Pricing decisions and remanufacturing strategies considering consumer recycling behavior

Abstract

This paper incorporates consumers’ recycling behavior in remanufacturing decisions and strategies. We first empirically demonstrate that both firms’ monetary incentives and consumers’ environmental awareness positively influence consumers’ recycling behavior, then construct theoretical models to incorporate such consumers’ recycling behavior in two common remanufacturing strategies: remanufacturing by the manufacturer itself (self-remanufacturing) and by the authorized remanufacturer (authorization remanufacturing). We find that: first, when consumers are of high environmental awareness, the recycling amount driven by environmental awareness is enough to support the optimal production plan. Thus, there is no necessity for firms to implement monetary incentives. When consumers’ environmental awareness becomes lower, firms make a tradeoff between collection cost and profit improvement by increasing collection and finally decide to implement monetary incentives only when consumers’environmental awareness is low. Second, except for the new products’ price under the self-remanufacturing strategy, firms’ decisions under each strategy, such as the new products’ price, remanufactured products’ price, and the license fee, will change with consumers’ recycling behavior when consumers’ environmental awareness is not very high. As a result, the manufacturer’s profit increases with consumers’ environmental awareness no matter which remanufacturing strategy it adopts. However, the remanufacturer’s profit (under the authorization remanufacturing strategy) may decrease with consumers’ environmental awareness. Third, consumers’ recycling behavior is the determining factor for the manufacturer’s remanufacturing strategy selection. Our results reveal that ignoring consumers’ recycling behavior will lead to tremendous decision and strategy deviation in remanufacturing.

Appendices

Appendix A: Questionnaire about consumers’ recycling behavior

Dear madam/Sir:

Hello!

We are from School of Management and Engineering, Nanjing University and we are conducting a survey on green recycling. This questionnaire is anonymous, and we will keep your answers strictly confidential. Please choose according to your actual situation. Thank you for your support!

PART A: BASIC INFORMATION

• 1. Your gender

• 2. Your age

• 3. Your highest education level

PART B: INVESTIGATION ON GREEN RECYCLING BEHAVIOR

This questionnaire is designed to understand the influencing factors of green recycling behavior of Chinese consumers.

To set up the consciousness of green development, follow the path of circular economy development and reduce the consumption of resources and energy is the inevitable requirement and fundamental way out of the sustainable development of Chinese industry. Recycling of waste electronic products is an important part of remanufacturing, which is of great significance to the coordinated and sustainable development of economy and society.

Please read carefully and tick the corresponding choice truthfully. In this part, 1 $=$ strongly agree, 2 $=$ agree, 3 $=$ not sure, 4 $=$ disagree,5 $=$ strongly disagree. Thank you!

• 1. I once participated in the recycling of waste electronic products.

• 2. I will participate in the recycling of waste electronic products in the future.

• 3. I will encourage my relatives and friends to take part in the recycling of used electronic products.

• 4. Recycling used electronic products can reduce environmental pollution and resource consumption.

• 5. I will contribute to reducing environmental pollution after participating in the recycling of waste electronic products.

• 6. If recycling used electronic products has a greater impact on the environment than before, I will definitely take part in recycling activities.

• 7. Participation in the recycling of waste electronic products should be rewarded to a certain extent.

• 8. I really care about the recycling price of used electronic products.

• 9. If the market can provide a higher recycling price, I am more willing to participate in the recycling of used electronic products.

Appendix B: Proofs

1.1 Proof of Table 4

Under strategy S, the manufacturer’s problem is

$$\begin{aligned}&\max _{p_r, p_n, p_{cr}} \quad \pi _M^S=(p_n-c_n)d_n+(p_r-c_r)d_r -p_{cr} \lambda (k e_r+\beta p_{cr})\\&\begin{array}{rr@{}ll} s.t.&{}d_r \le \lambda (k e_r+\beta p_{cr}), \quad p_{cr} \ge 0. \\ \end{array} \end{aligned}$$

Let $\gamma _1$, $\gamma _2$ denote the Lagrangian multiplier, then we have the Lagrangian function of optimization problem, $L(p_r, p_n, p_{cr}, \gamma _1, \gamma _2)= \pi _M^S- \gamma _1 \left( \lambda (k e_r+\beta p_{cr})-d_r \right) -\gamma _2 p_{cr}$, with the first-order conditions $\frac{\partial L\left( p_n,p_r,p_{\text {cr}}\right) }{\partial p_n}=0$, $\frac{\partial L\left( p_n,p_r,p_{\text {cr}}\right) }{\partial p_r}=0$, $\frac{\partial L\left( p_n,p_r,p_{\text {cr}}\right) }{\partial p_{cr}}=0$, and the constraints $\gamma _2 p_{cr} =0$, $\gamma _1 \left( \lambda (k e_r + \beta p_{cr}-d_r) \right) =0$. According to the values of $\gamma _1$ and $\gamma _2$, we have four cases.

Case S-I: $\gamma _1=0$, $\gamma _2=0$. The optimal solutions are $p_n^*=\frac{c_n+1}{2}$, $p_{r} ^*=\frac{c_r+\alpha }{2}$, and $p_{cr}^*=\frac{-k e_r}{2 \beta }$. Since $p_{cr}>0$ is not satisfied, this case is ommitted.

Case S-II: $\gamma _1=0$, $\gamma _2>0$. The optimal solutions are $p_n^*=\frac{c_n+1}{2}$, $p_{r} ^*=\frac{c_r+\alpha }{2}$, $p_{cr}^*=0$ and $\gamma _2=\lambda k e_r$. To ensure $\lambda (k e_r+\beta p_{cr})-d_r>0$, there must be $k>\frac{\alpha c_n-c_r}{\left( 2 \alpha -2 \alpha ^2\right) e_r \lambda }$.

