Online retailer’s optimal financing strategy in an online marketplace

Abstract

This paper studies an e-commerce supply chain, where a capital-constrained online retailer sells products through an e-commerce platform and may finance from the commercial bank, the upstream supplier as well as the e-commerce platform to relieve its capital distress. In the presence of bankruptcy risk for the online retailer, we establish a three-echelon Stackelberg game model, and characterise the optimal decision for each supply chain player. We examine the retailer’s financing strategy by comparing profits under three different financing schemes. Compared with existing literature, we find that in contrast to trade credit and e-commerce platform financing, the retailer has an incentive to finance from the bank. Also, the e-commerce platform is not always willing to provide financing service to the retailer when the commission rate is high. Furthermore, we investigate the impacts of the commission rate and the retailer’s initial capital on the equilibrium results. The results show that the interest rates of the bank and the e-commerce platform decrease with the commission rate. In contrast to the bank, the e-commerce platform generally charges a lower interest rate. In particular, when the commission rate exceeds a certain threshold, the e-commerce platform will provide free financing services to the retailer.

Appendices

Appendix 1

Proof of Lemma 1

The retailer’s profit can be rewritten as $\mathop {\max }\limits_{{q^{B} }} \pi_{R}^{B} = \left( {1 - \lambda } \right)\int_{{\Delta_{1} }}^{{q^{B} }} {\overline{F}\left( D \right)dD - K}$. Therefore, we have $\frac{{\partial \pi_{R}^{B} }}{{\partial q^{B} }} = \left( {1 - \lambda } \right)\left[ {\overline{F}\left( {q^{B} } \right) - \overline{F}\left( {{\Delta }_{1} } \right)\frac{{\partial {\Delta }_{1} }}{{\partial q^{B} }}} \right]$, where $\frac{{\partial {\Delta }_{1} }}{{\partial q^{B} }} = \frac{{w^{B} \left( {1 + r^{B} } \right)}}{1 - \lambda }$. According to the first order condition, we obtain $\overline{F}\left( {q^{B*} } \right) - {\Omega }_{1} \overline{F}\left( {{\Delta }_{1} } \right) = 0$, where ${\Omega }_{1} = \frac{{w^{B} \left( {1 + r^{B} } \right)}}{1 - \lambda }$. Further, $\left. {\frac{{\partial^{2} \pi_{R}^{B} }}{{\partial (q^{B} )^{2} }}} \right|_{{q^{B} = q^{B*} }} = \left( {1 - \lambda } \right)\Omega_{1} \overline{F}\left( {\Delta_{1} } \right)\left[ {\Omega_{1} h\left( {\Delta_{1} } \right) - h\left( {q^{B*} } \right)} \right]$. Since ${\Omega }_{1} < 1$, ${\Delta }_{1} < q^{B*}$ and $h\left( \cdot \right)$ is an increasing function, then $\left. {\frac{{\partial^{2} \pi_{R}^{B} }}{{\partial (q^{B} )^{2} }}} \right|_{{q^{B} = q^{B*} }} < 0$. Therefore, $q^{B*}$ is the unique optimal solution. $\square$.

Appendix 2

Proof of Proposition 1

• (i) From Lemma 1, we have $\frac{\partial {q}^{B*}}{\partial K}=\frac{h\left({\Delta }_{1}\right)\left(1+{r}^{B}\right)}{{\Omega }_{1}h\left({\Delta }_{1}\right)-h\left({q}^{B*}\right)}$. Since ${\Omega }_{1}=\frac{{w}^{B}{q}^{B*}\left(1+{r}^{B}\right)}{1-\lambda }<1$, ${\Delta }_{1}=\frac{({w}^{B}{q}^{B*}-K)(1+{r}^{B})}{1-\lambda }<{q}^{B*}$ and $h\left(\bullet \right)$ is an increasing function, then $\frac{\partial {q}^{B*}}{\partial K}<0$.

