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Carbon emission reduction and pricing policies of a supply chain considering reciprocal preferences in cap-and-trade system

  • S.I.: BOM in Social Networks
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Abstract

The traditional self-interest hypothesis is far from perfect. Social preference has a significant impact on every firm’s decision making. This paper incorporates reciprocal preferences and consumers’ low-carbon awareness (CLA) into the dyadic supply chain in which a single manufacturer plays a Stackelberg-like game with a single retailer. This research intends to investigate how reciprocity and CLA may affect the decisions and performances of the supply chain members and the system’s efficiency. In this study, the following two scenarios are discussed: (1) both the manufacturer and the retailer have no reciprocal preferences and (2) both of them have reciprocal preferences. We derive equilibriums under both scenarios and present a numerical analysis. We demonstrate that reciprocal preferences and CLA significantly affect the equilibrium and firms’ profits and utilities. First, the optimal retail price increases with CLA, while it decreases with the reciprocity of the retailer and the manufacturer; the optimal wholesale price increases with CLA and the retailer’s reciprocity, while it decreases with the manufacturer’s reciprocity. The optimal emission reduction level increases with CLA and the reciprocity of both the manufacturer and the retailer. Second, the optimal profits of the participants and the supply chain increase with CLA, the participants’ optimal profits are concave in their own reciprocity and increase with their co-operators’ reciprocity. Third, the participants’ optimal utilities increase with CLA and their reciprocity. Finally, the supply chain efficiency increases with the participants’ reciprocity, while the efficiency decreases with CLA.

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Acknowledgements

We thank the anonymous reviewers and the editor for their helpful comments for the revision of this manuscript. This work is partially supported by the NSFC Grant (Nos. 71502123, 71302115, 71301115 and 71540030), Tianjin science and technology plan Projects (15ZLZLZF00990), and the Training Program for Innovation Teams of Universities in Tianjin (No. TD12-5051).

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Correspondence to Juanjuan Qin.

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Appendices

Appendix A: Problem-solving process of max \(\pi _{m}\) with K–T condition

We can derive the manufacturer’s optimal solution with the K–T condition. The process can be described as the following:

$$\begin{aligned} \mathop {\max }\limits _{0\le e<1,w>c} \pi _m (w,e)=(w-c)(a-bp+re)+p_c [E-(1-e)(a-bp+re)]-\frac{1}{2}ue^{2} \end{aligned}$$
(1)

This problem is equal to the following problem:

$$\begin{aligned} \left\{ {\begin{array}{l} \max \;\;\pi _m (w,e)=(w-c)(a-bp+re)+p_c [E-(1-e)(a-bp+re)]-{}\frac{1}{2}ue^{2} \\ s.t.\;\;{}{}{}{}{}{}{}{}{}g_1 (w,e)=\;e-1<0 \\ \;\;\;\;\;\;\;{}{}{}{}g_2 (w,e)=c-w<0 \\ \;\;\;\;\;{}\;{}{}{}{}{}{}g_3 (w,e)=e\ge 0 \\ \;\;\;\;\;{}{}{}{}{}{}{}{}{}{}g_4 (w,e)=w>0 \\ \end{array}} \right. \end{aligned}$$
(2)

In Eqs. (1) and (2), \(p=\frac{a+re}{2b}+\frac{w}{2}\).

Since \(\nabla \pi _m (w,e)\hbox {=}\left( {\begin{array}{l} \frac{\partial \pi _m (w,e)}{\partial w} \\ \frac{\partial \pi _m (w,e)}{\partial e} \\ \end{array}} \right) \), \(\nabla g_1 (w,e)=\left( {\begin{array}{l} 0 \\ 1 \\ \end{array}} \right) \), \(\nabla g_2 (w,e)=\left( {\begin{array}{l} -1 \\ 0 \\ \end{array}} \right) \), \(\nabla g_3 (w,e)=\left( {\begin{array}{l} 0 \\ 1 \\ \end{array}} \right) \) and \(\nabla g_4 (w,e)=\left( {\begin{array}{l} 1 \\ 0 \\ \end{array}} \right) \), the K–T condition of Eq. (2) can be described as follows:

