Advance selling under deposit expansion and consumer’s valuation change

Abstract

Motivated by the emerging practices of China's major e-commerce platforms, we investigate the effect of consumer’s valuation change, deposit expansion and return policy on retailer’s advance booking decisions. We establish the retailer’s expected profit based on a two-stage newsvendor model and derive his optimal deposit and order quantity decisions under different scenarios. Our conclusions show that: (1) the retailer gains further profit under the advance selling strategy with a deposit, due to the reduction of inventory risk and exploitation of consumer’s valuation uncertainty; (2) the retailer benefits more from advance selling when he has a lower profit margin or the consumers have low-level prior valuations; (3) the effectiveness of advance selling on exploiting consumer’s valuation uncertainty will be weakened if the retailer allows customers to return products; (4) a lower deposit expansion rate leads to higher deposit and expected profit.

Appendices

Appendix

Proof of Proposition 1

If $0\le b<\frac{p-\overline{{v }_{0}}+M}{2}$ (Case III), we have ${D}_{c}-{D}_{r}=-\frac{{N}_{1}{\left(2b-M-p+\overline{{v }_{0}}\right)}^{2}}{4\overline{{v }_{0}}M}<0$. Hence, we have ${D}_{c}<{D}_{r}$. For case I and case II, Table 2 shows that ${D}_{c}={D}_{r}$.

Proof of Lemma 1

Since ${\pi }_{2}^{I}\left(\left.{Q}^{I}\right|{N}_{1},b\right)=\left(p-w\right){Q}^{I}-\frac{p\left(\overline{{v }_{0}}-p\right)}{\overline{{v }_{0}}}{\int }_{0}^{\frac{\overline{{v }_{0}}{Q}^{I}}{\overline{{v }_{0}}-p}}\Phi \left(n\right)dn+\frac{{N}_{1}\left(p-2b-w\right)\left(\overline{{v }_{0}}-p+b\right)}{\overline{{v }_{0}}}$, then the first order and the second order with respect to ${Q}^{I}$ are $\frac{d{\pi }_{2}^{I}\left(\left.{Q}^{I}\right|{N}_{1},b\right)}{d{Q}^{I}}=p-w-p\Phi \left(\frac{\overline{{v }_{0}}{Q}^{I}}{\overline{{v }_{0}}-p}\right),$ and $\frac{{d}^{2}{\pi }_{2}^{I}\left(\left.{Q}^{I}\right|{N}_{1},b\right)}{d{{Q}^{I}}^{2}}=-\frac{\overline{{v }_{0}}p}{\overline{{v }_{0}}-p}\phi \left(\frac{\overline{{v }_{0}}{Q}^{I}}{\overline{{v }_{0}}-p}\right)<0,$ respectively. Therefore, ${\pi }_{2}^{I}\left(\left.{Q}^{I}\right|{N}_{1},b\right)$ is a concave function of ${Q}^{I}$ and the optimal order quantity is ${Q}^{I}=\frac{\overline{{v }_{0}}-p}{\overline{{v }_{0}}}{\Phi }^{-1}\left(\frac{p-w}{p}\right).$ Hence, the total order quantity, ${Q}^{I}+\frac{{N}_{1}(\overline{{v }_{0}}-p+b)}{\overline{{v }_{0}}}$, is $\frac{\overline{{v }_{0}}-p}{\overline{{v }_{0}}}{\Phi }^{-1}\left(\frac{p-w}{p}\right)+\frac{{N}_{1}(\overline{{v }_{0}}-p+b)}{\overline{{v }_{0}}}.$

Proof of Proposition 2

Given ${\pi }_{1}^{I}\left(b|{N}_{1}\right)=\frac{E({N}_{1})(\overline{{v }_{0}}-p+b)(p-b-w)}{\overline{{v }_{0}}}+\left(p-w\right)\frac{\overline{{v }_{0}}-p}{\overline{{v }_{0}}}{\Phi }^{-1}\left(\frac{p-w}{p}\right)-\frac{p\left(\overline{{v }_{0}}-p\right)}{\overline{{v }_{0}}}{\int }_{0}^{\frac{\overline{{v }_{0}}{Q}^{I}}{\overline{{v }_{0}}-p}} \Phi \left(n\right)dn$, then the first order and the second order with respect to $b$ is

