The effect of supplier capacity on the supply chain profit

Abstract

In this paper, we study the role of capacity on the efficiency of a two-tier supply chain with two suppliers (leaders, first tier) and one retailer (follower, second tier). The suppliers compete via pricing (Bertrand competition) and, as one would expect in practice, are faced with production capacity. We consider a model with differentiated substitutable products where the suppliers are symmetric differing only by their production capacity. We characterize the prices, production amounts and profits in three cases: (1) the suppliers compete in a decentralized Nash equilibrium game, (2) the suppliers “cooperate” to optimize the total suppliers’ profit, and (3) the two tiers of the supply chain are centrally coordinated. We show that in a decentralized setting, the supplier with a lower capacity may benefit from restricting her capacity even when additional capacity is available at no cost. We also show that the loss of total profit due to decentralization cannot exceed 25 % of the centralized chain profits. Nevertheless, the loss of total profit is not a monotonic function of the “degree of asymmetry” of the suppliers’ capacities. Furthermore, we provide an upper bound on the supplier profit loss at equilibrium (compared with the cooperation setting) that depends on the “market power” of the suppliers as well as their market size. We show that there is less supplier profit loss as the asymmetry (in terms of their capacities) increases between the two suppliers. The worst case arises when the two suppliers are completely symmetric.

Appendices

Appendix 1: Notation

• wholesale price of supplier $k$: $p^k$ (subscript c for cooperative optimum, d for Nash equilibrium, cc for centrally coordinated)

• price of retailer for product of supplier $k$: $\bar{p}^k$

• production level of supplier $k$: $u^k$ (same subscripts)

• profit of supplier $k$: $J^k$ (same subscripts)

• profit of retailer: $J^R$ (same subscripts)

• overall supplier profit: $J=J^A+J^B$ (same subscripts)

• overall supply chain profit: $\Pi =J^A+J^B+J^R$ (same subscripts)

• demand level of supplier $k$: $d^k(\mathbf {p}) = \alpha -\beta p^k + \beta ^{\prime } p^{-k}$

• retail price for product of supplier $k$: $\bar{p}^k(\mathbf {D}) = \bar{\alpha }-\bar{\beta } d^k - \bar{\beta }^{\prime } d^{-k}$

• term of the supplier’s demand that is independent of prices: $\alpha$

• price elasticity of the demand of a supplier with respect to her own price: $\beta$

• price elasticity of the demand of a supplier with respect to her competitor’s price: $\beta ^{\prime }$

• $r=\frac{\beta ^{\prime }}{\beta }$

• production capacity of supplier $k$: $K^k$

• coefficient of quadratic supplier production cost: $\gamma$

• $t=\gamma \beta$

• $b=2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta )$

• $d=\sqrt{2\beta ^2(1+\gamma \beta )-\beta ^{'^2}(1+2\gamma \beta )}$

• capacity threshold in a monopoly setting: $K_1 = \frac{\alpha }{2(1+\gamma \beta )}$

• capacity threshold between regimes $a$ and $b$ (decentralized case): $l_a = \frac{\alpha \beta }{b}$

• capacity threshold between regimes $b$ and $c$ (decentralized case): $l_b(K^A) = \frac{\alpha \beta (\beta +\beta ^{\prime })-d^2 K^A}{\beta \beta ^{\prime }}$

• local maximum of $J^B$ as a function of $K^B$ in regime $b$: $K_0 =\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))}{2(2\beta (1+\gamma \beta )^2-\gamma \beta ^{'2}(1+2\gamma \beta ))}$

• local maximum of $J^B$ as a function of $K^B$ in regime $c$: $\bar{K}_0 =\frac{\alpha (\beta +\beta ^{\prime })-\beta K^A}{(\beta +\beta ^{\prime }) (1+2\gamma (\beta -\beta ^{\prime }))}$

• $K^* = \mathrm{Arg}\max _{K^B} (J^A_d+J^B_d)(K^B)$

• $\bar{K}_1 = \mathrm{Arg}\max _{K^B} J^B_d(K^B)$

• $\bar{K}=\frac{\beta K^B+\beta ^{\prime } K^A}{\beta +\beta ^{\prime }}$

• capacity threshold between regimes $a^{\prime }$ and $b^{\prime }$ (cooperative case): $l'_a = \frac{\frac{\alpha }{2}}{1+\gamma (\beta -\beta ^{\prime })}$

• capacity threshold between regimes $b^{\prime }$ and $c^{\prime }$ (cooperative case): $l'_b(K^A) = \frac{(\beta +\beta ^{\prime })\frac{\alpha }{2})-K^A(\beta +\gamma (\beta ^2-\beta ^{'^2}))}{\beta ^{\prime }}$

• profit ratio: $\rho = \frac{J^A_d+J^B_d}{J^A_c+J^B_c}$

• profit ratio for supplier $k$: $\rho ^k = \frac{J^k_d}{J^k_c}$

• $d_1 =4\beta (1+\gamma \beta )^2-2\gamma \beta ^{'^2}(1+2\gamma \beta )$

• $A=\beta (\beta +\beta ^{\prime })d_1-\beta \beta ^{\prime }(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))$

• $B=d_1(2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta ))$

Appendix 2: Optimal solution in a monopoly setting

$$\begin{aligned} J(K)=\max _{u,p}&p(\alpha -\beta p) - \gamma u^2\\ \mathrm{such\, that}&u=\alpha -\beta p\\&0\le p\le \frac{\alpha }{\beta }\\&0\le u \le K \end{aligned}$$

We obtain the optimal solution:

• if $\frac{\alpha }{2}-K(1+\gamma \beta )\le 0$, then $u=\frac{\alpha }{2(1+\gamma \beta )},\,p=\frac{\alpha }{2\beta }+\frac{\alpha }{2(\beta +\frac{1}{\gamma })}=\frac{1}{1+\gamma \beta }\Big ( \frac{\alpha }{2\beta }(1+2\gamma \beta )\Big )$.

• if $0 <\frac{\alpha }{2}-K(1+\gamma \beta )$, then $u=K,\,p=\frac{\alpha -K}{\beta }$.

$$\begin{aligned} J(K)=\left\{ \begin{array}{l@{\quad }l} \frac{\alpha ^2}{4\beta }-\frac{\gamma \alpha ^2}{4(1+\gamma \beta )} \,= \frac{\alpha ^2}{4\beta (1+\gamma \beta )},&{} \frac{\alpha }{2}-K(1+\gamma \beta )\le 0\\ \frac{K}{\beta }(\alpha -K)-\gamma K^2,&{} 0 <\frac{\alpha }{2}-K(1+\gamma \beta )\end{array}\right. \end{aligned}$$

In particular, $J(K)$ depends on $K$ only when the demand is high enough so that the production is at full capacity, i.e. when $0<\frac{\alpha }{2}-K(1+\gamma \beta )$, i.e. $K<K_1$, where $K_1\equiv \frac{\alpha }{2(1+\gamma \beta )}$.

We observe that on the domain $0<K<K_1$, $J(K)$ is quadratic and increasing. It reaches a maximum at $K=K_1$, while for $K>K_1$, $J(K)$ is independent of $K$, i.e. remains at the maximum value $J(K_1)$. In other words, beyond $K_1$, the supplier does not produce at full capacity and therefore any change of capacity above that value has no impact on the optimal strategy and thus the profits.

Appendix 3: Equilibrium solution derivation

Proposition 6

Due to Corollary 1, the equilibrium point and equilibrium profits are as follows:

• if $\,\alpha \beta -b K^B\le 0$,

  $$\begin{aligned}&u^A=u^{B}=\frac{\alpha \beta }{b},\,\, p^A=p^{B}=\frac{\alpha (1+2\gamma \beta )}{b}\\&J^A(K^A,K^B)=J^B(K^A,K^B)=\frac{\beta (1+\gamma \beta )\alpha ^2}{b^2} \end{aligned}$$

• if $\alpha \beta -b K^B>0$ and $\alpha \beta (\beta +\beta ^{\prime })-K^B\beta \beta ^{\prime }-d^2 K^A<0$,

  $$\begin{aligned}&p^A=(1+2\gamma \beta )\frac{\alpha (\beta +\beta ^{\prime })-K^B \beta ^{\prime }}{d^2},\\&p^B=\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-2K^B \beta (1+\gamma \beta )}{d^2}\\&u^A=\frac{\alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime }K^B}{d^2},\,\,u^B=K^B\\&J^A(K^A,K^B)=\beta (1+\gamma \beta )\left( \frac{\alpha (\beta +\beta ^{\prime })-\beta ^{\prime }K^B}{d^2}\right) ^2\\&J^B(K^A,K^B)\!=\!K^B\left( \frac{\alpha (2\beta (1+\gamma \beta )\!+\!\beta ^{\prime }(1+2\gamma \beta )) -2K^B \beta (1\!+\!\gamma \beta )}{2\beta ^2(1+\gamma \beta )\!-\!\beta ^{'2}(1+2\gamma \beta )}\right) -\gamma (K^B)^2 \end{aligned}$$

• if $\alpha \beta (\beta +\beta ^{\prime })-K^B\beta \beta ^{\prime }-d^2 K^A\ge 0$ (which implies $\,\alpha \beta -b K^B>0$),

  $$\begin{aligned}&u^A=K^A,\,\,u^{B}=K^B,\,\,p^A=\frac{\alpha -\frac{\beta K^A+\beta ^{\prime } K^B}{\beta +\beta ^{\prime }}}{\beta -\beta ^{\prime }},\,\, p^B=\frac{\alpha -\frac{\beta K^B+\beta ^{\prime } K^A}{\beta +\beta ^{\prime }}}{\beta -\beta ^{\prime }}\\&J^A(K^A,K^B)=\frac{K^A}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^A+\beta ^{\prime }K^B}{\beta +\beta ^{\prime }}\right) -\gamma (K^A)^2\\&J^B(K^A,K^B)=\frac{K^B}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^B+\beta ^{\prime }K^A}{\beta +\beta ^{\prime }}\right) -\gamma (K^B)^2 \end{aligned}$$

Proof

The best response solution can be derived using Lagrangian duality or using results from the monopoly case (Appendix 2):

• Case 1

  $$\begin{aligned}&\frac{\alpha +\beta ^{\prime } p^{-k}}{2}-K^k(1+\gamma \beta )\le 0\, \Rightarrow \\&u^k=\frac{\alpha +\beta ^{\prime } p^{-k}}{2(1+\gamma \beta )},\,p^k=\frac{\alpha +\beta ^{\prime }p^{-k}}{2\beta }+\ \frac{\gamma (\alpha + \beta ^{\prime } p^{-k})}{2(1+\gamma \beta )}=\frac{(\alpha +\beta ^{\prime }p^{-k})(1+2\gamma \beta )}{2\beta (1+\gamma \beta )}. \end{aligned}$$

• Case 2

  $$\begin{aligned} 0 <\frac{\alpha +\beta ^{\prime } p^{-k}}{2}-K^k(1+\gamma \beta )\,\Rightarrow \, u^k=K^k,\,\, p^k=\frac{\alpha +\beta ^{\prime } p^{-k}-K^k}{\beta }. \end{aligned}$$

To determine the equilibrium, we must consider three possibilities.