$$\begin{aligned} \left\{ {\begin{array}{l} \frac{\partial \pi _m (w,e)}{\partial w}-\lambda _2 +\lambda _4 =0 \\ \frac{\partial \pi _m (w,e)}{\partial e}+\lambda _1 +\lambda _3 =0 \\ \lambda _1 (e-1)=0 \\ \lambda _2 (c-w)=0 \\ \lambda _3 e=0 \\ \lambda _4 w=0 \\ \end{array}} \right. \;\; \end{aligned}$$
(3)

Since \(e~-~1~<~0\) and \(w~>~c, \lambda _{1}~=~ \lambda _{2}~=~\lambda _{4}~=~0\). Thus, we can solve the problem in two kinds of case.

  1. (i)
    $$\begin{aligned} \left\{ {\begin{array}{l} e=0 \\ \lambda _3 >0 \\ \frac{\partial \pi _m (w,e)}{\partial w}=0 \\ \frac{\partial \pi _m (w,e)}{\partial e}+\lambda _3 =0 \\ \end{array}} \right. \end{aligned}$$
    (4)

    The manufacturer’s optimal solution in this case is

    $$\begin{aligned} \left\{ {\begin{array}{l} e=0 \\ w_1 =\frac{a+b(c+p_c )}{2b} \\ \lambda _3 = {[2b(bp_c -r)w_1 -2abp_c -2br(c+p_c )]}/4 \\ \end{array}} \right. \end{aligned}$$
    (5)
  2. (ii)
    $$\begin{aligned} {}\left\{ {\begin{array}{l} e>0 \\ \lambda _3 =0 \\ \frac{\partial \pi _m (w,e)}{\partial w}=0 \\ \frac{\partial \pi _m (w,e)}{\partial e}+\lambda _3 =0 \\ \end{array}} \right. \end{aligned}$$
    (6)

The manufacturer’s optimal solution in this case is

$$\begin{aligned} \left\{ {\begin{array}{l} w^{{*}}=\frac{a+b(c+p_c )}{2b}+\frac{(r-bp_c )(a-bc-bp_c )(r+bp_c )}{2b[4bu-(r+bp_c )^{2}]} \\ e^{{*}}=\frac{(a-bc-bp_c )(r+bp_c )}{4bu-(r+bp_c )^{2}} \\ \frac{\partial \pi _m }{\partial \lambda _1 }=1-e\ge 0 \\ \frac{\partial \pi _m }{\partial \lambda _2 }=w-c\ge 0 \\ \end{array}} \right. . \end{aligned}$$
(7)

By comparing the optimal solutions above, we know the optimal solution of case (i) is a special case of case (ii) when \((a~-~ bc~-~ bp_{c})(r~+~ bp_{c})~=~0\). Therefore, we do not need to compare the extreme point and the maximum point on the boundaries of the feasible region.

Substituting Eq. (7) into\( p=\frac{a+re}{2b}+\frac{w}{2}\), we can describe the optimal solution of the game as follows:

$$\begin{aligned} \left\{ {\begin{array}{l} e^{{*}}=\frac{(a-bp_c -bc)(r+bp_c )}{4bu-(r+bp_c )^{2}} \\ w^{{*}}=\frac{2u(a+bc+bp_c )-(ap_c +rp_c +cr)(r+bp_c )}{4bu-(r+bp_c )^{2}}{}{}{}{}{}{} \\ p^{{*}}=\frac{u(3a+bc+bp_c )-(r+bp_c )(ap_c +rp_c +cr)}{4bu-(r+bp_c )^{2}} \\ \end{array}} \right. . \end{aligned}$$
(8)

