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Information advantage and payment disadvantage when selling goods through a powerful retailer

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Abstract

In practice, selling goods through a powerful retailer such as Wal-Mart enables the supplier to access the retailer’s ERP for accurate demand information (e.g., Wal-Mart’s Retail Link). However, in the recent years, we observe the suppliers are suffering from longer and longer average account period when they contract with powerful retailers. Therefore, whether partnering with a powerful retailer at the cost of a longer account period becomes the supplier’s strategic decision. In this paper, we formulate the supplier’s tradeoffs among the information advantage, payment disadvantage, and channel competition when it makes retailing decisions. We study the supplier’s two representative strategies: (1) relying on a small retailer that does not accumulate much information but can settle accounts immediately (referred to as Real-time Payment Retailing) or (2) relying on a powerful retailer that shares accurate demand information but incurs deferred payment (referred to as Deferred Payment Retailing). We built game-theoretical models and found that, interestingly, the supplier will prefer Deferred Payment Retailing when the supplier’s cash opportunity cost is high. We identify three interactive forces, namely, the pricing power effect, the demand size effect, and the information value, to interpret the rationality of the supplier’s preferences over Real-time and Deferred Payment Retailing strategies.

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Acknowledgements

The authors are grateful to the editors and reviewers for their helpful comments. This work was supported by NSFC Excellent Young Scientists Fund (No. 71822202), Chang Jiang Scholars Program (Niu Baozhuang 2017), and the Fundamental Research Funds for the Central Universities.

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Appendices

Appendix 1: The derivation of outcomes in each strategies

Under Strategy R, the supplier relies on the small retailer for product selling. There is a chain-to-chain competition where the supplier does not have demand information. We solve this game using backward induction. Since the powerful retailer has big data and perfect demand forecast, she determines her optimal quantity according to the perfectly forecasted ε. The small retailer has no big data, so it determines its optimal quantity according to E[ε]. The expected profit of the small retailer is

$$ \begin{aligned} E\left[ {\pi_{SR} } \right] & = E\left\{ {\left[ {a + \varepsilon - q_{S} - bq_{PR} - \left( {1 + r} \right)w} \right]q_{S} } \right\} \\ & = \left[ {a - E\left[ {q_{S} } \right] - bq_{PR} - \left( {1 + r} \right)w} \right]E\left[ {q_{S} } \right] \\ \end{aligned} $$
(1)

The profit of the powerful retailer is

$$ \pi_{PR} = \left[ {a + \varepsilon - q_{PR} - bq_{S} - \left( {1 + r} \right)c} \right]q_{PR} $$
(2)

For any given ε, the first-order conditions of \( \frac{{\partial E\left[ {\pi_{SR} } \right]}}{{\partial q_{SR} }} \) and \( \frac{{\partial \pi_{PR} }}{{\partial q_{PR} }} \) yield

$$ E\left[ {q_{S} \left( {w,q_{PR} } \right)} \right] = \frac{{a - \left( {1 + r} \right)w - bq_{PR} }}{2} $$
(3)
$$ q_{PR} \left( {w,E\left[ {q_{S} } \right]} \right) = \frac{{a + \varepsilon - \left( {1 + r} \right)c - bE\left[ {q_{S} } \right]}}{2} $$
(4)

Solving the equation sets, we have

$$ q_{S} = \frac{{a\left( {2 - b} \right) - \left( {1 + r} \right)\left( {2w - bc} \right)}}{{4 - b^{2} }} $$
(5)
$$ q_{PR} = \frac{{2a\left( {2 - b} \right) - 4c\left( {1 + r} \right) + 2b\left( {1 + r} \right)w + \left( {4 - b^{2} } \right)\epsilon}}{{2\left( {4 - b^{2} } \right)}} $$
(6)

The supplier determines his expected wholesale price to maximize his expected profit. Substituting Eq. (4) into Eq. (7) yields

$$ E\left[ {\pi_{S} \left] { = E} \right[\left( {1 + r} \right)\left( {w - c} \right)q_{S} \left( w \right)} \right] $$
(7)

The first-order condition \( \frac{{\partial E\left[ {\pi_{S} } \right]}}{\partial w} \) yields his optimal wholesale price

$$ w = \frac{{a\left( {2 - b} \right) + \left( {2 + b} \right)c\left( {1 + r} \right)}}{{4\left( {1 + r} \right)}} $$
(8)

Substituting Eq. (8) to the response functions (5) and (6) and profit functions (1), (2) and (7), we obtain the equilibrium outcomes under Strategy R in Table 9.