A and B in case 1 We observe that, only the production capacity differentiates the suppliers, and therefore the problem is symmetric if none of them produces at full capacity. By symmetry, $u^k=u^{-k}=u,\,\,p^k=p^{-k}=p$ and $u=\frac{\alpha +\beta ^{\prime }p}{2(1+\gamma \beta )},\,\,p=\frac{(\alpha +\beta ^{\prime } p)(1+2\gamma \beta )}{2\beta (1+\gamma \beta )}.$

$$\begin{aligned} p&= (\alpha +\beta ^{\prime } p)\frac{1+2\gamma \beta }{2\beta (1+\gamma \beta )}\\ p\left( 1-\beta ^{\prime }\frac{1+2\gamma \beta }{2\beta (1+\gamma \beta )}\right)&= \alpha \frac{1+2\gamma \beta }{2\beta (1+\gamma \beta )}\\ p&= \frac{\alpha +2\gamma \beta \alpha }{2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta )} \end{aligned}$$

and thus

$$\begin{aligned} u=\frac{\alpha \beta }{2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta )}. \end{aligned}$$

For this case to hold, we must verify

$$\begin{aligned}&0\le \alpha +\beta ^{\prime } p\le 2 K^k (1+\gamma \beta )\\ \Leftrightarrow&0\le \alpha +\beta ^{\prime }\frac{\alpha (1+2\gamma \beta )}{2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta )} \le 2 K^k (1+\gamma \beta )\\ \Leftrightarrow&(\alpha -2 K^k (1+\gamma \beta ))(2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta ))+\beta ^{\prime }\alpha (1+2\gamma \beta )\le 0\\ \Leftrightarrow&\alpha \beta -K^k(2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta ))\le 0,\,\,k=A,\,B\\ \Leftrightarrow&\alpha \beta -b K^k\le 0,\,\,k=A,\,B \end{aligned}$$

Since $K^B\le K^A$, this case holds if

$$\begin{aligned} \alpha \beta -b K^B\le 0. \end{aligned}$$

A and B in case 2 $u^k=K^k,\,\,p^A=\frac{\alpha +\beta ^{\prime }p^{B}-K^A}{\beta },\,\,p^B=\frac{\alpha +\beta ^{\prime }p^{A}-K^B}{\beta }$, therefore

$$\begin{aligned} p^A&= \frac{\alpha -K^A}{\beta }+\frac{\beta ^{\prime }}{\beta }\frac{\alpha +\beta ^{\prime }p^{A}-K^B}{\beta }\\ p^A\left( 1-\frac{\beta ^{'^2}}{\beta ^2}\right)&= \frac{\beta \alpha -\beta K^A+\beta ^{\prime }(\alpha -K^B)}{\beta ^2}\\ p^A&= \frac{\alpha -\frac{\beta K^A+\beta ^{\prime } K^B}{\beta +\beta ^{\prime }}}{\beta -\beta ^{\prime }} \end{aligned}$$

and by symmetry

$$\begin{aligned} p^B=\frac{\alpha -\frac{\beta K^B+\beta ^{\prime } K^A}{\beta +\beta ^{\prime }}}{\beta -\beta ^{\prime }}. \end{aligned}$$

For this case to hold, we must verify

$$\begin{aligned} 0&< \frac{\alpha }{2}-K^k(1+\gamma \beta )+\frac{\beta ^{\prime }}{2}\frac{\alpha -\frac{\beta K^{-k}+\beta ^{\prime }K^k}{\beta +\beta ^{\prime }}}{\beta -\beta ^{\prime }},\,\,k=A,\,B\\ 0&< \alpha \beta -2K^k(\beta -\beta ^{\prime })(1+\gamma \beta )-\beta ^{\prime }\frac{\beta K^{-k}+\beta ^{\prime }K^k}{\beta +\beta ^{\prime }},\,\,k=A,\,B\\ \Leftrightarrow \,\,0&< \alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime } K^{-k}-K^k(2\beta ^2(1+\gamma \beta )-\beta ^{'2} (1+2\gamma \beta )),\,\,k=A,\,B\\ \Leftrightarrow \,\,0&< \alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime } K^{-k}-d^2 K^k,\,\,k=A,\,B \end{aligned}$$

Lemma 2

$0<\alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime } K^{B}-d^2 K^A$ implies $0<\alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime } K^{A}-d^2 K^B$

Proof

To prove the lemma, it is sufficient to show that $\beta \beta ^{\prime } K^{B}+d^2 K^A\ge \beta \beta ^{\prime } K^{A}+d^2 K^B$.

$$\begin{aligned} \beta \beta ^{\prime } K^{B}+d^2 K^A- \beta \beta ^{\prime } K^{A}-d^2 K^B&= (K^A-K^B)(d^2-\beta \beta ^{\prime })\\&= (K^A-K^B)(-\beta \beta ^{\prime }+2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta )\\&= (K^A-K^B)((2\beta +\beta ^{\prime })(\beta -\beta ^{\prime })+2\gamma \beta (\beta ^2-\beta ^{'2}))\\&= (K^A-K^B)(\beta -\beta ^{\prime })(2\beta +\beta ^{\prime }+2\gamma \beta (\beta +\beta ^{\prime }))\ge 0 \end{aligned}$$

$\square$

Therefore, this case holds for

$$\begin{aligned} 0<\alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime } K^{B}-d^2 K^A \end{aligned}$$

Remark The inequality above implies $\,\alpha \beta -b K^B>0$. Indeed, since $K^A\ge K^B$, this inequality implies

$$\begin{aligned} 0\!<\!\alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime } K^{B}\!- d^2 K^A \le \alpha \beta (\beta +\beta ^{\prime })-(\beta \beta ^{\prime }\!+\!d^2)K^B\!=\!(\alpha \beta -b K^B)(\beta +\beta ^{\prime }). \end{aligned}$$

A in case 1, B in case 2

$$\begin{aligned} u^A=\frac{\alpha +\beta ^{\prime } p^B}{2(1+\gamma \beta )},\,u^B=K^B,\,\, p^A=\frac{(\alpha +\beta ^{\prime }p^{B})(1+2\gamma \beta )}{2\beta (1+\gamma \beta )},\,\, p^B=\frac{\alpha +\beta ^{\prime } p^{A}-K^B}{\beta } \end{aligned}$$

thus

$$\begin{aligned} p^A&= \frac{\alpha (1+2\gamma \beta )}{2\beta (1+\gamma \beta )}+\frac{\beta ^{\prime }(1+2\gamma \beta )}{2\beta (1+\gamma \beta )}\frac{\alpha +\beta ^{\prime } p^{A}-K^B}{\beta }\\ p^A&= \frac{\alpha (1+2\gamma \beta )(\beta +\beta ^{\prime })-K^B \beta ^{\prime }(1+2\gamma \beta ))}{2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta )}\\ p^B&= \frac{\alpha -K^B}{\beta }+\frac{\beta ^{\prime }}{\beta }(\frac{\alpha (1+2\gamma \beta )(\beta +\beta ^{\prime })-K^B \beta ^{\prime }(1+2\gamma \beta )}{2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta )})\\&= \frac{\alpha \beta (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-2K^B \beta ^2(1+\gamma \beta )}{\beta (2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta ))} \end{aligned}$$

For this case to hold, we must verify

$$\begin{aligned} 0&\ge \frac{\alpha +\beta ^{\prime } p^B}{2}-K^A(1+\gamma \beta )\\&= \beta \frac{\alpha (\beta +\beta ^{\prime })(1+\gamma \beta )-K^B\beta ^{\prime } (1+\gamma \beta )}{2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta )}-K^A(1+\gamma \beta )\\ \Leftrightarrow 0&\ge \alpha \beta (\beta +\beta ^{\prime })-K^B\beta \beta ^{\prime }-K^A(2\beta ^2(1+\gamma \beta )-\beta ^{'2} (1+2\gamma \beta ))\\ \Leftrightarrow 0&\ge \alpha \beta (\beta +\beta ^{\prime })-K^B\beta \beta ^{\prime }-d^2 K^A \end{aligned}$$

and

$$\begin{aligned} 0&\le \frac{\alpha +\beta ^{\prime } p^A}{2}-K^B(1+\gamma \beta )\\&= \frac{\alpha \beta (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))- K^B \beta ^{'2}(1+2\gamma \beta )}{4\beta ^2(1+\gamma \beta )-2\beta ^{'2}(1+2\gamma \beta )}-K^B(1+\gamma \beta )\\&= \frac{\alpha \beta (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))- K^B( 4\beta ^{2}(1+\gamma \beta )^2-\beta ^{'2}(1+2\gamma \beta )^2)}{4\beta ^2(1+\gamma \beta )-2\beta ^{'2}(1+2\gamma \beta )}\\ \Leftrightarrow 0&\le \alpha \beta (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-K^B( 4\beta ^{2} (1+\gamma \beta )^2-\beta ^{'2}(1+2\gamma \beta )^2)\\&= \alpha \beta -K^B(2\beta (1+\gamma \beta )-\beta ^{\prime }(1+2\gamma \beta ))\\&= \alpha \beta -b K^B \end{aligned}$$

Remark It is impossible to have A in case 2 and B in case 1. Indeed, for this case to hold, we would need $0 \ge \alpha \beta (\beta +\beta ^{\prime })-K^A\beta \beta ^{\prime }-d^2 K^B$ and $0<\alpha \beta -b K^A$, but $0 \ge \alpha \beta (\beta +\beta ^{\prime })-K^A\beta \beta ^{\prime }-d^2 K^B$ implies

$$\begin{aligned} 0 \ge \alpha \beta (\beta +\beta ^{\prime })-K^A\beta \beta ^{\prime }-d^2 K^A=(\beta +\beta ^{\prime })(\alpha \beta -K^Ab). \end{aligned}$$

