Contract manufacturer encroachment and its impact on OEM’s sales mode and social welfare in a platform-based supply chain

Abstract

Recently, contract manufacturers are increasingly establishing their private labels to compete with the national brands of original equipment manufacturers (OEMs) on the online platform (i.e., contract manufacturer encroachment). Yet, the extant literature rarely concerns the contract manufacturer’s encroachment strategy (non-encroachment, encroachment through the reselling or agency-selling modes) under different sales modes of the OEM and their strategic interplay, although this structure is common in practice. Considering a three-tier supply chain consisting of a contract manufacturer, an OEM and an online platform, we develop a game-theoretical model to examine the interactions of the contract manufacturer’s encroachment strategy and the OEM’s sales mode choice. We show that the OEM using the agency mode encourages the contract manufacturer to encroach through the agency mode, while the contract manufacturer encroaching via agency selling increases the probability that the OEM adopts the reselling mode. Surprisingly, to deal with the contract manufacturer encroachment, the OEM may utilize the reselling mode instead of the agency mode. Furthermore, we find that in most cases, contract manufacturer encroachment can increase the profits of the contract manufacturer and the platform but reduce the OEM’s profit, whereas in a few cases, the encroachment can achieve a win-win-win situation for the three firms. Moreover, contract manufacturer encroachment always increases consumer welfare but may reduce social welfare. In addition, the extended model investigates the influence of the game sequence and shows that there is a first-mover advantage for the OEM and a second-mover advantage for the contract manufacturer.

Appendix

Proof of Lemma 1

In scenario AN, using the equation $\pi ^{AN}_{OEM}=d^{\overline{E}}_1(1-r) p_1-d^{\overline{E}}_1 w_0$, we have $\frac{\partial ^2\pi ^{AN}_{OEM}}{\partial p_1^2}=-2(1-r)<0$ which indicates that $\pi ^{AN}_{OEM}$ is concave in $p_1$. From $\frac{\partial \pi ^{AN}_{OEM}}{\partial p_1}=0$, we have $p_1=\frac{1 - r+ w_0}{2(1-r)}$. Then, due to $\frac{\partial ^2\pi ^{AN}_{CM}}{\partial w_0^2}=\frac{-1}{1-r}<0$ which indicates that $\pi ^{AN}_{CM}$ is concave in $w_0$, from $\frac{\partial \pi ^{AN}_{CM}}{\partial w_0}=0$, we can get $w_{0}^{AN^*}=\frac{1 - r}{2}$. Substituting $w_{0}^{AN^*}=\frac{1 - r}{2}$ into $p_1=\frac{1 - r+ w_0}{2(1-r)}$, we can obtain the equilibrium result of scenarios AN.

In addition, we can utilize the same method to obtain the optimal outcomes of scenario RN as follows: $p_{1}^{RN^*}=\frac{7}{8}$, $w_{0}^{RN^*}=\frac{1}{2}$, and $w_{1}^{RN^*}=\frac{3 }{4}$. $\square$

