Optimal carbon capture and storage contracts using historical CO$_2$ emissions levels

Abstract

In an effort to reduce carbon dioxide (CO$_2$) emissions to the atmosphere, carbon capture and storage (CCS) technology has been developed to collect CO$_2$ from emissions generators and store it underground. Recent proposed legislation would limit the volume of emissions generated from power sources, effectively requiring some sources to participate in CCS. Both emissions sources and storage operators require incentives to enter into contracts to capture excess emissions at the source, and transport and store the CO$_2$ underground. As the level of emissions from power plants is stochastic and carryover into future time periods is expensive, we develop a newsvendor model to determine the optimal price and volume of these contracts to maximize the expected profit of the storage operator and encourage the participation of multiple emissions sources. Because the storage operator has a limit on the amount of CO$_2$ that can be injected each month, this limit affects the allocation of the optimal contract amounts between the emitters. The distribution of emissions and relative costs of transportation also influence the optimal policy. In addition to analytical solutions, we present data-driven methods for using correlated emissions data to determine the optimal price and volume of these contracts.

Appendices

Appendix A: Proof of Proposition 1

Proof of Proposition 1

To solve the constrained optimization problem $\mathbf {(P2)}$, we apply the Lagrangian method. Let $\mathcal {L}(q_1, q_2; \lambda )=E\Pi ^D(q_1, q_2 | p) - \lambda \cdot (q_1+q_2 - Q).$ The Karush–Kuhn–Tucker conditions for the optimal solution are:

$$\begin{aligned}&\frac{\partial \mathcal {L}(q_1, q_2; \lambda )}{\partial q_1} = \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_1}-\lambda =0 \end{aligned}$$

(11)

$$\begin{aligned}&\frac{\partial \mathcal {L}(q_1, q_2; \lambda )}{\partial q_2} = \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_2}-\lambda =0 \end{aligned}$$

(12)

$$\begin{aligned}&\lambda \cdot (q_1+q_2-Q)=0 \end{aligned}$$

(13)

$$\begin{aligned}&q_1+q_2-Q \le 0\end{aligned}$$

(14)

$$\begin{aligned}&q_1, q_2, \lambda \ge 0 \end{aligned}$$

(15)

To compute $\frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_1}$ and $\frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_2}$, we first rewrite the profit function according to Fig. 1 as

$$\begin{aligned}&E\Pi ^D(q_1,q_2 | p) =-K-\alpha _1 \cdot q_1-\alpha _2 \cdot q_2 + \int _{E_1=0}^{q_1}\int _{E_2=0}^{q_2} (p-c) \cdot (E_1+E_2) dF_2(E_2) dF_1(E_1) \nonumber \\&\quad + \int _{E_2=0}^{q_2} \int _{E_1=q_1}^{Q-E_2} (p-c) \cdot (E_1+E_2)-\beta _1(E_1-q_1) dF_1(E_1) dF_2(E_2) \nonumber \\&\quad + \int _{E_2=0}^{q_2} \int _{E_1=Q-E_2}^{\infty } (p-c) \cdot Q-\beta _1(Q-E_2-q_1)dF_1(E_1)dF_2(E_2) \nonumber \\&\quad + \int _{E_1=0}^{q_1} \int _{E_2=q_2}^{Q-E_1} (p-c) \cdot (E_1+E_2)-\beta _2(E_2-q_2) dF_2(E_2) dF_1(E_1) \nonumber \\&\quad + \int _{E_1=0}^{q_1}\int _{E_2=Q-E_1}^{\infty } (p-c) \cdot Q-\beta _2(Q-E_1-q_2) dF_2(E_2) dF_1(E_1) \nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\int _{E_1=q_1}^{Q-E_2} (p-c) \cdot (E_1+E_2)-\beta _1(E_1-q_1)- \beta _2(E_2-q_2) dF_1(E_1) dF_2(E_2)\nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\int _{E_1=Q-E_2}^{\infty } (p-c) \cdot Q-\beta _1(Q-E_2-q_1) - \beta _2(E_2-q_2) dF_1(E_1) dF_2(E_2) \nonumber \\&\quad + \int _{E_1=q_1}^{\infty }\int _{E_2=Q-q_1}^{\infty } (p-c) \cdot Q-\beta _2(Q-q_1-q_2) dF_2(E_2) dF_1(E_1). \end{aligned}$$

