Waste converting through by-product synergy: an insight from three-echelon supply chain

Abstract

By-product synergy (BPS) is an innovative method to convert waste into valuable by-products effectively. Based on a three-echelon supply chain composed of an upstream manufacturer, a processing plant with limited processing capacity, and a downstream manufacturer, this study derives the production quantity and waste disposal decisions of the upstream and downstream manufacturers as well as the optimal transfer price decision of the processing plant. Moreover, we assess the environmental performance of BPS. Analytical results suggest that the upstream manufacturer’s production quantity and waste disposal decisions and the processing plant’s transfer price decision are threshold dependent on the processing plant’s capacity, whereas the downstream manufacturer’s production quantity decision is threshold dependent on the processing plant’s capacity and price of raw materials. BPS is beneficial for all members of the supply chain to increase profit. The production promotion and cost-saving effects ensure that the supply chain members maximize their profit. However, BPS does not always have a positive effect on the environment; when the processing plant’s capacity and price of raw materials are below the threshold, implementing BPS results in a win-win situation of economic and environmental benefits.

Appendix proof

Proof of Lemma 1

Taking first and second derivatives of Q_A, we get $\frac{\partial {\prod}_U}{\partial {Q}_A}=a-2{Q}_A-{c}_A-{rc}_d$, $\frac{\partial^2{\prod}_U}{\partial {Q_A}^2}=-2$, therefore, ∏_U is concave in Q_A. Solving $\frac{\partial {\prod}_U}{\partial {Q}_A}=0$, we can obtain a maximum value ${Q}_A^N=\frac{a-{c}_A-{rc}_d}{2}$. Thus, the optimal profit is ${\Pi}_U^N=\frac{{\left(a-{c}_A-{rc}_d\right)}^2}{4}$. Similarly, we can obtain downstream manufacturer’s optimal production quantity is ${Q}_B^N=\frac{b-{c}_B-{c}_r}{2}$, optimal profit is ${\Pi}_D^N=\frac{{\left(b-{c}_B-{c}_r\right)}^2}{4}$.

Proof of Proposition 1

Depending on rQ_A > K or rQ_A ≤ K, we propose two sub-optimization problems (Sub-problem 1 and Sub-problem 2). The optimal solution for the original problem is thus the maximum of these two sub-problems.

• Sub-problem 1: When rQ_A > K, upstream manufacturer’s profit is:

$$ {\displaystyle \begin{array}{l}{\Pi}_U\left(\left.{Q}_A\right|{Q}_A,w\right)=\left({P}_A-{c}_A\right){Q}_A- wK-{c}_d\left({rQ}_A-K\right)\\ {}s.t\ {rQ}_A>K\end{array}} $$

The unconstraint solution is ${Q}_A(w)=\frac{a-{c}_A-{rc}_d}{2}$. Substituting this unconstraint solution into the constraint identifies a cutoff value of $\frac{r\left(a-{c}_A-{rc}_d\right)}{2}$. Due to $\frac{\partial^2{\Pi}_U\left(\left.{Q}_A\right|{Q}_A,w\right)}{\partial {Q_A}^2}<0$, Π_U(Q_A|Q_A, w) is concave in Q_A. Thus, when $K<\frac{r\left(a-{c}_A-{rc}_d\right)}{2}$, the unconstraint solution is optimal for sub-problem 1, and upstream manufacturer’s optimal production quantity and profit are ${Q}_{A1}(w)=\frac{a-{c}_A-{rc}_d}{2}$, ${\prod}_{U1}(w)=\frac{{\left(a-{c}_A-{rc}_d\right)}^2}{4}+K\left({c}_d-w\right)$. When $K\ge \frac{r\left(a-{c}_A-{rc}_d\right)}{2}$, upstream manufacturer’s optimal production quantity and profit are ${Q}_{A2}(w)=\frac{K}{r}$, ${\prod}_{U2}(w)=\frac{-{K}^2+ rK\left(a-{c}_A- rw\right)}{r^2}$.

