Strategic information sharing in online retailing under a consignment contract with revenue sharing

Abstract

This work develops a general model of a two-echelon supply chain in which a dominant retailer interacts with a manufacturer via a consignment contract with revenue sharing. The manufacturer’s cost function is known only to him, whereas the retailer has only an estimation of this function, which is based on common knowledge. We formulate the interaction between the parties as a Stackelberg game in which the less informed party (the retailer) moves first. We investigate a strategic information-sharing policy of the manufacturer under general formulations of (i) the supply chain’s revenue and cost functions, and (ii) the manufacturer’s decision functions. Two models are considered: (i) a point-estimation model—the retailer relies on a single-valued estimation of the manufacturer’s cost function, based on her “best belief”; and (ii) an interval-estimation model—the retailer faces uncertainty with regard to the cost function and thus estimates its parameter values within intervals. We find a condition that distinguishes between a case in which it is optimal for both parties for the manufacturer to share his exact cost function and a case in which such information-sharing is not optimal for the manufacturer but is optimal for the retailer. In the interval-estimation model, equilibrium is obtained using a normative (probabilistic) approach as well as behavioral-decision criteria (max–max, max–min and regret minimization). Under a normative approach both hidden and known superiority of the manufacturer are considered. Finally, we use our model to analyze a supply chain of a mobile application.

Appendix

Proof of Lemma 1

The total profit is $R(p,q) - C(q) = p\left( {a - \alpha p + \beta q} \right) - \gamma {{q^{2} } \mathord{\left/ {\vphantom {{q^{2} } 2}} \right. \kern-0pt} 2}$. The gradient of this function is $\left( {a - 2\alpha p + \beta q,\beta p - \gamma q} \right)\,$, and the Hessian is $\,\,\left( {\begin{array}{*{20}c} { - 2\alpha } & \beta \\ \beta & { - \gamma } \\ \end{array} } \right)$. It is easy to verify that when $\gamma > 0.5\beta^{2} /\alpha$, the function is concave with finite maxima. □

Proof of Proposition 3

The gradient and the Hessian of $\varPi_{m} (p,q,\eta ) = (1 - \eta )p(a - \alpha p + \beta q) - \gamma {{q^{2} } \mathord{\left/ {\vphantom {{q^{2} } 2}} \right. \kern-0pt} 2}$ with respect to p and q are

$$ \nabla \varPi_{m} (p,q,\eta ) = \left( {(1 - \eta )\left( {a + \beta q - 2\alpha p} \right)\,\,,\,\,(1 - \eta )p\beta - \gamma q} \right) $$

(and)

$$ H\left( {\varPi_{m} (p,q,\eta )} \right) = \left( {\begin{array}{*{20}c} { - 2\alpha (1 - \eta )} & {\beta (1 - \eta )} \\ {\beta (1 - \eta )} & { - \gamma } \\ \end{array} } \right). $$

It is easy to verify that when $\gamma > 0.5\beta^{2} /\alpha$, the profit function is concave. Solving $\nabla \varPi_{m} (p,q|\eta ) = 0$, we obtain the manufacturer’s best response for $\eta < 1$:

$$ p(\eta ) = a\gamma /(2\alpha \gamma - (1 - \eta )\beta^{2} ) $$

(A.1)

and

$$ q(\eta ) = a\beta (1 - \eta )/(2\alpha \gamma - (1 - \eta )\beta^{2} ). $$

(A.2)

The derivative of $\varPi_{r} (\eta ,p(n),q(n)) = \eta p(n)(a - \alpha p(n) + \beta q(n))$ with respect to $\eta$ is

$$ \frac{{d\varPi_{r} (\eta ,p(\eta ),q(\eta ))}}{d\eta } = \frac{{a^{2} \alpha \gamma^{2} }}{{(2\alpha \gamma - (1 - \eta )\beta^{2} )^{3} }}\left[ {2\alpha \gamma - (1 + \eta )\beta^{2} } \right]. $$

$\frac{{a^{2} \alpha \gamma^{2} }}{{(2\alpha \gamma - (1 - \eta )\beta^{2} )^{3} }}$ is positive because $\gamma > 0.5\beta^{2} /\alpha$. Therefore, the sign of the derivative is dictated by $2\alpha \gamma - (1 + \eta )\beta^{2}$.