(16)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_1}&= -\alpha _1+f_1(q_1) \int _{E_2=0}^{q_2} (p-c) \cdot (q_1+E_2) dF_2(E_2)\nonumber \\&\quad + \int _{E_1=q_1}^{Q-E_2}\left( \int _{E_2=0}^{q_2}\beta _1dF_2(E_2)\right) dF_1(E_1)\nonumber \\&\quad - f_1(q_1) \int _{E_2=0}^{q_2}(p-c) \cdot (q_1+E_2)dF_2(E_2)\nonumber \\&\quad +\int _{E_1=Q-E_2}^{\infty }\left( \int _{E_2=0}^{q_2} \beta _1 dF_2(E_2) \right) dF_1(E_1)\nonumber \\&\quad + f_1(q_1) \int _{E_2=q_2}^{Q-q_1} (p-c) \cdot (q_1+E_2)-\beta _2(E_2-q_2) dF_2(E_2) \nonumber \\&\quad + f_1(q_1) \int _{E_2=Q-q_1}^{\infty }(p-c) Q-\beta _2(Q-q_1-q_2) dF_2(E_2) \nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\left( \int _{E_1=q_1}^{Q-E_2}\beta _1 dF_1(E_1) \!-\! f_1(q_1) \cdot \left[ (p\!-\!c) \cdot (q_1\!+\!E_2)\!-\!\beta _2(E_2\!-\!q_2)\right] \right) dF_2(E_2)\nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\left( \int _{E_1=Q-E_2}^{\infty }\beta _1 dF_1(E_1)\right) dF_2(E_2) \nonumber \\&\quad + f_2(Q-q_1) \int _{E_1=q_1}^{\infty } (p-c) Q-\beta _2(Q-q_1-q_2) dF_{1}(E_{1})\nonumber \\&\quad + \int _{E_1=q_1}^{\infty }\!\!\left( \int _{E_2=Q-q_1}^{\infty } \beta _2 dF_2(E_2) \!-\!f_2(Q\!-\!q_1) \cdot \left[ (p\!-\!c)\cdot Q\!-\!\beta _2(Q\!-\!q_1\!-\!q_2)\right] \right) dF_1(E_1)\nonumber \\&\quad - f_1(q_1) \int _{E_2=Q-q_1}^{\infty } (p-c) Q-\beta _2(Q-q_1-q_2) dF_2(E_2) \end{aligned}$$

(17)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_1}&= -\alpha _1+\int _{E_1=q_1}^{\infty }\int _{E_2=0}^{q_2}\beta _1dF_2(E_2) dF_1(E_1)\nonumber \\&\quad +\int _{E_2=q_2}^{Q-q_1}\int _{E_1=q_1}^{\infty }\beta _1 dF_1(E_1) dF_2(E_2)\nonumber \\&\quad + \int _{E_1=q_1}^{\infty }\int _{E_2=Q-q_1}^{\infty } \beta _2 dF_2(E_2) dF_1(E_1) \end{aligned}$$

(18)

$$\begin{aligned}&\frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_1}\nonumber \\&\quad = -\alpha _1+\int _{E_1=q_1}^{\infty }\left( \int _{E_2=0}^{Q-q_1}\beta _1dF_2(E_2) + \int _{E_2=Q-q_1}^{\infty } \beta _2 dF_2(E_2)\right) dF_1(E_1)\nonumber \\ \end{aligned}$$

(19)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_1} = -\alpha _1+\overline{F_1}(q_1)\left( \beta _1 F_2(Q-q_1)+ \beta _2 \overline{F_2}(Q-q_1)\right) \end{aligned}$$

(20)

Thus, (11) becomes $-\alpha _1+\overline{F_1}(q_1)(\beta _1 F_2(Q-q_1)+ \beta _2 \overline{F_2}(Q-q_1))-\lambda =0$, i.e., $q_1^D({\lambda })=F_1^{-1}(1-\frac{\alpha _1+\lambda }{(\beta _1-\beta _2) \cdot F_2(Q-q_1^D({\lambda }))+\beta _2})$.

