Optimal financing decision with financial constraints for a manufacturer in a low-carbon supply chain

Abstract

In this paper, bank financing (BF) and trade credit financing (TCF) are viable. We investigate the financing choice problem for an emission-dependent manufacturer with capital constraints. Each supply chain member pursues its profit maximization. In the literature on the financing supply chain, enterprises and consumers become increasingly aware of environmental protection. A growing number of manufacturers produce low-carbon products, such as environmentally friendly bags, through a green supply chain system. We use the Stackelberg game to study the equilibrium financing choice and optimal decisions. We also perform numerical analysis to verify the impact of certain parameters on financing decisions. The results show no direct relationship between the degree of carbon reduction and the total amount of carbon emissions as defined by the government. In addition, when the trade credit interest rate is higher than the bank interest rate, the manufacturer chooses bank financing. When the credit interest rate is lower than a certain threshold, the retailer provides trade credit financing. Our study also provides useful insights for managers to understand and make financing decisions in a low-carbon supply chain with a capital-constrained manufacturer.

Appendices

Appendix 1

Proof of Lemma 1

Let us find the first derivative of Eqs. (12) and (13) with respect to $\theta$, and the following results can be obtained.

$$\begin{array}{c}\frac{\partial {\pi }_{Bm}^{*}}{\partial \theta }=\frac{k(b{\theta }_{e}+\theta +b{p}_{e})(1+{r}_{B}){(-a+2bc+be{p}_{e}+bc{r}_{B})}^{2}}{{(A+4bk{r}_{B})}^{2}}>0\\ \frac{\partial {\pi }_{Br}^{*}}{\partial \theta }=\frac{4b{k}^{2}(b{\theta }_{e}+\theta +b{p}_{e}){(1+{r}_{B})}^{2}{(-a+2bc+be{p}_{e}+bc{r}_{B})}^{2}}{{(A+4bk{r}_{B})}^{3}}>0\end{array}$$

Proof of Lemma 2

Let us find the first derivative of Eqs. (19) and (20) with respect to $\theta$, and the following results can be obtained.

$$\begin{array}{c}\frac{\partial {\pi }_{Tr}^{*}}{\partial \theta }=\frac{1}{2}(1+{r}_{T})\frac{12b{k}^{2}{(-a+bc+be{p}_{e})}^{2}[2\theta +2b({\theta }_{e}+{p}_{e})](1+{r}_{T})}{{(A+4bk{r}_{T})}^{3}}>0\\ \frac{\partial {\pi }_{Tm}^{*}}{\partial \theta }=\frac{k(b{\theta }_{e}+\theta +b{p}_{e}){(-a+bc+be{p}_{e})}^{2}(1+{r}_{T})}{{(A+4bk{r}_{T})}^{2}}>0\end{array}$$

Proof 4.1

Under the BF, it can be deduced from Eq. (7) that the Hessian matrix is

$$H=\left\{\begin{array}{cc}-b& \frac{\theta }{2}-\frac{b{p}_{e}}{2}-\frac{b{\theta }_{e}}{2}\\ \frac{\theta }{2}-\frac{b{p}_{e}}{2}-\frac{b{\theta }_{e}}{2}& \theta {p}_{e}-k(1+{r}_{B})+\theta {\theta }_{e}\end{array}\right.$$

Then, $|{H}_{1}|=\frac{1}{4}(A+4bk{r}_{B})>0.$

The Hessian matrix is negative.

Based on the conditions, we identify the only solution that satisfies all conditions:

$${p}_{B}^{*}=\frac{a+bw+\Delta \theta }{2b}$$

Inserting ${p}_{B}^{*}$ into the manufacturer’s profit function produces:

$$\begin{array}{c}\frac{\partial {\pi }_{Bm}}{\partial w}=\frac{1}{2}[a+2bc-2bw+\Delta e\theta +b(e-\Delta e){p}_{e}+bc{r}_{B}-b\Delta e{\theta }_{e}]=0\\ \frac{\partial {\pi }_{Bm}}{\partial \Delta e}=\frac{1}{2}[-2k\Delta e-2c\theta +w\theta +(a-bw-e\theta +2\Delta e\theta ){p}_{e}-(2k\Delta e+c\theta ){r}_{B}+a{\theta }_{e}-bw{\theta }_{e}+2\Delta e\theta {\theta }_{e}]=0\end{array}$$

