from desdeo_problem.problem.Variable import Variable
from desdeo_problem.problem.Objective import ScalarObjective
from desdeo_problem.problem.Problem import MOProblem, ProblemBase
import numpy as np
[docs]def river_pollution_problem(five_obj: bool = True, var_iv: np.array = np.array([0.5, 0.5])) -> MOProblem:
"""The river pollution problem with 4 or 5 objectives.
NARULA, S. C. & WEISTROFFER, H. R. (1989). A flexible method for
nonlinear multicriteria decisionmaking problems. IEEE transactions on
systems, man, and cybernetics, 19(4), 883-887.
Arguments:
five_obj (bool): If true utilize five objectives version and four objectives
version if false. Default is true.
var_iv (np.array): Optional, initial variable values. Must be between 0.3 and 1.0.
Defaults are 0.5 and 0.5.
Returns:
MOProblem: a problem object.
"""
if np.any(1 < var_iv) or np.any(var_iv < 0.3):
raise ValueError("Initial variable values need to be between lower and upper bounds")
def f_1(x: np.ndarray) -> np.ndarray:
x = np.atleast_2d(x)
return -4.07 - 2.27*x[:, 0]
def f_2(x: np.ndarray) -> np.ndarray:
x = np.atleast_2d(x)
return -2.60 - 0.03*x[:, 0] - 0.02*x[:, 1] - 0.01 / (1.39 - x[:, 0]**2) - 0.30 / (1.39 - x[:, 1]**2)
def f_3(x: np.ndarray) -> np.ndarray:
x = np.atleast_2d(x)
return -8.21 + 0.71 / (1.09 - x[:, 0]**2)
def f_4(x: np.ndarray) -> np.ndarray:
x = np.atleast_2d(x)
return -0.96 + 0.96 / (1.09 - x[:, 1]**2)
objective_1 = ScalarObjective(name="the DO level in the city", evaluator=f_1, maximize=[True])
objective_2 = ScalarObjective(name="the DO level at the municipality border", evaluator=f_2, maximize=[True])
objective_3 = ScalarObjective(name="the percent return on investment at the fishery", evaluator=f_3, maximize=[True])
objective_4 = ScalarObjective(name="the addition to the tax rate of city", evaluator=f_4, maximize=[True])
if five_obj:
def f_5(x: np.ndarray) -> np.ndarray:
return np.max([np.abs(x[:, 0] - 0.65), np.abs(x[:, 1] - 0.65)], axis=0)
objective_5 = ScalarObjective(name="BOD removed form the water close to the ideal value of 0.65", evaluator=f_5, maximize=[False])
objectives = [objective_1, objective_2, objective_3, objective_4, objective_5]
else:
# If five_obj is false, then problem is with 4 objectives.
objectives = [objective_1, objective_2, objective_3, objective_4]
x_1 = Variable("the proportionate amount of BOD removed from water at the fishery", var_iv[0], 0.3, 1.0)
x_2 = Variable("the proportionate amount of BOD removed from water at the city", var_iv[1], 0.3, 1.0)
variables = [x_1, x_2]
problem = MOProblem(variables=variables, objectives=objectives)
return problem
