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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.

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Acknowledgements

The authors thank the anonymous referees for their helpful comments and suggestions. This research is supported in part by National Natural Science Foundation of China (71771146). This paper is also a part of the project funded by Shanghai Pujiang Talent Project (16PJC059).

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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. (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. (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. (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.

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Zhang, Q., Cheng, X., Tsao, YC. et al. Advance selling under deposit expansion and consumer’s valuation change. Oper Res Int J (2021). https://doi.org/10.1007/s12351-021-00676-9

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Keywords

  • Advance selling
  • Deposit
  • Valuation change
  • Order cancellation
  • Return