Information sharing under agency selling in an e-commerce supply chain with competing OEMs

Abstract

With the rapid development of e-commerce platforms, more and more attention has been paid to the information asymmetry between the platform and upstream firms. This paper investigates the information-sharing strategy for an e-commerce platform on which two competing original equipment manufacturers outsource their manufacturing services to a common contract manufacturer and sell substitutable products directly to consumers. A game-theoretic model is employed to examine six information-sharing scenarios, and we derive the following results. First, the platform always voluntarily offers its demand information to other chain members when the contract manufacturer is the leader in the market. In particular, the platform obtains the most profit when it shares the information only with the contract manufacturer. Second, both the contract manufacturer and the original equipment manufacturers can benefit from information sharing. Information sharing can also help to increase the profit for the whole supply chain. Contrary to intuition, the whole supply chain is most profitable when the information is shared only with the contract manufacturer. Third, if the contract manufacturer gets the information, the profit for each member will increase with the demand forecasting accuracy. In addition, this paper explores the impacts of different leader–follower relationships on information sharing in the extension section. It is shown that when the original equipment manufacturers are the leaders and the contract manufacturer is the follower, the platform does not always share the information with others, and its information-sharing strategies change significantly.

Appendices

Appendix

Scenario IN

Using the equation $\mathop {Max}\limits _{p_1}E(\pi _{O1}^{IN})=E(d_1 (p_1 - r p_1) - d_1 w_1)$, we show $\frac{\partial ^2E(\pi _{O1}^{IN})}{\partial p_1^2}=2r-2<0$, which indicates that $E(\pi _{O1}^{IN})$ is concave in $p_1$. Similarly, using the equation $\mathop {Max}\limits _{p_2}E(\pi _{O2}^{IN})=E(d_2 (p_2 - r p_2) - d_2 w_2)$, we have $\frac{\partial ^2E(\pi _{O2}^{IN})}{\partial p_2^2}=2r-2<0$, which indicates that $E(\pi _{O2}^{IN})$ is concave in $p_2$. Then, from $\frac{\partial E(\pi _{O1}^{IN})}{\partial p_1}=0$ and $\frac{\partial E(\pi _{O2}^{IN})}{\partial p_1}=0$, we have $p_{1}^{IN}=\frac{2 a_0 + a_0 b - 2 a_0 r - a_0 b r + 2 w_1 + b w_2}{(4 - b^2) (1 - r)}$ and $p_{2}^{IN}=\frac{2 a_0 + a_0 b - 2 a_0 r - a_0 b r + b w_1 + 2 w_2}{(4 - b^2) (1- r)}$.

Note that $E(a|f)=A,E(a)=a_0$. Substituting $p_{1}^{IN}$ and $p_{2}^{IN}$ into equation $E(\pi _{M}^{IN})$, the CM’s problem can be rewriten as follows:

$\mathop {Max}\limits _{w_1,w_2}E(\pi _{M}^{IN})=\frac{( b^2-2 ) w_1^2 + 2 b w_1 w_2 + ( b^2-2 ) w_2^2 - a_0 (2 + b) (r-1 ) (w_1 + w_2)}{(2 - b) (2 + b) (1 - r)}.$

Then we have the first- and second- order derivatives of $E(\pi _{M}^{IN})$ regarding $w_1$ and $w_2$ as follows:

$\frac{\partial E(\pi _{M}^{IN})}{\partial w_1}=\frac{2 a_0 + a_0 b - 2 a_0 r - a_0 b r - 4 w_1 + 2 b^2 w_1 + 2 b w_2}{(2 - b) (2 + b) (1 - r)}$,

$\frac{\partial E(\pi _{M}^{IN})}{\partial w_2}=\frac{2 a_0 + a_0 b - 2 a_0 r - a_0 b r + 2 b w_1 - 4 w_2 + 2 b^2 w_2}{(2 - b) (2 + b) (1 - r)}$,

$\frac{\partial ^2E(\pi _{M}^{IN})}{\partial w_1^2}=\frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)}<0$,

$\frac{\partial ^2E(\pi _{M}^{IN})}{\partial w_1 \partial w_2}=\frac{\partial ^2E(\pi _{M}^{IN})}{\partial w_2 \partial w_1}=\frac{2 b}{(2 - b) (2 + b) (1 - r)}$,

$\frac{\partial ^2E(\pi _{M}^{IN})}{\partial w_2^2}=\frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)}<0$.