• Sub-problem 2:When rQ_A ≤ K, upstream manufacturer’s profit is:

$$ {\displaystyle \begin{array}{l}{\Pi}_U\left(\left.{Q}_A\right|{Q}_A,w\right)=\left({P}_A-{c}_A\right){Q}_A-{wrQ}_A\\ {}s.t\ {rQ}_A\le K\end{array}} $$

We solve this sub-problem similar to Sub-problem 1. When $K\ge \frac{r\left(a-{c}_A- rw\right)}{2}$, upstream manufacturer’s optimal production quantity and profit are ${Q}_{A3}(w)=\frac{a-{c}_A- rw}{2}{\prod}_{U3}(w)=\frac{{\left(a-{c}_A- rw\right)}^2}{4}$. When $K<\frac{r\left(a-{c}_A- rw\right)}{2}$, upstream manufacturer’s optimal production quantity and profit are ${Q}_{A4}(w)=\frac{K}{r}$, ${\prod}_{U4}(w)=\frac{-{K}^2+ rK\left(a-{c}_A- rw\right)}{r^2}$.

We compare the two sub-problems and derive the optimal solution for the original problem.

• (1) When $K<\frac{r\left(a-{c}_A-{rc}_d\right)}{2}$, the optimal production quantity is ${Q}_{A1}(w)=\frac{a-{c}_A-{rc}_d}{2}$, the optimal profit is ${\prod}_{U1}(w)=\frac{{\left(a-{c}_A-{rc}_d\right)}^2}{4}+K\left({c}_d-w\right)$;

• (2) When $K\in \left[\frac{r\left(a-{c}_A-{rc}_d\right)}{2},\frac{r\left(a-{c}_A- rw\right)}{2}\right]$, we need to compare ∏_U2(w) and ∏_U4(w). Because ∏_U2(w) − ∏_U4(w) = 0, the optimal production quantity is ${Q}_{A2}(w)={Q}_{A4}(w)=\frac{K}{r}$; the optimal profit is ${\prod}_{U2}(w)={\prod}_{U4}(w)=\frac{-{K}^2+ rK\left(a-{c}_A- rw\right)}{r^2}$.

• (3) When $K>\frac{r\left(a-{c}_A- rw\right)}{2}$, the optimal production quantity is ${Q}_{A3}(w)=\frac{a-{c}_A- rw}{2}$, the optimal profit is ${\prod}_{U3}(w)=\frac{{\left(a-{c}_A- rw\right)}^2}{4}$.

Therefore, Proposition 1 is proved

Proof of Proposition 2

According to Proposition 1, we propose three sub-optimization problems (Sub-problem 1, Sub-problem 2 and Sub-problem 3). The optimal solution for the original problem is thus the maximum of these three sub-problems.

• Sub-problem 1: When $K<\frac{r\left(a-{c}_A-{rc}_d\right)}{2}$, The plant’s profit function is ∏_p1 = (c_b + w − c_m)K, which is increasing in w. Due to w < c_d, Then the optimal transfer price and profit are w₁ = c_d, ∏_P1 = (c_b + c_d − c_m)K.

• Sub-problem 2: When $K\in \left[\frac{r\left(a-{c}_A-{rc}_d\right)}{2},\frac{r\left(a-{c}_A- rw\right)}{2}\right]$, The plant’s profit function is ∏_p2 = (c_b + w − c_m)K which is increasing in w. Due to $K<\frac{r\left(a-{c}_A- rw\right)}{2}\Rightarrow w<\frac{r\left(a-{c}_A\right)-2K}{r^2}$. Hence, the optimal transfer price and profit are ${w}_2=\frac{r\left(a-{c}_A\right)-2K}{r^2}$,${\prod}_{P2}=\frac{-2{K}^2+ rK\left[\left(a-{c}_A\right)+r\left({c}_b-{c}_m\right)\right]}{r^2}$.

• Sub-problem 3:When $K>\frac{r\left(a-{c}_A- rw\right)}{2}$, thus,$w>\frac{r\left(a-{c}_A\right)-2K}{r^2}$,The plant’s profit function is:

$$ {\displaystyle \begin{array}{l}{\prod}_P(w)=\left({c}_b+w-{c}_m\right){rQ}_A=\left({c}_b+w-{c}_m\right)\frac{r\left(a-{c}_A- rw\right)}{2}\\ {}s.t\ w>\frac{r\left(a-{c}_A\right)-2K}{r^2}\end{array}} $$