$\square$

Appendix 4: Optimal solution under supplier cooperation

While the capacity constraint is not tight for either supplier, by symmetry, the solutions are identical for each supplier, and they are a solution to the monopoly problem with price sensitivity $\beta -\beta ^{\prime }$, i.e.:

• if $\frac{\alpha }{2}-K^B(1+\gamma (\beta -\beta ^{\prime }))\le 0$, then

  $$\begin{aligned}&u^A=u^B=\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))},\,\,p^A=p^B= \frac{\alpha (1+2\gamma (\beta -\beta ^{\prime }))}{2(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))}.\\&J^A=J^B=\frac{\alpha ^2}{4(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))} \end{aligned}$$

Note that we can show

$$\begin{aligned} \frac{\alpha }{2}-K^A(1+\gamma (\beta -\beta ^{\prime }))+\frac{\beta ^{\prime }}{\beta +\beta ^{\prime }}(K^A-K^B)&= \frac{\alpha }{2}-\gamma K^A(\beta -\beta ^{\prime })- \frac{\beta K^A + \beta ^{\prime } K^B}{\beta +\beta ^{\prime }}\\&<\frac{\alpha }{2}-K^B(1+\gamma (\beta -\beta ^{\prime })). \end{aligned}$$

Some calculations involving minimizing the Lagrangian similarly to Appendix 3 lead to the additional cases:

• if $\frac{\alpha }{2}-K^A(1+\gamma (\beta -\beta ^{\prime }))+\frac{\beta ^{\prime }(K^A-K^B)}{\beta +\beta ^{\prime }}< 0 < \frac{\alpha }{2}-K^B(1+\gamma (\beta -\beta ^{\prime }))$, then

  $$\begin{aligned}&u^A = \frac{\frac{\alpha }{2}(\beta +\beta ^{\prime })-\beta ^{\prime } K^B}{\beta +\gamma (\beta ^2-\beta ^{'2})},\,\,u^B=K^B,\\&p^A=\frac{\alpha }{2(\beta -\beta ^{\prime })}+\frac{\frac{\alpha }{2}\gamma (\beta +\beta ^{\prime })-\gamma \beta ^{\prime } K^B}{\beta +\gamma (\beta ^2-\beta ^{'2})},\\&p^B =\frac{\alpha }{2(\beta -\beta ^{\prime })}+\frac{\frac{\alpha }{2}(1+\gamma (\beta +\beta ^{\prime }))- K^B(1+\gamma \beta )}{\beta +\gamma (\beta ^2-\beta ^{'2})}\\&J^A=\frac{\alpha \left( \frac{\alpha }{2}(\beta +\beta ^{\prime })-\beta ^{\prime } K^B\right) }{2(\beta -\beta ^{\prime })(\beta +\gamma (\beta ^2-\beta ^{'2}))},\\&J^B=-\gamma (K^B)^2+K^B\left( \frac{\alpha }{2(\beta -\beta ^{\prime })}+\frac{\frac{\alpha }{2}(1+\gamma (\beta +\beta ^{\prime }))- K^B(1+\gamma \beta )}{\beta +\gamma (\beta ^2-\beta ^{'2})}\right) \end{aligned}$$

• if $0\le \frac{\alpha }{2}-K^A(1+\gamma (\beta -\beta ^{\prime }))+\frac{\beta ^{\prime }(K^A-K^B)}{\beta +\beta ^{\prime }}$, then

  $$\begin{aligned}&u^B=K^B,\,u^A = K^A,\\&p^A =\frac{1}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^A + \beta ^{\prime } K^B}{\beta +\beta ^{\prime }}\right) ,\, p^B =\frac{1}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^B + \beta ^{\prime } K^A}{\beta +\beta ^{\prime }}\right) \\&J^A =-\gamma (K^A)^2 +\frac{K^A}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^A + \beta ^{\prime } K^B}{\beta +\beta ^{\prime }}\right) ,\,\, J^B =-\gamma (K^B)^2 +\frac{K^B}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^B + \beta ^{\prime } K^A}{\beta +\beta ^{\prime }}\right) \end{aligned}$$

Appendix 5: The centrally coordinated optimal solution

Proposition 7

The centrally coordinated optimal solution to the central planner’s optimization problem is:

• if $K^B\ge \frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}$, then

  $$\begin{aligned}&u^A=u^B=\frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}\equiv u_0,\,\,p^A=p^B= \frac{2\gamma \alpha }{1+2\gamma (\beta -\beta ^{\prime })}\equiv p_0\\&\Pi _{cc} =\frac{\alpha ^2}{(\beta -\beta ^{\prime })(1+2\gamma (\beta -\beta ^{\prime }))} \end{aligned}$$

• if $\frac{\alpha (\beta +\beta ^{\prime })-K^A(\beta +2\gamma (\beta ^2 - \beta ^{'2}))}{\beta ^{\prime }}\le K^B< \frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}$ then

  $$\begin{aligned}&u^A = u_0 + \frac{\beta ^{\prime } (u_0-K^B)}{\beta +2\gamma (\beta ^2 - \beta ^{'2})} = \frac{\alpha (\beta +\beta ^{\prime })-\beta ^{\prime } K^B}{\beta +2\gamma (\beta ^2 - \beta ^{'2})},\,\,u^B=K^B,\\&p^A = p_0 +\frac{2\gamma \beta ^{\prime } (u_0-K^B)}{\beta +2\gamma (\beta ^2 - \beta ^{'2})},\,\,\, p^B =p_0 +\frac{(1+2\gamma \beta ) (u_0-K^B)}{\beta +2\gamma (\beta ^2 - \beta ^{'2})} \end{aligned}$$

• if $K^B<\frac{\alpha (\beta +\beta ^{\prime })-K^A(\beta +2\gamma (\beta ^2 - \beta {'2}))}{\beta ^{\prime }}$, then

  $$\begin{aligned}&u^B=K^B,\,u^A = K^A,\\&p^A=\frac{\alpha (\beta +\beta ^{\prime })-\beta K^A -\beta ^{\prime } K^B}{\beta ^2 - \beta ^{'2}},\quad p^B =\frac{\alpha (\beta +\beta ^{\prime })-\beta ^{\prime } K^A -\beta K^B}{\beta ^2 - \beta ^{'2}} \end{aligned}$$

The proof uses Lagrangian duality and is straightforward; it is thus omitted.

Appendix 6: Figures

See Figs. 8, 9, 10, 11, 12.

Fig. 8

Cooperative and decentralized solution on two examples

Fig. 9

Example: total profit ratio $\frac{\Pi _{d}}{\Pi _{cc}}$ as a function of $K^B$ varying from 0 to $K^A$ with inputs (a) as in Fig. 8a; b as in Fig. 8b; c $\beta ^{\prime }=0.95$, $\beta =1$, $\alpha =6$, $K^A=7$, $\gamma =1$

Fig. 10

Example for asymmetric suppliers: $\Pi _{cc}$, $\Pi _{d}$, $\Pi _{c}$ and $\Pi _{cc}\Big (1-\frac{1}{4(1+\gamma (\beta -\beta ^{\prime }))^2}\Big )$ as a function of $K^B$ varying from 0 to $K^A$ with inputs as in Figs. 2, 8a, b, and 9c (In the first plot, $\Pi _{cc}$ and $\Pi _{d}$ overlap)

Fig. 11

Example: profit ratio as a function of $K^B$ varying from 0 to $K^A$ with inputs as in a Fig. 2; b Fig. 8a; c Fig. 8b; d Fig. 9c

Fig. 12

Illustration of an agreement between the suppliers to determine supplier B’s capacity

Appendix 7: Proofs

1.1 Proof of Lemma 1

(i) We can show that

$$\begin{aligned} l_b(K^A)-l'_b(K^A) = \frac{\beta +\beta ^{\prime }}{\beta \beta ^{\prime }}\left( \frac{\alpha }{2}\beta -(\beta -\beta ^{\prime })K^A(1+\gamma \beta )\right) \end{aligned}$$

Regime $c^{\prime }$ implies $l'_b(K^A)>0$, that is

$$\begin{aligned} K^A(\beta (1+\gamma \beta )-\gamma \beta ^{'2}) < (\beta +\beta ^{\prime })\frac{\alpha }{2}. \end{aligned}$$

Moreover, we can show that

$$\begin{aligned} \frac{\beta +\beta ^{\prime }}{\beta (1+\gamma \beta )-\gamma \beta ^{'2}}< \frac{1}{1+\gamma \beta }\frac{\beta }{\beta -\beta ^{\prime }} \end{aligned}$$

as follows:

$$\begin{aligned}&\frac{\beta +\beta ^{\prime }}{\beta (1+\gamma \beta )-\gamma \beta ^{'2}}< \frac{1}{1+\gamma \beta }\frac{\beta }{\beta -\beta ^{\prime }}\\&\Leftrightarrow (1+\gamma \beta )(\beta ^2-\beta ^{'2})<\beta (\beta (1+\gamma \beta )-\gamma \beta ^{'2})\\&\Leftrightarrow -\beta ^{'2}(1+\gamma \beta )<-\gamma \beta \beta ^{'2} \end{aligned}$$

which clearly holds.

Therefore, $l'_b(K^A)>0$ implies

$$\begin{aligned} K^A< \frac{(\beta +\beta ^{\prime })\frac{\alpha }{2}}{\beta (1+\gamma \beta )-\gamma \beta ^{'2}}< \frac{1}{1+\gamma \beta }\frac{\alpha }{2}\frac{\beta }{\beta -\beta ^{\prime }} \end{aligned}$$

which in turn implies $l_b(K^A)-l'_b(K^A)>0$. Finally, if regime $c^{\prime }$ holds, then $K^B<l'_b(K^A)<l_b(K^A)$ and therefore, in a decentralized setting, regime $c$ holds.

(ii) In this proof, the subscript $d$ (resp. $c$) refers to the decentralized (resp. cooperative) solution.

As seen in Section 2.3, regime a implies regime a$^{\prime }$ and regime c$^{\prime }$ implies regime c. Therefore, the inputs may correspond to one of 7 possible situations: regimes a and a$^{\prime }$, b and a$^{\prime }$, b and b$^{\prime }$, c and a$^{\prime }$, c and b$^{\prime }$, c and c$^{\prime }$, for respectively the decentralized and centralized settings.

• Regimes ($a,a^{\prime }$): Since the optimal and equilibrium (resp.) production quantities are given by $u^A_c = u^B_c =\frac{\alpha }{2(1+\gamma \beta )-2\gamma \beta ^{\prime }}$ and $u^A_d = u^B_d =\frac{\alpha }{2(1+\gamma \beta )-\frac{\beta ^{\prime }}{\beta }-2\gamma \beta ^{\prime }}$, it is clear that $u^A_c = u^B_c < u^A_d = u^B_d$.

