Inventory sharing of professional optics product supply chain with equal power agents

Abstract

This study investigates the inventory sharing policy of a professional optics product supply chain based on a real case study. The supply chain sells a product via two sales agents with equal power. The manufacturer receives a single order from each agent at the beginning of the selling season. Stock is kept in the manufacturer’s central warehouse. Thus, the agents hold virtual inventory only. To obtain greater flexibility and earn more profit, the agents implement an inventory sharing policy between themselves, under which they can trade the excess product by negotiation. This study seeks to answer three questions regarding the implementation of inventory sharing: (1) Does inventory sharing always benefit all members of the supply chain? (2) Should the manufacturer charge agents service fees and if so, how much? (3) What is the mechanism for achieving an all-win situation? By formulating the problem as a mixed Stackelberg and Nash-bargaining game, this study observes that the current negotiation mechanism does not always benefit the supply chain and the manufacturer, because it gives too much flexibility to agents. An all-win leading inventory sharing mechanism is then proposed, and managerial insights are generated.

A Proofs

Proof of Proposition 1

If $\delta > r-v$, then $v+\delta _p\le t \le r-\delta _r$ ($\Rightarrow \delta \le r-v$) does not hold, namely, at least one of $A_1$ and $A_2$ do not agree to trade the excess order if $\delta \ge r-v$. As $v+\delta _p<t <r-\delta _r$, $G_{\text {II}}(k,t)$ is strictly increasing in k. Therefore $k^*_{\text {II}}=\min (E_2,S_1)$. Then by considering $\partial G_{\text {II}}(k,t)/ \partial t = k^2 (r+v-\delta _r+ \delta _p-2t)$, and $\partial ^2 G_{\text {II}}(k,t)/ \partial t^2 = -2k^2t$, $t^*_{\text {II}}=t^*_{\text {III}}= \frac{1}{2}(r+v-\delta _r+\delta _p)$ and $t^*_{\text {II}}=t^*_{\text {III}}$ is unique. If $\delta \le r-v$, then $t^*_{\text {II}} \in (v+\delta _p, r-\delta _r)$ then by putting $t^*_{\text {II}}$ and $k^*_{\text {II}}$ in to $I_{A_i,AN,\text {II}}$, $i=1,2$, we obtain (3). $\square$

Proof of Proposition 2

The proof of Proposition 2 is similar to the proof of Proposition 1. $\square$

Proof of Proposition 3

By taking the first and second order partial derivatives of $\pi _{A_i,AN}(q_1,q_2)$ with respect to $q_i$, $i=1,2$, we obtain $\partial \pi _{A_i,AN}(q_1,q_2) / \partial q_i = (r-w)- \frac{1}{2}[(r-v+\delta )F_i(q_i) + (r-v-\delta )W(Q)]$ and $\partial \pi _{A_i,AN}(q_1,q_2)^2/ \partial ^2 dq_i = -\frac{1}{2}\{(r-v+\delta )f_i(q_i) + (r-v-\delta ) \int _0^Q f_i(Q-x_i)dF_i(x_i)\}<0$. Therefore, for any fixed $q_2\ge 0$, $\pi _{A_1,AN}(q_1,q_2)$ is strictly concave in $q_1$, and for any fixed $q_1\ge 0$, $\pi _{A_2,AN}(q_1,q_2)$ is strictly concave in $q_2$. Therefore, $q_{A_i,AN}^*$, $\forall i=1,2$ satisfy (7). By solving the simultaneous equations (7), we obtain $F_1(q_1^*)=F_2(q_2^*)$. By putting $F_1(q_1^*)=F_2(q_2^*)$ into (7), (7) becomes $2(r-w)= G(q_1^*)$, where $G(q_1^*) =(r-v+\delta )F_i(q_1^*) + (r-v-\delta )W(q_1^*+F^{-1}_2(F_1(q_1^*)))$. As W(Q) and $F_i(q_i)$, $i=1,2$, are strictly increasing in Q and $q_i$, respectively, $G(q_1^*)$ is strictly increasing in $q_1^*$. Moreover, as $Q(0)=F_i(0)=0$, for all $i=1,2$, $G(0)=0$. Therefore, there is a unique pair of ($q_1^*,q_2^*$) which satisfy $F_1(q_1^*)=F_2(q_2^*)$ and the simultaneous equations (7). Finally, by (7), $W(Q^*_A) \lesseqgtr F_i(q^*_{A_i}) \Leftrightarrow W(Q^*_A) \lesseqgtr (r-w)/(r-v) \lesseqgtr F_i(q^*_{A_i})$, $\forall i=1,2$. $\square$