$$\frac{d{\pi }_{1}^{I}\left(b\right)}{db}=\frac{E\left({N}_{1}\right)\left[2p-w-\overline{{v }_{0}}-2b\right]}{\overline{{v }_{0}}}$$

and

$$\frac{{d}^{2}{\pi }_{1}^{I}\left(b\right)}{d{b}^{2}}=-\frac{2E\left({N}_{1}\right)}{\overline{{v }_{0}}}<0,$$

respectively. Therefore, the first-order condition is $b=\frac{2p-w-\overline{{v }_{0}}}{2}.$ If$\frac{2p-w-\overline{{v }_{0}}}{2}<0$, then $\frac{d{\pi }_{1}^{I}\left(b\right)}{d b}<0$ for all $b\ge 0$ which means${b}^{*}=M$; if$\frac{2p-w-\overline{{v }_{0}}}{2}\ge 0$, then the optimal deposit is $b=\frac{2p-w-\overline{{v }_{0}}}{2}.$ Therefore, the optimal booking is $b_{1}^{*} = \left\{ {\begin{array}{*{20}c} {\frac{{2p - w - \overline{{v_{0} }} }}{2},} & { if\;\frac{{2p - w - \overline{{v_{0} }} }}{2} > M} \\ {M,} & {otherwise} \\ \end{array} } \right..$

Proof of Lemma 2

The proof is the same as Lemma 1, and we omit it.

Proof of Proposition 3

For case II, differentiate ${\pi }_{1}^{II}\left(b\right)$ with respect to $b$ yields

$$\frac{d{\pi }_{1}^{II}\left(b\right)}{db}=\frac{E\left({N}_{1}\right)}{4\overline{{v }_{0}}M}\left(6{b}^{2}-16Mb+8Mp-4M\overline{{V }_{0}}-4Mw+2{M}^{2}\right).$$

Further, we have

$$\frac{{d}^{2}{\pi }_{1}^{II}\left(b\right)}{d{b}^{2}}=\frac{E\left({N}_{1}\right)}{4\overline{{v }_{0}}M}(12b-16M)<0,$$

which proves the concavity of ${E}_{{N}_{1}}\left[{\pi }_{1}^{II}\left(b\right)\right]$. Hence the first order with respect to $b$ is the optimal decision. The optimal solution is $b=\frac{4M}{3}\pm \frac{\sqrt{M(13M-12p+6\overline{{v }_{0}}+6w)}}{3}$. Since $b<M$ we have

$${b}_{II}^{*}=\frac{4M}{3}-\frac{\sqrt{M\left(13M-12p+6\overline{{v }_{0}}+6w\right)}}{3}.$$

Recall that the optimal value of $b$ is derived with the condition, $\frac{p-\overline{{v }_{0}}+M}{2}\le b<M$, the optimal value of $b$ is

$$ b_{II}^{*} = \left\{ {\begin{array}{*{20}l} {\frac{{p - \overline{{v_{0} }} + M}}{2},} \hfill & {if\; \frac{4M}{3} - \frac{{\sqrt {M\left( {13M - 12p + 6\overline{{v_{0} }} + 6w} \right)} }}{3} < \frac{{p - \overline{{v_{0} }} + M}}{2};} \hfill \\ {M,} \hfill & { if\; \frac{4M}{3} - \frac{{\sqrt {M\left( {13M - 12p + 6\overline{{v_{0} }} + 6w} \right)} }}{3} > M;} \hfill \\ {\frac{4M}{3} - \frac{{\sqrt {M\left( {13M - 12p + 6\overline{{v_{0} }} + 6w} \right)} }}{3}} \hfill & {if\;{\text{max}}\left\{ {0,\frac{{p - \overline{v}_{0} + M}}{2}} \right\} \le \frac{4M}{3} - \frac{{\sqrt {M\left( {13M - 12p + 6\overline{v}_{0} + 6w} \right)} }}{3} \le M} \hfill \\ {a{\text{rgmax}}\left( {\pi_{1}^{III} \left( M \right),\pi_{1}^{III} \left( {\frac{{p - \overline{v}_{0} + M}}{2}} \right)} \right),} \hfill & {otherwise} \hfill \\ \end{array} } \right.. $$