The unconstraint solution is $w=\frac{\left(a-{c}_A\right)-r\left({c}_b-{c}_m\right)}{2r}$. Substituting this unconstraint solution into the constraint identifies a cutoff value of $\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$. Due to $\frac{\partial^2{\Pi}_P(w)}{\partial {w}^2}<0$,Π_P(w) is concave in w. Thus, when $K>\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, the unconstraint solution is optimal for Sub-problem 3, and then the optimal transfer price and profit are ${w}_3=\frac{\left(a-{c}_A\right)-r\left({c}_b-{c}_m\right)}{2r}$, ${\prod}_{P3}=\frac{{\left[a-{c}_A+r\left({c}_b-{c}_m\right)\right]}^2}{8}$. When $K\le \frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, the optimal transfer price and profit are ${w}_4=\frac{r\left(a-{c}_A\right)-2K}{r^2}$, ${\prod}_{P4}=\frac{-2{K}^2+ rK\left[\left(a-{c}_A\right)+r\left({c}_b-{c}_m\right)\right]}{r^2}$.

We compare the three sub-problems and derive the optimal solution for the original problem.

• (1) When $K<\frac{r\left(a-{c}_A-{rc}_d\right)}{2}$, the optimal transfer price is w₁ = c_d, the optimal profit is ∏_P1 = (c_b + c_d − c_m)K;

• (2) when $K\in \left[\frac{r\left(a-{c}_A-{rc}_d\right)}{2},\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right]$, we need to compare ∏_p2(w) and ∏_p4(w). Because ∏_p2(w) − ∏_p4(w) = 0, the optimal production quantity is ${w}_2={w}_4=\frac{r\left(a-{c}_A\right)-2K}{r^2}$, the optimal profit is ${\prod}_{P2}={\prod}_{P4}=\frac{-2{K}^2+ rK\left[\left(a-{c}_A\right)+r\left({c}_b-{c}_m\right)\right]}{r^2}$.

• (3) When $K>\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, the optimal transfer price is ${w}_3=\frac{\left(a-{c}_A\right)-r\left({c}_b-{c}_m\right)}{2r}$, the optimal profit is ${\prod}_{P3}=\frac{{\left[a-{c}_A+r\left({c}_b-{c}_m\right)\right]}^2}{8}$.

Therefore, Proposition 2 is proved.

Proof of Corollary 1

According to the processing capacity, we can obtain three cases.

• Case 1: When K ∈ (0, K₁), ${Q}_A^{\ast }={Q}_{A1}\frac{a-{c}_A-{rc}_d}{2}={Q}_A^N$, ${\prod}_U^{\ast }=\frac{{\left(a-{c}_A-{rc}_d\right)}^2}{4}={\prod}_U^N$;

• Case 2:When K ∈ [K₁, K₂], ${Q}_A^{\ast }={Q}_{A2}>\frac{K_1}{r}=\frac{a-{c}_A-{rc}_d}{2}={Q}_A^N$, ${\prod}_U^{\ast }={\prod}_{U2}>\frac{{\left({K}_1\right)}^2}{r^2}=\frac{{\left(a-{c}_A-{rc}_d\right)}^2}{4}{\prod}_U^{\ast }>{\prod}_U^N$;

• Case 3:When K > K₂, ${Q}_A^{\ast }-{Q}_A^N={Q}_{A3}-{Q}_A^N=\frac{2{c}_d+r\left({c}_b-{c}_m\right)-\left(a-{c}_A\right)}{4}=\frac{c_d-{rw}_3}{2}$. Because of c_d > w₃ and r < 1, hence, ${Q}_A^{\ast }>{Q}_A^N$. Due to ${\prod}_U^{\ast }-{\prod}_U^N={\left({Q}_A^{\ast}\right)}^2-{\left({Q}_A^N\right)}^2>0\Rightarrow {\prod}_U^{\ast }>{\prod}_U^N$. In summary, Corollary 1 is proved.

Proof of Proposition 3

Depending on the supply of by-product, we propose two sub-optimization problems (Sub-problem 1 and Sub-problem 2).

