Trade credit insurance, capital constraint, and the behavior of manufacturers and banks

Abstract

The manufacturer who is a supplier of trade credit may face non-payment risk from customers and a capital shortage problem simultaneously. Trade credit insurance, as one of the most important risk management tools, has been widely used in companies’ daily operation. In this study, the manufacturer who allows customers to delay payment for goods already delivered purchases trade credit insurance to transfer and reduce non-payment risk and borrows money from a bank to accommodate the capital constraint problem. The Stackelberg game and loss-averse theory are used to establish a newsboy model including trade credit insurance, and the optimal insurance coverage and total sales of the manufacturer are thereby investigated. Subsequently, the interest rate decision of the bank under different risk-averse situations is also characterized. We find that the interest rate set by a loss-averse bank is equal to or greater than that given by a risk-neutral bank. The use of trade credit insurance can help the manufacturer expand sales and dramatically reduce its default risk. Both the bank and the manufacturer are better off due to the use of trade credit insurance, but contrary to what one might expect, the bank prefers giving a higher interest rate to the manufacturer when the premium rate is in a reasonable region, which indicates that the manufacturer cannot use the insurance to negotiate better financing terms.

Appendix

Proof of Theorem 1

The decision problem (1) can be rewritten as

$$\begin{aligned} MinG(y,Q)=B(1+r)+\eta +hy-y-pQ\int \limits _{\frac{y}{pQ}}^1 {F\hbox {(x)}dx} \end{aligned}$$

$$\begin{aligned} \begin{array}{l} s.t. \qquad g_1 (Q,y)=y\ge 0 \\ g_2 (Q,y)=pQ-y\ge 0 \\ \end{array} \end{aligned}$$

Differentiating $G(y,Q),\,g_1 (Q,y)$ and $g_2 (Q,y)$ with respect to $y$ and $Q$, respectively, yields

$$\begin{aligned} \frac{\partial G}{\partial y}&= h+F\left( \frac{y}{pQ}\right) -1, \quad \frac{\partial G}{\partial Q}=2cQ(1+r)+p\int \limits _{\frac{y}{pQ}}^1 {xf(x)dx} -p,\\ \frac{\partial g_1 }{\partial y}&= 1, \quad \frac{\partial g_1 }{\partial Q}=0, \quad \frac{\partial g_2 }{\partial y}=-1, \quad \frac{\partial g_2 }{\partial Q}=p. \end{aligned}$$

$\gamma ^{*}_1$ and $\gamma ^{*}_2$ are Lagrange multipliers. $(Q^{*},y^{*})$ is K-T point and then Kuhn—Tucker conditions are given by

$$\begin{aligned} \left\{ {\begin{array}{l} h+F(\frac{y^{*}}{pQ^{*}})-1-\gamma _1 ^{*}+\gamma _2 ^{*}=0 \\ 2cQ^{*}(1+r)+p\int _{\frac{y^{*}}{pQ^{*}}}^1 {xf(x)dx} -p-p\gamma _2 ^{*}=0 \\ \gamma _1 ^{*}y^{*}=0 \\ \gamma _2 ^{*}(pQ^{*}-y^{*})=0 \\ \gamma _1 ^{*},\gamma _2 ^{*}\ge 0 \\ \\ \end{array}} \right. \end{aligned}$$

To solve these equations, we investigate several cases:

Case 1: $\gamma _1 ^{*}\ne 0,\,\gamma _2 ^{*}\ne 0$. $y^{*}=0$. $Q^{*}=0$. No solution.

Case 2: $\gamma _1 ^{*}\ne 0,\,\gamma _2 ^{*}=0$. $y^{*}=0,\,Q^{*}=\frac{p(1-E(\alpha ))}{2c(1+r)}$, but $\gamma _1 ^{*}=h-1<0$, so it is not K-T point.

Case 3: $\gamma _1 ^{*}=0,\,\gamma _2 ^{*}\ne 0$. $y^{*}=pQ^{*},\,Q^{*}=\frac{p(1-h)}{2c(1+r)}$, but $\gamma _2 ^{*}=-h<0$, so it is not K-T point.

