Cooperative advertising models under different channel power structure

Abstract

Cooperative (co-op) advertising is an arrangement between manufacturer and retailer, where two members initiate their respective advertising to help develop brand, motivate sales, and other purposes. This paper focuses on cooperative advertising in a supply chain consisting of one manufacturer and one retailer. As two basic elements of a supply chain, the channel power structure and information structure affect the firm’s pricing and advertising decisions. We attempt to investigate how they optimally decide their marginal returns and advertising investments. Utilizing the Stackelberg game theory, we derive the equilibrium of five scenarios in two different cases, i.e., (1) either the manufacturer or the retailer is acting as the Stackelberg leader and (2) whether the Stackelberg leader knows the advertising investment of his follower. Then we make a comparison and obtain some managerial implications as follows: (1) the supply chain power structure and the relative advertising effectiveness coefficient exert great influence on the optimal supply chain decisions, (2) the supply chain members’ profit is not parallel to their corresponding power structure, if the retailer’s advertising effectiveness is greater than the manufacturer’s, the Manufacturer-dominated Stackelberg game is a win–win strategy, and vice versa, (3) the dominant member can improve his and the overall performance of the supply chain by having the advertising information of the other.

Appendix

Proof of Proposition 2.1

Setting the first derivation $\partial \prod\nolimits_{s} /\partial u_{s} = 0,\partial \prod\nolimits_{s} /\partial a = 0,\partial \prod\nolimits_{s} /\partial A = 0$,we get $u_{s} = M_{c} /2$, $\sqrt a = k_{r} /8M_{c}^{2}$, $\sqrt A = k_{m} M_{c}^{2} /8$

The Hesse matrix of $\prod\nolimits_{s}$ is

$$ H = \left( {\begin{array}{*{20}c} { - 2\sqrt A k_{m} - 2\sqrt a k_{r} } & {\frac{{k_{r} (M_{c} - u_{s} )}}{2\sqrt a } - \frac{{k_{r} u_{s} }}{2\sqrt a }} & {\frac{{k_{m} (M_{c} - u_{s} )}}{2\sqrt A } - \frac{{k_{m} u_{s} }}{2\sqrt A }} \\ {\frac{{k_{r} (M_{c} - u_{s} )}}{2\sqrt a } - \frac{{k_{r} u_{s} }}{2\sqrt a }} & { - \frac{{k_{r} (M_{c} - u_{s} )u_{s} }}{{4a^{3/2} }}} & 0 \\ {\frac{{k_{m} (M_{c} - u_{s} )}}{2\sqrt A } - \frac{{k_{m} u_{s} }}{2\sqrt A }} & 0 & { - \frac{{k_{m} (M_{c} - u_{s} )u_{s} }}{{4A^{3/2} }}} \\ \end{array} } \right) $$

One easy proves that if $u_{s} = M_{c} /2$, then

$$ \left| {H_{1} } \right| = - 2[\sqrt A k_{m} + \sqrt a k_{r} ] < 0,\quad\left| {H_{2} } \right| = k_{r} M_{c}^{2} \frac{{k_{m} \sqrt A + k_{r} \sqrt a }}{{8a^{3/2} }} > 0,\quad\left| H \right| = \frac{{ - k_{m} k_{r} M_{c}^{4} (k_{m} \sqrt A + k\sqrt a_{r} )}}{{128a^{3/2} A^{3/2} }} < 0. $$

$$ \pi_{s} = (u_{m} + u_{r} )D - A - a = u_{s} (M_{c} - u_{s} )k_{r} (\sqrt a + \sqrt k \sqrt A ) - A - a $$

We set ${{\partial \pi_{s} } \mathord{\left/ {\vphantom {{\partial \pi_{s} } {\partial u_{s} }}} \right. \kern-0pt} {\partial u_{s} }} = k_{r} M_{c} (\sqrt a + \sqrt k \sqrt A ) - 2k_{r} (\sqrt a + \sqrt k \sqrt A )u_{s} = 0$

$$ {{\partial \pi_{s} } \mathord{\left/ {\vphantom {{\partial \pi_{s} } {\partial \sqrt a }}} \right. \kern-0pt} {\partial \sqrt a }} = u_{s} (M_{c} - u_{s} )k_{r} - 2\sqrt a = 0 $$

$$ {{\partial \pi_{s} } \mathord{\left/ {\vphantom {{\partial \pi_{s} } {\partial \sqrt A }}} \right. \kern-0pt} {\partial \sqrt A }} = u_{s} (M_{c} - u_{s} )k_{r} \sqrt k - 2\sqrt A = 0 $$