• Sub-problem 1: When Q_b = K, according to the Table 1,the processing capacity constraint is $K\le {K}_2=\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$. Under this condition, downstream manufacturer’s profit is:

$$ {\displaystyle \begin{array}{l}{\prod}_D^L\left({Q}_B\right)=\left({P}_B-{c}_B\right){Q}_B-{c}_bK-{c}_r\left({Q}_B-K\right)\\ {}s.t\ {Q}_B>K\end{array}} $$

The unconstraint solution is ${Q}_B=\frac{b-{c}_B-{c}_r}{2}$. Substituting this unconstraint solution into the constraint identifies a cutoff value of b − c_B − 2K. Due to $\frac{\partial^2{\Pi}_D^L\left({Q}_B\right)}{\partial {Q_B}^2}<0$, ${\Pi}_D^L\left({Q}_B\right)$ is concave in Q_B. Let c_r1 = b − c_B − 2K, thus when c_r < c_r1, the unconstraint solution is optimal for Sub-problem 1, and then downstream manufacturer’s optimal production quantity and profit are ${Q}_{B1}^L=\frac{b-{c}_B-{c}_r}{2}$, ${\prod}_{D1}^L=\frac{{\left(b-{c}_B-{c}_r\right)}^2}{4}+K\left({c}_r-{c}_b\right)$. c_r < b − c_B to ensure downstream manufacturer’s is positive. When c_r1 ≤ c_r < b − c_B, downstream manufacturer’s optimal production quantity and profit are ${Q}_{B2}^L=K$, ${\prod}_{D2}^L=-{K}^2+\left(b-{c}_B-{c}_b\right)K$.

• Sub-problem 2: When Q_b = rQ_A, according to the Table 1,the processing capacity constraint is $K>{K}_2=\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, Under this condition, downstream manufacturer’s profit is:

$$ {\displaystyle \begin{array}{l}{\prod}_D^H\left({Q}_B\right)=\left({P}_B-{c}_B\right){Q}_B-{c}_b{rQ}_A-{c}_r\left({Q}_B-{rQ}_A\right)\\ {}s.t\ {Q}_B>{rQ}_A\end{array}} $$

We solve this sub-problem similar to Sub-problem 1. Let ${c}_{r2}=b-{c}_B-\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{2}$, When c_r < c_r2, downstream manufacturer’s optimal production quantity and profit are ${Q}_{B1}^H=\frac{b-{c}_B-{c}_r}{2},$, ${\prod}_{D1}^H=\frac{{\left(b-{c}_B-{c}_r\right)}^2}{4}+\left({c}_r-{c}_b\right)\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$. When c_r2 ≤ c_r < b − c_B, downstream manufacturer’s optimal production quantity and profit are ${Q}_{B2}^H=\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, ${\prod}_{D2}^H=-{\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right]}^2+\left(b-{c}_B-{c}_b\right)\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$

Therefore, Proposition 3 is proved.

• Proof of Corollary 2

Obviously, in scenario L, ${Q}_{B1}^L>{Q}_{B2}^L$, ${\prod}_{D1}^L>{\prod}_{D2}^L$; in scenario H, ${Q}_{B1}^H>{Q}_{B2}^H$, ${\prod}_{D1}^H>{\prod}_{D2}^H$.

c_r1 = b − c_B − 2K, and $K\le \frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, thus ${c}_{r1}\ge b-{c}_B-\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{2}={c}_{r2}$.

Hence, the production quantity and profit under different conditions are shown as follow in Appendix Fig 7. We propose two sub-optimization problems (Sub-problem 1and Sub-problem 2).

Fig 7

production quantity and profit under different conditions

Sub-problem 1: Production quantity

According to the price of raw material, we can obtain three cases.

• Case 1: When c_r < c_r2, we need to compare ${Q}_{B1}^L$ with ${Q}_{B1}^H$. However, ${Q}_{B1}^L={Q}_{B1}^H=\frac{b-{c}_B-{c}_r}{2}$.

• Case 2: When c_r ∈ [c_r2, c_r1),we need to compare ${Q}_{B1}^L$ with ${Q}_{B2}^H$. Let $f\left[\Delta Q\left({c}_r\right)\right]={Q}_{B1}^L-{Q}_{B2}^H=\frac{b-{c}_B-{c}_r}{2}-\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, and the domain is c_r ∈ [c_r2, c_r1).f[ΔQ(c_r)] is decreasing in the domain, and the maximum value f[ΔQ(c_r2)] = 0, thus ${Q}_{B1}^L<{Q}_{B2}^H$。

• Case 3: When c_r ∈ [c_r1, b − c_B), we need to compare ${Q}_{B2}^L$ with ${Q}_{B2}^H$, ${Q}_{B2}^L=K\le \frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$, thus ${Q}_{B2}^L<{Q}_{B2}^H$.