Appendix B: Proof of Proposition 1

Proof

From Eq. (5) we can obtain the following equations:

$$\begin{aligned} \frac{de^{{*}}}{dr}= & {} \frac{(a-bp_c -bc)[4bu+(r+bp_c )^{2}]}{[4bu-(r+bp_c )^{2}]^{2}},\\ \frac{d\pi _m^{*} }{dr}= & {} \frac{u(r+bp_c )(a-bp_c -bc)^{2}}{[4bu-(r+bp_c )^{2}]^{2}},\\ \frac{d\pi _r^{*} }{dr}= & {} \frac{4bu^{2}(r+bp_c )(a-bp_c -bc)^{2}}{[4bu-(r+bp_c )^{2}]^{3}},\\ \frac{d^{2}e^{{*}}}{dr^{_2 }}= & {} 2(r+bp_c )(a-bp_c -bc)\frac{[4bu-(r+bp_c )^{2}]^{2}+2[4bu+(r+bp_c )^{2}]}{[4bu-(r+bp_c )^{2}]^{4}},\\ \frac{d^{2}\pi _m^{*} }{dr^{2}}= & {} u(a-bp_c -bc)^{2}\frac{[4bu-(r+bp_c )^{2}]^{2}+4(r+bp_c )^{2}}{[4bu-(r+bp_c )^{2}]^{2}},\\ \frac{d^{2}\pi _r^{*} }{dr^{2}}= & {} 4bu^{2}(a-bp_c -bc)^{2}\frac{[4bu-(r+bp_c )^{2}]^{3}+6(r+bp_c )^{2}}{[4bu-(r+bp_c )^{2}]^{3}}. \end{aligned}$$

\(\square \)

Given that the basic market demand (a) is sufficiently large and is significantly greater than the other parameters of the model, we assume \(a~-~{ bc}~-~{ bp}_{c}~>~0\). Considering 4 \(bu~-~(r~+~{ bp}_{c})^{2}~>~0\), we obtain \(\frac{de^{{*}}}{dr}>0, \frac{d^{2}e^{{*}}}{dr^{_2 }}>0, \frac{d\pi _m^{*} }{dr}>0, \frac{d^{2}\pi _m^{*} }{dr^{2}}>0, \frac{d\pi _r^{*} }{dr}>0\), and \(\frac{d^{2}\pi _r^{*} }{dr^{2}}>0\).

Proposition 1 is proven.

Appendix C: Problem-solving process of \(\hbox {max}~U_{m}\) with K–T Condition

We can derive the manufacturer’s optimal solution with the K–T condition. The process can be described as follows:

$$\begin{aligned} \begin{array}{l} \mathop {\max }\limits _{w,0\le e<1} U_m (w,e)=(w-c)(a-bp+re)+p_c [E-(1-e)(a-bp+re)]-{}\frac{1}{2}ue^{2}\\ \quad + \theta _m (p-w)(a-bp+re) \\ \end{array}. \end{aligned}$$
(1)

This problem is equal to the following problem:

$$\begin{aligned} \left\{ {\begin{array}{l} \max \;\;U_m (w,e)=(w-c)(a-bp+re)+p_c [E-(1-e)(a-bp+re)]-{}\frac{1}{2}ue^{2}\\ \qquad \qquad \qquad \qquad \quad \,+{}{}\theta _m (p-w)(a-bp+re) \\ s.t.\;\;{}{}{}{}{}{}{}{}{}g_1 (w,e)=\;e-1<0 \\ \;\;\;\;\;\;\;{}{}{}{}g_2 (w,e)=c-w<0 \\ \;\;\;\;\;{}\;{}{}{}{}{}{}g_3 (w,e)=e\ge 0 \\ \;\;\;\;\;{}{}{}{}{}{}{}{}{}{}g_4 (w,e)=w>0 \\ \end{array}} \right. . \end{aligned}$$
(2)

In Eqs. (1) and (2), \(p=\frac{(a+re+bw)+b\theta _r [(c-w)+p_c (1-e)]}{2b}\).

