The optimal advertising strategy with differentiated targeted effect consumers

Abstract

In this paper, we investigate the optimal advertising strategy for two heterogeneous retailers. To propagandize their products, they have to take the traditional advertising or the targeted one. Then a question appears, what are competitive retailers’ optimal equilibrium advertising strategy when they face with differentiated targeted effect consumers? For that, we design four mutually exclusive models: the traditional advertising for two retailers (the ‘nn’-model); the targeted advertising for one retailer and the traditional advertising for another (the ‘tn’ and the ‘nt’ models); the targeted advertising for both retailers (the ‘tt’-model). First, two retailers engage in Nash game to carry out price competitive under the above four models; then they participate in Stackelberg game to carry out advertising strategy competitive under two different game sequences: Retailer 1 and Retailer 2 as the decision leader respectively. By comparison and analysis, some interesting findings are obtained: when consumers’ expected targeted effect is positive, both retailers benefit from the targeted advertising, and the weaker retailer benefits more than the stronger; the stronger retailer is better off when he is the decision leader than when his competitor is; when consumers’ expected targeted effect is negative and near to zero, its optimal for the dominant retailer to take the traditional advertising and the weaker retailer to take the targeted one; two players get win-win when consumers’ expected targeted effect is low or high enough, and they may fall into lose-lose when consumers’ expected targeted effect is moderate; consumers and social get higher (lower) welfare under ‘tt’-model than under ‘nn’-model when consumers’ expected targeted effect is positive (negative).

Appendices

Appendix A

Proof A1

For ‘nn’-model, by solving $\left\{ \begin{array}{c}\frac{\partial \pi _1^{nn}}{\partial p_1}=0\\ \frac{\partial \pi _2^{nn}}{\partial p_2}=0\end{array}\right. \Rightarrow \left\{ \begin{array}{c}p_1^{nn}=\frac{2-2\theta }{4-\theta }\\ p_2^{nn}=\frac{\theta (1-\theta )}{4-\theta }\end{array}\right.$. Substituting them into (4.1), we get $s_1^{nn}$ and $s_2^{nn}$. Substituting them into $s^{nn}=s_1^{nn}+s_2^{nn}$, we otain the total market sale. Substituting $p_1^{nn},~ p_2^{nn},~s_1^{nn},~ s_2^{nn}$ into (4.2) and (4.3), we get $\pi _1^{nn}$ and $\pi _2^{nn}$.

For ‘tn’-model, by solving $\left\{ \begin{array}{c}\frac{\partial \pi _1^{tn}}{\partial p_1}=0\\ \frac{\partial \pi _2^{tn}}{\partial p_2}=0\end{array}\right. \Rightarrow \left\{ \begin{array}{c}p_1^{tn}=\frac{(2-\theta )(\phi (\alpha +\beta )-\beta )+2(1+f-\theta )}{4-\theta }=\frac{(2-\theta )E+2(1+f-\theta )}{4-\theta } \\ p_2^{tn}=\theta \frac{-\phi (\alpha +\beta )+\beta +1+f-\theta }{4-\theta }=\theta \frac{-E+1+f-\theta }{4-\theta }\end{array}\right.$.

Substituting them into (4.4), we get $s_1^{tn}$ and $s_2^{tn}$. Substituting $p_1^{tn},~ p_2^{tn}$ into (4.5) and (4.6), we get $\pi _1^{tn}$ and $\pi _2^{tn}$.

For ‘nt’-model, by solving $\left\{ \begin{array}{c}\frac{\partial \pi _1^{nt}}{\partial p_1}=0\\ \frac{\partial \pi _2^{nt}}{\partial p_2}=0\end{array}\right. \Rightarrow \left\{ \begin{array}{c}p_1^{nt}=\frac{-\phi (\alpha +\beta )+\beta +f-2\theta +2}{4-\theta }=\frac{-E+f-2\theta +2}{4-\theta } \\ p_2^{nt}=\frac{(\phi (\alpha +\beta )-\beta )(2-\theta )-\theta ^2+2f+\theta }{4-\theta }=\frac{E(2-\theta )-\theta ^2+2f+\theta }{4-\theta }\end{array}\right.$.

