Strategic interplay between store brand strategy and selling mode choice

Abstract

With the development of e-commerce, online retailers has increasingly developed their own store brand to compete with the national brand. It is necessary for the manufacturers who sell national brand products through these online retailers to consider strategically when making the selling mode decisions. Motivated by such practice issue, we consider a theoretic model comprised of a national brand manufacturer and an online retailer. The retailer chooses to introduce an average store brand, a premium store brand or neither, and the manufacturer sells national brand products through the retailer under reselling or agency selling mode. Comparing and analyzing the six possible scenarios, we investigate the interaction between the retailer’s store brand strategy and the manufacturer selling mode selection. We find that the unit production cost and the quality of the national brand affect the retailer’s store brand strategy. Under the reselling format, the retailer prefers to introduce a premium store brand when the national brand cost is relatively low. While under the agency selling format, the retailer prefers to introduce a premium store brand when the cost is low but the quality is high. Interestingly, we further find that the manufacturer may benefit from the premium store brand introduction. Whether for the reselling or agency selling, there exists the “win–win” situation for the manufacturer and the retailer with a higher national brand quality. Overall, our results provide several management insights for the online retailer’s store brand introduction and the manufacturer’s selling mode selection.

Appendices

Appendix A

Under the scenario RN, the profit functions of the retailer and the manufacturer are given in Eqs. (3) and (4). By backward induction, we can obtain the optimal prices, product demand and profits of the manufacturer and the retailer as follows

$$\begin{aligned}&p_n^{RN*}=\frac{3\gamma _{n}+c}{4}, w_n^{RN*}=\frac{\gamma _{n}+c}{2},&\\&\pi _R^{RN*}=\frac{(c-\gamma _{n})^2}{16\gamma _{n}}, \pi _M^{RN*}=\frac{(c-\gamma _{n})^2}{8\gamma _{n}}.&\end{aligned}$$

Under the scenario RA, the profit functions of the retailer and the manufacturer are given in Eqs. (5) and (6). By backward induction, we can obtain the retailer and the manufacturer’s optimal decisions as follows

$$\begin{aligned}&p_n^{RA*}=\frac{3\gamma _{n}-\gamma _{s}+2c}{4}, p_s^{RA*}=\frac{c+\gamma _{s}}{2},&\\&w_n^{RA*}=\frac{2c+\gamma _{n}-\gamma _{s}}{2},&\\&\pi _R^{RA*}=\frac{4(c-\gamma _{s})^2+\gamma _{s}(\gamma _{n}-\gamma _{s})}{16\gamma _{s}}, \pi _M^{RA*}=\frac{\gamma _{n}-\gamma _{s}}{8}.&\end{aligned}$$

Under the scenario RP, the profit functions of the retailer and the manufacturer are given in Eqs. (7) and (8). Similarly, we can obtain the optimal prices and profits as follows

$$\begin{aligned}&p_n^{RP*}=\frac{2\gamma _{n}+c+c_h\gamma _{n}}{4}, p_h^{RP*}=\frac{c_h+1}{2},&\\&w_n^{RP*}=\frac{c_h\gamma _{n}+c}{2},&\\&\pi _R^{RP*}=\frac{(c-c_h\gamma _{n})^2+4\gamma _{n}(1-\gamma _{n})(1-c_h)^2}{16\gamma _{n}(1-\gamma _{n})},&\\&\pi _M^{RP*}=\frac{(c-c_h\gamma _{n})^2}{8\gamma _{n}(1-\gamma _{n})}.&\end{aligned}$$

Under the scenario AN, the profit functions of the retailer and the manufacturer are given in Eqs. (9) and (10). By backward induction, we can obtain the optimal prices and profits as follows

$$\begin{aligned}&p_n^{AN*}=\frac{c+\gamma _{n}}{2},&\\&\pi _R^{AN*}=\frac{\alpha (c-\gamma _{n})^2}{4\gamma _{n}}, \pi _M^{AN*}=\frac{(1-\alpha )(\gamma _{n}-c)^2}{4\gamma _{n}}.&\end{aligned}$$

