Incentive mechanism to prevent moral hazard in online supply chain finance

Abstract

With e-commerce developing rapidly, banks have begun to cooperate with online platform operators to finance small and medium-sized enterprises (SMEs). However, this process engenders its own unique financial risks. This study highlights and investigates the risks in a four-party supply chain that include a third-party logistics provider, a bank, a B2B platform operator, and SMEs. In an asymmetric information setting, the collusion mechanisms in this four-party online supply chain are also explored. Subsequently, a two-part incentive contract is designed that can reduce the moral hazard faced by the banks while addressing the trade-off between the payments to the platform operator for better credit rating information and the payments to the third-party logistics provider for supervising collateral storage. For further confirmation, a numerical analysis is presented. The results indicate that based on a suitable capital coefficient, the two-part incentive contract may prevent moral hazard in online supply chains. Furthermore, when the line of credit is high, the bank must increase the incentives for the B2B platform operator to avoid default risk and decrease the incentives for 3PL.

Appendix 1: Proofs

Proof of Lemma 1

$$\begin{aligned} E(e^{\gamma \varepsilon } ) & = \frac{1}{{\sqrt {2\pi } \sigma }}\int_{ - \infty }^{ + \infty } {e^{\gamma \varepsilon } } e^{{ - \frac{{\varepsilon^{2} }}{{2\sigma^{2} }}}} {\text{d}}\varepsilon \\ & = \frac{1}{{\sqrt {2\pi } \sigma }}\int_{ - \infty }^{ + \infty } {e^{{ - \frac{{\varepsilon^{2} - \gamma \varepsilon^{2} \sigma^{2} }}{{2\sigma^{2} }}}} } {\text{d}}\varepsilon \\ & = \frac{1}{{\sqrt {2\pi } \sigma }}\int_{ - \infty }^{ + \infty } {e^{{ - \frac{{(\varepsilon^{2} - \gamma \sigma^{2} )^{2} - \gamma^{2} \sigma^{4} }}{{2\sigma^{2} }}}} } {\text{d}}\varepsilon \\ & = e^{{\frac{{\gamma^{2} \sigma^{2} }}{2}}} \frac{1}{{\sqrt {2\pi } \sigma }}\int_{ - \infty }^{ + \infty } {e^{{ - \frac{{(\varepsilon^{2} - \gamma \sigma^{2} )^{2} }}{{2\sigma^{2} }}}} } {\text{d}}\varepsilon \\ \end{aligned}$$

In the formula, $\frac{1}{{\sqrt {2\pi } \sigma }}\int_{ - \infty }^{ + \infty } {e^{{ - \frac{{(\varepsilon^{2} - \gamma \sigma^{2} )^{2} }}{{2\sigma^{2} }}}} } d\varepsilon$ is the area when the normal distribution mean is $\gamma \sigma^{2}$, and the variance is $\sigma^{2}$, the area is 1.

We can conclude $E(e^{\gamma \varepsilon } ) = e^{{\frac{{\gamma^{2} \sigma^{2} }}{2}}}$.

Proof of Proposition 1

Combining constraints (13) and (14) into the expressions $t_{1}$ and $t_{ 2}$, and replacing the bank’s objective function, we can have

$$\hbox{max} EU_{B1} = L(r + 1)A_{1} a_{1} + LA_{2} a_{2} - LA_{1} a_{1} A_{2} a_{2} - \frac{1}{2}c_{1} a_{1}^{2} - \frac{1}{2}c_{2} a_{2}^{2} - \frac{{\eta_{1} v_{1}^{2} L^{2} r^{2} \sigma_{1}^{2} }}{2} - \frac{{\eta_{2} v_{2}^{2} L^{2} r^{2} \sigma_{2}^{2} }}{2}$$

(26)

Deriving the Eq. (26) for $v_{1}$ and $v_{2}$, we have

$$\frac{{\partial EU_{B1} }}{{\partial v_{1} }} = - \eta_{1} L^{2} r^{2} \sigma_{1}^{2} v_{1} = 0$$

(27)

$$\frac{{\partial EU_{B1} }}{{\partial v_{2} }} = - \eta_{2} L^{2} r^{2} \sigma_{2}^{2} v_{2} = 0$$

(28)