To set simultaneous equations, we get

$$ u_{s}^{Co} = M_{c} /2,\quad\sqrt {a^{Co} } = M_{c}^{2} k_{r} /8,\quad\sqrt {A^{Co} } = M_{c}^{2} k_{m} /8,\quad \prod\nolimits_{s}^{Co} = M_{c}^{4} k_{r}^{2} (1 + k)/64. $$

Proof of Proposition 2.2

It’s easy to differentiate

$$ \frac{{\partial u_{s}^{Co} }}{\partial k} = 0;\quad\frac{{\partial a^{Co} }}{{\partial k_{r} }} = \frac{{M_{c}^{4} k_{r} }}{32} > 0;\quad\frac{{\partial a^{Co} }}{{\partial k_{m} }} = 0;\quad\frac{{\partial A^{Co} }}{{\partial k_{m} }} = \frac{{M_{c}^{4} k_{m} }}{32} > 0;\quad\frac{{\partial A^{Co} }}{{\partial k_{r} }} = 0. $$

Proof of Proposition 2.3

Because $\partial \prod\nolimits_{m}^{Ms} /\partial A^{Ms} = u_{m}^{Ms} k_{m} (M_{c} - u_{m}^{Ms} )/4\beta \sqrt A - 1$ and

$$ \frac{{\partial \prod\nolimits_{m}^{Ms} }}{{\partial u_{m}^{Ms} }} = \frac{{M_{c} - 2u_{m}^{Ms} }}{2} \cdot \left[\frac{{k_{r}^{2} (M_{c} - u_{m}^{Ms} )^{2} }}{8} + k_{m} \sqrt {A^{Ms} } \right] - \frac{{u_{m}^{Ms} (M_{c} - u_{m}^{Ms} )}}{2} \cdot \left[\frac{{k_{r}^{2} (M_{c} - u_{m}^{Ms} )}}{4}\right], $$

we utilize the first-order condition with respect to $u_{m}^{Ms}$ and obtain the Eq. (A1)

$$ (M_{c} - u_{m}^{Ms} )(M_{c} - 4u_{m}^{Ms} ) + 2ku_{m}^{Ms} (M_{c} - 2u_{m}^{Ms} ) = 0 $$

(A1)

The solution (A1) is $u_{m}^{Ms} = \Theta_{ \pm }^{MS} (k) \cdot M_{c}$ where $\Theta_{ \pm }^{Ms} (k) = \frac{{(5 - 2k) \pm \sqrt {(2k - 1)^{2} + 8} }}{8(1 - k)}$. According to (A1), then $u_{m}^{Ms}$ must satisfy $M_{c} /4 < u_{m}^{Ms} < M_{c} /2$, that is to say $1/4 < \Theta_{ \pm }^{Ms} (k) < 1/2$. Because of $\Theta_{ + }^{Ms} (k) - \frac{1}{2} = \frac{{(1 + 2k) + \sqrt {(2k - 1)^{2} + 8} }}{8(1 - k)}$ and $\Theta_{ + }^{Ms} (k) - \frac{1}{4} = \frac{{3 + \sqrt {(2k - 1)^{2} + 8} }}{8(1 - k)}$, we can easily prove that $\Theta_{ + }^{Ms} (k) > 1/2$ or $\Theta_{ + }^{Ms} (k) < 1/4$. So $\Theta_{ + }^{Ms} (k)$ isn’t the result of (A1).

Due to

$$ \Theta_{ - }^{Ms} (k) - \frac{1}{2} = - \frac{1}{{(1 + 2k) + \sqrt {(2k - 1)^{2} + 8} }}\Theta_{ - }^{Ms} (k) - \frac{1}{4} = \frac{1}{{2[3 + \sqrt {(2k - 1)^{2} + 8} ]}} $$

so $\Theta_{ - }^{Ms} (k)$ is the result of the Eq. (A1).

The Hesse matrix of $\prod\nolimits_{m}^{Ms}$ is

$$ H = \left( \begin{aligned} &- \frac{{(M_{c} - u_{m} )u_{m} k_{m} }}{{8A^{3/2} }} \, \frac{{(M_{c} - 2u_{m} )k_{m} }}{4\sqrt A } \hfill \\ &\quad\frac{{(M_{c} - 2u_{m} )k_{m} }}{4\sqrt A } \, - \sqrt A k_{m} - \frac{{3(M_{c} - u_{m} )^{2} k_{r}^{2} }}{8} + \frac{{3(M_{c} - u_{m} )u_{m} k_{r}^{2} }}{8} \hfill \\ \end{aligned} \right) $$

$$ \left| {H_{1} } \right| = - \frac{{(M_{c} - u_{m} )u_{m} k_{m} }}{{8A^{3/2} }},\quad\left| {H_{2} } \right| = \frac{{M_{c}^{2} k_{r}^{2} }}{4}[ - (1 - 6\theta + 6\theta^{2} )k + (1 - 2\theta )(1 - \theta )]. $$