Table 9 Outcomes under Strategy R

Next, we analyze the outcomes under Strategy D where the supplier sells products through the powerful retailer. Under Strategy D, the supplier has perfect demand forecast because of the sharing of the powerful retailer (e.g. Wal-Mart’s Retail Link). Their decisions are based on accurate forecast \( \epsilon \). Thus, the supplier and the powerful retailer’s profit functions are

$$ \pi_{S} = \left[ {w - \left( {1 + r} \right)c} \right]q_{S} $$
(9)
$$ \pi_{PR} = \left[ {p_{PR} - \left( {1 + r} \right)c} \right]q_{PR} + \left( {p_{S} - w} \right)q_{S} $$
(10)

The powerful retailer’s profit function is strictly concave in qS and qPR, so the first-order conditions yield the best response functions

$$ q_{S} = \frac{{a - w + \epsilon - 2bq_{PR} }}{2} $$
(11)
$$ q_{PR} = \frac{{a - c\left( {1 + r} \right) + \epsilon - 2bq_{S} }}{2} $$
(12)

Solving Eqs. (11) and (12), we have

$$ q_{S} = \frac{{\left( {a + \epsilon} \right)\left( {1 - b} \right) + bc\left( {1 + r} \right) - w}}{{2\left( {1 - b^{2} } \right)}} $$
(13)
$$ q_{PR} = \frac{{\left( {a + \epsilon} \right)\left( {1 - b} \right) - c\left( {1 + r} \right) + bw}}{{2\left( {1 - b^{2} } \right)}} $$
(14)

Substituting Eqs. (13) and (14) into Eq. (9), the supplier’s optimal wholesale price can be derived by letting \( \frac{{\partial \pi_{S} }}{\partial w} = 0. \) We have

$$ w = \frac{{\left( {a + \epsilon} \right)\left( {1 - b} \right) + \left( {1 + b} \right)c\left( {1 + r} \right)}}{2} $$
(15)

Substituting Eq. (15) to Eqs. (13) and (14), we obtain the outcomes under Strategy D in Table 10.

Table 10 Outcomes under Strategy D

Next, we prove the propositions and lemmas. Note that all the discussions are in the feasible set of the parameters \( \left\{ {\left( {a,b,c,r, \sigma} \right) \left |c < a< {2c,c} > 0, 0 < b< {1,\sigma}\,\right. \,> 0,0 < r < \frac{a - c}{c}} \right\} \).

Appendix 2: Proofs of propositions

Proof of Proposition 1

It’s easy to show that \( \frac{{\partial w^{R} }}{\partial r} = - \frac{{a\left( {2 - b} \right)}}{{4\left( {1 + r} \right)^{2} }} < 0 \) and \( \left\{ {\left( {a,b,c,r, \sigma}\right) \left |c < a< {2c,c} > 0, 0 < b< {1,\sigma}\,\right. \,> 0,0 < r < \frac{a - c}{c}} \right\} \).

Therefore, the difference between the wD and wR is

$$ w^{D} - w^{R} = \frac{{2c\left( {1 + b} \right)r^{2} + \left( {2a - 2ab + 2c + 3bc} \right)r + bc - ab}}{{4\left( {1 + r} \right)}} $$

The condition that \( w^{D} > w^{R} \) equals to the condition that \( f_{1} \left( r \right) = 2c\left( {1 + b} \right)r^{2} + \left( {2a - 2ab + 2c + 3bc} \right)r + bc - ab > 0 \). Next, we analyze \( f_{1} \left( r \right) \).

The discriminant of the quadratic function \( f_{1} \left( r \right) \) is \( \Delta_{1} = 8b\left( {1 + b} \right)\left( {a - c} \right)c + \left( { - 2a\left( { - 1 + b} \right) + \left( {2 + 3b} \right)c} \right)^{2} > 0 \), which indicates \( f_{1} \left( r \right) \) has two different roots.

Solving \( f_{1} \left( r \right) = 0 \), we have

$$ r_{w} = \frac{{\sqrt {\Delta_{1} } - 2a\left( {1 - b} \right) - c\left( {2 + 3b} \right)}}{{4\left( {1 + b} \right)c}},\quad r_{v} = \frac{{ - \sqrt {\Delta_{1} } - 2a\left( {1 - b} \right) - c\left( {2 + 3b} \right)}}{{4\left( {1 + b} \right)c}} $$

It becomes immediate that \( \frac{a - c}{c} > r_{w} > 0 \) and \( r_{v} < 0 \).