Case S-III: $\gamma _1>0$, $\gamma _2=0$. The optimal solutions are $p_n^*=\frac{c_n+1}{2}$, $p_r^*= \frac{\alpha \left( -c_n+(\alpha -1) ((c_r+\alpha ) \beta -k e_r) \lambda -1\right) }{2 (\alpha -1) \alpha \beta \lambda -2}$, $p_{cr}^*=\frac{-\alpha \beta c_n+\beta c_r+k (1-2 (\alpha -1) \alpha \lambda \beta ) e_r}{2 \beta ((\alpha -1) \alpha \lambda \beta -1)}$, and $\gamma _1=\frac{c_r-\alpha (c_n+(\alpha -1) e_r k \lambda )}{(\alpha -1) \alpha \beta \lambda -1}$. To ensure $p_{cr}>0$, there must be $k<\frac{\beta (\alpha c_n-c_r)}{e_r (2 (1-\alpha ) \alpha \beta \lambda +1)}$.

Case S-IV: $\gamma _1>0$, $\gamma _2>0$. The optimal results are $p_n^*=\frac{c_n+1}{2}$, $p_r^*=\frac{1}{2} \alpha (c_n+2 (\alpha -1) e_r k \lambda +1)$, $p_{cr}^*=0$, $\gamma _1=\alpha (c_n+2 (\alpha -1) e_r k \lambda )-c_r$, and $\gamma _2=\lambda (\beta (c_r-\alpha c_n)+e_r k (1-2 (\alpha -1) \alpha \beta \lambda ))$. To ensure $\gamma _1>0$, $\gamma _2>0$, there must be $\frac{\beta (\alpha c_n-c_r)}{e_r (2 (1-\alpha ) \alpha \beta \lambda +1)}<k<\frac{\alpha c_n-c_r}{\left( 2 \alpha -2 \alpha ^2\right) e_r \lambda }$.

Let $\widehat{k_1}=\frac{\alpha c_n- c_r}{2 \alpha \lambda (1-\alpha ) e_r}$, $\widehat{k_2}=\frac{\beta (\alpha c_n- c_r)}{e_r (1+2 \alpha (1-\alpha ) \lambda \beta ) }$ and summarize the results in cases S-I—S-IV, we can obtain the results shown in Table 4.

1.2 Proof of Propositions 1–3

According to Table 4, (i) when $k \le \widehat{k_2}$, $d_r^{S*}-\lambda k e_r=\frac{\lambda (\beta (c_r-\alpha c_n)+ e_r k (1-2 (\alpha -1) \alpha \beta \lambda ))}{2 (\alpha -1) \alpha \beta \lambda -2}>0$, $p_{cr}^{S*}>0$; (ii) when $\widehat{k_2}<k<\widehat{k_1}$, $d_{r}^{S*}=\lambda k e_r$ and $p_{cr}^{S*}=0$; (iii) when $k \ge \widehat{k_1}$, $d_{r}^{S*}-\lambda k e_r<0$ and $p_{cr}^{S*}=0$. Thus, Proposition 1 is obtained.

According to Table 4, no matter the valur of k, $p_n^{S*}$ always equals to $\frac{c_n+1}{2}$, which keeps unchanged as k increases. When $k=\widehat{k_2}$, there are $\frac{\alpha \left( -c_n+(\alpha -1) ((c_r+\alpha ) \beta -k e_r) \lambda -1\right) }{2 (\alpha -1) \alpha \beta \lambda -2}= \frac{1}{2} \alpha \left( c_n+2 (\alpha -1) \lambda k e_r+1\right)$ and $\frac{-\alpha \beta c_n+\beta c_r+k (1-2 (\alpha -1) \alpha \lambda \beta ) e_r}{2 \beta ((\alpha -1) \alpha \lambda \beta -1)}=0$. Thus, $p_r^{S*}$, $p_{cr}^{S*}$, $d_n^{S*}$ and $d_r^{S*}$ are continuous in $\widehat{k_2}$. Similarly, we can show all of the decisions and outcomes are continuous in $\widehat{k_1}$. When $k<\widehat{k_2}$, $\frac{\partial p_r^{S*}}{\partial k}=- \frac{(\alpha -1) \alpha e_r \lambda }{2 (\alpha -1) \alpha \beta \lambda -2}<0$, $\frac{\partial p_{cr}^{S*}}{\partial k}=\frac{(1-2 (\alpha -1) \alpha \lambda \beta ) e_r}{2 \beta ((\alpha -1) \alpha \lambda \beta -1)}<0$, $\frac{\partial d_n^{S*}}{\partial k}=\frac{\alpha \lambda e_r}{2 (\alpha -1) \alpha \lambda \beta -2}<0$, $\frac{\partial d_r^{S*}}{\partial k}=\frac{\lambda e_r}{2-2 (\alpha -1) \alpha \beta \lambda }>0$; when $\widehat{k_2}<k<\widehat{k_1}$, $\frac{\partial p_r^{S*}}{\partial k}= \alpha (\alpha -1) e_r \lambda$, $p_{cr}^{S*}=0$, $\frac{\partial d_n^{S*}}{\partial k}=-\alpha e_r<0$, $\frac{\partial d_r^{S*}}{\partial k}=\lambda e_r>0$. Thus, $p_r^{S*}$ decreases with k, $p_{cr}^{S*}$ decreases until to be zero, $p_n^{S*}$ decreases with k, and $d_r^{S*}$ increases with k. Proposition 2 is proved.

