Cooperative Advertising with Accrual Rate in a Dynamic Supply Chain

Abstract

We consider a supply chain in which a manufacturer stimulates his retailers investing more local advertising expenditures through a cooperative advertising program (Co-op). Co-op advertising programs usually involve two important contractual terms, a participation rate and an accrual rate. While previous literatures have discussed the participation rate excessively, they seldom study the role of the accrual rate in cooperative advertising. To investigate the impact of the accrual rate on cooperative advertising decisions, we develop a dynamic cooperative advertising model for a single manufacturer–single retailer supply chain that incorporates the participation rate and the accrual rate simultaneously. We derive the equilibrium co-op decisions of two channel members, including the manufacturer’s national advertising efforts and his participation rate, as well as the retailer’s local advertising expenditure. Our analysis of the equilibrium solutions shows that an increase in participation rate will not always increase the retailer’s advertising efforts because of the accrual rate. Also, both the manufacturer and retailer can benefit from a high accrual rate.

Appendix

1.1 Proof of Proposition 1

With the determined cooperative advertising program of the manufacturer, the optimization problem of the retailer can be specified as

$$\begin{aligned} V_r (M) = \mathop {{\text {max}}}\limits _{a( \cdot ) \ge 0} \left\{ {\int _0^\infty {e^{ - \phi t} \pi _r (t)} {\text {d}}t} \right\} \end{aligned}$$

(20)

s.t.

$$\begin{aligned} \frac{d}{{{\text {d}}t}}M(t) = u(t) - \delta M(t), \end{aligned}$$

where

$$\begin{aligned} \pi _r \left( t \right) = \left\{ {\begin{array}{*{20}c} {{\rho _{r}}\left( {a^\gamma M^\eta } \right) - \left( {1 - x} \right) a} &{} {{\text {if}}} &{} {xa \le y{\rho _{m}}S} \\ {\left( {{\rho _{r}} + {\rho _{m}}y} \right) \left( {a^\gamma M^\eta } \right) - a} &{} {{\text {if}}} &{} {xa > y{\rho _{m}}S}. \\ \end{array}} \right. \end{aligned}$$

(21)

The Hamilton–Jacobi–Bellman (HJB) equation is given by

$$\begin{aligned} \phi V_r = \mathop {{\text {max}}}\limits _{a( \cdot )} \left\{ {\pi _r + \frac{{\partial V_r }}{{\partial M}}(u - \delta M)} \right\} . \end{aligned}$$

(22)

The first-order condition for the maximization in (22) leads to

$$\begin{aligned} a(M) = \frac{{\partial \pi _r }}{{\partial a}}. \end{aligned}$$

(23)

Recall the retailer’s profit function at time t in Eq. (4)

$$\begin{aligned} \pi _r (t) = \left\{ {\begin{array}{l@{\quad }l@{\quad }l} {{\rho _{r}}a^\gamma M^\eta - (1 - x)a}, &{} {{\text {if}}} &{} {a \le \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}} } \\ {({\rho _{r}} + {\rho _{m}}y)(a^\gamma M^\eta ) - a}, &{} {{\text {if}}} &{} {a > \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}} }, \\ \end{array}} \right. \end{aligned}$$

(24)

where $\pi _r (t)$ is continuous when $a = \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}}$.

We solve the first derivative of $\pi _r$ with respect to a as

$$\begin{aligned} \frac{{\partial \pi _r }}{{\partial a}} = \left\{ {\begin{array}{l@{\quad }l@{\quad }l} {\gamma {\rho _{r}}a^{\gamma - 1} M^\eta - (1 - x)}, &{} {{\text {if}}} &{} {a \le \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}} } \\ {\gamma ({\rho _{r}} + {\rho _{m}}y)(a^{\gamma - 1} M^\eta ) - 1}, &{} {{\text {if}}} &{} {a > \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}} }. \\ \end{array}} \right. \end{aligned}$$

(25)

To solve the best response in local advertising investment for the retailer, we seek for the left derivative of $\pi _r(t)$ when $a = \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}}$ is

$$\begin{aligned} \frac{{\partial \pi _r }}{{\partial a}}\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^ - } } \right. = \frac{{\gamma {\rho _{r}}x}}{{{\rho _{m}}y}} - 1 + x \end{aligned}$$