Since \( 2c\left( {1 + b} \right) > 0 \), the quadratic function \( f_{1} \left( r \right) \) opens up. Therefore, \( f_{1} \left( r \right) > 0 \) equals to \( r > r_{w} \). Proposition 1 is proven.

Proof of Proposition 2

  1. (a)

    It is straightforward to show that \( q_{S}^{D} - q_{S}^{R} = - \frac{{b\left( {a - c - cr} \right)}}{{4\left( {1 + b} \right)\left( {2 + b} \right)}} < 0 \).

  2. (b)

    Obviously, we have \( \frac{{\partial q_{S}^{D} }}{\partial r} = - \frac{c}{{4\left( {1 + b} \right)}} < 0,\frac{{\partial q_{S}^{R} }}{\partial r} = - \frac{c}{{2\left( {2 + b} \right)}} < 0,\frac{{\partial q_{S}^{D} }}{\partial r} - \frac{{\partial q_{S}^{R} }}{\partial r} = \frac{bc}{{4\left( {2 + 3b + b^{2} } \right)}} > 0 \).

  3. (c)

    The difference between \( \frac{{q_{S}^{D} }}{{q_{S}^{D} + q_{PR}^{D} }} \) and \( \frac{{q_{S}^{R} }}{{q_{S}^{R} + q_{PR}^{R} }} \) is \( \frac{{q_{S}^{D} }}{{q_{S}^{D} + q_{PR}^{D} }} - \frac{{q_{S}^{R} }}{{q_{S}^{R} + q_{PR}^{R} }} = - \frac{b}{{18 + 9b + b^{2} }} < 0 \).

  4. (d)

    The difference between \( \frac{{q_{PR}^{D} }}{{q_{PR}^{D} + q_{S}^{D} }} \) and \( \frac{{q_{PR}^{R} }}{{q_{PR}^{R} + q_{S}^{R} }} \) is \( \frac{{q_{PR}^{D} }}{{q_{PR}^{D} + q_{S}^{D} }} - \frac{{q_{PR}^{R} }}{{q_{PR}^{R} + q_{S}^{R} }} = \frac{b}{{18 + 9b + b^{2} }} > 0 \).

Hereby, we prove Proposition 2.

Proof of Proposition 3

The difference between \( E\left[ {\pi_{S}^{D} } \right] \) and \( E\left[ {\pi_{S}^{R} } \right] \) is

$$ E\left[ {\pi_{S}^{D} } \right] - E\left[ {\pi_{S}^{R} } \right] = \frac{{ - 2bc^{2} r^{2} + 4bc\left( {a - c} \right)r - 2b\left( {a - c} \right)^{2} + \left( {2 - b - b^{2} } \right)\sigma^{2} }}{{8\left( {1 + b} \right)\left( {2 + b} \right)}} $$

The condition that \( E\left[ {\pi_{S}^{D} } \right] > E\left[ {\pi_{S}^{R} } \right] \) equals to the condition \( f_{2} \left( r \right) = - 2bc^{2} r^{2} + 4bc\left( {a - c} \right)r - 2b\left( {a - c} \right)^{2} + \left( {2 - b - b^{2} } \right)\sigma^{2} \). Next, we analyze \( f_{2} \left( r \right) \).

The discriminant of the quadratic function \( f_{2} \left( r \right) \) is \( \Delta_{2} = 8b\left( {2 - b - b^{2} } \right)c^{2} \sigma^{2} > 0 \), which indicates \( f_{2} \left( r \right) \) has two different roots.

Solving \( f_{2} \left( r \right) = 0 \), we have

$$ r_{1} = \frac{a - c}{c} - \frac{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }}{{c\sqrt {2b} }}\sigma ,\quad r_{a} = \frac{a - c}{c} + \frac{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }}{{c\sqrt {2b} }}\sigma $$

Then we can show that \( \frac{a - c}{c} > r_{1} \) and \( r_{a} > \frac{a - c}{c} \).

  1. 1.