Case 4: $\gamma _1 ^{*}=0,\,\gamma _2 ^{*}=0$. $(Q^{*},y^{*})$ satisfies the following equations:

$$\begin{aligned}&\displaystyle h+F\left( \frac{y^{*}}{pQ^{*}}\right) -1=0,\\&\displaystyle 2cQ^{*}(1+r)+p\int \limits _{\frac{y^{*}}{pQ^{*}}}^1 {xf(x)dx} -p=0. \end{aligned}$$

Since $\frac{\partial ^{2}G}{\partial y^{2}}=f\left( \frac{y}{pQ}\right) \frac{1}{pQ},\,\frac{\partial ^{2}G}{\partial Q^{2}}=2c(1+r)+f\left( \frac{y}{pQ}\right) \frac{y^{2}}{pQ^{3}}$,

$$\begin{aligned} \frac{\partial ^{2}G}{\partial y\partial Q}=\frac{\partial ^{2}G}{\partial Q\partial y}=-f\left( \frac{y}{pQ}\right) \frac{y}{pQ^{2}}, \end{aligned}$$

Hesse matrix is

$$\begin{aligned} H=\frac{\partial ^{2}G}{\partial Q^{2}}\frac{\partial ^{2}G}{\partial y^{2}}-\frac{\partial ^{2}G}{\partial y\partial Q}\frac{\partial ^{2}G}{\partial Q\partial y}=\frac{2c(1+r)}{pQ}f\left( \frac{y}{pQ}\right) >0. \end{aligned}$$

which is a positive matrix. Thus the objective is a convex function and the two constraints are linear functions so this is a convex programming. The optimal amount of the credit sales and optimal insurance coverage exist, given by

$$\begin{aligned} h+F\left( \frac{y^{*}}{pQ^{*}}\right) -1=0, \end{aligned}$$

and $2cQ^{*}(1+r)+p\int _{\frac{y^{*}}{pQ^{*}}}^1 {xf(x)dx} -p=0$.

Let $\frac{y^{*}}{pQ^{*}}=\phi ^{*}$, then $p\phi ^{*}F(\phi ^{*})-2cQ^{*}(1+r)+p\int _{\phi ^{*}}^1 {F(x)dx} =0$,

i.e., $Q^{*}=\frac{p(1-h)\phi ^{*}+p\int _{\phi ^{*}}^1 {F(x)dx} }{2c(1+r)}$. And $y^{*}=\phi ^{*}pQ^{*}$, where $\phi ^{*}=F^{-1}(1-h)$.$\square$

Proof of Lemma 1

Differentiating $Q^{*}=\frac{p(1-h)\phi ^{*}+p\int \nolimits _{\phi ^{*}}^1 {F(x)dx} }{2c(1+r)}$ with respect to $h$ yields $\frac{\partial Q^{*}}{\partial h}=\frac{p(1-h)\frac{\partial \phi ^{*}}{\partial h}-p\phi ^{*}-pF(\phi ^{*})\frac{\partial \phi ^{*}}{\partial h}}{2c(1+r)}$.

Since $F(\phi ^{*})=1-h,\,\frac{\partial Q^{*}}{\partial h}=\frac{-p\phi ^{*}}{2c(1+r)}<0$, i.e., the amount of the credit sales decreases as the premium rate increases.

Differentiating $F(\phi ^{*})=1-h$ with respect to $h$, we have

$$\begin{aligned} \frac{\partial y^{*}}{\partial h}=\frac{-1}{f(\phi ^{*})}pQ^{*}+p\phi ^{*}\frac{\partial Q^{*}}{\partial h}. \end{aligned}$$

Since $\frac{\partial Q^{*}}{\partial h}<0$, we have $\frac{\partial y^{*}}{\partial h}<0$, i.e., the insurance coverage is a decreasing function of the premium rate.$\square$

Proof of Theorem 2

When the manufacturer does not purchase the trade credit insurance, the objective function is $E\pi _s (Q,0)=pQ\int _0^1 {F\hbox {(x)}dx} -B(1+r)-\eta$.

Differentiating $E\pi _s (Q,0)$with respect to $Q$, we obtain

$$\begin{aligned} \frac{d\pi _s (Q,0)}{dQ}=(1-E(\alpha ))p-2cQ(1+r). \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \frac{d^{2}\pi _s (Q,0)}{dQ^{2}}=-2c(1+r)<0. \end{aligned}$$

$E\pi _s (Q,0)$ is a strictly concave function in $Q$ and the optimal amount of the credit sales is existent and unique. Let $Q_0 ^{*}$ be the solution to $\frac{d\pi _s (Q,0)}{dQ}=0$, then $Q_0 ^{*}=\frac{p(1-E(\alpha ))}{2c(1+r)}$.$\square$