Under the scenario AA, the profit functions of the retailer and the manufacturer are given in Eqs. (11) and (12). Similarly, we can obtain the optimal prices and profits of the retailer and the manufacturer’s as follows

$$\begin{aligned}&p_n^{AA*}=\frac{2\gamma _{n}^2-(2\gamma _{s}-3c)\gamma _{n}-\alpha c\gamma _{s}}{4\gamma _{n}-(1+\alpha )\gamma _{s}},&\\&p_s^{AA*}=\frac{(\gamma _{n}-\gamma _{s})(\alpha \gamma _{s}+\gamma _{s}+2c)+c\gamma _{s}(3-\alpha )}{4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s}},&\\&\pi _R^{AA*}=\frac{(\gamma _{n}-\gamma _{s})[(4\gamma _{n}-\alpha \gamma _{s})(\alpha \gamma _{n}\gamma _{s}-c\alpha \gamma _{s}-c\gamma _{s}+c^2)]}{\gamma _{s}(4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s})^2}\\&\qquad \quad \ +\frac{(\gamma _{n}-\gamma _{s})(\gamma _{n}\gamma _{s}^2)}{\gamma _{s}(4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s})^2},\\&\pi _M^{AA*}=\frac{(2\gamma _{n}-c)^2(\gamma _{n}-\gamma _{s})(1-\alpha )}{(4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s})^2}.&\end{aligned}$$

Under the scenario AP, the profit functions of the retailer and the manufacturer are given in Eqs. (13) and (14). Similarly, we can obtain the optimal prices and profits as follows

$$\begin{aligned}&p_n^{AP*}=\frac{c_h\gamma _n+\gamma _n+2c-\gamma _n^2-\alpha c\gamma _{n}}{4-(1+\alpha \gamma _{n})},&\\&p_h^{AP*}=\frac{2c_h+c+2-2\gamma _{n}-\alpha c}{4-(1+\alpha \gamma _{n})},&\\&\pi _R^{AP*}=\frac{(c-2c_h-2\gamma _{n}-\alpha c+c_h\gamma _{n}+\alpha c_h\gamma _{n}+2)\sigma _2}{(1-\gamma _{n})(4-\gamma _{n}-\alpha \gamma _{n})}&\\&\qquad \quad \ +\frac{\alpha \sigma _1^2}{\gamma _{n}(1-\gamma _{n})},&\\&\pi _M^{AP*}=\frac{(1-\alpha )\sigma _1^2}{\gamma _{n}(1-\gamma _{n})}.&\end{aligned}$$

Appendix B

1.1 B.1 Proof of Proposition 2

• (1) $\pi _R^{RA}-\pi _R^{RN}=\frac{4(c-\gamma _{s})^2+\gamma _{s}(\gamma _{n}-\gamma _{s})}{16\gamma _{s}}-\frac{(c-\gamma _{n})^2}{16\gamma _{n}}=\frac{4\gamma _{n}c_2-6c\gamma _{n}\gamma _s+3\gamma _n\gamma _s ^2-c^2\gamma _{s}}{16\gamma _{s}\gamma _n}.$ Since $4\gamma _{n}c_2-6c\gamma _{n}\gamma _s+3\gamma _n\gamma _s ^2-c^2\gamma _{s}=3\gamma _n(c-\gamma _{s})^2+c^2(\gamma _{n}-\gamma _{s})$, we can obtain that $\pi _R^{RA}-\pi _R^{RN}>0$.

• (2) $\pi _R^{RN}-\pi _R^{RP}=\frac{(c-\gamma _{n})^2}{16\gamma _{n}}-\frac{(c-c_h\gamma _{n})^2}{16\gamma _{n}(1-\gamma _{n})}-\frac{4\gamma _{n}(1-\gamma _{n})(1-c_h)^2}{16\gamma _{n}(1-\gamma _{n})}$. Since $c<c_h<1$, $\frac{(c-\gamma _{n})^2}{16\gamma _{n}}<\frac{(c-c_h\gamma _{n})^2}{16\gamma _{n}(1-\gamma _{n})}$. We can further obtain $\pi _R^{RN}<\pi _R^{RP}$.