Also, $\mathcal {L}^2_{q_1q_2}\equiv \frac{\partial ^2 \mathcal {L}(q_1, q_2; \lambda )}{\partial q_1 \partial q_2}=0$ and $\mathcal {L}^2_{q_1^2}\equiv \frac{\partial ^2 \mathcal {L}(q_1, q_2; \lambda )}{\partial q_1^2}$ $=-f_1(q_1)(\beta _1 F_2(Q-q_1)+ \beta _2 \overline{F_2}(Q-q_1))-\overline{F_1}(q_1)\cdot f_2(Q-q_1) \cdot (\beta _1-\beta _2)<0.$

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_2}&= -\alpha _2+f_2(q_2) \int _{E_1=0}^{q_1} (p-c)(E_1+q_2) dF_1(E_1) \nonumber \\&\quad + f_2(q_2)\int _{E_1=q_1}^{Q-q_2}(p-c)\cdot (E_1+q_2)-\beta _1(E_1-q_1) dF_1(E_1)\nonumber \\&\quad + f_2(q_2)\int _{E_1=Q-q_2}^{\infty }(p-c)\cdot Q-\beta _1(Q-q_1-q_2) dF_1(E_1) \nonumber \\&\quad + \int _{E_1=0}^{q_1} \left( \int _{E_2=Q-E_1}^{\infty } \beta _2 dF_2(E_2) \right) dF_1(E_1)\nonumber \\&\quad + \int _{E_1=0}^{q_1}\left( \int _{E_2=q_2}^{Q-E_1} \beta _2 dF_2(E_2) - f_2(q_2) \cdot (p-c)\cdot (E_1+q_2)\right) dF_1(E_1)\nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\left( \int _{E_1=q_1}^{Q-E_2} \beta _2 dF_1(E_1) \right) dF_2(E_2)\nonumber \\&\quad - f_2(q_2)\cdot \int _{E_1=q_1}^{Q-q_2} (p-c)\cdot (E_1+q_2)-\beta _1(E_1-q_1) dF_1(E_1)\nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\left( \int _{E_1=Q-E_2}^{\infty }\beta _2 dF_1(E_1) \right) dF_2(E_2) \nonumber \\&\quad - f_2(q_2)\cdot \int _{E_1=Q-q_2}^{\infty } (p-c)Q-\beta _1(Q-q_1-q_2) dF_1(E_1)\nonumber \\&\quad + \int _{E_1=q_1}^{\infty }\left( \int _{E_2=Q-q_1}^{\infty } \beta _2 dF_2(E_2)\right) dF_1(E_1) \end{aligned}$$

(21)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_2}&= -\alpha _2+\int _{E_1=0}^{q_1}\int _{E_2=q_2}^{Q-E_1} \beta _2 dF_2(E_2) dF_1(E_1)\nonumber \\ {}&\quad +\int _{E_1=0}^{q_1}\int _{E_2=Q-E_1}^{\infty } \beta _2 dF_2(E_2) dF_1(E_1)\nonumber \\&\quad + \int _{E_2=q_2}^{Q-q_1}\int _{E_1=q_1}^{Q-E_2} \beta _2 dF_1(E_1) dF_2(E_2)\nonumber \\&\quad +\int _{E_2=q_2}^{Q-q_1}\int _{E_1=Q-E_2}^{\infty }\beta _2 dF_1(E_1) dF_2(E_2) \nonumber \\&\quad + \int _{E_1=q_1}^{\infty }\int _{E_2=Q-q_1}^{\infty } \beta _2 dF_2(E_2)dF_1(E_1) \end{aligned}$$

(22)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1,q_2 | p)}{\partial q_2}&= -\alpha _2+\int _{E_1=0}^{q_1}\int _{E_2=q_2}^{\infty } \beta _2 dF_2(E_2) dF_1(E_1)\nonumber \\&\quad +\int _{E_1=q_1}^{\infty }\int _{E_2=q_2}^{\infty } \beta _2 dF_2(E_2)dF_1(E_1)\nonumber \\&= -\alpha _2+\beta _2 \overline{F_2}(q_2) \end{aligned}$$

(23)

Thus, (12) becomes $-\alpha _2+\beta _2 \overline{F_2}(q_2)-\lambda =0$, i.e., $q_2^D(\lambda )=F_2^{-1}(1-\frac{\alpha _2+\lambda }{\beta _2})$.

Also, $\mathcal {L}^2_{q_2 q_1}\equiv \frac{\partial ^2 \mathcal {L}(q_1, q_2; \lambda )}{\partial q_2 \partial q_1}=0$ and $\mathcal {L}^2_{q_2^2}\equiv \frac{\partial ^2 \mathcal {L}(q_1, q_2; \lambda )}{\partial q_2^2}=\frac{\partial ^2 E\Pi ^D(q_1,q_2 | p)}{\partial q_2^2}=-f_2(q_2)\cdot \beta _2 <0.$

We first compute $Q^c$ as the solution for $Q$ in $q_1^D(0)+q_2^D(0)=Q$, i.e.,

$$\begin{aligned} F_1^{-1}\left( 1-\frac{\alpha _1}{(\beta _1-\beta _2) \cdot F_2(Q^c-q_1^D(0))+\beta _2}\right) + F_2^{-1}\left( 1-\frac{\alpha _2}{\beta _2}\right) =Q^c. \end{aligned}$$

To find the optimal solution, we consider two cases.