Substituting them into (4.7), we get $s_1^{nt}$ and $s_2^{nt}$. Substituting $p_1^{nt},~ p_2^{nt},s_1^{nt},s_2^{nt}$ into (4.8), we get $\pi _1^{nt}$ and $\pi _2^{nt}$.

For ‘tt’-model, by solving $\left\{ \begin{array}{c}\frac{\partial \pi _1^{tt}}{\partial p_1}=0\\ \frac{\partial \pi _2^{tt}}{\partial p_2}=0\end{array}\right. \Rightarrow \left\{ \begin{array}{c}p_1^{tt}=\frac{(E+2)(1-\theta )+3f}{4-\theta }; \\ p_2^{tt}=\frac{(1-\theta )(2E+\theta )+f(\theta +2)}{4-\theta }.\end{array}\right.$. Substituting them into (4.9), (4.10), we get $s_1^{tt}$ and $s_2^{tt}$. Substituting $p_1^{tt},~ p_2^{tt},s_1^{tt},s_2^{tt}$ into (4.11), we get $\pi _1^{tt}$ and $\pi _2^{tt}$.

Proof A2

 

$$\begin{aligned}&\frac{\partial \pi _1^{tn}}{\partial f}=\frac{2(-2+\theta )((E-f)(2-\theta )+2(1-\theta ))}{(4-\theta )^2(1-\theta )}<0,\\&\frac{\partial \pi _1^{tn}}{\partial E}=\frac{2(\theta -2)[(2-\theta )(f-E)+2\theta -2]}{(4-\theta )^2(1-\theta )}>0,\\&\frac{\partial \pi _2^{tn}}{\partial f}=\frac{2\theta (E-f)+(1-\theta )(2-\theta )(8-\theta )}{(4-\theta )^2(\theta -1)}<0,~ \frac{\partial \pi _2^{tn}}{\partial E}=\frac{2\theta (E-f+\theta -1)}{(4-\theta )^2(1-\theta )}<0.\\&\frac{\partial \pi _1^{nt}}{\partial E}=\frac{2(E-f)-4(1-\theta )}{(1-\theta )(4-\theta )^2}<0; ~\frac{\partial \pi _1^{nt}}{\partial f}=\frac{2(f-E)-(1-\theta )(2-\theta )(6-\theta )}{(4-\theta )^2(1-\theta )}<0;\\&\frac{\partial \pi _2^{nt}}{\partial E}=\frac{2(\theta -2)((E-f)(\theta -2)-\theta +\theta ^2)}{\theta (1-\theta )(4-\theta )^2}>0,~\\&\quad \frac{\partial \pi _2^{nt}}{\partial f}=\frac{2(2-\theta )((E-f)(\theta -2)-\theta +\theta ^2)}{\theta (1-\theta )(4-\theta )^2}<0.\\&\frac{\partial \pi _1^{tt}}{\partial E}=\frac{2(1-\theta )(E-f+2)}{(4-\theta )^2}>0;~\frac{\partial \pi _1^{tt}}{\partial f}=\frac{-2(1-\theta )(E-f+2)}{(4-\theta )^2}<0. \end{aligned}$$

Others are easy to be checked.

Appendix B

Proof B1

 

• (1) When Retailer 1 takes the traditional advertising, $E\in (\underline{E^{nt}},\overline{E^{nt}})$, Retailer 2 needs to choose from the ‘nt’ and the ‘nn’ models. Because $\frac{\partial (\pi ^{nt}_2-\pi ^{nn}_2)}{\partial E}>0$, and the zero point of $\pi ^{nt}_2-\pi ^{nn}_2$ is $E_3=f+\frac{\theta (1-\theta )}{-2+\theta }+\frac{\sqrt{-f\theta (1-\theta )(4-\theta )^2+\theta ^2(1-\theta )^2}}{2-\theta }\in (\underline{E^{nt}},\overline{E^{nt}})$. Hence, when $E>E_3$, we have $\pi ^{nt}_2>\pi ^{nn}_2$, then Retailer 2 takes the targeted advertising; otherwise, she takes the traditional advertising.

