Pricing and financing strategies of a dual-channel supply chain with a capital-constrained manufacturer

Abstract

This study explores the operation decisions of a dual-channel supply chain consisting of a capital-constrained manufacturer, an e-commerce platform (ECP), and a third-party logistics company (3PL). This study first proposes two game models to obtain the equilibrium solutions of the supply chain members. Then, it compares and analyzes the equilibrium strategies under the two financing modes. This study has obtained the following interesting findings. The increase of the ECP interest rate will reduce the profit of the unit product in the distribution channel. However, the total income of the ECP will increase with the increase in the financing income. Under the 3PL financing mode, if the transportation fee of the direct sales channel is lower than that of the distribution channel, the ECP will reduce the sales price to compete with the direct sales channel of the manufacturer. Under the same conditions, the ECP financing mode is more competitive than the 3PL financing mode, and ECP financing services are the dominant strategy for the manufacturer, ECP, and consumers. If a 3PL company wants to win in the financing service, lowering interest rate or implementing differentiated transportation charges are important measures.

Appendices

Appendix

Proof of Proposition 1

After substituting $p_{{1}}^{E} - \hat{\omega } - t_{1} = n$ into the profit function $\pi_{m}^{E}$, the first and second derivative functions of $\pi_{m}^{E}$ with respect to $p_{2}^{E}$ and $\hat{\omega }$ are.

$\frac{{\partial \pi_{m}^{E} }}{{\partial p_{2}^{E} }}{ = }\alpha + \beta p_{1}^{E} - \beta (c - \hat{\omega }) + (c + t_{1} - 2p_{2}^{E} ) + c({1} - \beta )(i^{E} + 1)$, $\frac{{\partial^{2} \pi_{m}^{E} }}{{\partial (p_{2}^{E} )^{2} }}{ = } - {2}$,

$\frac{{\partial \pi_{m}^{E} }}{{\partial \hat{\omega }}}{ = }\alpha + c - p_{{1}}^{E} - \hat{\omega } + \beta p_{2}^{E} - \beta [c + t_{1} - p_{2}^{E} ] + c({1} - \beta )(i^{E} + 1)$, $\frac{{\partial^{2} \pi_{m}^{E} }}{{\partial \hat{\omega }^{2} }} = - 2$.

Given that $0 < \beta < 1$, the Hessian matrix $H(p_{2}^{E} ,\hat{\omega }) = \left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} { - 2} \\ {2\beta } \\ \end{array} } & {\begin{array}{*{20}c} {2\beta } \\ { - 2} \\ \end{array} } \\ \end{array} } \right]$ is a second-order negative definite matrix. That is, a unique set of optimal solutions ($p_{2}^{E}$ and $\hat{\omega }$) can maximize the manufacturer’s profit. After setting $\frac{{\partial \pi_{m}^{E} }}{{\partial p_{2}^{E} }}{ = 0}$ and $\frac{{\partial \pi_{m}^{E} }}{{\partial \hat{\omega }}}{ = 0}$, we obtain two equations: $p_{2}^{E} = \frac{{\beta (p_{1}^{E} + \hat{\omega }) + \alpha + c(1 - \beta )(i^{E} + 2) + t_{1} }}{{2}}$ and $\hat{\omega } = {2}\beta p_{2}^{E} - p_{1}^{E} + \alpha + c(1 - \beta )(i^{E} + 2) - \beta t_{1}$. According to the two equations, we can obtain $p_{2}^{E*} = \frac{{t_{1} + c(i^{E} + 2)}}{2} + \frac{\alpha }{{{2}(1 - \beta )}}$ and $\hat{\omega }^{ - } = \frac{{\alpha + c(1 - \beta )(i^{E} + 2)}}{2(1 - \beta )} - \frac{{t_{1} + n}}{2}$.