Proof of Lemma 2

For $Q^{*}=\frac{p(1-h)\phi ^{*}+p\int _{\phi ^{*}}^{1} {F(x)dx} }{2c(1+r)}$ and $F(\phi ^{*})=1-h$, we have $Q^{*}=\frac{p(1-\int _{\phi ^{*}}^{1} {xf(x)dx} )}{2c(1+r)}$. Since $E(\alpha )>\int _{\phi ^{*}}^{1} {xf(x)dx}$, we get $Q_0 ^{*}<Q^{*}$.$\square$

Proof of Theorem 3

Since $F\left( \frac{y^{*}}{pQ^{*}}\right) =1-h$, and $\frac{y^{*}}{Q^{*}}F\left( \frac{y^{*}}{pQ^{*}}\right) -2cQ^{*}(1+r)+p\int _{\frac{y^{*}}{pQ^{*}}}^{1} {F(x)dx} =0$, we can get

$$\begin{aligned} \frac{\partial y^{*}}{\partial r}=\frac{-y^{*}}{1+r}, \quad \frac{\partial Q^{*}}{\partial r}=\frac{-Q^{*}}{1+r},\text { and } \frac{\partial z}{\partial r}=\frac{2\eta }{pQ^{*}}. \end{aligned}$$

Case 1: The non-risk operation, i.e., $y^{*}\ge 2B(Q^{*})$ or $B(Q^{*})<y^{*}<2B(Q^{*})$ and $r\le \frac{y^{*}}{B(Q^{*})}-1$.

The expected profit of the bank is $E\pi _b (r)=B(Q^{*})r$.

Differentiating $E\pi _b (r)$ with respect to $r$, we have

$$\begin{aligned} \frac{dE\pi _b (r)}{dr}=B(Q^{*})-\frac{2cQ^{2}}{1+r}r. \end{aligned}$$

Furthermore, we obtain

$$\begin{aligned} \frac{d^{2}E\pi _b (r)}{dr^{2}}=-\frac{2cQ^{2}(2-r)}{(1+r)^{2}}<0. \end{aligned}$$

So the expected profit of the bank is a strictly concave function of $r$. The optimal interest rate is existent and unique. Let $r^{*}$ satisfy $\frac{dE\pi _b (r)}{dr}=0$, then we have $r^{*}=\frac{B(Q^{*})}{cQ^{^{*}{2}}+\eta }$. If in the case of $B(Q^{*})<y^{*}<2B(Q^{*})$ and $r\in \left[ {0,\frac{y^{*}}{B(Q^{*})}-1} \right]$, then the optimal interest rate is $\min \left\{ {\frac{y^{*}}{B(Q^{*})}-1,\frac{B(Q^{*})}{cQ^{^{*}{2}}+\eta }}\right\}$.

Case 2: The risk operation, i.e.,$y^{*}<B(Q^{*})$ or $B(Q^{*})<y^{*}<2B(Q^{*})$ and $r>\frac{y^{*}}{B(Q^{*})}-1$.

The expected profit of the bank is $E\pi _b (r)=y^{*}-B+pQ^{*}\int _z^1 {F(x)dx}$, where $z=\frac{pQ^{*}+y^{*}-B(Q^{*})(1+r)}{pQ^{*}}$.

Differentiating $E\pi _b (r)$ with respect to $r$, we have

$$\begin{aligned} \frac{dE\pi _b (r)}{dr}=\frac{2cQ^{^{*}{2}}-y^{*}}{1+r}-\frac{pQ^{*}}{1+r}\int \limits _z^1 {F(x)dx} -2\eta F(z), \end{aligned}$$

and then we obtain

$$\begin{aligned} \frac{d^{2}E\pi _b (r)}{dr^{2}}&= \frac{{2}y^{*}-{6}cQ^{^{*}{2}}}{{(1}+r)^{{2}}}+\frac{{2}pQ^{*}}{{(1}+r)^{{2}}}\int \limits _z^1 {F(x)dx} +\frac{2\eta }{{1}+r}F(z)-\frac{{4}\eta ^{{2}}}{pQ^{*}}f\hbox {(z)} \\&< \frac{2{y}^{*}{-6}cQ^{^{*}{2}}}{(1+r)^{{2}}}+\frac{{2}pQ^{*}}{(1+r)^{{2}}}(1-z)+\frac{2\eta }{{1}+r}F(z)-\frac{{4}\eta ^{{2}}}{pQ^{*}}f\hbox {(z)} \\&= \frac{2cQ^{^{*}{2}}(r-2)}{{(1}+r)^{{2}}}-\frac{{4}\eta ^{{2}}}{pQ^{*}}f\hbox {(z)<0}. \end{aligned}$$

Because $\frac{d^{{2}}E\pi _b (r)}{dr^{{2}}}<0$, the expected profit of the bank is a strictly concave function of $r$. The optimal interest rate is existent and unique. Let $r^{**}$satisfy $\frac{dE\pi _b (r)}{dr}=0$, so we have $\frac{2cQ^{^{*}{2}}-y^{*}}{1+r^{**}}-\frac{pQ^{*}}{{1}+r^{**}}\int _z^1 {F(x)dx} -2\eta F(z)=0$.