It is easy to prove that $\left| {H_{1} } \right| < 0,\left| {H_{2} } \right| > 0$ when $u_{m} = \theta^{Ms} M_{c} ,\sqrt A = k_{m} \theta^{Ms} (1 - \theta^{Ms} )M_{c}^{2} /4$. That is to say, $\Theta_{ - }^{Ms} (k)$ is the optimal solution for $\prod\nolimits_{m}^{Ms}$.The equilibrium solution (Ms) can easily be found based on $u_{m}^{Ms} = \theta^{Ms} M_{c}$ .

Proof of Proposition 2.4

Because $\frac{{\partial \theta^{Ms} }}{\partial k} = \frac{{4[\sqrt {(2k - 1)^{2} + 8} - (2k - 1)]}}{{[5 - 2k + \sqrt {(2k - 1)^{2} + 8} ]^{2} \sqrt {(2k - 1)^{2} + 8} }}$, we have $\partial \theta^{Ms} /\partial k > 0$. Based on this property, it is easy to prove that $\partial u_{m}^{Ms} /\partial k > 0,\partial u_{r}^{Ms} /\partial k < 0$ and $\partial A^{Ms} /\partial k > 0,\partial a^{Ms} /\partial k < 0$.

From Proposition 2.3, we have

$$ \frac{{\partial \psi^{Ms} }}{\partial k} = \frac{{ - 32(3 - 2k + \sqrt {9 - 4k + 4k^{2} } )}}{{(1 + k)^{2} \sqrt {9 - 4k + 4k^{2} } (5 - 2k + \sqrt {9 - 4k + 4k^{2} } )^{5} }} \cdot [h_{1} (k) + h_{2} (k)], $$

where

$$ \begin{aligned} & h_{1} (k) = (1 - k)\sqrt {9 - 4k + 4k^{2} } (27 - 12k + 18k^{2} + 4k^{3} ) \\ & h_{2} (k) = 81 - 139k + 140k^{2} - 66k^{3} + 24k^{4} + 8k^{5} . \\ \end{aligned} $$

It is easy to prove $27 - 12k + 18k^{2} + 4k^{3} > 0$ for $k \ge 0$.

If $k < 1$, because $h_{1} (k) > 2(1 - k)(27 - 12k + 18k^{2} + 4k^{3} )$, then

$$ h_{1} (k) + h_{2} (k) > (135 - 217k + 200k^{2} ) + ( - 94k^{3} + 16k^{4} + 8k^{5} ). $$

Since $\partial ( - 94k^{3} + 16k^{4} + 8k^{5} )/\partial k = 2k^{2} ( - 141 + 32k + 20k^{2} ) < 0$, we have $- 94k^{3} + 16k^{4} + 8k^{5} > - 70$ and $h_{1} (k) + h_{2} (k) > 65 - 217k + 200k^{2}$, therefore $h_{1} (k) + h_{2} (k) > 0$.

If $k \ge 1$, we have $h_{1} (k) < 0$.

Since $\partial h_{2} (k)/\partial k > 141k - 198k^{2} + 96k^{3} + 40k^{4} > 0$, so $h_{2} (k) > h_{2} (1) = 48 > 0$.

Because $[h_{2} (k)]^{2} - [h_{1} (k)]^{2} = 8k( - 81 + 299k - 95k^{2} + 113k^{3} + 48k^{4} + 4k^{5} ) > 0$, we can prove that $h_{1} (k) + h_{2} (k) > 0$ for $k \ge 1$.

Proof of Proposition 2.5

$$ \frac{{\partial \prod\nolimits_{m}^{Mst} }}{{\partial A^{Mst} }} = \frac{{k_{m} u_{m}^{Mst} (M_{c} - u_{m}^{Mst} )}}{{4\sqrt {A^{Mst} } }} - 1 = 0 $$

(A2)

$$ \frac{{\partial \prod\nolimits_{m}^{Mst} }}{\partial t} = \frac{{k_{r}^{2} (M_{c} - u_{m}^{Mst} )^{3} }}{{64t^{3} }}[ - (M_{c} + 3u_{m}^{Mst} )t + 2(M_{c} - u_{m}^{Mstt} )] = 0 $$