After substituting $p_{2}^{E*}$ and $\hat{\omega }^{ - }$ into the profit function $\pi_{p}^{E}$, the first and second derivative functions of $\pi_{p}^{E}$ with respect to $n$ are $\frac{{\partial^{2} \pi_{p}^{E} }}{{\partial n^{2} }} = - {1 < 0}$ and $\frac{{\partial \pi_{p}^{E} }}{\partial n} = \frac{{\alpha - 2n - (2c + 2ci^{E} + t_{1} )(1 - \beta )}}{2}$. After setting $\frac{{\partial \pi_{p}^{E} }}{\partial n} = {0}$, the optimal sales revenue per unit product in the ECP distribution channel is $n^{*} = \frac{{\alpha - 2c(1 + i^{E} )(1 - \beta ) - t_{1} (1 - \beta )}}{2}$. By substituting $n^{*}$ into $\hat{\omega }^{ - }$, we obtain $\hat{\omega }^{*} = \frac{{2c(2 - \beta )i^{E} - (1 + \beta )t_{1} }}{4} + M$. According to $p_{{1}}^{E} - \hat{\omega } - t_{1} = n$ and $\hat{\omega }^{*}$, we can obtain the optimal selling price of the ECP distribution channel, that is, $p_{1}^{E*} = \frac{{2\beta ci^{E} + t_{1} + \beta t_{{1}} }}{4} + N$, where $M = \frac{6c - 2c\beta }{4} + \frac{\alpha (1 + \beta )}{{4(1 - \beta )}}$ and $N = \frac{{\alpha {(3} - \beta )}}{4(1 - \beta )} + \frac{{2c({1} + \beta )}}{4}$.

Proof of Proposition 2

After substituting $p_{{1}}^{L} - \omega - t_{{1}} = n$ into the profit function $\pi_{m}^{L}$, the first and second derivative functions of $\pi_{m}^{L}$ with respect to $p_{2}^{E}$ and $\hat{\omega }$ could be obtained. By setting $\frac{{\partial \pi_{m}^{L} }}{{\partial p_{2}^{L} }}{ = 0}$ and $\frac{{\partial \pi_{m}^{L} }}{\partial \omega }{ = 0}$, we obtain two equations: $p_{{2}^{L-}} = \frac{{\alpha + \beta ({p_{1}^{L}} + \omega ) + c(1 - \beta )({i^{L}} + 2) + t_{2} }}{{2}}$ and ${\omega^{ - }} = \alpha - {p_{1}^{L}} + {2}\beta {p_{2}^{L}} + c(1 - \beta )({i^{L}} + 2) - \beta t_{2}$. According to the two equations, we can obtain $p_{2}^{L*} = \frac{\alpha }{{{2}(1 - \beta )}} + \frac{{t_{2} + c(i^{L} + 2)}}{2}$ and $\omega^{ - } = \frac{{\alpha + c(1 - \beta )(i^{L} + 2)}}{2(1 - \beta )} - \frac{{t_{1} + n}}{2}$.

After substituting $p_{2}^{L*}$ and $\omega^{ - }$ into the profit function $\pi_{p}^{L}$, the first and second derivative functions of $\pi_{p}^{E}$ with respect to $n$ are $\frac{{\partial^{2} \pi_{p}^{L} }}{{\partial n^{2} }} = - {1 < 0}$, $\frac{{\partial \pi_{p}^{L} }}{\partial n} = \frac{{\alpha - 2n - (2c + ci^{L} )(1 - \beta ) - t_{1} + \beta t_{2} }}{2}$. According to $\frac{{\partial \pi_{p}^{L} }}{\partial n}{ = 0}$, $p_{{1}}^{L} - \omega - t_{{1}} = n$, and $\omega^{ - } = \frac{{\alpha + c(1 - \beta )(i^{L} + 2)}}{2(1 - \beta )} - \frac{{t_{1} + n}}{2}$, we obtain $\omega^{*} = \frac{{c({3} - \beta )i^{L} - t_{{1}} - \beta t_{2} }}{4} + M$, where $M = \frac{6c - 2c\beta }{4} + \frac{\alpha (1 + \beta )}{{4(1 - \beta )}}$ and $N = \frac{{\alpha {(3} - \beta )}}{4(1 - \beta )} + \frac{{2c({1} + \beta )}}{4}$.