    If \( r_{1} > 0 \), we have \( \sigma \le \sigma_{1} = \frac{{\left( {a - c} \right)\sqrt {2b} }}{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }} \). Since \( - 2bc^{2} < 0 \), the quadratic function \( f_{2} \left( r \right) \) opens down, therefore, we find that \( f_{2} \left( r \right) > 0 \) equals to \( r > r_{1} \). It then can be shown that \( \sigma \le \sigma_{1} = \frac{{\left( {a - c} \right)\sqrt {2b} }}{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }} \) and \( r > r_{1} = \frac{a - c}{c} - \frac{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }}{{c\sqrt {2b} }}\sigma \).

  2. 2.

    If \( r_{1} < 0 \), we have \( \sigma \ge \sigma_{1} = \frac{{\left( {a - c} \right)\sqrt {2b} }}{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }} \). It can be shown that \( f_{2} \left( r \right) > 0 \) holds in the interval \( \frac{a - c}{c} > r > 0 \). Therefore, we have \( \sigma > \sigma_{1} = \frac{{\left( {a - c} \right)\sqrt {2b} }}{{\sqrt {\left( {1 - b} \right)\left( {2 + b} \right)} }} \).

Proposition 3 is proven.

Proof of Proposition 4

The difference between \( E\left[ {\pi^{D} } \right] \) and \( E\left[ {\pi^{R} } \right] \) is

$$ E\left[ {\pi^{D} } \right] - E\left[ {\pi^{R} } \right] = \frac{{ - 2b\left( {2 - 2b - b^{2} } \right)c^{2} r^{2} + 4b\left( {2 - 2b - b^{2} } \right)\left( {a - c} \right)cr - 2b\left( {2 - 2b - b^{2} } \right)\left( {a - c} \right)^{2} + 3\left( {1 - b} \right)\left( {2 + b} \right)^{2} \sigma^{2} }}{{16\left( {1 + b} \right)\left( {2 + b} \right)^{2} }} $$

The condition that \( E\left[ {\pi^{D} } \right] > E\left[ {\pi^{R} } \right] \) equals to \( f_{3} \left( r \right) = - 2b\left( {2 - 2b - b^{2} } \right)c^{2} r^{2} + 4b\left( {2 - 2b - b^{2} } \right)\left( {a - c} \right)cr - 2b\left( {2 - 2b - b^{2} } \right)\left( {a - c} \right)^{2} + 3\left( {1 - b} \right)\left( {2 + b} \right)^{2} \sigma^{2} > 0 \), which indicates \( f_{3} \left( r \right) \) has two different roots.

Solving the equation \( f_{3} \left( r \right) = 0 \), we have

$$ r_{2} = \frac{a - c}{c} - \frac{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }}{{c\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}\sigma ,\quad r_{b} = \frac{a - c}{c} + \frac{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }}{{c\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}\sigma . $$

It can be shown that \( r_{2} < \frac{a - c}{c} \) and \( r_{b} > \frac{a - c}{c} \).

  1. 1.

    If \( r_{2} > 0, \) we have \( \sigma \le \sigma_{2} = \frac{{\left( {a - c} \right)\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }} \). Since \( - 2b\left( {2 - 2b - b^{2} } \right)c^{2} < 0 \), the quadratic function \( f_{3} \left( r \right) \) opens down. Therefore, it can be shown that \( f_{3} \left( r \right) > 0 \) equals to \( r > r_{2} \). Thus, we find that \( \sigma \le \sigma_{2} = \frac{{\left( {a - c} \right)\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }} \) and \( r > r_{2} = \frac{a - c}{c} - \frac{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }}{{c\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}\sigma \).

  2. 2.

    If \( r_{2} < 0, \) we have \( \sigma \ge \sigma_{2} = \frac{{\left( {a - c} \right)\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }} \). It can be shown that \( f_{3} \left( r \right) > 0 \) holds in the interval \( \frac{a - c}{c} > r > 0 \). So, we find that \( \sigma \ge \sigma_{2} = \frac{{\left( {a - c} \right)\sqrt {2b\left( {2 - 2b - b^{2} } \right)} }}{{\left( {2 + b} \right)\sqrt {3\left( {1 - b} \right)} }} \).

Proposition 4 is proven.