(A3)

$$ \begin{aligned} &\frac{{\partial \prod\nolimits_{m}^{Mst} }}{{\partial u_{m}^{Mst} }} = \frac{{k_{r}^{2} (M_{c} - u_{m}^{Mst} )^{2} }}{16t}(M_{c} - 4u_{m}^{Mst} ) + k_{m} \frac{{M_{c} - 2u_{m}^{Mst} }}{2} \cdot \sqrt {A^{Mst} } \hfill \\ & + (1 - t^{Mst} )\frac{{k_{r}^{2} (M_{c} - u_{m}^{Mst} )^{3} }}{{16t^{2} }} = 0 \hfill \\ \end{aligned} $$

(A4)

Substitute the Eqs. (A2) and (A3) into A4, we have

$$ (M_{c} + 3u_{m}^{Mst} )(M_{c} - 3u_{m}^{Mst} ) + 8ku_{m}^{Mst} (M_{c} - 2u_{m}^{Mst} ) = 0. $$

(A5)

Thus, under the condition $M_{c} - u_{m}^{Mst} > 0$, we have $M_{c} /3 < u_{m}^{Mst} < M_{c} /2$.

Solving the Eq. (A5), we have $u_{m}^{Mst} = \Theta_{ \pm }^{Mst} (k)M_{c}$, where $\Theta_{ \pm }^{Mst} = \frac{{4k \pm \sqrt {16k^{2} + 16k + 9} }}{(9 + 16k)}$.

Since $\Theta_{ - }^{Mst} (k) = \frac{ - 16k - 9}{{(9 + 16k)(4k + \sqrt {16k^{2} + 16k + 9} )}} < 0$, $\Theta_{ - }^{Mst} (k)$ isn’t feasible solution of the Eq. (A5).

Because

$$ \Theta_{ + }^{Mst} - \frac{1}{3} = \frac{{\sqrt {144k^{2} + 144k + 81} - \sqrt {16k^{2} + 72k + 81} }}{3(9 + 16k)},\quad\Theta_{ + }^{Mst} - \frac{1}{2} = \frac{{\sqrt {64k^{2} + 64k + 36} - \sqrt {64k^{2} + 144k + 81} }}{2(9 + 16k)} $$

so $1/3 < \Theta_{ + }^{Mst} < 1/2$.

We denote $\theta^{Mst}$ as $\Theta_{ + }^{Mst} (k)$, and can get $1 - t = (5\theta^{Mst} - 1)/(1 + 3\theta^{Mst} ),\quad u_{m} = \theta^{Mst} M_{c} ,\quad\sqrt A = k_{m} \theta^{Mst} (1 - \theta^{Mst} )M_{c}^{2} /4$.

The Hesse matrix of $\prod\nolimits_{m}^{Mst}$ is

$$ \begin{aligned} H_{3} = \hfill \\ \left( {\begin{array}{*{20}c} { - (M_{c} - u_{m} )k_{r}^{2} \frac{{3(1 - t)(M_{c} - u_{m} ) + 2t(M_{c} + 2u_{m} )}}{{16t^{2} }} - \sqrt A k_{m} } & {\frac{{(M_{c} - 2u_{m} )k_{m} }}{4\sqrt A }} & { - (M_{c} - u_{m} )^{2} k_{r}^{2} \frac{{(1 - t)(M_{c} - u_{m} ) + t(2M_{c} - 5u_{m} )}}{{16t^{3} }}} \\ {\frac{{(M_{c} - 2u_{m} )k_{m} }}{4\sqrt A }} & { - \frac{{(M_{c} - u_{m} )u_{m} k_{m} }}{{8A^{3/2} }}} & 0 \\ { - k_{r}^{2} (M_{c} - u_{m} )^{2} \frac{{2(1 - t)(M_{c} - u_{m} ) + t(2M_{c} - 5u_{m} )}}{{16t^{3} }}} & 0 & { - (M_{c} - u_{m} )^{3} k_{r}^{2} \frac{{3(1 - t)(M_{c} - u_{m} ) + 2t(M_{c} - 3u_{m} )}}{{32t^{4} }}} \\ \end{array} } \right) \hfill \\ \end{aligned} $$

