River pollution problem

The problem is to improve the dissolved oxygen concentration in the city \((f_1)\) and at the municipality border \((f_2)\), maximize the percent return on investment at the fishery \((f_3)\), and minimize addition to the city tax \((f_4)\). Here is optional fifth objective [1] to keeping to proportional amount of biomechanical oxygen demanding material (BOD) removed from the water close to the ideal value \((f_5)\). More details about the test problem can be found in [2].

Definition

\[\begin{split}\min \; f_1(x) = & -4.07 - 2.27x_1 \\[2mm] \min \; f_2(x) = & -2.60 - 0.03x_1 - 0.02x_2 - \frac{0.01}{1.39 - x_1^2} - \frac{0.30}{1.39-x_2^2} \\[2mm] \min \; f_3(x) = & -8.21 + \frac{0.71}{1.09 - x_1^2} \\[2mm] \min \; f_4(x) = & -0.96 + \frac{0.96}{1.09 - x_2^2} \\ \\ \text{Optional fifth objective:}\\[2mm] \min \; f_5(x) = & \max \{ |x_1 - 0.65|, |x_2 - 0.65| \} \\\end{split}\]

The two variables represent the proportionate amount of BOD removed from water discharge at the fishery \((x_1)\) and at the city \((x_2)\).

Variable bounds are given as follows:

\[0.3 \leq x_1 \leq 1.0 \quad \quad 0.3 \leq x_2 \leq 1.0\]