benchmarkfcns.multiobjective.cf3

benchmarkfcns.multiobjective.cf3(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 CF3 constrained multi-objective benchmark function. SCORES = multiobjective.cf3(X) computes the value of the CF3 function at point X. multiobjective.cf3 accepts a matrix of size M-by-N and returns a matrix SCORES of size M-by-2. If return_constraints is True, returns an M-by-3 matrix where the last column contains the constraint violation (values > 0 are violations). Properties:

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

Mathematical Definition

\[\begin{aligned}\]

f_1(textbf{x}) &= x_1 + frac{2}{|J_1|} [4 sum_{j in J_1} y_j^2 - 2 prod_{j in J_1} cos(frac{20y_jpi}{sqrt{j}}) + 2] \ f_2(textbf{x}) &= 1 - x_1^2 + frac{2}{|J_2|} [4 sum_{j in J_2} y_j^2 - 2 prod_{j in J_2} cos(frac{20y_jpi}{sqrt{j}}) + 2] \ y_j &= x_j - sin(6pi x_1 + frac{jpi}{n}) \ text{Subject to: } & f_2 + f_1^2 - a sin(Npi(f_1^2 - f_2 + 1)) - 1 ge 0 \ & x_1 in [0, 1], x_j in [-2, 2], quad N=2, a=1 end{aligned}

Visualization

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