• Regimes ($b,a^{\prime }$): The equilibrium production quantity for B is $u_d^B=K^B$, so by feasibility it follows that $u_c^B\le u^B_d$. Moreover, for supplier A, $u_d^A=\frac{\alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime }K^B}{2\beta ^2(1+\gamma \beta )-\beta ^{'2} (1+2\gamma \beta )},\,\,u_c^A=\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}$. Some calculations lead to

  $$\begin{aligned} u_d^A\ge u_c^A&\Leftrightarrow \left( \alpha \beta (\beta +\beta ^{\prime })-\beta \beta ^{\prime }K^B\right) \left( 2(1+\gamma (\beta -\beta ^{\prime }))\right) \\&<\alpha \left( 2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta )\right) \\&\Leftrightarrow \alpha (2\beta +\beta ^{\prime })\ge 2\beta K^B(1+\gamma (\beta -\beta ^{\prime })) \end{aligned}$$

  Under regime b, we have $\alpha \beta \ge K^B(2\beta (1+\gamma \beta )-\beta ^{'}(1+2\gamma \beta ))$, and therefore

  $$\begin{aligned} \alpha (2\beta +\beta ^{\prime })\ge \frac{2\beta +\beta ^{\prime }}{\beta }\left( K^B(2\beta (1+\gamma \beta )-\beta ^{'}(1+2\gamma \beta ))\right) . \end{aligned}$$

  As a result, to show $u_d^A\ge u_c^A$, it is sufficient to show that the right hand side in the inequality above is greater than or equal to $2\beta K^B(1+\gamma (\beta -\beta ^{\prime }))$. We observe that

  $$\begin{aligned}&\frac{2\beta +\beta ^{\prime }}{\beta }\left( K^B(2\beta (1+\gamma \beta )-\beta ^{'}(1+2\gamma \beta ))\right) \\&\quad - 2\beta K^B(1+\gamma (\beta -\beta ^{\prime }))= K^B(\frac{2\beta ^2-\beta ^{'2}}{\beta }+2\gamma (\beta ^2-\beta ^{'2}))>0 \end{aligned}$$

  and the result follows.

• Regimes ($b,b'$): The optimal and equilibrium (resp.) production quantities are given by $u_d^A=\frac{\alpha (\beta +\beta ^{\prime })-\beta ^{\prime }K^B}{2\beta (1+\gamma \beta )- \frac{\beta ^{'2}}{\beta }-2\gamma \beta ^{'2}},\,\, u_c^A = \frac{\alpha (\beta +\beta ^{\prime })-2\beta ^{\prime } K^B}{2\beta (1+\gamma \beta )-2\gamma \beta ^{'2}},\;u_d^B=u_c^B=K^B$, thus it is clear that $u_d^A>u_c^A$.

• Regimes ($c,a^{\prime }$), ($c,b^{\prime }$) and ($c,c^{\prime }$): The equilibrium production quantities are given by $u^A_d = K^A,\,\,u^B_d =K^B$ so by feasibility $u_c^A\le u^A_d$ and $u_c^B\le u^B_d$ and the result follows.

$\square$

1.2 Proof of Theorem 3

Part 1

It is straightforward to derive that if $K^A=K^B\equiv K$, then

$$\begin{aligned} J^R_d&= \left\{ \begin{array}{l@{\quad }l}\frac{\alpha ^2\beta ^2}{b^2(\beta -\beta ^{\prime })} &{} \mathrm{if }\, \alpha \beta -bK\le 0\\ \frac{K^2}{\beta -\beta ^{\prime }} &{} \mathrm{else}\end{array}\right. \\ J^R_c&= \left\{ \begin{array}{l@{\quad }l}\frac{\alpha ^2}{4(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))^2} &{} \mathrm{if }\, \frac{1}{2}\alpha -K(1+\gamma (\beta -\beta ^{\prime }))\le 0\\ \frac{K^2}{\beta -\beta ^{\prime }} &{} \mathrm{else}\end{array}\right. \end{aligned}$$

In particular,

$$\begin{aligned} \frac{J^R_d}{J^R_c} =\left\{ \begin{array}{l@{\quad }l}\frac{4\beta ^2(1+\gamma (\beta -\beta ^{\prime }))^2}{b^2}&{} \mathrm{if }\, K\ge \alpha \beta /b\\ \frac{4K^2(1+\gamma (\beta -\beta ^{\prime }))^2}{\alpha ^2}&{} \mathrm{if }\, \frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}\le K< \alpha \beta /b \\ 1&{} \mathrm{else}\end{array}\right. \end{aligned}$$

We have $b^2 =(2\beta -\beta ^{\prime }+2\gamma \beta (\beta -\beta ^{\prime }))^2<(2\beta +2\gamma \beta (\beta -\beta ^{\prime }))^2 = 4\beta ^2(1+\gamma (\beta -\beta ^{\prime }))^2$. Moreover, if $\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}\le K$, then it is clear that $4K^2(1+\gamma (\beta -\beta ^{\prime }))^2\ge \alpha ^2$. As a result, it follows that $J^R_d/J^R_c\ge 1$.

Part 2

It is straightforward to derive that if $K^A=K^B\equiv K$, then

$$\begin{aligned} \Pi _d&= \left\{ \begin{array}{l@{\quad }l}\frac{\alpha ^2\beta (3\beta -2\beta ^{\prime }+2\gamma \beta (\beta -\beta ^{\prime }))}{b^2(\beta -\beta ^{\prime })} &{} \mathrm{if }\, \alpha \beta -bK\le 0\\ \frac{K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))}{\beta -\beta ^{\prime }} &{} \mathrm{else}\end{array}\right. \\ \Pi _c&= \left\{ \begin{array}{l@{\quad }l}\frac{\alpha ^2(3+2\gamma (\beta -\beta ^{\prime }))}{4(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))^2} &{} \mathrm{if }\, \frac{1}{2}\alpha -K(1+\gamma (\beta -\beta ^{\prime }))\le 0\\ \frac{K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))}{\beta -\beta ^{\prime }} &{} \mathrm{else}\end{array}\right. \\ \Pi _{cc}&= \left\{ \begin{array}{l@{\quad }l}\frac{\alpha ^2}{(\beta -\beta ^{\prime })(1+2\gamma (\beta -\beta ^{\prime }))} &{} \mathrm{if }\, K\ge \frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}\\ \frac{K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))}{\beta -\beta ^{\prime }} &{} \mathrm{else}\end{array}\right. \\ \end{aligned}$$

In particular,

$$\begin{aligned} \frac{\Pi _d}{\Pi _c} =\left\{ \begin{array}{l@{\quad }l}\frac{4\beta (3\beta -2\beta ^{\prime }+2\gamma \beta (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))^2}{b^2(3+2\gamma (\beta -\beta ^{\prime }))}&{} \mathrm{if }\, K\ge \alpha \beta /b\\ \frac{4K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))(1+\gamma (\beta -\beta ^{\prime }))^2}{\alpha ^2(3+2\gamma (\beta -\beta ^{\prime }))}&{} \mathrm{if }\, \frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}\le K< \alpha \beta /b\\ 1&{} \mathrm{else}\end{array}\right. \end{aligned}$$

We obtain after simplification:

$$\begin{aligned}&4\beta (3\beta -2\beta ^{\prime }+2\gamma \beta (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))^2-b^2(3+2\gamma (\beta -\beta ^{\prime }))\\&\quad \quad \quad = \beta ^{\prime }(4\beta -3\beta ^{\prime }) +\gamma \beta ^{\prime }(\beta -\beta ^{\prime })(4\beta -2\beta ^{\prime })>0 \end{aligned}$$

Moreover, we find that

$$\begin{aligned}&4K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))(1+\gamma (\beta -\beta ^{\prime }))^2-\alpha ^2(3+2\gamma (\beta -\beta ^{\prime }))\\&= -4(1+2\gamma (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))^2\left( K-\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))} \right) \\&\left( K - \frac{\alpha (3+2\gamma (\beta -\beta ^{\prime }))}{2(1+2\gamma (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))}\right) \end{aligned}$$

Clearly, for $\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}\le K$ we have $-4(1+2\gamma (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))^2 \left( K-\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))} \right) \le 0$. In addition, it is easy to obtain that

$$\begin{aligned} \frac{\beta }{b}\le \frac{3+2\gamma (\beta -\beta ^{\prime })}{2(1+2\gamma (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))}, \end{aligned}$$

therefore, for $K< \alpha \beta /b$, $K - \frac{\alpha (3+2\gamma (\beta -\beta ^{\prime }))}{2(1+2\gamma (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))}<0$. It follows that

$$\begin{aligned} 4K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))(1+\gamma (\beta -\beta ^{\prime }))^2-\alpha ^2(3+2\gamma (\beta -\beta ^{\prime }))<0 \end{aligned}$$

for $\frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}\le K< \alpha \beta /b$, and hence $\Pi _d\le \Pi _c$.

Part 3

Similarly,

$$\begin{aligned} \frac{\Pi _c}{\Pi _{cc}} =\left\{ \begin{array}{l@{\quad }l}\frac{(3+2\gamma (\beta -\beta ^{\prime }))(1+2\gamma (\beta -\beta ^{\prime }))}{4(1+\gamma (\beta -\beta ^{\prime }))^2} &{} \mathrm{if }\, K\ge \frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}\\ \frac{\alpha ^2(3+2\gamma (\beta -\beta ^{\prime }))}{4K(2\alpha -K(1+2\gamma (\beta -\beta ^{\prime })))(1+\gamma (\beta -\beta ^{\prime }))^2}&{} \mathrm{if }\, \frac{\alpha }{2(1+\gamma (\beta -\beta ^{\prime }))}\le K< \frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}\\ 1&{} \mathrm{else}\end{array}\right. \end{aligned}$$

We notice that

$$\begin{aligned} \frac{(3+2\gamma (\beta -\beta ^{\prime }))(1+2\gamma (\beta -\beta ^{\prime }))}{4(1+\gamma (\beta -\beta ^{\prime }))^2} = 1-\frac{1}{4(1+\gamma (\beta -\beta ^{\prime }))^2} \end{aligned}$$

Moreover, in the second expression, the denominator is clearly a increasing function of $K$ for $K< \frac{\alpha }{1+2\gamma (\beta -\beta ^{\prime })}$ so the ratio is decreasing with $K$. For the largest possible value of $K$ in the range, the ratio (at its minimum value) equals the ratio of the first case, which we showed above cannot go below $1-\frac{1}{4(1+\gamma (\beta -\beta ^{\prime }))^2}$. Therefore,

$$\begin{aligned} \frac{\Pi _c}{\Pi _{cc}} \ge 1-\frac{1}{4(1+\gamma (\beta -\beta ^{\prime }))^2}\ge \frac{3}{4} \end{aligned}$$

where the first inequality is tight for large capacity level and the second inequality is tight in the case of no production cost ($\gamma =0$) or non differentiated suppliers ($\beta =\beta ^{\prime }$).