If $B(Q^{*})<y^{*}<2B(Q^{*})$and$r>\frac{y^{*}}{B(Q^{*})}-1$, then the optimal interest rate is $\max \left\{ {\frac{y^{*}}{B(Q^{*})}-1,r^{**}} \right\}$ $\square$

Proof of Theorem 4

Differentiating $EU(\pi _b )$ with respect to $r$ yields

$$\begin{aligned} \frac{\partial EU(\pi _b )}{\partial r}&= \left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] \frac{\partial H}{\partial r}+\left( \frac{\partial B(Q^{*})}{\partial r}r+B\right) F(z)-pT\frac{\partial Q}{\partial r} \\&= -\left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] \frac{pQ^{*}+y^{*}-2cQ^{^{*}{2}}}{1+r}\\&\quad +\left( -\frac{2cQ^{2}}{1+r}r+B\right) F(z)+\frac{pTQ^{*}}{1+r}. \end{aligned}$$

Let $r^{*}$ satisfies $\frac{\partial EU(\pi _b )}{\partial r}=0$, then

$$\begin{aligned} \left[ {B(1+r)-2cQ^{2}r} \right] F(z)+pTQ^{*}=\left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] \left[ {pQ^{*}+y^{*}-2cQ^{^{*}{2}}} \right] . \end{aligned}$$

Furthermore, we obtain

$$\begin{aligned} \frac{\partial ^{2}EU(\pi _b )}{\partial r^{2}}&= \left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] \frac{\partial ^{2}H}{\partial r^{2}}+\left[ {-f(z)\frac{\partial z}{\partial r}+(1-\lambda )f(z_0 )\frac{\partial z_0 }{\partial r}} \right] \frac{\partial H}{\partial r} \\&+\left( \frac{\partial ^{2}B(Q^{*})}{\partial r^{2}}r+2\frac{\partial B(Q^{*})}{\partial r}\right) F(z)+\left( \frac{\partial B(Q^{*})}{\partial r}r+B\right) f(z)\frac{\partial z}{\partial r}\\&-\,pT\frac{\partial ^{2}Q}{\partial r^{2}}-p\frac{\partial T}{\partial r}\frac{\partial Q}{\partial r} \\&= 2\left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] \frac{pQ^{*}+y^{*}-3cQ^{^{*}{2}}}{(1+r)^{2}}-pT\frac{2Q^{*}}{(1+r)^{2}}\\&+\,2\frac{r-2}{(1+r)^{2}}cQ^{2}F(z)-f(z)\frac{4\eta ^{2}}{pQ^{*}}+f(z_0 )\frac{(cQ^{^{*}{2}}+\eta )^{2}}{pQ(1+r)^{2}}(1-\lambda ) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial ^{2}EU(\pi _b )}{\partial r^{2}}(1+r)^{2}\left| {_{r=r^{*}} } \right.&= -2\left[ {\eta (1+r)+cQ^{2}} \right] F(z)\\&\quad -\,2\left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] cQ^{^{*}{2}} \\&\quad -\,f(z)\frac{4\eta ^{2}}{pQ^{*}}(1+r)^{2}+f(z_0 )\frac{(cQ^{^{*}{2}}+\eta )^{2}}{pQ}(1-\lambda ) \end{aligned}$$

Since $z_0 >z$, we get $F(z_0 )>F(z)$.

So $\lambda -F(z)+(1-\lambda )F(z_0 )=\lambda \bar{{F}}(z_0 )+F(z_0 )-F(z)>0$.

We also know $\lambda \ge 1$, then$\frac{d^{2}EU(\pi _b )}{dr^{2}}(1+r)^{2}\left| {_{r=r^{*}} } \right. <0$ and the sufficient condition is satisfied. The optimal interest rate $r^{*}$satisfies the following equation:

$$\begin{aligned} \left[ {B(1+r)-2cQ^{2}r} \right] F(z)+pTQ^{*} =\left[ {\lambda -F(z)+(1-\lambda )F(z_0 )} \right] \left[ {pQ^{*}+y^{*}-2cQ^{^{*}{2}}} \right] . \end{aligned}$$

$\square$