Port-based supply chain decisions considering governmental pollution tax

Abstract

Considering the adverse impacts of pollutant emissions from port areas on the environment, an analytical model is developed to investigate optimal decisions for two port-based supply chains with or without governmental pollution tax. When the governmental pollution tax is introduced, the impacts of local governments independently determining pollution tax and jointly determining pollution tax on the environment and transport chains are investigated, respectively. The results show that the enterprises will not pay attention to the pollution if the governmental pollution tax is not introduced, which will result in the higher pollution level than the optimal level. It is interesting to note that negative pollution tax (i.e., subsidies) is more likely to occur when local governments set pollution tax separately. When local governments cooperate with each other to determine pollution tax, the pollution spillover problem can be solved, and optimal social welfare can be achieved. Thus, cooperation between local governments is more effective. Furthermore, a numerical example is given to analyze the influences of the local pollution damage coefficient and the spillover pollution damage coefficient on the optimal strategies of all participants.

Appendix

Proof of Corollary 1.

It can be easily obtained by derivation that

$$\begin{gathered} \frac{{\partial w_{i}^{*} \left( {t_{1} ,t_{2} } \right)}}{{\partial t_{i} }} = \frac{{n\left( {b^{2} - 8} \right)}}{{16 - b^{2} }} < 0,\quad \frac{{\partial w_{i}^{*} \left( {t_{1} ,t_{2} } \right)}}{{\partial t_{j} }} = \frac{2bn}{{16 - b^{2} }} > 0; \hfill \\ \frac{{\partial q_{i}^{*} \left( {t_{1} ,t_{2} } \right)}}{{\partial t_{i} }} = \frac{{2n\left( {b^{2} - 8} \right)}}{{b^{4} - 20b^{2} + 64}} < 0,\quad \frac{{\partial q_{i}^{*} \left( {t_{1} ,t_{2} } \right)}}{{\partial t_{j} }} = \frac{4bn}{{b^{4} - 20b^{2} + 64}} > 0. \hfill \\ \end{gathered}$$

Proof of Corollary 2 .

Because $\in \left( {0,1} \right)$, then we get

$$\begin{gathered} \frac{{\partial t_{i}^{L*} }}{\partial x} = \frac{{ - b^{4} + 2b^{3} + 16b^{2} - 16b - 64}}{{b^{3} + 4b^{2} - 8b - 16}} > 0,\quad \frac{{\partial w_{i}^{L*} }}{\partial x} = \frac{{n\left( {b^{2} - 8} \right)\left( {b^{2} - 4} \right)}}{{b^{3} + 4b^{2} - 8b - 16}} < 0; \hfill \\ \frac{{\partial t_{i}^{L*} }}{\partial y} = \frac{{ - 2b^{3} + 4b^{2} + 16b}}{{b^{3} + 4b^{2} - 8b - 16}} < 0,\quad \frac{{\partial q_{i}^{L*} }}{\partial x} = - \frac{{2n\left( {b^{2} - 8} \right)}}{{b^{3} + 4b^{2} - 8b - 16}} < 0; \hfill \\ \frac{{\partial w_{i}^{L*} }}{\partial y} = \frac{{2nb\left( {b^{2} - 4} \right)}}{{b^{3} + 4b^{2} - 8b - 16}} > 0,\quad \frac{{\partial q_{i}^{L*} }}{\partial y} = - \frac{4nb}{{b^{3} + 4b^{2} - 8b - 16}} > 0. \hfill \\ \end{gathered}$$

Proof of Corollary 3 .

Similar to the proof of Corollary 1, we can easily get

$$\begin{gathered} \frac{{\partial t_{i}^{C*} }}{\partial x} = \frac{{ - b^{2} + 2b + 8}}{2b + 2} > 0,\quad \frac{{\partial w_{i}^{C*} }}{\partial x} = \frac{{n\left( {b^{2} - 4} \right)}}{2b + 2} < 0,\quad \frac{{\partial q_{i}^{C*} }}{\partial x} = - \frac{n}{b + 1} < 0; \hfill \\ \frac{{\partial t_{i}^{C*} }}{\partial y} = \frac{{ - b^{2} + 2b + 8}}{2b + 2} > 0,\quad \frac{{\partial w_{i}^{C*} }}{\partial y} = \frac{{n\left( {b^{2} - 4} \right)}}{2b + 2} < 0,\quad \frac{{\partial q_{i}^{C*} }}{\partial y} = - \frac{n}{b + 1} < 0. \hfill \\ \end{gathered}$$

Proof of Corollary 4 .

