Optimal replenishment and pricing decisions under the collect-on-delivery payment scheme

Abstract

In this paper, we study how the timing of payments affects joint replenishment and pricing decisions. Specifically, we consider a periodic review system in which consumers pay upon receiving their goods, referred to as the collect- on-delivery scheme, as opposed to paying when placing their orders, referred to as the instant payment scheme commonly assumed in the literature. We show that a base-stock list price policy is optimal without a fixed ordering cost, and provide a sufficient condition for an $(s, S, p)$ policy to be optimal with a fixed ordering cost. We also perform both analytical and numerical studies to compare the two schemes.

Appendix A

1.1 A.1 Proof of Lemma 4

Proof

Note that,

$$\begin{aligned} \frac{dH(z,p_y(z))}{dz}&= \left\{ \frac{\partial H(z,p)}{\partial z}+\frac{\partial H(z,p)}{\partial p}\left( -\frac{1}{d^{\prime }(p)}\right) \right\} |_{p=p_y(z)}\nonumber \\&= -[p_y(z)-\alpha c+h]+[(1-\alpha )p_y(z)+h+b]\bar{\Phi }(z) \nonumber \\&\quad -\frac{d(p_y(z))+\mu -(1-\alpha )\mathbf{\mathsf E}(-z+\epsilon _t)^{+}}{d^{\prime }(p_y(z))},\\ \frac{d^2H(z,p_y(z))}{dz^2}&= \frac{2-2(1-\alpha )\bar{\Phi }(z)}{d^{\prime }(p_y(z))}-[(1-\alpha )p_y(z)+h+b]\phi (z)\nonumber \\&\quad -\frac{[d(p_y(z))+\mu -(1-\alpha )\mathbf{\mathsf E}(-z+\epsilon _t)^{+}]d^{\prime \prime }(p_y(z))}{[d^{\prime }(p_y(z))]^3}.\nonumber \end{aligned}$$

(9)

If $z\le 0$, then

$$\begin{aligned} d(p_y(z))+\mu -(1-\alpha )\mathbf{\mathsf E}(-z+\epsilon _t)^{+}=y-z+\mu -(1-\alpha )(-z+\mu )>0; \end{aligned}$$

otherwise if $z>0$, then

$$\begin{aligned} d(p_y(z))+\mu -(1-\alpha )\mathbf{\mathsf E}(-z+\epsilon _t)^{+}\ge d(p_y(z))+\mu -(1-\alpha )[\mu -z\bar{\Phi }(z)]>0. \end{aligned}$$

Therefore, $H(z,p_y(z))$ is concave in $z$ and hence, there exists a unique maximizer $z(y)$ and $p(y)=p_y(z(y))$. As $H(z,p)$ is jointly continuous in $z$ and $p$, and $p_y(z)$ is continuous in $z$ for any given $y\ge 0$, $z(y)$ is continuous for $y\in [0,+\infty )$. Then, $p(y)$ is also continuous for $y\in [0,+\infty )$.

Note that $\lim \nolimits _{y\rightarrow +\infty }z(y)=\lim \nolimits _{y\rightarrow +\infty }(y-d(p(y)))=+\infty$. If $z(y)$ is non-increasing in some interval within $[0,+\infty )$, then there must exist $y_1<y_2$ such that $z(y_1)=z(y_2)=z_0$. Thus, $p_{y_1}(z_0)>p_{y_2}(z_0)$. Moreover, the derivative of (9) with respect to $p_y(z)$ is

$$\begin{aligned} \left. \left\{ -1+(1-\alpha )\bar{\Phi }(z)-1+\frac{d(p)+\mu -(1-\alpha )\mathbf{\mathsf E}(-z+\epsilon _t)^{+}}{(d^{\prime }(p))^2}d^{\prime \prime }(p)\right\} \right| _{p=p_y(z)}\le -1. \end{aligned}$$

Therefore, $\left. \frac{dH(z,p_{y_2}(z))}{dz}\right| _{z=z_0}>\left. \frac{dH(z,p_{y_1}(z))}{dz}\right| _{z=z_0}$ and hence, $\left. \frac{dH(z,p_{y_2}(z))}{dz}\right| _{z=z_0}>0$ or,

$\left. \frac{dH(z,p_{y_1}(z))}{dz}\right| _{z=z_0}<0$. The former implies $z_0=y_2$ and the latter implies $z_0=y_1-d(c)$, any of which contradicts $z(y_1)=z(y_2)=z_0$. Therefore, $z(y)$ increases in $y$ for $y\in [0,+\infty )$. $\square$