Proof of Proposition 3

The first and second derivative functions of $\pi_{p}^{E}$ with respect to $i^{E}$ are $\frac{{\partial \pi_{p}^{E} }}{{\partial i^{E} }} = \frac{{c(1 + \beta )[2\alpha - (2K + t_{1} + 2c(1 + i^{E} ))(1 - \beta )]}}{2}$ and $\frac{{\partial^{{2}} \pi_{p}^{E} }}{{\partial (i^{E} )^{2} }} = c^{2} (\beta^{2} - 1) < 0$. We conclude that $\pi_{p}^{E}$ is a strictly concave function with respect to $i^{E}$. By setting $\frac{{\partial \pi_{p}^{E} }}{{\partial i^{E} }} = {0}$, we obtain $i^{E*} = \frac{{2\alpha - (2K + 2c + t_{1} )(1 - \beta )}}{2c(1 - \beta )}$.

Proof of Corollary 3

• (1) The first derivative of $i^{E*}$ with respect to $t_{1}$ is $\frac{{\partial i^{E*} }}{{\partial t_{1} }} = - \frac{{t_{1} }}{2c} < 0$. Then, we could conclude that $i^{E*}$ is always negatively related to $t_{1}$.

• (2) The first derivative of $i^{E*}$ with respect to $c$ is $\frac{{\partial i^{E*} }}{\partial c} = \frac{{\alpha - (1 - \beta )t_{1} }}{{2c^{2} (\beta - 1)}}$. Given that $2c^{2} (\beta - 1) < 0$, the sign of $\frac{{\partial i^{E*} }}{\partial c}$ depends on $\alpha - (1 - \beta )t_{1}$. According to $\alpha > (1 - \beta )(2c + t_{1} )$, we obtain $\frac{{\partial i^{E*} }}{\partial c} < 0$ holds.

Proof of Proposition 4

According to $\frac{{\partial \pi_{t}^{L} }}{{\partial i^{L} }} = \frac{{c[(3 + \beta )\alpha - 2c(1 - \beta )(3 + \beta )(1 + i^{L} ) - 2t_{1} (1 - \beta ) - 2t_{2} (1 - \beta )(2 + \beta )]}}{4}$ and $\frac{{\partial^{{2}} \pi_{t}^{L} }}{{\partial (i^{L} )^{2} }} = \frac{{c^{2} (\beta^{2} + 2\beta - 3)}}{2} < 0$, we conclude that $\pi_{t}^{L}$ is a strictly concave function with respect to $i^{L}$. By setting $\frac{{\partial \pi_{t}^{L} }}{{\partial i^{L} }} = {0}$, we obtain $i^{L*} = \frac{\alpha }{2c(1 - \beta )} - \frac{{t_{1} + (2 + \beta )t_{2} }}{c(3 + \beta )} - 1$. Given that $0 < i^{L*} < 1$, we have $0 < \alpha (3 + \beta ) - 2(1 - \beta )(t_{1} + (2 + \beta )t_{2} + c(3 + \beta )) < 2c(1 - \beta )(3 + \beta )$. Solving the inequality, we can obtain the requirement that $\frac{{2(1 - \beta )(t_{1} + (2 + \beta )t_{2} + c(3 + \beta ))}}{3 + \beta } < \alpha < \frac{{2(1 - \beta )(t_{1} + (2 + \beta )t_{2} + 2c(3 + \beta ))}}{3 + \beta }$.

