benchmarkfcns.multiobjective.cf10

benchmarkfcns.multiobjective.cf10(x: Annotated[numpy.typing.NDArray[numpy.float64], '[m, n]', 'flags.c_contiguous'], return_constraints: bool = False) → Annotated[numpy.typing.NDArray[numpy.float64], '[m, n]']

Computes the value of the CEC 2009 CF10 constrained multi-objective benchmark function. SCORES = multiobjective.cf10(X) computes the value of the CF10 function at point X. multiobjective.cf10 accepts a matrix of size M-by-N and returns a matrix SCORES of size M-by-3. If return_constraints is True, returns an M-by-4 matrix where the last column contains the constraint violation (values > 0 are violations). Properties:

• Recommended domain: x1, x2 in [0, 1], xj in [-2, 2] for j=3..N

Mathematical Definition

\[\begin{aligned}\]

f_1(textbf{x}) &= cos(0.5x_1pi) cos(0.5x_2pi) + frac{2}{|J_1|} sum_{j in J_1} [4y_j^2 - 2cos(frac{20pi y_j}{sqrt{j}}) + 2] \ f_2(textbf{x}) &= cos(0.5x_1pi) sin(0.5x_2pi) + frac{2}{|J_2|} sum_{j in J_2} [4y_j^2 - 2cos(frac{20pi y_j}{sqrt{j}}) + 2] \ f_3(textbf{x}) &= sin(0.5x_1pi) + frac{2}{|J_3|} sum_{j in J_3} [4y_j^2 - 2cos(frac{20pi y_j}{sqrt{j}}) + 2] \ y_j &= x_j - 2x_2 sin(2pi x_1 + frac{jpi}{n}) \ text{Subject to: } & 1.0 - (f_1^2 + f_2^2 + f_3^2) + sin(2pi (f_1^2 + f_2^2 + f_3^2)) ge 0 \ & x_1, x_2 in [0, 1], x_j in [-2, 2] end{aligned}

Visualization

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