Substitute the equilbrium pricing decisions and demands in Table 4 to $\pi _{Mn}=(p_n-c_n)d_n$, $\pi _{Mr}=(p_r-c_r)d_r$, and $C_{r}=p_{cr} \lambda q_r$ and $\pi _{M}=\pi _{Mn}+\pi _{Mr}-C_r$, then we can obtain that when $k<\widehat{k_2}$, $\frac{\partial \pi _{Mn}^S}{\partial k}=\frac{\alpha (c_n-1) e_r \lambda }{4 (1-\alpha ) \alpha \beta \lambda +4}<0$, $\frac{\partial \pi _{Mr}^S}{\partial k}=\frac{e_r \lambda (\alpha ((\alpha -1) \lambda (\alpha \beta (c_n-1)+2 e_r k)+c_n+1)-2 c_r)}{4 ((\alpha -1) \alpha \beta \lambda -1)^2}>0$, $\frac{\partial C_r^S}{\partial k}=\frac{e_r \lambda ((\alpha -1) \alpha \beta \lambda (\alpha \beta c_n-\beta c_r+2 e_r k)-e_r k)}{2 \beta ((\alpha -1) \alpha \beta \lambda -1)^2}<0$, $\frac{\partial \left( \pi _{Mr}^S-C_r^S \right) }{\partial k}=-\frac{e_r \lambda (\beta (\alpha +\alpha c_n-2 c_r)+2 e_r k)}{4 \beta ((\alpha -1) \alpha \beta \lambda -1)}>0$, and $\frac{\partial \pi _{M}^S}{\partial k}=\frac{e_r \lambda (\alpha \beta c_n-\beta c_r+e_r k)}{2 \beta ((1-\alpha ) \alpha \beta \lambda +1)}>0$; When $\widehat{k_2}<k<\widehat{k_1}$, $\frac{\partial \pi _{Mn}^S}{\partial k}=\frac{1}{2} \alpha (c_n-1) e_r \lambda <0$, $C_r^S=0$, $\frac{\partial \left( \pi _{Mr}^S-C_r^S \right) }{\partial k}=\frac{\partial \pi _{Mr}^S}{\partial k}=\frac{1}{2} e_r \lambda (\alpha (c_n+4 (\alpha -1) e_r k \lambda +1)-2 c_r)>0$,, and $\frac{\partial \pi _{M}^S}{\partial k}=e_r \lambda (\alpha (c_n+2 (\alpha -1) e_r k \lambda )-c_r)>0$. Since $\pi _{Mn}^S$, $\pi _{Mr}^S$, $C^S_r$, $\pi _{Mr}^S-C^S_r$, and $\pi _{M}^S$ are continuous functions of k, as shown in Proposition 2. Then, when $k<\widehat{k_1}$, as k increaes, (i) $\pi _{Mn}^S$ decreases and $\pi _{Mr}^S$ increases; (ii) $C^S_r$ decreases until it is zero, and $\pi _{Mr}^S-C^S_r$ increases; (iii) $\pi _{M}^S$ increases. That is, Proposition 3 is proved.

1.3 Proof of Table 5

Under strategy A, given $t^A$, the manufacturer’s problem is $\max _{p_n} \pi _M^A=(p_n-c_n)d_n +t^A d_r$, and the remanufacturer’s problem is

$$\begin{aligned}&\max _{p_r, p_{cr}} \quad \pi _R^A=(p_r-c_r)d_r -t^A d_r- p_{cr} (k e_r+\beta p_{cr})\\&\begin{array}{rr@{}ll} s.t.&{}d_r \le k e_r+\beta p_{cr}, \quad p_{cr} \ge 0. \\ \end{array} \end{aligned}$$

Let $\gamma _3$, $\gamma _4$ denote the Lagrangian multiplier, then we have the Lagrangian function of optimization problem $L(p_r, p_{cr}, \gamma _3, \gamma _4)= \pi _R^A- \gamma _3 (k e_r+\beta p_{cr}-d_r)-\gamma _4 p_{cr}$, with the first-order conditions $\frac{\partial L\left( p_r,p_{\text {cr}}\right) }{\partial p_r}=0$, $\frac{\partial L\left( p_r,p_{\text {cr}}\right) }{\partial p_{cr}}=0$, and the constraints $\gamma _3 (k e_r + \beta p_{cr}-d_r) =0$, $\gamma _4 p_{cr} =0$. The optimization problem of the manufacturer is to satisfy the firsr-order condition $\frac{\partial \pi _M^A}{\partial p_n}=0$. According to the values of $\gamma _3$ and $\gamma _4$, we have four cases.

Case A-I: $\gamma _3=0$, $\gamma _4=0$. Under this case, we have $q_r>d_r$ and $p_{cr}>0$, which conflict with each other. Thus, similar to the case S-I, this case is omitted.

Case A-II: $\gamma _3=0$, $\gamma _4>0$. The optimal solutions are $p_n^{A}=\frac{-2 \alpha +2 c_n+c_r+3 t^A+2}{4-\alpha }$, $p_r^A=\frac{\alpha ^2-\alpha (c_n+t^A+1)-2 (c_r+t^A)}{\alpha -4}$, $p_{cr}^A=0$, and $\gamma _4=k e_r$. To ensure $q_r>d_r$, there must be $t^A>\frac{c_r (-2+\alpha )+\alpha (c_n+1-\alpha -k e_r (4-\alpha )(1-\alpha ))}{2 (1-\alpha )}$.

Case A-III: $\gamma _3>0$, $\gamma _4=0$. The optimal solutions are $p_n^{A}=\frac{t^A (2+3 \alpha \beta (1-\alpha ))+N_1}{4 - 2 \alpha + (4 - \alpha ) (1 - \alpha ) \alpha \beta }$, $p_r^{A}=\frac{t^A \alpha (2+(2-\alpha -\alpha ^2) \beta ) + N_2}{4 - 2 \alpha + (4 - \alpha ) (1 - \alpha ) \alpha \beta }$, $p_{cr}^A=\frac{-2 \beta t^A (1-\alpha )+N_3 }{\beta (4 - 2 \alpha + (4 - \alpha ) (1 - \alpha ) \alpha \beta )}$, and $\gamma _3=\frac{2 \alpha c_n+2 (\alpha -2) c_r-(\alpha -1) \alpha ((\alpha -4) e_r k+2)+4 (\alpha -1) t^A}{(\alpha -4) (\alpha -1) \alpha \beta -2 \alpha +4}$. To ensure $p_{cr}^A>0$ and $\gamma _3>0$, there must be $t^A<\frac{\alpha \beta (\alpha -c_n-c_r-1)+2 \beta c_r+e_r k ((\alpha -4) (\alpha -1) \alpha \beta -\alpha +2)}{2 (\alpha -1) \beta }$. Where $N_1= c_n (2 + 2 (1 - \alpha ) \alpha \beta ) + (1 - \alpha ) (2 - \alpha k e_r + \alpha (2 + c_r - 2 \alpha ) \beta )$, $N_2=\alpha (c_n (2 + (1 - \alpha ) \alpha \beta ) + (1 - \alpha ) (2 - 2 k e_r + 2 c_r \beta + (1 - \alpha ) \alpha \beta ))$, $N_3=\alpha \beta (c_n+c_r-(\alpha -1) ((\alpha -4) k e_r+1))-2 \beta c_r+(\alpha -2) k e_r$.

