Cooperative advertising, pricing strategy and firm performance in the e-marketing age

Abstract

Cooperative advertising plays a strategically important role in marketing programs. In this paper, we use a game theoretical model to study not only cooperative advertising but also pricing strategy in a manufacturer—e-retailer supply chain with the consideration of product categories. First, two cooperative advertising models (the leader-follower Stackelberg and the strategic alliance) are established and analyzed. We then compare the two models to develop some important theories and managerial insights. Furthermore, we utilize a bargaining model to implement profit sharing and determine the manufacturer’s participation rate for cooperative advertising in the channel coordination of strategic alliance. Based on our results, we derive optimal market strategies and identify probable paths of future research.

Appendices

Appendix 1

$$ D = \left( {\theta - p} \right)\left( {\lambda r\sqrt A + k\sqrt Q } \right) $$

(A1)

The profit for the manufacturer is as follows:

$$ {\pi_m} = wD - tA - Q $$

(A2)

Similarly, for the e-retailer, the profit is as follows:

$$ {\pi_r} = \left( {p - w} \right)D - \left( {1 - t} \right)A $$

(A3)

The total profit for the whole supply chain is as follows:

$$ {\pi_T} = {\pi_m} + {\pi_r} = pD - A - Q $$

(A4)

Taking the derivative of (A3) with respect to p and A, respectively, and letting $\left( {{{\partial {\pi_r}} \mathord{\left/{\vphantom {{\partial {\pi_r}} {\partial p}}} \right.} {\partial p}}} \right) = 0$ and $\left( {{{\partial {\pi_r}} \mathord{\left/{\vphantom {{\partial {\pi_r}} {\partial A}}} \right.} {\partial A}}} \right) = 0$ yields:

$$ {p^S} = \frac{1}{2}\left( {w + \theta } \right) $$

(A5)

$$ {A^S} = \frac{{{\lambda^2}{r^2}{{\left( {\theta - w} \right)}^4}}}{{64{{\left( {1 - t} \right)}^2}}} $$

(A6)

Substituting (A5) and (A6) into (A2), then by the differential of π _m on w, Q and t, respectively, and letting $\left( {{{\partial {\pi_m}} \mathord{\left/{\vphantom {{\partial {\pi_m}} {\partial w}}} \right.} {\partial w}}} \right) = 0$, $\left( {{{\partial {\pi_m}} \mathord{\left/{\vphantom {{\partial {\pi_m}} {\partial Q}}} \right.} {\partial Q}}} \right) = 0$ and $\left( {{{\partial {\pi_m}} \mathord{\left/{\vphantom {{\partial {\pi_m}} {\partial t}}} \right.} {\partial t}}} \right) = 0$, we obtain:

$$ {w_S} = \frac{{4{k^2}\theta + {\lambda^2}{r^2}\theta \sqrt {9 + \frac{{16{k^2}\left( {{k^2} + {\lambda^2}{r^2}} \right)}}{{{\lambda^4}{r^4}}}} }}{{9{\lambda^2}{r^2} + 16{k^2}}} $$

(A7)

$$ {t^S} = \frac{{5{w_S} - \theta }}{{3{w_S} + \theta }} $$

(A8)

$$ {Q^S} = \frac{{{k^2}w_S^2{{\left( {\theta - {w_S}} \right)}^2}}}{{16}} $$

(A9)

Substituting (A7), (A8) and (A9) into all of the functions, we then obtain all of results listed in Table 2.

