Energy conservation strategy and coordination of a closed-loop supply chain with risk-averse members under carbon tax regulation

Abstract

This paper considers a closed-loop supply chain including a risk-averse manufacturer and a risk-averse retailer under carbon tax regulation, where the manufacturer takes back used products collected from customers by the retailer. The stochastic demand for the product is linearly dependent on the sales price and level of energy saving equipment. Under the mean-variance framework, two manufacturer-led decentralized systems are modeled and compared to reveal whether the risk-averse manufacturer should invest in energy saving equipment. A two-part tariff contract is then proposed to coordinate the decentralized system with equipment investment. The conditions of realizing a win-win outcome for the system agents are further derived. The developed models are illustrated by numerical examples, and a sensitivity analysis is conducted to identify the effects of major parameters on optimal decisions of the closed-loop supply chain. Some managerial implications are also discussed.

Appendix A

Proof of Theorem 1

With the Stackelberg game approach, we first solve and simplify $\frac{\partial U(\Pi _{r})}{\partial p}$ and $\frac{\partial U(\Pi _{r})}{\partial \tau }$ for any values of w and s fromEq. (6).

$$\begin{aligned} \frac{\partial U(\Pi _{r})}{\partial p}=(d-\alpha p+\beta s)-(\alpha +\lambda _{r}\sigma ^{2})[(b-a)\tau +p-w] \end{aligned}$$

(A1)

and

$$\begin{aligned} \frac{\partial U(\Pi _{r})}{\partial \tau }=(b-a)(d-\alpha p+\beta s)-2 c_{L} \tau -(b-a)\lambda _{r}\sigma ^{2}[(b-a)\tau +p-w].\nonumber \\ \end{aligned}$$

(A2)

Taking the second partial derivatives of $U(\Pi _{r})$ with respect to p and $\tau$, we obtain that $\frac{\partial ^{2} U(\Pi _{r})}{\partial p^{2}}=-(2\alpha +\lambda _{r}\sigma ^{2})$, $\frac{\partial ^{2} U(\Pi _{r})}{\partial p \partial \tau }=-(b-a)(\alpha +\lambda _{r}\sigma ^{2})$, and $\frac{\partial ^{2} U(\Pi _{r})}{\partial \tau ^{2}}=-2c_{L}-\lambda _{r}\sigma ^{2}(b-a)^{2}$. Using the assumption that $c_{n}\ge c_{r}+b+\delta _{e}c_{t}$, we have $\delta _{c}-\delta _{e} c_{t}\ge b-a$. From $b>a$ and $\underline{c_{L}}>\frac{\alpha ^{2}(b-a)^{2}}{2(2\alpha +\lambda _{r}\sigma ^{2})}$, we have $c_{L}\ge \underline{c_{L}}>\frac{\alpha ^{2}(b-a)^{2}}{2(2\alpha +\lambda _{r}\sigma ^{2})}$ and further have $\frac{\partial ^{2} U(\Pi _{r})}{\partial \tau ^{2}}\cdot \frac{\partial ^{2} U(\Pi _{r})}{\partial p^{2}}-(\frac{\partial ^{2} U(\Pi _{r})}{\partial \tau \partial p})^{2}=2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}>0$, which means that $U(\Pi _{r})$ is jointly concave with respect to p and $\tau$.

By solving $\frac{\partial U(\Pi _{r})}{\partial p}=0$ and $\frac{\partial U(\Pi _{r})}{\partial \tau }=0$ and from Eqs.(A1) and (A2), we have the expressions for $p^{D-H}$ and $\tau ^{D-H}$ for any given w and s:

$$\begin{aligned} p^{D-H}=w+\frac{[2c_{L}-\alpha (b-a)^{2}](d-\alpha w+\beta s)}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}} \end{aligned}$$

(A3)

and

$$\begin{aligned} \tau ^{D-H}=\frac{\alpha (b-a)(d-\alpha w+\beta s)}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}}. \end{aligned}$$

(A4)

From Eqs. (A3) and (A4), we further have

$$\begin{aligned} d-\alpha p^{D-H}+\beta s=\frac{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})(d-\alpha w+\beta s)}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}} \end{aligned}$$

(A5)

and

$$\begin{aligned} w-b \tau ^{D-H} - c-c_{t}(e_{0}-s) &=w-(c_{n}+c_{t}e_{n})+c_{t}s\nonumber \\&\quad +\frac{\alpha (b-a)(c_{n}-c_{r}-b-\delta _{e} c_{t})(d-\alpha w+\beta s)}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}}. \end{aligned}$$

(A6)

Substituting Eqs. (A5) and (A6) into Eq. (7), and solving and simplifying $\frac{\partial U(\Pi _{m})}{\partial w}$ and $\frac{\partial U(\Pi _{m})}{\partial s}$, we have

$$\begin{aligned} \frac{\partial U(\Pi _{m})}{\partial w}= & {} \frac{1}{A_2} \left\{ A_{1}(d-\alpha p^{D-H}+\beta s)-[2\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+A_{1}\lambda _{m}\sigma ^{2}]\cdot \right. \nonumber \\{} & {} \left. [w-b \tau ^{D-H} - c-c_{t}(e_{0}-s)]\right\} \end{aligned}$$

