dual simplex method calculator with code examples

The dual simplex method is one of the most widely used optimization algorithms in linear programming. It is a variant of the simplex method – an algorithm that maximizes or minimizes a linear objective function subject to a set of linear constraints. The dual simplex method is used when the primal problem is infeasible or unbounded, or when the objective function is maximization. It is also used to solve problems with multiple optimal solutions.

A dual simplex method calculator is a tool that implements the dual simplex algorithm to solve linear programming problems. The calculator typically takes inputs in the form of coefficients of the objective function and the constraints, and outputs the optimal solution to the problem.

The algorithm of the dual simplex method involves two phases. The first phase is the primal feasibility phase, where the slack variables are added to convert the problem into a feasible form. The second phase is the dual simplex method, where the algorithm iteratively moves from one feasible basis to another until the optimal solution is found.

Let's look at an example of solving a linear programming problem using the dual simplex method calculator and code.

Example:
Maximize z = 3×1 + 2×2
Subject to:
x1 + x2 <= 4
2×1 + x2 <= 5
x1, x2 >= 0

Using the dual simplex method calculator, we can input the coefficients of the objective function and the constraints as follows:

Objective function: 3×1 + 2×2
Constraints:
x1 + x2 <= 4
2×1 + x2 <= 5

The calculator then outputs the optimal solution to the problem as z = 10/3, x1 = 2/3, x2 = 10/3.

Let's now look at an example code implementation of the dual simplex method algorithm in Python:

import numpy as np

def dual_simplex(c, A, b):
    """
    Solve a linear programming problem using the dual simplex method.
    Maximize c.T * x
    Subject to:
        A * x <= b
        x >= 0
    Returns the optimal solution x and the optimal value of the objective function.
    """
    m, n = A.shape

    # Convert the problem to a maximization problem by negating the objective function
    c = -c

    # Phase 1: Primal feasibility phase
    # Add slack variables to convert the problem into a feasible form
    slack_A = np.hstack([A, np.eye(m)])
    c = np.hstack([c, np.zeros(m)])
    basis = np.arange(n, n + m)
    x = np.zeros(n + m)
    z = 0

    while True:
        # Calculate the simplex multipliers
        B = slack_A[:, basis]
        cB = c[basis]
        pi = np.linalg.solve(B.T, cB)

        # Calculate the reduced costs
        reduced_c = c - slack_A.T @ pi

        # Check if the basis is optimal
        if np.all(reduced_c[basis] >= 0):
            break

        # Choose the entering variable with the most negative reduced cost
        entering = np.argmin(reduced_c)

        # Calculate the leaving variable
        B_inv = np.linalg.inv(B)
        d = B_inv @ slack_A[:, entering]

        if np.all(d <= 0):
            return None, -np.inf

        leaving = basis[np.argmin(x[basis] / d)]

        # Update the basis
        basis = np.setdiff1d(basis, leaving)
        basis = np.append(basis, entering)

        # Update the solution and the objective function value
        x = np.zeros(n + m)
        x[basis] = np.linalg.solve(B, b)
        z = -c @ x

    # Phase 2: Dual simplex method
    while True:
        # Calculate the simplex multipliers
        B = slack_A[:, basis]
        cB = c[basis]
        pi = np.linalg.solve(B.T, cB)

        # Calculate the reduced costs
        reduced_c = c - slack_A.T @ pi

        # Check if the basis is optimal
        if np.all(reduced_c[basis] >= 0):
            break

        # Choose the leaving variable with the smallest x value
        leaving = basis[np.argmin(x[basis])]

        # Calculate the entering variable
        d = np.zeros(n + m)
        d[basis] = np.linalg.solve(B, slack_A[:, leaving])
        d[leaving] = -1

        entering = np.argmin(reduced_c / d)

        # Update the basis
        basis = np.setdiff1d(basis, leaving)
        basis = np.append(basis, entering)

        # Update the solution and the objective function value
        x = np.zeros(n + m)
        x[basis] = np.linalg.solve(B, b)
        z = -c @ x

    return x[:n], z

# Test the function with the example problem
c = np.array([3, 2])
A = np.array([[1, 1], [2, 1]])
b = np.array([4, 5])
x, z = dual_simplex(c, A, b)
print("Optimal solution:", x)
print("Optimal value:", z)

The code implementation works as follows:

The input parameters to the function are the coefficients of the objective function c, the matrix of constraints A, and the vector of constraint values b.
The function first converts the problem to a maximization problem by negating the objective function.
The function then proceeds with the primal feasibility phase, where slack variables are added to convert the problem into a feasible form. The slack variables are added to the constraints in the form of a slack matrix slack_A, and the new objective function coefficients are added in the form of a new objective function vector c.
The function initializes the basis to be the indices of the slack variables, and sets the starting solution x and objective function value z to be zero.
The function then enters a loop that iteratively performs the following steps:
- Calculate the simplex multipliers pi using the current basis.
- Calculate the reduced costs reduced_c using the current simplex multipliers.
- Check if the current basis is optimal by checking if all the reduced costs with respect to the basis are non-negative. If they are, the function exits the loop and proceeds to phase 2. Otherwise, it proceeds with finding the entering and leaving variables.
- The function chooses the entering variable with the most negative reduced cost, and calculates the leaving variable by finding the variable in the basis with the smallest ratio of current solution value to the value of the corresponding element in the direction vector.
- The function updates the basis, solution, and objective function value using the new entering and leaving variables.
The function then proceeds with phase 2, the dual simplex method. This phase is similar to the previous phase, except that the function uses the current solution as the starting point and proceeds by iteratively finding the entering and leaving variables. The function terminates when the basis becomes optimal.
The function returns the optimal solution x and optimal value z.

In conclusion, the dual simplex method is a powerful algorithm for solving linear programming problems with multiple optimal solutions or with an unbounded or infeasible primal problem. The dual simplex method calculator is a useful tool for implementing this algorithm and obtaining optimal solutions to linear programming problems. The code example provided here demonstrates how to implement the dual simplex method algorithm in Python.

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dual simplex method calculator with code examples

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Surarchith Midhunakula

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Surarchith Midhunakula

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