Given that \(\nabla U_m (w,e)\hbox {=}\left( {\begin{array}{l} \frac{\partial U_m (w,e)}{\partial w} \\ \frac{\partial U_m (w,e)}{\partial e} \\ \end{array}} \right) \), \(\nabla g_1 (w,e)=\left( {\begin{array}{l} 0 \\ 1 \\ \end{array}} \right) \), \(\nabla g_2 (w,e)=\left( {\begin{array}{l} -1 \\ 0 \\ \end{array}} \right) \), \(\nabla g_3(w,e)=\left( {\begin{array}{l} 0 \\ 1 \\ \end{array}} \right) \), and \(\nabla g_4 (w,e)=\left( {\begin{array}{l} 1 \\ 0 \\ \end{array}} \right) \), the K–T condition of Eq. (2) can be described as

$$\begin{aligned} \left\{ {\begin{array}{l} \frac{\partial U_m (w,e)}{\partial w}-\lambda _2 +\lambda _4 =0 \\ \frac{\partial U_m (w,e)}{\partial e}+\lambda _1 +\lambda _3 =0 \\ \lambda _1 (e-1)=0 \\ \lambda _2 (c-w)=0 \\ \lambda _3 e=0 \\ \lambda _4 w=0 \\ \end{array}} \right. \;\;\quad . \end{aligned}$$
(3)

Given that \(e-1<~0_{ }\) and \(w > c\), \({\lambda }_{1}=\lambda _{2}={\lambda }_{4}=0\). Thus, we can solve the problem in two cases.

  1. (i)
    $$\begin{aligned} \left\{ {\begin{array}{l} e=0 \\ \lambda _3 >0 \\ \frac{\partial U_m (w,e)}{\partial w}=0 \\ \frac{\partial U_m (w,e)}{\partial e}+\lambda _3 =0 \\ \end{array}} \right. \end{aligned}$$
    (4)

    The manufacturer’s optimal solution in this case is

    $$\begin{aligned} \left\{ {\begin{array}{l} e=0 \\ w_1 =\frac{a(1-\theta _m )+b(c+p_c )(\theta _r ^{2}\theta _m -2\theta _r +1)}{b(1-\theta _r )(2-\theta _m -\theta _r \theta _m )} \\ \lambda _3 \hbox {=}{\left\{ {\begin{array}{l} -b[(\theta _r -1)(2bp_c +r\theta _m -bp_c \theta _m \theta _r )+(r+bp_c \theta _r )(2-\theta _r \theta _m -\theta _m )]w_1 - \\ 2bp_c (a-2bc\theta _r -2bp_c \theta _r )\hbox {+}2ar\theta _m +2b(bp_c \theta _m \theta _r ^{2}-r)(c+p_c ) \\ \end{array}} \right\} }/4 \\ \end{array}} \right. . \end{aligned}$$
    (5)
  2. (ii)
    $$\begin{aligned} {}\left\{ {\begin{array}{l} e>0 \\ \lambda _3 =0 \\ \frac{\partial U_m (w,e)}{\partial w}=0 \\ \frac{\partial U_m (w,e)}{\partial e}+\lambda _3 =0 \\ \end{array}} \right. \end{aligned}$$
    (6)

The manufacturer’s optimal solution in this case is

$$\begin{aligned} \left\{ {\begin{array}{l} w_h^{*} =\frac{a(1-\theta _m )+b(c+p_c )(\theta _r ^{2}\theta _m -2\theta _r +1)}{b(1-\theta _r )(2-\theta _m -\theta _m \theta _r )}+ \\ {}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}\frac{[r(1-\theta _m )-bp_c (\theta _r ^{2}\theta _m -2\theta _r +1)](a-bc-bp_c )(r+bp_c )(1-\theta _r \theta _m )^{2}}{b(1-\theta _r )(2-\theta _m -\theta _r \theta _m )[2bu(1-\theta _r )(2-\theta _m -\theta _r \theta _m )-(r+bp_c )^{2}(1-\theta _r \theta _m )^{2}]} \\ e_h^{*} =\frac{(a-bc-bp_c )(r+bp_c )(1-\theta _r \theta _m )^{2}}{2bu(1-\theta _r )(2-\theta _m -\theta _r \theta _m )-(r+bp_c )^{2}(1-\theta _r \theta _m )^{2}} \\ \frac{\partial U_m }{\partial \lambda _1 }=1-e\ge 0 \\ \frac{\partial U_m }{\partial \lambda _2 }=w-c\ge 0 \\ \end{array}} \right. . \end{aligned}$$
(7)