Then we can get the determinant of the Hessian matrix $H_{1}$ of $E(\pi _{M}^{IN})$ to $w_1$ and $w_2$ as follows.

$$\begin{aligned} |H_{1}|=\left| \begin{array}{cc} \frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)} &{} \frac{2 b}{(2 - b) (2 + b) (1 - r)}\\ \frac{2 b}{(2 - b) (2 + b) (1 - r)} &{}\frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)} \\ \end{array} \right| =\frac{4 (1 - b) (1 + b)}{(2 - b) (2 + b) (1 - r)^2}>0. \end{aligned}$$

Therefore, $E(\pi _{M}^{IN})$ is strictly concave in $w_1$ and $w_2$. Then from $\frac{\partial E(\pi _{M}^{IN})}{\partial w_1}=0$ and $\frac{\partial E(\pi _{M}^{IN})}{\partial w_2}=0$, we have $w_1^{IN*}=\frac{(2 a_0 - a_0 - a_0 b + a_0 b) (1 - r)}{2 (1 - b)}$ and $w_2^{IN*}=\frac{(2 a_0 - a_0 - a_0 b + a_0 b) (1 - r)}{2 (1 - b)}$. Substituting $w_1^{IN*}$ and $w_1^{IN*}$ into $p_{1}^{IN}$ and $p_{2}^{IN}$, we can obtain the optimal solutions of Scenario IN. $\square$

Scenario IS-M

Using the equation $\mathop {Max}\limits _{p_1}E(\pi _{O1}^{IS-M})=E(d_1 (p_1 - r p_1) - d_1 w_1)$, we have $\frac{\partial ^2E(\pi _{O1}^{IS-M})}{\partial p_1^2}=2r-2<0$ for $0<r<1$, which indicates that $E(\pi _{O1}^{IS-M})$ is concave in $p_1$. Similarly, by using the equation $\mathop {Max}\limits _{p_2}E(\pi _{O2}^{IS-M})=E(d_2 (p_2 - r p_2) - d_2 w_2)$, we can obtain $\frac{\partial ^2E(\pi _{O2}^{IS-M})}{\partial p_2^2}=2r-2<0$ for $0<r<1$, which indicates that $E(\pi _{O2}^{IS-M})$ is concave in $p_2$. Then from $\frac{\partial E(\pi _{O1}^{IS-M})}{\partial p_1}=0$ and $\frac{\partial E(\pi _{O2}^{IS-M})}{\partial p_2}=0$, we have $p_{1}^{IS-M}=\frac{2 a_0 + a_0 b - 2 a_0 r - a_0 b r + 2 w_1 + b w_2}{(4 - b^2) (1 - r)}$ and $p_{2}^{IS-M}=\frac{2 a_0 + a_0 b - 2 a_0 r - a_0 b r + 2 w_1 + b w_2}{(4 - b^2) (1 - r)}$.

Note that $E(a|f)=A,E(a)=a_0$. Substituting $p_{1}^{IS-M}$ and $p_{2}^{IS-M}$ into equation $E(\pi _{M}^{IS-M})$, the CM’s problem can be rewriten as follows:

$\mathop {Max}\limits _{w_1,w_2}E(\pi _{M}^{IS-M})=\frac{ b^2 w_1^2-2 w_1^2 + 2 b w_1 w_2 - 2 w_2^2 + b^2 w_2^2 + A ( b^2-4) ( r-1) (w_1 + w_2) - a_0 ( b + b^2-2) ( r-1) (w_1 + w_2)}{(2 - b) (2 + b) (1 - r)}.$

Then we have the first- and second- order derivatives of $E(\pi _{M}^{IS-M})$ regarding $w_1$ and $w_2$ as follows:

$\frac{\partial E(\pi _{M}^{IS-M})}{\partial w_1}=\frac{4 A - 2 a_0 + a_0 b - A b^2 + a_0 b^2 - 4 A r + 2 a_0 r - a_0 b r + A b^2 r - a_0 b^2 r - 4 w_1 + 2 b^2 w_1 + 2 b w_2}{(2 - b) (2 + b) (1 - r)}$,

$\frac{\partial E(\pi _{M}^{IS-M})}{\partial w_2}=\frac{4 A - 2 a_0 + a_0 b - A b^2 + a_0 b^2 - 4 A r + 2 a_0 r - a_0 b r + A b^2 r - a_0 b^2 r + 2 b w_1 - 4 w_2 + 2 b^2 w_2}{(2 - b) (2 + b) (1 - r)}$,

$\frac{\partial ^2E(\pi _{M}^{IS-M})}{\partial w_1^2}=\frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)}<0$,

$\frac{\partial ^2E(\pi _{M}^{IS-M})}{\partial w_1 \partial w_2}=\frac{\partial ^2E(\pi _{M}^{IS-M})}{\partial w_2 \partial w_1}=\frac{2 b}{(2 - b) (2 + b) (1 - r)}$,

$\frac{\partial ^2E(\pi _{M}^{IS-M})}{\partial w_2^2}=\frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)}<0$.