• (2) When Retailer 1 takes the targeted advertising, because $\underline{E^{tt}}<\underline{E^{tn}}<\overline{E^{tn}}<\overline{E^{tt}}$, when $E\in (\underline{E^{tn}},\overline{E^{tn}})$, Retailer 2 needs to choose from the ‘tn’ and the ‘tt’ models. Because $\frac{\partial (\pi ^{tn}_2-\pi ^{tt}_2)}{\partial E}<0$, and the zero point of $\pi ^{tn}_2-\pi ^{tt}_2$ is $E_0=\frac{-2 f + \theta + 3 f \theta - \theta ^2}{-2 + 3 \theta } - \sqrt{\frac{ 32 f \theta - 2 \theta ^2 - 96 f \theta ^2 + 5 \theta ^3 + 90 f \theta ^3 - 4 \theta ^4 - 29 f \theta ^4 + \theta ^5 + 3 f \theta ^5}{(-2 + \theta ) (-2 + 3 \theta )^2}}\in (\underline{E^{tn}},\overline{E^{tn}})$. Hence, when $E\in (\underline{E^{tn}},E_0)$, $\pi ^{tn}_2>\pi ^{tt}_2$, then Retailer 2 takes the traditional advertising; otherwise, when $E\in (\underline{E^{tt}},\underline{E^{tn}})\cup (E_0,\overline{E^{tt}})$, Retailer 2 takes the targeted advertising.

Proof B2

Because $(-1,\underline{E^{tt}})$ is out of the feasible zone of E under the ‘nt’, ‘tn’ and ‘tt’ models. Hence, Retailer 1 and Retailer 2’s optimal advertising strategy is the ‘nn’-model.

When $E\in (\underline{E^{tt}},E_3)$, Retailer 1’s needs to choose from ‘nn’ and the ‘tt’-models. Because $\frac{\partial (\pi _1^{tt}-\pi _1^{nn})}{\partial E}>0$, and the zero point of $\pi _1^{tt}=\pi _1^{nn}$ is $E_2=-2+f+\sqrt{4+\frac{f(4-\theta )^2}{-1+\theta }}\in (\underline{E^{tt}},E_3)$. Hence, when $E\in (E_2,E_3)$, $\pi ^{tt}_1>\pi ^{nn}_1,$ then Retailer 1 takes the ‘tt’ model. When $E\in (\underline{E^{tt}},E_2)$, Retailer 1 takes the ‘nn’-model.

When $E\in (E_3,\underline{E^{tn}})$, Retailer 1 needs to choose from ‘tt’ and ‘nt’-models. It is easy to check that $\pi _1^{nt}>\pi _1^{tt}$, hence Retailer 1 takes the ‘nt’-model.

When $E\in (\underline{E^{tn}},E_0)$, Retailer 1 needs to choose from ‘tn’ and ‘nt’-models. It is easy to check that $\pi _1^{nt}>\pi _1^{tn}$, hence Retailer 1 takes the ‘nt’-model.

When $E\in (E_0,\overline{E^{nt}})$, Retailer 1 needs to choose from ‘tt’ and ‘nt’-models. Because $\frac{\partial (\pi _1^{nt}-\pi _1^{tt})}{\partial E}<0$, and the zero point of $\pi _1^{nt}-\pi _1^{tt}$ is $E_1=-2+f+\frac{2}{\theta }-\frac{1}{\theta }\sqrt{\frac{(1-\theta )^2(2-\theta )+f\theta (4-\theta )^2(\theta -1)}{-2+\theta }}\in (E_0,\overline{E^{nt}})$. Hence, when $E\in (E_0,E_1)$, $\pi _1^{nt}>\pi _1^{tt}$, then Retailer 1 takes the ‘nt’-model. When $E\in (E_1,\overline{E^{nt}})$, Retailer 1 takes the ‘tt’-model.

When $E\in (\overline{E^{nt}},\overline{E^{tt}})$, Retailer 1 needs to choose from ‘tt’ and ‘nn’-models. It is easy to check that $\pi _1^{tt}>\pi _1^{nn}$, hence Retailer 1 takes the ‘tt’-model.