• (ii) According to Lemma 1, we have $\frac{\partial {q}^{B*}}{\partial {w}^{B}}=\frac{1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)}{{w}^{B}\left[{\Omega }_{1}h\left({\Delta }_{1}\right)-h\left({q}^{B*}\right)\right]}$. Note that $\frac{\partial (\overline{F}\left(Q\right)Q)}{\partial Q}=\overline{F}\left(Q\right)[1-H(Q)]$. Since $H(Q)$ increases with $Q$, then $\overline{F}\left(Q\right)Q$ is a concave function in $Q$ and ${Q}^{*}$ maximizes $\overline{F}\left(Q\right)Q$, where ${Q}^{*}$ holds $1-H\left({Q}^{*}\right)=0$. Since $\overline{F}\left({q}^{B*}\right)={\Omega }_{1}\overline{F}\left({\Delta }_{1}({q}^{B*})\right)>{\Omega }_{1}\overline{F}({\Omega }_{1}{q}^{B*})$, then we have $\overline{F}\left({q}^{B*}\right){q}^{B*}>{\Omega }_{1}{q}^{B*}\overline{F}({\Omega }_{1}{q}^{B*})$. Therefore, ${\Omega }_{1}{q}^{B*}<{Q}^{*}$ and $1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)>1-H\left({\Omega }_{1}{q}^{B*}\right)>1-H\left({Q}^{*}\right)=0$. Since ${\Omega }_{1}h\left({\Delta }_{1}\right)<h\left({q}^{B*}\right)$, then $\frac{\partial {q}^{B*}}{\partial {w}^{B}}<0$.

• (iii) According to Lemma 1, we have $\frac{\partial {q}^{B*}}{\partial {r}^{B}}=\frac{{w}^{B}-{\Omega }_{1}\left({w}^{B}{q}^{B*}-K\right)h\left({\Delta }_{1}\right)}{\left(1-\lambda \right){\Omega }_{1}[{\Omega }_{1}h\left({\Delta }_{1}\right)-h\left({q}^{B*}\right)]}$. Since ${w}^{B}-$ ${\Omega }_{1}\left({w}^{B}{q}^{B*}-K\right)h\left({\Delta }_{1}\right)>{w}^{B}\left[1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)\right]>0$ and ${\Omega }_{1}h\left({\Delta }_{1}\right)-h\left({q}^{B*}\right)<0$, we have $\frac{\partial {q}^{B*}}{\partial {r}^{B}}$ $<0$. $\square$

Appendix 3

Proof of Proposition 2

We prove this by contradiction. Suppose ${q}^{N*}\ge {q}^{B*}$ for any $K$, then we have ${q}^{N*}\ge {q}^{B*}{|}_{K=0}=\widehat{q}$. Therefore, $\overline{F}\left(\widehat{q}\right)\widehat{q}={\Omega }_{1}\widehat{q}\overline{F}({\Omega }_{1}\widehat{q})$. According to the Proof of Proposition 1, we obtain ${\Omega }_{1}\widehat{q}<{Q}^{*}<\widehat{q}$ and $1-H\left({\Omega }_{1}\widehat{q}\right)>1-H\left({Q}^{*}\right)=0>1-H\left(\widehat{q}\right)$. As ${q}^{N*}$ holds $\overline{F}\left({q}^{N*}\right)-f\left({q}^{N*}\right){q}^{N*}-\frac{c}{1-\lambda }=0$, we obtain $1-H\left({q}^{N*}\right)=\frac{c}{\left(1-\lambda \right)\overline{F}\left({q}^{N*}\right)}>0$. Therefore, $H\left(\widehat{q}\right)>H\left({q}^{N*}\right)$, which implies $\widehat{q}>{q}^{N*}$. This is a contradiction to ${q}^{N*}\ge \widehat{q}$. Hence, we obtain ${q}^{N*}<{q}^{B*}$. $\square$

Appendix 4

Proof of Proposition 3

• (i) Since ${w}^{B}{q}^{B*}>K$, then ${q}^{B*}>\frac{K}{{w}^{B}}={q}^{NF*}$.

• (ii) $\frac{\partial {\pi }_{R}^{B*}}{\partial {r}^{B}}=\left(1-\lambda \right)[\overline{F}\left({q}^{B*}\right)\frac{\partial {q}^{B*}}{\partial {r}^{B}}-\overline{F}\left({\Delta }_{1}\right)\frac{\partial {\Delta }_{1}}{\partial {r}^{B}}]=-\left(1-\lambda \right)\overline{F}\left({\Delta }_{1}\right)\Delta ({q}^{B*})<0$.