Case 1: $Q\ge Q^c$, then $q_1^D(0)+q_2^D(0) \le Q$ is satisfied with $\lambda =0$. The optimal contract amounts solve the following equations: $F_1(q_1^D)=1-\frac{\alpha _1}{(\beta _1-\beta _2) \cdot F_2(Q-q_1^D)+\beta _2}$ and $F_2(q_2^D)=1-\frac{\alpha _2}{\beta _2}$.

Case 2: $Q<Q^c$, $q_1^D+q_2^D\le Q$ is violated. We thus need to solve for $\lambda$ using $q_1^D(\lambda )+q_2^D(\lambda )=Q$. Because $\mathcal {L}^2_{q_1^2}<0$ and $\mathcal {L}^2_{q_1^2}\mathcal {L}^2_{q_2^2}-\mathcal {L}^2_{q_1 q_2}\mathcal {L}^2_{q_2 q_1}>0$, $\mathcal {L}$ is concave. Thus, the optimal solution satisfies all KKT conditions and meets the second-order sufficient conditions as well. $\square$

Appendix B: Derivation of the results for the special case

Single-emitter case. First, we compute $q_i^*$.

$$\begin{aligned} \overline{F_i}(q_i^*)=e^{-\gamma q_i^*}=\frac{\alpha _2}{\beta _2} \quad \Rightarrow q_i^*=\frac{1}{\gamma }\ln \left( \frac{\beta _i}{\alpha _i}\right) . \end{aligned}$$

(24)

$$\begin{aligned} \text {Equation }(2) \Rightarrow E\Pi (q_i^*|p_i)= & {} -K-\frac{\alpha _i}{\gamma } \ln (\rho )-\frac{\beta _i}{\gamma } \left( \frac{1}{\rho }-e^{-\gamma Q}\right) \nonumber \\&+\frac{p-c}{\gamma }(1-e^{-\gamma Q}), \end{aligned}$$

(25)

$$\begin{aligned}&\frac{t-p_i^S-(\mu -\delta )}{2\delta } \left( 1+\frac{1}{\gamma }(1-e^{-\gamma Q})\right) =\frac{1}{2\delta } E\Pi (q_i^*|p_i) \nonumber \\&\quad \Rightarrow p_i^S=\left( t-(\mu -\delta )+K+\frac{\alpha _i}{\gamma } \ln (\rho )+\frac{\beta _i}{\gamma } \left( \frac{1}{\rho }-e^{-\gamma Q}\right) +\frac{c}{\gamma }(1-e^{-\gamma Q})\right) \Bigg /\nonumber \\&\quad \qquad \qquad \quad \left( 1+\frac{1}{\gamma }(1-e^{-\gamma Q})\right) . \end{aligned}$$

(26)

Dual-emitter case. First, we compute $q_1^D, q_2^D$, and $E\Pi ^D(q_1^D, q_2^D|p)$.

$$\begin{aligned} \overline{F_2}(q_2)&=e^{-\gamma q_2}=\frac{\alpha _2}{\beta _2} \quad \Rightarrow q_2^D=\frac{1}{\gamma }\ln \left( \frac{\beta _2}{\alpha _2}\right) . \end{aligned}$$

(27)

$$\begin{aligned} \overline{F_1}(q_1)&=e^{-\gamma q_1}=\frac{\alpha _1}{(1-e^{-\gamma (Q-q_1)})(\beta _1-\beta _2)+\beta _2} \Rightarrow \nonumber \\ q_1^D&=\frac{1}{\gamma }\ln \left( \frac{\beta _1}{\alpha _1+e^{-\gamma Q}(\beta _1-\beta _2)}\right) . \end{aligned}$$

(28)

$$\begin{aligned} \text {Let }\overline{\rho }&=\frac{\beta _1}{\alpha _1+e^{-\gamma Q}(\beta _1-\beta _2)}, \nonumber \\ E\Pi ^D(q_1^D, q_2^D|p)&=-K-\frac{\alpha _1}{\gamma } \ln (\overline{\rho })-\frac{\alpha _2}{\gamma } \ln (\rho )\nonumber \\&\quad +(p-c)\left( \frac{2}{\gamma }(1-e^{-\gamma Q})-Q e^{-\gamma Q}\right) \nonumber \\&\quad - \beta _2 \left( \frac{1}{\gamma }\left( \frac{1}{\rho }-e^{-\gamma Q}\right) -e^{-\gamma Q} q_1^D\right) \nonumber \\&\quad - \beta _1 \left( \frac{1}{\gamma }\left( \frac{1}{\overline{\rho }}-e^{-\gamma Q}\right) -e^{-\gamma Q} (Q-q_1^D)\right) . \end{aligned}$$