For case III, we have:

$$\frac{{d}^{2}{\pi }_{1}^{III}\left(b\right)}{d{b}^{2}}=\frac{E\left({N}_{1}\right)}{\overline{{v }_{0}}M}\left(6p-9b-4\overline{{v }_{0}}-2w\right).$$

It is not determined whether the second-order is less than zero. For the first order with respect to $b$, we have:

$$\frac{d{\pi }_{1}^{III}\left(b\right)}{db}=\frac{E\left({N}_{1}\right)}{2\overline{{v }_{0}}M}\left(-9{b}^{2}+12bp-8b\overline{{v }_{0}}-4wb-3{p}^{2}+4p\overline{{v }_{0}}+2wp-{\overline{{v }_{0}}}^{2}-2w\overline{{v }_{0}}\right).$$

Then the first-order condition is:

$${b}_{III}^{i*},{b}_{III}^{j*}=\frac{6p-4\overline{{v }_{0}}-2w}{9}\pm \frac{\sqrt{9{p}^{2}-12p\overline{{v }_{0}}-6pw+7{\overline{{v }_{0}}}^{2}-2\overline{{v }_{0}}w+4{w}^{2}}}{9}.$$

We derive the optimal solution by comparing the function value of the first order and the boundary points. That is

$$ b_{III}^{*} = \arg max\left( {\pi_{1}^{III} \left( {b_{III}^{i*} } \right),\pi_{1}^{III} \left( {b_{III}^{j*} } \right),\pi_{1}^{III} \left( 0 \right),\pi_{1}^{III} \left( {\frac{{p - \overline{{v_{0} }} + M}}{2}} \right)} \right),\quad if b_{III}^{i*} \,or\, b_{III}^{j*} \in I_{III} $$

where,

$$ b_{III}^{i*} ,b_{III}^{j*} = \frac{{5p - 3\overline{{v_{0} }} - 2w}}{7} \pm \frac{{\sqrt {4p^{2} - 2p\overline{{v_{0} }} - 6pw + 2\overline{{v_{0} }}^{2} - 2\overline{{v_{0} }} w + 4w^{2} } }}{7},\quad i,j \in \{ 1,2\} $$

$${I}_{III}=\left\{b|0\le b<\frac{p-{\overline{v} }_{0}+M}{2}\right\}.$$

Proof of Proposition 5

The proof is similar to Proposition 3. In particular, we can easily derive the retailer’s profit at the first stage is

• (1) if $M\le b$;

  $$\underset{b}{\mathit{max}}E\left[{\pi }_{1}^{I}\left(b\right)\right]=\left(p-w\right){Q}_{r}^{I}-\frac{p\left(\overline{{v }_{0}}-p\right)}{\overline{{v }_{0}}}{\int }_{0}^{{\Phi }^{-1}\left(\frac{p-w}{p}\right)} \Phi \left(n\right)dn+\frac{E\left({N}_{1}\right)\left(p-2b-w\right)\left(\overline{{v }_{0}}-p+b\right)}{\overline{{v }_{0}}}+\frac{bE\left({N}_{1}\right)M}{4\overline{{v }_{0}}}.$$

• (2) if $p-\overline{{v }_{0}}+M\le b<M$;

  $$ \begin{aligned} \mathop {\max }\limits_{b} E\left[ {\pi_{1}^{II} \left( b \right)} \right] & = \left( {p - w} \right)Q_{r}^{II} - \frac{{p\left( {\overline{{v_{0} }} - p} \right)}}{{\overline{{v_{0} }} }}\mathop \int \limits_{0}^{{{\Phi }^{ - 1} \left( {\frac{p - w}{p}} \right)}} \Phi \left( n \right)dn \\ & \quad + E\left( {N_{1} } \right)\left( {p - 2b - w} \right)\frac{{\left( {\overline{{v_{0} }} - p + b} \right)}}{{\overline{{V_{0} }} }} + \frac{{\left( {p - w} \right)E\left( {N_{1} } \right)\left( {M - b} \right)^{2} }}{{4\overline{{v_{0} }} M}} + \frac{{bE\left( {N_{1} } \right)M}}{{4\overline{{v_{0} }} }}. \\ \end{aligned} $$

