## 4 Methods for Generating a Near-Optimal Subset of Pareto Optimal Solutions
### Revised \(\epsilon\)-Constraint Method  
Considering that numerous weakly efficient solutions may be generated by the basic \(\epsilon\)-constraint method, a revised \(\epsilon\)-constraint method is introduced to address the drawback of the basic version. We solve the following single-objective optimization models by varying \(k\) from 0, 1,..., all the way to \(K:=(f_{1}^{\max}-f_{1}^{\min})/\epsilon\):  
\[\text{[M2($k$)]: Objective (2)}\]  
subject to  
Constraints (3)-(4),  
and let \(\mathbf{\hat{x}}(k)\) be an optimal solution obtained. We then solve the following model:  
\[\text{[M3($k$)]:}\ \min f_{1}(\mathbf{x}) \tag{5}\]  
subject to  
\[\text{Constraint}\quad\eqref{constraint}\quad\] \[f_{2}(\mathbf{x}) \leq f_{2}(\mathbf{\hat{x}}(k)), \tag{6}\]and let \(\mathbf{x}^{*}(k)\) be an optimal solution obtained. The following lemmas, propositions, and corollaries are proposed. **Lemma 1**. : \(f_{1}(\mathbf{x}^{*}(k))\leq f_{1}(\mathbf{\hat{x}}(k))\) _and \(f_{2}(\mathbf{x}^{*}(k))\leq f_{2}(\mathbf{\hat{x}}(k))\)._  
Proof. : Since \(f_{2}(\mathbf{\hat{x}}(k))\leq f_{2}(\mathbf{\hat{x}}(k))\), \(\mathbf{\hat{x}}(k)\) is feasible to [M3(\(k\))], and therefore, \(f_{1}(\mathbf{x}^{*}(k))\leq f_{1}(\mathbf{\hat{x}}(k))\) due to the optimality of \(\mathbf{x}^{*}(k)\) to [M3(\(k\))]. Since \(\mathbf{x}^{*}(k)\) is optimal, and thus, feasible to [M3(\(k\))], it satisfies the constraint \(f_{2}(\mathbf{x}^{*}(k))\leq f_{2}(\mathbf{\hat{x}}(k))\). According to Lemma 1, Corollary 1 is derived. **Corollary 1**. : \(\mathbf{x}^{*}(k)\) _is an optimal solution to [M2(\(k\))]._  
**Proposition 1**. : \(\mathbf{x}^{*}(k)\) _is a Pareto optimal solution._  
Proof. : Suppose there is an \(\mathbf{x}^{\#}\in X\) such that \(f_{1}(\mathbf{x}^{\#})<f_{1}(\mathbf{x}^{*}(k))\) and \(f_{2}(\mathbf{x}^{\#})\leq f_{2}(\mathbf{x}^{*}(k))\). Then, this contradicts the optimality of \(\mathbf{x}^{*}(k)\) to [M3(\(k\))]. Suppose there is an \(\mathbf{x}^{\#}\in X\) such that \(f_{1}(\mathbf{x}^{\#})\leq f_{1}(\mathbf{x}^{*}(k))\) and \(f_{2}(\mathbf{x}^{\#})<f_{2}(\mathbf{x}^{*}(k))\). Combined with Lemma 1, \(f_{1}(\mathbf{x}^{\#})\leq f_{1}(\mathbf{\hat{x}}(k))\) and \(f_{2}(\mathbf{x}^{\#})<f_{2}(\mathbf{\hat{x}}(k))\). Then, this contradicts the optimality of \(\mathbf{\hat{x}}(k)\) to [M2(\(k\))]. Define a multiset \(\Omega:=\{\mathbf{x}^{*}(k);k=0,\ldots,K\}\). If there are multiple solutions in \(\Omega\) that have the same objective vector, keep only one of these solutions. **Proposition 2**. : _Let \(\Phi\) be the full set of Pareto optimal solutions. Then, \(\Omega\) is a \"good representation\" of \(\Phi\) in the sense that for any \(\mathbf{\tilde{x}}\in\Phi\), there exists an \(\mathbf{x}^{*}\in\Omega\) satisfying \(f_{1}(\mathbf{x}^{*})<f_{1}(\mathbf{\tilde{x}})+\epsilon\) and \(f_{2}(\mathbf{x}^{*})\leq f_{2}(\mathbf{\tilde{x}})\)._  
Proof. : For any \(\mathbf{\tilde{x}}\in\Phi\), let \(k\) be an integer such that \(f_{1}^{\min}+(k-1)\epsilon<f_{1}(\mathbf{\tilde{x}})\leq f_{1}^{\min}+k\epsilon\); evidently, \(k\in\{0,\ldots,K\}\). We now prove that \(\mathbf{x}^{*}(k)\) satisfies \(f_{1}(\mathbf{x}^{*}(k))<f_{1}(\mathbf{\tilde{x}})+\epsilon\) and \(f_{2}(\mathbf{x}^{*}(k))\leq f_{2}(\mathbf{\tilde{x}})\). Lemma 1 states that \(f_{1}(\mathbf{x}^{*}(k))\leq f_{1}(\mathbf{\hat{x}}(k))\). Since \(\mathbf{\hat{x}}(k)\) is optimal, and thus, feasible to [M2(\(k\))]\(f_{1}(\mathbf{\hat{x}}(k))\leq f_{1}^{\min}+k\epsilon\). Therefore, \(f_{1}(\mathbf{x}^{*}(k))\leq f_{1}^{\min}+k\epsilon\). Combined with \(f_{1}^{\min}+(k-1)\epsilon<f_{1}(\mathbf{\tilde{x}})\), we have \(f_{1}(\mathbf{x}^{*}(k))<f_{1}(\mathbf{\hat{x}})+\epsilon\). Evidently, \(\mathbf{\tilde{x}}\) is feasible to [M2(\(k\))]. Since \(\mathbf{\hat{x}}(k)\) is optimal to [M2(\(k\))], \(f_{2}(\mathbf{\hat{x}}(k))\leq f_{2}(\mathbf{\tilde{x}})\). Lemma 1 states that \(f_{2}(\mathbf{x}^{*}(k))\leq f_{2}(\mathbf{\hat{x}}(k))\). Therefore, \(f_{2}(\mathbf{x}^{*}(k))\leq f_{2}(\mathbf{\tilde{x}})\). Proposition 2 immediately implies Corollary 2 as follows.