Proof of Corollary 4

Given that $\frac{{\partial i^{L*} }}{{\partial t_{1} }} = \frac{\beta - 1}{{c(3 - \beta^{2} - 2\beta )}} < 0$, $i^{L*}$ is always negatively related to $t_{1}$.

Given that $\frac{{\partial i^{L*} }}{{\partial t_{{2}} }} = \frac{{\beta + \beta^{2} - 2}}{{c(3 - \beta^{2} - 2\beta )}} < 0$, $i^{L*}$ is always negatively related to $t_{{2}}$.

The first derivative of $i^{L*}$ with respect to $c$, namely, $\frac{{\partial i^{L*} }}{\partial c} = - \frac{{\alpha (3 + \beta ) - 2(1 - \beta )(t_{1} + (2 + \beta )t_{2} )}}{{2c^{2} (1 - \beta )(\beta + 3)}} < 0$.

Proof of Corollary 5

• (i) On the basis of Sect. 4.1, we have $\hat{\omega }^{*} - \omega^{*} = \frac{{2c(2 - \beta )i^{E} - c({3} - \beta )i^{L} - \beta (t_{1} - t_{2} )}}{4}$. Therefore, if $\phi_{1} > \Delta t$, then we have $\hat{\omega }^{*} > \omega^{*}$; if $\phi_{1} \le \Delta t$, then we have $\hat{\omega }^{*} \le \omega^{*}$, where $\phi_{{1}} { = }\frac{{({4} - {2}\beta )ci^{E} - (3 - \beta )ci^{L} }}{\beta }$

• (ii) $p_{1}^{E*} - p_{1}^{L*} = \frac{{2\beta ci^{E} - (1 + \beta )ci^{L} + \beta (t_{1} - t_{2} )}}{4}$. Therefore, if $\phi_{{2}} < \Delta t$, then we have $p_{1}^{E*} > p_{1}^{L*}$; if $\phi_{{2}} \ge \Delta t$, then we have $p_{1}^{E*} \le p_{1}^{L*}$, where $\phi_{{2}} { = }\frac{{(1 + \beta )ci^{L} - 2\beta ci^{E} }}{\beta }$.

• (iii) Similarly, $p_{2}^{E*} - p_{2}^{L*} = \frac{{c(i^{E} - i^{L} ) + (t_{1} - t_{2} )}}{2}$. Therefore, if $\phi_{{3}} < \Delta t$, then we have $p_{2}^{E*} > p_{2}^{L*}$; if $\phi_{{3}} \ge \Delta t$, then we have $p_{2}^{E*} \le p_{2}^{L*}$, where $\phi_{{3}} { = }c(i^{L} - i^{E} )$.

Proof of Corollary 6

The proof of Corollary 6 is similar to that of Corollary 5.

Proof of Corollary 7

In the formula $i^{E*} - i^{L*} = \frac{{2t_{2} (2 + \beta ) - t_{1} (1 + \beta )}}{2c(3 + \beta )}$, the denominator is positive, namely, ${2}c(3 + \beta ) > 0$. Therefore, if the numerator is greater than zero, then $i^{E*} - i^{L*} > 0$ holds. The inequality (the numerator is greater than zero) can be transformed into $0 < t_{1} < \frac{{{2}(\beta + 2)}}{1 + \beta }t_{2}$. Therefore, if $t_{1} > \frac{{{2}(2 + \beta )}}{1 + \beta }t_{2}$, then $i^{E*} < i^{L*}$ holds; if $0 < t_{1} < \frac{{{2}(2 + \beta )}}{1 + \beta }t_{2}$, then $i^{E*} > i^{L*}$ holds. By setting $t_{1} = t_{2}$, we find that $i^{E*} - i^{L*} > 0$ always holds.