We can obtain $\left\{\begin{array}{c}{w}_{B}^{*},\Delta {e}_{B}^{*},{p}_{B}^{*}\\ {\pi }_{Bm}^{*},{\pi }_{Br}^{*}\end{array}\right.$

Proof 4.2

Under TCF, it can be deduced from Eq. (9) that the Hessian matrix is ${H}_{2}=\left(\begin{array}{cc}-b& \frac{\theta }{2}-\frac{b{p}_{e}}{2}-\frac{b{\theta }_{e}}{2}\\ \frac{\theta }{2}-\frac{b{p}_{e}}{2}-\frac{b{\theta }_{e}}{2}& \theta {p}_{e}-k(1+{r}_{T})+\theta {\theta }_{e}\end{array}\right)$.

Then, $|{H}_{2}|=\frac{1}{4}(A+4bk{r}_{T})>0$.

The Hessian matrix is negative.

Based on the conditions, we identify the only solution that satisfies all conditions:

$${p}_{T}^{*}=\frac{a-bc+bw+\Delta e\theta -bc{r}_{T}}{2b}$$

Inserting ${p}_{T}^{*}$ into the manufacturer’s profit function produces

$$\begin{array}{c}\frac{\partial {\pi }_{Tm}}{\partial w}=\frac{1}{2}[a+2bc-2bw+\Delta e\theta +b(e-\Delta e){p}_{e}+bc{r}_{T}-b\Delta e{\theta }_{e}]=0\\ \frac{\partial {\pi }_{Tm}}{\partial \Delta e}=\frac{1}{2}[-2k\Delta e(1+{r}_{T})+(-2c+w-c{r}_{T})\theta +{p}_{e}(a+bc-bw-e\theta +2\Delta e\theta +bc{r}_{T})+(a+bc-bw+2\Delta e\theta +bc{r}_{T}){\theta }_{e}]=0\end{array}$$

We can obtain $\left\{\begin{array}{c}{w}_{T}^{*},\Delta {e}_{T}^{*},{p}_{T}^{*}\\ {\pi }_{Tm}^{*},{\pi }_{Tr}^{*}\end{array}\right.$

Proof of Proposition 1

• ① We perform sensitivity analysis on the optimal wholesale price under the benchmark model.

$$\begin{array}{c}\frac{\partial {w}^{*}}{\partial e}=\frac{{p}_{e}[2bk-\theta (\theta +b{p}_{e}+b{\theta }_{e})]}{A}>0;\\ \frac{\partial {w}^{*}}{\partial {\theta }_{e}}=\frac{(a-bc-be{p}_{e})[{\theta }^{3}+b({p}_{e}+{\theta }_{e})(-4bk+2{\theta }^{2}+b\theta ({p}_{e}+{\theta }_{e}))]}{{A}^{2}}<0;\\ \frac{\partial {w}^{*}}{\partial {p}_{e}}=\frac{\begin{array}{c}2b(\theta +b({p}_{e}+{\theta }_{e}))((a({p}_{e}+{\theta }_{e})-\theta (c+e{p}_{e}))(\theta +b({p}_{e}+{\theta }_{e}))+2(a+bc+be{p}_{e})(k-\theta ({p}_{e}+{\theta }_{e})))\\ +[-2bek+\theta (a+3bc+e\theta )-2b(a-3e\theta ){p}_{e}+b(-2a+3e\theta ){\theta }_{e}]A\end{array}}{{A}^{2}}>0.\end{array}$$

• ② We perform sensitivity analysis on the optimal level of carbon emission reduction under the benchmark model.

$$\begin{array}{c}\frac{\partial \Delta {e}^{*}}{\partial e}=\frac{b{p}_{e}(\theta +b{p}_{e}+b{\theta }_{e})}{-A}<0;\\ \frac{\partial \Delta {e}^{*}}{\partial {\theta }_{e}}=\frac{b(a-bc-be{p}_{e})(4bk+{\theta }^{2}+b({p}_{e}+{\theta }_{e})(2\theta +b({p}_{e}+{\theta }_{e})))}{{A}^{2}}>0;\\ \frac{\partial \Delta {e}^{*}}{\partial {p}_{e}}=\frac{b(8bk(a-bc-be{p}_{e})-A(a-bc+e\theta +be{\theta }_{e}))}{{A}^{2}}>0.\end{array}$$