(i) When $\gamma < \beta^{2} /\alpha$, $2\alpha \gamma - (1 + \eta )\beta^{2}$ crosses 0 only once and it is from above; therefore, $\varPi_{r} (\eta ,p(\eta ),q(\eta ))$ is quasi-concave with a maximum point $\eta^{*} = (2\alpha \gamma - \beta^{2} )/\beta^{2}$,$\eta^{*} \in (0,1)$. Substituting $\eta^{*}$ into (A.1) and (A.2) yields the equilibrium price and quality level. The retailer’s and manufacturer’s profits at equilibrium are obtained by substituting $\eta^{*} ,p^{*}$ and q^* into (7) and (8), respectively.

(ii) Otherwise, if $\alpha \gamma - \beta^{2} \ge 0$, then $\varPi_{r} (\eta ,p(\eta ),q(\eta ))$ monotonically increases and hence is maximized at $\eta = 1$. □

Proof of Proposition 4

(i) Similarly to the proof of Proposition 3, by substituting the expressions of (7) and (8) into optimization problem (4), we obtain $\hat{\eta }^{*} = \frac{{2\alpha \hat{\gamma } - \beta^{2} }}{{\beta^{2} }}$ for $\frac{{\beta^{2} }}{2\alpha } < \hat{\gamma } < \frac{{\beta^{2} }}{\alpha }$. Substituting $\hat{\eta }^{*}$ into (A.1) and (A.2) yields the equilibrium price and quality level. The retailer’s and manufacturer’s profits at equilibrium are obtained by substituting $\hat{\eta }^{*} ,\hat{p}^{*}$ and $\hat{q}^{*}$ into (7) and (8), respectively.

(ii) If $\hat{\gamma } \ge \beta^{2} /\alpha$, then $\varPi_{r} (\eta ,p(\eta ),q(\eta ))$ monotonically increases and hence is maximized at $\eta = 1$. □

Proof of Corollary 4

(i) We prove the claim using simple calculus. (ii) Straightforward from Proposition 1. (iii) By Proposition 2, $\varPi_{m} (p^{*} ,q^{*} ,\eta^{*} ) \ge \varPi_{m} (\hat{p}^{*} ,\hat{q}^{*} ,\hat{\eta }^{*} )$ if and only if $\eta^{*} = \frac{{2\alpha \gamma - \beta^{2} }}{{\beta^{2} }} < \frac{{2\alpha \hat{\gamma } - \beta^{2} }}{{\beta^{2} }} = \hat{\eta }^{*}$, that is, if and only if $\gamma < \hat{\gamma }$. □

Proof of Lemma 2

Set $\varOmega = \left\{ {x|0.5\beta^{2} /\alpha < x < \beta^{2} /\alpha } \right\}$ according to the conditions of Proposition 3(i). Following (5), the retailer, based on her belief, sets her equilibrium revenue share $\eta_{f}^{*}$ by solving

$$ \begin{aligned} & \mathop {\hbox{max} }\limits_{\eta } \quad \left\{ {E\left[ {\varPi_{r} (\eta ,p(\eta ,x),q(\eta ,x)) = \int_{x \in \varOmega } {\eta p(\eta ,x)(a + \alpha q(\eta ,x) - \beta p(\eta ,x))f(x)dx} } \right]} \right\} \\ & s.t.\quad \left( {p(\eta ,x),q(\eta ,x)} \right) \equiv \mathop {\arg \hbox{max} }\limits_{p,q} \left\{ {\varPi_{m} (p,q,\eta ,x) \equiv (1 - \eta )p(a + \alpha q - \beta p) - x{{q^{2} } \mathord{\left/ {\vphantom {{q^{2} } 2}} \right. \kern-0pt} 2}} \right\}\, \\ & \quad \quad (1 - \eta )R(p(\eta ,x),q(\eta ,x)) - C_{x} (p(\eta ,x),q(\eta ,x)) \ge 0 \\ & \quad \quad 0 \le \eta \le 1. \\ \end{aligned} $$

(A.3)

Following (A.1) and (A.2),

$$ p(\eta ,x) = \frac{ax}{{2\alpha x - (1 - \eta )\beta^{2} }}\quad {\text{and}}\quad q(\eta ,x) = \frac{a\beta (1 - \eta )}{{2\alpha x - (1 - \eta )\beta^{2} }}. $$

(A.4)

The claim is proved by substituting the last two expressions into the objective function of (A.3). □

Proof of Lemma 3

As stated in Sect. 3.2.1, the optimal $\eta_{b}^{*}$ solves $\max_{\eta } \left\{ {\min_{x \in \varOmega } \left[ {\varPi_{r} (\eta ,p(\eta ,x),q(\eta ,x))} \right]} \right\}$. Substituting expression (A.4) into the retailer’s profit (7), we obtain

$$ \varPi_{r} (\eta ,p(\eta ,x),q(\eta ,x)) = \frac{{a^{2} x^{2} \alpha \eta }}{{(2\alpha x - \beta^{2} (1 - \eta ))^{2} }}, $$