(26)

and the right derivative is

$$\begin{aligned} \frac{{\partial \pi _r }}{{\partial a}}\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^+ } } \right. = \frac{{\gamma {\rho _{r}}x}}{{{\rho _{m}}y}} - 1 + \gamma x. \end{aligned}$$

(27)

According to $0 < \gamma < 1$, we get

$$\begin{aligned} \partial \pi _r /\partial a\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}} ]^ - } } \right. > \partial \pi _r /\partial a\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}} ]^ + } } \right. , \end{aligned}$$

which leads to the following three possible cases as well as the corresponding best response in advertising investment.

(i) If ${\rho _{r}} < {\rho _{m}}y\left( {1 - x} \right) /(\gamma x)$, i.e.,

$$\begin{aligned} \partial \pi _r /\partial a\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^ + } } \right. < \partial \pi _r /\partial a\left| {_{a =[ ({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^- } } \right. < 0, \end{aligned}$$

the retailer will get his best response in local advertising investment when $a < \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}}$. Utilizing (21), we can obtain the retailer’s best response in local advertising investment is $a\left( M \right) \mathrm{{ = }}\left[ {\gamma {\rho _{r}}M^\eta /(1 - x)} \right] ^{\frac{1}{{^{1 - \gamma } }}}$.

(ii) If ${\rho _{m}}y\left( {1 - x} \right) /(\gamma x) \le {\rho _{r}} \le {\rho _{m}}y\left( {1 - \gamma x} \right) /(\gamma x)$, i.e.,

$$\begin{aligned} \partial \pi _r /\partial a\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^+ } } \right. \le 0 \le \partial \pi _r /\partial a\left| {_{a = [({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^ - } } \right. , \end{aligned}$$

the retailer’s best response in advertising investment is achieved when $a\left( M \right) = \left( {{\rho _{m}}yM^\eta /x} \right) ^{\frac{1}{{1 - \gamma }}}$.

(iii) If ${\rho _{r}} > {\rho _{m}}y\left( {1 - \gamma x} \right) /(\gamma x)$, i.e.,

$$\begin{aligned} \partial \pi _r /\partial a\left| {_{a =[ ({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}} ]^- } } \right. > \partial \pi _r /\partial a\left| {_{a =[ ({\rho _{m}}yM^\eta /x)^{\frac{1}{{1 - \gamma }}}]^+ } } \right. > 0, \end{aligned}$$

the retailer gets his optimal local advertising investment when $a > ( \frac{{\rho _{m}}yM^\eta }{x})^{\frac{1}{{1 - \gamma }}}$ and we obtain that the retailer’s best response in local advertising investment is $a = \left( {\gamma ({\rho _{r}} + {\rho _{m}}y)M^\eta } \right) ^{\frac{1}{{1 - \gamma }}}$ by utilizing (26).

1.2 Proof of Proposition 2

Under the assumption that $\gamma + \eta /2\mathrm{{ = }}1$, the optimization problem of the manufacturer can be specified as

$$\begin{aligned} \mathop {\max }\limits _{u(.)} \int _0^\infty {e^{ - \phi t} (\varDelta _i M^2 (t) - \frac{1}{2}u^2 (t)){\text {d}}t} \end{aligned}$$

s.t.

$$\begin{aligned} \frac{{dM(t)}}{{{\text {d}}t}} = u(t) - \delta M(t),M(0) = M_0. \end{aligned}$$

The Hamilton–Jacobi–Bellman (HJB) equation is given by

$$\begin{aligned} \phi V_m = \mathop {\max }\limits _u \left\{ {\varDelta (x)M^2 - \frac{1}{2}u^2 + \frac{{\partial V_m }}{{\partial M}}(u - \delta M)}, \right\} \end{aligned}$$

(28)

where $\varDelta (x)$ is given by Eq. (9).

