Equilibrium analysis of marketing strategies in supply chain with marketing efforts induced demand considering free riding

Abstract

Retailers can make marketing efforts to increase the market demand, but the results from their activities are generally uncertain and influenced by free riding. This paper considers marketing strategies in a two-echelon supply chain under free riding, where a manufacturer sells products through two competitive retailers who have different powers. The dominant retailer will decide whether to make marketing efforts, and the following retailer will choose whether to follow the decision of the dominant retailer. We establish our demand functions relying on the price and marketing efforts, and then build six decentralized game models to examine how marketing strategies and power structures (manufacturer-dominant and retailer-dominant) affect supply chain members’ performances. It is found that, for the dominant retailer, he will make marketing efforts if free riding is not severe. As for the following retailer, in retailer-dominant structure, he will also make marketing efforts if the dominant retailer makes that, while his strategy varies with the degree of free riding in manufacturer-dominant structure. We also show that if the dominant retailer wants to make marketing efforts, he will make the same level of marketing efforts regardless of his market base and competitor’s decision.

Appendix

Proof of Proposition 2

Referring to Tables 2 and 3, it can be easily obtained that if the two retailers both make marketing efforts, we have

$$\begin{aligned} \pi _1^{\mathrm{MS}\text {-}{1}}-\pi _1^{\mathrm{RS}\text {-}{1}}= & {} \frac{(8 \lambda -3) r^2}{32 \eta },\\ \pi _2^{\mathrm{MS}\text {-}{1}}-\pi _2^{\mathrm{RS}\text {-}{1}}= & {} \frac{(8 \lambda -3) r^2}{32 \eta }. \end{aligned}$$

If $\frac{3}{8}<\lambda \le$, then we have $\pi _1^{\mathrm{MS}\text {-}{1}}>\pi _1^{\mathrm{RS}\text {-}{1}}$, $\pi _2^{\mathrm{MS}\text {-}{1}}>\pi _2^{\mathrm{RS}\text {-}{1}}$.

If only the dominant retailer makes marketing efforts, we have

$$\begin{aligned} \pi _1^{\mathrm{MS}\text {-}{2}}-\pi _1^{\mathrm{RS}\text {-}{2}}= & {} -\frac{r^2}{32 \eta }<0, \\ \pi _2^{\mathrm{MS}\text {-}{2}}-\pi _2^{\mathrm{RS}\text {-}{2}}= & {} \frac{(4 \lambda -1) r^2}{16 \eta }. \end{aligned}$$

If $0\le \lambda <\frac{1}{4}$, then we have

$$\begin{aligned} \pi _2^{\mathrm{MS}\text {-}{2}}<\pi _2^{\mathrm{RS}\text {-}{2}}. \end{aligned}$$

Therefore, Proposition 2 is proved.

Proof of Proposition 4

From Table 2, we can find that

$$\begin{aligned} \pi _1^{\mathrm{MS}\text {-}{1}}-\pi _1^{\mathrm{MS}\text {-}{3}}= & {} -\frac{(\lambda -1) \lambda r^2}{2 \eta }> 0, \\ \pi _2^{\mathrm{MS}\text {-}{1}}-\pi _2^{\mathrm{MS}\text {-}{3}}= & {} -\frac{(\lambda -1) \lambda r^2}{2 \eta }> 0, \\ \pi _1^{\mathrm{MS}\text {-}{1}}-\pi _1^{\mathrm{MS}\text {-}{2}}= & {} -\frac{\left( 4 \lambda ^2-5 \lambda +1\right) r^2}{4 \eta }, \\ \pi _2^{\mathrm{MS}\text {-}{2}}-\pi _2^{\mathrm{MS}\text {-}{3}}= & {} -\frac{\left( 4 \lambda ^2-5 \lambda +1\right) r^2}{4 \eta }. \end{aligned}$$

If $\frac{1}{4}<\lambda \le 1$, then we have

$$\begin{aligned} \pi _1^{\mathrm{MS}\text {-}{1}}>\pi _1^{\mathrm{MS}\text {-}{2}},\quad \pi _2^{\mathrm{MS}\text {-}{2}}>\pi _2^{\mathrm{MS}\text {-}{3}}. \end{aligned}$$

