Optimize Your Assembly Line Scheduling Using Dynamic Programming in C

Optimize Your Assembly Line Scheduling Using Dynamic Programming in C

Subtitle: Streamline your production process with an efficient dynamic programming solution for assembly line scheduling.

In manufacturing, the assembly line scheduling problem is crucial for optimizing production efficiency. Using dynamic programming, we can determine the fastest way to process tasks through two assembly lines, minimizing the total time required. In this blog, we will explore a C language implementation of this solution.

Problem Explanation

In an assembly line scheduling problem, we have two assembly lines with \(n\) stations each. Tasks can switch between lines but incur a switching cost. The goal is to find the optimal sequence that minimizes the total processing time.

C Implementation

Here’s a step-by-step C implementation of the assembly line scheduling using dynamic programming:

#include <stdio.h>

int n, i, fe, le;
int a[2][10], t[2][9], e[2], x[2], f1[10], f2[10], l[2][10];

void display() {
    int j, i = le;
    printf("\nThe Optimal path is:\n");
    for (j = 1; j < n; ++j) {
        i = l[i-1][j];
        printf("line %d ", i);
        printf("station %d \n", j);
    }
    i = le;
    printf("line %d ", i);
    printf("station %d \n", n);
}

void fastest_way() {
    f1[0] = e[0] + a[0][0];
    f2[0] = e[1] + a[1][0];
    int j;

    for (j = 1; j < n; ++j) {
        if ((f1[j-1] + a[0][j]) <= (f2[j-1] + t[1][j-1] + a[0][j])) {
            f1[j] = f1[j-1] + a[0][j];
            l[0][j] = 1;
        } else {
            f1[j] = f2[j-1] + t[1][j-1] + a[0][j];
            l[0][j] = 2;
        }
        if ((f2[j-1] + a[1][j]) <= (f1[j-1] + t[0][j-1] + a[1][j])) {
            f2[j] = f2[j-1] + a[1][j];
            l[1][j] = 2;
        } else {
            f2[j] = f1[j-1] + t[0][j-1] + a[1][j];
            l[1][j] = 1;
        }
    }

    if ((f1[n-1] + x[0]) <= (f2[n-1] + x[1])) {
        fe = f1[n-1] + x[0];
        le = 1;
    } else {
        fe = f2[n-1] + x[1];
        le = 2;
    }

    display();
}

void main() {
    printf("Dynamic Programming: Assembly Line Scheduling >>\n\n");
    printf("Enter the number of nodes: ");
    scanf("%d", &n);
    printf("\nEnter The Entry Chassis:\n");
    for (i = 0; i < 2; i++)
        scanf("%d", &e[i]);

    printf("Enter The Exit Chassis:\n");
    for (i = 0; i < 2; i++)
        scanf("%d", &x[i]);
    printf("\nEnter The Assembly Time at Line S1:\n");
    for (i = 0; i < n; ++i)
        scanf("%d", &a[0][i]);
    printf("\nEnter The Assembly Time at Line S2:\n");
    for (i = 0; i < n; ++i)
        scanf("%d", &a[1][i]);
    printf("\nEnter Time Required to Change from line S1 to S2:\n");
    for (i = 0; i < n-1; ++i)
        scanf("%d", &t[0][i]);
    printf("\nEnter Time Required to Change from line S2 to S1:\n");
    for (i = 0; i < n-1; ++i)
        scanf("%d", &t[1][i]);
    printf("\n");
    fastest_way();
    printf("\nOptimal Path Costs are:\n");
    i = 1;
    printf("For Path[%d] ->\t", i);
    for (int j = 0; j < n; j++)
        printf("S[%d][%d]=%d\t", i, j+1, f1[j]);
    printf("Total cost: %d", fe);
    printf("\n");
    i = 2;
    printf("For Path[%d] ->\t", i);
    for (int j = 0; j < n; j++)
        printf("S[%d][%d]=%d\t", i, j+1, f2[j]);
    fe = f2[n-1] + x[1];
    printf("Total cost: %d", fe);
}

Explanation:

1. Initialization:
— Arrays `a` and `t` store the assembly and transition times.
— Arrays `e` and `x` store entry and exit times for the assembly lines.
— Arrays `f1` and `f2` store the minimum time to reach each station on both lines.
— Array `l` keeps track of the line choices.

2. Input Handling:
— The number of stations and times are taken as input from the user.

3. Dynamic Programming Calculation:
— The minimum time to reach each station is calculated by considering both staying on the same line and switching lines.
— The results are stored in `f1` and `f2`, with the path tracked in `l`.

4. Display:
— The optimal path and its cost are printed.

Conclusion

Using dynamic programming for assembly line scheduling allows for efficient calculation of the optimal production path, minimizing the total processing time. The provided C implementation illustrates this approach clearly, making it easier to understand and apply to real-world problems. Experiment with different inputs to see how the algorithm adapts to various scenarios. Happy coding!

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