$\square$

1.3 Proof of Proposition 4

We consider the 7 possible combinations of regimes in the decentralized and cooperative settings.

• Regimes $(a,a')$: Let’s first note that

  $$\begin{aligned} b^2-4\beta (\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))(1+\gamma \beta )=\beta ^{'2}. \end{aligned}$$

  Some calculations then lead to

  $$\begin{aligned} J_c^B-J_d^B=\frac{\frac{\alpha ^2}{4(\beta -\beta ^{\prime })}}{1+\gamma (\beta -\beta ^{\prime })}- \frac{\beta (1+\gamma \beta )\alpha ^2}{b^2} =\frac{\beta ^{'2}\alpha ^2}{4b^2(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))}\ge 0. \end{aligned}$$

• Regimes $(b,a')$: Some calculations then lead to

  $$\begin{aligned} J_c^B-J_d^B&= \frac{\frac{\alpha ^2}{4(\beta -\beta ^{\prime })}}{1+\gamma (\beta -\beta ^{\prime })}+\gamma (K^B)^2\\&-K^B\left( \frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1\!+\!2\gamma \beta ))-2K^B \beta (1\!+\!\gamma \beta )}{d^2}\right) \\&= \frac{-\alpha K^B(2\beta (1\!+\!\gamma \beta )\!+\!\beta ^{\prime }(1\!+\!2\gamma \beta ))\!+\!(K^B)^2( 2\beta (1\!+\!\gamma \beta )^2\!-\!\gamma \beta ^{'2}(1\!+\!2\gamma \beta ))}{d^2}\\&+\ \frac{\alpha ^2}{4(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))} \end{aligned}$$

  We observe that this is a polynomial of degree 2 in $\alpha$ with concavity turned up. Straightforward calculations enable to find that this polynomial reaches a minimum at

  $$\begin{aligned} \frac{2(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))K^B(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta )) }{d^2} \end{aligned}$$

  and thus the minimum value of the polynomial is

  $$\begin{aligned}&\frac{(K^B)^2}{d^4}\left( -(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))^2\right. \\&\quad \quad \left. +\ d^2(2\beta (1+\gamma \beta )^2-\gamma \beta ^{'2}(1+2\gamma \beta ))\right) \\&= \frac{(K^B)^2}{d^4}\left( \beta ^{'3}(1+2\gamma \beta )^3+\beta \beta ^{'2}(1+\gamma \beta )(1+2\gamma \beta )^2-\gamma \beta ^2\beta ^{'2}(1+\gamma \beta )^2\right. \\&\quad \quad \left. -\ 4\gamma \beta \beta ^{'3}(1+\gamma \beta )(1+2\gamma \beta )\right) \\&= \frac{(K^B)^2}{d^4}\left( \beta ^{'3}(1+2\gamma \beta )+\beta \beta ^{'2}(1+\gamma \beta )(1+3\gamma \beta +3\gamma ^2\beta ^2)\right) >0 \end{aligned}$$

• Regimes $(b,b^{\prime })$:

  $$\begin{aligned} J_c^B-J_d^B&= K^B\left( \frac{\alpha }{2(\beta -\beta ^{\prime })}+\frac{\frac{\alpha }{2}(1+\gamma (\beta +\beta ^{\prime }))- K^B(1+\gamma \beta )}{\beta +\gamma (\beta ^2-\beta ^{'2})}\right. \\&\left. -\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-2K^B \beta (1+\gamma \beta )}{d^2}\right) \end{aligned}$$

  thus $J^B_c\ge J_d^B$ iff

  $$\begin{aligned}&\frac{\alpha }{2(\beta -\beta ^{\prime })}+\frac{\frac{\alpha }{2}(1+\gamma (\beta +\beta ^{\prime }))- K^B(1+\gamma \beta )}{\beta +\gamma (\beta ^2-\beta ^{'2})}\\&\qquad -\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-2K^B \beta (1+\gamma \beta )}{d^2}>0 \\&\Leftrightarrow \alpha \left( \frac{2\beta -\beta ^{\prime }+2\gamma (\beta ^2-\beta ^{'2})}{2(\beta -\beta ^{\prime })(\beta +\gamma (\beta ^2-\beta ^{'2}))}- \frac{2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta )}{d^2}\right) \\&\qquad +\frac{K^B\beta ^{'2}(1+\gamma \beta )}{d^2(\beta +\gamma (\beta ^2-\beta ^{'2}))}>0 \end{aligned}$$

  It is sufficient to show that the slope of $\alpha$ above is positive. Since we observe that $(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))(\beta -\beta ^{\prime }) = d^2-\beta \beta ^{\prime }$, the slope of $\alpha$ is positive iff

  $$\begin{aligned} d^2(2(\beta +\gamma (\beta ^2-\beta ^{'2}))-\beta ^{\prime }) -2(\beta +\gamma (\beta ^2-\beta ^{'2}))(d^2-\beta \beta ^{\prime })>0 \end{aligned}$$

  which simplifies to $\beta ^{'3}>0$. Therefore, $J^B_c\ge J_d^B$.

• Regimes $(c,a')$:

  $$\begin{aligned} J_c^B-J_d^B&= \frac{\frac{\alpha ^2}{4(\beta -\beta ^{\prime })}}{1+\gamma (\beta -\beta ^{\prime })}- \frac{K^B}{\beta -\beta ^{\prime }}\left( \alpha -\bar{K}\right) +\gamma (K^B)^2\\&= \frac{\alpha ^2}{4(\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))}-\frac{\alpha K^B}{\beta -\beta ^{\prime }} +\frac{K^B\bar{K}}{\beta -\beta ^{\prime }}+\gamma (K^B)^2 \end{aligned}$$

  where $\bar{K}=\frac{\beta K^B+\beta ^{\prime }K^A}{\beta +\beta ^{\prime }}>K^B$. We observe that $J_c^B-J_d^B$ is quadratic in $\alpha$ with the concavity turned up. Therefore, it if necessary and sufficient to check the sign of $J_c^B-J_d^B$ at the point where the polynomial reaches its minimum: $2 K^B(1+\gamma (\beta -\beta ^{\prime }))$, where the polynomial takes value

  $$\begin{aligned} \frac{K^B\bar{K}}{\beta -\beta ^{\prime }}+ \gamma (K^B)^2-\frac{(K^B)^2(1+\gamma (\beta -\beta ^{\prime }))}{\beta -\beta ^{\prime }} =\frac{K^B(\bar{K}-K^B}{\beta -\beta ^{\prime }}>0 \end{aligned}$$

  Therefore, $J^B_c\ge J_d^B$.

• Regimes $(c,b^{\prime })$:

  $$\begin{aligned} J_c^B-J_d^B\!&= \!K^B\left( -\frac{\alpha }{2(\beta -\beta ^{\prime })}\!+\!\frac{\frac{\alpha }{2}(1\!+\!\gamma (\beta \!+\!\beta ^{\prime }))\!-\! K^B(1\!+\!\gamma \beta )}{\beta +\gamma (\beta ^2-\beta ^{'2})} +\frac{\beta K^B\!+\!\beta ^{\prime }K^A}{\beta ^2\!-\!\beta ^{'2}}\right) \\&= K^B\left( \frac{-\alpha \beta ^{\prime }}{2(\beta -\beta ^{\prime })(\beta +\gamma (\beta ^2-\beta ^{'2}))}\right. \\&\left. \qquad +\ \frac{-K^B(1+\gamma \beta )(\beta ^2-\beta ^{'2})+(\beta K^B+\beta ^{\prime }K^A)(\beta +\gamma (\beta ^2-\beta ^{'2}))}{(\beta +\gamma (\beta ^2-\beta ^{'2}))(\beta ^2-\beta ^{'2})} \right) \\&= K^B\beta ^{\prime }\left( \frac{-\alpha }{2(\beta -\beta ^{\prime })(\beta +\gamma (\beta ^2-\beta ^{'2}))}+\frac{K^B\beta ^{'}+K^A(\beta +\gamma (\beta ^2-\beta ^{'2}))}{(\beta +\gamma (\beta ^2-\beta ^{'2}))(\beta ^2-\beta ^{'2})} \right) \end{aligned}$$

  Since in regime $b^{\prime }$, $\frac{\alpha }{2}<K^A(1+\gamma (\beta -\beta ^{\prime }))-\frac{\beta ^{\prime }(K^A-K^B)}{\beta +\beta ^{\prime }}$, we have

  $$\begin{aligned} J_c^B-J_d^B&> K^B\beta ^{\prime }\frac{-K^A(\beta +\beta ^{\prime }+\gamma (\beta ^2-\beta ^{'2}))+\beta ^{\prime }(K^A-K^B)+K^B\beta ^{'}+ K^A(\beta +\gamma (\beta ^2-\beta ^{'2}))}{(\beta +\gamma (\beta ^2-\beta ^{'2}))(\beta ^2-\beta ^{'2})}\\&= 0 \end{aligned}$$

  which proves that on the valid domain $J^B_c\ge J_d^B$.

• Regimes $(c,c^{\prime })$: $J_c^B=J_d^B=\frac{K^B}{\beta -\beta ^{\prime }}\left( \alpha -\frac{\beta K^B+\beta ^{\prime }K^A}{\beta +\beta ^{\prime }}\right)$ (the cooperative solution and the decentralized equilibrium coincide).

$\square$

1.4 Proof of Proposition 5 and Theorem 4

Three regimes are possible: $(a,a')$, $(c,a')$ and $(c,c^{\prime })$.

• Regimes $(c,c^{\prime })$: In this regime, the optimal solution under cooperation is identical to the equilibrium, therefore $\rho =1$. The condition to be in this regime is $0\le \frac{\alpha }{2}-K(1+\gamma (\beta -\beta ^{\prime }))$.