(A.5)

which can be shown to be a decreasing function of x. Thus, its minimal value is attained at $x = \beta^{2} /\alpha$, which according to Proposition 3(ii), yields $\eta = 1$. □

Proof of Lemma 4

According to Sect. 3.2.1, $\eta_{b}^{*} = \arg \min_{\eta } \left\{ {\max_{x \in \varOmega } \left[ {G(\eta ,x)} \right]} \right\}$, where $G(\eta ,x) \equiv \max_{\eta } \left[ {\eta R(s(\eta ,x))} \right] - \eta R(s(\eta ,x))$. By Proposition 3, for any $x \in \varOmega$, $\max_{\eta } \left[ {\eta R(s(\eta ,x))} \right] = \varPi_{r} (\eta^{*} (x),p^{*} (x),q^{*} (x)) = \frac{{a^{2} x^{2} \alpha }}{{4\beta^{2} (2\alpha x - \beta^{2} )}}$, which is the retailer’s maximal profit for a given manufacturer’s efficiency. Using (A.5), $G(\eta ,x) = \frac{{a^{2} x^{2} \alpha }}{{4\beta^{2} (2\alpha x - \beta^{2} )}} - \frac{{a^{2} x^{2} \alpha \eta }}{{(2\alpha x - \beta^{2} (1 - \eta ))^{2} }}$. To find the maximum value of $G(\eta ,x)$, consider the first derivative of $G(\eta ,x)$ with respect to x over $\varOmega = \left\{ {x|0.5\beta^{2} /\alpha < x < \beta^{2} /\alpha } \right\}$:

$$ \frac{\partial }{\partial x}G(\eta ,x) = \frac{{a^{2} x\alpha (2\alpha x - \beta^{2} (1 + \eta ))\varLambda (x)}}{{2\beta^{2} (2\alpha x - \beta^{2} )^{2} (2\alpha x - \beta^{2} (1 - \eta ))^{3} }}, $$

where $\varLambda (x) \equiv 4\alpha^{3} x^{3} - 8\alpha^{2} \beta^{2} (1 - \eta )x^{2} + \alpha \beta^{4} (1 - \eta )(5 + \eta )x - \beta^{6} (1 - \eta^{2} )$. The polynomial equation $\varLambda (x) = 0$ has only one real root, and $\varLambda (0.5\beta^{2} /\alpha ) = 0.125\beta^{6} \eta^{2} > 0$ and $\varLambda (\beta^{2} /\alpha ) = \beta^{6} \eta > 0$. In addition, $\frac{\partial }{\partial x}G(\eta ,x) = 0$ at $x = 0.5\beta^{2} (1 + \eta )/\alpha$ and $\left. {\frac{{\partial^{2} }}{{\partial x^{2} }}G(\eta ,x)} \right|_{{x = \frac{{\beta^{2} (1 + \eta )}}{2\alpha }}} = \frac{{a^{2} \alpha (1 + \eta )^{2} }}{{8\beta^{4} \eta^{3} }} > 0.$ Consequently, over $\varOmega = \left\{ {\gamma |0.5\beta^{2} /\alpha < \gamma < \beta^{2} /\alpha } \right\}$, $G(\eta ,x)$ has a unique minimum, which implies $\max_{x \in \varOmega } \left[ {G(\eta ,x)} \right] = \hbox{max} \left( {\mathop {\lim }\limits_{{x \to 0.5\beta^{2} /\alpha }} G(\eta ,x),\mathop {\lim }\limits_{{x \to \beta^{2} /\alpha }} G(\eta ,x)} \right)$. Because $\mathop {\lim }\limits_{{x \to 0.5\beta^{2} /\alpha }} G(\eta ,x) = \infty$ and $\mathop {\lim }\limits_{{x \to \beta^{2} /\alpha }} G(\eta ,x) = \frac{{a^{2} (1 - \eta )^{2} }}{{4\alpha (1 + \eta )^{2} }}$, for any $0 \le \eta \le 1$,$\max_{x \in \varOmega } \left[ {G(\eta ,x)} \right] = \mathop {\lim }\limits_{{x \to 0.5\beta^{2} /\alpha }} G(\eta ,x)$. Because $\frac{\partial }{\partial \eta }G(\eta ,x) = \frac{{a^{2} }}{{4\alpha \eta^{2} }} > 0$, $\eta_{b}^{*} = 0$ solves $\min_{\eta } \left\{ {\max_{x \in \varOmega } \left[ {G(\eta ,x)} \right]} \right\}.$ □