(29)

Next, we compute $Q^c$ and $\lambda$.

$$\begin{aligned} Q^c&= q_1^D+q_2^D=\frac{1}{\gamma }\ln \left( \frac{(\alpha _1+e^{-\gamma Q^c}(\beta _1-\beta _2))\alpha _2}{\beta _1\beta _2}\right) \nonumber \\&\Rightarrow Q^c=\frac{1}{\gamma }\ln \left( \frac{\beta _1\beta _2-(\beta _1-\beta _2)\alpha _2}{\alpha _1\alpha _2}\right) .\end{aligned}$$

(30)

$$\begin{aligned} e^{-\gamma q_1}&=\frac{\alpha _1+\lambda }{(1-e^{-\gamma (Q-q_1)})(\beta _1-\beta _2)+\beta _2}\nonumber \\&\Rightarrow q_1^D(\lambda )=\frac{1}{\gamma }\ln \left( \frac{\beta _1}{\alpha _1+\lambda +e^{-\gamma Q}(\beta _1-\beta _2)}\right) , \end{aligned}$$

(31)

$$\begin{aligned} e^{-\gamma q_2}&=\frac{\alpha _2+\lambda }{\beta _2} \quad \Rightarrow q_2^D(\lambda )=\frac{1}{\gamma }\ln \left( \frac{\beta _2}{\alpha _2+\lambda }\right) . \end{aligned}$$

(32)

$$\begin{aligned} q_1^D(\lambda )+q_2^D(\lambda )&=Q \Rightarrow (\alpha _1+\lambda +(\beta _1-\beta _2)e^{-\gamma Q})(\alpha _2+\lambda ) -\beta _1\beta _2e^{-\gamma Q}=0 \nonumber \\&\Rightarrow \lambda =\frac{1}{2}\sqrt{(\alpha _1-\alpha _2+e^{-\gamma Q}(\beta _1-\beta _2))^2+4\beta _1\beta _2e^{-\gamma Q}} \nonumber \\&\quad -\frac{1}{2}(\alpha _1+\alpha _2+e^{-\gamma Q}(\beta _1-\beta _2)). \end{aligned}$$

(33)

We also need to show that $\lambda >0$ as long as $\beta _2>\alpha _1$, and $\frac{\partial \lambda }{\partial Q}<0$. $Q<Q^c \Rightarrow e^{-\gamma Q}>e^{-\gamma Q^c}=\frac{\beta _1\beta _2-(\beta _1-\beta _2)\alpha _2}{\alpha _1\alpha _2}$. Thus,

$$\begin{aligned}&(\alpha _1-\alpha _2+e^{-\gamma Q}(\beta _1-\beta _2))^2+4\beta _1\beta _2e^{-\gamma Q}-(\alpha _1+\alpha _2+e^{-\gamma Q}(\beta _1-\beta _2))^2\\&\quad = 4(\beta _1\beta _2e^{-\gamma Q}-\alpha _1\alpha _2-\alpha _2e^{-\gamma Q}(\beta _1-\beta _2))\\&\quad =4(e^{-\gamma Q}\beta _1(\beta _2-\alpha _2)+\alpha _2(e^{-\gamma Q}\beta _2-\alpha _1))\\&\quad \displaystyle >e^{-\gamma Q}\beta _1(\beta _2-\alpha _2)+\alpha _2\left( \frac{\beta _1\beta _2-(\beta _1-\beta _2)\alpha _2}{\alpha _1\alpha _2}\beta _2-\alpha _1\right) \\&\quad \displaystyle =e^{-\gamma Q}\beta _1(\beta _2-\alpha _2)+\frac{1}{\alpha _1}(\beta _1\beta _2(\beta _2-\alpha _2)+\alpha _2(\beta _2^2-\alpha _1^2))>0\quad \mathrm{if} \quad \beta _2>\alpha _1. \end{aligned}$$

To verify $\frac{\partial \lambda }{\partial Q}<0$, first showing $D(Q) = S(Q)-U(Q)<0$, where $S(Q)\equiv (\alpha _1-\alpha _2+e^{-\gamma Q}(\beta _1-\beta _2))^2+4\beta _1\beta _2e^{-\gamma Q}$ and $U(Q) \equiv (\alpha _1-\alpha _2+e^{-\gamma Q}(\beta _1+\beta _2))^2$.

