) - (Arrival Time) Also, Waiting Time (WT) = (Turn Around Time) - (Burst Time)
Therefore,
Output:
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
Algorithm:
- Step-1: Create a structure of process containing all necessary fields like AT (Arrival Time), BT(Burst Time), CT(Completion Time), TAT(Turn Around Time), WT(Waiting Time).
- Step-2: Sort according to the AT;
- Step-3: Find the process having Largest Burst Time and execute for each single unit. Increase the total time by 1 and reduce the Burst Time of that process with 1.
- Step-4: When any process have 0 BT left, then update the CT(Completion Time of that process CT will be Total Time at that time).
- Step-2: After calculating the CT for each process, find TAT and WT.
(TAT = CT - AT) (WT = TAT - BT)
Implementation of Algorithm3s
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 structure 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; // sorting 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; } |
Java
// Java Program to implement // longest remaining time first import java.util.*; class GFG { // creating a class of a process static class process { int processno; int AT; int BT; // for backup purpose to print in last int BTbackup; int WT; int TAT; int CT; } static process[] p = new process[ 4 ]; // variable to find the total time static int totaltime = 0 ; static int prefinaltotal = 0 ; // finding the largest Arrival Time among all the // available process at that time static 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 static void findCT() { int index; int flag = 0 ; int i = p[ 0 ].AT; while ( true ) { if (i <= 4 ) { index = findlargest(i); } else index = findlargest( 4 ); System.out.print( "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; System.out.println( " Process P" + p[index].processno + " is completed at " + totaltime); } System.out.println(); // loop termination condition if (totaltime == prefinaltotal) break ; } } public static void main(String[] args) { int i; // initializing the process number for (i = 0 ; i < 4 ; i++) { p[i] = new process(); p[i].processno = i + 1 ; } for (i = 0 ; i < 4 ; i++) // taking AT { p[i].AT = i + 1 ; } 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 System.out.print( "PNo\tAT\tBT\n" ); for (i = 0 ; i < 4 ; i++) { System.out.print(p[i].processno + "\t" ); System.out.print(p[i].AT + "\t" ); System.out.println(p[i].BT + "\t" ); } System.out.println(); Arrays.sort(p, (process p1, process p2) -> { return p1.AT - p2.AT; }); // 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; } System.out.print( "After execution of all processes ... \n" ); // after all process executes System.out.print( "PNo\tAT\tBT\tCT\tTAT\tWT\n" ); for (i = 0 ; i < 4 ; i++) { System.out.print(p[i].processno + "\t" ); System.out.print(p[i].AT + "\t" ); System.out.print(p[i].BTbackup + "\t" ); System.out.print(p[i].CT + "\t" ); System.out.print(p[i].TAT + "\t" ); System.out.println(p[i].WT + "\t" ); } System.out.println(); System.out.println( "Total TAT = " + totalTAT); System.out.println( "Average TAT = " + (totalTAT / 4.0 )); System.out.println( "Total WT = " + totalWT); System.out.println( "Average WT = " + totalWT / 4.0 ); } } // This code is contributed by Karandeep Singh |
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 () # sorting 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) |
Output:
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
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