Proofs of Lemmas 2-5

In scenario AA, using the equation $\pi ^{AA}_{OEM}=d^{E}_{1} (1-r)p_1-d^{E}_{1} w_0$, we can get $\frac{\partial ^2\pi ^{AA}_{OEM}}{\partial p_1^2}=\frac{2(r-1)}{1-\theta }<0$, which indicates that $\pi ^{AA}_{OEM}$ is concave in $p_1$. Similarly, using the equation $\pi ^{AA}_{CM}=d^{E}_{1} w_0+d^{E}_{2} (1-r)p_2$, we can get $\frac{\partial ^2\pi ^{AA}_{CM}}{\partial p_2^2}=\frac{2(r-1)}{\theta (1-\theta )}<0$, which indicates that $\pi ^{AA}_{CM}$ is concave in $p_2$. Then from $\frac{\partial \pi ^{AA}_{CM}}{\partial p_2}=0$ and $\frac{\partial \pi ^{AA}_{OEM}}{\partial p_1}=0$, we have $p_1=\frac{2 - 2 r + 2 w_0 - 2 \theta + 2 r \theta + w_0 \theta }{(1 - r) (4 - \theta )}$ and $p_2=\frac{\theta (1 - r + 3 w_0 - \theta + r \theta )}{(1 - r) (4 - \theta )}$. Subsequently, due to $\frac{\partial ^2\pi ^{AA}_{CM}}{\partial w_0^2}=\frac{2 (8 + \theta )}{(-1 + r) (4 - \theta )^2}<0$ which indicates that $\pi ^{AA}_{CM}$ is concave in $w_0$, from $\frac{\partial \pi ^{AA}_{CM}}{\partial w_0}=0$, we have $w_{0}^{AA^*}=\frac{(1 - r) (8 + \theta ^2)}{2 (8 + \theta )}$. Substituting $w_{0}^{AA^*}=\frac{(1 - r) (8 + \theta ^2)}{2 (8 + \theta )}$ into $p_1=\frac{2 - 2 r + 2 w_0 - 2 \theta + 2 r \theta + w_0 \theta }{(1 - r) (4 - \theta )}$ and $p_2=\frac{\theta (1 - r + 3 w_0 - \theta + r \theta )}{(1 - r) (4 - \theta )}$, we can obtain the Lemma 2.

Furthermore, we can use the same method to get the optimal outcomes of scenarios AR, RA and RR. Here, the proof process is omitted due to simplicity. $\square$

Proof of Proposition 1

Using Lemma 1, (i) we have $p_{1}^{RN^*}-p_{1}^{AN^*}=\frac{1}{8}>0$ and $w_{0}^{RN^*}-w_{0}^{AN^*}=\frac{( r) \theta }{2}>0$. Hence, $p_{1}^{RN^*}>p_{1}^{AN^*}$ and $w_{0}^{RN^*}>w_{0}^{AN^*}$. (ii) From $\pi _{OEM}^{RN^*}-\pi _{OEM}^{AN^*}=\frac{1-2r}{32}=0$, we can get the solution, that is $r=\frac{1}{2}$. Similarly, from $\pi _{CM}^{RN^*}-\pi _{CM}^{AN^*}=\frac{1-2r}{16}=0$, we can get the solution, that is $r=\frac{1}{2}$. Thus, when $r<\frac{1}{2}$, $\pi _{CM}^{AN^*}>\pi _{CM}^{RN^*}$ and $\pi _{OEM}^{AN^*}>\pi _{OEM}^{RN^*}$; when $r>\frac{1}{2}$, $\pi _{CM}^{AN^*}<\pi _{CM}^{RN^*}$ and $\pi _{OEM}^{AN^*}<\pi _{OEM}^{RN^*}$.(iii) From $\pi _{P}^{RN^*}-\pi _{P}^{AN^*}=\frac{-1+12r}{64}=0$, we can get the solution, that is $r=\frac{1}{12}$. Thus we have when $r<\frac{1}{12}$, $\pi _{P}^{RN^*}>\pi _{P}^{AN^*}$; when $r>\frac{1}{12}$, $\pi _{P}^{RN^*}<\pi _{P}^{AN^*}$. $\square$

Proof of Propositions 2–10

Similar to the Proof of Proposition 1, the proofs of Propositions 2–5 are omitted here. Note that $r_1$ ($\theta _1$) is the only meaningful root of the equation $\pi _{CM}^{RA^*}-\pi _{CM}^{RN^*}=0$ ($\pi _{OEM}^{RA^*}-\pi _{OEM}^{RN^*}=0$). Moreover, we can get two meaningful roots $r_2$ and $r_3$ ($r_2<r_3$) from the equation $\pi _{P}^{RA^*}-\pi _{P}^{RN^*}=0$.

The proofs of Propositions 6–10 can be easily obtained; hence, they are also omitted. Note that we can obtain thresholds $r_4$, $r_5$ and $r_6$ from the equation $\pi _{CM}^{RA^*}-\pi _{CM}^{RR^*}=0$, $\pi _{OEM}^{AA^*}-\pi _{OEM}^{RA^*}=0$ and $\pi _{OEM}^{AR^*}-\pi _{OEM}^{RR^*}=0$, respectively. $\square$