$$\begin{aligned} D(Q)&=4e^{-\gamma Q}\beta _2(\beta _1-e^{-\gamma Q}\beta _1-(\alpha _1-\alpha _2))\nonumber \\&<4e^{-\gamma Q}\beta _2 \left( \beta _1-\frac{\beta _1\beta _2-(\beta _1-\beta _2)\alpha _2}{\alpha _1\alpha _2}\beta _1-(\alpha _1-\alpha _2)\right) \nonumber \\&=4e^{-\gamma Q}\frac{\beta _2}{\alpha _1\alpha _2}(\beta _1\alpha _2(\alpha _1-\beta _2)+\beta _1^2(\alpha _2-\beta _2)+\alpha _1\alpha _2(\alpha _2-\alpha _1))<0. \end{aligned}$$

(34)

$$\begin{aligned} \frac{\partial \lambda }{\partial Q}&= \frac{1}{4}(S(Q))^{-\frac{1}{2}}\cdot \frac{\partial S(Q)}{\partial Q}+\frac{1}{2}\gamma e^{-\gamma Q}(\beta _1-\beta _2) \nonumber \\&=\frac{-1}{2}\gamma e^{-\gamma Q}(S(Q))^{-\frac{1}{2}}((\alpha _1-\alpha _2+e^{-\gamma Q}(\beta _1-\beta _2))(\beta _1-\beta _2)\nonumber \\&\quad +2\beta _1\beta _2 -(\beta _1-\beta _2)(S(Q))^{\frac{1}{2}}) \nonumber \\&<\frac{-1}{2}\gamma e^{-\gamma Q}(S(Q))^{-\frac{1}{2}}(e^{-\gamma Q}(\beta _1-\beta _2)^2+2\beta _1\beta _2 -e^{-\gamma Q}(\beta _1-\beta _2)(\beta _1\!+\!\beta _2)) \nonumber \\&=-\gamma e^{-\gamma Q}(S(Q))^{-\frac{1}{2}}\beta _2((1-e^{-\gamma Q})\beta _1+e^{-\gamma Q}\beta _2)<0. \end{aligned}$$

(35)

Last, we compute the optimal price $p^D$ by solving for it from Eq. (8):

$$\begin{aligned} E\Pi ^D-E\Pi _1^S-E\Pi _2^S-\frac{1}{\gamma }(1-e^{-\gamma Q})&=K+\frac{\alpha _1}{\gamma }(\ln (\rho )-\ln (\overline{\rho }))\nonumber \\&\quad + \frac{\beta _1}{\gamma }\left( \frac{1}{\rho }-\frac{1}{\overline{\rho }}\right) -(p-c-\beta _1)Qe^{-\gamma Q}\nonumber \\&\quad - \frac{e^{-\gamma Q}}{\gamma }(\beta _1-\beta _2)\ln (\overline{\rho }) \nonumber \\&\quad - \frac{1}{\gamma }\left( 1-e^{-\gamma Q}\right) \equiv T_1-(p-c)Qe^{-\gamma Q} \end{aligned}$$

(36)

$$\begin{aligned} E\Pi _1^S+E\Pi _2^S&=(p-c)\frac{2}{\gamma }(1-e^{-\gamma Q})-2K-(\alpha _1+\alpha _2)\frac{1}{\gamma }\ln (\rho )\nonumber \\&\quad - (\beta _1+\beta _2)\frac{1}{\gamma }\left( \frac{1}{\rho }-e^{-\gamma Q}\right) \equiv (p-c)\frac{2}{\gamma }(1-e^{-\gamma Q})-T2\quad \end{aligned}$$

(37)

$$\begin{aligned}&G(t-p)^2 (Qe^{-\gamma Q})+2G(t-p)\left( E\Pi ^D-E\Pi _1^S-E\Pi _2^S-\frac{1}{\gamma }(1-e^{-\gamma Q})\right) \nonumber \\&\quad + E\Pi _1^S+E\Pi _2^S=0. \end{aligned}$$

(38)

Let $\overline{t}=t-c-\mu +\delta ,$

$$\begin{aligned}&\left( \frac{\overline{t}-(p-c)}{2\delta }\right) ^2 (Qe^{-\gamma Q})+\frac{(\overline{t}-(p-c))}{\delta }(T_1-(p-c)Qe^{-\gamma Q}) \nonumber \\&\quad + (p-c)\frac{2}{\gamma } (1-e^{-\gamma Q})-T_2=0. \end{aligned}$$

