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) )。
示例：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，
%20CPU%20Scheduling%20Program_0.jpg)
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) 所以，
输出：
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 –
- 步骤1：创建一个流程结构，其中包含所有必填字段，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)。
- 步骤2：根据AT进行排序；
- 步骤3：找到爆发时间最大的进程，然后对每个单元执行一次。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CT的完成时间将是该时间的总时间)。
- 步骤2：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
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：
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