Enhance financing for small- and medium-sized suppliers with reverse factoring: a game theoretical analysis

Abstract

Reverse factoring, a financial scheme in which established retailers facilitate financing for suppliers, is becoming an increasingly important tool in the industry. Normally, an SME supplier, a core retailer and a bank participate in the reverse factoring scheme. A three-level Stackelberg game is proposed in this study to investigate the interaction of the participants. The closed-form equilibria of the retailer’s replenishment decision, the supplier’s payment term decision and the bank’s financing decision are derived from the theoretical model. To our knowledge, this study is the first attempt which takes banks into account and endogenises their interest rates in the modelling of reverse factoring. The reverse factoring scheme is compared with commercial loans and traditional factoring. Compared to commercial loans, the introduction of factoring can lower credit risk, but fraud risk still exists. Reverse factoring solves this fraud problem and further decreases the financing cost for the supplier. Consequently, reverse factoring benefits the retailer through a significantly increased payment extension granted by the supplier. The numerical results also indicate that the utility of the bank significantly improves by 8–50% under varying levels of default risk compared with traditional factoring. Our study provides incentives and guidelines for supply chain participants to adopt such schemes when faced with capital constraints and the credit risk of the supplier.

Appendices

Appendices

Appendix A1: Proof of Proposition 1.

We use the method of backward induction to obtain the optimal solution. First, the retailer decides its own order quantity ${q}_{B}$ based on the existing payment term ${T}_{B}$. As f(x) follows a uniform distribution U[a, b], $f(x)=\frac{1}{b-a}$. At the same time, the retail price per unit p is normalised to 1, that is, p = 1.

Equation (1) can be rewritten as

$$ \begin{array}{*{20}c} {Max} \\ {q_{B} } \\ \end{array} U(\pi_{B}^{r} ) = \pi_{B}^{r} = \frac{{q_{B}^{2} - a^{2} }}{{2\left( {b - a} \right)}} + \frac{{\left( {b - q_{B} } \right)q_{B} }}{b - a} - wq_{B} + wq_{B} T_{B} \theta $$

Taking the first derivative of ${q}_{B}$ in Eq. (1), we have:

$$\frac{{d\pi }_{B}^{r}}{{dq}_{B}}=\frac{{q}_{B}}{b-a}+\frac{b-2{q}_{B}}{b-a}-w+w{T}_{B}\theta $$

(A.1)

Taking the second derivative of ${q}_{B}$ in Eq. (1), we have:

$$\frac{{d}^{2}{\pi }_{B}^{r}}{d{q}_{B}^{2}}=-\frac{1}{b-a}$$

(A.2)

As $\frac{{d}^{2}{\pi }_{B}^{r}}{d{q}_{B}^{2}}=-\frac{1}{b-a}$ <0, the objective function ${\pi }_{B}^{r}$ is concave in ${q}_{B}$.

• A. When a supplier has no financial constraint, we can obtain the optimal solution of ${q}_{B}$

by solving $\frac{{d\pi }_{B}^{r}}{{dq}_{B}}=0$:

$$q_B^* = {q_{BU}} = b - w\left( {1 - {T_B}\theta } \right)(b - a)$$

(A.3)

To make the problem nontrivial, we suppose T≧ 0. Under this condition, the minimum order quantity of the retailer is q_b^min=b–w(b–a)≦ b–w(1–T_Bθ)(b–a). The retailer will choose the minimum order quantity when the supplier does not grant it any payment term, that is, ${T}_{B}$=0.

• B. When the supplier has a financial constraint but ${cq}_{B}^{min}$<L, i.e., c(b–w(b–a)) < L, the supplier is able to meet the minimum order quantity of the retailer. The supplier can still set up a ${T}_{B}$>0 to stimulate the demand from the retailer. The optimal solution is ${q}_{B}^{*}$=${q}_{BL}=$ b–w(1–${T}_{B}$ θ)(b–a), obtained by solving $\frac{{d\pi }_{B}^{r}}{{dq}_{B}}=0$.

• C. When a supplier has a financial constraint and ${cq}_{B}^{min}$>L, that is, c(b–w(b–a)) > L, the supplier is unable to meet the minimum order quantity ${q}_{B}^{min}$ of the retailer. The order quantity that the supplier can meet is no more than L/c. It is obvious that the retailer’s profit is increasing in [0, L/c]. Hence, the order quantity of the retailer in this circumstance is ${q}_{B}^{*}$=${q}_{BLL}$=L/c, and ${q}_{BLL}$<${q}_{B}^{min}$<${q}_{BL}$.

