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Installation

You can install pymoors directly from PyPi doing

pip install pymoors

Example: The Multi-Objective Knapsack Problem

The multi-objective knapsack problem is a classic example in optimization where we aim to select items, each with its own benefits and costs, subject to certain constraints (e.g., weight capacity). In the multi-objective version, we want to optimize more than one objective function simultaneously—often, maximizing multiple benefits or qualities at once.

Mathematical Formulation

Suppose we have \(n\) items. Each item \(i\) has: - A profit \(p_i\). - A quality \(q_i\). - A weight \(w_i\).

Let \(x_i\) be a binary decision variable where \(x_i = 1\) if item \(i\) is selected and \(x_i = 0\) otherwise. We define a knapsack capacity \(C\). A common multi-objective formulation for this problem is:

\[ \begin{aligned} &\text{Maximize} && f_1(x) = \sum_{i=1}^{n} p_i x_i \\ &\text{Maximize} && f_2(x) = \sum_{i=1}^{n} q_i x_i \\ &\text{subject to} && \sum_{i=1}^{n} w_i x_i \leq C,\\ & && x_i \in \{0, 1\}, \quad i = 1,\dots,n. \end{aligned} \]

Solving it with pymoors

import numpy as np

from pymoors import (
    Nsga2,
    RandomSamplingBinary,
    BitFlipMutation,
    SinglePointBinaryCrossover,
    ExactDuplicatesCleaner,
)
from pymoors.typing import TwoDArray


PROFITS = np.array([2, 3, 6, 1, 4])
QUALITIES = np.array([5, 2, 1, 6, 4])
WEIGHTS = np.array([2, 3, 6, 2, 3])
CAPACITY = 7


def knapsack_fitness(genes: TwoDArray) -> TwoDArray:
    # Calculate total profit
    profit_sum = np.sum(PROFITS * genes, axis=1, keepdims=True)
    # Calculate total quality
    quality_sum = np.sum(QUALITIES * genes, axis=1, keepdims=True)

    # We want to maximize profit and quality,
    # so in pymoors we minimize the negative values
    f1 = -profit_sum
    f2 = -quality_sum
    return np.column_stack([f1, f2])


def knapsack_constraint(genes: TwoDArray) -> TwoDArray:
    # Calculate total weight
    weight_sum = np.sum(WEIGHTS * genes, axis=1, keepdims=True)
    # Inequality constraint: weight_sum <= capacity
    return weight_sum - CAPACITY


algorithm = Nsga2(
    sampler=RandomSamplingBinary(),
    crossover=SinglePointBinaryCrossover(),
    mutation=BitFlipMutation(gene_mutation_rate=0.5),
    fitness_fn=knapsack_fitness,
    constraints_fn=knapsack_constraint,
    duplicates_cleaner=ExactDuplicatesCleaner(),
    n_vars=5,
    population_size=32,
    n_offsprings=32,
    num_iterations=100,
    mutation_rate=0.1,
    crossover_rate=0.9,
    keep_infeasible=False,
)

algorithm.run()

In this small example, the algorithm finds a single solution on the Pareto front: selecting the items (A, D, E), with a profit of 7 and a quality of 15. This means there is no other combination that can match or exceed both objectives without exceeding the knapsack capacity (7).

Once the algorithm finishes, it stores a population attribute that contains all the individuals evaluated during the search.

pop = algorithm.population
# Get genes
>>> pop.genes
array([[1., 0., 0., 1., 1.],
       [0., 1., 0., 0., 1.],
       [1., 1., 0., 1., 0.],
       [0., 0., 0., 1., 1.],
       [1., 0., 0., 0., 1.],
       [1., 0., 0., 1., 0.],
       [1., 1., 0., 0., 0.],
       [0., 0., 1., 0., 0.],
       [0., 1., 0., 1., 0.],
       [1., 0., 0., 0., 0.],
       [0., 0., 0., 1., 0.],
       [0., 0., 0., 0., 1.],
       [0., 1., 0., 0., 0.],
       [0., 0., 0., 0., 0.]])
# Get fitness
>>> pop.fitness
array([[ -7., -15.],
       [ -7.,  -6.],
       [ -6., -13.],
       [ -5., -10.],
       [ -6.,  -9.],
       [ -3., -11.],
       [ -5.,  -7.],
       [ -6.,  -1.],
       [ -4.,  -8.],
       [ -2.,  -5.],
       [ -1.,  -6.],
       [ -4.,  -4.],
       [ -3.,  -2.],
       [ -0.,  -0.]])
# Get constraints evaluation
>>> pop.constraints
array([[ 0.],
       [-1.],
       [ 0.],
       [-2.],
       [-2.],
       [-3.],
       [-2.],
       [-1.],
       [-2.],
       [-5.],
       [-5.],
       [-4.],
       [-4.],
       [-7.]])
# Get rank
>>> pop.rank
array([0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6], dtype=uint64)

Note that in this example there is just one individual with rank 0, i.e Pareto optimal. Algorithms in pymoors store all individuals with rank 0 in a special attribute best, which is list of pymoors.schemas.Individual objects

# Get best individuals
best = pop.best
>>> best
[<pymoors.schemas.Individual object at 0x11b8ec110>]
# In this small exmaple as mentioned, best is just one single individual (A, D, E)
>>> best[0].genes
array([1., 0., 0., 1., 1.])
>>> best[0].fitness
array([ -7., -15.])
>>> best[0].constraints
array([0.])