• (3) if $0\le b<p-\overline{{v }_{0}}+M$;

  $$ \begin{aligned} \mathop {\max }\limits_{b} E\left[ {\pi_{1}^{II} \left( b \right)} \right] & = \left( {p - w} \right)Q_{r}^{II} - \frac{{p\left( {\overline{{v_{0} }} - p} \right)}}{{\overline{{v_{0} }} }}\mathop \int \limits_{0}^{{{\Phi }^{ - 1} \left( {\frac{p - w}{p}} \right)}} \Phi \left( n \right)dn + E\left( {N_{1} } \right)\left( {p - 2b - w} \right)\frac{{\left( {\overline{{v_{0} }} - p + b} \right)}}{{\overline{{V_{0} }} }} \\ & \quad + \frac{{\left( {p - w} \right)E\left( {N_{1} } \right)\left( {M - b} \right)^{2} }}{{4\overline{{v_{0} }} M}} + b \frac{{3\left( {\overline{{v_{0} }} - p + b} \right)^{2} + 2\left( {\overline{{v_{0} }} - p + b} \right)M}}{{4M\overline{{v_{0} }} }} \\ \end{aligned} $$

Solve the above problem yields Proposition 5. Further, for the proof of ${b}_{rI}^{*}<{b}_{I}^{*}$. since $\frac{2p-w-\overline{{v }_{0}}}{2}>M$, we have $-\frac{p}{4}+\frac{w}{4}+\frac{M}{16}<\frac{7w-6p-\overline{{v }_{0}}}{32}<0$. Hence, we have ${b}_{rI}^{*}<{b}_{I}^{*}$.

This completes the proof.

Proof of Lemma 4

It is easy to prove that

$$ N_{1} \left( {\overline{{v_{0} }} - p + \alpha b} \right)(p - \left( {2 + \alpha } \right)b + 2M - \overline{{v_{0} }} - N_{1} \left( {M - b} \right)^{2} = - \left( {b\left( {1 + \alpha } \right) - M - p + \overline{v}_{0} } \right)^{2} < 0, $$

and ${N}_{1}{(M-\alpha b)}^{2}-{N}_{1}{\left(M-b\right)}^{2}>0,$ when $0<{\alpha }<1$.

Therefore, we have:

$$ N_{1} \left( {\overline{{v_{0} }} - p + \alpha b} \right)(p - \left( {2 + \alpha } \right)b + 2M - \overline{{v_{0} }} < N_{1} \left( {M - b} \right)^{2} < N_{1} \left( {M - \alpha b} \right)^{2} , $$

Recall that: ${D}_{c}=\frac{{N}_{1}\left(\overline{{v }_{0}}-p+\alpha b\right)(p-(2+\alpha )b+2M-\overline{{v }_{0}})}{4\overline{{v }_{0}}M}$, ${D}_{r}=\frac{{N}_{1}{(M-\alpha b)}^{2}}{4\overline{{v }_{0}}M}$. Thus, we have ${\mathrm{D}}_{\mathrm{c}}<{\mathrm{D}}_{\mathrm{r}}$.

Proof of Lemma 5

The proof is similar to the proof of Lemma 1, and we omit it.

Proof of Proposition 6

Notice that when $0<\alpha <1$ & $b>\frac{M}{\alpha }$ and ${\alpha }\ge 1$ & $b>M$, from $\frac{d{\pi }_{1}^{I}}{d\alpha }=0$, $\frac{d{\pi }_{1}^{I}}{db}=0$, we derive:

$$2\alpha b=2p-w-{\overline{v} }_{0}$$

When $0<\alpha <1$ and $M\le b<\frac{M}{\alpha }$, from $\frac{d{\pi }_{1}^{I}}{d\alpha }=0$, $\frac{d{\pi }_{1}^{I}}{db}=0$, we derive:

$$(4M-p+w)\alpha b=M(3p-w-2{\overline{v} }_{0})$$

Thus, under the above cases, the retailer actually jointly decides expansion rate and deposit. Namely, since the optimal final discount $\alpha b$ is a constant, a lower expansion rate will lead to a higher-level deposit.

Other parts of the proof are similar to the proof of Lemma 1, so we omit it.