Combine case 1,2 and 3, when c_r < c_r2, ${Q}_B^L={Q}_B^H$; when c_r ∈ [c_r2, b − c_B), ${Q}_B^L<{Q}_B^H$.

Sub-problem 2: Profit

According to the price of raw material, we can obtain three cases.

• Case 1: when c_r < c_r2, we need to compare ${\prod}_{D1}^L$ with ${\prod}_{D1}^H$.${\prod}_{D1}^H-{\prod}_{D1}^L=\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}-K\right]\left({c}_r-{c}_b\right)>0$, thus ${\prod}_{D3}^H>{\prod}_{D1}^L$.

• Case 2:When c_r ∈ [c_r2, c_r1), we need to compare ${\prod}_{D1}^L$ with ${\prod}_{D2}^H$. Let

$$ f\left[\varDelta {\prod}_D\left({c}_r\right)\right]={\prod}_{D1}^L-{\prod}_{D2}^H=\frac{{\left(b-{c}_B-{c}_r\right)}^2}{4}+K\left({c}_r-{c}_b\right)+{\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right]}^2-\left(b-{c}_B-{c}_b\right)\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4} $$

and the domain is c_r ∈ [c_r2, c_r1), and $K\le \frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$. The first and second derivatives of c_r are $\frac{\partial f\left[\varDelta {\prod}_D\left({c}_r\right)\right]}{\partial {c}_r}=\frac{c_B-b+{c}_r+2k}{2}$,$\frac{\partial^2f\left[\varDelta {\prod}_D\left({c}_r\right)\right]}{\partial {c_r}^2}=\frac{1}{2}$. Hence f[Δ∏_D(c_r)] decreasing in the domain, and the maximum value$f\left[\varDelta {\prod}_D\left({c}_{r2}\right)\right]=K\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}-{c}_b\right]-\left(b-{c}_B-{c}_b\right)\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right]\le -\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\left[b-{c}_B-\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right]<0$.

Hence, ${\prod}_{D1}^L<{\prod}_{D2}^H$

• Case 3:When c_r ∈ [c_r1, b − c_B), we need to compare ${\prod}_{D2}^L$ with ${\prod}_{D2}^H$. Let

$f\left[\varDelta {\prod}_D(K)\right]={\prod}_{D2}^H-{\prod}_{D2}^L={K}^2-\left(b-{c}_B-{c}_b\right)K-{\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right]}^2+\left(b-{c}_B-{c}_b\right)\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$and the domain is $K\le \frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$. The first and second derivatives of K are $\frac{\partial f\left[\varDelta {\prod}_D(K)\right]}{\partial K}=2K-b+{c}_B+{c}_b$, $\frac{\partial^2f\left[\varDelta {\prod}_D\left({c}_r\right)\right]}{\partial {c_r}^2}=2$. In addition, when c_r ∈ [c_r1, b − c_B), ${Q}_{B1}^L>{Q}_{B2}^L\Rightarrow \frac{b-{c}_B-{c}_b}{2}>\frac{b-{c}_B-{c}_r}{2}>\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}$. Therefore, f[Δ∏_D(K)] decreasing in the domain, and the maximum value $f\left[\varDelta {\prod}_D\left(\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{4}\right)\right]=0$, thus ${\prod}_{D2}^H\ge {\prod}_{D2}^L$.

Combine case 1,2 and 3, ${\prod}_D^L<{\prod}_D^H$.

In summary, Corollary 1 is proved.

Proof of Corollary 3

Sub-problem 1: Scenario L

According to the price of raw material, we can obtain two cases.

• Case 1: When c_r < c_r1, ${Q}_B^N={Q}_{B1}^L$, and ${\prod}_D^N-{\prod}_{D1}^L=-K\left({c}_r-{c}_b\right)<0\Rightarrow {\prod}_{D1}^L>{\prod}_D^N$

