## Appendix A  
_Appendix A.1. Proofs of Theorems 1 and 2_  
**Proof of Theorem 1. ** We demonstrate this via a synthetic counterexample. Assume that there is only one candidate route \(r_{1}\), which excludes U.S. ports and needs three ships for operation, i.e., \(N_{r_{1}}=3\). There is only one OD pair \((o,d)\), and its weekly transport demand is 2000 TEUs, i.e., \(D_{od}=2000\) TEUs. The revenue earned per TEU for this OD pair is set as \(p_{od}=800\) USD. Consider that the company owns three ships of the same type, \(s_{1}\) (\(|S|=1\)), i.e., \(A_{s_{1}}=3\). The transport capacity of each ship is 6000 TEUs, i.e., \(Q_{s_{1}}=6000\) TEUs. The transportation cost incurred by each ship deployed on route \(r_{1}\) consists of the following components: the fuel cost \(F_{s_{1}r_{1}}=0.6\) million USD, the berthing cost \(B_{s_{1}r_{1}}=0.6\) million USD, and the extra penalty cost of \(B_{s_{1}r_{1}}^{\text{extra}}=0\), since the route does not include any U.S. ports. The weekly cost for chartering in a ship of type \(s_{1}\) is set as \(c_{s_{1}}^{\text{in}}=0.5\) million USD, and the weekly revenue from chartering out one idle ship is set as \(c_{s_{1}}^{\text{out}}=0.4\) million USD. Then, for the original model, there will be two possible route operation and ship assignment solutions: (1) Operating no route, i.e., \(x_{r_{1}}=0\). This solution is feasible, and the company can charter out all three ships to earn revenue. The weekly profit in this case is \(c_{s_{1}}^{\text{out}}n_{s_{1}}^{\text{out}}=0.4\times 3=1.2\) million USD, which is the first objective of the model. The total weekly transport volume for the second objective in this case is \(z_{adr_{1}}=0\). (2) Operating route \(r_{1}\), i.e., \(x_{r_{1}}=1\). In this case, the company deploys all three owned ships on the route, i.e., \(n_{s_{1}r_{1}}=3\), and no ship is chartered in or out, i.e., \(n_{s_{1}}^{\text{in}}=n_{s_{1}}^{\text{out}}=0\). Since the demand is 2000 TEUs and each ship has a capacity of 6000 TEUs, the demand is fully satisfied. Then, the weekly profit in this case can be calculated as follows:  
\[ \begin{array}{l}p_{od}z_{adr_{1}}-n_{s_{1}r_{1}}/N_{r_{1}}(F_{s_{1}r_{1}}+B_ {s_{1}r_{1}})-n_{s_{1}r_{1}}/N_{r_{1}}\cdot B_{s_{1}r_{1}}^{\text{extra}}\\ =800/10^{6}\times 2000-(0.6+0.6)-0=0.4\text{ million USD}.\end{array} \]  
The total weekly transport volume in this case is 2000 TEUs. It is notable that strategy (1) yields a higher weekly profit (1.2 million USD), but the transport volume is zero. In contrast, strategy (2) results in a lower weekly profit (0.4 million USD), yet it fully satisfies the demand, achieving a transport volume of 2000 TEUs. These two solutions represent trade-offs between the two conflicting objectives of the model. Therefore, both solutions constitute a set of Pareto-optimal solutions under the original MIP model. However, if we relax the binary variable \(x_{r}\) to be continuous, \(x_{r_{1}}=1/3\) will also satisfy the weekly operation requirement, as the number of required ships becomes \(N_{r_{1}}x_{r_{1}}=3\times 1/3=1\). The company can assign one owned ship to route \(r_{1}\), i.e., \(n_{s_{1}r_{1}}=1\), and charter out the remaining two idle ships, i.e., \(n_{s_{1}}^{\text{out}}=2\). Since the ship capacity is 6000 TEUs and the weekly transport demand is 2000 TEUs, \(D_{od}=Q_{s_{1}}n_{s_{1}r_{1}}/N_{r_{1}}\). Therefore, the demand of the OD pair can be fully transported, i.e., \(z_{adr_{1}}=2000\) TEUs, and thus constraint (1-7) can be satisfied. The first objective function value, then, can be calculated as follows:  
\[ \begin{array}{l}p_{od}z_{odr_{1}}-n_{s_{1}r_{1}}/N_{r_{1}}(F_{s_{1}r_{1}}+B_ {s_{1}r_{1}})-n_{s_{1}r_{1}}/N_{r_{1}}\cdot B_{s_{1}r_{1}}^{\text{extra}}+c_{s_ {1}}^{\text{out}}n_{s_{1}}^{\text{out}}\\ =800/10^{6}\times 2000-\frac{1}{3}(0.6+0.6)-0+0.4\times 2=2.0\text{ million USD}.\end{array} \]The second objective is 2000 TEUs.