Proof of Proposition 4

By (8), we have $d q^*_1 / dq^*_2 = f_2(q^*_2) / f_1(q^*_1)>0$ as $f_1(q_2^*) >0$ and $f_2(q_2^*) >0$. Therefore, we obtain part (a). Then by (7), for other parameters being fixed, we obtain

$$\begin{aligned} dq_1^* / dv= & {} \frac{1}{\Lambda _1} \left\{ f_2(q^*_2)\left[ F_1(q^*_1) + \int _0^{Q^*_A}F_2(Q^*_A-x_1)dF_1(x_1)\right] \right\} ,\\ dq_1^*/dr= & {} -dq_1^*/dw = 2f_2(q^*_2)/\Lambda _1,\\ dq_1^* / d \delta= & {} \frac{1}{\Lambda _1} f_2(q^*_2)\left[ F_1(q^*_1) - \int _0^{Q^*_A}F_2(Q^*_A-x_1)dF_1(x_1)\right] , \end{aligned}$$

where

As $r-v > \delta$ , $\Lambda _1>0$, $dq_1^*/dr >0$, $dq_1^*/dv >0$, $dq_1^*/dw <0$, $dq_1^*/d \delta >0$ if and only if $W(Q^*_A) > F_1(q^*_1)$, $dq_1^*/d \delta =0$ if and only if $W(Q^*_A) = F_1(q^*_1)$ and $dq_1^*/d\delta <0$ if and only if $W(Q^*_A) < F_1(q^*_1)$. Similar results are obtain for $q_2^*$. Then, by Proposition 3(b), we obtain the results of part(d).$\square$

Proof of Proposition 5

For all $i=1,2$, $W(Q_A^*) \lesseqgtr (r-w)/(r-v)\lesseqgtr F_i(q_i^*) \Leftrightarrow$$F_i(q_i^o)= (r-w)/(r-v)\lesseqgtr F_i(q_i^*) \Leftrightarrow$$q_i^o \lesseqgtr q_i^*$ (because $F_i(q)$ is a strictly increasing function of q) $\Leftrightarrow$$Q_A^* \lesseqgtr Q_A^o$ (because $Q_A^* = q_1^*+q_2^*$ and $Q_A^o = q_1^o+q_2^o$). $\square$

Proof of Proposition 6

Let $\Pi _{A_i,AN}(q_1,q_2,x_1,x_2)$ and $\Pi _{A_i,B}(q_1,q_2,x_1,x_2)$ be the profit of $A_i$ for Model AN and Model B, respectively, for any given $q_i\ge 0$, and $x_i\ge 0$, for $i=1,2$. For any fixed $q_i > 0$, $i=1,2$, $\Pi _{A_i,AN}(q_1,q_2,x_1,x_2) \ge \Pi _{A_i,B}(q_1,q_2,x_1,x_2)$, $\forall x_1\ge 0$ and $\forall x_2\ge 0$. Moreover, $\Pi _{A_i,AN}(q_1,q_2,x_1,x_2) > \Pi _{A_i,B}(q_1,q_2,x_1,x_2)$ for $q_1 < x_2$ and $q_2 > x_1$. Therefore, $\pi _{A_i,AN}(q_1,q_2) > \pi _{A_i,B}(q_1,q_2)$ for all $\forall q_i > 0$ and $i=1,2$. $\square$