(A7)

and

$$\begin{aligned} \frac{\partial U(\Pi _{m})}{\partial s}&= \frac{1}{A_2}\big \{[A_{2}c_t+\alpha \beta (b-a)(c_{n}-c_{r}-b-\delta _{e} c_{t})]\cdot \nonumber \\{} & {} [d-\alpha p^{D-H}+\beta s+\lambda _m\sigma ^2 (w-b\tau ^{D-H}-c-c_{t}(e_0-s))]-\nonumber \\&\quad - [w-b \tau ^{D-H}-c-c_{t}(e_{0}-s)]2\beta c_{L}(\alpha +\lambda _{r}\sigma ^{2})\big \}-\eta s. \end{aligned}$$

(A8)

From Eqs. (A7) and (A8), we can reduce the second partial derivatives of $U(\Pi _m)$ as follows. $\frac{\partial ^{2} U(\Pi _{m})}{\partial w^{2}}=-\frac{A_{1}}{A^{2}_{2}}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]$, $\frac{\partial ^{2} U(\Pi _{m})}{\partial s^{2}}=\frac{4\beta c_L(\alpha +\lambda _r\sigma ^2)}{A^2_2}[A_2 c_t+\alpha \beta (b-a)(c_n-c_r-b-\delta _e c_t)]-\frac{\lambda _m\sigma ^2}{A_2^2}[A_{2}c_{t}+\alpha \beta (b-a)(c_{n}-c_{r}-b-\delta _{e} c_t)]^2-\eta$, and $\frac{\partial ^{2} U(\Pi _{m})}{\partial w \partial s}=\frac{A_1}{A_2^2}2\beta c_{L}(\alpha +\lambda _{r}\sigma ^{2})-\frac{2\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+A_{1}\lambda _{m}\sigma ^{2}}{A_2^2}[A_{2}c_{t}+\alpha \beta (b-a)(c_{n}-c_{r}-b-\delta _{e} c_{t})]$. Using $\eta \ge \underline{\eta }$, we have $\eta >\frac{4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}{A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}$ and further have $\frac{\partial ^{2} U(\Pi _{m})}{\partial w^{2}}\cdot \frac{\partial ^{2} U(\Pi _{m})}{\partial s^{2}}-(\frac{\partial ^{2} U(\Pi _{m})}{\partial w \partial s})^{2}=\frac{\eta A_1}{A_2^2}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]-\frac{4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}{A_2^2}>0$, which means that $U(\Pi _{m})$ is jointly concave with respect to w and s.

Solving $\frac{\partial U(\Pi _{m})}{\partial w}=0$ and $\frac{\partial U(\Pi _{m})}{\partial s}=0$ and from Eqs. (A7) and (A8), we have the expressions for $w^{D-H}$ and $s^{D-H}$:

$$\begin{aligned} w^{D-H}=\frac{d}{\alpha }-\left( \frac{\eta [2\alpha c_{L}(\alpha +\lambda _{r} \sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]A_{2}}{4\alpha (\beta +\alpha c_{t})c^{2}_{L}(\alpha +\lambda _{r} \sigma ^{2})^{2}}-\frac{\beta }{\alpha }\right) s^{D-H} \end{aligned}$$

(A9)

and

$$\begin{aligned} s^{D-H}=\frac{4(\beta +\alpha c_{t})c^{2}_{L}(\alpha +\lambda _{r} \sigma ^{2})^{2}[d-\alpha (c_{n}+c_{t}e_{n})]}{\eta A_{1}[4\alpha c_{L}(\alpha +\lambda _{r} \sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]-4c^{2}_{L}(\alpha +\lambda _{r} \sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}. \end{aligned}$$

(A10)

Substituting the above two Equations into Eqs.(A3) and (A4) and after simplification, we obtain the optimal selling price $p^{D-H}$ and collection rate $\tau ^{D-H}$:

$$\begin{aligned} p^{D-H}=\frac{d}{\alpha }- \frac{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})}{\alpha ^{2}(b-a)} \left( 1-\frac{\beta (\beta +\alpha c_{t})}{\eta [2\alpha c_{L}(\alpha +\lambda _{r} \sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}\right) \tau ^{D-H}\nonumber \\ \end{aligned}$$

(A11)

and

$$\begin{aligned} \tau ^{D-H}=\frac{\alpha (b-a)\eta [2\alpha c_{L}(\alpha +\lambda _{r} \sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}][d-\alpha (c_{n}+c_{t}e_{n})]}{\eta A_{1}[4\alpha c_{L}(\alpha +\lambda _{r} \sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]-4c^{2}_{L}(\alpha +\lambda _{r} \sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}. \end{aligned}$$

(A12)

Using Eq. (A12), we rearrange Eq. (A10) as follows:

$$\begin{aligned} s^{D-H}\nonumber & =\frac{2 c_L \left( \beta +\alpha c_t\right) \left( \alpha + \lambda _r\sigma ^2\right) }{2 c_L \left( \beta +\alpha c_t\right) ^2 \left( \alpha + \lambda _r\sigma ^2\right) -\alpha A_1\eta } \left[ \frac{A_1\tau ^{D-H}}{\alpha (b-a)}-d+\alpha (c_t e_n+c_n)\right] .\nonumber \\ \end{aligned}$$