Proof of Proposition 5

The difference between \( \frac{{E\left[ {\pi_{S}^{D} } \right]}}{{E\left[ {\pi^{D} } \right]}} \) and \( \frac{{E\left[ {\pi_{S}^{R} } \right]}}{{E\left[ {\pi^{R} } \right]}} \) is

$$ \frac{{E\left[ {\pi_{S}^{D} } \right]}}{{E\left[ {\pi^{D} } \right]}} - \frac{{E\left[ {\pi_{S}^{R} } \right]}}{{E\left[ {\pi^{R} } \right]}} = \frac{{ - \left( {4 - b} \right)b\left( {3 + b} \right)c^{2} r^{2} + 2\left( {4 - b} \right)b\left( {3 + b} \right)c\left( {a - c} \right)r + 8\sigma^{2} - b\left( {3 + b} \right)\left[ {\left( {4 - b} \right)\left( {a - c} \right)^{2} + 2b\sigma^{2} } \right]}}{{\left( {7 + b} \right)\left( {\left( {28 + 8b - b^{2} } \right)\left( {a - c - cr} \right)^{2} + 4\left( {2 + b} \right)^{2} \sigma^{2} } \right)}} $$

We find that the condition that \( \frac{{E\left[ {\pi_{S}^{D} } \right]}}{{E\left[ {\pi^{D} } \right]}} > \frac{{E\left[ {\pi_{S}^{R} } \right]}}{{E\left[ {\pi^{R} } \right]}} \) equals to \( f_{4} \left( r \right) = - \left( {4 - b} \right)b\left( {3 + b} \right)c^{2} r^{2} + 2\left( {4 - b} \right)b\left( {3 + b} \right)c\left( {a - c} \right)r + 8\sigma^{2} - b\left( {3 + b} \right)\left[ {\left( {4 - b} \right)\left( {a - c} \right)^{2} + 2b\sigma^{2} } \right] > 0 \), indicating \( f_{4} \left( r \right) \) has two different roots.

Solving the equation \( f_{4} \left( r \right) = 0 \), we have\( r_{3} = \frac{a - c}{c} - \frac{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }}{{c\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}\sigma ,\quad r_{c} = \frac{a - c}{c} + \frac{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }}{{c\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}\sigma . \).Therefore, it is shown that \( r_{3} < \frac{a - c}{c} \) and \( r_{c} > \frac{a - c}{c} \).

  1. 1.

    If \( r_{3} > 0, \) we have \( \sigma \le \sigma_{3} = \frac{{\left( {a - c} \right)\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }} \). Since \( - \left( {4 - b} \right)b\left( {3 + b} \right)c^{2} < 0 \), the quadratic function \( f_{4} \left( r \right) \) opens down. Therefore, it can be shown that \( f_{4} \left( r \right) > 0 \) equals to \( r > r_{3} \). We find that \( \sigma \le \sigma_{3} = \frac{{\left( {a - c} \right)\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }} \) and \( r > r_{3} = \frac{a - c}{c} - \frac{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }}{{c\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}\sigma \).

  2. 2.

    If \( r_{3} < 0, \) we have \( \sigma \ge \sigma_{3} = \frac{{\left( {a - c} \right)\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }} \). It is obviously that \( f_{4} \left( r \right) > 0 \) holds in the region \( \frac{a - c}{c} > r > 0 \). So, we show that \( \sigma \ge \sigma_{3} = \frac{{\left( {a - c} \right)\sqrt {b\left( {4 - b} \right)\left( {3 + b} \right)} }}{{\left( {2 + b} \right)\sqrt {2\left( {1 - b} \right)} }} \).

Proposition 5 becomes immediate.

Appendix 3: Powerful retailer has no incentive to take the first-mover advantage

In this subsection, we investigate the powerful retailer’s incentive to take the first-mover advantage under Strategy D, which means the powerful retailer determines her order quantity qPR before the supplier determines the wholesale price w. Therefore, the event sequence under Strategy D is revised. See Fig. 12.

Fig. 12
figure 12

Illustration of event sequence and payment cycle under Strategy D

We first derive the equilibrium outcomes when the powerful retailer is the first-mover under Strategy D, and then investigate whether she has incentives to be the first-mover.