By comparing the optimal solutions above, we know that the optimal solution of case (i) is a special case of case (ii). We need not compare the extreme point and maximum point on the boundaries of the feasible region. Substituting Eq. (7) into \(p=\frac{(a+re+bw)+b\theta _r [(c-w)+p_c (1-e)]}{2b}\), we can describe the optimal solution of the game as

$$\begin{aligned} \left\{ {\begin{array}{l} e_h^{*} =\frac{(a-bc-bp_c )(r+bp_c )(1-\theta _r \theta _m )^{2}}{2bu(1-\theta _r )(2-\theta _m -\theta _r \theta _m )-(r+bp_c )^{2}(1-\theta _r \theta _m )^{2}} \\ w_h^{*} =\frac{a(1-\theta _m )+b(c+p_c )(\theta _r ^{2}\theta _m -2\theta _r +1)}{b(1-\theta _r )(2-\theta _m -\theta _m \theta _r )}+\frac{B[r(1-\theta _m )-bp_c (\theta _r ^{2}\theta _m -2\theta _r +1)]}{Ab(1-\theta _r )(2-\theta _m -\theta _r \theta _m )}{}{}{}{}{}{} \\ p_h^{*} =\frac{2a(1-\theta _m )+(a+bc+bp_c )(1-\theta _m \theta _r )}{2b(2-\theta _m -\theta _r \theta _m )}+{}{}{}\frac{B(3r-2r\theta _m +bp_c \theta _r \theta _m -r\theta _r \theta _m -bp_c )}{2Ab(2-\theta _m -\theta _r \theta _m )} \\ \end{array}} \right. . \end{aligned}$$
(8)

Appendix D: Proof of Proposition 2

Proof

From \(e_h^{*} =\frac{(a-bc-bp_c )(r+bp_c )(1-\theta _r \theta _m )^{2}}{2bu(1-\theta _r )(2-\theta _m -\theta _r \theta _m )-(r+bp_c )^{2}(1-\theta _r \theta _m )^{2}}\), we can obtain \(\frac{\partial e_{\mathrm{h}} ^{{*}}}{\partial r}\) \( =\frac{(a-bc-bp_c )(1-\theta _r \theta _m )^{2}}{A}\),\(\frac{\partial e_{\mathrm{h}} ^{{*}}}{\partial \theta _m }=\frac{2buB(1-\theta _r )(1-3\theta _r +\theta _r \theta _m +\theta _r ^{2}\theta _m )}{(1-\theta _r \theta _m )A^{2}}\), and \(\frac{\partial e_{\mathrm{h}} ^{*}}{\partial \theta _r }=\frac{4buB(1-\theta _m )^{2}}{(1-\theta _r \theta _m )A^{2}}\). Given that \(A~>~0, B~>~0, -1\le \theta _{\mathrm{m}} \le 1\), and \(-1\le \theta _r \le \sqrt{5}-2\), we obtain \(\frac{\partial e_{\mathrm{h}} ^{{*}}}{\partial r}>0,\frac{\partial e_{\mathrm{h}} ^{{*}}}{\partial \theta _m }>0\), and\(\frac{\partial e_{\mathrm{h}} ^{*}}{\partial \theta _r }>0\). Similarly, \(\frac{\partial ^{2}e_{\mathrm{h}} ^{{*}}}{\partial r^{2}}=\frac{2(r+bp_c )(1-\theta _r \theta _m )^{4}(a-bc-bp_c )}{A^{2}}>0\) is evident. \(\square \)