When $1 - t = (5\theta - 1)/(1 + 3\theta ),u_{m} = \theta M_{c} ,\sqrt A = k_{m} \theta (1 - \theta )M_{c}^{2} /4$,we have

$$ \begin{aligned} & \left| {H_{1} } \right| = \frac{{\partial^{2} \prod\nolimits_{m} }}{{\partial u_{m}^{2} }} = \frac{{ - 64\sqrt A k_{m} - 9M_{m}^{2} (1 + 2\theta - 3\theta^{2} )k_{r}^{2} }}{64} \\ & \left| {H_{2} } \right| = \frac{{\partial^{2} \prod\nolimits_{m} }}{{\partial u_{m}^{2} }}\frac{{\partial^{2} \prod\nolimits_{m} }}{{\partial A_{{}}^{2} }} - \left( {\frac{{\partial^{2} \prod\nolimits_{m} }}{\partial u\partial A}} \right)^{2} = \frac{{ - 8(1 - 6\theta + 6\theta^{2} )k + 9(1 + 3\theta )(1 - \theta )}}{{8M^{2} (1 - \theta )^{2} \theta^{2} k}} \\ & \left| {H_{3} } \right| = - \frac{{(1 + 3\theta )^{4} k_{r}^{4} M^{4} k_{m} [8( - 1 + 7\theta - 12\theta^{2} + 6\theta^{3} )k + (1 + 18\theta - 27\theta^{2} )(1 - \theta )]}}{{4096(1 - \theta )^{3} \theta k}}. \\ \end{aligned} $$

In the interval $1/3 \le \theta \le 1/2$, it is easy to derive that $1 + 2\theta - 3\theta^{2} > 0,1 - 6\theta + 6\theta^{2} < 0,1 + 18\theta - 27\theta^{2} > 0$, and $- 1 + 7\theta - 12\theta^{2} + 6\theta^{3} > 0$,

Therefore, $\left| {H_{1} } \right| < 0,\left| {H_{2} } \right| > 0,\left| {H_{3} } \right| < 0$.

Since $\frac{{\partial t^{Mst} }}{{\partial \theta^{Mst} }} = - 8/(1 + 3\theta^{Mst} )^{2} < 0$ and $1/3 \le \theta^{Mst} \le 1/2$, we can get $2/5 < t^{Mst} < 2/3$.

Proof of Proposition 2.6

For $1/3 < \theta^{Mst} < 1/2$, because $\partial \theta^{Mst} /\partial k = \frac{20}{{\sqrt {16k^{2} + 16k + 9} \cdot (18 - 4k + 9\sqrt {16k^{2} + 16k + 9} )}}$, we have $\partial \theta^{{Ms{\text{t}}}} /\partial k > 0$. Based on this property, it is easy to prove that:$\partial u_{m}^{Mst} /\partial k > 0,\partial u_{r}^{Mst} /\partial k < 0$, $\partial A^{Mst} /\partial k > 0,\partial a^{Mst} /\partial k < 0,$$\partial (1 - t^{Mst} )/\partial k < 0$.

Through a complex algebraic operation, we have

$$ \partial \psi^{Mst} /\partial k = \frac{\begin{array}{ll} &2[ - 24300 - 223128k - 550848k^{2} - 651392k^{3} - 81920k^{5} - 12(243 + 7000k^{2} \hfill \\ &\quad+ 846k + 13376k^{3} + 7168k^{4} )\sqrt {9 + 16k + 16k^{2} } ] \hfill \\ \end{array} }{{[(\sqrt {9 + 16k + 16k^{2} } + 9 + 12k)(1 + k)(9 + 16k)^{4} \sqrt {9 + 16k + 16k^{2} } ]}} < 0. $$

Proof of Proposition 2.7

According to the retailer’s profit, it is easy to get

$$ \begin{aligned} &\frac{{\partial \prod\nolimits_{r}^{Rs} }}{{\partial a^{Rs} }} = k_{r} \frac{{u_{r}^{Rs} (M_{c} - u_{r}^{Rs} )}}{{4\beta \sqrt {a^{Rs} } }} - 1 \hfill \\ &\frac{{\partial \prod\nolimits_{r}^{Rs} }}{{\partial u_{r}^{Rs} }} = \frac{{(M_{c} - 2u_{r}^{Rs} )}}{2\beta }[k_{r} \sqrt {a^{Rs} } + k_{m}^{2} \frac{{(M_{c} - u_{r}^{Rs} )^{2} }}{8\beta } - \frac{{u_{r}^{Rs} }}{2\beta }[k_{m}^{2} \frac{{(M_{c} - u_{r}^{Rs} )^{2} }}{4\beta }]. \hfill \\ \end{aligned} $$

Set $\partial \prod\nolimits_{r}^{Rs} /\partial a^{Rs} = 0,\partial \prod\nolimits_{r}^{Rs} /\partial u_{r}^{Rs} = 0$, we have

$$ 2u_{r}^{Rs} (M_{c} - 2u_{r}^{Rs} ) + k(M_{c} - 4u_{r}^{Rs} )(M_{c} - u_{r}^{Rs} ) = 0. $$

(A6)

So we can obtain $u_{r}^{Rs} = \Theta_{ \pm }^{Rs} (k)M_{c}$, where $\Theta_{ \pm }^{Rs} (k) = \frac{{(5k - 2) \pm \sqrt {9k^{2} - 4k + 4} }}{8(k - 1)}$.

