Trade credit contract with limited liability in the supply chain with budget constraints

Abstract

The paper investigates the impacts of trade credit (i.e., open account payment) and limited liability on the performance of a two-level supply chain with budget constraints. It shows that limited liability accounts for the reason why the retailer with a lower initial budget initiates a higher ordering level under trade credit contract. Furthermore, this paper finds that trade credit contract could create value in a supply chain with budget constraints, and partly coordinate the supply chain.

Appendix

Proof of Proposition 1

Part 1. Let $\pi^{R}(Q)\equiv \mathbb{E} [(\min[D,Q]-(wQ-B))^{+}-B ]$. Then

$$\pi^R(Q)=\int_{(wQ-B)}^Q\bigl[D-(wQ-B)\bigr]dF(D)+\int_Q^\infty \bigl[Q-(wQ-B)\bigr]dF(D)-B$$

Differentiating π ^R with respect to Q and collecting terms, we have

$$\frac{d\pi^R(Q)}{dQ}=\bar{F}(Q)-w\bar{F}(wQ-B)$$

The retailer’s optimal ordering level, Q ^∗, must satisfy the first-order condition:

$$\bar{F}\bigl(Q^*\bigr)=w\bar{F}\bigl(wQ^*-B\bigr).$$

We further have

Since h is an increasing function by Assumption A1, we have h(Q ^∗)>h(wQ ^∗−B). Therefore, $\frac{d^{2}\pi^{R}(Q)}{d(Q)^{2}}|_{Q=Q^{*}}<0$ and Q ^∗ is thus unique.

Part 2. Inspection of $\bar{F}(Q^{*})=w\bar{F}(wQ^{*}-B)$ readily reveals that

$$\frac{dQ^*}{dB}=-\frac{h(wQ^*-B)}{h(Q^*)-wh(wQ^*-B)}<0.$$

Then, we can show that Q ^∗ is decreasing in B. We further show that $\frac{dQ^{*}}{dw}<0$. Let

$$G\bigl(Q^*,w,B\bigr)=\bar{F}\bigl(Q^*\bigr)-w\bar{F}\bigl(wQ^*-B\bigr).$$

Because of $\frac{dQ^{*}}{dw}=-\frac{dG(Q^{*},w,B)}{dw}/\frac{dG(Q^{*},w,B)}{dQ^{*}}$ and $\frac{dwQ^{*}}{dw}=Q^{*}+w\frac{dQ^{*}}{dw}$, we obtain

Suppose $\frac{dQ^{*}}{dw}\geq 0$. Then wQ ^∗ and Q ^∗ h(Q ^∗) are increasing in w. Let w ₀ satisfy 1−Q ^∗(w ₀)h(Q ^∗(w ₀))=0. We have $\frac{dQ^{*}}{dw}<0$ when w<w ₀ and $\frac{dQ^{*}}{dw}>0$ when w>w ₀. That means that wQ achieves its minimum at w ₀. Recall that under trade credit contract the feasible region of wholesale price is (c,1). The remaining proof splits into three cases.

• Case A: w ₀≥1. In this case,

  $$Q^*(w)h\bigl(Q^*(w)\bigr)\leq Q^*(1)h\bigl(Q^*(1)\bigr)\leq Q^*(w_0)h\bigl(Q^*(w_0)\bigr)=1$$

  which implies that $\frac{dwQ^{*}}{dw}\leq 0$. A contradiction.

• Case B: c<w ₀<1. We have wQ ^∗ being decreasing in w over (c,w ₀). Again, a contradiction.

• Case C: w ₀=c. In this case, we have

  $$wQ^*(c)h\bigl(wQ^*(c)-B\bigr)<Q^*(c)h\bigl(Q^*(c)\bigr)=1$$

  because w<1 and B≥0 by assumption. Since we suppose $\frac{dQ^{*}}{dw}\geq 0$, we also obtain wQ ^∗ h(wQ ^∗−B)≥1, a contradiction to the inequality above.

We have reached a contradiction in each of the three cases. Therefore, we must have $\frac{dQ^{*}}{dw}<0$. □

Proof of Proposition 2

First, we need to proof Π^s(w) is a unimodal function of w∈[c,p], where p=1.

In (9), we can show that $\frac{d\Pi^{S}}{dw}=\frac{[1-wQ^{*}h(wQ^{*}-B)]}{-w[h(Q^{*})-wh(wQ^{*}-B)]}<0$.

Let $\varepsilon(w)=\frac{\bar{F}(Q^{*})[1-H(Q^{*})]}{1-wQ^{*}h(wQ^{*}-B)}$, then the sign of $\frac{d\Pi^{S}}{dw}$ depends on that of [ε(w)−c]. Let w ₀ satisfies 1−H(Q ^∗)=0. When w<w ₀, we have ε(w)<0 and thus $\frac{d\Pi^{S}}{dw}>0$. When w>w ₀, we have ε>0.

Next, we aim to show that ε(w) is an increasing function when w>w ₀. We have

where Q≡Q ^∗, $H'(D)\equiv\frac{dH(D)}{dD}>0$, $Q'\equiv \frac{dQ^{*}}{dw}$, to simplify notation.

Therefore, ε(w) is an increasing function when w>w ₀. Let $\hat{w}$ solves ε(w)=c. This further implies that $\frac{d\Pi^{S}}{dw}>0$ when $w<\hat{w}$, $\frac{d\Pi^{S}}{dw}=0$ when $w=\hat{w}$ and $\frac{d\Pi^{S}}{dw}<0$ when $w>\hat{w}$. And, w=c, we can get ε<c, and $\frac{d\Pi^{S}}{dw}>0$; Alternatively, w=c, we can get ε>c, and $\frac{d\Pi^{S}}{dw}<0$. Therefore, we can conclude that Π^S(w) is a unimodal function of w∈[c,1]. Given the feasible region for $w\in[\underline{w},\bar{w}]$, as a consequence, we can show that the optimal solution for the supplier, w ^∗, is equivalent to $\min[\max[\underline{w},\hat{w}],\bar{w}]$. Then, the proof is well done. □