Accelerating the diffusion of innovations under mixed word of mouth through marketing–operations interaction

Abstract

In this paper, an extension of the Bass model is suggested that accounts for the influence of conformance quality on mixed (i.e., positive and negative) word-of-mouth in the diffusion of a new product. A primary goal is to determine how an active operational policy seeking to continuously improve conformance quality affects the optimal leveraging of marketing instruments used to diffuse new products, and the resulting sales and profits. To do so, an optimal tradeoff by a monopolistic firm between advertising effort and price, on the one hand, and conformance quality, on the other hand, is analyzed, along with the implications for word of mouth effectiveness. Our results can be summarized as follows. Price and advertising levels are respectively lower and higher under an operations–marketing policy than under a marketing policy only. As a result, the market potential and the innovation effect are higher under an operations–marketing policy than under a marketing policy only, as is the imitation effect due to conformance quality improvements over time. Also, greater cumulative sales and cumulative profits are obtained. However, higher design quality results in a lower price and greater advertising effort under an operations–marketing policy than under a marketing policy only. Finally, for lower design quality, the two policies result in different patterns (non-monotonic vs. monotonic) for price and advertising yet cumulative sales and profits are of quite similar magnitude.

Appendices

Appendix

A1

Under an operations–marketing policy, the corresponding Hamiltonian writes:

$$\begin{aligned} H= & {} \left[ {Q\left( {p+\phi } \right) -\frac{dq_0 }{1+X}+\lambda } \right] \left( {\alpha -\beta p-X} \right) \left[ {aq_0 +\left( {b_1 +b_2 } \right) Y-b_2 X+cu} \right] \nonumber \\&-\frac{u^{2}}{2}-\frac{v^{2}}{2}+\mu v\left( {1-Q} \right) \end{aligned}$$

(A1.1)

where $\lambda (t)$, $\mu (t)$ and $\varphi (t)$ are the current-value costate variables associated with X(t), Q(t) and Y(t), respectively, that are given by the equations:

$$\begin{aligned} \dot{\lambda }= & {} r\lambda +\left[ {Q\left( {p+\phi } \right) -\frac{dq_0 }{1+X}+\lambda } \right] \left[ {aq_0 +\left( {b_1 +b_2 } \right) Y+b_2 \left( {\alpha -\beta p-2X} \right) +cu} \right] \nonumber \\&-\frac{dq_0 \left( {\alpha -\beta p-X} \right) ^{2}}{\left( {1+X} \right) ^{2}}\left[ {aq_0 +\left( {b_1 +b_2 } \right) Y-b_2 X+cu} \right] ^{2} \end{aligned}$$

(A1.2)

$$\begin{aligned} \dot{\mu }= & {} \mu \left( {r+v} \right) -\left( {p+\varphi } \right) \left( {\alpha -\beta p-X} \right) \left[ {aq_0 +\left( {b_1 +b_2 } \right) Y-b_2 X+cu} \right] \end{aligned}$$

(A1.3)

$$\begin{aligned} \dot{\varphi }= & {} r\varphi -\left( {b_1 +b_2 } \right) \left[ {Q\left( {p+\varphi } \right) -\frac{dq_0 }{1+X}+\lambda } \right] \left( {\alpha -\beta p-X} \right) \end{aligned}$$

(A1.4)

along with their respective transversality conditions, that is:

$$\begin{aligned} \lambda (T)= & {} 0\\ \mu (T)= & {} 0\\ \varphi (T)= & {} 0 \end{aligned}$$

Necessary conditions for optimality are:

$$\begin{aligned} H_p= & {} 0\Rightarrow p=\frac{1}{2\beta Q}\left[ {Q\left( {\alpha -\beta \phi -X} \right) +\beta \left( {\frac{dq_0 }{1+X}-\lambda } \right) } \right] \end{aligned}$$

(A1.5)

$$\begin{aligned} H_u= & {} 0\Rightarrow u=c\left[ {Q\left( {p+\phi } \right) -\frac{dq_0 }{1+X}+\lambda } \right] \left( {\alpha -\beta p-X} \right) \end{aligned}$$

(A1.6)

$$\begin{aligned} H_v= & {} 0\Rightarrow v=\mu \left( {1-Q} \right) \end{aligned}$$

(A1.7)

It can be shown that the Legendre-Clebsch condition is fulfilled as the Hessian matrix of the Hamiltonian in terms of the control variables, that is:

$$\begin{aligned} H=\left[ {{\begin{array}{ccc} {-2\beta Q\left[ {aq_0 +\left( {b_1 +b_2 } \right) Y-b_2 X+cu} \right] }&{} 0&{} 0 \\ 0&{} {-1}&{} 0 \\ 0&{} 0&{} {-1} \\ \end{array} }} \right] \end{aligned}$$

is definite negative. Therefore, the Hamiltonian is concave with respect to the control vector $\left( {p,u,v} \right)$, which is a maximizer of the Hamiltonian.

