Demand–supply dynamics in FMCG business: exploration of customers’ herd behavior

Abstract

Nowadays, social interactions among customers play an increasingly important role in product demand. However, the existing literature mostly focusses on the social influence of other customers’ stated preference while disregarding the social influence of revealed preference. To bridge the gap, this paper incorporates the latter, referred to as herd behavior of customers, into a dynamic demand–supply system of fast moving consumer goods, whose sales to some extent depend on social interaction. Equilibrium points and their local stability are analyzed first. Then, we solve the optimal supply strategy and analyze the evolution characteristics of this demand–supply dynamics. Results indicate that there may be more than one equilibrium point in this system. Provider has two optimal supply policies if initial point is the indifference point, which separates state space if two steady states exist. In addition, the demand–supply evolution is periodic as the provider alternates between cultivating and exploiting the herd effect.

Appendices

Appendix A

Proof of Proposition 2

As $\mu ' = 1 - \frac{{\Omega '(\Lambda )\phi (p) - \Omega (\Lambda )\phi '(p)h'(\Lambda )}}{{{\phi ^2}(p)}}$, for any equilibrium point $({\Lambda ^*},{p^*};\mu )$ in a decreasing interval of $f( \Lambda )$, ${\mathrm{P}}({\mathrm{1}}){\mathrm{= 1 + }}v(1 - \phi + \Omega ') - (1 - \phi + v + \Omega ' + (\mu - \Lambda )(1 - v)h'\phi ')=({\mathrm{1}} - v)[\phi ({p^*}) - \Omega '({\Lambda ^*}) - (\mu - {\Lambda ^*})h'({\Lambda ^*})\phi '({p^*})]=({\mathrm{1}} - v)\phi ({p^*})\mu ' < 0$ holds, so this equilibrium point is always unstable. For the equilibrium point in an increasing interval, ${\mathrm{P(1)}}>0$ and ${\mathrm{P}}( - {\mathrm{1}}){\mathrm{= 1 + }}v(1 - \phi + \Omega ') + (1 - \phi + \Omega ' + v + (\mu - \Lambda )(1 - v)h'\phi ')=(1 + v)(2 - \phi + \Omega ') + (\mu - \Lambda )(1 - v)h'\phi ' \ge (1 + v)(2 - \mu '\phi ({p^*})) > 0$ when $\mu '({\Lambda ^{\mathrm{*}}})\phi ({p^{\mathrm{*}}}) < 2$. In addition, $\det (J)<1$ always holds if $\Omega '({\Lambda ^{\mathrm{*}}}) < \phi ({p^*})$, so Jury stability condition is satisfied and the equilibrium point is asymptotically stable. However, if $\Omega '({\Lambda ^{\mathrm{*}}}) >\phi ({p^*})$, Jury stability condition will be violated due to $\det (J)>1$ with the increase in v.

This is to say, when v crosses ${v^{\mathrm{0}}} = \frac{1}{{1 - \phi ({p^*}) + \Omega '({\Lambda ^*})}}$ from the left side, a pair of complex-conjugate eigenvalues crosses the unit circle [28]. As a consequence, the equilibrium point becomes unstable. $\square$

Proof of Proposition 3

For a given ${\chi _t}$, $\frac{{\partial {\mathcal {H}}}}{{\partial {q_t}}} = w{\Lambda _t} - 2\alpha {q_t} + {\chi _t}\frac{{{{\mathrm{e}}^{w{\Lambda _t} + \alpha {q_t}}}\alpha ({\Lambda _t} - \mu )}}{{{{({{\mathrm{e}}^{w{\Lambda _t}}} + {{\mathrm{e}}^{\alpha {q_t}}})}^2}}}$. As $\frac{{\partial {\mathcal {H}}}}{{\partial {q_t}}} >0$ when ${q_t}=0$ and $\frac{{\partial {\mathcal {H}}}}{{\partial {q_t}}}<0$ when ${q_t} \rightarrow \infty$, so $\frac{{\partial {\mathcal {H}}}}{{\partial {q_t}}} =0$ must have solution(s) in ${q_t} \in (0,\infty )$. The second-order derivative satisfies $\frac{{{\partial ^2}{\mathcal {H}}}}{{{\partial }{q_t}^2}} = \frac{{{{\mathrm{e}}^{w{\Lambda _t} + \alpha {q_t}}}({{\mathrm{e}}^{w{\Lambda _t}}} - {{\mathrm{e}}^{\alpha {q_t}}})}}{{{{({{\mathrm{e}}^{w{\Lambda _t}}} + {{\mathrm{e}}^{\alpha {q_t}}})}^3}}}$${\alpha ^2}({\Lambda _t} - \mu ){\chi _t} - 2\alpha$. Any solution of $\frac{{\partial {\mathcal {H}}}}{{\partial {q_t}}} =0$ also satisfies $\frac{{{\partial ^2}\mathcal{H}}}{{\partial {q_t}^2}} = \alpha \frac{{{{\mathrm{e}}^{w{\Lambda _t}}} - {{\mathrm{e}}^{\alpha {q_t}}}}}{{{{\mathrm{e}}^{w{\Lambda _t}}} + {{\mathrm{e}}^{\alpha {q_t}}}}}( 2\alpha {q_t}-w{\Lambda _t}) - 2\alpha < 0$. Therefore, this solution must be a local maximum. If there are two local maximums, there must be a local minimum between them. This contradicts with $\frac{{{\partial ^2}\mathcal{H}}}{{\partial {q_t}^2}} < 0$. So $\frac{{\partial \mathcal{L}}}{{\partial {q_t}}} =0$ has a unique solution. Assume the unique solution of $\frac{{\partial {\mathcal {H}}}}{{\partial {q_t}}} =0$ is ${{\hat{q}}_t}$. Then, if ${{\hat{q}}_t}$ satisfies ${{\hat{q}}_t}<w\Lambda _t$, ${{\hat{q}}_t}$ also is the unique solution of $\frac{{\partial \mathcal{L}}}{{\partial {q_t}}} =0$ with ${\eta _t} =0$. If ${{\hat{q}}_t}>w\Lambda _t$, $\frac{{\partial \mathcal{L}}}{{\partial {q_t}}} =0$ has a unique solution ${q_t} = \frac{{w{\Lambda _t}}}{\alpha }$ with ${\eta _t} = {\chi _t}\frac{{({\Lambda _t} - \mu )}}{4} - \frac{w}{\alpha }{\Lambda _t}$. What is more, Legendre–Clebsch condition is satisfied for the interior solution. $\square$

Appendix B

Step 1 Initialize decision variable ${q_t}$ and the slack variable ${\eta _t} =0$, $t=1,2,\ldots ,T$,

Step 2 Solve state variable ${\Lambda _t}$ (${\Lambda _t}$ and ${r_t}$) in Eq. 8 (Eq. 7) forward in time,

Step 3 Solve adjoint variable ${\chi _t}$ (${\chi _t}$ and ${\kappa _t}$) via Eq. 10 (Eq. 12 and Eq. 13) backward in time, subject to transversality condition ${\chi _T}=0$ (${\chi _T}=0$ and ${\kappa _T}=0$),

Step 4 Solve decision variable ${q_t}$ via Proposition 3 (Proposition 4),

Step 5 Check convergence. If $\left\| \mathbf{q - \mathbf q _{old}} \right\| > {10^{ - 4}}$, go to Step 2. Otherwise, stop and output ${q_t}$ and the corresponding profit.