• Regimes $(a,a')$: This regime holds if $\alpha \beta -bK\le 0$, i.e. $\alpha -K(2-r+2\gamma (\beta -\beta ^{\prime }))\le 0$. Since $b^2-4\beta (\beta -\beta ^{\prime })(1+\gamma (\beta -\beta ^{\prime }))(1+\gamma \beta )=\beta ^{'2},$

  $$\begin{aligned} \rho&= \frac{4\beta (\beta -\beta ^{\prime })(1+\gamma \beta )(1+\gamma (\beta -\beta ^{\prime }))}{b^2}=\frac{b^2-r^2\beta ^{2}}{b^2} = 1-\frac{r^2\beta ^{2}}{b^2} \end{aligned}$$

  Moreover, $\frac{b}{\beta }=2-r+2\gamma (\beta -\beta ^{\prime })>2-r$ so $\frac{r^2\beta ^{2}}{b^2}<\frac{r^2}{(2-r)^2}$. Therefore,

  $$\begin{aligned} \rho > 1-\frac{r^2}{(2-r)^2}. \end{aligned}$$

• Regimes $(c,a')$: This regime holds if $\alpha -K(2-r+2\gamma (\beta -\beta ^{\prime }))\ge 0$ and $0\ge \frac{\alpha }{2}-K(1+\gamma (\beta -\beta ^{\prime }))$, i.e.

  $$\begin{aligned} \frac{1}{2(1+\gamma (\beta -\beta ^{\prime }))}\le \frac{K}{\alpha }\le \frac{1}{2-r+2\gamma (\beta -\beta ^{\prime })}. \end{aligned}$$
  $$\begin{aligned} \rho&= 4(1+\gamma (\beta -\beta ^{\prime }))\frac{K(\alpha -K)-\gamma (\beta -\beta ^{\prime })K^2}{\alpha ^2}\\&= 4(1+\gamma (\beta -\beta ^{\prime }))\frac{K}{\alpha }\left( 1-\frac{K}{\alpha }(1+\gamma (\beta -\beta ^{\prime })) \right) \end{aligned}$$

  Denoting $X=(1+\gamma (\beta -\beta ^{\prime }))\frac{K}{\alpha }$, we have $\rho = 4X (1-X)$, where $\frac{1}{2}\le X\le X_0 \equiv \frac{1+\gamma (\beta -\beta ^{\prime })}{2-r+2\gamma (\beta -\beta ^{\prime })}$. On this domain, $\rho$ reaches its minimum at $X_0$, i.e.

  $$\begin{aligned} \rho&\ge 4X_0 (1-X_0) \\&= \frac{4(1+\gamma (\beta -\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime })-r)}{(2+2\gamma (\beta -\beta ^{\prime })-r)^2}\\&= 1-\frac{r^2}{(2+2\gamma (\beta -\beta ^{\prime })-r)^2}\\&\ge 1-\frac{r^2}{(2-r)^2} \end{aligned}$$

It follows that $1-\rho \le \frac{r^2}{(2-r)^2}$. We now show that $1-\rho \le r^2 \frac{K^2}{\alpha ^2}$. In regimes $(c,c^{\prime })$, $\rho =1$ so the bound clearly holds.

In regimes $(a,a')$, $\frac{b}{\beta }=2-r+2\gamma (\beta -\beta ^{\prime })\ge \frac{\alpha }{K}$ and thus $\rho =1-\frac{r^2\beta ^{2}}{b^2}\ge 1- r^2 \frac{K^2}{\alpha ^2}$.

In regimes $(c,a')$, for brevity let us denote $u \equiv 1 +\gamma (\beta -\beta ^{\prime })$. We have (as seen in Proof of Proposition 5 and Theorem 4 of Appendix 7) $\rho = 1-(1-2u\frac{K}{\alpha })^2$ and the valid domain in this regime is $2u-r\le \frac{\alpha }{K}\le 2u$. Therefore, we have in particular $\frac{K}{\alpha }(2u-r)\le 1$, which implies $0\le 2u\frac{K}{\alpha }-1\le r\frac{K}{\alpha }$. Therefore, $\left( 2u\frac{K}{\alpha }-1\right) ^2\le r^2\frac{K^2}{\alpha ^2}$, so we have $\rho = 1-(2u\frac{K}{\alpha }-1)^2\ge 1-r^2\frac{K^2}{\alpha ^2}.$ $\square$

1.5 Proof of Theorem 5

The derivative of $\rho$ with respect to $K^B$ is

$$\begin{aligned} \frac{\partial \rho }{\partial K^B} = \frac{\partial }{\partial K^B}\left( \frac{J^A_d+J^B_d}{J^A_c+J^B_c}\right) \end{aligned}$$

therefore $\rho$ is non increasing with $K^B$ iff

$$\begin{aligned} (J^A_d+J^B_d)\frac{\partial (J^A_c+J^B_c)}{\partial K^B}-(J^A_c+J^B_c)\frac{\partial (J^A_d+J^B_d)}{\partial K^B}\ge 0 \end{aligned}$$

(5)

• Regimes $(c,c^{\prime })$: $\rho = 1$ is independent of $K^B$.

• Regimes $(a,a^{\prime })$: $J^A_d,\,J^B_d,\, J^A_c$ and $J^B_c$ are independent of $K^B$, so $\rho$ is independent of $K^B$.

• Regimes $(c,a^{\prime })$: We have $\frac{\partial (J^A_c+J^B_c)}{\partial K^B}=0$ and $J^A_c+J^B_c\ge 0$, therefore (5) holds iff $\frac{\partial (J^A_d+J^B_d)}{\partial K^B}\le 0$. We have

  $$\begin{aligned} \frac{\partial (J^A_d+J^B_d)}{\partial K^B}&= -\frac{\beta ^{\prime }}{\beta +\beta ^{\prime }}\frac{K^A}{\beta -\beta ^{\prime }}-2\gamma K^B {+}\frac{\alpha }{\beta -\beta ^{\prime }} -\frac{\beta K^B +\beta ^{\prime } K^A}{\beta ^2-\beta ^{'2}}{-}\frac{\beta }{\beta +\beta ^{\prime }}\frac{K^B}{\beta -\beta ^{\prime }}\\&= \frac{\alpha }{\beta -\beta ^{\prime }}-2\gamma K^B-2\frac{\beta K^B +\beta ^{\prime } K^A}{\beta ^2-\beta ^{'2}}\\&\le \frac{\alpha -2\gamma K^B(\beta -\beta ^{\prime })-2K^B}{\beta -\beta ^{\prime }} \end{aligned}$$

  where the last inequality follows from the observation that $\frac{\beta K^B +\beta ^{\prime } K^A}{\beta +\beta ^{'}}\ge K^B$ for $K^A\ge K^B$. In regime $a'$, $\alpha -2K^B(1+\gamma (\beta -\beta ^{\prime })<0$, so the result follows.

• Regimes $(b,a^{\prime })$: We have $\frac{\partial (J^A_c+J^B_c)}{\partial K^B}=0$ and $J^A_c+J^B_c\ge 0$, therefore (5) holds iff

  $$\begin{aligned} \frac{\partial (J^A_d+J^B_d)}{\partial K^B}\le 0 \end{aligned}$$

  (6)

  We observe that $\frac{\partial J^A_d}{\partial K^B}$ and $\frac{\partial J^B_d}{\partial K^B}$ are linear in $K^B$, with a slope respectively equal to $\frac{2\beta \beta ^{'2}(1+\gamma \beta )}{d^4}>0$ and $-2\gamma -\frac{4\beta (1+\gamma \beta )}{d^2}<0$. Therefore, $\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is linear in $K^B$ with slope

  $$\begin{aligned} \frac{2}{d^4}\left( \beta \beta ^{'2}(1+\gamma \beta )-\gamma d^4-2\beta (1+\gamma \beta )d^2\right) <0 \end{aligned}$$

  since $d^2 \ge 2\beta ^2 -\beta ^{'2}$ implies $\beta ^{'2}-2d^2\le -4 \beta ^2+\beta ^{'2}<0$. Therefore, (6) holds iff $\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is non positive at the lowest value allowed for $K^B$ in regimes $(b,a')$, i.e.

  $$\begin{aligned} \max \{\frac{\alpha \beta (\beta +\beta ^{\prime })-d^2 K^A }{\beta \beta ^{\prime }} ,\, \frac{\frac{\alpha }{2} }{1+\gamma (\beta -\beta ^{\prime })}\}. \end{aligned}$$

  We will prove the sufficient condition that $\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is non positive for $K^B=\frac{\frac{\alpha }{2} }{1+\gamma (\beta -\beta ^{\prime })}$.

$\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is linear decreasing in $K^B$, and some calculations lead to showing that it takes value 0 for

$$\begin{aligned} K^B = \frac{\alpha }{2}\frac{d^2(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-2\beta \beta ^{\prime }(1+\gamma \beta )(\beta +\beta ^{\prime })}{-\beta \beta ^{'2}(1+\gamma \beta )+\gamma d^4+2\beta (1+\gamma \beta )d^2}. \end{aligned}$$

Therefore $\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is non positive for $K^B=\frac{\frac{\alpha }{2}}{1+\gamma (\beta -\beta ^{\prime })}$ iff

$$\begin{aligned} \frac{d^2(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-2\beta \beta ^{\prime }(1+\gamma \beta )(\beta +\beta ^{\prime })}{-\beta \beta ^{'2}(1+\gamma \beta )+\gamma d^4+2\beta (1+\gamma \beta )d^2}\le 1, \end{aligned}$$

or equivalently,

$$\begin{aligned} \gamma d^4-d^2 \beta ^{\prime }(1+2\gamma \beta )+\beta \beta ^{\prime }(1+\gamma \beta )(2\beta +\beta ^{\prime })\ge 0. \end{aligned}$$

Denoting $r=\frac{\beta ^{\prime }}{\beta }\in [0,1]$ and $t=\gamma \beta >0$, we have $\frac{d^2}{\beta ^2}=(2-r^2)+2t(1-r^2)$, so we observe that the inequality above is equivalent to

$$\begin{aligned} t \big ((2-r^2)+2t(1-r^2)\big )^2- \big ((2-r^2)+2t(1-r^2)\big )r(1+2t)+r(1+t)(2+r) \ge 0 \end{aligned}$$

which, as illustrated in Fig. 13, holds for any $r$ in $[0,1]$ and $t>0$. This ends the proof.

Fig. 13

Plot of $t \big ((2-r^2)+2t(1-r^2)\big )^2- \big ((2-r^2)+2t(1-r^2)\big )r(1+2t)+r(1+t)(2+r)$ as a function of $r$ and $t$, where $r=\frac{\beta ^{\prime }}{\beta }$ and $t=\gamma \beta$

Lemma 3

Consider two polynomials of degree 2: $P_1(X)=aX^2 + bX + c$ and $P_2(X)=a'X^2 + b^{\prime }X + c^{\prime }$. Then $P_1(X) P_2'(X)-P_1'(X) P_2(X)$ is a polynomial of degree 2 given by $(a^{\prime }b-ab^{\prime })X^2 +2(a^{\prime }c-ac^{\prime })X+(cb^{\prime }-c^{\prime }b)$.

