Cooperative Advertising in a Dynamic Retail Market Oligopoly

Abstract

Cooperative advertising is a key incentive offered by a manufacturer to influence retailers’ promotional decisions. We study cooperative advertising in a dynamic retail oligopoly where a manufacturer sells his product through N competing retailers. We model the problem as a Stackelberg differential game in which the manufacturer announces his shares of advertising costs of the N retailers or his subsidy rates, and the retailers in response play a Nash differential game in choosing their optimal advertising efforts over time. We obtain the feedback equilibrium solution consisting of the optimal advertising policies of the retailers and manufacturer’s subsidy rates. We identify key drivers that influence the optimal subsidy rates and, in particular, obtain the conditions under which the manufacturer will not support the retailers. For the special case of two retailers we obtain insights on some key supply chain issues. First, we analyze its impact on profits of channel members and the extent to which it can coordinate the channel. Second, we investigate the case of an anti-discrimination act which restricts the manufacturer to offer equal subsidy rates to the two retailers.

Appendices

Appendix A: Derivation of P _i ,i=1,2,…,N, for N Identical Retailers

For N identical retailers, with $\theta^{*}_{i} = 0, \forall i = 1, 2,\ldots, N$, by subtracting (17) from (16), we get $\gamma_{ij} = \beta_{i} - \frac{m}{(r+\delta)}, i = 1, 2, \ldots, N, j = i = 1, 2, \ldots, N, j \neq i$. Since the retailers are identical, it is not surprising that γ _ij is same for all j’s, where i=1,2,…,N,j=i=1,2,…,N,j≠i. Using this relation in (16), we can rewrite (16) as

$$ 4(r+\delta) \beta_i = 4m - \rho^2 \beta^2_i - 2\rho^2 \sum_{k\neq i}{\beta_k \biggl[\beta_i - \frac{m}{(r+\delta)} \biggr]}, \quad i = 1, 2,\ldots, N. $$

From the above equation we can say that β ₁=β ₂=⋯=β _N =β, which is intuitive as the retailers are identical. With this we get a quadratic equation in β which gives two roots, one negative and one positive. Clearly it makes sense to have the value function of retailer i, i.e., V _i (X), increase with his own market share x _i . Therefore, we expect β _i =∂V _i (X)/∂x _i >0. Thus, we ignore the negative root and get for i=1,2,…N,

$$ \beta_i = \beta= \frac{(n-1)m\rho^2 - 2(r+\delta)^2 + \sqrt {4(r+\delta)^4 + 4nm\rho^2(r+\delta)^2 + (n-1)^4m^2\rho^4}}{(2n-1)\rho^2(r+\delta)^2}. $$

(43)

Using (43) in (19) we solve for B _i ,i=1,2,…N, and then using (21) we get (22).

Appendix B: Uniqueness of Optimal Solution in the Case of Two Symmetric Retailers

The uniqueness of an optimal solution to the problem defined by (6), (7), (9)–(11) is guaranteed by a unique solution of the system of equations (15)–(20). It appears to be difficult to prove the uniqueness in the general case. However, in the special case of symmetric retailers (M ₁=M ₂=M,m ₁=m ₂=m,δ ₁=δ ₂=δ and ρ ₁=ρ ₂), we can establish the result as follows.

We first look at the signs of α _i ,β _i and γ _i ,i=1,2. It is expected that β _i >0. Now consider γ _i , which can be expressed in terms of β ₁ and β ₂ by using Eq. (17):

$$\gamma_i = \frac{\beta_i^2(-1+\theta^*_{3-i})\rho_i^2}{2(-1+\theta^*_i)((r+\delta_{3-i})(-1+\theta^*_{3-i}) - \beta_{3-i} \rho_{3-i}^2)}, \quad i = 1,2. $$

Since β _i >0 and $\theta^{*}_{i} < 1, i = 1,2$, we have $\gamma_{i} = V_{ix_{3-i}} < 0$, as intuition would suggest on account of the competition between the retailers. We can also use (16) and (17) to write

$$ \gamma_i = \frac{-m_i + \beta_i(r+\delta_i)}{(r+\delta_{3-i})}. $$

(44)

Since γ _i <0, we must have

$$ \beta_i < \frac{m_i}{(r+\delta_i)}. $$

(45)