Then we can get the determinant of the Hessian matrix $H_{2}$ of $E(\pi _{M}^{IS-M})$ to $w_1$ and $w_2$ as follows.

$$\begin{aligned} |H_{2}|=\left| \begin{array}{cc} \frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)} &{} \frac{2 b}{(2 - b) (2 + b) (1 - r)}\\ \frac{2 b}{(2 - b) (2 + b) (1 - r)} &{}\frac{-2 (2 - b^2)}{(2 - b) (2 + b) (1 - r)} \\ \end{array} \right| =\frac{4 (1 - b) (1 + b)}{(2 - b) (2 + b) (1 - r)^2}>0. \end{aligned}$$

Therefore, $E(\pi _{M}^{IS-M})$ is strictly concave in $w_1$ and $w_2$. Then from $\frac{\partial E(\pi _{M}^{IS-M})}{\partial w_1}=0$ and $\frac{\partial E(\pi _{M}^{IS-M})}{\partial w_2}=0$, we have $w_1^{IS-M*}=\frac{(2 A - a_0 - A b + a_0 b) (1 - r)}{2 (1 - b)}$ and $w_2^{IS-M*}=\frac{(2 A - a_0 - A b + a_0 b) (1 - r)}{2 (1 - b)}$. Substituting $w_1^{IS-M*}$ and $w_1^{IS-M*}$ into $p_{1}^{IS-M}$ and $p_{2}^{IS-M}$, we can obtain the optimal solutions of Scenario IS-M. $\square$

Proof of Scenarios IS-O1, IS-MO1, IS-O12, and IS-MO12

The proof of Scenarios IS-O1, IS-MO1, IS-O12, and IS-MO12 is omitted due to similarity. Moreover, the optimal expected profits for the firms are given as follows (Table 4).$\square$

Table 4 Optimal expected profits for chain members

Proof of Proposition 1

If A > a₀, we have:

$$\begin{aligned} w^{IS-M^*}_1-w^{IS-MO12^*}_1= & {} \frac{(A - a_0) (1 - r)}{2}>0\\ w^{IS-MO12^*}_1-w^{IS-MO1^*}_1= & {} \frac{(A - a_0) b (1 - r)}{2 (2 - b)}>0\\ w^{IS-MO1^*}_1-w^{IN^*}_1= & {} \frac{(A - a_0) (2 - 2 b + b^2) (1 - r)}{2 (2 - b) (1 - b)}>0\\ w^{IN^*}_1-w^{IS-O1^*}_1= & {} 0\\ w^{IS-O1^*}_1-w^{IS-O12^*}_1= & {} 0 \end{aligned}$$

Hence, $w^{IS-M^*}_1>w^{IS-MO12^*}_1>w^{IS-MO1^*}_1>w^{IN^*}_1=w^{IS-O1^*}_1=w^{IS-O12^*}_1$. simialerly, we can get $w^{IS-MO1^*}_2>w^{IS-M^*}_2>w^{IS-MO12^*}_2>w^{IN^*}_2=w^{IS-O1^*}_2=w^{IS-O12^*}_2$; $p^{IS-MO12^*}_1=p^{IS-MO1^*}_1>p^{IS-M^*}_1>p^{IS-O1^*}_1=p^{IS-O12^*}_1>p^{IN^*}_1$; $p^{IS-MO12^*}_2>p^{IS-MO1^*}_2=p^{IS-M^*}_2>p^{IS-O12^*}_2>p^{IN^*}_2=p^{IS-O1^*}_2$. In addition, if $A<a_0$ holds, there will be the opposite situation. $\square$

Proof of Proposition 2

By comparison, we can know:

$$\begin{aligned} E(\pi ^{IS-M^*}_{M})-E(\pi ^{IS-MO12^*}_{M})= & {} \frac{(3 - b) (1 - r) t v}{2 (2 - b)}>0\\ E(\pi ^{IS-MO1^*}_{M})-E(\pi ^{IS-MO12^*}_{M})= & {} \frac{(6 - 2 b + b^2) (1 - r) t v}{4 (2- b)^2}>0\\ E(\pi ^{IS-MO12^*}_{M})-E(\pi ^{IN^*}_{M})= & {} \frac{(1 - r) t v}{2 (2 - b) (1 - b)}>0\\ E(\pi ^{IN^*}_{M})-E(\pi ^{IS-O1^*}_{M})= & {} 0\\ E(\pi ^{IS-O1^*}_{M})-E(\pi ^{IS-O12^*}_{M})= & {} 0\\ E(\pi ^{IS-M^*}_{M})-E(\pi ^{IS-MO1^*}_{M})= & {} \frac{(6 - 8 b + b^2) (1 - r) t v}{4 (2 - b)^2} \end{aligned}$$