(39)

$$\begin{aligned} p^D&=c+\frac{-B+\sqrt{B^2-4AC}}{2A}, \text { where } A=Qe^{-\gamma Q}\left( \frac{1}{4\delta ^2}+\frac{1}{\delta }\right) , \nonumber \\ B&=\frac{2}{\gamma }(1-e^{-\gamma Q})-\frac{2\overline{t}}{\delta }Qe^{-\gamma Q}-\frac{T_1}{\delta }, \nonumber \\ \text { and } C&=\frac{\overline{t}^2}{4\delta ^2}Qe^{-\gamma Q}+\frac{\overline{t}}{\delta }T_1-T_2. \end{aligned}$$

(40)

As $Q$ increases, $T_1$ increases while $T_2$ decreases. Thus, both $A$ and $C$ increase, and $B$ decreases. As a result, $p^D$ decreases in $Q$.

Appendix C: Proof of Proposition 2

We can use Fig. 1 to construct the derivative as the sum of three integrals:

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1^D,q_2^D | p)}{\partial p}&= \int _{E_1=0}^{Q}\int _{E_2=0}^{Q-E_1}(E_1+E_2)dF_2(E_2)dF_1(E_1)\nonumber \\&\quad +\int _{E_1=0}^{Q}\int _{E_2=Q-E_1}^{\infty }QdF_2(E_2)dF_1(E_1)\nonumber \\&\quad + \int _{E_1=Q}^{\infty }\int _{E_2=0}^{\infty }QdF_2(E_2)dF_1(E_1) \end{aligned}$$

(41)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1^D,q_2^D | p)}{\partial p}&= \int _{E_1=0}^{Q}E_1 F_2(Q-E_1)dF_1(E_1)\nonumber \\ {}&\quad + \int _{E_1=0}^{Q}\int _{E_2=0}^{Q-E_1}E_2dF_2(E_2)dF_1(E_1)\nonumber \\&\quad + Q\int _{E_1=0}^Q\overline{F_2}(Q-E_1)dF_1(E_1)+Q\overline{F_1}(Q) \end{aligned}$$

(42)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1^D,q_2^D | p)}{\partial p}&= \int _{E_1=0}^Q \left( E_1 F_2(Q-E_1)+Q\overline{F_2}(Q-E_1)\right. \nonumber \\ {}&\quad \left. +\int _{E_2=0}^{Q-E_1}E_2dF_2(E_2)\right) dF_1(E_1)+Q(1-F_1(Q)) \end{aligned}$$

(43)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1^D,q_2^D | p)}{\partial p}&= \int _{E_1=0}^Q \bigg (E_1 F_2(Q-E_1)+Q\overline{F_2}(Q-E_1)+(Q-E_1)F_2(Q-E_1)\nonumber \\ {}&\quad -\int _{E_2=0}^{Q-E_1}F_2(E_2)dE_2\bigg )dF_1(E_1)+ Q-QF_1(Q) \end{aligned}$$

(44)

$$\begin{aligned} \frac{\partial E\Pi ^D(q_1^D,q_2^D | p)}{\partial p}= & {} Q-QF_1(Q)+QF_1(Q)\nonumber \\&-\int _{E_1=0}^{Q}\left( \int _{E_2=0}^{Q-E_1}F_2(E_2)dE_2\right) dF_1(E_1). \end{aligned}$$

(45)

Using integration by parts and Leibniz’ rule on the double integral leads to

$$\begin{aligned}&=Q-\left( F_1(E_1)\int _{E_2=0}^{Q-E_1}F_2(E_2)dE_2\right) \bigg |_{E_1=0}^Q+\int _{E_1=0}^QF_1(E_1)F_2(Q-E_1)dE_1\end{aligned}$$

(46)

$$\begin{aligned}&= Q-\int _{E_1=0}^Q F_1(E_1)F_2(Q-E_1)dE_1. \end{aligned}$$

(47)