(A13)

$\square$

Proof of Corollary 2

• (i) From Eqs. (8), (9), (10), and (11), we have

  $$\begin{aligned} d-\alpha p^{D-H}+\beta s^{D-H}=\frac{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})\tau ^{D-H}}{\alpha (b-a)} \end{aligned}$$

  (A14)

  and

  $$\begin{aligned} p^{D-H}-w^{D-H}+(b-a)\tau ^{D-H}=\frac{2c_{L}\tau ^{D-H}}{\alpha (b-a)} \end{aligned}$$

  (A15)

  Substituting above two equations into Eq. (6), we have

  $$\begin{aligned} U^{D-H}(\Pi _{r})=\frac{c_{L}[2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}](\tau ^{D-H})^{2}}{\alpha ^{2}(b-a)^{2}}. \end{aligned}$$

  (A16)

  Similarly, from Eqs. (9), (10), and (11), we have

  $$\begin{aligned}{} & {} w^{D-H}-b\tau ^{D-H}-c-c_{t}(e_{0}-s^{D-H})\nonumber \\{} & {} =w^{D-H}-(c_{n}+c_{t}e_{n})+c_{t}s^{D-H}+(c_{n}-c_{r}-b-\delta _{e} c_{t})\tau ^{D-H}\nonumber \\{} & {} =\frac{\eta A_{1}s^{D-H}}{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})(\beta +\alpha c_{t})}. \end{aligned}$$

  (A17)

  The following utility of the manufacturer in model D-H is obtained from Eqs. (A14), (A17) andEq. (7):

  $$\begin{aligned} U^{D-H}(\Pi _{m})=\eta \left( \frac{\eta A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}{8c_L^2(\alpha +\lambda _r\sigma ^2)^2 (\beta +\alpha c_t)^2}-\frac{1}{2}\right) (s^{D-H})^{2}. \end{aligned}$$

  (A18)

• (ii) From Eqs. (8) and (11), we have

  $$\begin{aligned} d-\alpha p^{D-H}+\beta s^{D-H}=\frac{\eta [2\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2} A_{1}]s^{D-H}}{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})(\beta +\alpha c_{t})}. \end{aligned}$$

  (A19)

  Using $e_{0}=e_{r}\tau +e_{n}(1-\tau )$ and $\delta _{e}=e_{r}-e_{n}$, and from Eqs. (A19) and (3), we have

  $$\begin{aligned} E^{D-H}(J_{m})&= (e_{0}-s^{D-H})(d-\alpha p^{D-H}+\beta s^{D-H})\nonumber \\&= (\delta _{e} \tau ^{D-H}+e_{n}-s^{D-H})(d-\alpha p^{D-H}+\beta s^{D-H})\nonumber \\&= \frac{\eta [2\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}](e_{n}-s^{D-H}+\delta _{e} \tau ^{D-H})s^{D-H}}{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})(\beta +\alpha c_{t})}.\nonumber \\ \end{aligned}$$

  (A20)

$\square$

Proof of Corollary 3

• (i) From Table 2, we take the first partial derivative of $\tau ^{D-N}$ with respect to $\lambda _{m}$ and have

  $$\begin{aligned} \frac{\partial \tau ^{D-N}}{\partial \lambda _{m}}=\frac{2(b-a)c_{L}[d-\alpha (c_{n}+c_{t}e_{n})](\alpha +\lambda _{r}\sigma ^{2})\alpha ^{2}\sigma ^{2}}{[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]^{2}} \end{aligned}$$

  (A21)

  From Eq. (A21), we have $\frac{\partial \tau ^{D-N}}{\partial \lambda _{m}}>0$. Similarly, from Table 2, we also have

  $$\begin{aligned} \frac{\partial w^{D-N}}{\partial \lambda _{m}}=\frac{-2c_{L}[d-\alpha (c_{n}+c_{t}e_{n})]A_{2}(\alpha +\lambda _{r}\sigma ^{2})\sigma ^{2}}{[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]^{2}} \end{aligned}$$

  (A22)

  and

  $$\begin{aligned} \frac{\partial p^{D-N}}{\partial \lambda _{m}}=\frac{-4[d-\alpha (c_{n}+c_{t}e_{n})]c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}\sigma ^{2}}{[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]^{2}} \end{aligned}$$

  (A23)

  From Eqs. (A22) and (A23), we have $\frac{\partial w^{D-N}}{\partial \lambda _{m}}<0$ and $\frac{\partial p^{D-N}}{\partial \lambda _{m}}<0$.