• Regimes $(c,b^{\prime })$: $J^A_c+J^B_c$ and $J^A_d+J^B_d$ are quadratic in $K^B$, so, using the Lemma above, inequality (5) is a quadratic expression in $K^B$, which we will denote polynomial $P(K^B)=s K^{B^2} + t_0 K^B + v$. The coefficient of $K^{B^2}$ is, after calculations, equal to

  $$\begin{aligned} s&= \frac{(1+\gamma (\beta +\beta ^{\prime }))\alpha }{\beta ^2-\beta ^{'2}}-\frac{\alpha -2K^A\frac{\beta ^{\prime }}{\beta +\beta ^{\prime }}}{\beta -\beta ^{\prime }} \Big (\gamma +\frac{1+\gamma \beta }{\beta +\gamma (\beta ^2-\beta ^{'2})}\Big )\\&= \frac{\beta ^{\prime }(1+\gamma (\beta +\beta ^{\prime }))}{(\beta ^2-\beta ^{'2})(\beta +\gamma (\beta ^2-\beta ^{'2}))}(-\alpha +2K^A(1+\gamma (\beta -\beta ^{\prime }))) \end{aligned}$$

  This coefficient is positive iff

  $$\begin{aligned} \alpha <2K^A(1+\gamma (\beta -\beta ^{\prime })) \end{aligned}$$

  Regime $b^{\prime }$ implies $(\beta +\beta ^{\prime })\alpha -2K^A(\beta +\gamma (\beta ^2-\beta ^{'2}))< 2\beta ^{\prime } K^B$. Since $K^B\le K^A$, we have in this regime

  $$\begin{aligned} (\beta +\beta ^{\prime })\alpha <2K^A(\beta +\beta ^{\prime }+\gamma (\beta ^2-\beta ^{'2})), \end{aligned}$$

  i.e. $\alpha <2K^A(1+\gamma (\beta -\beta ^{\prime }))$. This proves that $(J^A_d+J^B_d)\frac{\partial (J^A_c+J^B_c)}{\partial K^B}-(J^A_c+J^B_c)\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is quadratic in $K^B$ with a coefficient of $(K^B)^2$ that is positive. To show that polynomial $P(K^B)$ is non negative on the valid domain for $K^B$ in regimes $(c,b^{\prime })$, we will show that at the minimum allowed value in this regime $\max (0, K_1^B)$, $P(K^B)$ and its first derivative are non negative, where $l'_b(K^A)= \frac{(\beta +\beta ^{\prime })\frac{\alpha }{2}-K^A(\beta +\gamma (\beta ^2-\beta ^{'2}))}{\beta ^{\prime }}$. First, it is clear that $J^A_c, J^B_c, J^A_d$ and $J^B_d$ are continuous in $K^B$. Furthermore, we observe that $\frac{\partial (J^A_c+J^B_c)}{\partial K^B}$ is continuous in $K^B$ at the threshold $l'_b(K^A)$ between regimes $b^{\prime }$ and $c^{\prime }$. Indeed, after calculations, to the left (regime $b^{\prime }$), it is equal to

  $$\begin{aligned} \frac{\alpha (1+\gamma (\beta +\beta ^{\prime }))}{\beta +\gamma (\beta ^2-\beta ^{'2})}- 2l'_b(K^A)\frac{(1+\gamma (\beta +\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))}{\beta +\gamma (\beta ^2-\beta ^{'2})} \end{aligned}$$

  while to the right (regime $c^{\prime }$), it is equal to

  $$\begin{aligned} \frac{\alpha -2\frac{\beta ^{\prime }}{\beta +\beta ^{\prime }}K^A}{\beta -\beta ^{\prime }}-2 l'_b(K^A)\frac{\beta +\gamma (\beta ^2-\beta ^{'2})}{\beta ^2-\beta ^{'2}} \end{aligned}$$

  Straightforward simplifications show that these two quantities are equal. Since $l'_b(K^A)$ is not a threshold between two regimes for $J_d$, $\frac{\partial (J^A_d+J^B_d)}{\partial K^B}$ is continuous in $K^B$ at $l'_b(K^A)$. Therefore, the first derivative of $\rho$ with respect to $K^B$ is continuous at $l'_b(K^A)$. To the left of $l'_b(K^A)$, regimes $(c,b^{\prime })$ become regimes $(c,c^{\prime })$ where $\rho =1$ and in particular its first derivative is equal to zero. Therefore, by continuity, polynomial $P(K^B)$ takes value zero at $l'_b(K^A)$. If $l'_b(K^A)$ is the larger root (i.e. $l'_b(K^A)>-\frac{t_0}{2s}$), then in regime $(c,b^{\prime })$, $K^B>\max (0, l'_b(K^A))\ge l'_b(K^A)$ and thus $P(K^B)$ is non negative and the result is shown. Assume $l'_b(K^A)$ is the smaller root (i.e. $l'_b(K^A)<-\frac{t_0}{2s}$); we need to show that $t_0$ the derivative of polynomial $P(K^B)$ at zero is non negative. It will then follow that $l'_b(K^A)<0$ (because we showed $s>0$), hence the result. The coefficient of the linear term in $P(K^B)$ is:

  $$\begin{aligned} t_0&= -\frac{2}{\beta -\beta ^{\prime }}(1+\gamma (\beta +\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))\Big (\frac{K^A\alpha }{\beta +\gamma (\beta ^2-\beta ^{'2})}- \frac{(K^A)^2}{(\beta +\beta ^{'})}\Big )\\&+\,\frac{2}{\beta -\beta ^{\prime }}\Big (\frac{\alpha ^2}{4(\beta -\beta ^{'})}\Big )\\&= \frac{2}{\beta -\beta ^{\prime }}\Big ( \frac{\alpha ^2}{4(\beta -\beta ^{'})}-(1+\gamma (\beta +\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))\\&\times \,\Big (\frac{K^A\alpha }{\beta +\gamma (\beta ^2-\beta ^{'2})}- \frac{(K^A)^2}{\beta +\beta ^{'}}\Big )\Big ) \end{aligned}$$

  It is easy to check (using for example the Excel solver) that $t_0$ above is minimized for $\beta ^{\prime } = 0$ and then equals $\frac{2}{\beta ^2}\Big ((\frac{\alpha }{2}-K^A)^2+\gamma (K^A)^2\Big )\ge 0$. Therefore $t_0\ge 0$, which ends the proof.

• Regimes $(b,b^{\prime })$: $J^A_c+J^B_c$ and $J^A_d+J^B_d$ are quadratic in $K^B$, so, using the Lemma above, inequality (5) is a quadratic expression in $K^B$: $P_1(K^B)=s_1 (K^B)^2 + t_1 K^B + v_1$. First, let’s show that the coefficient $s_1$ of $(K^B)^2$ is positive. This coefficient is equal to

  $$\begin{aligned} s_1&= -\frac{(1+\gamma (\beta +\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))}{\beta +\gamma (\beta ^2-\beta ^{'2})}\Big (-\frac{2\beta \beta ^{\prime }(1+\gamma \beta )}{d^4}( \alpha (\beta +\beta ^{\prime }))\\&+\,\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))}{d^2} \Big )\\&-\,\Big (\big (-\gamma +\frac{\beta \beta ^{'2}(1+\gamma \beta )}{d^4}-\frac{2\beta (1+\gamma \beta )}{d^2}\big )\frac{\alpha -(1+\gamma (\beta +\beta ^{\prime }))}{\beta +\gamma (\beta ^2-\beta ^{'2})}\Big ) \end{aligned}$$

  It is easy to check (using for example the Excel solver) that this expression is minimized for $\beta ^{\prime }=0$, in which case it equals

  $$\begin{aligned} \frac{\alpha }{\beta }\frac{1+\gamma \beta }{\beta }+(\gamma +\frac{1}{\beta })\frac{\alpha }{\beta }=0. \end{aligned}$$

  Therefore $s_1$ is non negative. If polynomial $P_1(K^B)$ had less than 2 roots, then it would always be non negative and the result follows. Let’s assume that it has 2 roots. Moreover, we have that the coefficient of the linear term in $P_1(K^B)$ is given by

  $$\begin{aligned} t_1&= -2\frac{(1+\gamma (\beta +\beta ^{\prime }))(1+\gamma (\beta -\beta ^{\prime }))}{\beta +\gamma (\beta ^2-\beta ^{'2})}\Big ( \frac{\beta (1+\gamma \beta )}{d^4}\big (\alpha (\beta +\beta ^{\prime }))\big )^2\Big )\\&-\Big (-\gamma +\frac{\beta \beta ^{'2}(1+\gamma \beta )}{d^4}-\frac{2\beta (1+\gamma \beta )}{d^2}\Big ) \frac{\frac{\alpha ^2}{2}(\beta +\beta ^{\prime })}{(\beta -\beta ^{\prime })(\beta +\gamma (\beta ^2-\beta ^{'2}))} \end{aligned}$$

  It is easy to check (using for example the Excel solver) that $t_1$ above is minimized for $\beta ^{\prime } = 0$ and then equals

  $$\begin{aligned} -\frac{2}{\beta }(1+\gamma \beta )\Big (\frac{\alpha ^2}{4\beta (1+\gamma \beta )}\Big ) +\frac{\alpha ^2}{2\beta ^2}=\frac{-\frac{\alpha )^2}{2}+\frac{\alpha ^2}{2} }{\beta ^2}=0 \end{aligned}$$

  Therefore $t_1\ge 0$. In other words, polynomial $P_1(K^B)$ has a derivative at 0 that is non negative. Finally, we have to show that $P_1(0)\ge 0$ to conclude that both roots are negative, and as a result on the valid domain for $K^B$, we have $P_1(K^B)\ge 0$. We have

  $$\begin{aligned}&v_1=\frac{\alpha (1+\gamma (\beta +\beta ^{\prime }))}{\beta +\gamma (\beta ^2-\beta ^{'2})}\Big ( \frac{\beta (1+\gamma \beta )}{d^4}\big (\alpha (\beta +\beta ^{\prime }))\big )^2\Big ) -\frac{\frac{\alpha ^2}{2}(\beta +\beta ^{\prime })}{2(\beta -\beta ^{\prime })(\beta +\gamma (\beta ^2-\beta ^{'2}))}\\&\quad \qquad +\Big (-\frac{2\beta \beta ^{\prime }(1+\gamma \beta )}{d^4}( \alpha (\beta +\beta ^{\prime }))\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))}{d^2} \Big ) \end{aligned}$$

  It is easy to check (using for example the Excel solver) that $v_1$ above is minimized for $\beta ^{\prime } = 0$ and then equals

  $$\begin{aligned} \frac{\alpha \!-\!2I_0}{\beta }\Big (\!-\!\gamma I_0^2\!+\!\frac{(\alpha \!+\!2\gamma \beta I_0)^2}{4\beta (1\!+\!\gamma \beta )}\!+\!\frac{I_0(\alpha \!-\!I_0)}{\beta }\Big ) \!-\!\frac{\frac{\alpha ^2}{4}\!+\!I_0(\alpha \!-\!I_0)(1\!+\!2\gamma \beta )}{\beta (1\!+\!\gamma \beta )}\frac{\alpha \!-\!2I_0}{\beta }\!=\!0 \end{aligned}$$

  Therefore $v_1\ge 0$.