From \(e_h^{*} =\frac{(a-bc-bp_c )(r+bp_c )(1-\theta _r \theta _m )^{2}}{2bu(1-\theta _r )(2-\theta _m -\theta _r \theta _m )-(r+bp_c )^{2}(1-\theta _r \theta _m )^{2}}\), we derive

$$\begin{aligned} \begin{array}{l} \frac{\partial ^{2}e_{\mathrm{h}} ^{{*}}}{\partial \theta _m^2 }=2bu(1-\theta _r )(a-bc-bp_c )(r+bp_c )[\frac{2\theta _r ^{2}(2-\theta _m -\theta _r \theta _m )}{A^{2}}\\ \qquad \qquad + {}{}{}\frac{(1-\theta _r \theta _m )[4bu(1-\theta _r ^{2})-4\theta _r (r+bp_c )^{2}](1-3\theta _r +\theta _r \theta _m +\theta _r ^{2}\theta _m )}{A^{4}}]>0 \\ \end{array} \end{aligned}$$

and

$$\begin{aligned}&\frac{\partial ^{2}e_{\mathrm{h}} ^{*}}{\partial \theta _r ^{2}}=\frac{4bu(a-bc-bp_c )(r+bp_c )(1-\theta _m )^{2}(1-\theta _r \theta _m )}{A^{3}}\\&\qquad \quad \quad \,\,\times \left\{ {\begin{array}{l} -2bu\theta _m (1-\theta _r )(2-\theta _m -\theta _r \theta _m )+ \\ (1-\theta _r \theta _m )^{2}[8bu-3\theta _m (r+bp_c )^{2}] \\ \end{array}} \right\} . \end{aligned}$$

Considering \(-1\le \theta _{\mathrm{m}} \le 1, -1\le \theta _r \le \sqrt{5}-2,A~>~0, B~>~0, 4{bu}~-~(r~+~{bp}_{c})^{2}~>~0\), and \(\frac{\partial ^{2}U_m }{\partial w^{2}}\frac{\partial ^{2}U_m }{\partial e^{2}}-\frac{\partial ^{2}U_m }{\partial w\partial e}\frac{\partial ^{2}U_m }{\partial e\partial w}\ge 0\), we derive

$$\begin{aligned} 4bu(1-\theta _r )(1+\theta _r )-4\theta _r (r+bp_c )^{2}>(r+bp_c )^{2}[(1-\theta _r )(1+\theta _r )-4\theta _r ]>0 \end{aligned}$$

and

$$\begin{aligned} (1~-~\theta _{r}\theta _{m})^{2}[8{ bu}~-~3 \theta _{m}(r~+~{bp}_{c})^{2}]~-~2 bu\theta _{m}(1~-~\theta _{r})(2~-~ \theta _{m}~-~\theta _{r} \theta _{m})~>~0. \end{aligned}$$

Thus, \(\frac{\partial ^{2}e_{\mathrm{h}} ^{{*}}}{\partial \theta _m^2 }>0\) and \(\frac{\partial ^{2}e_{\mathrm{h}} ^{{*}}}{\partial \theta _r^2 }>0\).

Proposition 2 is proven.

Appendix E: Proof of Proposition 3

Proof

Let \(\theta _{m}\) = 1; we obtain the corresponding optimal decisions, utilities, and profits as follows:

$$\begin{aligned} \left\{ {\begin{array}{ll} \pi _{{\mathrm{h}r}}^{*} =\frac{bu^{2}\left( {a-bc-bp_c } \right) ^{2}}{\left[ {2bu-\left( {r+bp_c } \right) ^{2}} \right] ^{2}} \\ \pi _{\mathrm{h}m}^{*} =p_c E-\frac{u(r+bp_c )^{2}(a-bc-bp_c )^{2}}{2[2bu-(r+bp_c )^{2}]^{2}} \\ \end{array}} \right. ,\\ \left\{ {\begin{array}{ll} U_{hm}^{*} =\frac{u\left( {a-bc-bp_c } \right) ^{2}}{2\left[ {2bu-\left( {r+bp_c } \right) ^{2}} \right] }+p_c E \\ U_{{\mathrm{h}} _r }^{*} =\frac{u\left( {a-bc-bp_c } \right) ^{2}\left[ {2bu-\theta _r \left( {r+bp_c } \right) ^{2}} \right] }{2\left[ {2bu-\left( {r+bp_c } \right) ^{2}} \right] ^{2}}+\theta _r p_c E \\ \end{array}} \right. ,\\ \left\{ {\begin{array}{ll} e_h^{*} =\frac{(r+bp_c )(a-bc-ap_c )}{2bu-(r+bp_c )^{2}} \\ w_h^{*} =c-p_c -{}{}\frac{p_c (r+bp_c )(a-bc-ap_c )}{2bu-(r+bp_c )^{2}} \\ p_h^{*} =\frac{a+bc+bp_c }{2b}+\frac{(r-bp_c )(r+bp_c )(a-bc-ap_c )}{2b[2bu-(r+bp_c )^{2}]} \\ \end{array}} \right. . \end{aligned}$$

\(\square \)

The optimal decisions and corresponding profits of the manufacturer and retailer and the optimal utility of the manufacturer have nothing to do with \(\theta _{r}\).

In the case of \(\theta _{m}~=~1\), we obtain \(\frac{\partial U_{\mathrm{h}r} ^{*} }{\partial \theta _r }=p_c E-\frac{u(r+bp_c )^{2}(a-bc-bp_c )^{2}}{2[2bu-(r+bp_c )^{2}]^{2}}=\pi _{\mathrm{h}m} ^{*} \). Given that \(\pi _{\mathrm{h}m} ^{*}>0, \frac{\partial U_{\mathrm{h}r} ^{*} }{\partial \theta _r }>0\). The optimal utility of the retailer is non-decreasing with \(\theta _{r}\). Thus, the retailer will not be mean to the manufacturer, i.e., \(\theta _{r} >~0.\)

In the case of \(\theta _{m}~=~1\), we obtain \(\pi _{\mathrm{h}m} ^{*} +\pi _{\mathrm{h}r}^{*} -\pi _m^{*} -\pi _r^{*} =\frac{2b^{2}u^{3}(a-bc-bp_c )^{2}}{[2bu-(r+bp_c )^{2}][4bu-(r+bp_c )^{2}]^{2}}\) and \(\pi _{\mathrm{h}r} ^{*} -\pi _r^{*} =\frac{4b^{2}u^{3}(a-bc-bp_c )^{2}[3bu-(r+bp_c )^{2}]}{[2bu-(r+bp_c )^{2}]^{2}[4bu-(r+bp_c )^{2}]^{2}}\). We previously assume that \(2 bu(1 - \theta _{r})(2 - \theta _{m}~-~ \theta _{r} \theta _{m})~-~(r~+~bp_{c})^{2}(1~-~ \theta _{r}\theta _{m})^{2}~>~0\); therefore, we derive \((1~-~\theta _{r})^{2}[2 bu~-~(r~+~{bp}_{c})^{2}]~>~0\) in case of \(\theta _{m}~=~1, \hbox {then } 2{bu}~-~(r~+~{bp}_{c})^{2}~>~0\). We obtain \(\pi _{\mathrm{h}r} ^{*} >\pi _r^{*} \) and \(\pi _{\mathrm{h}m} ^{*} +\pi _{\mathrm{h}r} ^{*} >\pi _m^{*} +\pi _r^{*} \).

Proposition 3 is proven.

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Xia, L., Guo, T., Qin, J. et al. Carbon emission reduction and pricing policies of a supply chain considering reciprocal preferences in cap-and-trade system. Ann Oper Res 268, 149–175 (2018). https://doi.org/10.1007/s10479-017-2657-2

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