• 1. When $c_{1} = c_{2}$,

  $$t_{i}^{C*} - t_{i}^{L*} = \frac{{\left( {b - 4} \right)^{2} \left( {b + 2} \right)^{2} \left[ {\left( {a - c_{1} } \right)b + \left( {by - bx + 2y} \right)n} \right]}}{{2n\left( {b + 1} \right)\left( {16 + 8b - 4b^{2} - b^{3} } \right)}}$$

  Because ${ }b \in \left( {0,1} \right)$ and $16 + 8b - 4b^{2} - b^{3} > 0$, we can verify that if $a - c_{1} > \frac{{\left( {bx - by - 2y} \right)n}}{b}$, then $t_{i}^{C*} > t_{i}^{L*}$; otherwise $t_{i}^{C*} < t_{i}^{L*}$.

• 2. When $c_{1} = c_{2}$,

  $$w_{i}^{C*} - w_{i}^{L*} = - \frac{{\left( {b + 2} \right)^{2} \left( {b - 2} \right)\left( {b - 4} \right)\left[ {\left( {a - c_{1} } \right)b + \left( {by - bx + 2y} \right)n} \right]}}{{2\left( {b + 1} \right)\left( {16 + 8b - 4b^{2} - b^{3} } \right)}}$$

  Similarly, if $a - c_{1} > \frac{{\left( {bx - by - 2y} \right)n}}{b}$, then $w_{i}^{C*} < w_{i}^{L*}$; otherwise $w_{i}^{C*} > w_{i}^{L*}$.

• 3. When $c_{1} = c_{2}$,

  $$q_{i}^{C*} - q_{i}^{L*} = \frac{{\left( {b + 2} \right)\left( {b - 4} \right)\left[ {\left( {a - c_{1} } \right)b + \left( {by - bx + 2y} \right)n} \right]}}{{\left( {b + 1} \right)\left( {16 + 8b - 4b^{2} - b^{3} } \right)}}$$

  If $a - c_{1} > \frac{{\left( {bx - by - 2y} \right)n}}{b}$, we get $q_{i}^{C*} < q_{i}^{L*}$; otherwise $q_{i}^{C*} > q_{i}^{L*}$.

• 4. Substituting $q_{i}^{C*}$ and $q_{i}^{L*}$ into Eq. (1), (2) respectively, we can get $p_{i}^{C*}$ and $p_{i}^{L*}$. When $c_{1} = c_{2}$,

  $$p_{i}^{C*} - p_{i}^{L*} = - \frac{{\left( {b + 2} \right)\left( {b - 4} \right)\left[ {\left( {a - c_{1} } \right)b + \left( {by - bx + 2y} \right)n} \right]}}{{16 + 8b - 4b^{2} - b^{3} }}$$

If $a - c_{1} > \frac{{\left( {bx - by - 2y} \right)n}}{b}$, then $p_{i}^{C*} > p_{i}^{L*}$; otherwise $p_{i}^{C*} < p_{i}^{L*}$.

Proof of Corollary 5 .

• 1. When $c_{1} = c_{2}$, the profit of the port in different scenarios are: $\pi_{{P_{i} }}^{C*} = w_{i}^{C*} q_{i}^{C*}$, $\pi_{{P_{i} }}^{L*} = w_{i}^{L*} q_{i}^{L*}$,

  From the previous proof, we know that if $a - c_{1} > \frac{{\left( {bx - by - 2y} \right)n}}{b}$, then $w_{i}^{C*} < w_{i}^{L*} ,q_{i}^{C*} < q_{i}^{L*}$, so $w_{i}^{C*} q_{i}^{C*} < w_{i}^{L*} q_{i}^{L*}$, i.e. $\pi_{{P_{i} }}^{C*} < \pi_{{P_{i} }}^{L*}$; otherwise $\pi_{{P_{i} }}^{C*} > \pi_{{P_{i} }}^{L*}$.

• 2. When $c_{1} = c_{2}$,

  $$sw^{C*} - \left( {sw_{1}^{L*} + sw_{2}^{L*} } \right) = \frac{{\left( {b - 4} \right)^{2} \left( {b + 2} \right)^{2} \left[ {\left( {a - c_{1} } \right)b + \left( {by - bx + 2y} \right)n} \right]^{2} }}{{\left( {b + 1} \right)\left( {16 + 8b - 4b^{2} - b^{3} } \right)^{2} }}$$

We have that $sw^{C*} > \left( {sw_{1}^{L*} + sw_{2}^{L*} } \right)$.

Proof of Corollary 6 .

When $c_{1} = c_{2}$,

$$D_{i}^{C*} - D_{i}^{L*} = \frac{{n\left( {b + 2} \right)\left( {b - 4} \right)\left( {x + y} \right)\left[ {\left( {a - c_{1} } \right)b + \left( {by - bx + 2y} \right)n} \right]}}{{\left( {b + 1} \right)\left( {16 + 8b - 4b^{2} - b^{3} } \right)}}$$

If $a - c_{1} > \frac{{\left( {bx - by - 2y} \right)n}}{b}$, then $D_{i}^{C*} < D_{i}^{L*}$; otherwise $D_{i}^{C*} > D_{i}^{L*}$.