If $k > 1$, $\Theta_{ + }^{Rs} (k) > \frac{5k - 2 + 3k - 2}{8(k - 1)} > 1$; if $k < 1$, $\Theta_{ + }^{Rs} (k) = \frac{{(5k - 2) + \sqrt {(5k - 2)^{2} - 16(k - 1)k} }}{8(k - 1)} < 0$, thus $u_{r}^{Rs} = \Theta_{ + }^{Rs} (k)M_{c}$ isn’t feasible solution.

Because

$$ \begin{aligned} \Theta_{ - }^{Rs} (k) - \frac{1}{2} = \frac{ - k}{{k + 2 + \sqrt {9k^{2} - 4k + 4} }},\quad\Theta_{ - }^{Rs} (k) - \frac{1}{4} = \frac{1}{{6k + 2\sqrt {9k^{2} - 4k + 4} }} \hfill \\ \end{aligned} $$

thus $1/4 \le \Theta_{ - }^{Rs} (k) \le 1/2$.

The Hesse matrix of $\prod\nolimits_{r}^{Rs}$ is

$$ H_{2} = \left( {\begin{array}{*{20}c} { - \frac{{3(M - u)(M - 2u)k_{m}^{2} }}{8} - \frac{{\sqrt a k_{r} }}{{}}} & {\frac{{(M - 2u)k_{r} }}{4\sqrt a }} \\ {\frac{{(M - 2u)k_{r} }}{4\sqrt a }} & { - \frac{{(M - u)uk_{r} }}{{8a^{3/2} }}} \\ \end{array} } \right). $$

$$ \left| {H_{1} } \right| = - \frac{{3(M - u)(M - 2u)k_{m}^{2} }}{8} - \sqrt a k_{r} ,\quad \left| {H_{2} } \right| = \frac{{k_{r}^{3} M^{4} (1 - \theta )\theta (3(1 - 2\theta )(1 - \theta )k - (1 - 6\theta + 6\theta^{2} ))}}{{64a^{3/2} }} $$

In the interval $1/4 \le \theta \le 1/2$, it is easy to derive that $1 - 6\theta + 6\theta^{2} < 0$ and $\left| {H_{1} } \right| < 0,\left| {H_{2} } \right| > 0$.

Therefore $\Theta_{ - }^{Rs} (k)$ is optimal solution for $\prod\nolimits_{r}^{Rs}$. We denote $\theta^{Rs}$ as $\Theta_{ - }^{Rs} (k)$. The equilibrium solution (Rs) can easily be derived based on $u_{r}^{Rs} = \theta_{{}}^{Rs} (k)M_{c}$ .

Proof of Proposition 2.10

Because $\partial \theta^{Rs} /\partial k = - \frac{{4( - 2 + k + \sqrt {4 - 4k + 9k^{2} } )}}{{\sqrt {4 - 4k + 9k^{2} } ( - 2 + 5k + \sqrt {4 - 4k + 9k^{2} } )^{2} }}$, we have $\partial \theta^{Rs} /\partial k > 0$ and $1/4 \le \theta^{Rs} \le 1/2$.Based on this property, it is easy to prove that : $\partial u_{r}^{Rs} /\partial k < 0,\partial u_{m}^{Rs} /\partial k > 0$, $\partial a^{Rs} /\partial k < 0,\partial A^{Rs} /\partial k > 0.$ We can get

$$ \partial \psi^{Rs} /\partial k = \frac{{32(3k - 2 + \sqrt {4 - 4k + 9k^{2} } )}}{{(1 + k)^{2} \sqrt {4 - 4k + 9k^{2} } (5k - 2 + \sqrt {4 - 4k + 9k^{2} } )^{5} }}h_{3} (k), $$

where $h_{3} (k) = h_{31} (k) + h_{32} (k)$ and $h_{31} (k) = (k - 1)\sqrt {4 - 4k + 9k^{2} } (4 + 18k - 12k^{2} + 27k^{3} )$, $h_{32} (k) = (8 + 24k - 66k^{2} + 140k^{3} - 139k^{4} + 81k^{5} )$.