• (ii) From Table 2, we take the first partial derivative of $U^{D-N}(\Pi _{r})$ and $E^{D-N}(J_{m})$ with respect to $\lambda _{m}$. We have

  $$\begin{aligned} \frac{\partial U^{D-N}(\Pi _{r})}{\partial \lambda _{m}}=\frac{2c_{L}A_{2}\tau ^{D-N}}{\alpha ^{2}(b-a)^{2}}\cdot \frac{\partial \tau ^{D-N}}{\partial \lambda _{m}}, \end{aligned}$$

  (A24)

  and

  $$\begin{aligned} \frac{\partial E^{D-N}(J_{m})}{\partial \lambda _{m}}=\frac{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})(e_{n}+2\delta _{e}\tau ^{D-N}) }{\alpha (b-a)}\cdot \frac{\partial \tau ^{D-N}}{\partial \lambda _{m}}. \end{aligned}$$

  (A25)

  Using Eq. (A21) and from Eqs. (A24) and (A25), we have $\frac{\partial U^{D-N}(\Pi _{r})}{\partial \lambda _{m}}>0$ and $\frac{\partial E^{D-N}(J_{m})}{\partial \lambda _{m}}>0$. From Table 2, we also simplify $U^{D-N}(\Pi _{m})$ as

  $$\begin{aligned} U^{D-N}(\Pi _{m})=\frac{2c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}[d-\alpha (c_{n}+c_{t}e_{n})]^{2}}{A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}. \end{aligned}$$

  (A26)

  From Eq. (A26), we take the first partial derivative of $U^{D-N}(\Pi _{m})$ with respect to $\lambda _{m}$ and have

  $$\begin{aligned} \frac{\partial U^{D-N}(\Pi _{m})}{\partial \lambda _{m}}=\frac{-2c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}\sigma ^{2}[d-\alpha (c_{n}+c_{t}e_{n})]^{2}}{[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}. \end{aligned}$$

  (A27)

  Equation (A27) yields $\frac{\partial U^{D-N}(\Pi _{m})}{\partial \lambda _{m}}<0$.

$\square$

Proof of Theorem 4

• (i) According to Table 3 andEq. (9), the comparison between $\tau ^{D-H}$ and $\tau ^{D-N}$ yields

  $$\begin{aligned} \frac{\tau ^{D-N}}{\tau ^{D-H}}=1-\frac{4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}{\eta A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}. \end{aligned}$$

  (A28)

  Using $\eta >\frac{4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}{A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}$, we easily have $\tau ^{D-H}>\tau ^{D-N}$.

• (ii) With Eq. (8) and Table 3, we compare $p^{D-H}$ with $p^{D-N}$ and have

  $$\begin{aligned} p^{D-H}-p^{D-N}=\frac{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})}{\alpha ^{2}(b-a)}(\tau ^{D-N}-\tau ^{D-H}\Phi _{1}), \end{aligned}$$

  (A29)

  where $\Phi _{1}=1-\frac{2c_{L}(\alpha +\lambda _{r}\sigma ^{2})\beta (\beta +\alpha c_{t})}{\eta [2\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}$. From Eq. (A29), we have that if $\tau ^{D-H}\Phi _{1}<\tau ^{D-N}$, then $p^{D-H}>p^{D-N}$; otherwise, $p^{D-H}\le p^{D-N}$.

• (iii) WithEq. (10) and Table 3, we compare $w^{D-H}$ with $w^{D-N}$ and have

  $$\begin{aligned} w^{D-H}-w^{D-N}=\frac{A_{2}}{\alpha ^{2}(b-a)}(\tau ^{D-N}-\tau ^{D-H}\Phi _{2}), \end{aligned}$$

  (A30)

  where $\Phi _{2}=1-\frac{4\beta (\beta +\alpha c_{t})c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}}{\eta A_{2}[2\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}$. From Eq. (A30), we have that if $\tau ^{D-H}\Phi _{2}<\tau ^{D-N}$, then $w^{D-H}>w^{D-N}$; Otherwise, $w^{D-H}\le w^{D-N}$.

$\square$

Proof of Theorem 5

(i) According to Eqs.(12), and (15), and Table 3, the comparison between $U^{D-H}_{r}$ and $U^{D-N}_{r}$ yields

$$\begin{aligned} U^{D-H}_{r}-U^{D-N}_{r}=\frac{c_{L}[2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}]}{\alpha ^{2}(b-a)^{2}} [(\tau ^{D-H})^{2}-(\tau ^{D-N})^{2}]. \end{aligned}$$

Using Theorem 4(i), we have $U^{D-H}_{r}>U^{D-N}_{r}$.

Let $y=\eta A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]-4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}$. Using Eqs. (13) and (15), and Table 3, we compare $U^{D-H}_{m}$ with $U^{D-N}_{m}$ and have

$$\begin{aligned}{} & {} U^{D-H}_{m}-U^{D-N}_{m}\nonumber \\{} & {} \quad =\frac{8c^{4}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{4}(\beta +\alpha c_{t})^{2}A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}](\tau ^{D-N})^{2}}{y\alpha ^{2}(b-a)^{2}[2\alpha (\alpha +\lambda _{r}\sigma ^{2}) c_{L}+\lambda _{m}\sigma ^{2}A_{1}]^{2}}\nonumber \\{} & {} \quad >0. \end{aligned}$$

(A31)

The last equation in Eq. (A31) holds because $\eta >\frac{4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}}{A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}$. Therefore, from Eq. (A31), we have $U^{D-H}_{m}>U^{D-N}_{m}$.