Proof of Proposition 7

Part (a): By Proposition 5, $q_i^* \ge q_i^o$, for all $i=1,2$, if $W(Q^*_A)\ge (r-w)/(r-v)$. Therefore, $\pi _{M,B}(q_1^o,q_2^o) = (w-c) (q^o_1 + q^o_2) \le (w-c) Q^*_R$. In addition, as $\sum _{i=1}^2 \int _0^{q_i} F_i(x_i) dx_i > \int _0^Q F_1(x_1)F_2(Q-x_1)dx_1$, $\forall q_1>0$ and $\forall q_2>0$, $\pi _{M,B}(q_1^o,q_2^o) \le (w-c) Q^*_A + \delta [\sum _{i=1}^2 \int _0^{q_i^*} F_i(x_i) dx_i - \int _0^{Q_A^*} F_1(x_1)F_2(Q-x_1)dx_1] = \pi _{M,AN}(q_1^*,q_2^*)$. As $q_i^* > q_i^o$, for all $i=1,2$, if $W(Q^*_A) > (r-w)/(r-v)$, inequality (11) becomes strict. Finally, by (11) and Proposition 6, we obtain (12).

Part (b): According to Proposition 5, $q^o_1 + q^O_2 > Q^*_A$ if $W(Q^*_A) < (r-w)/(r-v)$. Then by considering $\pi _{M,AN}(q^*_1,q^*_2)>\pi _{M,B}(q^o_1,q^o_2)$ and $\pi _{SC,AN}(q^*_1,q^*_2)>\pi _{SC,B}(q^o_1,q^o_2)$, we obtain (13) and (14), respectively. $\square$

Proof of Proposition 8

If $t < v$, then the provider suffers loss in the trading of product. If $T = t +\delta > v$, then the purchaser suffers loss in the trading. Therefore, $k^e=0$ if $t < v$ or $t +\delta > r$. If $v \le t \le r-\delta$, then both the provider and the purchaser are profitable by trading any unit of product. Therefore, the provider wish to provide all the excess inventory, and the purchaser wish to receive as much as possible to fulfill the demand. For $q_1 >x_1$, $A_1$ has $E_1$ excess stock, and $A_2$ has $S_2$ amount of shortage. Similarly, for $q_2 >x_2$, $A_2$ has $E_2$ excess stock, and $A_1$ has $S_1$ amount of shortage. For $q_1< x_1$ and $q_2< x_2$, both $A_1$ and $A_2$ have shortage. Therefore, $k^e=0$ and no trading occur. $\square$

Proof of Proposition 9

By considering the first order optimality condition of $\pi _{A_i,MC}(q_1,q_2)$ with respect to $q_i$, for $i=1,2$, we obtain (15). By considering the second order optimality condition $\pi _{A_1,MC}(q_1,q_2)$ with respect to $q_1$, we obtain $\partial ^2 \pi _{A_1,MC}(q_1,q_2)/ \partial q_1^2 = -(r-t-\delta ) \{ f_1(q_1) \bar{F}_2(q_2) + \int _{q_1}^Q f_2(Q-x_1)dF_1(x_1)\}-(t-v)\{f_1(q_1)F_2(q_2) + \int _0^{q_1} f_2(Q-x_1)dF_1(x_1)\}-\delta f_1(x_1) <0$. Similarly, we obtain $\partial ^2 \pi _{A_2,MC}(q_1,q_2) \partial q_2^2 <0$. Therefore, for any fixed $q_i \ge 0$, (15) is the necessary and sufficient optimality condition of $A_i$, for $i=1,2$.

Next, we prove that there exists a unique solution of the simultaneous equations (15). Equation (15) contains two equations. We use (15-1) to represent (15) for $i=1$ and (15-2) to represent (15) for $i=2$. Let $(\hat{q}_1,\hat{q}_2)$ be the pair of $(q_1, q_2)$ that satisfies (15-1), and $(\tilde{q}_1,\tilde{q}_2)$ be the pair of $(q_1, q_2)$ that satisfies (15-2). The simultaneous equations (15) has a unique solution if and only if there are unique $(\hat{q}_1,\hat{q}_2)$ and unique $(\tilde{q}_1,\tilde{q}_2)$ such that $(\hat{q}_1,\hat{q}_2)=(\tilde{q}_1,\tilde{q}_2)$.

