Datasets:

Monketoo
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math-docs-dataset

$\begin{align*} \text{Minimize: } & y \tag{7} \\ \text{subject to: } & w_i(f_i(x,z) - f_i^L) \le y, \quad (j=1,\dots,k) \\ & g_j(d,z) < 0, \quad (j=1,\dots,m) \\ & x_i^L \le x_i \le x_i^U, \quad (i=1,\dots,n_x) \end{align*}$

Step 5: If the objective function values are satisfactory, the search is finished. Otherwise, update the aspiration level $f_i^A$ and return to Step 2.

The weight coefficient $w_i$ plays an important role in obtaining the Pareto solution in the direction of the aspiration level, which is directly related to the designer's preference. As shown in Fig. 4, the Pareto optimal solution is usually located on the line connecting the ideal point and the aspiration level in the objective function space, regardless of whether or not the aspiration level lies in the feasible region. When Eq. (7) is solved using a nonlinear programming method, an accurate Pareto optimal solution is obtained.

An accurate Pareto set is obtained by parametrically changing the aspiration level. On the other hand, the designers can investigate only the desired region in detail by arranging the aspiration level properly without obtaining the full Pareto set.

4. Numerical Examples

4.1 Example 1: two-bar truss structure

Consider the following problem involving the two-bar truss structure shown in Fig. 5 [9]. The original design problem is to find the nominal diameter for member $x_1$ and the height of structure $x_2$ to minimize the element stress under the constraints of the total volume $g_1(x)$ and buckling stress $g_2(x)$. The external force $F$ is set at 150 kN, the thickness of the circular tube $T$ is set at 2.5 mm, the width of structure $B$ is set at 750 mm, and the elastic modulus $E$ is set at 210 GPa. The original optimization problem is formulated as follows:

$\text{Minimize:} \quad f(x) = \frac{F\sqrt{B^2 + x_2^2}}{2\pi T x_1 x_2} \qquad (8)$

Fig. 5. Two-bar truss design problem

$\begin{array}{ll} \text{subject to:} & g_1(\mathbf{x}) = \frac{2\pi T x_1 \sqrt{B^2 + x_2^2}}{70000} \le 1 \\ & g_2(\mathbf{x}) = \frac{4F_1(B^2 + x_2^2)^{3/2}}{E\pi^3 T x_1 x_2(T^2 + x_1^2)} \le 1 \\ & 1 \le x_1 \le 100 \\ & 1 \le x_2 \le 1000 \end{array}$

The robust design problem is considered when the member diameter and height have uncertainties, with ($Δx_1$, $Δx_2$) used to denote random variables, where the mean value of $x_1$ and $x_2$ are treated as design variables. In this example, the standard deviation of a random variable is assumed to be $Δx_i/3$ [9]. The robust multiobjective design problem is formulated as follows:

$\begin{align} \text{Minimize:} \quad & f_1(\mathbf{x}) = f(E[\mathbf{x}]) = \frac{F\sqrt{B^2 + x_2^2}}{2\pi T x_1 x_2} \tag{9} \\ & f_2(\mathbf{x}, \Delta\mathbf{x}) = \sqrt{\mathrm{Var}[f(\mathbf{x}, \Delta\mathbf{x})]} \nonumber \\ & = \frac{1}{3}\sqrt{\left(\frac{\partial f}{\partial x_1}\right)^2 \Delta x_1^2 + \left(\frac{\partial f}{\partial x_2}\right)^2 \Delta x_2^2} \nonumber \\ \text{subject to:} \quad & \hat{g}_1(\mathbf{x}) = g_1(E[\mathbf{x}]) + \sqrt{\mathrm{Var}[g_1(\mathbf{x})]} \le 1 \nonumber \\ & \hat{g}_2(\mathbf{x}) = g_2(E[\mathbf{x}]) + \sqrt{\mathrm{Var}[g_2(\mathbf{x})]} \le 1 \nonumber \\ & 1 + \Delta x_1 \le x_1 \le 10 - \Delta x_1 \nonumber \\ & 10 + \Delta x_2 \le x_2 \le 100 - \Delta x_2 \nonumber \end{align}$

where $\hat{g}_j(x)$ are the constraints for the considered variations of the design variables evaluated as the first-order approximation [8]. The constraints mean the volume and buckling load limits should be satisfied under variations.

The Pareto front under several values of $\Delta x$ is obtained by parametrically changing the aspiration level, as shown in Fig. 6. The ideal points are set by solving each single-objective function problem, as listed in Table 1. Fig. 6 (a) shows the Pareto set when changing $\Delta x_1$ to 1.0, 1.5, and 2.0, where $\Delta x_2$ is constant at $\Delta x_2 = 5.0$. On the other hand, Fig. 6 (b) corresponds to the Pareto set when changing $\Delta x_2$ to 5.0, 7.0, and 10.0, where $\Delta x_1$ is constant at $\Delta x_1 = 1.0$. Note that the blue