Case A-IV: $\gamma _3>0$, $\gamma _4>0$. The optimal solutions are $p_n^A=\frac{c_n+(1- \alpha ) (1-\alpha e_r k)+t^A}{2-\alpha }$, $p_r^A=\frac{\alpha (-\alpha +c_n+2 (\alpha -1) e_r k+t^A+1)}{2-\alpha }$, $p_{cr}^A=0$, $\gamma _3=\frac{\alpha (c_n-(\alpha -1) ((\alpha -4) e_ k+1))+(\alpha -2) c_r+2 (\alpha -1) t^A}{2-\alpha }$, and $\gamma _4=\frac{2 \beta (c_r+t^A)}{2-\alpha }+\frac{\alpha \beta (-\alpha +c_n+c_r+2 t^A+1)+e_r k (-(\alpha -4) (\alpha -1) \alpha \beta +\alpha -2)}{\alpha -2}$. To ensure $\gamma _3>0$ and $\gamma _4>0$, there must be $\frac{\alpha \beta (\alpha -c_n-c_r-1)+2 \beta c_r+e_r k ((\alpha -4) (\alpha -1) \alpha \beta -\alpha +2)}{2 (\alpha -1) \beta }<t^A<\frac{\alpha (c_n-(\alpha -1) ((\alpha -4) e_r k+1))+(\alpha -2) c_r}{2 (1-\alpha )}$.

Substituting $p_n^A$ and $p_r^A$ into $d_n=\frac{1-\alpha -p_n+p_r}{1-\alpha }$ and $d_r=\frac{\alpha p_n-p_r}{\alpha (1-\alpha )}$, we can easily derive the values of $d)n^A$ and $d_r^A$ under each case. Then, let $\widehat{t_1}=\frac{\alpha (c_n-(\alpha -1) ((\alpha -4) e_r k+1))+(\alpha -2) c_r}{2 (1-\alpha )}$ and $\widehat{t_2}=\frac{\alpha \beta (\alpha -c_n-c_r-1)+2 \beta c_r+e_r k ((\alpha -4) (\alpha -1) \alpha \beta -\alpha +2)}{2 (\alpha -1) \beta }$, and summarize the results in cases A-I—A-IV, Table 5 is obtained.

1.4 Proof of Lemmas 1–2

According to Table 5, we can obtain that when $t^A \le \widehat{t_2}$, $\frac{\partial p_n^A}{\partial t^A}=\frac{2+3 (1-\alpha ) \alpha \beta }{(\alpha -4) (\alpha -1) \alpha \beta -2 \alpha +4}>0$, $\frac{\partial p_r^A}{\partial t^A}=\frac{\alpha \left( 2-\left( \alpha ^2+\alpha -2\right) \beta \right) }{(\alpha -4) (\alpha -1) \alpha \beta -2 \alpha +4}>0$, $\frac{\partial p_{cr}^A}{\partial t^A}=\frac{2 (\alpha -1)}{(\alpha -4) (\alpha -1) \alpha \beta -2 \alpha +4}<0$, $\frac{\partial d_n^A}{\partial t^A}=\frac{(\alpha -1) \alpha \beta -2}{(\alpha -4) (\alpha -1) \alpha \beta -2 \alpha +4}<0$, and $\frac{\partial d_r^A}{\partial t^A}=\frac{2 (\alpha -1) \beta }{(\alpha -4) (\alpha -1) \alpha \beta -2 \alpha +4}<0$; When $\widehat{t_2}<t^A \le \widehat{t_1}$, $\frac{\partial p_n^A}{\partial t^A}=\frac{1}{2-\alpha }>0$, $\frac{\partial p_r^A}{\partial t^A}=\frac{\alpha }{2-\alpha }>0$, $p_{cr}^A=0$, $\frac{\partial d_n^A}{\partial t^A}=-\frac{1}{2-\alpha }<0$, and $\frac{\partial d_r^A}{\partial t^A}=0$; When $t^A<\widehat{t_1}$, $\frac{\partial p_n^A}{\partial t^A}=\frac{3}{4-\alpha }>0$, $\frac{\partial p_r^A}{\partial t^A}=\frac{2+\alpha }{4-\alpha }>0$, $p_{cr}^A=0$, $\frac{\partial d_n^A}{\partial t^A}=\frac{\alpha -1}{(4-\alpha )(1-\alpha )}<0$, and $\frac{\partial d_r^A}{\partial t^A}=\frac{2(\alpha -1)}{\alpha (4-\alpha )(1-\alpha )}<0$. In all, there are i) $\partial p_n^{A}/ \partial t^{A} >0$, $\partial p_r^{A}/ \partial t^{A} >0$, $\partial p_{cr}^{A}/ \partial t^{A} \le 0$; ii) $\partial d_n^{A}/ \partial t^{A} <0$, $\partial d_r^{A}/ \partial t^{A} \le 0$. Lemma 1 is proved.

Similarly, differentiating $p_n^A$, $p_r^A$, $p_{cr}^A$, $d_n^A$, and $d_r^A$ in Table 5 with respect to k and summarize the results, we can easily obtain Lemma 2.