Now consider α _i , which is retailer i’s value function when the initial market is zero for both retailers. We now show that this value is positive. By adding (15) and (17), we can conclude that α _i =−γ _i (r+δ _3−i )/r, which is positive since γ _i <0. Thus,

$$ \alpha_i = \frac{m_i - \beta_i(r+\delta_i)}{r} > 0. $$

(46)

Moreover, using (44) in Eq. (15), we can write α _i in terms of β ₁ and β ₂, and then rewrite (46) as

$$ \alpha_i = \frac{1}{4r} \biggl[\frac{2\beta_{3-i}(m_i-\beta_i(r+\delta_i))\rho_{3-i}^2}{(r+\delta_{3-i})(-1+\theta_{3-i})} - \frac{\beta_i^2\rho_i^2}{(-1+\theta_i)} \biggr] > 0, \quad i = 1, 2. $$

(47)

Clearly, in the symmetric case, we will have α ₁=α ₂=α,β ₁=β ₂=β, γ ₁=γ ₂=γ, B ₁=B ₂=B, and hence $\theta^{*}_{1} = \theta^{*}_{2} = \theta^{*}$. We can thus rewrite (47) as

$$\frac{\beta(2m-3\beta(r+\delta))\rho^2}{2r(r+\delta)(-1+\theta)} > 0. $$

This, along with β>0 and θ<1, gives us

$$ \beta> \frac{2m}{3(r+\delta)}. $$

(48)

From (45) and (48), we have

$$ \frac{2m}{3(r+\delta)} < \beta< \frac{m}{(r+\delta)}. $$

(49)

To prove a unique solution to Eqs. (15)–(20), we reduce them into one equation of a single variable β, and then aim for the unique solution of β. We will separately consider the cases of a cooperative equilibrium where θ ^∗>0 and a non-cooperative equilibrium where θ ^∗=0.

Case I: Cooperative equilibrium (θ ^∗>0)

Since β ₁=β ₂=β and B ₁=B ₂=B in the symmetric case, (20) reduces to θ ^∗=(2B−β)/(2B+β). Using this, (44), and (46), we can reduce Eqs. (15)–(20) to two equations in variables β and B, i.e.,

(50)

(51)

Using (50), (51) and (14) in (21) and setting P ₁=P ₂=P on account of the case being symmetric, we obtain the participation threshold function

$$ P = 2B - \beta= \frac{8(r+\delta)(-m+2M+(\beta-2B)\delta) - (\beta +2B)(2m-(\beta-4B)(r+\delta))\rho^2}{8r(r+\delta)}. $$

(52)

Using (50), we can write B in terms of β as follows:

$$ B = \frac{-2m(4(r+\delta)+\beta\rho^2) + \beta(r+\delta)(8(r+\delta)+3\beta\rho^2)}{2(2m-3\beta(r+\delta ))\rho^2}. $$

(53)

Now using (52) and (53), we can rewrite P in terms of β only, and then write the condition of the cooperative equilibrium as

$$ P = \frac{-4m((r+\delta)^2+\beta\rho^2) + 2\beta(r+\delta)(4(r+\delta)+3\beta\rho^2)}{(2m-3\beta(r+\delta ))\rho^2} > 0. $$

(54)

Next, we find the values of β for which the inequality in (54) holds. In order to see how P varies with β, we first find the roots of the equation P=0. The numerator is quadratic in β with the roots denoted as η ₁ and η ₂:

Clearly, η ₁>0 and η ₂<0. Also, the denominator (2m−3β(r+δ))ρ ² of (54) changes sign at β=2m/3(r+δ); the denominator is strictly positive when β<2m/3(r+δ) and strictly negative when β>2m/3(r+δ).