Consider $\hat{q}_1^0$ and $\hat{q}_2^0$ which satisfy $r-w=(t-v)\int _0^{\hat{q}_1^0} F_2(\hat{q}_1^0-x_1)dF_1(x_1) + (r-t) F_1(\hat{q}_1^0)$ and $r-w=(r-t-\delta )\int _0^{\hat{q}_2^0} F_2(\hat{q}_2^0-x_1)dF_1(x_1)$, respectively. Moreover, consider $\tilde{q}_1^0$ and $\tilde{q}_2^0$ which satisfy $r-w=(r-t-\delta )\int _0^{\tilde{q}_1^0} F_1(\tilde{q}_1^0-x_2)dF_2(x_2)$ and $r-w=(t-v)\int _0^{\tilde{q}_2^0} F_1(\tilde{q}_2^0-x_2)dF_2(x_2) + (r-t) F_2(\tilde{q}_2^0)$, respectively. Notice that $(\hat{q}_1^0,0)$ and $(0,\hat{q}_2^0)$ are solutions of (15-1) and $(\tilde{q}_1^0,0)$ and $(0,\tilde{q}_2^0)$ are solutions of (15-2). Next, we explore the relationships between $\hat{q}_1^0$, $\hat{q}_2^0$, $\tilde{q}_1^0$, and $\tilde{q}_2^0$.

As $\int _0^{q} F_1(q^0-x_2)dF_2(x_2) = \int _0^{q} F_2(q-x_1)dF_1(x_1)$ , $\forall q\ge 0$, $\hat{q}_1^0 = \tilde{q}_2^0$. By considering $\hat{q}_1^0$ and $\tilde{q}_2^0$, we have

$$\begin{aligned}&(t-v)\left[ \int _0^{\tilde{q}_2^0}F_1(\tilde{q}_2^0-x_2)dF_2(x_2) - \int _0^{\hat{q}_1^0}F_2(\hat{q}_1^0-x_1)dF_1(x_1)\right] \nonumber \\&\qquad \qquad \quad = (r-t)[F_1(\hat{q}_1^0)-F_2(\tilde{q}_2^0)]. \end{aligned}$$

(22)

For $\hat{q}_1^0>F_2\tilde{q}_2^0$, L.H.S. of (22) is greater than 0 but R.H.S. of (22) is smaller than 0. For $\hat{q}_1^0<F_2\tilde{q}_2^0$, L.H.S. of (22) is smaller than 0 but R.H.S. of (22) is bigger than 0. Therefore, $\hat{q}_1^0=\tilde{q}_2^0$. Next, if $\hat{q}_2^0 = \tilde{q}_1^0$, then

$$\begin{aligned} \frac{r-t}{r-2t+v-\delta } = \frac{\int _0^{\tilde{q}_1^0} F_1(\tilde{q}_1^0-x_2)dF_2(x_2)}{F_1(\hat{q}_2^0)}. \end{aligned}$$

(23)

As $\int _0^{\tilde{q}_1^0} F_1(\tilde{q}_1^0-x_2)dF_2(x_2)<F_1(\hat{q}_2^0)$ and $t> v-\delta$. The L.H.S of (23) is bigger than 1 but the R.H.S. of (23) is smaller than 1, and it is a contradiction. Hence, $\hat{q}_2^0 \ne \tilde{q}_1^0$. In summary the above analysis implies that either (1) $\hat{q}_1^0 > \tilde{q}_1^0$ and $\hat{q}_2^0 < \tilde{q}_2^0$ hold, or (2) $\hat{q}_1^0 < \tilde{q}_1^0$ and $\hat{q}_2^0 > \tilde{q}_2^0$ hold.