1.5 Proof of Table 6

Table 5 shows that when $t^A$ is in three kinds of different ranges, different equilibrium decisions and results will appear. We can derive the manufacturer’s profits under different cases. For convenience, we denote the profits under $t \le \widehat{t_2}$, $\widehat{t_2}<t \le \widehat{t_1}$, and $t > \widehat{t_1}$ as $\pi _M^\textrm{I}$, $\pi _M^\textrm{II}$, and $\pi _M^\textrm{III}$ respectively. We can show that all of $\pi _M^\textrm{I}$, $\pi _M^\textrm{II}$, and $\pi _M^\textrm{III}$ are concave functions of $t^A$. $\frac{\partial {\pi _M^\textrm{I}}}{\partial t^A}=0$ when $t^A=t^{A\textrm{I}*}=\frac{8 e_r k - 4 (-1 + c_n + 2 e_r k) \alpha + 4 (-1 + \alpha ) (2 c_r + \alpha (-2 + 2 e_r k (-1 + \alpha ) + (-1 + c_n) \alpha )) \beta }{8 + 2 \beta (8 - 4 \alpha (1 + \alpha ) + (-1 + \alpha )^2 \alpha (8 + \alpha ) \beta )}-\frac{ (1 - \alpha )^2 \alpha (8 c_r + \alpha (-8 + (-1 + c_n) \alpha )) \beta ^2}{8 + 2 \beta (8 - 4 \alpha (1 + \alpha ) + (-1 + \alpha )^2 \alpha (8 + \alpha ) \beta )}$, $\frac{\partial {\pi _M^\textrm{II}}}{\partial t^A}=0$ when $t^A=t^{A\textrm{II}*}=2 e_r k +\frac{\alpha }{2}{ (1 - c_n - 4 e_r k)}$, and $\frac{\partial {\pi _M^\textrm{III}}}{\partial t^A}=0$ when $t^{A}=t^{A\textrm{III}*}=\frac{-8 c_r + \alpha (8 + \alpha - c_n \alpha )}{2 (8 + \alpha )}$.

Comparing the value of $t^{A\textrm{I}*}$, $t^{A\textrm{II}*}$, $\widehat{t_1}$, $\widehat{t_2}$, and $t^{A\textrm{III}*}$, we can obtain the following results:

• (i) When $k>\widehat{k_3}$, there are $t^{A\textrm{II}*}>\widehat{t_1}$, $t^{A\textrm{III}*}>\widehat{t_1}$, and $t^{A\textrm{I}*}>\widehat{t_2}$. Thus, $\pi _M^\textrm{I}$ increases in $(0, \widehat{t_2})$, $\pi _M^\textrm{II}$ increases in $(\widehat{t_2},\widehat{t_1})$, $\pi _M^\textrm{III}$ first increases and then decreases in $t>\widehat{t_1}$. Hence, the optimal license fee is $t^{A\textrm{III}*}$.

• (ii) When $\widehat{k_4}<k<\widehat{k_3}$, there $t^{A\textrm{II}*}>\widehat{t_1}$ and $t^{A\textrm{III}*}<\widehat{t_1}$. Thus, $\pi _M^\textrm{II}$ increases in $(\widehat{t_2},\widehat{t_1})$, $\pi _M^\textrm{III}$ decreases in $t>\widehat{t_1}$. To obtain the trend in $t<\widehat{t_2}$, we need to compare the values of $\widehat{t_2}$ and $t^{A\textrm{I}*}$. The result is that when $k<\widehat{k_9}$ ($\widehat{k_9}=\frac{\beta \left( \left( \alpha ^2+\alpha -2\right) \beta -2\right) (c_r-\alpha c_n)}{e_r \left( \beta \left( \alpha (\alpha +8) (\alpha -1)^2 \beta -3 \alpha (\alpha +1)+6\right) +2\right) }$), $t^{A\textrm{I}*}<\widehat{t_2}$. Thus, if $\widehat{k_4}>\widehat{k_8}$, i.e., $\beta <\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}$, $t^{A\textrm{I}*}>\widehat{t_2}$ holds all the times under this condition. Thus, $\pi _M^\textrm{I}$ decreases in $t<\widehat{t_2}$ when $\beta <\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}$. Then, the optimal license fee is $\widehat{t_1}$. Otherwise if $\beta >\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}$, there are $t^{A\textrm{I}*}<\widehat{t_2}$ when $k<\widehat{k_9}$, and $t^{A\textrm{I}*}>\widehat{t_2}$ when $\widehat{k_8}<k<\widehat{k_3}$. Thus, when $\widehat{k_9}<k<\widehat{k_3}$, $\pi _M^\textrm{I}$ decreases in $t<\widehat{t_2}$. Thus, the optimal license fee is $\widehat{t_1}$; when $\widehat{k_4}<k<\widehat{k_9}$, $\pi _M^\textrm{I}$ first increases and then decreases in $t<\widehat{t_2}$. Thus, we need to compare the profits in $\widehat{t_1}$ and $t^{A\textrm{I}*}$. Then, we can find that (a) when $\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}<\beta <\widehat{\beta }_1$, the profit in $\widehat{t_1}$ is always larger than that in $t^{A\textrm{I}*}$ if $\widehat{k_4}<k<\widehat{k_9}$; (b) when $\beta >\frac{3}{2 \alpha ^2-4 \alpha +2}$, the profit in $\widehat{t_1}$ is larger (smaller) than that in $t^{A\textrm{I}*}$ if $\widehat{k_6}<k<\widehat{k_9}$ ($\widehat{k_4}<k<\widehat{k_6}$). Thus, the optimal license fee is $\widehat{t_1}$ and $t^{A\textrm{I}*}$ when $\widehat{k_6}<k<\widehat{k_9}$ and $\widehat{k_4}<k<\widehat{k_6}$, respectively.