Proof of Corollary 8

• (1) According to the subtraction of the two profit functions, we obtain $\pi_{m}^{E*} - \pi_{m}^{L*} = \frac{ci(1 - \beta )(2\alpha - (4c + 2t + ci)(1 - \beta ))}{{16}}$. When $i^{E} = i^{L} = i$ and $t_{1} = t_{2} = t$, then according to $d_{{1}}^{E*} = \frac{{\alpha - (2c + t_{1} )(1 - \beta )}}{4} > 0$ and $d_{{2}}^{E*} = \frac{{\alpha (2 + \beta ) - (2c + t_{1} + ci^{E} )(1 - \beta )(2 + \beta ) - ci^{E} (1 - \beta )\beta }}{4} > 0$, we can obtain $\alpha > ({2}c + t)(1 - \beta )$ and $\alpha > ({2}c + t + ci)(1 - \beta )$. Therefore, if $2\alpha > (4c + 2t + ci)(1 - \beta )$, then $\pi_{m}^{E*} - \pi_{m}^{L*} > 0$ holds.

• (2) The first derivative of $\pi_{m}^{E*} - \pi_{m}^{L*}$ with respect to $t$ is $\frac{{\partial (\pi_{m}^{E*} - \pi_{m}^{L*} )}}{\partial t} = \frac{{ - (c(\beta - 1)(4i^{E} - 5i^{L} + 4\beta i^{E} - 3\beta i^{L} ))}}{8}$. If $\frac{{i^{E} }}{{i^{L} }} > \frac{(5 + 3\beta )}{{4(1 + \beta )}}$, then we obtain $\frac{{\partial (\pi_{m}^{E*} - \pi_{m}^{L*} )}}{\partial t} > 0$; if $\frac{{i^{E} }}{{i^{L} }} < \frac{(5 + 3\beta )}{{4(1 + \beta )}}$, then we obtain $\frac{{\partial (\pi_{m}^{E*} - \pi_{m}^{L*} )}}{\partial t} < 0$. By setting $\pi_{m}^{E*} - \pi_{m}^{L*} = 0$, we obtain $t = t^{\prime}$. Therefore, when $\frac{{i^{E} }}{{i^{L} }} > \frac{(5 + 3\beta )}{{4(1 + \beta )}}$, if $0 < t^{\prime} < t$, then $\pi_{m}^{E*} > \pi_{m}^{L*}$ holds; if $0 < t < t^{\prime}$, then $\pi_{m}^{E*} < \pi_{m}^{L*}$ holds. When $\frac{{i^{E} }}{{i^{L} }} < \frac{(5 + 3\beta )}{{4(1 + \beta )}}$, if $0 < t^{\prime} < t$, then $\pi_{m}^{E*} < \pi_{m}^{L*}$ holds; if $0 < t < t^{\prime}$, then $\pi_{m}^{E*} > \pi_{m}^{L*}$ holds.

• (3) After setting $\pi_{m}^{E*} - \pi_{m}^{L*} = 0$, we can obtain the two roots of the equation, namely, $i_{1}$ and $i_{{2}}$. By solving $\frac{{\partial (\pi_{m}^{E*} - \pi_{m}^{L*} )}}{\partial i} > 0$, we obtain $i < \tilde{i} = \frac{{\alpha - 2c(1 - \beta ) + (3 + 4\beta )t_{1} - (4 + 3\beta )t_{2} }}{c(1 - \beta )}$. That is, if $i < \tilde{i}$, then $\pi_{m}^{E*} - \pi_{m}^{L*}$ increases as $i$ increases; if $i > \tilde{i}$, then $\pi_{m}^{E*} - \pi_{m}^{L*}$ decreases as $i$ increases. Given that $\pi_{m}^{E*} - \pi_{m}^{L*}$ is a quadratic function with respect to $i$, if $i_{1} < i < i_{2}$, then $\pi_{m}^{E*} > \pi_{m}^{L*}$ holds; if $i < i_{1}$ or $i > i_{{2}}$, then $\pi_{m}^{E*} < \pi_{m}^{L*}$ holds.

Proof of Corollary 9

The proof of Corollary 9 is similar to that of Corollary 8.