Second, the supplier decides its payment term for the retailer while considering the impact of its decision on the retailer.

Taking the first derivative of ${T}_{B}$ in Eq. (2), we have:

$$\frac{{d\pi }_{B}^{s}}{{dT}_{B}}=(w-c)*\frac{{dq}_{B}}{{dT}_{B}}-w\alpha (\frac{{dq}_{B}}{{dT}_{B}}*{T}_{B}+{q}_{B})$$

(A.4)

Taking the second derivative of ${T}_{B}$ in Eq. (2), we have:

$$\frac{{d}^{2}{\pi }_{B}^{s}}{d{T}_{B}^{2}}=(w-c)\frac{{d}^{2}{q}_{B}}{{dT}_{B}^{2}}-w\alpha (\frac{{d}^{2}{q}_{B}}{{dT}_{B}^{2}}*{T}_{B}+\frac{2{dq}_{B}}{{dT}_{B}})=\frac{{dq}_{B}}{{dT}_{B}}*(-2w\alpha )$$

(A.5)

As $\frac{{d}^{2}{\pi }_{B}^{s}}{d{T}_{B}^{2}}=\frac{{dq}_{B}}{{dT}_{B}}*(-2w\alpha )<0$, the objective function ${\pi }_{B}^{s}$ is concave in ${T}_{B}$.

• A. When there is no financial constraint, solving $\frac{{d\pi }_{B}^{s}}{{dT}_{B}}=0$ gives us:

$${{T}_{B}^{*}=T}_{BU}=\frac{w-c}{2w\alpha }+\frac{1}{2\theta }-\frac{b}{(b-a)*2w\theta }$$

(A.6)

At the same time, the order quantity of the retailer is $q_{B}^{*} = q_{BU} = b - w\left( {1 - T_{BU} \theta } \right)\left( {b - a} \right)$.

For ${T}_{B}^{*}\geqq 0$, the condition $\alpha \leqq \frac{(b-a)\theta (w-c)}{b-(b-a)w}$ should be met. This implies that the interest rate on commercial loans cannot be too high; otherwise, it will be too costly for the supplier to set up any credit term.

• B. When there is a financial constraint but c ${q}_{B}^{min}$<L, i.e., c(b–w(b–a)) < L, the optimal solution remains as ${q}_{B}^{*}$=${q}_{BL}$=b–w(1–${T}_{B}$ θ)(b–a); however, the supplier does not have enough funds to afford the original optimal payment term ${T}_{BU}=\frac{w-c}{2w\alpha }+\frac{1}{2\theta }-\frac{b}{(b-a)*2w\theta }$. We denote the maximum payment term that the supplier can afford in this circumstance as T_BL. In this case, ${T}_{BL}<{T}_{BU}$. It is obvious that the supplier’s profit is increasing in [0, T_BL]. Hence, the optimal payment term ${T}_{B}^{*}$ in this circumstance would be T_BL.

The production cost of the supplier is equal to cq_BL + wq_BLT_BLα. By solving cq_BL + wq_BLT_BLα = L, we have:

$$\begin{aligned}&T_B^* = {T_{BL}} \\ &\quad= \frac{{ - b\alpha w {-} a\alpha {w^2} {+} b\alpha {w^2} {+} acw\theta {-} bcw\theta {+} \sqrt {{{\left( {b\alpha w {+} a\alpha {w^2} {-} b\alpha {w^2} {-} acw\theta {+} bcw\theta } \right)}^2} {-} 4\left( {bc {-} L {+} acw {-} bcw} \right)\left( { - a\alpha {w^2}\theta {+} b\alpha {w^2}\theta } \right)} }}{{2\left( { - a\alpha {w^2}\theta { +} b\alpha {w^2}\theta } \right)}}\end{aligned}$$

(A.7)

At the same time, the order quantity of the retailer is ${q}_{B}^{*}$=q_BL=b–w(1–${T}_{BL}$ θ)(b–a).

In addition, ${q}_{B}^{*}$=q_BL=b–w(1–${T}_{BL}$ θ)(b–a) < b–w(1–${T}_{BU}$ θ)(b–a) = q_BU.