The first-order condition for the maximization in (28) leads to

$$\begin{aligned} u(M) = \frac{{\partial V_m }}{{\partial M}}. \end{aligned}$$

(29)

Substituting (29) into (28), one has

$$\begin{aligned} \phi V_m = \varDelta (x)M^2 + \frac{1}{2}\left( \frac{{\partial V_m }}{{\partial M}}\right) ^2 - \delta M\frac{{\partial V_m }}{{\partial M}}. \end{aligned}$$

(30)

Conjecture the following form of the value function

$$\begin{aligned} V_m = a_0 + a_1 M + a_2 M^2, \end{aligned}$$

(31)

where the coefficients $a_i(i=0,1,2)$ are to be determined.

Substituting (31) into (30) renders

$$\begin{aligned} \phi a_2= & {} \varDelta (x) + 2a_2 ^2 - 2\delta a_2, \end{aligned}$$

(32)

$$\begin{aligned} \phi a_1= & {} 2a_1 a_2 - a_1 \delta , \end{aligned}$$

(33)

$$\begin{aligned} \phi a_0= & {} \frac{1}{2}a_1 ^2. \end{aligned}$$

(34)

Solving Eqs. (32)–(34) and for the sake of stability, we have

$$\begin{aligned} a_0 = a_1 = 0, a_2 = \frac{1}{4}(2\delta + \phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta (x)} ). \end{aligned}$$

Then we get the feedback strategy of the manufacturer’s national advertising

$$\begin{aligned} u(M) = (\delta + \frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta (x)} }}{2})M. \end{aligned}$$

(35)

1.3 Proof of Proposition 3

Substituting the feedback national advertising of the manufacturer in Proposition 2 into the goodwill dynamic equation, we have

$$\begin{aligned} \frac{d}{{{\text {d}}t}}M(t) = \frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta (x)} }}{2}M(t). \end{aligned}$$

(36)

Solving Eq. (36), we get the trajectory of the goodwill level is

$$\begin{aligned} M(t) = M_0 e^{\frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta } }}{2}t}. \end{aligned}$$

(37)

Substituting (37) into a(M) and u(M) in Propositions  1 and 2, respectively, we get the best response in local advertising expenditure is

$$\begin{aligned} a(t)\mathrm{{ = }}\left\{ {\begin{array}{l@{\quad }l@{\quad }l} {M_0 \left( {\frac{{\gamma {\rho _{r}}}}{{1 - x}}} \right) ^{\frac{1}{{^{1 - \gamma } }}} e^{\frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta _1 } }}{2}t} } &{} {{\text {if}}} &{} {x < A} \\ {M_0 \left( {\frac{{{\rho _{m}}y}}{x}} \right) ^{\frac{1}{{1 - \gamma }}} e^{\frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta _2 } }}{2}t} } &{} {{\text {if}}} &{} {A \le x \le B} \\ {M_0 \left( {\gamma {\rho _{r}} + \gamma {\rho _{m}}y} \right) ^{\frac{1}{{1 - \gamma }}} e^{\frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta _3 } }}{2}t} } &{} {{\text {if}}} &{} {x > B,} \\ \end{array}} \right. \end{aligned}$$

(38)

and the best response in national advertising effort is

$$\begin{aligned} u(t) = \frac{1}{2}M_0 \left( {2\delta + \phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta } } \right) e^{\frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\varDelta } }}{2}t}. \end{aligned}$$

(39)

1.4 Proof of Proposition 4

The manufacturer makes decisions on participation rate by solving his maximization problem given by Eq. (8). The HJB equation is given by

$$\begin{aligned} \phi V_m = \mathop {\max }\limits _{0 \le x( \cdot ) \le 1} \left\{ {\varDelta (x)M^{\frac{\eta }{{1 - \gamma }}} - \frac{1}{2}u^2 + \frac{{\partial V_m }}{{\partial M}}(u - \delta M)} \right\} , \end{aligned}$$