• (iii) When $\frac{\beta (c_r-\alpha c_n)}{e_r \left( \left( \alpha ^2+\alpha -2\right) \beta -1\right) }<k<\widehat{k_4}$, there are $\widehat{t_2}<t^{A\textrm{II}*}<\widehat{t_1}$ and $t^{A\textrm{III}*}<\widehat{t_1}$. Hence, $\pi _M^\textrm{II}$ first increases and then decreases in $(\widehat{t_2},\widehat{t_1})$, $\pi _M^\textrm{III}$ decreases in $t>\widehat{t_1}$. According to (ii), when $k<\widehat{k_9}$, $t^{A\textrm{I}*}<\widehat{t_2}$. Since $\widehat{k_9}>\frac{\beta (c_r-\alpha c_n)}{e_r \left( \left( \alpha ^2+\alpha -2\right) \beta -1\right) }$, then if $\beta <\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}$, there is $\widehat{k_4}>\widehat{k_9}$. Thus, $\pi _M^\textrm{I}$ first increases and then decreases in $t<\widehat{t_2}$ when $\frac{\beta (c_r-\alpha c_n)}{e_r \left( \left( \alpha ^2+\alpha -2\right) \beta -1\right) }<k<\widehat{k_9}$, and always increases when $\widehat{k_9}<k<\widehat{k_4}$. Hence, when $\widehat{k_9}<k<\widehat{k_4}$, the optimal license fee is $t^{A\textrm{II}*}$. When $\frac{\beta (c_r-\alpha c_n)}{e_r \left( \left( \alpha ^2+\alpha -2\right) \beta -1\right) }<k<\widehat{k_9}$, we need to compare the profits in $t^{A\textrm{I}*}$ and $t^{A\textrm{II}*}$. Then, we can obtain that when $k>\widehat{k_5}$ ($k<\widehat{k_5}$), the profit in $t^{A\textrm{II}*}$ is larger (smaller) than $t^{A\textrm{I}*}$. If $\beta >\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}$, there is $\widehat{k_9}>\widehat{k_4}$. Thus, we also need to compare the profits in $t^{A\textrm{I}*}$ and $t^{A\textrm{II}*}$. When $\beta >\widehat{\beta }_1$, there is $\widehat{k_5}>\widehat{k_4}$. Then the profit in $t^{A\textrm{II}*}$ is always smaller than $t^{A\textrm{I}*}$ because $k<\widehat{k_5}$. Thus, the optimal license fee is $t^{A\textrm{I}*}$. Otherwise when $\frac{1}{8} \sqrt{\frac{\alpha ^2-28 \alpha +36}{(\alpha -1)^4}}+\frac{\alpha +2}{8 (\alpha -1)^2}<\beta <\widehat{\beta }_1$, there is $\widehat{k_5}<\widehat{k_4}$. Then, the profit in $t^{A\textrm{II}*}$ is larger (smaller) than $t^{A\textrm{I}*}$ when $k>\widehat{k_5}$ ($k<\widehat{k_5}$). Thus, the optimal license fee is $t^{A\textrm{II}*}$ and $t^{A\textrm{I}*}$ when $k>\widehat{k_5}$ and $k<\widehat{k_5}$, respectively.

• (iv) When $k<\frac{\beta (c_r-\alpha c_n)}{e_r \left( \left( \alpha ^2+\alpha -2\right) \beta -1\right) }$, there are $t^{A\textrm{II}*}<\widehat{t_2}$, $t^{A\textrm{III}*}<\widehat{t_1}$, and $t^{A\textrm{I}*}<\widehat{t_2}$. Thus, $\pi _M^\textrm{I}$ first increases and then decreases in $(0, \widehat{t_2})$, $\pi _M^\textrm{II}$ decreases in $(\widehat{t_2},\widehat{t_1})$, $\pi _M^\textrm{III}$ decreases in $t>\widehat{t_1}$. Hence, the optimal license fee is $t^{A\textrm{I}*}$.

Summarizing the above results, we can obtain the optimal license fee decision under each condition. Then, according to Table 5, we can easily show the optimal pricing decisions under each condition. Thus, Table 6 is proved.

1.6 Proof of Propositions 4–7 and Corollary 1

According to Table 6, we can obtain the values of $d_n^{A*}$ and $d_r^{A*}$ under different cases based on the equations of $d_n=\frac{1-\alpha -p_n+p_r}{1-\alpha }$ and $d_r=\frac{\alpha p_n-p_r}{\alpha (1-\alpha )}$. Then, (i) when $\beta \le \widehat{\beta _1}$ and $k \le \widehat{k_5}$ or when $\beta > \widehat{\beta _1}$ and $k \le \widehat{k_6}$, there are $p_{cr}^{A*}>0$ and $d^{A*}_r-k e_r=\beta p_{cr}^{A*}>0$; (ii) when $\beta \le \widehat{\beta _1}$ and $\widehat{k_5}<k \le \widehat{k_3}$ or when $\beta > \widehat{\beta _1}$ and $\widehat{k_6}<k \le \widehat{k_3}$, there are $p_{cr}^{A*}=0$ and $d_{r}^{A*}-k e_r=0$; (iii) when $k > \widehat{k_3}$, there are $p_{cr}^{A*}=0$ and $d_{r}^{A*}-k e_r=\frac{(\alpha +2) c_r-\alpha ((\alpha +2) c_n+(\alpha -1) (\alpha +8) e_r k)}{(\alpha -1) \alpha (\alpha +8)}<0$. Then, Proposition 4 is proved.

When $\beta \le \widehat{\beta _1}$ and $k=\widehat{k_5}$ or when $\beta > \widehat{\beta _1}$ and $k=\widehat{k_6}$, $t^{A*}_\textrm{I} \ne t^{A*}_\textrm{II}$. Thus, $t^{A*}$ is not continuous at this point. Corresponding, $p_n^{A*}$, $p_r^{A*}$, $p_{cr}^{A*}$, $d_n^{A*}$, $d_r^{A*}$, and $\pi _R^{A*}$ are not continuous at this point. For the manufacturer’s profit $\pi _M^{A*}$, it is a continuous function because the two options at this point ($t^{A*}_\textrm{I}$ and $t^{A*}_\textrm{II}$) have no difference in terms of its profit. Differentiating the pricing decisions and all the outcomes with respect to k, we can have the results as shown in Table 7, from which Propositions 5–7 and Corollary 1 can be easily derived.