The supplier and the powerful retailer’s profit functions are the same as Eqs. (9) and (10). We solve the game by backward induction. First, given qPR and w, we derive the best response function of the powerful retailer with respect to qS as

$$ q_{S} \left( {q_{PR} ,w} \right) = \frac{{a + \epsilon - 2bq_{PR} - w}}{2} $$
(16)

Then, substituting Eq. (16) into Eq. (9), we obtain the supplier’s optimal wholesale price

$$ w\left( {q_{PR} } \right) = \frac{{a + c\left( {1 + r} \right) + \epsilon - 2bq_{PR} }}{2} $$
(17)

Substituting Eq. (17) into Eq. (16), we have

$$ q_{S} \left( {q_{PR} } \right) = \frac{{a - c\left( {1 + r} \right) + \epsilon - 2bq_{PR} }}{4} $$
(18)

Then the powerful retailer’s optimal quantity of her self-branded product qPR can be derived as

$$ q_{PR} = \frac{{\left( {4 - b} \right)\left[ {a - c\left( {1 + r} \right)} \right]}}{{2\left( {4 - b^{2} } \right)}} + \frac{{\left( {4 - b} \right)\epsilon}}{{2\left( {4 - b^{2} } \right)}} $$
(19)

Substituting Eq. (19) back to Eqs. (17) and (18), we obtain the supplier’s equilibrium wholesale price w and the powerful retailer’s equilibrium order quantity qS

$$ w = \frac{{2a\left( {1 - b} \right) + \left( {2 + 2b - b^{2} } \right)c\left( {1 + r} \right)}}{{4 - b^{2} }} + \frac{{2\left( {1 - b} \right)\epsilon}}{{4 - b^{2} }} $$
(20)
$$ q_{S} = \frac{{\left( {1 - b} \right)\left[ {a - c\left( {1 + r} \right)} \right]}}{{4 - b^{2} }} + \frac{{\left( {1 - b} \right)\epsilon}}{{4 - b^{2} }} $$
(21)

Similarly, we obtain the expected profits of the supplier and the powerful retailer. All the outcomes under Strategy D are in Table 11. For ease of expression, we use Strategy D3 to denote the case in which the powerful retailer is the first-mover.

Table 11 Expected equilibrium outcomes under Strategy D3

Next, we investigate the powerful retailer’s incentive to be the first-mover by comparing her profits under Strategy D and Strategy D3. Note that all the discussions require the parameters satisfy \( \left\{ {\left( {c,a,b,\sigma ,r} \right) |c < a\left\langle {2c,c} \right\rangle 0, 0 < b\left\langle {1,\sigma } \right\rangle 0,0 < r < \frac{a - c}{c}} \right\} \).

Proposition 9

The powerful retailer prefers Strategy D for any feasible parameters (i.e., \( E\left[ {\pi_{PR}^{{D_{3} }} } \right] < E\left[ {\pi_{PR}^{D} } \right] \)).

Intuitively, a player who moves first can acquire the first-mover advantage Cournot competition (Lus and Muriel 2009). However, Proposition 9 demonstrates that the powerful retailer has no incentive to be the first-mover. To explain the interesting finding, we derive the following three lemmas.

Lemma 1

If the powerful retailer is the first-mover, the supplier will determine a higher wholesale price (i.e., \( w^{{D_{3} }} > w^{D} \)).

Lemma 2

If the powerful retailer is the first-mover, the powerful retailer will determine a larger quantity of her self-branded product and the supplier will receive a smaller order quantity from the powerful retailer (i.e., \( q_{PR}^{{D_{3} }} > q_{PR}^{D} \) and \( q_{S}^{{D_{3} }} < q_{S}^{D} \)).

Lemma 3

If the powerful retailer is the first-mover, she will obtain a higher marginal profit from her self-branded product and a lower marginal profit from supplier’s product (i.e., \( E\left[ {m_{PR}^{{D_{3} }} } \right] > E\left[ {m_{PR}^{D} } \right] \) and \( E\left[ {m_{S}^{{D_{3} }} } \right] < E\left[ {m_{S}^{D} } \right] \)). In addition, the marginal profit of the powerful retailer is hurt more when she is the first-mover (i.e., \( E\left[ {m_{PR}^{{D_{3} }} } \right] - E\left[ {m_{PR}^{D} } \right] < E\left[ {m_{S}^{D} } \right] - E\left[ {m_{S}^{{D_{3} }} } \right] \)).