Proof B3

It is easy to prove $\frac{\partial E_1}{\partial f}=\frac{1}{2} (2 - \frac{(-4 + \theta )^2 (-1 + \theta )}{(-2 +\theta ) \theta \sqrt{\frac{(-1 + \theta ) (8 + 4 (-3 + 4 f) \theta + (4 - 8 f) \theta ^2 + f\theta ^3)}{(-2 + \theta ) \theta ^2}}})>0$, $\frac{\partial E_2}{\partial f}=1 + \frac{(-4 + \theta )^2}{ 2 \sqrt{\frac{f (-4 + \theta )^2 + 4 (-1 + \theta )}{-1 + \theta }} (-1 + \theta )}>0$, $\frac{\partial E_3}{\partial f}=1+\frac{(-4+\theta )^2(-1+\theta )\theta }{2(2-\theta )\sqrt{(-1+\theta )\theta (f(-4+\theta )^2+(-1+\theta )\theta )}}>0$. Because $E_1$, $E_2$ and $E_3$ are all negative numbers, then their absolute values decrease along with the increasing of f.

$$\begin{aligned}&\frac{\partial (E_1-E_3)}{\partial f}\\&\quad =((-4 + \theta )^2 (-1 + \theta )\\&\qquad \theta (-(1/ \sqrt{(((-1 + \theta ) \theta (f (-4 + \theta )^2 + (-1 + \ \theta ) \theta ))/(-2 + \theta )^2)}) + ( 2 - \theta )/(\theta ^2\\&\quad \quad \sqrt{((-1 + \theta ) (8 + 4 (-3 + 4 f) \theta + (4 - 8 f) \theta ^2 + f \theta ^3)}{((-2 + \theta ) \theta ^2)})))/(2 (-2 + \ \theta )^2)>0,\\&\frac{\partial (E_3-E_2)}{\partial f}\\&\quad =\frac{1}{2} (-((-4 + \theta )^2/( \sqrt{(f (-4 + \theta )^2 + 4 (-1 + \theta ))/(-1 + \theta )} (-1 + \theta ))) \\&\qquad + ((-4 + \ \theta )^2 (-1 + \theta ) \theta )/((-2 + \theta )^2 \sqrt{((-1 + \ \theta ) \theta (f (-4 + \theta )^2 + (-1 + \theta ) \ \theta ))/(-2 + \theta )^2}))\\&\qquad \left\{ \begin{array}{cc}>0,~~0<f<f_2\\<0,~~,f_2<f<f_0\end{array}\right. \end{aligned}$$

indicate when $0<f<f_2$, along with f, the speed of move tawards right of $E_2$, $E_3$, $E_1$ are slower and slower; when $f_2<f<f_0$, the speed of $E_3$ moving tawards right is the fastest of the three.

Proof B4

We give the proof of $CS^{1}_{13}$, $CS^{1}_{22}$others can be checked through Mathematica. First, according to the definition of $CS^{1}_{13}$, it is east to be checked that

$$\begin{aligned}&CS^{1}_{13}\\&\quad =\frac{[2+f+4\beta +3E-(2+\beta +E)\theta ][\beta (-4+\theta )+f(-3+2\theta )-(1+\theta )(-2+E+2\theta )](1-\phi ) +[2+f+3E+\alpha (-4+\theta )-(2+E)\theta ][2+4\alpha -E+f(-3+2\theta )-\theta (\alpha +E+2\theta )]\phi }{2(1-\theta )^2(4-\theta )^2}\\&\quad =\frac{4(1-\theta )^2(1+\theta )+(E-f)^2(-3+2\theta )+E^2(4-\theta )^2+(1-\theta )(2-3\theta )2(E-f)}{2(1-\theta )^2(4-\theta )^2}\\&\qquad -\frac{\beta ^2(1-\phi )+\alpha ^2\phi }{2(1-\theta )^2}\\&\quad =\frac{4(1-\theta )^2(1+\theta )+(E-f)^2(-3+2\theta )+E^2(4-\theta )^2+(1-\theta )(2-3\theta )2(E-f)}{2(1-\theta )^2(4-\theta )^2}\\&\qquad -\frac{\sigma ^2+E^2}{2(1-\theta )^2}. \end{aligned}$$