• C. When there is a financial constraint and ${cq}_{B}^{min}$>L, i.e., c(b–w(b–a)) > L, the supplier is unable to meet the minimum order quantity of the retailer and ${q}_{B}^{*}$=q_BLL < ${q}_{B}^{min}$<q_BL<q_BU. There is no reason for the supplier to set up any payment term to stimulate the order quantity. Hence, we have $T_{B}^{*} = T_{BLL} = 0,$ and ${T}_{BLL}$<${T}_{BL}$<${T}_{BU}$.

Appendix A2: Proof of Proposition 2

We use the method of backward induction to obtain the optimal solution. First, the retailer decides on its order quantity based on the payment term ${T}_{TF}$. In this circumstance, there is no financial constraint. Similar to the proof of proposition 1, by solving $\frac{{d\pi }_{TF}^{r}}{{dq}_{TF}}$=0, we obtain the optimal solution q^*_TF(T_TF)=b–w(1–T_TFθ)(b–a).

Next, by solving $\frac{{d\pi }_{TF}^{s}}{{dT}_{TF}}=0$, we have: $T_{TF}^{*} = \frac{w - c}{{2wr_{TF}^{*} }} + \frac{1}{2\theta } - \frac{b}{{\left( {b - a} \right)*2w\theta }}$.

Since we assume $T_{TF}^{*}$≧0, the condition $r_{TF} { \leqq }\frac{{\left( {b - a} \right)\theta \left( {w - c} \right)}}{{b - \left( {b - a} \right)w}}$ should be met.

Finally, the bank decides on ${r}_{TF}$. Taking the first derivative of Eq. (3), we have:

$$ \frac{{d lnU( \pi_{TF}^{b} )}}{dr} = \frac{1}{{q_{TF} }}*\frac{{dq_{TF} }}{{dr_{TF} }} + \frac{1}{{T_{TF} }}*\frac{{dT_{TF} }}{{dr_{TF} }} + \frac{\eta }{{\eta r_{TF} - \left( {\eta + \lambda - \lambda \eta } \right)\theta }} $$

$$ \begin{aligned} \frac{{d lnU( \pi_{TF}^{b} )}}{dr} &= \frac{1}{{q_{TF} }}*\frac{{dq_{TF} }}{{dr_{TF} }} + \frac{1}{{T_{TF} }}*\frac{{dT_{TF} }}{{dr_{TF} }} + \frac{\eta }{{\eta r_{TF} - \left( {\eta + \lambda - \lambda \eta } \right)\theta }} \\ &= \frac{{\left( {b - a} \right)\theta \left( {w - c} \right)}}{{rb - r\left( {b - a} \right)w + \left( {w - c} \right)\left( {b - a} \right)\theta }}*\frac{1}{ - r} \\ &+ \frac{{\left( {b - a} \right)\theta \left( {w - c} \right)}}{{r\left( {b - a} \right)w + \left( {w - c} \right)\left( {b - a} \right)\theta - br}}*\frac{1}{ - r} + \frac{\eta }{{\eta r - \left( {\eta + \lambda - \lambda \eta } \right)\theta }} \\ \end{aligned} $$

(A.8)

It can be proved that there exists one point x. $\frac{d{ lnU( \pi }_{TF}^{b})}{dr}$>0 when ${r}_{TF}$ is inside [0, x], $\frac{d{ lnU( \pi }_{TF}^{b})}{dr}$<0 when ${r}_{TF}$ is inside [x,$\frac{(b-a)\theta (w-c)}{b-(b-a)w}$], under the condition of $\frac{(b-a)(w-c){/[\eta /(\eta +\lambda -\lambda \eta )]}^\frac{1}{2}}{b-(b-a)w}>1$.

By solving $\frac{d{ lnU( \pi }_{TF}^{b})}{dr}=$ 0, we have

$$\begin{array}{ccccc} r_{TF}^* = & \sqrt[3]{{\frac{{{u^2}\theta }}{{\eta {k^2}/\left( {\eta + \lambda - \lambda \eta } \right)}} + \sqrt {\frac{{{u^4}{\theta ^2}}}{{{\eta ^2}{k^4}/{{\left( {\eta + \lambda - \lambda \eta } \right)}^2}}} + \frac{{{u^6}}}{{27{k^6}}}} }}\\ \sqrt[3]{{\frac{{{u^2}\theta }}{{\eta {k^2}/\left( {\eta + \lambda - \lambda \eta } \right)}} - \sqrt {\frac{{{u^4}{\theta ^2}}}{{{\eta ^2}{k^4}/{{\left( {\eta + \lambda - \lambda \eta } \right)}^2}}} + \frac{{{u^6}}}{{27{k^6}}}} }} \end{array}$$

(A.9)

where.

