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Longest Remaining Time First (LRTF) CPU Scheduling Program
• Difficulty Level : Hard
• Last Updated : 14 Aug, 2019

Prerequisite – CPU Scheduling | Longest Remaining Time First (LRTF) algorithm
We have given some process 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.

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, complietion time (CT) can be directly determined by Gantt chart, and

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

## C++

 #include   using namespace std;  // creating a structure of a processstruct 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 processesstruct process p[4];  // variable to find the total timeint 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 timeint 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 processint 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 processesp = []for i in range(4):    p.append([0, 0, 0, 0, 0, 0, 0])  # variable to find the total time totaltime = 0prefinaltotal = 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)

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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