• (iii) Since $\frac{\partial ({\pi }_{R}^{B*}-{\pi }_{R}^{NF*})}{\partial {r}^{B}}=-\left(1-\lambda \right)\overline{F}\left({\Delta }_{1}\right)\Delta ({q}^{B*})<0$, then there exists a unique $\overline{r}$ such that ${\pi }_{R}^{B*}\ge {\pi }_{R}^{NF*}$ if ${r}^{B}\le \overline{r}$, where $\overline{r}$ satisfies ${\pi }_{R}^{B*}-{\pi }_{R}^{NF*}=0$, i.e., ${\int }_{{\Delta }_{1} }^{{q}^{B*}(\overline{r}) }\overline{F}(D)dD-$ ${\int }_{0 }^{\frac{K}{{w}^{B}}}\overline{F}\left(D\right)dD=0$.$\square$

Appendix 5

Proof of Lemma 2

The bank’s profit function can be rewritten as ${\pi }_{B}^{B}=\left(1-\lambda \right){\int }_{0 }^{{\Delta }_{1}}\overline{F}\left(D\right)dD-\left({w}^{B}{q}^{B*}-K\right)$, then $\frac{\partial {\pi }_{B}^{B}}{\partial {r}^{B}}=\overline{F}\left({\Delta }_{1}\right){w}^{B}\left(1+{r}^{B}\right)\frac{{w}^{B}-\left({w}^{B}{q}^{B*}-K\right)h\left({q}^{B*}\right)}{\left(1-\lambda \right){\Omega }_{1}[{\Omega }_{1}h\left({\Delta }_{1}\right)-h\left({q}^{B*}\right)]}-{w}^{B}\frac{\partial {q}^{B*}}{\partial {r}^{B}}={w}^{B}\frac{\partial {q}^{B*}}{\partial {r}^{B}}[\overline{F}\left({\Delta }_{1}\right){w}^{B}\left(1+{r}^{B}\right){\xi }_{1}\left({r}^{B}\right)-1]$, where ${\xi }_{1}\left({r}^{B}\right)=\frac{{w}^{B}-\left({w}^{B}{q}^{B*}-K\right)h\left({q}^{B*}\right)}{{w}^{B}-{\Omega }_{1}\left({w}^{B}{q}^{B*}-K\right)h\left({\Delta }_{1}\right)}<1$. From Proposition 1, we have $\frac{\partial {q}^{B*}}{\partial {r}^{B}}<0$. Therefore, the signal of $\frac{\partial {\pi }_{B}^{B}}{\partial {r}^{B}}$ is related to $\overline{F}\left({\Delta }_{1}\right){w}^{B}\left(1+{r}^{B}\right){\xi }_{1}\left({r}^{B}\right)-1$. Based on the form of $\frac{\partial {\pi }_{B}^{B}}{\partial {r}^{B}}$, as in Jing et al. [35] and Cao et al. [62], we show that $\overline{F}\left({\Delta }_{1}\right){w}^{B}\left(1+{r}^{B}\right){\xi }_{1}\left({r}^{B}\right)-1$ increases with ${r}^{B}$, and we derive that ${\pi }_{B}^{B}$ is a monotone or unimodal function of ${r}^{B}$ for ${r}^{B}\in [0,\overline{r}]$. Since ${w}^{B}\le p=1$, then $\overline{F}\left({\Delta }_{1}\right){w}^{B}\left(1+{r}^{B}\right){\xi }_{1}\left({r}^{B}\right)-1=\overline{F}\left({\Delta }_{1}\right){w}^{B}{\xi }_{1}\left({r}^{B}\right)-1<0$ if ${r}^{B}=0$, which indicates ${r}^{B*}>0$. Therefore, if ${r}_{1}\in \left(0,\overline{r}\right)$, then ${r}^{B*}={r}_{1}$; if ${r}_{1}\ge \overline{r}$, then ${r}^{B*}=\overline{r}$. $\square$