Appendix 2

Because ${w_S} = \frac{{4{k^2}\theta + {\lambda^2}{r^2}\theta \sqrt {9 + \frac{{16{k^2}\left( {{k^2} + {\lambda^2}{r^2}} \right)}}{{{\lambda^4}{r^4}}}} }}{{9{\lambda^2}{r^2} + 16{k^2}}}$ and ${Q^S} = \frac{{{k^2}w_S^2{{\left( {\theta - {w_S}} \right)}^2}}}{{16}}$, thus it is easy to prove that ${{\partial {Q^S}} \mathord{\left/{\vphantom {{\partial {Q^S}} {\partial k > 0}}} \right.} {\partial k > 0}}$, ${{\partial {Q^S}} \mathord{\left/{\vphantom {{\partial {Q^S}} {\partial \theta > 0}}} \right.} {\partial \theta > 0}}$ and ${{\partial {Q^S}} \mathord{\left/{\vphantom {{\partial {Q^S}} {\partial \lambda\kern1.5pt<\kern1.5pt0}}} \right.} {\partial \lambda\kern1.5pt<\kern1.5pt0}}$. Similarly, we can easily prove that ${{\partial {A^S}} \mathord{\left/{\vphantom {{\partial {A^S}} {\partial k\kern1.5pt<\kern1.5pt0}}} \right.} {\partial k\kern1.5pt<\kern1.5pt0}}$, ${{\partial {A^S}} \mathord{\left/{\vphantom {{\partial {A^S}} {\partial \lambda > 0}}} \right.} {\partial \lambda > 0}}$, ${{\partial {A^S}} \mathord{\left/{\vphantom {{\partial {A^S}} {\partial r > 0}}} \right.} {\partial r > 0}}$, ${{\partial {A^S}} \mathord{\left/{\vphantom {{\partial {A^S}} {\partial {t^S} > 0}}} \right.} {\partial {t^S} > 0}}$, ${{\partial {A^S}} \mathord{\left/{\vphantom {{\partial {A^S}} {\partial \theta > 0}}} \right.} {\partial \theta > 0}}$; ${{\partial {p^S}} \mathord{\left/{\vphantom {{\partial {p^S}} {\partial \theta > 0}}} \right.} {\partial \theta > 0}}$,${{\partial {w_S}} \mathord{\left/{\vphantom {{\partial {w_S}} {\partial \theta > 0}}} \right.} {\partial \theta > 0}}$, ${{\partial {p^S}} \mathord{\left/{\vphantom {{\partial {p^S}} {\partial k > 0}}} \right.} {\partial k > 0}}$, ${{\partial {p^S}} \mathord{\left/{\vphantom {{\partial {p^S}} {\partial \lambda\kern1.5pt<\kern1.5pt0}}} \right.} {\partial \lambda\kern1.5pt<\kern1.5pt0}}$, ${{\partial {w_S}} \mathord{\left/{\vphantom {{\partial {w_S}} {\partial k > 0}}} \right.} {\partial k > 0}}$, ${{\partial {w^S}} \mathord{\left/{\vphantom {{\partial {w^S}} {\partial \lambda\kern1.5pt<\kern1.5pt0}}} \right.} {\partial \lambda\kern1.5pt<\kern1.5pt0}}$, ${{\partial {w_S}} \mathord{\left/{\vphantom {{\partial {w_S}} {\partial \theta > 0}}} \right.} {\partial \theta > 0}}$, ${{\partial {t^S}} \mathord{\left/{\vphantom {{\partial {t^S}} {\partial k > 0}}} \right.} {\partial k > 0}}$, ${{\partial {t^S}} \mathord{\left/{\vphantom {{\partial {t^S}} {\partial \lambda > 0}}} \right.} {\partial \lambda > 0}}$ and ${{\partial {t^S}} \mathord{\left/{\vphantom {{\partial {t^S}} {\partial \theta = 0}}} \right.} {\partial \theta = 0}}$.

Appendix 3

$$ {\text{Because}}{\pi_A} = pD - A - Q $$

(A10)

Thus taking the derivative of π _A on p, Q and A, respectively, and letting $({{\partial {\pi_A}} \mathord{\left/{\vphantom {{\partial {\pi_A}} {\partial p}}} \right.} {\partial p}}) = 0$, $\left( {{{\partial {\pi_A}} \mathord{\left/{\vphantom {{\partial {\pi_A}} {\partial Q}}} \right.} {\partial Q}}} \right) = 0$ and $\left( {{{\partial {\pi_A}} \mathord{\left/{\vphantom {{\partial {\pi_A}} {\partial A}}} \right.} {\partial A}}} \right) = 0$, we then obtain:

$$ {Q^A} = \frac{{{k^2}{\theta^4}}}{{64}},{p^A} = \frac{\theta }{2},{A^A} = \frac{{{\lambda^2}{r^2}{\theta^4}}}{{64}} $$

(A11)

Substituting (A11) into all of the functions, then we obtain all of results listed in Table 3.