• Case 2: When c_r ≥ c_r1, let $f\left[\Delta Q\left({c}_r\right)\right]={Q}_B^N-{Q}_{B2}^L=\frac{b-{c}_B-{c}_r}{2}-K$, and the domain isc_r ∈ [c_r1, b − c_B). f[ΔQ(c_r)] is decreasing in the domain, and the maximum value f[ΔQ(c_r1)] = 0, hence ${Q}_B^N<{Q}_{B2}^L$. Let $f\left[\varDelta \prod \left({c}_r\right)\right]={\prod}_D^N-{\prod}_{D2}^L=\frac{{\left(b-{c}_B-{c}_r\right)}^2}{4}+{K}^2-\left(b-{c}_B-{c}_b\right)K$, and the domain is c_r ∈ [c_r1, b − c_B) The first and second derivatives of c_r are $\frac{\partial f\left[\varDelta \prod \left({c}_r\right)\right]}{\partial {c}_r}=\frac{c_B+{c}_r-b}{8}$,$\frac{\partial^2f\left[\varDelta \prod \left({c}_r\right)\right]}{\partial {c_r}^2}=\frac{1}{8}$. Therefore, f[Δ ∏ (c_r)] is decreasing in the domain, and the maximum value $f\left[\varDelta \prod \left({c}_{r1}\right)\right]=2K\left(K-\frac{b-{c}_B-{c}_b}{2}\right)<0$, hence, ${\prod}_D^N<{\prod}_{D2}^L$

Combine case 1,2 and 3, when c_r < c_r1, ${Q}_B^{\ast }={Q}_B^N$, ${\prod}_D^{\ast }>{\prod}_D^N$; when c_r ≥ c_r1, ${Q}_B^{\ast }>{Q}_B^N$, ${\prod}_D^{\ast }>{\prod}_D^N$.

Sub-problem 2:Scenario H

Similar to the Sub- problem 1, we can get the similar conclusions.

Therefore, Corollary 3 is proved.

Proof of Proposition 4

• Sub-problem 1:when K < K₁, the derivatives with respect to c_m:

$\frac{\partial {Q}_{A1}}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{U1}}{\partial {c}_m}=0$, $\frac{\partial {w}_1}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{P1}}{\partial {c}_m}=-K$, $\frac{\partial {Q}_{B1}^L}{\partial {c}_m}=0$, $\frac{\partial {Q}_{B2}^L}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{D1}}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{D2}}{\partial {c}_m}=0$

• Sub-problem 2:when K ∈ [K₁, K₂], the derivatives with respect to c_m:

$\frac{\partial {Q}_{A2}}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{U2}}{\partial {c}_m}=0$, $\frac{\partial {w}_2}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{P2}}{\partial {c}_m}=-K$, $\frac{\partial {Q}_{B1}^L}{\partial {c}_m}=0$, $\frac{\partial {Q}_{B2}^L}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{D1}^L}{\partial {c}_m}=0$, $\frac{\partial {\prod}_{D2}^L}{\partial {c}_m}=0$

• Sub-problem 3:when K > K₁, the derivatives with respect to c_m

$\frac{\partial {Q}_{A3}}{\partial {c}_m}=-\frac{r}{4}$, $\frac{\partial {\prod}_{U3}}{\partial {c}_m}=\frac{r\left({c}_m-{c}_b\right)}{8}<0$, $\frac{\partial {w}_3}{\partial {c}_m}=\frac{1}{2}$, $\frac{\partial {\prod}_{P3}}{\partial {c}_m}=\frac{r\left({c}_m-{c}_b\right)}{4}<0$, $\frac{\partial {Q}_{B1}^H}{\partial {c}_m}=0$, $\frac{\partial {Q}_{B2}^H}{\partial {c}_m}=-\frac{r^2}{4}$,$\frac{\partial {\prod}_{D1}^H}{\partial {c}_m}=\frac{r^2\left({c}_b-{c}_r\right)}{4}<0$, $\frac{\partial {\prod}_{D2}^H}{\partial {c}_m}=-\frac{r^2}{4}\left(b-{c}_B-\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{2}-{c}_b\right)=-\frac{r^2}{4}\left({c}_{r2}-{c}_b\right)<0$

In summary, Proposition 2 is proved.

Proof of Corollary 4

According to the processing capacity, we can obtain three cases

• Case 1: When K ≤ K₁, ${E}_{U1}-{E}_U^N={E}_A\left({Q}_{A1}-{Q}_A^N\right)+{rE}_d\left({Q}_{A1}-{Q}_A^N\right)+\left({E}_b-{E}_d\right)K$, because of ${Q}_{A1}={Q}_A^N$ and E_b < E_d, hence ${E}_{U1}<{E}_U^N$.