• ③ We perform sensitivity analysis on the optimal retail price under the benchmark model.

$$\begin{array}{c}\frac{\partial {p}^{*}}{\partial e}=\frac{{p}_{e}(bk-{\theta }^{2}-2b\theta {p}_{e}-2b\theta {\theta }_{e})}{A}>;0\\ \frac{\partial {p}^{*}}{\partial {\theta }_{e}}=\frac{\begin{array}{c}2b[(3a+bc)k-c{\theta }^{2}+{p}_{e}(b(ek-2c\theta )-\theta (a+e\theta -2be{p}_{e}))-\\ \theta (a+2bc+2be{p}_{e}){\theta }_{e}][\theta +b({p}_{e}+{\theta }_{e})]-\theta (a+2bc+2be{p}_{e})A\end{array}}{{A}^{2}}<0;\\ \frac{\partial {p}^{*}}{\partial {p}_{e}}=\frac{\begin{array}{c}[(3a+bc)k-c{\theta }^{2}+{p}_{e}(b(ek-2c\theta )-\theta (a+e\theta +2be{p}_{e}))-\theta (a+2bc+2be{p}_{e}){\theta }_{e}]2b(\theta +b{p}_{e}+b{\theta }_{e})\\ +[b(ek-2c\theta )-\theta (a+e\theta )-2be\theta ({\theta }_{e}+2{p}_{e})]A\end{array}}{{A}^{2}}>0.\end{array}$$

Proof of Proposition 2

• ① We perform sensitivity analysis on the optimal wholesale price under the BF.

$$\begin{array}{c}\frac{\partial {w}_{B}^{*}}{\partial e}=\frac{{p}_{e}[2bk(1+{r}_{B})-{\theta }^{2}-b\theta {p}_{e}-b\theta {\theta }_{e}]}{A+4bk{r}_{B}}>0;\\ \frac{\partial {w}_{B}^{*}}{\partial {\theta }_{e}}=\frac{\{\eta [{\theta }^{3}+b({p}_{e}+{\theta }_{e})(-4bk+2{\theta }^{2}-4bk{r}_{B}+b\theta ({p}_{e}+{\theta }_{e}))]\}}{{(A+4bk{r}_{B})}^{2}}>0,when-2bc-b(e{p}_{e}+c{r}_{B})+a=\eta <0.\\ \begin{array}{c}\frac{\partial {w}_{B}^{*}}{\partial {p}_{e}}=\frac{\theta (a+e\theta )-2b(ek-c\theta )+b[2(a+e\theta ){p}_{e}+(-2ek+c\theta ){r}_{B}+(2a+e\theta ){\theta }_{e}]}{A+4bk{r}_{B}}-\\ \frac{2b[\theta +b({p}_{e}+{\theta }_{e})]\{[(-a+e\theta ){p}_{e}+c\theta (2+{r}_{B})-a{\theta }_{e}](\theta -b({p}_{e}+{\theta }_{e}))+2(a+2bc+be{p}_{e}+bc{r}_{B})[-k(1+{r}_{B})+\theta ({p}_{e}+{\theta }_{e})]\}}{{(A+4bk{r}_{B})}^{2}}>0;\end{array}\end{array}$$

According to the shrinkage method, $\frac{\partial {w}_{B}^{*}}{\partial {p}_{e}}=\frac{\sigma }{A}(\frac{[4b(\theta +b({p}_{e}+{\theta }_{e}))-A]\psi +A}{{A}^{2}})>0$where

$$\begin{array}{c}\sigma ={r}_{B}(2ak+6bck-c{\theta }^{2}-bc\theta {\theta }_{e})+{p}_{e}(2bek-a\theta -2bc\theta -e{\theta }^{2}+b(2ek-c\theta ){r}_{B}-b(2a+e\theta ){\theta }_{e})>0;\\ \psi =2ak+4bck-2c{\theta }^{2}-b(a+e\theta ){p}_{e}^{2}+2bck{r}_{B}^{2}-a\theta {\theta }_{e}-2bc\theta {\theta }_{e}-ab{\theta }_{e}^{2}>0\end{array}$$

• ② We perform sensitivity analysis on the optimal level of carbon emission reduction under the BF.