Since $(3k - 2) + \sqrt {(3k - 2)^{2} + 8k} > 0$, the sign of $\partial \psi^{Rs} /\partial k$ is same as $h_{3} (k)$. Because $\partial h_{32} (k)/\partial k = 24 - 132k + 200k^{2} + k^{2} (220 - 556k + 405k^{2} )$, so $\partial h_{32} (k)/\partial k > 0$. Therefore $h_{32} (k) \ge h_{32} (0) > 0$. Similarly, we have $18 - 12k + 27k^{2} \ge 0$, that means $\sqrt {4 - 4k + 9k^{2} } (4 + 18k - 12k^{2} + 27k^{3} ) \ge 0$. If $k \ge 1$, $h_{31} (k) \ge 0$, so that $h_{3} (k) \ge 0$. If $0 \le k < 1$, because $[h_{32} (k)]^{2} - [h_{31} (k)]^{2} = - 8k^{4} ( - 4 - 48k - 113k^{2} + 95k^{3} - 299k^{4} + 81k^{5} ) \ge 0$, so that $h_{3} (k) \ge 0$.With the above results, $\partial \psi^{Rs} /\partial k \ge 0$.

Proof of Proposition 2.11

Since $\prod\nolimits_{m}^{Rst} = u_{m}^{Rst} D^{Rst} /\beta - t^{Rst} A^{Rst} ,\prod\nolimits_{r}^{Rst} = u_{r}^{Rst} D^{Rst} /\beta - (1 - t^{Rst} )A^{Rst} - a^{Rst}$,

$$ \begin{aligned} & \partial \prod\nolimits_{m}^{Rst} /\partial u_{m}^{Rst} = (M_{c} - u_{r}^{Rst} - 2u_{m}^{Rst} )(k_{r} \sqrt {a^{Rst} } + k_{m} \sqrt {A^{Rst} } )/\beta \\ & \partial \prod\nolimits_{m}^{Rst} /\partial A^{Rst} = u_{m}^{Rst} (M_{c} - u_{s}^{Rst} )k_{m} /(2\beta \sqrt {A^{Rst} } ) - t^{Rst} , \\ \end{aligned} $$

The manufacture’s optimal profits are given by the following equations:

$$ u_{m}^{Rst} = \left( {M_{c} - u_{r}^{Rst} } \right)/2,\sqrt {A^{Rst} } = (M_{c} - u_{r}^{Rst} )^{2} k_{m} /(8\beta t^{Rst} ) $$

(A7)

So $\prod\nolimits_{r}^{Rst} = k_{r} u_{r}^{Rst} \frac{{(M_{c} - u_{r}^{Rst} )}}{2\beta }\sqrt {a^{Rst} } - a^{Rst} + k_{m}^{2} \frac{{(M_{c} - u_{r}^{Rst} )^{3} (M_{c} + 3u_{r}^{Rst} )}}{{64\beta^{2} t^{Rst} }} - k_{m}^{2} \frac{{(M_{c} - u_{r}^{Rst} )^{4} }}{{64\beta^{2} t^{Rst} t^{Rst} }}$.

Solving equation $\partial \prod\nolimits_{r}^{Rst} /\partial a^{Rst} = 0$, $\partial \prod\nolimits_{r}^{Rst} /\partial t^{Rst} = 0$, we have $\sqrt {a^{Rst} } = k_{r} u_{r}^{Rst} (M_{c} - u_{r}^{Rst} )/(4\beta )$, $t^{Rst} = 2(M_{c} - u_{r}^{Rst} )/(M_{c} + 3u_{r}^{Rst} )$. Substitute it into $\partial \prod\nolimits_{r}^{Rst} /\partial u_{r}^{Rst} = 0$, we have that

$$ k(M_{c} + 3u_{r}^{Rst} )(M_{c} - 3u_{r}^{Rst} ) + 8u_{r}^{Rst} (M_{c} - 2u_{r}^{Rst} ) = 0 $$

(A8)

So we have $u_{r}^{Rst} = \Theta_{ \pm }^{Rst} (k)M_{c}$, $\Theta_{ \pm }^{Rst} (k) = \frac{{4 \pm \sqrt {16 + k(9k + 16)} }}{(9k + 16)}$. Since $u_{r}^{Rst} > 0$, $u_{r}^{Rs} = \Theta_{ - }^{Rst} (k)M_{c}$ isn’t feasible solution.

Since

$$ \begin{aligned} \Theta_{ + }^{Rst} (k) - \frac{1}{3} &= \frac{{ - \sqrt {81k^{2} + 72k + 16} + \sqrt {81k^{2} + 144k + 144} }}{3(9k + 16)} \\ \Theta_{ + }^{Rst} (k) - \frac{1}{2} &= \frac{{ - \sqrt {81k^{2} + 144k + 64} + \sqrt {36k^{2} + 64k + 64} }}{2(9k + 16)} \\ \end{aligned} $$

so $u_{r}^{Rst} = \Theta_{ + }^{Rst} (k)M_{c}$ is optimal solution and $1/3 \le \Theta_{ - }^{Rst} (k) \le 1/2$. We denote $\theta^{Rst}$ as $\Theta_{ - }^{Rst} (k)$.

