Bargaining on supply chain networks with heterogeneous valuations

Abstract

In this paper, we generalize the two-sided bargaining model which is developed in Manea (Am Econ Rev 101(5):2042–2080, 2011) to analyse the trade dynamics in supply chains. No further restrictions on the network structure are imposed. Both buyers and sellers can assume the role of a proposer in the bargaining game. Any finite number of buyers and sellers is allowed. More importantly, valuations of the buyers are heterogeneous. This generalized model allows us to investigate the impact of network structure and the heterogeneous pie size (resulting from heterogeneous valuations) on the equilibrium in full generality which is lack in the existing supply chain literature.

Appendix

Proof of Lemma 1

Without loss of generality, suppose that $\max \{\omega _{1}, \omega _{2}, \omega _{3}, \omega _{4} \} = \omega _{1}$. Hence,

$$\begin{aligned} |\max \{ \omega _{1}, \omega _{2} \} - \max \{ \omega _{3}, \omega _{4} \} | \le \omega _{1} - \omega _{3} = |\omega _{1} - \omega _{3}| \le \max \{ |\omega _{1} - \omega _{3}|,|\omega _{2} - \omega _{4}| \}. \end{aligned}$$

$\square$

Proof of Lemma 2

By the definition of contraction mapping, we need to prove that

$$\begin{aligned} \forall u,w \in [0,1]^{n}: \, \Vert f^{\delta }(u) - f^{\delta }(w) \Vert \le \delta \Vert u - w \Vert , \end{aligned}$$

which means for all $k \in S \cup B$, $|f_{k}^{\delta }(u) - f_{k}^{\delta }(w)| \le \delta \Vert u - w \Vert$.

Without loss of generality, we prove the above inequality for all $s \in S$. This inequality can be easily shown for any buyer $b \in B$.

$$\begin{aligned}&\Vert f_{s}^{\delta }(u) - f_{s}^{\delta }(w)| \\&\quad = \Bigg |\Bigg (1-\sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2}\Bigg ) \delta (u_{s} - w_{s}) + \sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2} (\max \{ v_{b} - \delta u_{b} , \delta u_{s} \}\\&\qquad - \max \{ v_{b} - \delta w_{b} , \delta w_{s} \}) \Bigg | \\&\quad \le \Bigg |\Bigg (1-\sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2}\Bigg ) \delta (u_{s} - w_{s})| + \sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2} \delta (\max \{ | u_{b} - w_{b} |, | u_{s} - w_{s} |\}) \Bigg | \\&\quad \le \Bigg (1-\sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2}\Bigg ) \delta \Vert u-w\Vert + \sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2} \delta \Vert u-w\Vert \\&\quad = \delta \Vert u - w \Vert , \end{aligned}$$

implying that the function $f^{\delta }$ is a contraction mapping. Hence, the function has a unique fixed point. $\square$

Proof of Lemma 3

$(s,b) \in G^{*\delta }$ means that that s and b are connected and $\max \{ v_{b} - \delta u_{s}^{*\delta }, \delta u_{b}^{*\delta } \} = v_{b} - \delta u_{s}^{*\delta }$. Since $u^{*\delta }$ is the fixed point of $f^{\delta }$, $u^{*\delta }$ is the solution of the following linear equations system

$$\begin{aligned} u_{s}&= \Bigg (1-\sum \limits _{\{b|(s,b) \in G^{*\delta }\}} \dfrac{p_{sb}}{2}\Bigg ) \delta u_{s} + \sum \limits _{\{b|(s,b) \in G^{*\delta }\}} \dfrac{p_{sb}}{2} (v_{b} - \delta u_{b} ), \,\, \forall s \in S \\ u_{b}&= \Bigg (1-\sum \limits _{\{ s|(s,b) \in G^{*\delta }\}} \dfrac{p_{sb}}{2}\Bigg ) \delta u_{b} + \sum \limits _{\{s|(s,b) \in G^{*\delta }\}} \dfrac{p_{sb}}{2} (v_{b} - \delta u_{s} ), \,\, \forall b \in B. \end{aligned}$$

Take any non-empty subnetwork H of G and define a mapping $h^{\delta , H}: {\mathbb {R}}^{n} \rightarrow {\mathbb {R}}^{n}$ such that for each $s \in S$ and $b \in B$,

$$\begin{aligned} h_{s}^{\delta , H}(u)= & {} \Bigg (1-\sum \limits _{\{b|(s,b) \in H\}} \dfrac{p_{sb}}{2}\Bigg ) \delta u_{s} + \sum \limits _{\{b|(s,b) \in H\}} \dfrac{p_{sb}}{2} (v_{b} - \delta u_{b} ), \, \forall s \in S \nonumber \\ h_{b}^{\delta , H}(u)= & {} \Bigg (1-\sum \limits _{\{ s|(s,b) \in H\}} \dfrac{p_{sb}}{2}\Bigg ) \delta u_{b} + \sum \limits _{\{s|(s,b) \in H\}} \dfrac{p_{sb}}{2} (v_{b} - \delta u_{s} ), \, \forall b \in B. \end{aligned}$$