$$\begin{array}{c}\frac{\partial \Delta {e}_{B}^{*}}{\partial {p}_{e}}=\frac{\begin{array}{c}b[a-2bc-b(e{p}_{e}+c{r}_{B})]\\ -be[\theta +b({p}_{e}+{\theta }_{e})]\end{array}}{A+4bk{r}_{B}}+\frac{[a-2bc-b(e{p}_{e}+c{r}_{B})](2b\theta +2{b}^{2}{p}_{e}+2{b}^{2}{\theta }_{e})[\theta +b({p}_{e}+{\theta }_{e})]}{{(A+4bk{r}_{B})}^{2}}>0;\\ \frac{\partial \Delta {e}_{B}^{*}}{\partial {\theta }_{e}}=-\frac{b(-a+2bc+be{p}_{e}+bc{r}_{B})[4bk+{\theta }^{2}+4bk{r}_{B}+b({p}_{e}+{\theta }_{e})(2\theta +b({p}_{e}+{\theta }_{e}))]}{{(A+4bk{r}_{B})}^{2}}>0;\\ \frac{\partial \Delta {e}_{B}^{*}}{\partial e}=\frac{-b\theta {p}_{e}-{b}^{2}{p}_{e}^{2}-{b}^{2}{p}_{e}{\theta }_{e}}{A+4bk{r}_{B}}<0\end{array}$$

• ③ We perform sensitivity analysis on the optimal retail price under the BF.

  $$\begin{array}{c}\frac{\partial {p}_{B}^{*}}{\partial e}=\frac{{p}_{e}[bk(1+{r}_{B})-{\theta }^{2}-b\theta ({p}_{e}+{\theta }_{e})]}{A+4bk{r}_{B}}>0;\\ \frac{\partial {p}_{B}^{*}}{\partial {\theta }_{e}}=-\frac{{p}_{e}(2b\theta +2{b}^{2}{p}_{e}+2{b}^{2}{\theta }_{e})[-bek+a\theta +2bc\theta +e{\theta }^{2}+b(-ek+c\theta ){r}_{B}+b(2a+e\theta ){\theta }_{e}]}{{(A+4bk{r}_{B})}^{2}}-\frac{b(2a+e\theta ){p}_{e}}{A+4bk{r}_{B}}<0;\end{array}$$

  where $E=-bek+a\theta +2bc\theta +e{\theta }^{2}+b(-ek+c\theta ){r}_{B}+b(2a+e\theta ){\theta }_{e}$.

According to the shrinkage method, when $\frac{(2a+e\theta )A-\theta E}{{A}^{2}}<0,$ $\frac{\partial {p}_{B}^{*}}{\partial {\theta }_{e}}<0$.

$$\begin{array}{c}\frac{\partial {p}_{B}^{*}}{\partial {p}_{e}}=\frac{\begin{array}{c}2b[\theta +b({p}_{e}+{\theta }_{e})]\{3ak+2bck-2c{\theta }^{2}-b(a+e\theta ){p}_{e}^{2}+{r}_{B}(3(a+bc)k-c{\theta }^{2}+bck{r}_{B})\\ -\theta (a+2bc+bc{r}_{B}){\theta }_{e}-ab{\theta }_{e}^{2}+{p}_{e}[b(ek-2c\theta )-\theta (a+e\theta )+b(ek-c\theta ){r}_{B}-b(2a+e\theta ){\theta }_{e}]\}\end{array}}{{(A+4bk{r}_{B})}^{2}}\\ -\frac{2b(a+e\theta ){p}_{e}+E}{A+4bk{r}_{B}}>0.\end{array}$$

Proof of Proposition 3

• ① We perform sensitivity analysis on the optimal retail price under TCF.