According to Eqs. (27) and (28), the incentive coefficients of bank payment the B2B platform and 3PL can be obtained as

$$v_{1}^{*} = 0;\quad v_{2}^{*} = 0.$$

Proof of Proposition 2

Deriving Eq. (26) for $a_{1}$ and $a_{2}$, we have

$$\frac{{\partial EU_{B1} }}{{\partial a_{1} }} = L(r + 1)A_{1} - LA_{1} A_{2} a_{2} - c_{1} a_{1} = 0$$

(29)

$$\frac{{\partial EU_{B1} }}{{\partial a_{2} }} = LA_{2} - LA_{1} A_{2} a_{1} - c_{2} a_{2} = 0$$

(30)

According to Eqs. (29) and (30), the level of effort of the B2B platform and 3PL can be obtained

$$a_{1}^{*} = \frac{{LA_{1} \left[ {\left( {r + 1} \right)c_{2} - LA{}_{2}^{2} } \right]}}{{c_{1} c_{2} - L^{2} A{}_{1}^{2} A{}_{2}^{2} }}$$

(31)

$$a_{2}^{*} = \frac{{LA_{2} \left[ {c_{1} - LA{}_{1}^{2} \left( {r + 1} \right)} \right]}}{{c_{1} c_{2} - L^{2} A{}_{1}^{2} A{}_{2}^{2} }}$$

(32)

From $a_{1}^{*} > 0$, we can have

$$\left[ {\left( {r + 1} \right)c_{2} - LA{}_{2}^{2} } \right]\left( {c_{1} c_{2} - L^{2} A{}_{1}^{2} A{}_{2}^{2} } \right) > 0$$

(33)

Then we can have $L < \hbox{min} \{ \frac{{\left( {r + 1} \right)c_{2} }}{{A{}_{2}^{2} }},\frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}\}$ or $L > \hbox{max} \{ \frac{{\left( {r + 1} \right)c_{2} }}{{A{}_{2}^{2} }},\frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}\}$.

From $a_{2}^{*} > 0$, we can have

$$\left[ {c_{1} - LA{}_{1}^{2} \left( {r + 1} \right)} \right]\left( {c_{1} c_{2} - L^{2} A{}_{1}^{2} A{}_{2}^{2} } \right) > 0$$

(34)

Then we can obtain the constraint conditions

$$L < \hbox{min} \left\{ {\frac{{c_{1} }}{{A{}_{1}^{2} \left( {r + 1} \right)}},\frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}} \right\}\;{\text{or}}\;L > \hbox{max} \left\{ {\frac{{c_{1} }}{{A{}_{1}^{2} \left( {r + 1} \right)}},\frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}} \right\}.$$

As $c_{1} < c_{2}$; $A_{1} > A_{2}$, we can deduct

$$\frac{{\left( {r + 1} \right)c_{2} }}{{A{}_{2}^{2} }} > \frac{{c_{1} }}{{A{}_{1}^{2} \left( {r + 1} \right)}}$$

(35)

So it needs to satisfy: $L < \hbox{min} \{ \frac{{c_{1} }}{{A{}_{1}^{2} \left( {r + 1} \right)}},\frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}\}$ or $L > \hbox{max} \{ \frac{{\left( {r + 1} \right)c_{2} }}{{A{}_{2}^{2} }},\frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}\}$.

Proof of Corollary 2.

• (1) According to $a_{1}^{*} = \frac{{LA_{1} \left[ {\left( {r + 1} \right)c_{2} - LA{}_{2}^{2} } \right]}}{{c_{1} c_{2} - L^{2} A{}_{1}^{2} A{}_{2}^{2} }}$, numerator and denominator are divided by $A_{1}$, we can get $a_{1}^{*} = \frac{{L\left[ {\left( {r + 1} \right)c_{2} - LA{}_{2}^{2} } \right]}}{{{\raise0.7ex\hbox{${c_{1} c_{2} }$} \!\mathord{\left/ {\vphantom {{c_{1} c_{2} } {A_{1} }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${A_{1} }$}} - L^{2} A{}_{1}^{2} A{}_{2}^{2} }}$. From the equitation, $c_{1} c_{2} - L^{2} A{}_{1}^{2} A{}_{2}^{2} > 0$, which means $L < \frac{{\sqrt {c_{1} c_{2} } }}{{A_{1} A_{2} }}$, when credit rating capability coefficient $A_{1}$ is bigger, $a_{1}^{*}$ is bigger.

  Similarly for 3PL, it can be proved that the greater the coefficient, the greater $A_{2}$ is, and the greater $a_{2}^{*}$ is.