It can be proved that ${0\leqq r}_{TF}^{*}\leqq \frac{(b-a)\theta (w-c)}{b-(b-a)w}$ at the current condition. Therefore, ${r}_{TF}^{*}$ is the equilibrium result.

Similar to the proof of proposition 1, we can obtain:

$$\begin{array}{ccccc} T_{TF}^*\left( {r_{TF}^*} \right) = & \frac{{w - c}}{{2wr_{TF}^*}} + \frac{1}{{2\theta }} - \frac{b}{{\left( {b - a} \right)*2w\theta }};q_{TF}^*\left( {r_{TF}^*} \right)\\ = \frac{{r_{TF}^*b - \left( {b - a} \right)wr_{TF}^* + \left( {b - a} \right)\theta \left( {w - c} \right)}}{{2r_{TF}^*}} \end{array}$$

“Appendix A3”: Proof of Proposition 4

$$\begin{aligned}A.r_{TF}^* &= \sqrt[3]{{\frac{{{u^2}\theta }}{{\eta {k^2}/\left( {\eta + \lambda - \lambda \eta } \right)}} + \sqrt {\frac{{{u^4}{\theta ^2}}}{{{\eta ^2}{k^4}/{{\left( {\eta + \lambda - \lambda \eta } \right)}^2}}} + \frac{{{u^6}}}{{27{k^6}}}} }} \\ &\quad+ \sqrt[3]{{\frac{{{u^2}\theta }}{{\eta {k^2}/\left( {\eta + \lambda - \lambda \eta } \right)}} - \sqrt {\frac{{{u^4}{\theta ^2}}}{{{\eta ^2}{k^4}/{{\left( {\eta + \lambda - \lambda \eta } \right)}^2}}} + \frac{{{u^6}}}{{27{k^6}}}} }}\end{aligned}$$

$$ r_{RF}^{*} = \sqrt[3]{{\frac{{u^{2} \theta }}{{k^{2} }} + \sqrt {\frac{{u^{4} \theta^{2} }}{{k^{4} }} + \frac{{u^{6} }}{{27k^{6} }}} }} + \sqrt[3]{{\frac{{u^{2} \theta }}{{k^{2} }} - \sqrt {\frac{{u^{4} \theta^{2} }}{{k^{4} }} + \frac{{u^{6} }}{{27k^{6} }}} }} $$

As $\eta$<1 and $\lambda$<1, it is obvious that $\eta /(\eta +\lambda -\lambda \eta )$ <1.

As $\eta /(\eta +\lambda -\lambda \eta )$<1, it is obvious that ${r}_{TF}^{*}$>${r}_{RF}^{*}$.

$$\begin{array}{ccccc} B.As\,r_{TF}^* > r_{RF}^*,T_{TF}^* = & \frac{{w - c}}{{2wr_{TF}^*}} + \frac{1}{{2\theta }} - \frac{b}{{\left( {b - a} \right)*2w\theta }}\\ < \frac{{w - c}}{{2wr_{RF}^*}} + \frac{1}{{2\theta }} - \frac{b}{{\left( {b - a} \right)*2w\theta }}\\ = T_{RF}^* \end{array}$$

$$ C.\,AsT_{TF}^{*}\! <\! T_{RF}^{*} ,q_{TF}^{*}\!=\! b - w(1 - T_{TF} *\theta )(b - a) \!<\! b - w(1 - T_{RF} *\theta )(b \!-\! a) \!=\! q_{RF}^{*} $$

$$ \begin{aligned} D.\,\pi_{RF}^{b} \left( {r_{RF}^{*} } \right) &= wq_{RF} \left( {r_{RF}^{*} } \right)T_{RF} \left( {r_{RF}^{*} } \right)\left( {r_{RF}^{*} - \theta } \right) \\ &> \pi_{RF}^{b} \left( {r_{TF}^{*} } \right) = wq_{RF} \left( {r_{TF}^{*} } \right)T_{RF} \left( {r_{TF}^{*} } \right)\left( {r_{TF}^{*} - \theta } \right) \\ &> wq_{RF} \left( {r_{TF}^{*} } \right)T_{RF} \left( {r_{TF}^{*} } \right)[\eta r_{TF}^{*} - (\eta + \lambda - \lambda \eta )\theta ] \\ &= wq_{TF} \left( {r_{TF}^{*} } \right)T_{TF} \left( {r_{TF}^{*} } \right)[\eta r_{TF}^{*} - (\eta + \lambda - \lambda \eta )\theta ] \\ &= \pi_{TF}^{b} \left( {r_{TF}^{*} } \right) \\ \end{aligned} $$