$$\begin{array}{c}\frac{\partial {p}_{T}^{*}}{\partial e}=\frac{{p}_{e}[bk(1+{r}_{T})-{\theta }^{2}-b\theta ({p}_{e}+{\theta }_{e})]}{A+4bk{r}_{T}}>0;\\ \frac{\partial {p}_{T}^{*}}{\partial {\theta }_{e}}=\frac{(a-bc-be{p}_{e})[2bk\theta +{\theta }^{3}+b(b\theta {p}_{e}^{2}+2k{r}_{T}(\theta -b{\theta }_{e})+{\theta }_{e}(-2bk+2{\theta }^{2}+b\theta {\theta }_{e})+2{p}_{e}(-bk+{\theta }^{2}-bk{r}_{T}+b\theta {\theta }_{e}))]}{{(A+4bk{r}_{T})}^{2}}<0;\\ \begin{array}{c}\frac{\partial {p}_{T}^{*}}{\partial {p}_{e}}=-\frac{{p}_{e}[-bek+a\theta +bc\theta +e{\theta }^{2}-bek{r}_{T}+b(2a+e\theta ){\theta }_{e}][2{b}^{2}{p}_{e}+2b(\theta +b{\theta }_{e})]}{{(A+4bk{r}_{T})}^{2}}-\\ \frac{2b(a+e\theta ){p}_{e}-bek+a\theta +bc\theta +e{\theta }^{2}-bek{r}_{T}+b(2a+e\theta ){\theta }_{e}}{A+4bk{r}_{T}}\\ -\frac{[-3ak-bck+c{\theta }^{2}+b(a+e\theta ){p}_{e}^{2}-(3a+bc)k{r}_{T}+a\theta {\theta }_{e}+bc\theta {\theta }_{e}+ab{\theta }_{e}^{2}][2{b}^{2}{p}_{e}+2b(\theta +b{\theta }_{e})]}{{(A+4bk{r}_{T})}^{2}}>0.\end{array}\end{array}$$

• ② We perform sensitivity analysis on the optimal wholesale price under TCF.

$$\begin{array}{c}\frac{\partial {w}_{T}^{*}}{\partial e}=\frac{{p}_{e}[2bk-{\theta }^{2}+2bk{r}_{T}-b\theta ({p}_{e}+{\theta }_{e})]}{A+4bk{r}_{T}}>0;\\ \frac{\partial {w}_{T}^{*}}{\partial {\theta }_{e}}=\frac{(a-bc-be{p}_{e})({\theta }^{3}+b({p}_{e}+{\theta }_{e})[-4bk+2{\theta }^{2}-4bk{r}_{T}+b\theta ({p}_{e}+{\theta }_{e})])}{{(A+4bk{r}_{T})}^{2}}<0;\\ \begin{array}{c}\frac{\partial {w}_{T}^{*}}{\partial {p}_{e}}=\frac{b(2ek-3c\theta )-\theta (a+e\theta )-b(2a+2bc+e\theta ){\theta }_{e}-2b{r}_{T}(-ek+c\theta +bc{\theta }_{e})-2b{p}_{e}(a+bc+e\theta +bc{r}_{T})}{A+4bk{r}_{T}}+\\ \frac{2b(\theta +b{p}_{e}+b{\theta }_{e})(2ak+6bck-2c{\theta }^{2}+4bck{r}_{T}^{2}-a\theta {\theta }_{e}-3bc\theta {\theta }_{e}-ab{\theta }_{e}^{2}-{b}^{2}c{\theta }_{e}^{2})}{{(A+4bk{r}_{T})}^{2}}-\\ \frac{{p}_{e}(2b\theta +2{b}^{2}{p}_{e}+2{b}^{2}{\theta }_{e})[b(-2ek+3c\theta )+\theta (a+e\theta )+b(2a+2bc+e\theta ){\theta }_{e}+2b{r}_{T}(-ek+c\theta +bc{\theta }_{e})]}{{(A+4bk{r}_{T})}^{2}}\\ +\frac{2b(\theta +b{p}_{e}+b{\theta }_{e})[-b{p}_{e}^{2}(a+bc+e\theta +bc{r}_{T})+{r}_{T}(2(a+5bc)k-c{\theta }^{2}-bc{\theta }_{e}(2\theta +b{\theta }_{e}))]}{{(A+4bk{r}_{T})}^{2}}>0.\end{array}\end{array}$$

• ③ We perform sensitivity analysis on the optimal level of carbon emission reduction under TCF.