Using (A1.5), (A1.6) rewrites:

$$\begin{aligned} u=\frac{c\left\{ {\left( {1+X} \right) \left[ {Q\left( {\alpha +\beta \varphi -X} \right) +\beta \lambda } \right] -\beta dq_0 } \right\} ^{2}}{4\beta Q\left( {1+X} \right) ^{2}} \end{aligned}$$

(A1.8)

Using (A1.5), (A1.7), (A1.8) to rewrite (7b), (7c), (7d) and (A1.2)–(A1.4), we get the associated TPBVP (10)–(15). $\square$

A2

From (A1.5), we compute the partial derivative of price with respect to price, which gives $p_Q =-\frac{1}{2Q^{2}}\left( {\frac{dq_0 }{1+X}-\lambda } \right) <0$ whenever $\frac{dq_0 }{1+X}>\lambda$. In our context, the positive influence of sales on the objective function suggests that its corresponding costate variable $\lambda$ should be lower than the unit production cost $\frac{dq_0 }{1+X}$. In contrast, from (A1.5), the greater the cumulative sales, the lower the price because $p_X =-\frac{1}{2}\left[ {\frac{1}{\beta }+\frac{dq_0 }{Q\left( {1+X} \right) ^{2}}} \right] <0$. $\square$

A3

From (A1.6), we compute the partial derivative of advertising effort with respect to conformance quality, which gives $u_Q =\frac{c}{4\beta }\left[ {\left( {\alpha +\beta \phi -X} \right) ^{2}-\frac{\beta ^{2}}{Q^{2}}\left( {\frac{dq_0 }{1+X}-\lambda } \right) ^{2}} \right]$, which is strictly positive. Further, from (A1.6), the greater the cumulative sales, the lower the advertising effort because $u_X =-\frac{c}{2\beta }\left[ {Q\left( {\alpha +\beta \phi -X} \right) -\beta \left( {\frac{dq_0 }{1+X}-\lambda } \right) } \right] <0$. $\square$

A4

From (A1.7), the greater the conformance quality, the lower the quality improvement effort as $v_Q =-\mu$, which is negative for a non-negative quality improvement effort in A1.7. $\square$

A5

Under a marketing policy only, the corresponding Hamiltonian writes:

$$\begin{aligned} H= & {} \left[ {Q\left( {p+\phi } \right) -\frac{dq_0 }{1+X}+\lambda } \right] \left( {\alpha -\beta p-X} \right) \left[ {aq_0 +\left( {b_1 +b_2 } \right) Y-b_2 X+cu} \right] \nonumber \\&-\frac{u^{2}}{2}-\frac{v^{2}}{2}+\mu v\left( {1-Q} \right) \end{aligned}$$

(A5.1)

where $\eta (t)$ is the current-value costate variable associated with X(t), that is given by the equation:

$$\begin{aligned} \dot{\eta }=r\eta +\left[ {Q_0 p-\frac{dq_0 }{1+X}+\eta } \right] \left\{ {aq_0 +2\left[ {\left( {b_1 +b_2 } \right) Q_0 -b_2 } \right] X+cu-\left[ {\left( {b_1 +b_2 } \right) Q_0 -b_2 } \right] (\alpha -\beta p)} \right\} \nonumber \\ -\frac{dq_0 \left( {\alpha -\beta p-X} \right) }{\left( {1+X} \right) ^{2}}\left\{ {aq_0 +\left[ {\left( {b_1 +b_2 } \right) Q_0 -b_2 } \right] X+cu} \right\} \nonumber \\ \end{aligned}$$

(A5.2)

along with its transversality condition, that is:

$$\begin{aligned} \eta (T)=0 \end{aligned}$$

Necessary conditions for optimality are:

$$\begin{aligned} H_p =0\Rightarrow p= & {} \frac{1}{2\beta Q_0 }\left[ {Q_0 \left( {\alpha -X} \right) +\beta \left( {\frac{dq_0 }{1+X}-\eta } \right) } \right] \end{aligned}$$

(A5.3)

$$\begin{aligned} H_u =0\Rightarrow u= & {} c\left( {\alpha -\beta p-X} \right) \left[ {Q_0 p-\frac{dq_0 }{1+X}+\eta } \right] \end{aligned}$$

(A5.4)

The Legendre-Clebsch condition is also fulfilled in this case as the Hessian matrix of the Hamiltonian in terms of the control variables, that is:

$$\begin{aligned} H=\left[ {{\begin{array}{ll} {-2\beta Q_0 \left( {aq_0 +cu+[-b_2 +(b_1 +b_2 )Q_0 ]X} \right) }&{} 0 \\ 0&{} {-1} \\ \end{array} }} \right] \end{aligned}$$

is definite negative. Therefore, the Hamiltonian is concave with respect to the control vector$\left( {p,u} \right)$, which is a maximizer of the Hamiltonian.

Using (A5.3), (A5.4) rewrites:

$$\begin{aligned} u=\frac{c\left\{ {\left( {1+X} \right) \left[ {Q_0 \left( {\alpha -X} \right) +\beta \eta } \right] -\beta dq_0 } \right\} ^{2}}{4\beta Q_0 \left( {1+X} \right) ^{2}} \end{aligned}$$

(A5.5)

Using (A5.3)–(A5.5) to rewrite (8b) and (A5.2), we get the associated TPBVP (23)–(24).

$\square$

A6

From (A5.3) and (A5.4), we get:

$$\begin{aligned} u=\frac{cQ_0 }{\beta }\left( {\alpha -\beta p-X} \right) ^{2} \end{aligned}$$

(A6.1)

which gives $u_{Q_0 } =\frac{c}{\beta }\left( {\alpha -\beta p-X} \right) ^{2}\ge 0$, $u_{\left( {M-X} \right) } ={2cQ_0 \left( {\alpha -\beta p-X} \right) }/\beta \ge 0$, $u_p =-2cQ_0 \left( {\alpha -\beta p-X} \right) \le 0$ and $u_X =-{2cQ_0 \left( {\alpha -\beta p-X} \right) }/\beta \le 0$. $\square$