1 and n workpieces arrive at the same time;
2. Any workpiece can be processed on any machine;
3. The processing time of the workpiece is known;
Basic idea:
The short work priority algorithm is adopted to minimize the total processing time.
For a detailed description of this algorithm, please refer to any book on operating system principles.
Solution steps:
1, sort the workpieces according to the processing time, and the one that needs the least time ranks first;
2. Set an array U with the size of m to represent the elapsed time of each machine;
3. Set an array n with the size of m, indicating the time required for each machine to process the current workpiece;
4. Set an array r with the size of m to represent the running time of each machine;
4. Set all elements of the array u, r and n to 0;
5. Do the following for each element in the workpiece data:
Find the element x with the smallest value from the array n;
Taking the value of each element in the array n minus the value of the element x as its new value;
Adding the value of each element in the array R and the value of the element X as its new value;
Setting the value of element X as the processing time required by the workpiece;
Modify the values in the array u: U[x]+=R[x]+N[x];
6. Sum the values of the elements in the array U to get the shortest total time.
Good luck!
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The program is written in JAVA language, and the sorting operation is omitted, assuming that the number of machines is 3.
The public static void main(String[] arg) throws an exception.
{
int[] W=new int[]{ 1，2，3，55，56，77，77，88，99， 100 }； //Working time required for the workpiece
int[] U=new int[]{0，0，0 }； //The total time taken by each machine to process workpieces, including waiting time.
int[] R=new int[]{0，0，0 }； //Running time of each machine
int[] N=new int[]{0，0，0 }； //Time required for each machine to complete the current work.
for(int I = 0； I<w. Length; i++)
{
/* * Find the earliest available machine in the current machine, where m is the machine code and x is the time required for the machine to complete the current work */
int m = 0；
for(int j = 1； J<n. Length; j++)if(N[j]& lt； n[m])m = j；
int x = N[m]；
//Take the value of each element in array N minus the value of element X as its new value;
//Add the value of each element in the array R with the value of element X as its new value;
for(int j = 0； J<n. Length; j++)
{
n[j]-= x；
r[j]+= x；
}
//Set the value of element X (that is, M machine) as the processing time required by the workpiece;
n[m]= W[I]；
u[m]+=(R[m]+N[m])；
}
int total = 0；
for(int I = 0； I<u. Length; i++)total+= U[I]；
System.out.println ("total time equals:"+total ");
}