Appendix 4

Because ${w_S} = \frac{{4{k^2}\theta + {\lambda^2}{r^2}\theta \sqrt {9 + \frac{{16{k^2}\left( {{k^2} + {\lambda^2}{r^2}} \right)}}{{{\lambda^4}{r^4}}}} }}{{9{\lambda^2}{r^2} + 16{k^2}}}$, we let $g = \frac{k}{{\lambda r}}$, then we have ${w_S} = \frac{{4{g^2}\theta + \theta \sqrt {9 + 16{g^2}\left( {1 + {g^2}} \right)} }}{{9 + 16{g^2}}}$, ${{\partial {w_S}} \mathord{\left/{\vphantom {{\partial {w_S}} {\partial g > 0}}} \right.} {\partial g > 0}}$ and g > 0, thus we obtain $\frac{\theta }{3}{|_{g = 0}} \leqslant {w_S} \leqslant \frac{\theta }{2}{|_{g \to \infty }}$. Also, ${\pi_T} = \frac{{{{\left( {\theta - {w_S}} \right)}^2}\left( {2\left( {5 + 8{g^2}} \right){w_S}\theta + 3w_S^2 + 3{\theta^2}} \right)}}{{256}}$, thus we obtain ${{\partial {\pi_T}} \mathord{\left/{\vphantom {{\partial {\pi_T}} {\partial {w_S}\kern1.5pt<\kern1.5pt0}}} \right.} {\partial {w_S}\kern1.5pt<\kern1.5pt0}}$. ${\pi_T}\kern1.5pt<\kern1.5pt{\pi_T}{|_{{w_S} = \theta /3}} = \frac{{{\theta^4}\left( {5 + 4{g^2}} \right)}}{{1296}}$. Furthermore, ${\pi_A} = \frac{{{\theta^4}\left( {1 + {g^2}} \right)}}{{64}}$.

Thus, it is easy to prove that ${\pi_A} > {\pi_T}{|_{{w_S} = \theta /3}}$. Thus, we have ${\pi_A} > {\pi_T}$.

By the same way, we can prove that A ^A > A ^S, p ^A < p ^S and D ^A > D ^S.

Thus, Proposition 2 is proved.

Appendix 5

Let $f = \frac{{4{g^2} + \sqrt {9 + 16{g^2}\left( {1 + {g^2}} \right)} }}{{9 + 16{g^2}}}$, thus ${w_S} = f\theta$.

Thus, ${\pi_T} = \frac{{{\theta^4}{{\left( {1 - f} \right)}^2}\left( {2f\left( {5 + 8{g^2}} \right) + 3{f^2} + 3} \right)}}{{256}}$ and ${\pi_A} = \frac{{{\theta^4}\left( {1 + {g^2}} \right)}}{{64}}$,

So $\partial ({\pi_A} - {\pi_T})/\partial \theta = \frac{{{\theta^3}\left( {4\left( {1 + {g^2}} \right) - {{\left( {1 - f} \right)}^2}\left( {2f\left( {5 + 8{g^2}} \right) + 3{f^2} + 3} \right)} \right)}}{{64}} > 0.$

Thus, Proposition 3 is proved.

Appendix 6

If the profit sharing is successful, the acceptable schemes to the manufacturer and the e-retailer respectively are:

$$ \Delta {\pi_m} = \pi_m^P - {\pi_m} \geqslant 0 $$

$$ \Delta {\pi_r} = \pi_r^P - {\pi_r} \geqslant 0 $$

From Appendix 4, we know ${\pi_A} > {\pi_T} = \pi_m^S + \pi_r^S.$

Thus, we have $\pi_m^P + \pi_r^P = {\pi_A} > \pi_m^S + \pi_r^S.$

Therefore, Proposition 4 is proved.