(iii) Using Eqs. (11) and (14) and Table 3, we compare $E^{D-H}(J_{m})$ with $E^{D-N}(J_{m})$ and have

$$\begin{aligned} \frac{E^{D-H}(J_{m})}{E^{D-N}(J_{m})}=\frac{(e_{n}+\delta _{e} \tau ^{D-H}-s^{D-H})\tau ^{D-H}}{(e_{n}+\delta _{e} \tau ^{D-N})\tau ^{D-N}}. \end{aligned}$$

(A32)

From Eq. (15) and the expression for $\Delta _{0}$, we have $e_{n}+\delta _{e}(\tau ^{D-H}+\tau ^{D-N})=\Delta _{0}$. Therefore, when $s^{D-H}>s_{t}=\frac{4c^{2}_{L}(\alpha +\lambda _{r}\sigma ^{2})^{2}(\beta +\alpha c_{t})^{2}\Delta _{0}}{\eta A_{1}[4\alpha c_{L}(\alpha +\lambda _{r}\sigma ^{2})+\lambda _{m}\sigma ^{2}A_{1}]}$, using Eq. (15), we have $s^{D-H}>\frac{(\tau ^{D-H}-\tau ^{D-N})\Delta _{0}}{\tau ^{D-H}}=\frac{(\tau ^{D-H}-\tau ^{D-N})[e_{n}+\delta _{e}(\tau ^{D-H}+\tau ^{D-N})]}{\tau ^{D-H}}$. Substituting it into Eq. (A32), we have $E^{D-H}(J_{m})<E^{D-N}(J_{m})$. Similarly, when $s^{D-H}\le s_{t}$, using Eqs.(15) and (A32), we have $E^{D-H}(J_{m})\ge E^{D-N}(J_{m})$. $\square$

Proof of Theorem 6

• (i) Solving $\frac{\partial U(\Pi _{c})}{\partial p}$ and $\frac{\partial U(\Pi _{c})}{\partial \tau }$ and from Eq. (17), we have

  $$\begin{aligned} \frac{\partial U(\Pi _{c})}{\partial p}&= (d-\alpha p+\beta s)-(\alpha +\lambda _{c}\sigma ^{2})[p-(c_{n}+c_{t}e_{n})\nonumber \\&\quad +(\delta _{c}-\delta _{e} c_{t})\tau +c_{t}s], \end{aligned}$$

  (A33)
  $$\begin{aligned} \frac{\partial U(\Pi _{c})}{\partial \tau }= & {} (\delta _{c}-\delta _{e} c_{t})(d-\alpha p+\beta s)-2 c_{L} \tau \nonumber \\{} & {} -(\delta _{c}-\delta _{e} c_{t})\lambda _{c}\sigma ^{2}[p-(c_{n}+c_{t}e_{n})+(\delta _{c}-\delta _{e} c_{t})\tau +c_{t}s], \end{aligned}$$

  (A34)

  and

  $$\begin{aligned} \frac{\partial U(\Pi _{c})}{\partial s}= & {} c_{t}(d-\alpha p+\beta s)-\eta s \nonumber \\{} & {} +(\beta -c_{t}\lambda _{c}\sigma ^{2})[p-(c_{n}+c_{t}e_{n})+(\delta _{c}-\delta _{e} c_{t})\tau +c_{t}s]. \end{aligned}$$

  (A35)

  Taking the second partial derivatives of $U(\Pi _{c})$ with respect to three decision variables, we have $\frac{\partial ^{2} U(\Pi _{c})}{\partial p^{2}}=-(2\alpha +\lambda _{c}\sigma ^{2})$, $\frac{\partial ^{2} U(\Pi _{c})}{\partial \tau ^{2}}=-[2c_{L}+\lambda _{c}\sigma ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}]$, $\frac{\partial ^{2} U(\Pi _{c})}{\partial s^{2}}=2\beta c_{t}-\lambda _{c}\sigma ^{2}c^{2}_{t}-\eta$, $\frac{\partial ^{2} U(\Pi _{c})}{\partial p \partial \tau }=-(\alpha +\lambda _{c}\sigma ^{2})(\delta _{c}-\delta _{e} c_{t})$, $\frac{\partial ^{2} U(\Pi _{c})}{\partial p \partial s}=\beta -c_{t}(\alpha +\lambda _{c}\sigma ^{2})$, and $\frac{\partial ^{2} U(\Pi _{c})}{\partial s \partial \tau }=(\beta -\lambda _{c}\sigma ^{2} c_{t})(\delta _{c}-\delta _{e} c_{t})$. Using $c_{L}>\underline{c_{L}}$, we have $\frac{\partial ^{2} U(\Pi _{c})}{\partial \tau ^{2}}\cdot \frac{\partial ^{2} U(\Pi _{c})}{\partial p^{2}}-(\frac{\partial ^{2} U(\Pi _{c})}{\partial \tau \partial p})^{2}=2c_{L}(2\alpha +\lambda _{c}\sigma ^{2})-\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}>0$. Using $\eta >\underline{\eta }$, we have $\frac{\partial ^{2} U(\Pi _{c})}{\partial s^{2}}<0$ and $|\nabla ^{2}\Pi _{c}|<0$, where $\nabla ^{2}\Pi _{c}$ is the Hessian matrix of $U(\Pi _{c})$, and