We will now compare the value of η ₁ with m/(r+δ) and 2m/3(r+δ). We can see that the difference

$$\eta_1 - \frac{2m}{3(r+\delta)} = \frac{-2(r+\delta)^2 - m\rho^2 + \sqrt{4(r+\delta)^4 + 8m(r+\delta)^2\rho^2 + m^2\rho^4}}{3(r+\delta)\rho^2} > 0, $$

and thus η ₁>2m/3(r+δ). Furthermore,

$$\eta_1 - \frac{m}{(r+\delta)} = \frac{-2(r+\delta)^2 - 2m\rho^2 + \sqrt{4(r+\delta)^4 + 8m(r+\delta)^2\rho^2 + m^2\rho^4}}{3(r+\delta)\rho^2} < 0, $$

and thus we have

$$ \frac{2m}{3(r+\delta)} < \eta_1 < \frac{m}{(r+\delta)}. $$

(55)

We can then conclude from (54) that P>0 is satisfied when

$$ \beta\in(-\infty, -\eta_2) \quad \mbox{or} \quad \beta\in\biggl( \frac {2m}{3(r+\delta)}, \eta_1\biggr). $$

(56)

Therefore, the conditions (49), (55) and (56) along with the fact that β>0 give us the desirable range of the solution for β, i.e.,

(57)

Now using (53) in (51), we can write a single equation in β. After some steps of algebra, for the symmetric retailer case with positive cooperation, this single equation for β can be written as

$$ F(\beta) = \frac{8\beta(r+\delta)^3(m-\beta(r+\delta))-(2m-3\beta(r+\delta ))^2(2M+\beta(r+\delta))\rho^2}{(2m-3\beta(r+\delta))^2\rho^2} = 0. $$

Thus, a unique cooperative solution in the symmetric retailer case is guaranteed when exactly one root of the equation F(β)=0 lies in the range given by (57). The numerator of the above expression, denoted as N(β), is cubic in β. Thus, we can rewrite the equation for β as

$$ N(\beta) = a\beta^3 + b \beta^2 + c\beta+ d = 0, $$

(58)

where

Since the denominator of F(β) is positive for all values of β except 2m/3(r+δ), the sign of F(β) is the same as that of N(β). In what follows, we perform a simple sign analysis of N(β) to draw inference about the roots of (58). After a few steps of algebra with the help of Mathematica, the following observations can be made:

These observations make it clear that the equation N(β)=0 has three real roots in the following intervals:

$$(-\infty, 0),\quad \biggl(0, \frac{2m}{3(r+\delta)}\biggr) \quad \mbox{and}\quad \biggl(\frac{2m}{3(r+\delta)}, \frac{m}{(r+\delta)}\biggr). $$

Moreover, from (55) and (57), we see that there should be exactly one root in the desired interval $(\frac{2m}{3(r+\delta)}, \eta_{1})$ for there to be cooperation in the equilibrium solution. In fact, the location of the third root in the interval $(\frac{2m}{3(r+\delta)}, \frac{m}{(r+\delta)})$ w.r.t. η ₁ determines whether we will have a cooperative or non-cooperative equilibrium. Figure 14 shows the curve N(β) when β>0. This curve gives us an idea of when exactly one of the two positive roots of the equation N(β)=0 would be in the interval $(\frac{2m}{3(r+\delta)}, \eta_{1})$. Note that one root of this equation is negative and is not shown in the figure. It can be easily seen that to attain exactly one root in the interval $(\frac{2m}{3(r+\delta)}, \eta_{1})$ and thereby to have a cooperative equilibrium, we must have

$$ F(\beta)|_{\beta=\eta_1} < 0, $$

(59)

which, when using β=η ₁ in F(β), gives us

(60)

After a few steps of algebra, one can see that the condition (60) has just the opposite sign to the one that ensures a non-cooperative solution in the case of symmetric retailers, which can be obtained by simply using M ₁=M ₂=M in Proposition 3. In other words, when the parameters m,M,r,δ, and ρ are such that the inequality (60) is not satisfied, then the third root of the equation N(β)=0 will be greater than or equal to η ₁, and the optimal solution will be a non-cooperative one. Thus, a unique cooperative equilibrium is guaranteed when (60) is satisfied.

Fig. 14

Subsidy rates vs. M ₁

Case II: Non-cooperative Equilibrium

We now consider the non-cooperative equilibrium (θ ^∗=0) in the symmetric retailer case. As illustrated in Appendix A, the system of equations (15)–(20) can be solved explicitly in the non-cooperative case to get a unique positive solution of β ₁ and β ₂, given by (43). Since the symmetric retailer case is a further simplification of the case of identical retailers, i.e., with M ₁=M ₂=M, the solution of β is unique and is given by (43). Note that this value of β equals η ₁.