Table 7 Partial derivative with respect to k under strategy A

1.7 Proof of Propositions 8–9 and Corollary 2

Tables 4 and 6 have shown that under strategies S and A, there are three and four cases of equilibria, respectively. We can easily show that $\widehat{k_7}<\widehat{k_2}$ always holds. Thus, we can figure out eight types of comparison combinations of the equilibrium decisions under strategies S and A, as shown in Fig. 11. Then, according the results in Tables 4 and 6, we can easily show that in each region, $p_n^{S*}-p_n^{A*}<0$ and $p_{cr}^{S*}-p_{cr}^{A*} \ge 0$ always exist. Thus, Porposition 8(i) can be obtained. Table 8 lists the comparison results of $p_r^{S*}$ and $p_r^{A*}$. Summarizing the results in Table 8, we can obtain Proposition 8(ii).

Similar to Proposition 8 (ii), we show the conditions where the profit under strategy A is larger than that under strategy S in each region. The results are in Table 9. The value of the threshold points are presentes in Table 10.

To obtain the comprehensive conditions for $\pi _M^{A*}-\pi ^{S*}_M>0$, we next compare the thresholds in the adjacent regions. Then, we can find that $\widehat{k}_{10}=\widehat{k}_{14}=\widehat{k_2}$ in $\beta =\frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}$, $\widehat{k}_{11}=\widehat{k}_{12}$ in $\beta =\widehat{\beta }_7$, $\widehat{k}_{12}=\widehat{k}_{13}$ in $\beta =\widehat{\beta }_{10}$, and $\widehat{k}_{13}=\widehat{k}_{17}=\widehat{k_2}$ in $\beta =\frac{1}{2} \sqrt{\frac{\alpha +8}{(\alpha +4) \left( \alpha ^2-\alpha \right) ^2 \lambda ^2}}+\frac{1}{2 \left( \alpha ^2-\alpha \right) \lambda }$.

The value of $\lambda$ also impacts the existence of the conditions. Because $\widehat{\lambda }_1<\widehat{\lambda }_3<\widehat{\lambda }_4<\widehat{\lambda }_2$, we can consider the following scenarios:

(i) $\lambda <\widehat{\lambda }_1$. In Region I, $\pi _M^{A*}-\pi _M^{S*}>0$ if $\widehat{\beta }_2<\beta <\widehat{\beta }_3$. In Region VI, $\pi _M^{A*}-\pi _M^{S*}>0$ always holds because $\widehat{k}_{15}<\widehat{k_4}$ and $\widehat{k}_{16}>\widehat{k_3}$. In other regions, the lower bounds of the conditions in Regions V, II and III are connected end to end, forming a comprehensive lower bound for the condition in full region (i.e., $\underline{K_2}(\lambda , \beta )$). The upper bounds for the conditions in Region IV and VII are connected end to end, forming a comprehensive upper bound for the condition in full region (i.e., $\overline{K_2}(\lambda , \beta )$). According Table 9, we can know the upper bound under this condition is

$$\begin{aligned} \overline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{17}, &{} \beta<\frac{1}{2} \sqrt{\frac{\alpha +8}{(\alpha +4) \left( \alpha ^2-\alpha \right) ^2 \lambda ^2}}+\frac{1}{2 \left( \alpha ^2-\alpha \right) \lambda }; \\ \widehat{k}_{13}, &{}\frac{1}{2} \sqrt{\frac{\alpha +8}{(\alpha +4) \left( \alpha ^2-\alpha \right) ^2 \lambda ^2}}+\frac{1}{2 \left( \alpha ^2-\alpha \right) \lambda }<\beta<\widehat{\beta }_{10}; \\ \widehat{k}_{12}, &{} \widehat{\beta }_{10}<\beta <\widehat{\beta }_7. \\ \end{array}\right. } \end{aligned}$$

Fig. 11

Segements

Table 8 Conditions of $p_r^{S*}-p_r^{A*}>0$ in each region

Table 9 Conditions of $\pi _M^{A*}-\pi _M^{S*}>0$ in each region

Table 10 Values of the thresholds

However, the lower bound need to be further subdivided based on the relative value of $\widehat{\beta _1}$, $\widehat{\beta _2}$ and $\widehat{\beta _3}$. (i-a) If $\widehat{\beta _1}<\widehat{\beta _2}$, there are $\widehat{\beta }_4<\frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}$, $\widehat{\beta }_6<\frac{1}{2 \alpha ^2 \lambda -\alpha ^2-2 \alpha \lambda -\alpha +2}$ and $\widehat{\beta }_5<\widehat{\beta }_1$. Thus,

$$\begin{aligned} \underline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{14}, &{} \beta<\frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}; \\ \widehat{k}_{10}, &{} \frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}<\beta<\widehat{\beta }_5; \\ \widehat{k}_{11}, &{}\widehat{\beta }_5<\beta<\widehat{\beta }_2; \\ \widehat{k}_{6}, &{} \widehat{\beta }_2<\beta<\widehat{\beta }_3; \\ \widehat{k}_{11}, &{} \widehat{\beta }_3<\beta <\widehat{\beta }_7. \end{array}\right. } \end{aligned}$$

(i-b) If $\widehat{\beta _2}<\widehat{\beta _1}<\widehat{\beta _3}$, there are $\widehat{k}_{10}<\widehat{k}_{5}$ when $\widehat{\beta _2}<\beta <\widehat{\beta _1}$. Thus,

$$\begin{aligned} \underline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{14}, &{} \beta<\frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}; \\ \widehat{k}_{10}, &{} \frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}<\beta<\widehat{\beta }_2; \\ \widehat{k}_{5}, &{}\widehat{\beta }_2<\beta<\widehat{\beta }_1; \\ \widehat{k}_{6}, &{} \widehat{\beta }_1<\beta<\widehat{\beta }_3; \\ \widehat{k}_{11}, &{} \widehat{\beta }_3<\beta <\widehat{\beta }_7. \end{array}\right. } \end{aligned}$$