(9)

$h^{\delta , H}$ is a contraction mapping. All equations in the system (9) are linear functions of $\delta$, implying that for each $k \in S \cup B$, $u_{k}^{\delta , H}$ is given by the Cramer’s rule

$$\begin{aligned} u_{k}^{\delta , H} = \dfrac{P_{k}^{H}(\delta )}{Q_{k}^{H}(\delta )}, \end{aligned}$$

(10)

where $P_{k}^{H}$ and $Q_{k}^{H}$ are polynomials in $\delta$. Since the linear system in (10) is non-singular, $Q_{k}^{H}(\delta ) \ne 0$ for all $\delta \in (0,1)$ and for all non-empty subnetworks H of G.

Take any $s \in S$, $b \in B$ and any non-empty subnetwork H of G. $\delta (u_{s}^{\delta , H} + u_{b}^{\delta , H}) = v_{b}$ is equivalent to

$$\begin{aligned} v_{b} = \delta \left ( \dfrac{P_{s}^{H}(\delta )}{Q_{s}^{H}(\delta )} + \dfrac{P_{b}^{H}(\delta )}{Q_{b}^{H}(\delta )} \right ). \end{aligned}$$

Since the equation above is valid for all $\delta \in (0,1)$, it holds also for $\delta = 1/3$. Rewriting the equation for this specific value of $\delta$, we have

$$\begin{aligned} 3v_{b} = u_{s}^{1/3,H} + u_{b}^{1/3,H}, \end{aligned}$$

which contradicts with for all $k \in S \cup B$, $u_{k}^{1/3,H} \le v_{b}$. It follows that for all (s, b, H) the statement $\delta (u_{s}^{1/3,H} + u_{b}^{1/3,H}) = v_{b}$ holds for a finite set of solutions $\delta$, which concludes the proof. $\square$

Proof of Lemma 4

Take any link $(s,b) \in G$. $(s,b) \in G \backslash G^{*}$ implies that for all $\delta > {\underline{\delta }}$ (${\underline{\delta }}$ is determined in Theorem 2),

$$\begin{aligned} \delta (u_{s}^{*\delta } + u_{b}^{*\delta }) > v_{b}. \end{aligned}$$

(11)

If the link (s, b) is involved in the limit equilibrium agreement network $G^{*}$, then for all $\delta > {\underline{\delta }}$,

$$\begin{aligned} u_{s}^{*\delta }&= \Bigg (1-\sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2}\Bigg ) \delta u_{s}^{*\delta } + \sum \limits _{\{b|(s,b) \in G\}} \dfrac{p_{sb}}{2} ( v_{b} - \delta u_{b}^{*\delta } ) \\ u_{b}^{*\delta }&= \Bigg (1-\sum \limits _{\{s|(s,b) \in G\}} \dfrac{p_{sb}}{2}\Bigg ) \delta u_{b}^{*\delta } + \sum \limits _{\{s|(s,b) \in G\}} \dfrac{p_{sb}}{2} (v_{b} - \delta u_{s}^{*\delta } ). \end{aligned}$$

Since for all $k \ne s \in S$ with $(k,b) \in G^{*}$, $v_{b} - \delta u_{k}^{*\delta } \ge \delta u_{b}^{*\delta }$ and for all $l \ne b \in B$ with $(s,l) \in G^{*}$, $v_{l} - \delta u_{l}^{*\delta } \ge \delta u_{s}^{*\delta }$ we obtain

$$\begin{aligned} u_{s}^{*\delta }\ge & {} \Bigg (1-\dfrac{p_{sb}}{2}\Bigg ) \delta u_{s}^{*\delta } + \dfrac{p_{sb}}{2} ( v_{b} - \delta u_{b}^{*\delta } ) \nonumber \\ u_{b}^{*\delta }\ge & {} \Bigg (1-\dfrac{p_{sb}}{2}\Bigg ) \delta u_{b}^{*\delta } + \dfrac{p_{sb}}{2} (v_{b} - \delta u_{s}^{*\delta } ). \end{aligned}$$

(12)

As $\delta \rightarrow 1$, by (11), we have $u_{s}^{*} + u_{b}^{*} \ge v_{b}$ for all $(s,b) \in G \backslash G^{*}$. And from (11) and (12), for all $(s,b) \in G^{*}$, $u_{s}^{*} + u_{b}^{*} = v_{b}$. $\square$