We also find that

$$\begin{aligned} \pi _1^{\mathrm{MS}\text {-}{2}}-\pi _1^{\mathrm{MS}\text {-}{3}}= & {} \frac{\left( 2 \lambda ^2-3 \lambda +1\right) r^2}{4 \eta },\\ \pi _2^{\mathrm{MS}\text {-}{1}}-\pi _2^{\mathrm{MS}\text {-}{2}}= & {} \frac{\left( 2 \lambda ^2-3 \lambda +1\right) r^2}{4 \eta }. \end{aligned}$$

If $0\le \lambda <\frac{1}{2}$, then we have

$$\begin{aligned} \pi _1^{\mathrm{MS}\text {-}{2}}>\pi _1^{\mathrm{MS}\text {-}{3}},\quad \pi _2^{\mathrm{MS}\text {-}{1}}>\pi _2^{\mathrm{MS}\text {-}{2}}. \end{aligned}$$

Thus, Proposition 4 is proved.

Proof of Proposition 5

Referring to Tables 2 and 3, we have

$$\begin{aligned} e_1^{\mathrm{RS}\text {-}{1}}=e_1^{\mathrm{RS}\text {-}{2}}<e_1^{\mathrm{MS}\text {-}{1}}=e_1^{\mathrm{MS}\text {-}{2}}. \end{aligned}$$

We can also get that

$$\begin{aligned} w^{\mathrm{RS}\text {-}{1}}-w^{\mathrm{RS}\text {-}{2}}= & {} \frac{(3-4 \lambda ) r}{16 \eta }, \\ w^{\mathrm{RS}\text {-}{2}}-w^{\mathrm{RS}\text {-}{3}}= & {} \frac{(3-4 \lambda ) r}{8 \eta }, \\ w^{\mathrm{RS}\text {-}{1}}-w^{\mathrm{RS}\text {-}{3}}= & {} \frac{(3-4 \lambda ) r}{16 \eta }. \end{aligned}$$

If $0\le \lambda <\frac{3}{4}$, then we have

$$\begin{aligned} w^{\mathrm{RS}\text {-}{1}}>w^{\mathrm{RS}\text {-}{2}}>w^{\mathrm{RS}\text {-}{3}}. \end{aligned}$$

Thus, Proposition 5 is proved.

Proof of Proposition 6

Referring to Table 3, we get

$$\begin{aligned} \pi _1^{\mathrm{RS}\text {-}{2}}-\pi _1^{\mathrm{RS}\text {-}{3}}= & {} \frac{(3-4 \lambda )^2 r^2}{32 \eta }> 0,\\ \pi _2^{\mathrm{RS}\text {-}{1}}-\pi _2^{\mathrm{RS}\text {-}{2}}= & {} \frac{(3-4 \lambda )^2 r^2}{32 \eta }> 0,\\ \pi _1^{\mathrm{RS}\text {-}{1}}-\pi _1^{\mathrm{RS}\text {-}{2}}= & {} -\frac{\left( 16 \lambda ^2-16 \lambda +3\right) r^2}{16 \eta },\\ \pi _2^{\mathrm{RS}\text {-}{2}}-\pi _2^{\mathrm{RS}\text {-}{3}}= & {} -\frac{\left( 16 \lambda ^2-16 \lambda +3\right) r^2}{16 \eta }. \end{aligned}$$

If $\frac{1}{4}<\lambda <\frac{3}{4}$, then we have

$$\begin{aligned} \pi _1^{\mathrm{RS}\text {-}{1}}>\pi _1^{\mathrm{RS}\text {-}{2}},\quad \pi _2^{\mathrm{RS}\text {-}{2}}>\pi _2^{\mathrm{RS}\text {-}{3}}. \end{aligned}$$

Besides, we have

$$\begin{aligned} \pi _1^{\mathrm{RS}\text {-}{1}}-\pi _1^{\mathrm{RS}\text {-}{3}}= & {} \frac{\left( -16 \lambda ^2+8 \lambda +3\right) r^2}{32 \eta },\\ \pi _2^{\mathrm{RS}\text {-}{1}}-\pi _2^{\mathrm{RS}\text {-}{3}}= & {} \frac{\left( -16 \lambda ^2+8 \lambda +3\right) r^2}{32 \eta }. \end{aligned}$$

If $0\le \lambda <\frac{3}{4}$, then we have

$$\begin{aligned} \pi _1^{\mathrm{RS}\text {-}{1}}>\pi _1^{\mathrm{RS}\text {-}{3}},\quad \pi _2^{\mathrm{RS}\text {-}{1}}>\pi _2^{\mathrm{RS}\text {-}{3}}. \end{aligned}$$

Thus, Proposition 6 is proved.