$$ \begin{aligned} E.\pi_{RF}^{s} \left( {T_{RF}^{*} } \right) &= wq_{RF} \left( {T_{RF}^{*} } \right) - cq_{RF} \left( {T_{RF}^{*} } \right) - wq_{RF} \left( {T_{RF}^{*} } \right)T_{RF}^{*} r_{RF}^{*} \\ &> wq_{RF} \left( {T_{TF}^{*} } \right) - cq_{RF} \left( {T_{TF}^{*} } \right) - wq_{RF} \left( {T_{TF}^{*} } \right)T_{TF}^{*} r_{RF}^{*} \\ &> wq_{RF} \left( {T_{TF}^{*} } \right) - cq_{RF} \left( {T_{TF}^{*} } \right) - wq_{RF} \left( {T_{TF}^{*} } \right)T_{TF}^{*} r_{TF}^{*} \\ &= wq_{TF} \left( {T_{TF}^{*} } \right) - cq_{TF} \left( {T_{TF}^{*} } \right) - wq_{TF} \left( {T_{TF}^{*} } \right)T_{TF}^{*} r_{TF}^{*} = \pi_{TF}^{s} \left( {T_{TF}^{*} } \right) \\ \end{aligned} $$

Appendix A4: Proof of Proposition 5

q_BLL < q_BL < q_BU and T_BLL < T_BL < T_BU is already proved in proposition 1.

$$\begin{array}{ccccc} A.As\;\alpha > r_{RF}^*,{T_{BU}} = & \frac{{w - c}}{{2w\alpha }} + \frac{1}{{2\theta }} - \frac{b}{{\left( {b - a} \right)*2w\theta }}\\ < \frac{{w - c}}{{2wr_{RF}^*}} + \frac{1}{{2\theta }} - \frac{b}{{\left( {b - a} \right)*2w\theta }}\\ = T_{RF}^* \end{array}$$

B. As T_BU <  T_RF^*, q_BU = b–w(1–T_BUθ)(b–a) < b–w(1–T_RF^*θ)(b–a) =  q_RF^*.

As α>${r}_{RF}^{*},$ T_BU < ${T}_{RF}^{*}$, and q_BU${<q}_{RF}^{*}$, we have.

$$ \begin{aligned} C.\,\pi_{RF}^{s} \left( {T_{RF}^{*} } \right) & = wq_{RF} \left( {T_{RF}^{*} } \right) - cq_{RF} \left( {T_{RF}^{*} } \right) - wq_{RF} \left( {T_{RF}^{*} } \right)T_{RF}^{*} r_{RF}^{*} \\ &> wq_{RF} \left( {T_{BU} } \right) - cq_{RF} \left( {T_{BU} } \right) - wq_{RF} \left( {T_{BU} } \right)T_{BU} r_{RF}^{*} \\ &> wq_{RF} \left( {T_{BU} } \right) - cq_{RF} \left( {T_{BU} } \right) - wq_{RF} \left( {T_{BU} } \right)T_{BU} \alpha \\& = wq_{BU} \left( {T_{BU} } \right) - cq_{BU} \left( {T_{BU} } \right) - wq_{BU} \left( {T_{BU} } \right)T_{BU} \alpha = \pi_{BU}^{s} \left( {T_{BU} } \right) \\ \end{aligned} $$

It is obvious that ${\pi }_{B}^{s}$ is increasing in [0, ${T}_{BU}$]. Hence, ${\pi }_{BU}^{s*}$>${\pi }_{BL}^{s*}$>${\pi }_{BLL}^{s*}$.

It is obvious that ${\pi }_{B}^{r}$ is increasing in [0, ${q}_{BU}$]. Hence, ${\pi }_{BU}^{r*}$>${\pi }_{BL}^{r*}$>${\pi }_{BLL}^{r*}$.