1.2 A.2 Proof of Lemma 5

Proof

We first consider the case $0\le y_1\le y^*$. By Lemma 2, $z_0\le z^*$ and then by Lemma 3,

$$\begin{aligned} \left. \frac{dH(z,p_{y_1}(z))}{dz}\right| _{z=z_0}&= \left\{ \frac{\partial H(z,p)}{\partial z}+\frac{\partial H(z,p)}{\partial p}\left( -\frac{1}{d^{\prime }(p)}\right) \right\} _{z=z_0,p=p_0}\\&= \left. \frac{\partial H(z,p)}{\partial z}\right| _{z=z_0,p=p_0}=\left. \frac{dH(z,\hat{p}(z))}{dz}\right| _{z=z_0}\ge 0. \end{aligned}$$

Since $H(z,p_{y_1}(z))$ is concave in $z$ by Lemma 4, we have $z(y_1)\ge z_0$ and hence, $p(y_1)\ge p_0$. Similarly, for any $y_2>y^*$, $z_0>z^*$ and $\left. \frac{dH(z,p_{y_2}(z))}{dz}\right| _{z=z_0}<0$. Thus, $p(y_2)<p_0$. Hereby, we complete the proof of the first part.

Note that for any $y_1\in [0,y^*]$,

$$\begin{aligned}&\left. \frac{dH(z(y),p(y))}{dy}\right| _{y=y_1}\\&\quad =\left\{ \frac{\partial H(z,p)}{\partial z}z^{\prime }(y)+\frac{\partial H(z,p)}{\partial p}p^{\prime }(y)\right\} _{y=y_1,z=z(y_1),p=p(y_1)}\\&\quad =\left. \frac{\partial H(z,p)}{\partial z}\right| _{z=z(y_1),p=p(y_1)} - d^{\prime }(p)p^{\prime }(y)\left[ \frac{\partial H(z,p)}{\partial z}\right. \\&\quad \quad +\left. \left. \frac{\partial H(z,p)}{\partial p}\left( -\frac{1}{d^{\prime }(p)}\right) \right] \right| _{y=y_1,z=z(y_1),p=p(y_1)}\\&=\left. \frac{\partial H(z,p)}{\partial z}\right| _{z=z(y_1),p=p(y_1)}=\left. \frac{d G_t(z,p_{y_1}(z))}{dz}\right| _{z=z(y_1)}\\&\quad \quad -\left. \frac{\partial H(z,p)}{\partial p}\left( -\frac{1}{d^{\prime }(p)}\right) \right| _{z=z(y_1),p=p(y_1)}. \end{aligned}$$

Since $z(y_1)\ge z_0$, by Lemma 2 we have $z(y_1)+d(\hat{p}(z(y_1)))\ge z_0+d(\hat{p}(z_0))=y_1=z(y_1)+d(p(y_1))$ and hence, $p(y_1)\ge \hat{p}(z(y_1))$. Therefore, $\left. \frac{\partial H(z,p)}{\partial p}\right| _{z=z(y_1),p=p(y_1)}\le 0$. Furthermore, $\left. \frac{dH(z,p_{y_1}(z))}{dz}\right| _{z=z(y_1)}\ge 0$ as $p(y_1)\ge p_0>c$. Then,

$$\begin{aligned} \left. \frac{dH(z(y),p(y))}{dy}\right| _{y=y_1}&\ge 0. \end{aligned}$$

On the other hand, for any $y_2>y^*$, $\left. \frac{dH(z,p_{y_2}(z))}{dz}\right| _{z=z(y_1)}<0$ as $p(y_2)<p_0<P^u$. Similarly, $p(y_2)<\hat{p}(z(y_2))$ because $z(y_2)<z_0$, and then $\left. \frac{dH(z(y),p(y))}{dy}\right| _{y=y_2}<0$. Therefore, $H(z(y),p(y))$ is quasi-concave in $y$ and $y^*$ is its maximizer. $\square$

1.3 A.3 Proof of Lemma 6

Proof

We prove the three parts one by one.

• 1. When $y^*\le y_1<y_2$ and $p_2\in [c,P^u]$ is given, let $z_2=y_2-d(p_2)$ and $y^{\prime }=z_2+d(\hat{p}(z_2))$. By Lemma 5,

  $$\begin{aligned} G(y_1,p(y_1)) \!=\! H(z(y_1),p(y_1)) \!\ge \! H(z(y_2),p(y_2)) \!=\! G(y_2,p(y_2)) \ge G(y_2,p_2). \end{aligned}$$

  If $z(y_1)\le \max \{z_2,y^*\}$, $p_1=p(y_1)$ satisfies all the requirements. Otherwise, if $z(y_1)>\max \{z_2,y^*\}$, consider the following two cases.