• (3) Set $\Delta \pi =\pi _R^{RA}-\pi _R^{RP}=\frac{4(c-\gamma _{s})^2+\gamma _{s}(\gamma _{n}-\gamma _{s})}{16\gamma _{s}}-\frac{4\gamma _{n}(1-\gamma _{n})(1-c_h)^2}{16\gamma _{n}(1-\gamma _{n})}-\frac{4\gamma _{n}(1-\gamma _{n})(1-c_h)^2}{16\gamma _{n}(1-\gamma _{n})}$. c has two solutions $c_{RR1}=\frac{\gamma _{n}\gamma _{s}[(1-\gamma _{n})(4-\delta _1)-c_h]}{4\gamma _{n}(1-\gamma _{n})-\gamma _{s}}<0$ and $c_{RR2}=\frac{\gamma _{n}\gamma _{s}[4(1-\gamma _{n}-\delta _1(1+\gamma _{n}-c_h))]}{4\gamma _{n}(1-\gamma _{n})-\gamma _{s}}$ when $\Delta \pi =0$. $\frac{\partial \Delta \pi }{\partial c}\vert _{c=c_{RR2}}>0$, so $\Delta \pi$ is monotonically decreasing with respect to c under the constraint $\bar{c}>c>0$. Since $\Delta \pi \vert _{c=c_{RR2}}=0$, we can prove that $\pi _R^{RA}<\pi _R^{RP}$ when $0<c<c_{RR2}$ and $\pi _R^{RA}\ge \pi _R^{RP}$ when $c\ge c_{RR2}$. Set $c_{RR}^*=c_{RR2}$. (Note that $\delta _1=\frac{A}{\gamma _{n}\gamma _{s}(1-\gamma _{n})}$, where $A=-12c_h^2\gamma _{n}^2+16c_h^2\gamma _{n}-4c_h^2\gamma _{s}+32c_h\gamma _{n}^2-8c_h\gamma _{n}\gamma _{s}-32c_h\gamma _{n}+8c_h\gamma _{s}+4\gamma _{n}^3-4\gamma _{n}^2\gamma _{s}-20\gamma _{n}^2+5\gamma _{n}\gamma _{s}+16\gamma _{n}+3\gamma _{s}^2-4\gamma _{s}$)

• (4) $\pi _M^{RN}-\pi _M^{RA}=\frac{(c-\gamma _{n})^2}{8\gamma _{n}}-\frac{\gamma _{n}-\gamma _{s}}{8}=\frac{c(c-\gamma _{n})-\gamma _{n}(c-\gamma _{s})}{8\gamma _{n}}$. Since $\gamma _{n}>\gamma _{s}>c$, $\pi _M^{RN}>\pi _M^{RA}$.

• (5) $\pi _M^{RN}-\pi _M^{RP}=\frac{(c-\gamma _{n})^2}{8\gamma _{n}}-\frac{(c-c_h\gamma _{n})^2}{8\gamma _{n}(1-\gamma _{n})}=\frac{(c-\gamma _{n})^2(1-\gamma _{n})-(c-c_h\gamma _{n})^2}{8\gamma _{n}(1-\gamma _{n})}$. Since $(c-\gamma _{n})^2(1-\gamma _{n})-(c-c_h\gamma _{n})^2=\gamma _{n}^2-\gamma _{n}^3+2c\gamma ^2+2cc_h\gamma _{n}-2c\gamma _{n}-c^2\gamma _{n}>0$, we can prove that $\pi _M^{RN}>\pi _M^{RP}$.