$$\begin{array}{c}\frac{\partial \Delta {e}_{T}^{*}}{\partial e}=\frac{-b\theta {p}_{e}-{b}^{2}{p}_{e}^{2}-{b}^{2}{p}_{e}{\theta }_{e}}{A+4bk{r}_{T}}<0;\\ \begin{array}{c}\frac{\partial \Delta {e}_{T}^{*}}{\partial {p}_{e}}=\frac{b[a-bc-e\theta -be(2{p}_{e}+{\theta }_{e})](A+4bk{r}_{T})}{{(A+4bk{r}_{T})}^{2}}+\frac{2b(a-bc-be{p}_{e}){[\theta +b({p}_{e}+{\theta }_{e})]}^{2}}{{(A+4bk{r}_{T})}^{2}}>0;\\ \frac{\partial \Delta {e}_{T}^{*}}{\partial {\theta }_{e}}=\frac{b(a-bc-be{p}_{e})(4bk+{\theta }^{2}+4bk{r}_{T}+b({p}_{e}+{\theta }_{e})[2\theta +b({p}_{e}+{\theta }_{e})])}{{(A+4bk{r}_{T})}^{2}}>0\end{array}\end{array}$$

Appendix 2

Proof of Proposition 4

• (1) When ${r}_{B}<\frac{A}{[2c{\theta }^{2}+b(a+e\theta ){p}_{e}^{2}+a\theta {\theta }_{e}+2bc\theta {\theta }_{e}+ab{\theta }_{e}^{2}-3ak]}$, ${p}_{B}^{*}-{p}_{T}^{*}>0$.

• (2) $\Delta {e}_{B}^{*}-\Delta {e}_{T}^{*}=[\theta +b({p}_{e}+{\theta }_{e})](\frac{a-2bc-be{p}_{e}-bc{r}_{B}}{A+4bk{r}_{B}}+\frac{bc+be{p}_{e}-a}{A+4bk{r}_{T}})<0$.

• (3) When ${r}_{T}<\frac{a-2cb-eb{p}_{e}}{cb}$, ${w}_{B}^{*}-{w}_{T}^{*}<0$.

Proof of Proposition 5

• (1) According to the shrinkage method, ${\pi }_{Bm}^{*}-{\pi }_{Tm}^{*}=\frac{(2AB+kz)({r}_{T}-{r}_{B})}{2A}.$

When $0<{r}_{B}<{r}_{T}$, ${\pi }_{Bm}^{*}-{\pi }_{Tm}^{*}>0$; or ${r}_{B}>{r}_{T}>0$, ${\pi }_{Bm}^{*}-{\pi }_{Tm}^{*}<0$.

• (2) ${\pi }_{Br}^{*}-{\pi }_{Tr}^{*}=\frac{b{k}^{2}z{(1+{r}_{B})}^{2}}{{(A+4bk{r}_{B})}^{2}}-\frac{1}{2}(1+{r}_{T})(-2B-\frac{kz}{A+4bk{r}_{T}}+\frac{6b{k}^{2}z(1+{r}_{T})}{{(A+4bk{r}_{T})}^{2}})$

where

$$\begin{array}{c}z={\left(-a+2bc+be{p}_{e}+bc{r}_{B}\right)}^{2}\\ \begin{array}{c}m=\frac{2B{(A+4bk{r}_{T})}^{2}-12b{k}^{2}z+k(A+4bk{r}_{T})z+\frac{(A+4bk{r}_{T})l}{A+4bk{r}_{B}}}{12b{k}^{2}z},\\ \begin{array}{c}n=\frac{2B{(A+4bk{r}_{T})}^{2}-12b{k}^{2}z+k(A+4bk{r}_{T})z-\frac{(A+4bk{r}_{T})l}{A+4bk{r}_{B}}}{12b{k}^{2}z},\\ l=\sqrt{48{b}^{2}{k}^{4}{z}^{2}{r}_{B}(2+{r}_{B})+4B{(A+4bk{r}_{B})}^{2}(A+4bk{r}_{T})(AB+k+4bBk{r}_{T})+{k}^{2}{z}^{2}(48{b}^{2}{k}^{2}+k{(A+4bk{r}_{B})}^{2}+B(A+4bk{r}_{B})(A+4bk{r}_{T}))}.\end{array}\end{array}\end{array}$$

If ${r}_{T}>\mathrm{min}{\{m,n\}}^{+}$, we can obtain the result ${\pi }_{Br}^{*}>{\pi }_{Tr}^{*}$.

If $0<{r}_{T}<\mathrm{min}{\{m,n\}}^{+}$, we can obtain the result ${\pi }_{Br}^{*}<{\pi }_{Tr}^{*}$.