The comparison and complex analysis on dual-channel supply chain under different channel power structures and uncertain demand

Abstract

In this paper, a dual-channel supply chain composed of one manufacturer and one retailer is discussed. The manufacturer sells its products to its customers, becoming the rival of the retailers. In this case, we consider two scenarios in which the manufacturer holds either asymmetric or symmetric channel power as the retailer. According to the changing demand, we develop the dynamic game models for the two scenarios and analyze the models’ dynamic behavior. The corporations’ profits and system stability are compared under these two scenarios. We find that the manufacturer’s profit is greater with asymmetric channel power than that with symmetric channel power. For the retailer, it will get more profit if it holds a symmetric channel power compared with the manufacturer. The supply chain profit will increase as the internal competition increases. Besides, the system stability will be easier to maintain with asymmetric channel power and the market will be much easier to fall into chaos with symmetric channel power.

Appendix

Proof of Proposition 1

We make ${\Delta } {\pi _{\mathrm{m}}} = \pi _{\mathrm{m}}^{\mathrm{AS}} - \pi _{\mathrm{m}}^\mathrm{S}$ then we can obtain:

$$\begin{aligned} {\Delta } {\pi _{\mathrm{m}}}(\sigma ) = \frac{{f(\sigma )}}{{144{b_{\mathrm{r}}}\pi }} \end{aligned}$$

(30)

where,

$$\begin{aligned} f(\sigma )= & {} \left( 2{b_{\mathrm{r}}}^2{c^2}\pi + 2{c^2}{k^2}\pi + 2{a^2}\pi {\rho ^2}\right. \nonumber \\&\left. +\,2a\sqrt{2\pi } \rho \sigma + {\sigma ^2} + 2ck\left( 2a\pi \rho + \sqrt{2\pi } \sigma \right) \right. \nonumber \\&\left. -\,2{b_{\mathrm{r}}}c(2ck\pi + 2a\pi \rho + \sqrt{2\pi } \sigma )\right) \end{aligned}$$

(31)

Noticed that $144{b_{\mathrm{r}}}\pi > 0$, we can infer the value range of ${\Delta } {\pi _{\mathrm{m}}}$ by analyzing the curve shape of the function $f(\sigma )$. By solving the condition $\partial f(\sigma )/\partial \sigma = 0$, we can get $\sigma = ({b_{\mathrm{r}}}c - kc - a\rho )\sqrt{2\pi }$. Considering that second derivative ${\partial ^2}f(\sigma )/\partial {\sigma ^2} = 2 > 0$, the function can get the minimum value at extreme point.

When $({b_{\mathrm{r}}}c - kc - a\rho )\sqrt{2\pi } > 0$, the minimum value will be:

$$\begin{aligned} f\left( \sigma = {b_{\mathrm{r}}}c\sqrt{2\pi } - kc\sqrt{2\pi } - a\rho \sqrt{2\pi } \right) = 0 \end{aligned}$$

(32)

When $({b_{\mathrm{r}}}c - kc - a\rho )\sqrt{2\pi } < 0$, the minimum value will be:

$$\begin{aligned} \mathop {\lim }\limits _{\sigma \rightarrow 0} f(\sigma ) = 2\pi {\left( - {b_{\mathrm{r}}}c + ck + a\rho \right) ^2} \ge 0 \end{aligned}$$

(33)

By judging the value of the $f(\sigma )$, we can infer that the value of function ${\Delta } {\pi _{\mathrm{m}}}(\sigma )$ will never be less than 0. In other words, ${\Delta } {\pi _{\mathrm{m}}}(a) \ge 0$, or $\pi _{\mathrm{m}}^{\mathrm{AS}} \ge \pi _{\mathrm{m}}^\mathrm{S}$. This completes the proof. $\square$

Proof of Proposition 2

We make ${\Delta } {\pi _{\mathrm{r}}} = \pi _{\mathrm{r}}^{\mathrm{AS}} - \pi _{\mathrm{r}}^\mathrm{S}$ then we can obtain:

$$\begin{aligned} \Delta {\pi _\mathrm{r}}(\sigma ) = \frac{{ - 7f(\sigma )}}{{288{b_r}\pi }} \end{aligned}$$

(34)

Here, $f(\sigma )$ has been expressed in Eq. 32. In the Proof of Proposition 1, we have proved that $f(\sigma ) \ge 0$. So it is easy to check that $\Delta {\pi _r}(\sigma ) \le 0$. In other words, $\pi _r^{AS} \le \pi _r^S$. This completes the proof. $\square$

Proof of Proposition 3

We make ${\Delta } {\pi _\mathrm{SC}} = \pi _\mathrm{SC}^{\mathrm{AS}} - \pi _\mathrm{SC}^\mathrm{S}$ to denote the supply chain profit difference between model AS and model S. We can obtain:

$$\begin{aligned} \Delta {\pi _\mathrm{SC}}(\sigma ) = \Delta {\pi _\mathrm{m}} + \Delta {\pi _\mathrm{r}} = \frac{{ - 5f(\sigma )}}{{288{b_\mathrm{r}}\pi }} \end{aligned}$$

(35)

Here, $f(\sigma )$ has been expressed in Eq. 32. In the Proof of Proposition 1, we have proved that $f(\sigma ) \ge 0$. So it is easy to check that $\Delta {\pi _\mathrm{SC}}(\sigma ) \le 0$. In other words, $\pi _\mathrm{SC}^{\mathrm{AS}} \le \pi _\mathrm{SC}^\mathrm{S}$. This completes the proof. $\square$