  • (a) If $y^{\prime }\le y_1$, let $p_1=d^{-1}(y_1-z_2)$. Then, we have $d(\hat{p}(z_2))\le y_1-z_2=d(p_1)<y_2-z_2=d(p_2)$ and $\hat{p}(z_2)\ge p_1>p_2$. By Lemma 2,

    $$\begin{aligned} G(y_1,p_1) = H(z_2,p_1) \ge H(z_2,p_2) = G(y_2,p_2), \end{aligned}$$

    and $y_1-d(p_1)=z_2=y_2-d(p_2)$. Thus, $p_1$ satisfies all the requirements.

  • (b) If $y^{\prime }>y_1$, then $y^{\prime }>y_1\ge z(y_1)>y^*$. By Lemma 5, we have $p(y^{\prime })\le \hat{p}(z_2)$ and hence,

    $$\begin{aligned} z_2 = y^{\prime }-d(\hat{p}(z_2))\ge y^{\prime }-d(p(y^{\prime }))=z(y^{\prime }). \end{aligned}$$

    Therefore, $z(y^{\prime })\le z_2<z(y_1)$. However, this contradicts the monotonicity of $z(y)$ from Lemma 4 as $y^{\prime }>y_1$. So $y^{\prime }>y_1$ cannot happen.

• 2. When $y^*\ge y_1>y_2$ and $p_2\in [c,P^u]$ is given, let $z_2=y_2-d(p_2)$ and $y^{\prime }=z_2+d(\hat{p}(z_2))$. If $z_2\le z(y_1)$, let $p_1=p(y_1)$ and it satisfies all the requirements. Otherwise, if $z_2>z(y_1)$, consider the following two situations.

  • (a) If $y^{\prime }\ge y_1$, let $p_1=d^{-1}(y_1-z_2)$, then we have $d(p_2)=y_2-z_2<y_1-z_2=d(p_1)\le y^{\prime }-z_2=d(\hat{p}(z_2))$ and $\hat{p}(z_2)\le p_1<p_2$. By Lemma 2,

    $$\begin{aligned} G(y_1,p_1) = H(z_2,p_1) \ge H(z_2,p_2) = G(y_2,p_2), \end{aligned}$$

    and $y_1-d(p_1)=z_2=y_2-d(p_2)$. Thus, $p_1$ satisfies all the requirements.

  • (b) If $y^{\prime }<y_1$, by the monotonicity of $z(y)$ from Lemma 4, we have $z(y^{\prime })<z(y_1)<z_2$. Therefore,

    $$\begin{aligned} d(\hat{p}(z_2)) = y^{\prime }-z_2 < y^{\prime }-z(y^{\prime }) = d(p(y^{\prime })), \end{aligned}$$

    and then $\hat{p}(z_2)>p(y^{\prime })$. However, this contradicts Lemma 5 as $y^{\prime }<y_1\le y^*$. So this situation cannot happen.

• 3. When $t=T$, since $R_T(y)$ is quasi-concave in $y$, we know that $R_T(y_T^1)\ge R_T(y_T^2)$. Then, we consider two systems that starts with $y_t^1$ and $y_t^2$, $t<T$, and experience the same sample path of the random shocks. We further assume that the system starts with $y_t^2$ follows the optimal decisions to attain $R_t(y_t^2)$ and $p_t^2$ is the optimal price. If $y_t^2>y_t^1\ge y^*$, by part 1 of Lemma 6, there exists $p_t^1$ such that $G_t(y_t^1,p_t^1)\ge G_t(y_t^2,p_t^2)$ and $x_{t+1}^1\le \max \{x_{t+1}^2,y^*\}\le \max \{y_{t+1}^2,y^*\}$. Otherwise, if $y_t^2<y_t^1\le y^*$, since $\max _p G(y,p)$ is quasi-concave in $y$, there exists $p_t^1$ such that $G_t(y_t^1,p_t^1)\ge G_t(y_t^2,p_t^2)$ and $x_{t+1}^1<y_t^1\le y^*$. In both cases, it can be either $x_{t+1}^1\le y_{t+1}^2$ or $y_{t+1}^2<x_{t+1}^1\le y^*$. If $x_{t+1}^1\le y_{t+1}^2$, order up to $y_{t+1}^2$ in period $t+1$ and mimic the decisions of system $2$ in the following periods for system 1. If $y_{t+1}^2<x_{t+1}^1\le y^*$, no order is placed in period $t+1$ in system $1$. Still, by part 1 of Lemma 6, there exists $p_{t+1}^1$ such that $G_{t+1} (y_{t+1}^1,p_{t+1}^1)\ge G_{t+1}(y_{t+1}^2,p_{t+1}^2)$. We continue making decisions for system $1$ in this way until $x_s^1\le y_s^2$ for some $s>t$ or the end of the horizon is reached. Therefore, in either case, system $1$ is at most $\alpha K$ worse than system $2$. Since $R_t(y_t^1)$ is the maximum profit of system $1$ from period $t$ to the end of horizon, we obtain the desired result. $\square$