(40)

where

$$\begin{aligned} \varDelta (x)\mathrm{{ = }}\left\{ {\begin{array}{l@{\quad }l@{\quad }l} {\left( {\frac{{{\rho _{m}} - \left( {{\rho _{m}} + \gamma {\rho _{r}}} \right) x}}{{\gamma {\rho _{r}}}}} \right) \left( {\frac{{\gamma {\rho _{r}}}}{{(1 - x)}}} \right) ^{\frac{1}{{^{1 - \gamma } }}} } &{} {{\text {if}}} &{} {x < A} \\ {\left( {\frac{{x\left( {1 - y} \right) }}{y}} \right) (\frac{{{\rho _{m}}y}}{x})^{\frac{1}{{1 - \gamma }}} } &{} {{\text {if}}} &{} {A \le x \le B} \\ {\frac{{{\rho _{m}}\left( {1 - y} \right) }}{{\gamma ({\rho _{r}} + {\rho _{m}}y)}}\left( {\gamma ({\rho _{r}} + {\rho _{m}}y)} \right) ^{\frac{1}{{1 - \gamma }}} } &{} {{\text {if}}} &{} {x > B.} \\ \end{array}} \right. \end{aligned}$$

and $A = {\rho _{m}}y/(\gamma {\rho _{r}} + {\rho _{m}}y)$, $B = {\rho _{m}}y/\left( {\gamma \left( {{\rho _{r}} + {\rho _{m}}y} \right) } \right)$.

To get the optimal participation rate, we first examine the first-order condition for the maximization of $V_m$ which leads to

$$\begin{aligned} \frac{{\partial V_m (M)}}{{\partial x}} = \frac{{\partial \varDelta (x)}}{{\partial x}}M^{\frac{\eta }{{1 - \gamma }}} \end{aligned}$$

(41)

and discuss the optimal participation rate under three conditions sequentially:

• (i) When $x<A$, letting $\partial V_m (M)/\partial x = 0$, we get the solution $x^*$ denoted by

  $$\begin{aligned} x^\mathrm{{*}} = \frac{{\partial \varDelta (x)}}{{\partial x}}\mathrm{{ = }}\frac{{\gamma {\rho _{r}} + {\rho _{m}} - {\rho _{r}}}}{{\gamma {\rho _{r}} + {\rho _{m}}}}. \end{aligned}$$

  Furthermore, because

  $$\begin{aligned} \left. {\frac{{\partial ^2 V_m (M)}}{{\partial x^2 }}} \right| _{x = \frac{{\gamma {\rho _{r}} + {\rho _{m}} - {\rho _{r}}}}{{\gamma {\rho _{r}} + {\rho _{m}}}}} = - \frac{1}{{(1 - \gamma ){\rho _{r}}^2 }}(\gamma {\rho _{r}} + {\rho _{m}})^2 (\gamma ^2 {\rho _{r}} + \gamma {\rho _{m}})^{\frac{1}{{1 - \gamma }}} < 0, \end{aligned}$$

  we can conclude that the manufacturer gets his optimal participation rate at $x\mathrm{{ = }}(\gamma {\rho _{r}} + {\rho _{m}} - {\rho _{r}})/(\gamma {\rho _{r}}+ {\rho _{m}})$ when $x<A$.

• (ii) When $A \le x \le B$, we get the first derivative of $V_m(M)$ with respect to x as

  $$\begin{aligned} \frac{{\partial V_m (M)}}{{\partial x}} = - \frac{{(1 - y)\gamma }}{{y(1 - \gamma )}}(\frac{{{\rho _{m}}y}}{x})^{\frac{1}{{1 - \gamma }}}. \end{aligned}$$

  (42)

  From Eq. (38), we get $\partial V_m (M)/\partial x < 0$, which indicates the manufacturer’s value function is a monotone decreasing function of x. Thus the manufacturer gets the maximum of the present value of the profit at $x=A$ when $A \le x \le B$.

• (iii) We can eliminate the condition $x>B$ from consideration for obtaining the manufacturer’s optimal participation rate because we can prove that the manufacturer’s profit under this condition is always smaller than that under the second condition.