  $$\begin{aligned} |\nabla ^{2}\Pi _{c}|& = \begin{vmatrix} \frac{\partial ^{2} U(\Pi _{c})}{\partial p^{2}}&\frac{\partial ^{2} U(\Pi _{c})}{\partial p\partial \tau }&\frac{\partial ^{2} U(\Pi _{c})}{\partial p\partial s}\\ \frac{\partial ^{2} U(\Pi _{c})}{\partial \tau \partial p}&\frac{\partial ^{2} U(\Pi _{c})}{\partial \tau ^2}&\frac{\partial ^{2} U(\Pi _{c})}{\partial \tau \partial s}\\ \frac{\partial ^{2} U(\Pi _{c})}{\partial s \partial p}&\frac{\partial ^2 U(\Pi _c)}{\partial s \partial \tau }&\frac{\partial ^{2} U(\Pi _{c})}{\partial s^2} \end{vmatrix}_{3\times 3}\nonumber \\&= 2c_{L}(\beta +\alpha c_{t})^{2}-\eta [2c_{L}(2\alpha +\lambda _{c}\sigma ^{2})-\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}]<0, \end{aligned}$$

  (A36)

  which means that $U(\Pi _{c})$ is jointly concave with respect to p, $\tau$, and s. Based on Eqs.(A33)–(A35), we solve $\frac{\partial U(\Pi _{c})}{\partial p}=0$, $\frac{\partial U(\Pi _{c})}{\partial \tau }=0$, and $\frac{\partial U(\Pi _{c})}{\partial s}=0$, and have

  $$\begin{aligned} p^{C}= & {} \frac{d}{\alpha }-\frac{2c_{L}[(\alpha +\lambda _{c}\sigma ^{2})\eta -\beta (\beta +\alpha c_{t})]}{\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})\eta }\tau ^{C}, \end{aligned}$$

  (A37)
  $$\begin{aligned} \tau ^{C}= & {} \frac{\alpha \eta (\delta _{c}-\delta _{e} c_{t})[d-\alpha (c_{n}+c_{t}e_{n})]}{\eta [2c_{L}(2\alpha +\lambda _{c}\sigma ^{2})-\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}]-2c_{L}(\beta +\alpha c_{t})^{2}}, \end{aligned}$$

  (A38)

  and

  $$\begin{aligned} s^{C}=\frac{2c_{L}(\beta +\alpha c_{t})[d-\alpha (c_{n}+c_{t}e_{n})]}{\eta [2c_{L}(2\alpha +\lambda _{c}\sigma ^{2})-\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}]-2c_{L}(\beta +\alpha c_{t})^{2}}. \end{aligned}$$

  (A39)

• (ii) Using Eqs. (A37), (A38), and (A39) yields

  $$\begin{aligned} d-\alpha p^{C}+\beta s^{C}= & {} \frac{2c_{L}(\alpha +\lambda _{c}\sigma ^{2})\tau ^{C}}{\alpha (\delta _{c}-\delta _{e} c_{t})}, \end{aligned}$$

  (A40)
  $$\begin{aligned} s^{C}= & {} \frac{2c_{L}(\beta +\alpha c_{t})\tau ^{C}}{\alpha \eta (\delta _{c}-\delta _{e} c_{t})}, \end{aligned}$$

  (A41)

  and

  $$\begin{aligned} p^{C}-(c_{n}+e_{n}c_{t})+(\delta _{c}-\delta _{e} c_{t})\tau ^{C}+c_{t}s^{C}=\frac{2c_{L}\tau ^{C}}{\alpha (\delta _{c}-\delta _{e} c_{t})}. \end{aligned}$$

  (A42)

  Substituting Eqs. (A40), (A41), and (A42) into Eq. (17) and after simplification, we have

  $$\begin{aligned} U^{C}(\Pi _{c})= & {} \frac{2c^{2}_{L}[(2\alpha +\lambda _{c}\sigma ^{2})\eta -(\beta +\alpha c_{t})^{2}](\tau ^{C})^{2}}{\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}\eta }-c_{L}(\tau ^{C})^{2}\nonumber \\= & {} \frac{c_{L}\{[2c_{L}(2\alpha +\lambda _{c}\sigma ^{2})-\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}]\eta -2c_{L}(\beta +\alpha c_{t})^{2}\}(\tau ^{C})^{2}}{\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}\eta }.\nonumber \\ \end{aligned}$$

  (A43)

  Substituting Eqs. (A40) and (A41) into Eq. (3) and using $e_{0}=e_{r}\tau +e_{n}(1-\tau )$, we simplify the expected carbon emissions as

  $$\begin{aligned} E^{C}(J_{m})&= \frac{2_{L}(\alpha +\lambda _{c}\sigma ^{2})(e_{n}+\delta _{e} \tau ^{C}-s^{C})\tau ^{C}}{\alpha (\delta _{c}-\delta _{e} c_{t})}\nonumber \\ & = \frac{\eta (\alpha +\lambda _{c}\sigma ^{2})(e_{n}+\delta _{e} \tau ^{C}-s^{C})s^{C}}{\beta +\alpha c_{t}}. \end{aligned}$$