• (2) According to $a_{1}^{*} = \frac{{L\left[ {\left( {r + 1} \right)c_{2} - LA{}_{2}^{2} } \right]}}{{{\raise0.7ex\hbox{${c_{1} c_{2} }$} \!\mathord{\left/ {\vphantom {{c_{1} c_{2} } {A_{1} }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${A_{1} }$}} - L^{2} A{}_{1}^{2} A{}_{2}^{2} }}$, numerator and denominator are both divided by $L^{2}$, we can get $a_{1}^{*} = \frac{{A_{1} \left[ {{\raise0.7ex\hbox{${\left( {r + 1} \right)c_{2} }$} \!\mathord{\left/ {\vphantom {{\left( {r + 1} \right)c_{2} } L}}\right.\kern-0pt} \!\lower0.7ex\hbox{$L$}} - A{}_{2}^{2} } \right]}}{{{\raise0.7ex\hbox{${c_{1} c_{2} }$} \!\mathord{\left/ {\vphantom {{c_{1} c_{2} } {L^{2} }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${L^{2} }$}} - A_{1}^{2} A{}_{2}^{2} }}$. From the equitation, we can find that as loan amount $L$ increases, and $a_{1}^{*}$ also increases.

Proof of Lemma 2 The deterministic equivalence of the B2B platform is obtained from Eq. (10)

$$\hat{w}_{1} \left( {a_{1} } \right) = t_{1} + v_{1} rLA_{1} a_{1} - \frac{1}{2}c_{1} a_{1}^{2} - \frac{{\eta_{1} v_{1}^{2} L^{2} r^{2} \sigma_{1}^{2} }}{2}$$

(36)

Finding the derivative of $a_{1}$: $\frac{{\partial \hat{w}_{1} \left( {a_{1} } \right)}}{{\partial a_{1} }} = v_{1} LrA_{1} - c_{1} a_{1} = 0$, we can have $a_{1}^{**} = \frac{{v_{1} LrA_{1} }}{{c_{1} }}$.

Similarly, deriving 3PL’s deterministic equivalence of wealth $\hat{w}_{2} \left( {a_{2} } \right)$ for $a_{2}$, we can get $a_{2}^{**} = \frac{{v_{2} LrA_{2} }}{{c_{2} }}$.

Proof of Proposition 3 According to $(IR_{1} )$ and $(IR_{2} )$, we have

$$t_{1} = \bar{w}_{1} - v_{1} rLA_{1} a_{1} + \frac{1}{2}c_{1} a_{1}^{2} + \frac{{\eta_{1} v_{1}^{2} L^{2} r^{2} \sigma_{1}^{2} }}{2}$$

(37)

$$t_{2} = \bar{w}_{2} - v_{2} rLA_{2} a_{2} + \frac{1}{2}c_{2} a_{2}^{2} + \frac{{\eta_{2} v_{2}^{2} L^{2} r^{2} \sigma_{2}^{2} }}{2}$$

(38)

Substituting $a_{1}^{**}$; $a_{2}^{**}$; $t_{1}$; $t_{2}$ into the bank’s objective function, we have

$$\mathop {\hbox{max} }\limits_{{t_{1} ;t_{2} ;v_{1} ;v_{2} }} EU_{B1} = \frac{{L^{2} r(r + 1)A_{1}^{2} }}{{c_{1} }}v_{1} + \frac{{L^{2} A_{2}^{2} r}}{{c_{2} }}v_{2} - \frac{{L^{3} A_{1}^{2} A_{2}^{2} r^{2} }}{{c_{1} c_{2} }}v_{1} v_{2} - \left( {\frac{{L^{2} A_{1}^{2} r^{2} }}{{2c_{1} }} + \frac{{\eta_{1} L^{2} r^{2} \sigma_{1}^{2} }}{2}} \right)v_{1}^{2} - \left( {\frac{{L^{2} A_{2}^{2} r^{2} }}{{2c_{2} }} + \frac{{\eta_{2} L^{2} r^{2} \sigma_{2}^{2} }}{2}} \right)v_{2}^{2} - \overline{{w_{1} }} - \overline{{w_{2} }}$$

(39)

Solving the problem of maximizing the utility of the banks can be solved as follows:

• (1) For any given $v_{2}$, solved by the optimal trade-off about incentive, we have

  $$v_{1} = \frac{{A{}_{1}^{2} \left( {r + 1} \right)}}{{r(A{}_{1}^{2} + \eta_{1} \sigma_{1}^{2} c_{1} )}} - \frac{{LA{}_{1}^{2} A{}_{2}^{2} }}{{c_{2} (A{}_{1}^{2} + \eta_{1} \sigma_{1}^{2} c_{1} )}}v_{2}$$

  (40)

• (2) Substituting the Eq. (40) into the objective function of the bank, the solution is

  $$v_{2}^{**} = \frac{{c_{2} A{}_{2}^{2} [c_{1} F_{1} - LA_{1}^{4} \left( {r + 1} \right)]}}{{r(c_{1} c_{2} F_{1} F_{2} - L^{2} A{}_{1}^{4} A{}_{2}^{4} )}}$$

  (41)

Where $F_{1} = A{}_{1}^{2} + \eta_{1} \sigma_{1}^{2} c_{1}$, $F_{2} = A{}_{2}^{2} + \eta_{2} \sigma_{2}^{2} c_{2}$.

Substituting Eq. (41) into Eq. (40), we can have

$$v_{1}^{**} = \frac{{c_{1} A{}_{1}^{2} \left[ {c_{2} F_{2} \left( {r + 1} \right) - LA_{2}^{4} } \right]}}{{r\left( {c_{1} c_{2} F_{1} F_{2} - L^{2} A{}_{1}^{4} A{}_{2}^{4} } \right)}}$$

(42)

Proof of Corollary 3

Deriving Eqs. (15) and (16) respectively for the loan interest rates $r$, we have

$$\frac{{\partial a_{1}^{**} }}{\partial r} = \frac{{LA{}_{1}^{3} c_{2} F_{2} }}{{c_{1} c_{2} F_{1} F_{2} - L^{2} A{}_{1}^{4} A{}_{2}^{4} }}$$

(43)

$$\frac{{\partial a_{2}^{**} }}{\partial r} = - \frac{{L^{2} A{}_{1}^{4} A{}_{2}^{3} }}{{c_{1} c_{2} F_{1} F_{2} - L^{2} A{}_{1}^{4} A{}_{2}^{4} }}$$

(44)

For $c_{1} c_{2} F_{1} F_{2} - L^{2} A{}_{1}^{4} A{}_{2}^{4} > 0$, i.e.$L < \frac{{\sqrt {c_{1} c_{2} F_{1} F_{2} } }}{{A_{1}^{2} A_{2}^{2} }}$ according to the Eq. (43), we can have $\frac{{\partial a_{1}^{**} }}{\partial r} > 0$, that is, $a_{1}^{**}$ and $r$ are proportional. According to the Eq. (44), we can have $\frac{{\partial a_{2}^{**} }}{\partial r} < 0$, that is, $a_{2}^{**}$ and $r$ are inversely proportional.

Similarly, it can be proved for $L > \frac{{\sqrt {c_{1} c_{2} F_{1} F_{2} } }}{{A_{1}^{2} A_{2}^{2} }}$, $a_{1}^{**}$ and $r$ are inversely proportional, while $a_{2}^{**}$ and $r$ are proportional.

Proof of Proposition 4 The expected extra revenue that the B2B platform chooses collusion can obtain is

$$\vartriangle U_{E} = g(M_{1} - C_{{M_{1} }} ) + (1 - g)(M_{1} - C_{{M_{1} }} - K)$$

(45)

The extra expected gain of no collusion is 0. The B2B platform chooses collusion when the expected benefits of collusion bring about more utility than non-collusion. Therefore, according to the utility function of the B2B platform, the condition that the B2B platform can reject the collusion is

$$- e^{{ - \eta_{1} [g(M_{1} - C_{{M_{1} }} ) + (1 - g)(M_{1} - C_{{M_{1} }} - K)]}} < - e^{0}$$

(46)

From the above formula (46), the condition can be derived B2B platform to refuse collusion $K > \frac{{M_{1} - C_{{M_{1} }} }}{1 - g}$.

After observing that the B2B platform has accepted the collusion, 3PL refused to collude on the condition $- e^{{ - \eta_{2} (M_{2} - C_{{M_{2} }} )}} < - e^{{ - \eta_{2} R}}$, we have $R > M_{2} - C_{{M_{2} }}$.

Proofs for Sect. 4.3 are similar to those above, so we do not repeat them.