For any fixed r, w, v, t and $\delta$, let $D_1(\hat{q}_1,\hat{q}_2)=d\hat{q}_1/d\hat{q}_2$ and $D_2(\tilde{q}_1,\tilde{q}_2)=d\tilde{q}_1/d\tilde{q}_2$. Furthermore, for any given $q_i \ge 0$, $i=1,2$, let $B_1(q_1,q_2) = (r-t-\delta ) \int _{q_1}^{Q} f_2(Q -x_1) dF_1(x_1) + (t-v) \int _0^{q_1} f_2(Q -x_1) dF_1(x_1)$, $C_1(q_1,q_2) = B(q_1,q_2) + f_1(q_1)[(t-v+\delta ) F_2(q_2) + (r-t) \bar{F}_2(q_2)]$, $B_2(q_1,q_2) = (r-t-\delta ) \int _{q_2}^{Q} f_1(Q -x_2) dF_2(x_2) + (t-v) \int _0^{q_2} f_1(Q -x_2) dF_2(x_2)$, and $C_2(q_1,q_2) = E(q_1,q_2) + f_2(q_2)[(t-v+\delta ) F_1(q_1) + (r-t) \bar{F}_1(q_1)]$. Notice that $C_1(q_1,q_2)>B_1(q_1,q_2)>0$ and $C_2(q_1,q_2)>B_2(q_1,q_2)>0$, and hence $B_1(q_1,q_2)/C_1(q_1,q_2)<C_2(q_1,q_2)/B_2(q_1,q_2)$, $\forall q_1,q_2>0$.

By taking the first order derivative on both sides of (15-1), we obtain $D_1(\hat{q}_1,\hat{q}_2) = - B_1(\hat{q}_1,\hat{q}_2)/C_1(\hat{q}_1,\hat{q}_2)<0$. Similarly, for (15-2), we obtain $D_2(\tilde{q}_1,\tilde{q}_2) = - B_2(\tilde{q}_1,\tilde{q}_2)/C_2(\tilde{q}_1,\tilde{q}_2)<0$. As $B_1(q_1,q_2)/C_1(q_1,q_2)<C_2q_1,q_2)/B_2(q_1,q_2)$, $\forall q_1,q_2>0$, $D_1(\hat{q}_1,\hat{q}_2) < D_2(\tilde{q}_1,\tilde{q}_2)$, $\forall$$\hat{q}_1=\tilde{q}_1$ and $\hat{q}_2=\tilde{q}_2$.

Fig. 2

Unique equilibrium of Model MC

As (1) $\hat{q}_1^0 = \tilde{q}_2^0$ and $\hat{q}_2^0=\tilde{q}_1^0$ and (2) $D_1(\hat{q}_1,\hat{q}_2) < D_2(\tilde{q}_1,\tilde{q}_2)$, $\forall$$\hat{q}_1=\tilde{q}_1$ and $\hat{q}_2=\tilde{q}_2$, $\hat{q}_1^0 < \tilde{q}_1^0$ and $\hat{q}_2^0 > \tilde{q}_2^0$. Therefore, there exist a unique set of $(\hat{q}_1,\hat{q}_2)$ and $(\tilde{q}_1,\tilde{q}_2)$, such that $(\hat{q}_1,\hat{q}_2)=(\tilde{q}_1,\tilde{q}_2)$ (see Fig. 2). $\square$

Proof of Proposition 10

Remind that we consider the case for $v<t<r-\delta$ only. For any given $q_1\ge 0$ and $q_2 \ge 0$, let $E_1(q_1,q_2)=\int _0^Q F_2(Q-x_1)dF_1(x_1) - 2 \int _0^{q_1} F_2(Q-x_1)dF_1(x_1) dF_1(q_1) >0$ and $E_2(q_1,q_2)=\int _0^Q F_1(Q-x_2)dF_2(x_2) - 2 \int _0^{q_2} F_1(Q-x_2)dF_2(x_2) dF_2(q_2) >0$. For fixed $w>c$ and $\delta \ge 0$, by considering the first order derivative of (15-1) and (15-2), we obtain the systems of ordinary differential equations

$$\begin{aligned} \left\{ \begin{array}{l} dt = \hat{h}_{11} dQ^e_A + \hat{h}_{12} dq^e_1, \\ dt = \hat{h}_{21} dQ^e_A - \hat{h}_{22} dq^e_1, \\ \end{array} \right. \end{aligned}$$