(i-c) If $\widehat{\beta _1}>\widehat{\beta _3}$, there are $\widehat{\beta _5}<\widehat{\beta _1}$. Thus,

$$\begin{aligned} \underline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{14}, &{} \beta<\frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}; \\ \widehat{k}_{10}, &{} \frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}<\beta<\widehat{\beta }_2; \\ \widehat{k}_{5}, &{}\widehat{\beta }_2<\beta<\widehat{\beta }_3; \\ \widehat{k}_{10}, &{} \widehat{\beta }_3<\beta<\widehat{\beta }_5; \\ \widehat{k}_{11}, &{} \widehat{\beta }_5<\beta <\widehat{\beta }_7. \end{array}\right. } \end{aligned}$$

(ii) $\widehat{\lambda }_1<\lambda <\widehat{\lambda }_3$. Under this scenario, $\pi _M^{A*}$ is always lower than $\pi _M^{S*}$ when $k<\widehat{k_7}$. The upper bound keeps unchanged. Since $\widehat{k}_{10}>\widehat{k_5}$ and $\widehat{k}_{11}>\widehat{k_6}$ always hold under this condition, the lower bound becomes to be

$$\begin{aligned} \underline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{14}, &{} \beta<\frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}; \\ \widehat{k}_{10}, &{} \frac{\lambda }{\alpha ^2 \lambda ^2-\alpha \lambda ^2-\alpha +1}<\beta<\widehat{\beta }_5; \\ \widehat{k}_{11}, &{} \widehat{\beta }_5<\beta <\widehat{\beta }_7. \end{array}\right. } \end{aligned}$$

(iii) $\widehat{\lambda }_3<\lambda <\widehat{\lambda }_4$. Under this scenario, $\pi _M^{A*}$ is always lower than $\pi _M^{S*}$ in Regions I, II and V. The upper bound still keeps unchanged because the upper bound of the conditions in Region III, IV, VII do not change. But the lower bound changes due to $\pi _M^{A*}>\pi _M^{S*}$ no longer conditional holds in Regions II and V. Then, because $\widehat{k}_{11}>\widehat{k_4}$ and $\widehat{k}_{11}>\widehat{k_6}$ always hold, $\widehat{k}_{15}<widehat{k_4}$ and $\widehat{k}_{15}=\widehat{k}_{11}=\widehat{k}_2$ when $\beta =\frac{\alpha -2 \lambda +2}{(\alpha -1) \alpha \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) }-2 \sqrt{\frac{\alpha \lambda -\lambda ^2+2 \lambda -1}{(\alpha -1) \alpha ^2 \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) ^2}}$, the formulation of the lower bound is

$$\begin{aligned} \underline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{15}, &{} \beta<\frac{\alpha -2 \lambda +2}{(\alpha -1) \alpha \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) }-2 \sqrt{\frac{\alpha \lambda -\lambda ^2+2 \lambda -1}{(\alpha -1) \alpha ^2 \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) ^2}} \\ \widehat{k}_{11}, &{}\frac{\alpha -2 \lambda +2}{(\alpha -1) \alpha \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) }-2 \sqrt{\frac{\alpha \lambda -\lambda ^2+2 \lambda -1}{(\alpha -1) \alpha ^2 \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) ^2}}<\beta <\widehat{\beta }_7. \end{array}\right. } \end{aligned}$$

(iv) $\widehat{\lambda }_4<\lambda <\widehat{\lambda }_2$. Under this scenario, the lower bound is same with that in scenario (iii). But the form of the upper bound $\overline{K}_2(\lambda , \beta )$ has changed because $\pi _M^{A*}$ can no longer be larger than $\pi _M^{S*}$ in Region VII. Since $\widehat{k}_{16}=\widehat{k_12}=\widehat{k_2}$ when $\beta =2 \sqrt{\frac{\alpha \lambda -\lambda ^2+2 \lambda -1}{(\alpha -1) \alpha ^2 \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) ^2}}+\frac{\alpha -2 \lambda +2}{(\alpha -1) \alpha \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) }$, and $\widehat{k}_{16}<\widehat{k_3}$, the upper bound in this scenario is

$$\begin{aligned} \overline{K}_2(\lambda , \beta )= {\left\{ \begin{array}{ll} \widehat{k}_{16}, &{} \beta<2 \sqrt{\frac{\alpha \lambda -\lambda ^2+2 \lambda -1}{(\alpha -1) \alpha ^2 \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) ^2}}+\frac{\alpha -2 \lambda +2}{(\alpha -1) \alpha \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) }; \\ \widehat{k}_{12}, &{}2 \sqrt{\frac{\alpha \lambda -\lambda ^2+2 \lambda -1}{(\alpha -1) \alpha ^2 \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) ^2}}+\frac{\alpha -2 \lambda +2}{(\alpha -1) \alpha \left( 4 \alpha \lambda -\alpha -4 \lambda ^2+8 \lambda -8\right) }<\beta <\widehat{\beta }_7. \\ \end{array}\right. } \end{aligned}$$

(v) $\lambda <\widehat{\lambda }_2$. All the conditions can not hold. Thus, $\pi _M^{A*}$ is always lower than $\pi _M^{S*}$.

Summarizing the scenarios (i)–(v) (including (i-a), (i-b), (i-c)), we can obtain Proposition 9.

Then, according to Proposition 9, if we let $k=0$, there is $\pi _M^{A}>\pi _M^{S}$ only when $\lambda < \widehat{\lambda }_1$ and $\widehat{\beta }_2<\beta <\widehat{\beta }_3$. Thus, Corollary 2 is proved.