• (6) $\pi _M^{RA}-\pi _M^{RP}=\frac{\gamma _{n}-\gamma _{s}}{8}-\frac{(c-c_h\gamma _{n})^2}{8\gamma _{n}(1-\gamma _{n})}=\frac{(\gamma _{n}-\gamma _{s})\gamma _{n}(1-\gamma _{n})-(c-c_h\gamma _{n})^2}{8\gamma _n(1-\gamma _{n})}$. When $(\gamma _{n}-\gamma _{s})\gamma _{n}(1-\gamma _{n})-(c-c_h\gamma _{n})^2<0$, that is $0<c\le c_{h}\gamma _{n}-\sqrt{\gamma _{n}(1-\gamma _{n})(\gamma _{n}-\gamma _{s})}$, $\pi _M^{RA}<\pi _M^{RP}$. When $c\ge c_{h}\gamma _{n}-\sqrt{\gamma _{n}(1-\gamma _{n})(\gamma _{n}-\gamma _{s})}$, $\pi _M^{RA}\ge \pi _M^{RP}$. Set $c_{RM}^*=c_{h}\gamma _{n}-\sqrt{\gamma _{n}(1-\gamma _{n})(\gamma _{n}-\gamma _{s})}$.

1.2 B.2 Proof of Proposition 4

• (1) Set $\Delta \pi =\pi _R^{AN*}-\pi _R^{AA*}=\frac{\alpha (c-\gamma _{n})^2}{4\gamma _{n}}-\frac{(\gamma _{n}-\gamma _{s})[(4\gamma _{n}-\alpha \gamma _{s})(\alpha \gamma _{n}\gamma _{s}-c\alpha \gamma _{s}-c\gamma _{s}+c^2)]}{\gamma _{s}(4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s})^2}+\frac{(\gamma _{n}-\gamma _{s})(\gamma _{n}\gamma _{s}^2)}{\gamma _{s}(4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s})^2}$. c has two solutions $c_{AN1}$ and $c_{AN2}$ when $\Delta \pi =0$. $c_{AN2}=\frac{\gamma _{n}\gamma _{s}\sigma _3}{\sigma _4}$, where $\sigma _3=8\gamma _{n}\gamma _{s}+8\alpha \gamma _{n}^2-\alpha \gamma _{s}^2-8\gamma _{n}^2+8\gamma _{n}\sqrt{\alpha \gamma _{s}(\gamma _{n}-\gamma _{s})}-2\gamma _{s}\sqrt{\alpha \gamma _{s}(\gamma _{n}-\gamma _{s})}+\alpha ^3\gamma _{s}^2-2\alpha \gamma _{s}\sqrt{\alpha \gamma _{s}(\gamma _{n}-\gamma _{s})}+2\alpha \gamma _{n}\gamma _{s}-6\alpha ^2\gamma _{n}\gamma _{s}$ and $\sigma _4=\alpha ^3\gamma _{s}^3-8\alpha ^2\gamma _{n}\gamma _{s}^2+2\alpha ^2\gamma _{s}^3+20\alpha \gamma _{n}^2\gamma _{s}-12\alpha \gamma _{n}\gamma _{s}^2+\alpha \gamma _{s}^3-16\gamma _{n}^3+16\gamma _{n}^2\gamma _{s}$. We can obtain that $c_{AN2}<0$ and $c_{AN1}>\bar{c}$. $\Delta \pi \vert _{c=0}$, so $\Delta \pi >0$ under the constrain $0<c<\bar{c}$. We can prove that $\pi _R^{AN*}>\pi _R^{AA*}$.

• (2) $\pi _M^{AN*}-\pi _M^{AA*}=\frac{(1-\alpha )(\gamma _{n}-c)^2}{4\gamma _{n}}-\frac{(2\gamma _{n}-c)^2(\gamma _{n}-\gamma _{s})(1-\alpha )}{(4\gamma _{n}-\gamma _{s}-\alpha \gamma _{s})^2}$. We solve $\pi _M^{AN*}-\pi _M^{AA*}=0$ and obtain two solutions, $c_{MAN1}$ and $c_{MAN2}$. $\bar{c}<c_{MAN1}<c_{MAN2}$. We can obtain that $\pi _M^{AN*}>\pi _M^{AA*}$.