(24)

where $\hat{h}_{11} = B_1(q^e_1,q^e_2)/E_1(q^e_1,q^e_2)>0$, $\hat{h}_{12} = [C_1(q^e_1,q^e_2)-B_1(q^e_1,q^e_2)]/E_1(q^e_1,q^e_2)>0$, $\hat{h}_{21} = C_1(q^e_1,q^e_2)/E_2(q^e_1,q^e_2)>0$ and $\hat{h}_{22} = [C_2(q^e_1,q^e_2)-B_2(q^e_1,q^e_2)]/E_2(q^e_1,q^e_2)>0$. By solving the simultaneous equation (24) for $dQ^e_A/dt$ and $dq^e_1/dt$, we obtain $dQ^e_A/dt = (\hat{h}_{12}+\hat{h}_{22})/(\hat{h}_{11}\hat{h}_{22}+\hat{h}_{12}\hat{h}_{21})>0$. Therefore, $Q^e_A$ in strictly increasing in t, for $v< t < r-\delta$. By solving the simultaneous equation (24) for $dq^e_1/dt$, we obtain $dq^e_1/dt= (\hat{h}_{21}-\hat{h}_{11})/(\hat{h}_{11}\hat{h}_{22}+\hat{h}_{12}\hat{h}_{21})$. $dq^e_1/dt \lesseqgtr 0$ if and only if $\hat{h}_{21} \lesseqgtr \hat{h}_{11}$. Therefore, $q^e_1$ is non-monotonic with t. By following a similar approach, we obtain $dq^e_2/dt= (\hat{h}_{12}-\hat{h}_{22})/(\hat{h}_{11}\hat{h}_{22}+\hat{h}_{12}\hat{h}_{21})$. $dq^e_1/dt \lesseqgtr 0$ if and only if $\hat{h}_{12} \lesseqgtr \hat{h}_{22}$. Therefore, $q^e_2$ is non-monotonic with t too.

For fixed $w>c$ and $v<t<r$, by considering the first order derivative of (15-1) and (15-2), we obtain the systems of ordinary differential equations

$$\begin{aligned} \left\{ \begin{array}{l} d \delta = \tilde{h}_{11} dQ^e_R + \tilde{h}_{12} dq^e_1, \\ d \delta = \tilde{h}_{21} dQ^e_R - \tilde{h}_{22} dq^e_1, \\ \end{array} \right. \end{aligned}$$

(25)

where $\tilde{h}_{11} = B_1(q^e_1,q^e_2)/[W(Q^e_A)-W(q^e_1)]>0$, $\tilde{h}_{12} = [C_1(q^e_1,q^e_2)-B_1(q^e_1,q^e_2)]/[W(Q^e_A)-W(q^e_1)]>0$, $\tilde{h}_{21} = C_1(q^e_1,q^e_2)/[W(Q^e_A)-W(q^e_2)]>0$ and $\tilde{h}_{22} = [C_2(q^e_1,q^e_2)-B_2(q^e_1,q^e_2)]/[W(Q^e_A)-W(q^e_2)]>0$. By solving the simultaneous equation (25) for $dQ^e_A/d \delta$, we obtain $dQ^e_A/d \delta = (\tilde{h}_{12}+\tilde{h}_{22})/(\tilde{h}_{11}\tilde{h}_{22}+\tilde{h}_{12}\tilde{h}_{21})>0$. Therefore, $Q^e_A$ in strictly increasing in $\delta$, for $0< \delta < r-t$. By solving the simultaneous equation (25) for $dq^e_1/d \delta$, we obtain $dq^e_1/d \delta = (\tilde{h}_{21}-\tilde{h}_{11})/(\tilde{h}_{11}\tilde{h}_{22}+\tilde{h}_{12}\tilde{h}_{21})$. $dq^e_1/d \delta \lesseqgtr 0$ if and only if $\tilde{h}_{21} \lesseqgtr \tilde{h}_{11}$. Therefore, $q^e_1$ is non-monotonic with $\delta$. By following a similar approach, we obtain $dq^e_2/d \delta = (\tilde{h}_{12}-\tilde{h}_{22})/(\tilde{h}_{11}\tilde{h}_{22}+\tilde{h}_{12}\tilde{h}_{21})$. $dq^e_1/d \delta \lesseqgtr 0$ if and only if $\tilde{h}_{12} \lesseqgtr \tilde{h}_{22}$. Therefore, $q^e_2$ is non-monotonic with $\delta$ too. $\square$