Appendix 6

Proof of Lemma 3

First, $\frac{\partial {\pi }_{S}^{B}}{\partial {w}^{B}}=\frac{{w}^{B}\left[1-H\left({q}^{B*}\right)\right]-c\left[1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)\right]}{{w}^{B}\left[{\Omega }_{1}h\left({\Delta }_{1}\right)-h\left({q}^{B*}\right)\right]}=\frac{\partial {q}^{B*}}{\partial {w}^{B}}\{\frac{{w}^{B}\left[1-H\left({q}^{B*}\right)\right]}{1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)}-c\}$, where $\frac{\partial {q}^{B*}}{\partial {w}^{B}}<0$ (see Proposition 1). Let ${\eta }_{1}\left({w}^{B}\right)=\frac{1-H\left({q}^{B*}\right)}{1-\Omega {q}^{B*}h\left({\Delta }_{1}\right)}$, then the sign of $\frac{\partial {\pi }_{S}^{B}}{\partial {w}^{B}}$ is related to that of ${\eta }_{1}\left({w}^{B}\right)-c$. Based on the formulation of ${\eta }_{1}\left({w}^{B}\right)-c$, according to the proof of Proposition 6 in Jing et al. [35], we have $\frac{\partial {\eta }_{1}\left({w}^{B}\right)}{\partial {w}^{B}}>0$. Hence, $\frac{{w}^{B}[1-H({q}^{B*})]}{1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)}$ is also an increasing function of ${w}^{B}$. Assume that ${w}_{1}$ holds $\frac{{w}_{1}[1-H({q}^{B*})]}{1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)}-c=0$, then ${w}_{1}=\frac{c\left[1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)\right]}{1-H\left({q}^{B*}\right)}$. Since $1-{\Omega }_{1}{q}^{B*}h\left({\Delta }_{1}\right)>1-H\left({q}^{B*}\right)$, then we have ${w}_{1}>c$. Hence, if ${w}_{1}\ge \frac{1-\lambda }{1+\overline{r}}$, then ${\pi }_{S}^{B}$ is a decreasing function with respect to ${w}^{B}$ and the optimal wholesale price is ${w}^{B*}=\frac{1-\lambda }{1+\overline{r}}$; if ${w}_{1}<\frac{1-\lambda }{1+\overline{r}}$, then ${\pi }_{S}^{B}$ is concave in ${w}^{B}$, and the optimal wholesale price is ${w}^{B*}={w}_{1}$.$\square$

Appendix 7

Proof of Proposition 4

The proof of Proposition 4 is similar to that of Proposition 1, and we omit it. $\square$

Appendix 8

Proof of Lemma 4

The proof of Lemma 4 is similar to Zhan et al. [61], and hence we omit it. $\square$

Appendix 9

Proof of Proposition 5

The proof of Proposition 5 is similar to that of Proposition 1, and we omit it. $\square$

Appendix 10

Proof of Lemma 5

The e-commerce platform’s profit function can be written as $\pi_{E}^{E} = \lambda \int_{0 }^{{q^{E*} }} {\overline{F}\left( D \right)dD + \left( {1 - \lambda } \right)\int_{0 }^{{\Delta_{3} }} {\overline{F}\left( D \right)dD - \left( {w^{E} q^{E*} - K} \right)} }$, then we have $\frac{\partial {\pi }_{E}^{E}}{\partial {r}^{E}}={w}^{E}\frac{\partial {q}^{E*}}{\partial {r}^{E}}\left[\varphi \left({r}^{E}\right)-1\right]$ and$\varphi \left({r}^{E}\right)=\frac{\left[\lambda \overline{F}\left({q}^{E*}\right)+{w}^{E}\left(1+{r}^{E}\right)\overline{F}\left({\Delta }_{3}\right)\right]\left[1-H\left({\Delta }_{3}\right)\right]+\left(1-\lambda \right){\Delta }_{3}\overline{F}\left({\Delta }_{3}\right)\left[{\Omega }_{3}h\left({\Delta }_{3}\right)-h\left({q}^{E*}\right)\right]}{{w}^{E}\left[1-H\left({\Delta }_{3}\right)\right]}$. Following the proof of Proposition 1, we obtain$\frac{\partial {q}^{E*}}{\partial {r}^{E}}<0$. Therefore, the signal of $\frac{\partial {\pi }_{E}^{E}}{\partial {r}^{E}}$ depends on$\varphi \left({r}^{E}\right)-1$. Based on the form of$\frac{\partial {\pi }_{E}^{E}}{\partial {r}^{E}}$, as in [40, 41] and Cao et al. [62], we derive that ${\pi }_{E}^{E}$ is a monotone or unimodal function of ${r}^{E}$ for${r}^{E}\in \left[0,\widehat{r}\right]$. If${r}_{3}\le 0$, then${r}^{E*}=0$; if${r}_{3}\in \left(0,\widehat{r}\right)$, then${r}^{E*}={r}_{3}$; if${r}_{3}\ge \widehat{r}$, then${r}^{E*}=\widehat{r}$.

Appendix 11

Proof of Lemma 6

The proof of Lemma 6 is similar to that of Lemma 3, and hence we omit it.$\square$