Population Size and Duplicates

Note that although the specified population_size was 32, the final population ended up being 13 individuals, of which 1 had rank = 0. This is because we used the keep_infeasible=False argument, removing any individual that did not satisfy the constraints (in this case, the weight constraint). We also used a duplicate remover called ExactDuplicatesCleaner that eliminates all exact duplicates— meaning whenever genes1 == genes2 in every component.

Variable Types in pymoors

In pymoors, there is no strict enforcement of whether variables are integer, binary, or real. The core Rust implementation works with f64 ndarrays. To preserve a specific variable type—binary, integer, or real—you must ensure that the genetic operators themselves maintain it.

It is the user's responsibility to choose the appropriate genetic operators for the variable type in question. In the knapsack example, we use binary-style genetic operators, which is why the solutions are arrays of 0s and 1s.

Example: A Real-Valued Multi-Objective Optimization Problem

Below is a simple two-variable multi-objective problem to illustrate real-valued optimization with pymoors. We have two continuous decision variables, \( x_1 \) and \( x_2 \), both within a given range. We define two objective functions to be minimized simultaneously, and we solve this using the popular NSGA2 algorithm.

Mathematical Formulation

Let \(\mathbf{x} = (x_1, x_2)\) be our decision variables, each constrained to the interval \([-2, 2]\). We define the following objectives:

\[ \begin{aligned} \min_{\mathbf{x}} \quad &f_1(x_1, x_2) = x_1^2 + x_2^2 \\ \min_{\mathbf{x}} \quad &f_2(x_1, x_2) = (x_1 - 1)^2 + x_2^2 \\ \text{subject to} \quad & -2 \le x_1 \le 2, \\ & -2 \le x_2 \le 2. \end{aligned} \]

Interpretation

  1. \(f_1\) measures the distance of \(\mathbf{x}\) from the origin \((0,0)\) in the 2D plane.
  2. \(f_2\) measures the distance of \(\mathbf{x}\) from the point \((1,0)\).

Thus, \(\mathbf{x}\) must compromise between being close to \((0,0)\) and being close to \((1,0)\). There is no single point in \([-2,2]^2\) that simultaneously minimizes both distances perfectly (other than at the boundary of these trade-offs), so we end up with a Pareto front rather than a single best solution.

Solving it with pymoors

Below is a complete example in Python demonstrating how to set up this problem using and solve it with the NSGA2 genetic algorithm.

import numpy as np

from pymoors import (
    Nsga2,
    RandomSamplingFloat,
    GaussianMutation,
    ExponentialCrossover,
    CloseDuplicatesCleaner
)
from pymoors.typing import TwoDArray


def fitness(genes: TwoDArray) -> TwoDArray:
    x1 = genes[:, 0]
    x2 = genes[:, 1]
    # Objective 1: Distance to (0,0)
    f1 = x1**2 + x2**2
    # Objective 2: Distance to (1,0)
    f2 = (x1 - 1)**2 + x2**2
    # Combine the two objectives into a single array
    return np.column_stack([f1, f2])



algorithm = Nsga2(
    sampler=RandomSamplingFloat(min = -2, max = 2),
    crossover=ExponentialCrossover(exponential_crossover_rate = 0.75),
    mutation=GaussianMutation(gene_mutation_rate=0.1, sigma=0.01),
    fitness_fn=fitness,
    duplicates_cleaner=CloseDuplicatesCleaner(epsilon=1e-5),
    n_vars=2,
    population_size=200,
    n_offsprings=200,
    num_iterations=200,
    mutation_rate=0.1,
    crossover_rate=0.9,
    keep_infeasible=False,
    lower_bound=-2,
    upper_bound=2
)

algorithm.run()

This simple problem has a known Pareto Optimal

  • Decision-Space Pareto Front: \( \{ (x_1, 0) \mid x_1 \in [0,1] \} \)

  • Objective-Space Pareto Front: \( \{ (t^2, (t-1)^2) \mid t \in [0,1] \} \)

Each point on that curve represents a different trade-off between minimizing the distance to \((0,0)\) and to \((1,0)\).

pymoors best front