In summary, owing to $V_m$ is a continuous function, the optimal participation rate of the manufacturer is

$$\begin{aligned} \bar{x} = \left\{ {\begin{array}{l@{\quad }l@{\quad }l} 0 &{} {{\text {if}}} &{} {{} {\rho _{m}}/{\rho _{r}} < \left( {1 - \gamma } \right) } \\ {\frac{{{\rho _{m}}y}}{{\gamma {\rho _{r}} + {\rho _{m}}y}} } &{} {{\text {if}}} &{} {{\rho _{m}}/{\rho _{r}} \ge \left( {1 - \gamma } \right) \ {\text {and}}\ y < {\hat{y}}} \\ {\frac{{{\rho _{m}} + \gamma {\rho _{r}} - {\rho _{r}}}}{{{\rho _{m}} + \gamma {\rho _{r}}}} } &{} {{\text {if}}} &{} {{\rho _{m}}/{\rho _{r}} \ge \left( {1 - \gamma } \right) \ {\text {and}}\ y \ge {\hat{y}}}, \\ \end{array}} \right. \end{aligned}$$

(43)

where ${\hat{y}} = \gamma \left( {{\rho _{m}} - \left( {1 - \gamma } \right) {\rho _{r}}} \right) /{\rho _{m}}$.

Substituting the optimal participation rate into Eqs. (12)–(14), we can get the manufacturer’s equilibrium advertising efforts is

$$\begin{aligned} \overline{u} (t) = \frac{1}{2}\left( {2\delta + \phi - \sqrt{(2\delta + \phi )^2 - 8\overline{\varDelta }} } \right) M, \end{aligned}$$

(44)

the manufacturer’s goodwill stock along with time t is

$$\begin{aligned} \overline{M} (t) = M_0 e^{\frac{{\phi - \sqrt{(2\delta + \phi )^2 - 8\overline{\varDelta }} }}{2}t} \end{aligned}$$

(45)

and the retailer’s equilibrium local advertising expenditure is

$$\begin{aligned} \bar{a}(t)\mathrm{{ = }}\left\{ {\begin{array}{l@{\quad }l@{\quad }l} {\left( {\gamma {\rho _{r}}} \right) ^{\frac{1}{{^{1 - \gamma } }}} (\overline{M} (t))^2 } &{} {{\text {if}}} &{} {{\rho _{m}} < (1 - \gamma ){\rho _{r}}} \\ {(\gamma {\rho _{r}}+ {\rho _{m}}y)^{\frac{1}{{1 - \gamma }}} (\overline{M} (t))^2 } &{} {{\text {if}}} &{} {{\rho _{m}} \ge (1 - \gamma ){\rho _{r}}{} \mathrm{{ and }}{} y < {\hat{y}}} \\ {\left( {\gamma (\gamma {\rho _{r}} + {\rho _{m}})} \right) ^{\frac{1}{{1 - \gamma }}} (\overline{M} (t))^2 } &{} {{\text {if}}} &{} {{\rho _{m}} \ge (1 - \gamma ){\rho _{r}}{} \mathrm{{ and }}{} y \ge {\hat{y}}}, \\ \end{array}} \right. \end{aligned}$$

(46)

where

$$\begin{aligned} \overline{\varDelta }\mathrm{{ = }}\left\{ {\begin{array}{l@{\quad }l@{\quad }l} {{\rho _{m}}\left( {\gamma {\rho _{r}}} \right) ^{\frac{\gamma }{{^{1 - \gamma } }}} } &{} {{\text {if}}} &{} {{\rho _{m}} < (1 - \gamma ){\rho _{r}}} \\ {{\rho _{m}}(1 - y)(\gamma {\rho _{r}} + y{\rho _{m}})^{\frac{\gamma }{{1 - \gamma }}} } &{} {{\text {if}}} &{} {{\rho _{m}} \ge (1 - \gamma ){\rho _{r}}{} \mathrm{{ and }}{} y < {\hat{y}}} \\ {(\frac{1}{\gamma } - 1)(\gamma ^2 {\rho _{r}} + \gamma {\rho _{m}})^{\frac{1}{{1 - \gamma }}} } &{} {{\text {if}}} &{} {{\rho _{m}} \ge (1 - \gamma ){\rho _{r}}{} \mathrm{{ and }}{} y \ge {\hat{y}},} \\ \end{array}} \right. \end{aligned}$$

and ${\hat{y}} = \gamma \left( {\rho _{m} - \left( {1 - \gamma } \right) {\rho _{r}} } \right) /{\rho _{m}}$.