Proof of Proposition 11

Part (a): For $t=r$ and $\delta =0$, (15) becomes

$$\begin{aligned} \left\{ \begin{array}{l} (r-w)/(r-v) = \int _0^{q^e_1} F_2(Q^e_R-x_1)dF_1(x_1), \\ (r-w)/(r-v) = \int _0^{q^e_2} F_1(Q^e_R-x_2)dF_2(x_2). \\ \end{array} \right. \end{aligned}$$

As $F_1(q_1^o)=F_2(q_2^o)=(r-w)/(r-v)$, $F_1(q^e_1)>\int _0^{q^e_1} F_2(Q^e_A-x_1)dF_1(x_1)$ and $F_2(q^e_1)>\int _0^{q^e_2} F_1(Q^e_R-x_2)dF_2(x_2)$, we have

$$\begin{aligned} F_1(q_1^o)< F_1(q^e_1) \; \mathrm{and} \; F_2(q_2^o) < F_1(q^e_2). \end{aligned}$$

Therefore, $q_1^o<q^e_1$, $q_2^o<q^e_2$ and $Q_A^o<Q^e_A$.

Part (b) For $t=v$ and $\delta =0$, (15) becomes

$$\begin{aligned} \left\{ \begin{array}{l} (r-w)/(r-v) = \int _{q^e_1}^{Q^e_A} F_2(Q^e_A-x_1)dF_1(x_1)+ F_1(q^e_1), \\ (r-w)/(r-v) = \int _{q^e_2}^{Q^e_A} F_1(Q^e_A-x_2)dF_2(x_2)+ F_2(q^e_2). \\ \end{array} \right. \end{aligned}$$

As $F_1(q_1^o)=F_2(q_2^o)=(r-w)/(r-v)$, $\int _{q^e_1}^{Q^e_A} F_2(Q^e_A-x_1)dF_1(x_1) >0$ and $\int _{q^e_2}^{Q^e_A} F_2(Q^e_A-x_2)dF_2(x_2) >0$, we obtain

$$\begin{aligned} F_1(q_1^o)> F_1(q^e_1), \; \mathrm{and} F_2(q_2^o) > F_1(q^e_2). \end{aligned}$$

Therefore, $q_1^o>q^e_1$, $q_2^o>q^e_2$ and $Q_A^o>Q^e_A$.

Part (c): For $t= (r+v-\delta )/2$, (15) is equivalently to (7). Therefore, $q_1^*=q^e_1$, $q_2^*=q^e_2$ and $Q_R^*=Q^e_R$. $\square$

Proof of Proposition 12

Similar to the proof of Proposition 6. $\square$

Proof of Proposition 13

Part (a): For $(\delta ,t)\in \mathcal {Q}^+$, $Q^e_A > Q^o_A$. Therefore, $\pi _{M,B}(q_1^o,q_2^o) = (w-c) Q^o_A < (w-c) Q^e_A$. In addition, as $\sum _{i=1}^2 \int _0^{q_i} F_i(x_i) dx_i > \int _0^Q F_1(x_1)F_2(Q-x_1)dx_1$, $\forall q_1>0$ and $\forall q_1>0$, $\pi _{M,B}(q_1^o,q_2^o) < \pi _{M,MC}(q_1^e,q_2^e)$. By (16) and Proposition 12, we obtain (17). The proof of part (b) is similar to the proof of part (a), and we omit it here. Part (c): By considering $\pi _{M,MC}(q^e_1,q^e_2)>\pi _{M,B}(q^o_1,q^o_2)$ and $\pi _{SC,MC}(q^e_1,q^e_2)>\pi _{SC,B}(q^o_1,q^o_{A_2})$, we obtain (20) and (21), respectively. $\square$