The Hesse matrix of $\prod\nolimits_{m}^{Rst}$ is

$$ H_{3} = \left( {\begin{array}{*{20}c} { - \frac{{(M_{c} - u_{r} )u_{r} k_{r} }}{{8a^{3/2} }}} & {\frac{{(M_{c} - 2u_{m} )k_{r} }}{4\sqrt a }} & 0 \\ {\frac{{(M_{c} - 2u_{r} )k_{r} }}{4\sqrt a }} & { - \frac{{3(M_{c} - u_{r} )(M_{c} + M_{c} t - u_{r} - 3tu_{r} )k_{m}^{2} + 16\sqrt a t^{2} \beta k_{r} }}{{16t^{2} }}} & {\frac{{(M_{c} - u_{r} )^{2} [ - 2M_{c} + (2 + 3t)u_{r} ]k_{m}^{2} }}{{16t^{3} }}} \\ 0 & {\frac{{(M_{c} - u_{r} )^{2} [ - 2M + (2 + 3t)u_{r} ]k_{m}^{2} }}{{16t^{3} }}} & {\frac{{(M_{c} - u_{r} )^{3} [M_{c} ( - 3 + t) + 3(1 + t)u_{r} ]k_{m}^{2} }}{{32t^{4} }}} \\ \end{array} } \right) $$

and

$$ \left| {H_{1} } \right| = - u_{r} k_{r} \frac{{M_{c} - u_{r} }}{{8a^{3/2} }},\quad \left| {H_{2} } \right| = M_{c}^{2} k_{r} \frac{{9M_{c}^{2} ( - 1 + \theta )^{2} \theta (1 + 3\theta )k_{m}^{2} - 32\sqrt a \beta (1 - 6\theta + 6\theta^{2} )k_{r} }}{{512a^{3/2} }}, $$

$$ \left| {H_{3} } \right| = M_{c}^{8} (1 - \theta )\theta (1 + 3\theta )^{4} k_{m}^{2} k_{r} \frac{{( - 1 - 18\theta + 27\theta^{2} )k_{m}^{2} + 8(1 - 6\theta + 6\theta^{2} )k_{r}^{2} }}{{262144a^{3/2} }}. $$

Because in the interval $1/3 \le \theta \le 1/2$, then $1 - 6\theta + 6\theta^{2} < 0, - 1 - 18\theta + 27\theta^{2} < 0$.It is easy to prove that $\left| {H_{1} } \right| < 0,\left| {H_{2} } \right| > 0,\left| {H_{3} } \right| < 0$. Therefore, $u_{r} = \theta^{Rst} M_{c} ,\sqrt a = k_{r} M_{c}^{2} \frac{\theta (1 - \theta )}{4},t = 2\frac{1 - \theta }{1 + 3\theta }$ is optimal decision for the retailer.

Since $\frac{{\partial t^{Rst} }}{{\partial \theta^{Rst} }} = - 8/(1 + 3\theta^{Rst} )^{2} < 0$ and $1/3 \le \theta^{Rst} \le 1/2$, we can get $2/5 < t^{Rst} < 2/3$.

Proof of Proposition 2.12

Because $\frac{{\partial \theta^{Rst} }}{\partial k} = \frac{{144k - 72\sqrt {16 + k(9k + 16)} - 32}}{{2(9k + 16)^{2} \sqrt {16 + k(9k + 16)} }}$, we have $\partial \theta^{Rst} /\partial k < 0$ and $1/3 \le \theta^{Rst} \le 1/2$.Based on this property, it is easy to prove that $\partial u_{r}^{Rst} /\partial k < 0,\partial u_{m}^{Rst} /\partial k > 0,\quad \partial a^{Rst} /\partial k < 0,\partial A^{Rst} /\partial k > 0,\quad \partial (1 - t^{Rst} )/\partial k < 0.$

Through a complex algebraic operation, we have

$$ \partial \psi^{Rst} (k)/\partial k = \frac{\begin{array}{ll} &(12 + 9k - \sqrt {16 + 16k + 9k^{2} } )(81920 + 388096k + 651392k^{2} + 550848k^{3} + 223128k^{4} + 24300k^{5} \\ & \quad + 12\sqrt {16 + 16k + 9k^{2} } (7168 + 13376k + 7000k^{2} + 846k^{3} + 243k^{4} )) \hfill \\ \end{array} }{{4(1 + k)^{2} (16 + 9k)^{5} \sqrt {16 + 16k + 9k^{2} } }} > 0. $$