Appendix C: Proof of Proposition 7

As defined in Sect. 7, $V^{c}_{r}$ is the combined value function of the two retailers in the cooperative scenario and $V^{n}_{r}$ is the same in the non-cooperative scenario. We can write γ _i and α _i in terms of β _i from (44) and (46), respectively. Recall that in the case of symmetric retailers, α ₁=α ₂=α,β ₁=β ₂=β, and γ ₁=γ ₂=γ. Furthermore, when there is no cooperation, β=η, and its value is given by (43). Using β=η, (44), and (46), we can find $V^{n}_{r}$ by adding computing (12) for i=1,2, and adding the two. After a few steps of algebra, we get

$$ V^n_r = \frac{2(x_1+x_2)(m(r+2\delta)-2\delta(r+\delta)\eta_1)+(-1+x_1+x_2)\eta_1(2m-3(r+\delta)\eta_1)\rho^2}{2r(r+\delta)}. $$

(61)

In the case of symmetric retailers with cooperation, using (44) and (46) from Appendix B, and using the fact that α ₁=α ₂=α,β ₁=β ₂=β and γ ₁=γ ₂=γ, we get

$$ V^c_r=\frac{4(x_1+x_2)(m(r+2\delta) - 2\beta\delta(r+\delta ))-(\beta+2B)(1-x_1-x_2)(2m-3\beta(r+\delta))\rho^2}{4r(r+\delta)}. $$

(62)

Clearly, $V^{n}_{r}$ and $V^{c}_{r}$ depend only on the sum (x ₁+x ₂), and this proves the first statement of Proposition 7. Next, we define $\Delta V_{r} = V^{c}_{r}-V^{n}_{r}$, which can be computed as follows:

$$\mbox{\fontsize{9.0}{11.0}\selectfont$\displaystyle \Delta V_r \hspace{-0.5pt}= \hspace{-0.5pt} \frac{-8x\delta(r+\delta)(\beta-\eta_1)\hspace{-0.5pt}+\hspace{-0.5pt}(-1+x)(2m(\beta+2B-2\eta_1)\hspace{-0.5pt}-\hspace{-0.5pt}3(r+\delta)(\beta^2+2B\beta-2\eta_1^2))\rho^2}{4r(r+\delta)} $}, $$

where x=x ₁+x ₂. ΔV _r is linear in x, and we will write it as ΔV _r (x). We can see that ΔV _r (1)=2δ(η ₁−β)/r>0. This is because we know from (57) that to sustain a cooperative equilibrium, the parameters (m,M,r,δ,ρ) should be such that β<η ₁.

Now consider

$$\Delta V_r(0) = \frac{(-2m(\beta+2B-2\eta_1)+3(r+\delta)(\beta^2+2B\beta_2-2\eta_1^2))\rho^2}{4r(r+\delta)}. $$

Substituting the value of B in terms of β (from (53)) in the above expression, we can write

$$ \Delta V_r(0) = \frac{4(r+\delta)(m-\beta(r+\delta))+\eta(2m-3(r+\delta)\eta )}{2r(r+\delta)}. $$

(63)

It is clear by (63) that a decrease in the value of β (caused by changes in parameters) also decreases the value of ΔV _r (0). We know that for a cooperative equilibrium, β<η ₁, and so a lower bound for ΔV _r (0) can be obtained by using β=η ₁ in (63). This lower bound is

$$\frac{4m(r+\delta)+2m\eta\rho^2-(r+\delta)\eta(4(r+\delta)+3\eta \rho^2)}{2r(r+\delta)}. $$

By using the value of η ₁ from Appendix B, and after a few steps of algebra, we can see that the above expression reduces to zero. Therefore, ΔV _r (0)>0.

Because ΔV _r (x) is linear in x, $\Delta V_{r}(0) \hspace{-0.5pt}>\hspace{-0.5pt} 0$, and $\Delta V_{r}(1) \hspace{-0.5pt}>\hspace{-0.5pt} 0$, we can say that $\Delta V_{r}(x) \hspace{-0.5pt}>\hspace{-0.5pt} 0, \forall x \in [0,1]$. Thus, $V^{c}_{r}(x) > V^{n}_{r}(x)$, ∀x∈[0,1]. The equality holds when β=η ₁, i.e., when non-cooperation is optimal for the manufacturer.