Hence, if $0<b<4-\sqrt{10}$, $E(\pi ^{IS-M^*}_{M})>E(\pi ^{IS-MO1^*}_{M})>E(\pi ^{IS-MO12^*}_{M})>E(\pi ^{IN^*}_{M}) =E(\pi ^{IS-O1^*}_{M})=E(\pi ^{IS-O12^*}_{M})$; if $4-\sqrt{10}<b<1$, $E(\pi ^{IS-MO1^*}_{M})>E(\pi ^{IS-M^*}_{M})>E(\pi ^{IS-MO12^*}_{M})>E(\pi ^{IN^*}_{M}) =E(\pi ^{IS-O1^*}_{M})=E(\pi ^{IS-O12^*}_{M})$. $\square$

Proof of Proposition 3

By comparison, we can get the following relationship: when $0<b<0.75$, $E(\pi ^{IS-O12^*}_{O1})>E(\pi ^{IS-M^*}_{O1})>E(\pi ^{IS-O1^*}_{O1})> E(\pi ^{IS-MO12^*}_{O1})>E(\pi ^{IS-MO1^*}_{O1})>E(\pi ^{IN^*}_{O1})$; when $0.75<b<1$, $E(\pi ^{IS-O12^*}_{O1})>E(\pi ^{IS-M^*}_{O1})>E(\pi ^{IS-MO12^*}_{O1})> E(\pi ^{IS-O1^*}_{O1})>E(\pi ^{IS-MO1^*}_{O1})>E(\pi ^{IN^*}_{O1})$. Furthermore, we have: $E(\pi ^{IS-O12^*}_{O2})>E(\pi ^{IS-MO12^*}_{O2})>E(\pi ^{IS-M^*}_{O2})> E(\pi ^{IS-MO1^*}_{O2})>E(\pi ^{IS-O1^*}_{O2})=E(\pi ^{IN^*}_{O2})$. $\square$

Proof of Proposition 4

By comparison, we can easily get the following relationship: if $0<b<0.5$, $E(\pi ^{IS-M^*}_{P})>E(\pi ^{IS-O12^*}_{P})>E(\pi ^{IS-MO1^*}_{P})>E(\pi ^{IS-MO12^*}_{P})>E(\pi ^{IS-O1^*}_{P})>E(\pi ^{IN^*}_{P})$; if $0.5<b<1$, $E(\pi ^{IS-M^*}_{P})>E(\pi ^{IS-MO12^*}_{P})>E(\pi ^{IS-MO1^*}_{P})>E(\pi ^{IS-O12^*}_{P})>E(\pi ^{IS-O1^*}_{P})>E(\pi ^{IN^*}_{P})$. $\square$

Proof of Proposition 4

By comparison, we can easily get the following relationship: we have:

• (1) if $0<b<0.5$ and $\frac{3(3 -2b)}{4(2 -b)}<r<1$, $E(\pi ^{IS-M^*}_{SC})>E(\pi ^{IS-O12^*}_{SC})>E(\pi ^{IS-MO1^*}_{SC})>E(\pi ^{IS-MO12^*}_{SC})> E(\pi ^{IS-O1^*}_{SC})>E(\pi ^{IN^*}_{SC})$;

• (2) if $0<b<0.5$ and $0<r<\frac{3(3-2b)}{4(2-b)}$, $E(\pi ^{IS-M^*}_{SC})>E(\pi ^{IS-MO1^*}_{SC})>E(\pi ^{IS-O12^*}_{SC})>E(\pi ^{IS-MO12^*}_{SC})> E(\pi ^{IS-O1^*}_{SC})>E(\pi ^{IN^*}_{SC})$;

• (3) if $0.5<b<1$ and $\frac{7-11b+6b^2}{4(2 -3b+b^2)}<r<1$, $E(\pi ^{IS-M^*}_{SC})>E(\pi ^{IS-MO12^*}_{SC})>E(\pi ^{IS-MO1^*}_{SC})>E(\pi ^{IS-O12^*}_{SC})> E(\pi ^{IS-O1^*}_{SC})>E(\pi ^{IN^*}_{SC})$;

• (4) if $0.5<b<1$ and $0<r<\frac{7-11b+6b^2}{4(2 -3b+b^2)}$, $E(\pi ^{IS-M^*}_{SC})>E(\pi ^{IS-MO1^*}_{SC})>E(\pi ^{IS-MO12^*}_{SC})>E(\pi ^{IS-O12^*}_{SC})> E(\pi ^{IS-O1^*}_{SC})>E(\pi ^{IN^*}_{SC})$.

$\square$

Proof of Proposition 6–9

Following a similar logic in proving the base model, we can prove Propositions 6–9. $\square$