It is easy to get

$$\begin{aligned} CS^{1}_{22}= & {} \frac{[-2f-2E+\beta (-4+\theta )+\theta +E\theta ]^2(1-\phi )+[-2f-2E-\alpha (-4+\theta )+\theta +E\theta ]^2\phi }{2\theta (4-\theta )^2}\\= & {} \frac{(-2f-2E+\theta +E\theta )^2-2(-2f-2E+\theta +E\theta )E(-4+\theta )+(4-\theta )^2(\beta ^2(1-\phi )+\alpha ^2\phi )}{2\theta (4-\theta )^2}\\= & {} \frac{(-2f-2E+\theta +E\theta )^2-2(-2f-2E+\theta +E\theta )E(-4+\theta )+(4-\theta )^2(\sigma ^2+E^2)}{2\theta (4-\theta )^2}\\= & {} \frac{(-2f-2E+\theta +E\theta +E(4-\theta ))^2+(4-\theta )^2\sigma ^2}{2\theta (4-\theta )^2}\\= & {} \frac{(2E-2f+\theta )^2}{2\theta (4-\theta )^2}+\frac{\sigma ^2}{2\theta }; \end{aligned}$$

By the same way

$$\begin{aligned}&CS^{1}_{23}\\&\quad =\frac{[-2E+\beta (-4+\theta )+f(-2+\theta )+\theta -\theta ^2]^(1-\phi )+[2E+\alpha (-4+\theta )-f(-2+\theta )-\theta +\theta ^2]^2\phi }{2(4-\theta )^2(1-\theta )^2\theta }\\&\quad =\frac{(-2E+f(-2+\theta )+\theta +\theta ^2)^2+2(-2E+f(-2+\theta )+\theta +\theta ^2)E(4-\theta )+(4-\theta )^2(\beta ^2(1-\phi )+\alpha ^2\phi )}{2(4-\theta )^2(1-\theta )^2\theta }\\&\quad =\frac{(-2E+f(-2+\theta )+\theta +\theta ^2)^2+2(-2E+f(-2+\theta )+\theta +\theta ^2)E(4-\theta )+(4-\theta )^2(\sigma ^2+E^2)}{2(4-\theta )^2(1-\theta )^2\theta }\\&\quad =\frac{[-2E+f(-2+\theta )+\theta +\theta ^2+(4-\theta )E+(4-\theta )^2\sigma ^2]^2}{2(4-\theta )^2(1-\theta )^2\theta }\\&\quad =\frac{[(2-\theta )(E-f)+\theta -\theta ^2]^2+(4-\theta )^2\sigma ^2}{2(4-\theta )^2(1-\theta )^2\theta } \end{aligned}$$

Proof B5

Because $\frac{\partial (\pi _1^{tn}-\pi _1^{nn})}{\partial E}>0$ and the zero point of $\pi _1^{tn}-\pi _1^{nn}$ is $E_5=\frac{2 - 2 f - 2 \theta + f \theta }{-2 + \theta } + \sqrt{\frac{ 4 - 16 f - 8 \theta + 24 f \theta + 4 \theta ^2 - 9 f \theta ^2 + f \theta ^3}{(-2 + \theta )^2}}$, we get conclusion (1). It is easy to prove that $\frac{\partial (\pi _1^{tt}-\pi _1^{nt})}{\partial E}>0$ and the zero point of $\pi _1^{tt}-\pi _1^{nt}$ is $E_1=-2+f+\frac{2}{\theta }-\frac{1}{ \theta }\sqrt{\frac{-8 + 20 \theta - 16 f \theta - 16 \theta ^2 + 24 f \theta ^2 + 4 \theta ^3 - 9 f \theta ^3 + f \theta ^4}{-2+\theta }}$, we get conclusion (2). We $E\in (-1,\underline{E^{tt}})$, it is out of the feasible region of E for the ‘nt’, ‘tt’ and ‘tn’-models, retailers has to take the ‘nn’-model, we get conclusion (3).