  (A44)

$\square$

Proof of Theorem 7

Based on Eq. (23), we solve $\frac{\partial U(\Pi _{r/tpt})}{\partial p}=0$ and $\frac{\partial U(\Pi _{r/tpt})}{\partial \tau }=0$ and have the optimal sales price $p^{TPT}$ and collection rate $\tau ^{TPT}$. The corresponding expressions are given by

$$\begin{aligned} p^{TPT}=w^{TPT}+\frac{[2c_{L}-\alpha (b-a)^{2}](d-\alpha w^{TPT}+\beta s^{TPT})}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}} \end{aligned}$$

(A45)

and

$$\begin{aligned} \tau ^{TPT}=\frac{\alpha (b-a)(d-\alpha w^{TPT}+\beta s^{TPT})}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}}. \end{aligned}$$

(A46)

The TPT contract coordinates the CLSC system if and only if the two supply chain members in the manufacturer-led CLSC system make decisions in accordance with the centralized case. Using the coordination conditions $p^{TPT}=p^{C}$ and $\tau ^{TPT}=\tau ^{C}$ and comparing above two equations with Eqs. (18) and (19), we have

$$w^{{TPT}} = \frac{d}{\alpha } - \left[ {\frac{{2c_{L} (2\alpha + \lambda _{r} \sigma ^{2} ) - \alpha ^{2} (b - a)^{2} }}{{\alpha (b - a)}} - \frac{{2\beta c_{L} (\beta + \alpha c_{t} )}}{{\alpha \eta (\delta _{c} - \delta _{e} c_{t} )}}} \right]\tau ^{C} {\text{ }}$$

(A47)

and

$$\begin{aligned} s^{TPT}=\frac{\alpha ^{2}(b-a)p^{C}+2c_{L}(\alpha +\lambda _{r}\sigma ^{2})\tau ^{C}-\alpha (b-a)d}{\alpha \beta (b-a)}. \end{aligned}$$

(A48)

We compare Eq. (20) with Eq. (A48) and have that $s^{TPT}=s^{C}$ holds if and only if $\frac{\alpha +\lambda _{r}\sigma ^{2}}{b-a}=\frac{\alpha +\lambda _{c}\sigma ^{2}}{\delta _{c}-\delta _{e} c_{t}}$ holds. This also indicates that the TPT contract effectively coordinates the manufacturer-led CLSC system. $\square$

Proof of Theorem 8

• (i) From Theorem 7, we have that when the CLSC system is coordinated by the TPT contract, $\frac{\alpha +\lambda _{r}\sigma ^{2}}{b-a}=\frac{\alpha +\lambda _{c}\sigma ^{2}}{\delta _{c}-\delta _{e} c_{t}}$ holds. Recalling an assumption $c_{n}\ge c_{r}+b+\delta _{e} c_{t}$, we have $\delta _{c}-\delta _{e} c_{t}\ge b-a$. From Gan et al. (2004), we have that the coordination of a two-echelon system with two risk-averse members yields $\lambda _{c}<\lambda _{r}$. Therefore, $\frac{\alpha +\lambda _{r}\sigma ^{2}}{b-a}=\frac{\alpha +\lambda _{c}\sigma ^{2}}{\delta _{c}-\delta _{e} c_{t}}$ yields $\lambda _{r}=0$ and $c_{n}=c_{r}+b+\delta_{e} c_{t}.$

• (ii) From the proof of Theorem 7, we have that the coordination conditions $p^{TPT}=p^{C}$, $\tau ^{TPT}=\tau ^{C}$, and $s^{TPT}=s^{C}$ hold when the CLSC system is coordinated by the TPT contract. In this scenario, using Eqs.(23), (24), and (25), we have

  $$\begin{aligned} U^{TPT}(\Pi _{r/tpt})= & {} \frac{c_{L}[2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(b-a)^{2}](\tau ^{C})^{2}}{\alpha ^{2}(b-a)^{2}}-F\nonumber \\= & {} \frac{c_{L}[4c_{L}-\alpha (b-a)^{2}](\tau ^{C})^{2}}{\alpha (b-a)^{2}}-F, \end{aligned}$$

  (A49)

  and

  $$\begin{aligned} U^{TPT}(\Pi _{m/tpt})= & {} \frac{-2c^{2}_{L}(\beta +\alpha c_{t})^{2}(\tau ^{C})^{2}}{\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}\eta }+F. \end{aligned}$$

  (A50)

  The last terms of above two Equations hold because of Theorem 8(i). According to Eqs.(A49), (A50), and (21), we further have $U^{TPT}(\Pi _{m/tpt})+U^{TPT}(\Pi _{r/tpt})=U^{C}(\Pi _{c}).$ Similarly, using the coordination conditions mentioned above and from Eq. (22), we have $E^{TPT}(J_{m})=E^{C}(J_{m}).$

