Financing Schemes in Supply Chains with a Capital-Constrained Supplier: Coopetition and Risk

Abstract

In this paper, we consider a small capital-constrained supplier that can finance from a bank or a peer supplier or a downstream manufacturer in a supply chain. We use a game-theoretical model to analyze different financing schemes. Through analytical comparison, we find that the small supplier will always choose purchase order financing when the production cost is low. As the production cost increases, the internal financing (buyer direct financing and peer financing) dominates external financing.

Appendix: Proofs

Proof of Proposition 1

The firstand second-order condition of (1) is as follows:

$\frac{{\partial \pi_{s}^{BDF} }}{\partial e} = a + w_{1} - 2ek - c(1 + r_{b} )$ and $\frac{{\partial^{2} \pi_{s}^{BDF} }}{{\partial e^{2} }} = - 2k < 0$. Thus, we have $e(w_{1} ) = {{(a - c(1 + r_{b} ) + w_{1} )} \mathord{\left/ {\vphantom {{(a - c(1 + r_{b} ) + w_{1} )} {2k}}} \right. \kern-\nulldelimiterspace} {2k}}$.

For (6), its first and second-order condition is

$\frac{{\partial \pi_{m}^{POF} }}{{\partial w_{1} }} = \frac{{w_{2} - w_{1} - (a + w_{1} - c(1 + r_{b} ))}}{2k}$, $\frac{{\partial^{2} \pi_{m}^{POF} }}{{\partial w_{1}^{2} }} = - \frac{1}{k} < 0$. Then, we obtain $w_{1} (w_{2} ) = {{(w_{2} + c(1 + r_{b} ) - a)} \mathord{\left/ {\vphantom {{(w_{2} + c(1 + r_{b} ) - a)} 2}} \right. \kern-\nulldelimiterspace} 2}$. Considering the participation constraint $w_{1} \ge c(1 + r_{b} ) + 2\sqrt {ak} - a$, we have the manufacturer’s optimal decision: $w_{1} (w_{2} ) = {{(w_{2} + c(1 + r_{b} ) - a)} \mathord{\left/ {\vphantom {{(w_{2} + c(1 + r_{b} ) - a)} 2}} \right. \kern-\nulldelimiterspace} 2}$, if $\frac{{w_{2} + c(1 + r_{b} ) - a}}{2} \ge c(1 + r_{b} ) + 2\sqrt {ak} - a$; $w_{1} = c(1 + r_{b} ) + 2\sqrt {ak} - a$, otherwise. Therefore, we have the small supplier’s optimal decision in (8). As $0 < e < 1$, we have $a < k$. Inserting (8) into (3) we have.

$$\pi_{p}^{POF} = \mathop \{ \nolimits_{{(1 - \sqrt {{a \mathord{\left/ {\vphantom {a k}} \right. \kern-\nulldelimiterspace} k}} )(w_{2} - c),\;{\text{otherwise}}}}^{{\frac{{a - c(1 + r_{b} ) + w_{2} }}{4k}(w_{2} - c),w_{2} > c(1 + r_{b} ) + 4\sqrt {ak} - a}} {.}$$

If $w_{2} > c(1 + r_{b} ) + 4\sqrt {ak} - a$, we have

$\frac{{\partial \pi_{p}^{POF} }}{{\partial w_{2} }} = \frac{{4k - a + c(1 + r_{b} ) - 2w_{2} + c}}{4k}$,$\frac{{\partial^{2} \pi_{p}^{POF} }}{{\partial w_{2}^{2} }} = - \frac{1}{2k} < 0$, Thus $w_{2} = c + 2k + \frac{{cr_{b} - a}}{2}$. Solving the inequation $w_{2} = c + 2k + \frac{{cr_{b} - a}}{2} > c(1 + r_{b} ) + 4\sqrt {ak} - a$, we have $cr_{b} < a + 4k - 8\sqrt {ak}$. Then, if $0 < cr_{b} < a + 4k - 8\sqrt {ak}$, we have the equilibrium: $w_{1}^{POF*} = c + k + \frac{{3(cr_{b} - a)}}{4},$$e^{POF*} = \frac{{a + 4k - cr_{b} }}{8k},$$w_{2}^{POF*} = c + 2k + \frac{{cr_{b} - a}}{2}$; Otherwise, $\partial \pi_{p}^{POF} = 1 - \sqrt {\frac{a}{k}} > 0$, we have $\pi_{p}^{POF}$ increases as $w_{2}$. From $w_{2} \le c(1 + r_{b} ) + 4\sqrt {ak} - a$ we obtain $w_{2} = c(1 + r_{b} ) + 4\sqrt {ak} - a$, and further have the equilibrium:$w_{2}^{POF*} = c(1 + r_{b} ) + 4\sqrt {ak} - a,$$w_{1}^{POF*} = c(1 + r_{b} ) + 2\sqrt {ak} - a,$$e^{POF*} = \sqrt {{a \mathord{\left/ {\vphantom {a k}} \right. \kern-\nulldelimiterspace} k}}$.

Proof of Proposition 4

Suppose $\frac{{a + 4k - 8\sqrt {ak} }}{{r_{b} }} > \frac{{a + 2\sqrt {ak} }}{{1 + r_{m} }} > 8\sqrt {ak} - 4k$, we have $\frac{{1 + r_{m} }}{{r_{b} }} > \frac{{a + 2\sqrt {ak} }}{{a + 4k - 8\sqrt {ak} }}$ and $r_{m} < \frac{{a + 4k - 6\sqrt {ak} }}{{8\sqrt {ak} - 4k}}$.

• (i) When $c \in (0,\frac{{a + 2\sqrt {ak} }}{{1 + r_{m} }}),$ we have $\pi_{s}^{PF*} = \pi_{s}^{BDF*} = 0,$ $\pi_{s}^{POF*} = ((a - cr_{b} )^{2} - 56ak - 8kcr_{b} + 16k^{2} )/64k.$ Thus, we have $\pi_{s}^{POF*} > \pi_{s}^{PF*} = \pi_{s}^{BDF*} ;$

• (ii) when $c \in (8\sqrt {ak} - 4k,(a + 2\sqrt {ak} )/(1 + r_{m} )),$ we have $\pi_{s}^{PF*} = [c^{2} + 8ck + 16k(k - 4a)]/(64k).$ Further, we get $\pi_{s}^{PF*} - \pi_{s}^{POF*} = \,c^{2} + 8ck(1 + r_{b} ) - 8ak - (a - cr_{b} )^{2}$. From $c > a,$ it follows that $\pi_{s}^{PF*} - \pi_{s}^{POF*} > 0$. Then we have $\pi_{s}^{PF*} > \pi_{s}^{P0F*} > \pi_{s}^{BDF*} ;$

• (iii) otherwise, we have $\pi _{s}^{{BDF*}} = [c^{2} + 8ck + 16k(k - 4a)]\,/(64k)$, then we have $\pi_{s}^{PF*} \, = \,\pi_{s}^{BDF*} \, > \,\pi_{s}^{POF8} \,$.

For part (b) and (c), we could prove it similarly.