$\square$

1.6 Proof of Proposition 1

In regime $a$, $\frac{\partial J^A}{\partial K^B}(K^A,K^B)=0$.

In regime $c$,

$$\begin{aligned} \frac{\partial J^A}{\partial K^B}(K^A,K^B)=-\frac{\beta ^{\prime }K^A}{\beta ^2-\beta ^{'2}}<0. \end{aligned}$$

In regime $b$,

$$\begin{aligned} \frac{\partial J^A}{\partial K^B}(K^A,K^B)=-\frac{2\beta \beta ^{\prime }(1+\gamma \beta )}{d^4}(\alpha (\beta +\beta ^{\prime })-\beta ^{\prime } K^B) \end{aligned}$$

and in that regime,

$$\begin{aligned} \beta ^{\prime } K^B<\alpha \frac{\beta \beta ^{\prime }}{b}=\alpha (\beta +\beta ^{\prime }) -\alpha \frac{d^2}{b} \end{aligned}$$

therefore $\alpha (\beta +\beta ^{\prime })-\beta ^{\prime } K^B>\alpha \frac{\beta \beta ^{\prime }}{b}>0$ and thus $\frac{\partial J^A}{\partial K^B}(K^A,K^B)<0$. $\square$

1.7 Proof of Proposition 2

We first show that $\alpha \beta -b K_0>0$ and that if $\alpha \beta (\beta +\beta ^{\prime })-K_0\beta \beta ^{\prime }-d^2 K^A<0$, then $K_0\le K^A$, and thus $K^B=K_0$ corresponds to regime $b$.

Let $d_1 \equiv 4\beta (1+\gamma \beta )^2-2\gamma \beta ^{'2}(1+2\gamma \beta )$ so that $K_0=\frac{\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))}{d_1}$.

$$\begin{aligned} \alpha \beta -b K_0>0&\Leftrightarrow \beta d_1-b(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))>0\\&\Leftrightarrow \beta ^{'2}(1+2\gamma \beta )>0 \end{aligned}$$

which holds.

$$\begin{aligned} \alpha \beta (\beta +\beta ^{\prime })-K_0\beta \beta ^{\prime }-d^2 K^A<0&\Rightarrow \alpha \beta (\beta +\beta ^{\prime })d_1-\beta \beta ^{\prime }\alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))\\&\,\,-K^A d_1(2\beta ^2(1+\gamma \beta )- \beta ^{'2}(1+2\gamma \beta ))<0 \end{aligned}$$

Let

$$\begin{aligned} A&= \beta (\beta +\beta ^{\prime })d_1-\beta \beta ^{\prime }(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))\\ B&= d_1(2\beta ^2(1+\gamma \beta )-\beta ^{'2}(1+2\gamma \beta ))>0 \end{aligned}$$

so that the inequality above is: $A\alpha -B K^A<0$. We can show that $A>0$: since

$$\begin{aligned} d_1>2\beta ^{\prime }(2(1+2\gamma \beta +\gamma ^2\beta ^2)-\gamma \beta ^{'}(1+2\gamma \beta )>2\beta ^{\prime }(2+3\gamma \beta ) \end{aligned}$$

we have

$$\begin{aligned} A=\beta ^2(d_1-2\beta ^{\prime }(1+\gamma \beta ))+\beta ^{\prime }\beta (d_1-\beta ^{\prime }(1+2\gamma \beta ))>0 \end{aligned}$$

$K_0 \le K^A$ iff

$$\begin{aligned} \alpha (2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta )) -d_1 K^A\le 0. \end{aligned}$$

In order to show that $K_0 \le K^A$, it is sufficient to show

$$\begin{aligned} \frac{B}{A}<\frac{d_1}{2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta )}. \end{aligned}$$

After calculations,

$$\begin{aligned}&B(2\beta (1+\gamma \beta )+\beta ^{\prime }(1+2\gamma \beta ))-Ad_1=-d_1\beta ^{'2}(\beta +\beta ^{\prime })(1+2\gamma \beta )<0. \end{aligned}$$

This proves that if $\alpha \beta (\beta +\beta ^{\prime })-K_0\beta \beta ^{\prime }-d^2 K^A<0$, then $K_0\in [0, K^A]$ and $K^B=K_0$ is in regime $b$. Therefore, a simple calculation of derivatives (using the expression found in Proposition 6 leads to $\frac{\partial J^B}{\partial K^B}\Big \arrowvert _{K^B=K_0}=0$, which means that $K_0$ is a local maximum of $J^B$. (It is not a local minimum or a point of inflexion because we proved that the profit was quadratic concave in regime $b$.)

Similarly, it is easy to show that under conditions (2) and (3), if $\bar{K}_0\in [0, K^A]$, regime $c$ holds for $K^B=\bar{K}_0$ and $\frac{\partial J^B}{\partial K^B}\Big \arrowvert _{K^B=\bar{K}_0}=0$.

$\square$

1.8 Proof of Proposition 3

$J^A+J^B$ is independent of $K^B$ in regime $a$. Therefore if $K^*$ is in regime $a$, then $l_a= \frac{\alpha \beta }{b}$ the threshold between regime $a$ and $b$ is also a maximum for $J^A+J^B$, and $K^*\ge l_a$.

Assume that $\bar{K}_1$ the value of $K^B$ that maximizes $J^B(K^B)$ satisfies $\bar{K}_1<l_a$. Then $\bar{K}_1$ is in regime $b$ or $c$, and $J^B(\bar{K}_1)\ge J^B(l_a)$. Since $J^A$ is decreasing with $K^B$ in regimes $b$ and $c$, $J^A(\bar{K}_1)> J^A(l_a)$, and therefore $(J^A+J^B)(\bar{K}_1)> (J^A+J^B)(l_a)$, which contradicts that $l_a$ is a maximum for $J^A+J^B$.

Now consider the case when $K^*$ corresponds to either regime $b$ or $c$. Thus we have

$$\begin{aligned} \frac{\partial (J^A+J^B)(K^A,K_B)}{\partial K^B}\Big \arrowvert _{K^B=K^*}=0. \end{aligned}$$

As we proved above, $J^A$ is non decreasing with $K^B$, and in particular $\frac{\partial J^A}{\partial K^B}(K^A,K^B)\Big \arrowvert _{K^B=K^*}<0$. Therefore, $\frac{\partial J^B}{\partial K^B}(K^A,K^B)\Big \arrowvert _{K^B=K^*}>0$.

Moreover, in regimes $b$ and $c$ respectively, $\frac{\partial J^B}{\partial K^B}(K^A,K^B)$ is linear decreasing with $K^B$, possibly crossing the horizontal axis at respectively $K_0$ and $\bar{K}_0$. First suppose that $J^B$’s global maximum is $K_0$ (regime $b$). Then for $K^B\ge K_0$ (regime $a$ and rest of regime $b$), $\frac{\partial J^B}{\partial K^B}(K^A,K^B)\le 0$ and thus $K^*<K_0$. Now suppose the global maximum is $\bar{K}_0$ (regime $c$). Then for $K^B>\bar{K}_0$ and in regime $c$, $\frac{\partial J^B}{\partial K^B}(K^A,K^B)<0$, so if $K^*$ is in regime $c$, $K^*<\bar{K}_0$. To end the proof, we only need to show that if $J^B$’s global maximum is $\bar{K}_0$, then $K^*$ is in regime $c$ (not $b$).

Assume that $J^B$’s global maximum is $\bar{K}_0$ and $K^*$ is not in regime $c$ (and as a result, $\bar{K}_0<K^*$). Then $J^B(\bar{K}_0)>J^B(K^*)$. Moreover, $J^A$ is continuous with $K^B$ and non increasing, therefore $J^A(\bar{K}_0)\ge J^A(K^*)$, which implies $(J^A+J^B)(\bar{K}_0)>(J^A+J^B)(K^*)$ and is a contradiction. $\square$

1.9 Proof of Theorem 2

It is in supplier B’s best interests to choose her capacity level at $K_0$ or $\bar{K}_0$, whichever value maximizes her equilibrium profit (depending on which of the local maxima is a global maximum). Denote $\bar{K}_1$ the value that supplier B would select, i.e., $argmax_{K^B} J^B(K^B)$. From Proposition 3, it follows that $\bar{K}_1\ge K^*$. At that level $\bar{K}_1$, the system total profits are not maximized. In other words, by decreasing supplier B’s capacity level from $\bar{K}_1$ to $K^*$, supplier B’s profits decrease less than supplier A’s profits increase (see Fig. 12). Supplier A would benefit from paying supplier B a fee of $J^B(\bar{K}_1)-J^B(K^*) + \epsilon$ to change her capacity level from $\bar{K}_1$ to $K^*$. This would leave supplier B better off, and incurs for supplier A a profit of

$$\begin{aligned} (J^A+J^B)(K^*)-J^B(\bar{K}_1)-\epsilon \ge (J^A+J^B)(\bar{K}_1)-J^B(\bar{K}_1)-\epsilon =J^A(\bar{K}_1)-\epsilon . \end{aligned}$$

Thus, such a fee together with a change from supplier B capacity level from $\bar{K}_1$ to $K^*$ incurs a benefit of $\epsilon$ for supplier B and of $-\epsilon +(J^A+J^B)(K^*)-(J^A+J^B)(\bar{K}_1)$ for supplier A (supplier A benefits as long as the premium $\epsilon$ is no greater than $(J^A+J^B)(K^*)-(J^A+J^B)(\bar{K}_1)$). Notice that the setting is still decentralized, in the sense that both suppliers make decisions “selfishly” in order to optimize their own profits, not the system profits. $\square$