• (iii) From Theorem 8(i), we simplify the expression for $A_{1}$ shown in Theorem 1 as $A_{1}=\alpha [4c_{L}-\alpha (\delta _{c}-\delta _{e} c_{t})^{2}]=\alpha [4c_{L}-\alpha (b-a)^{2}]$In this scenario, comparing Eqs. (9) and (11) with Eqs.(19) and (20) yields

  $$\begin{aligned} \frac{\tau ^{C}}{\tau ^{D-H}}=\frac{\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1}) -4c^{2}_{L}\alpha ^{2}(\beta +\alpha c_{t})^{2}}{(2\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})[\eta A_{1}-2c_{L}(\beta +\alpha c_{t})^{2}]}, \end{aligned}$$

  (A51)

  and

  $$\begin{aligned} \frac{s^{C}}{s^{D-H}}=\frac{\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2} A_{1})-4c^{2}_{L}\alpha ^{2}(\beta +\alpha c_{t})^{2}}{2\eta c_{L}\alpha ^{2}A_{1}-4c^{2}_{L}\alpha ^{2}(\beta +\alpha c_{t})^{2}}. \end{aligned}$$

  (A52)

Using $\eta >\underline{\eta }$ and from Eq. (A51), we have $\tau ^{C}>\tau ^{D-H}$ and $s^{C}>s^{D-H}$. When the TPT contract is implemented by the two risk-averse members, the corresponding utilities are no less than those in the manufacturer-led system, i.e., $U^{TPT}(\Pi _{r/tpt})\ge U^{D-H}(\Pi _{r})$ and $U^{TPT}(\Pi _{m/tpt})\ge U^{D-H}(\Pi _{m})$. Using Eqs. (A51) and (A52), we compare Eq. (12) with Eq. (A49) and have

$$\begin{aligned} F\le & {} \frac{c_{L}A_{1}[(\tau ^{C})^{2}-(\tau ^{D-H})^{2}]}{\alpha ^{2}(b-a)^{2}}\nonumber \\= & {} \frac{\eta A_{1}(3\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-c_{L}(\beta +\alpha c_{t})^{2}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^2 A_{1})}{[\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-4c^{2}_{L}\alpha ^{2}(\beta +\alpha c_{t})^{2}]^2}\cdot \nonumber \\{} & {} \frac{4c^2_{L} A^2_{1}[\alpha ^{2}\eta +(\beta +\alpha c_{t})^{2}\lambda _{m}\sigma ^{2}](\tau ^{C})^{2}}{\alpha ^2(\delta _c-\delta _e c_t)^2} \nonumber \\= & {} \frac{\eta A_{1}(3\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-c_{L}(\beta +\alpha c_{t})^{2}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^2 A_{1})}{[\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-4c^{2}_{L}\alpha ^{2}(\beta +\alpha c_{t})^{2}]^2}\cdot \nonumber \\{} & {} \frac{\eta ^{2} A^{2}_{1}[\alpha ^{2}\eta +(\beta +\alpha c_{t})^{2}\lambda _{m}\sigma ^{2}](s^{C})^{2}}{(\beta +\alpha c_{t})^{2}}. \end{aligned}$$

(A53)

The last term of Eq. (A53) holds because of Eqs. (19) and (20).

Similarly, using Eq. (A52) and comparing Eq. (13) with Eq. (A50) yield

$$\begin{aligned} F\ge \frac{\eta ^{2}A^{2}_{1}[\alpha ^{2}\eta +(\beta +\alpha c_{t})^{2}\lambda _{m}\sigma ^{2}](s^{c})^{2}}{2(\beta +\alpha c_{t})^{2}[\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-4\alpha ^{2}c^{2}_{L}(\beta +\alpha c_{t})^{2}]}. \end{aligned}$$

(A54)

Let $F_{max}=\frac{[\eta A_{1}(3\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-c_{L}(\beta +\alpha c_{t})^{2}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^2 A_{1})][\alpha ^{2}\eta +(\beta +\alpha c_{t})^{2}\lambda _{m}\sigma ^{2}]}{[\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-4c^{2}_{L}\alpha ^{2}(\beta +\alpha c_{t})^{2}]^2}\cdot \frac{\eta ^{2} A^{2}_{1}(s^{C})^{2}}{(\beta +\alpha c_{t})^{2}}.$ and $F_{min}=\frac{\eta ^{2}A^{2}_{1}[\alpha ^{2}\eta +(\beta +\alpha c_{t})^{2}\lambda _{m}\sigma ^{2}](s^{c})^{2}}{2(\beta +\alpha c_{t})^{2}[\eta A_{1}(4\alpha ^{2}c_{L}+\lambda _{m}\sigma ^{2}A_{1})-4\alpha ^{2}c^{2}_{L}(\beta +\alpha c_{t})^{2}]}$. Using $\eta >\frac{2c_{L}(\beta +\alpha c_{t})^{2}}{2c_{L}(2\alpha +\lambda _{r}\sigma ^{2})-\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}}$, we have that $\eta >\frac{2c_{L}(\beta +\alpha c_{t})^{2}}{4c_{L}\alpha -\alpha ^{2}(\delta _{c}-\delta _{e} c_{t})^{2}}=\frac{2c_{L}(\beta +\alpha c_{t})^{2}}{A_{1}}$ holds for $\lambda _{r}=0$. Therefore, comparing $F_{max}$ with $F_{min}$, we have $F_{max}>F_{min}$. $\square$