• Case 2:When K ∈ [K₁, K₂] , ${E}_{U2}-{E}_U^N=\frac{K}{r}\left({E}_A+{rE}_b\right)-\frac{a-{c}_A-{rc}_d}{2}\left({E}_A+{rE}_d\right)$, because ofE_b < E_d and $K>\frac{r\left(a-{c}_A-{rc}_d\right)}{2}$, the positive or negative of ${E}_{U2}-{E}_U^N$ is indefinite. Let ${E}_{U2}-{E}_U^N=0\Rightarrow {K}_E=\frac{E_A+{rE}_d}{E_A+{rE}_b}{K}_1$ is the threshold, when K ≥ K_E, ${E}_{U2}>{E}_U^N$.

• Case 3: When K ≥ K₂, ${E}_{U3}-{E}_U^N=\left({E}_A+{rE}_b\right)\left[\frac{a-{c}_A+r\left({c}_b-{c}_m\right)}{4}\right]+\left({E}_A+{rE}_d\right)\frac{a-{c}_A-{rc}_d}{2}$. Similar to case 2, the positive or negative of ${E}_{U2}-{E}_U^N$ is indefinite.${E}_{U3}-{E}_U^N<\left({E}_A+{rE}_b\right)K+\left({E}_A+{rE}_d\right)\frac{a-{c}_A-{rc}_d}{2}\Rightarrow K<{K}_E$, ${E}_{U3}<{E}_U^N$. Thus, K_E is the threshold.

In summary, Corollary 4 is proved.

Proof of Corollary 5

Sub-problem 1: Scenario L

According to the price of raw material, we can obtain two cases.

• Case 1: When c_r < c_r1, ${E}_{D1}^L-{E}_D^N={E}_B\left({Q}_{B1}^L-{Q}_B^N\right)+{E}_r\left({Q}_{B1}^L-{Q}_B^N\right)-{E}_rK$. According to the Corollary 2, we have ${Q}_{B1}^L={Q}_B^N$. Therefore, ${E}_{D1}^L<{E}_D^N$.

• Case 2: When c_r ≥ c_r1, ${E}_{D2}^L-{E}_D^N={E}_B\left({Q}_{B2}^L-{Q}_B^N\right)+{E}_r\left({Q}_{B2}^L-{Q}_B^N\right)-{E}_rK$. According to the Corollary 2, we have ${Q}_{B2}^L>{Q}_B^N$. the positive or negative of ${E}_{D2}^L-{E}_D^N$ is indefinite. Let ${E}_{D2}^L-{E}_D^N=0\Rightarrow {c}_{r1E}={c}_{r1}+2K\frac{E_r}{E_B+{E}_r}$ is the threshold. When c_r1 ≥ c_r1E, ${E}_{D2}^L\ge {E}_D^N$.

Combine case 1 and case 2, when c_r1 < c_r1E, ${E}_D^L<{E}_D^N$; when c_r1 ≥ c_r1E, ${E}_D^L\ge {E}_D^N$.

Sub-problem 2: Scenario H

Similar to the Sub-problem 1, we can get the similar conclusions. ${c}_{r2E}={c}_{r2}+\left[\frac{r\left(a-{c}_A\right)+{r}^2\left({c}_b-{c}_m\right)}{2}\right]\frac{E_r}{E_B+{E}_r}$ is the threshold, when c_r1 < c_r1E, ${E}_D^L<{E}_D^N$; when c_r1 ≥ c_r1E, ${E}_D^L\ge {E}_D^N$.

Therefore, Corollary 5 is proved.

Proof of Corollary 6

Sub-problem 1: proof of (1)

• Case 1: According to Corollary 1, ${Q}_A^{\ast}\ge {Q}_A^N$, ${\prod}_U^{\ast}\ge {\prod}_U^N$. According to Corollary 4, when K < K_E, ${E}_U<{E}_U^N$, so the ${SW}_U-{SW}_U^N=\left({\Pi}_U-{\Pi}_U^N\right)+\left({CS}_U-{CS}_U^N\right)-v\left({E}_U-{E}_U^N\right)>0$ when K < K_E.

• Case 2: Similar to the Sub- problem 1, we can get the similar conclusions.

Sub-problem 2: proof of (2)

According to Proposition 1, when K > K₂, ${Q}_{A3}=\frac{\left(a-{c}_A\right)+r\left({c}_b-{c}_m\right)}{4}$, which K is not included. The value of Q_A will not change with K. So the value of Π_U, CS_U and E_U will not change with K either, SW_U stays the same while K increase.

Therefore, Corollary 6 is proved.