Lemma 1 points out that the supplier will determine a higher wholesale price if the powerful retailer acts as the first-mover, which increases the powerful retailer’s procurement cost and hence, lowers her order quantity under Strategy D3. This motivates the powerful retailer to increase the quantity of self-branded product, as Lemma 2 shows. Therefore, the powerful retailer’s profit from the reselling business is hurt (\( E\left[ {m_{S}^{{D_{3} }} } \right] < E\left[ {m_{S}^{D} } \right] \)) but that from the self-brand business is increased (\( E\left[ {m_{PR}^{{D_{3} }} } \right] > E\left[ {m_{PR}^{D} } \right] \)). In equilibrium, we find that the marginal profit of the powerful retailer is hurt more when she is the first-mover, as Lemma 3 reveals that \( E\left[ {m_{PR}^{{D_{3} }} } \right] - E\left[ {m_{PR}^{D} } \right] < E\left[ {m_{S}^{D} } \right] - E\left[ {m_{S}^{{D_{3} }} } \right] \). Consequently, the powerful retailer has no incentive to be the first-mover.

We provide the proofs of Proposition 9, Lemmas 1, 2 and 3 as follows.

Proof of Proposition 9

Proposition 9 can be derived by comparing \( E\left[ {\pi_{PR}^{D} } \right] \) with \( E\left[ {\pi_{PR}^{{D_{3} }} } \right] \). We have

$$ E\left[ {\pi_{PR}^{D} } \right] - E\left[ {\pi_{PR}^{{D_{3} }} } \right] = \frac{{3\left\{ {\left[ {a - c\left( {1 + r} \right)} \right]^{2} + \sigma^{2} } \right\}b^{2} }}{{16\left( {1 + b} \right)}}\frac{1 - b}{{4 - b^{2} }} $$

Clearly, we have \( 1 - b > 0 \) and \( 4 - b^{2} > 0 \) for any \( b \in \left( {0,1} \right) \). Therefore, it can be verified that \( E\left[ {\pi_{PR}^{D} } \right] - E\left[ {\pi_{PR}^{{D_{3} }} } \right] > 0 \) for any \( 0 < r < \frac{a - c}{c} \), \( 0 < b < 1 \) and \( \sigma > 0 \).

Thus, Proposition 9 is proven.

Proof of Lemma 1

Lemma 1 can be derived by comparing \( E\left[ {w^{{D_{3} }} } \right] \) with \( E\left[ {w^{D} } \right] \). We have

$$ w^{{D_{3} }} - w^{D} = \frac{{\left( {1 - b} \right)b^{2} \left[ {a - c\left( {1 + r} \right)} \right]}}{{2\left( {4 - b^{2} } \right)}} $$

It can be verified that \( w^{{D_{3} }} - w^{D} > 0 \) for any \( 0 < r < \frac{a - c}{c} \) and \( 0 < b < 1 \).

Lemma 1 is proven.

Proof of Lemma 2

The proof of Lemma 2 is similar to the proof of Lemma 1. We therefore omit the proof here.

Proof of Lemma 3

Lemma 3 can be derived by comparing \( m_{PR}^{{D_{3} }} \) with \( m_{PR}^{D} ,m_{S}^{{D_{3} }} \) with \( m_{S}^{D} \) and \( m_{PR}^{D} - m_{PR}^{{D_{3} }} \) with \( m_{S}^{{D_{3} }} - m_{S}^{D} \), respectively. We have

$$ \begin{aligned} m_{PR}^{D} - m_{PR}^{{D_{3} }} & = \frac{{\left( {1 - b} \right)b^{2} \left[ {a - c\left( {1 + r} \right)} \right]}}{{4\left( {4 - b^{2} } \right)}} > 0 \\ m_{S}^{{D_{3} }} - m_{S}^{D} & = - \frac{{\left( {1 - b} \right)b^{2} \left[ {a - c\left( {1 + r} \right)} \right]}}{{2\left( {4 - b^{2} } \right)}} < 0 \\ \left( {m_{PR}^{D} - m_{PR}^{{D_{3} }} } \right) - \left( {m_{S}^{{D_{3} }} - m_{S}^{D} } \right) & = - \frac{{\left( {1 - b} \right)b\left[ {a - c\left( {1 + r} \right)} \right]}}{{4\left( {2 - b} \right)}} < 0 \\ \end{aligned} $$

for any \( 0 < r < \frac{a - c}{c} \) and \( 0 < b < 1 \).

Thus, Lemma 3 is proven.

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Niu, B., Shen, Z. & Li, Q. Information advantage and payment disadvantage when selling goods through a powerful retailer. Ann Oper Res 331, 417–446 (2023). https://doi.org/10.1007/s10479-020-03889-x

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