Also, $\frac{\partial ^2 E\Pi ^D(q_1^D,q_2^D | p)}{\partial p^2}=0.$ We can write down the first-order condition on the optimal price:

$$\begin{aligned} \frac{\partial E\Pi ^D (q_1^D, q_2^D, p)}{\partial p}&=2G(t-p) \cdot (-g(t-p)) \cdot E\Pi ^D(q_1^D, q_2^D|p) \nonumber \\&\quad + G^{2}(t-p) \cdot \frac{\partial E\Pi ^D (q_1^D, q_2^D|p)}{\partial p}\nonumber \\&\quad + (1-2G(t-p)) \cdot (-g(t-p)) \cdot (E\Pi _1^S(q_1^S| p)+E\Pi _2^S(q_2^S| p))\nonumber \\&\quad + G(t-p) \cdot \overline{G}(t-p) \cdot \left( \frac{\partial E\Pi _1^S(q_1^S| p)}{\partial p}+\frac{\partial E\Pi _2^S(q_2^S| p)}{\partial p} \right) \end{aligned}$$

(48)

$$\begin{aligned} \frac{\partial E\Pi ^D (q_1^D, q_2^D, p)}{\partial p}&= -2 G(t-p) \cdot g(t-p) \cdot E\Pi ^D(q_1^D, q_2^D|p) \nonumber \\&\quad + G^{2}(t-p) \cdot \left( \int _{E_1=0}^{Q}(1-F_1(E_1) \cdot F_2(Q-E_1)) dE_1\right) \nonumber \\&\quad +(2G(t-p)-1) \cdot g(t-p) \cdot (E\Pi _1^S(q_1^S| p)+E\Pi _2^S(q_2^S| p))\nonumber \\&\quad + G(t-p) \cdot \overline{G}(t-p) \cdot \bigg (\int _{E_1=0}^{Q}\overline{F_1}(E_1) dE_1\nonumber \\ {}&\quad +\int _{E_2=0}^{Q}\overline{F_2}(E_2) dE_2\bigg )=0, \end{aligned}$$

(49)

which gives the value of the optimal price $p^D$. The second-order sufficient condition for $p^D$ is:

$$\begin{aligned}&\frac{\partial ^2 E\Pi ^D(q_1^D, q_2^D,p)]}{\partial p^2} \nonumber \\&\quad = 2[g^{'}(t-p) \cdot G(t-p) +g^2(t-p)] \cdot E\Pi ^D(q_1^D, q_2^D|p)\nonumber \\&\qquad - 2g(t-p) \cdot G(t-p) \cdot \frac{\partial E\Pi ^D(q_1^D, q_2^D|p)}{\partial p} \nonumber \\&\qquad - [(g^{'}(t-p)\cdot (2G(t-p)-1)+2g^2(t-p)] \cdot (E\Pi _1^S(q_1^S| p)+E\Pi _2^S(q_2^S| p))\nonumber \\&\qquad + [g(t-p)\cdot (2G(t-p)-1)] \cdot \left( \frac{\partial E\Pi _1^S(q_1^S| p)}{\partial p} +\frac{\partial E\Pi _2^S(q_2^S| p)}{\partial p} \right) \nonumber \\&\qquad + [-g(t-p) \cdot (1-G(t-p))+G(t-p)\cdot g(t-p)] \cdot \bigg (\frac{\partial E\Pi _1^S(q_1^S| p)}{\partial p}+\frac{\partial E\Pi _2^S(q_2^S| p)}{\partial p} \bigg ). \end{aligned}$$

(50)

Thus, $p^D$ must satisfy the second-order condition:

$$\begin{aligned}&2[g^{'}(t-p^D) \cdot G(t-p^D) +g^2(t-p^D)] \cdot E\Pi ^D(q_1^D, q_2^D|p^D) \nonumber \\&\quad - 2g(t-p^D) \cdot G(t-p^D) \cdot \left( \int _{E_1=0}^{Q}(1-F_1(E_1) \cdot F_2(Q-E_1)) dE_1\right) \nonumber \\&\quad - \left[ (g^{'}(t-p^D)\cdot (2G(t-p^D)-1)+2g^2(t-p^D)\right] \cdot \left( E\Pi _1^S(q_1^S| p^D)+E\Pi _2^S(q_2^S| p^D)\right) \nonumber \\&\quad + \left[ 4g(t-p^D)\cdot G(t-p^D)-2g(t-p^D)\right] \cdot \left( \int _{E_1=0}^{Q}\overline{F_1}(E_1) dE_1+\int _{E_2=0}^{Q}\overline{F_2}(E_2) dE_2\right) <0. \end{aligned}$$

(51)

Additionally, we need to impose the following conditions to ensure the expected profit of the storage operator is non-negative:

$$\begin{aligned} E\Pi ^D\left( q_1^D, q_2^D|p^D\right) , \, E\Pi _1^S\left( q_1^S| p^D\right) , \, \text{ and } \, E\Pi _2^S\left( q_2^S| p^D\right) \, \ge 0. \end{aligned}$$

(52)