Game Theory Analysis for Carbon Auction Market Through Electricity Market Coupling

Abstract

In this paper, we analyze Nash equilibria between electricity producers selling their production on an electricity market and buying CO₂ emission allowances on an auction carbon market. The producers’ strategies integrate the coupling of the two markets via the cost functions of the electricity production. We set out a clear Nash equilibrium on the power market that can be used to compute equilibrium prices on both markets as well as the related electricity produced and CO₂ emissions released.

Appendix

1.1 Proof of Proposition 1

1.1.1 A. First We Prove the Dominance Property (i)

Suppose that one producer, let us say producer 1, deviates and chooses C ₁ instead of s ₁. We have to show that its market share cannot be reduced by this deviation. By definition of the admissibility (see (2)) we have

$$\displaystyle{s_{1}(q) \geq C_{1}(q),\forall q \in [0,\kappa _{1}].}$$

Hence the offer functions defined by (3) satisfy $\mathcal{O}(s_{1};\cdot ) \leq \mathcal{O}(C_{1};\cdot ).$ By adding the unchanged offers of the other producers

$$\displaystyle{ \mathcal{O}((\mathbf{s}_{-1},s_{1});\cdot ) \leq \mathcal{O}((\mathbf{s}_{-1},C_{1});\cdot ), }$$

(A.1)

where (s ₋₁, C ₁) denotes the strategy profile that includes producer 1 deviation. The minimum market clearing price (5) for strategy profile s is

$$\displaystyle{ \underline{p}(\mathbf{s}) =\inf \{ p,\;\mathcal{O}(\mathbf{s};p) > D(p)\}. }$$

The minimum market clearing price (5) for strategy profile (s ₋₁, C ₁) is

$$\displaystyle{ \underline{p}(\mathbf{s}_{-1},C_{1}) =\inf \{ p,\;\mathcal{O}((\mathbf{s}_{-1},C_{1});p) > D(p)\}. }$$

The inequality (A.1) together with the fact that the demand D(⋅ ) is a non-increasing function imply that $\underline{p}(\mathbf{s}_{-1},C_{1}) \leq \underline{ p}(\mathbf{s})$, from which, with (6) we deduce that

$$\displaystyle{p^{\mathrm{elec}}(\mathbf{s}_{ -1},C_{1}) \leq p^{\mathrm{elec}}(\mathbf{s}).}$$

Now let us show that Producer 1 does not reduce its market share by deviating from s ₁(⋅ ) to C ₁(⋅ ), that is $\varphi _{1}(\mathbf{s}_{-1},C_{1}) \geq \varphi _{1}(\mathbf{s})$.

For the sake of clarity we adopt, in this paragraph, the following notation:

$$\displaystyle{ \begin{array}{ll} \underline{p}_{\mathbf{s}} &:=\underline{ p}(\mathbf{s}) \\ p_{\mathbf{s}}^{\mathrm{elec}} &:= p^{\mathrm{elec}}(\mathbf{s})\end{array} \quad \mbox{ and }\quad \begin{array}{ll} \underline{p}_{\mathbf{s}C} &:=\underline{ p}(\mathbf{s}_{-1},C_{1}) \\ p_{\mathbf{s}C}^{\mathrm{elec}} &:= p^{\mathrm{elec}}((\mathbf{s}_{-1},C_{1})).\end{array} }$$

We first consider the case where $\underline{p}_{\mathbf{s}C} <\underline{ p}_{\mathbf{s}}$ . By definition of the minimum clearing price $\underline{p}_{\mathbf{s}C}$, the fact that $D(\underline{p}_{\mathbf{s}}) \leq D(\underline{p}_{\mathbf{s}C})$ and the fact that $\mathcal{O}((\mathbf{s}_{-1},C_{1});\cdot )$ is non-decreasing, we have

$$\displaystyle{D(p_{\mathbf{s}}^{\mathrm{elec}}) \leq D(\underline{p}_{\mathbf{ s}}) \leq D(\underline{p}_{\mathbf{s}C}) \leq \mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}_{\mathbf{s}C}) \leq \mathcal{O}((\mathbf{s}_{-1},C_{1});p_{\mathbf{s}C}^{\mathrm{elec}}).}$$

Hence,

$$\displaystyle\begin{array}{rcl} \mathcal{O}((\mathbf{s}_{-1},s_{1}),p_{\mathbf{s}}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}}^{\mathrm{elec}})& \leq & \mathcal{O}((\mathbf{s}_{ -1},C_{1});p_{\mathbf{s}C}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}C}^{\mathrm{elec}}), {}\\ \mathcal{O}((\mathbf{s}_{-1},s_{1}),\underline{p}_{\mathbf{s}}) \wedge D(\underline{p}_{\mathbf{s}})& \leq & \mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}_{\mathbf{s}C}) \wedge D(\underline{p}_{\mathbf{s}C}). {}\\ \end{array}$$

From the market clearing (8) we get

$$\displaystyle\begin{array}{rcl} \varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1})&& {}\\ & =& \mathcal{O}((\mathbf{s}_{-1},s_{1}),p_{\mathbf{s}}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}}^{\mathrm{elec}}) -\mathcal{O}((\mathbf{s}_{ -1},C_{1});p_{\mathbf{s}C}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}C}^{\mathrm{elec}}) {}\\ & & +\sum _{j>1}\left (\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})\right ). {}\\ \end{array}$$

According to Definition 2, let us denote

$$\displaystyle{ \mathcal{E}(\underline{p}_{\mathbf{s}}) = \left \{j \in \{ 2,\ldots,J\}\mbox{ s.t. }\varDelta ^{-}\mathcal{O}(s_{ j};\underline{p}_{\mathbf{s}}) > 0\right \}. }$$

We have

$$\displaystyle\begin{array}{rcl} \varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1})&& {}\\ & =& \mathcal{O}((\mathbf{s}_{-1},s_{1});p_{\mathbf{s}}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}}^{\mathrm{elec}}) -\mathcal{O}((\mathbf{s}_{ -1},C_{1});p_{\mathbf{s}C}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}C}^{\mathrm{elec}}) {}\\ & & +\sum _{j>1,j\notin \mathcal{E}(\underline{p}_{\mathbf{s}})}\left (\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\mathcal{O}(s_{j};p_{\mathbf{s}}^{\mathrm{elec}})\right ) {}\\ & & +\sum _{j>1,j\in \mathcal{E}(\underline{p}_{\mathbf{s}})}(\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})) {}\\ & \leq & \mathcal{O}((\mathbf{s}_{-1},s_{1});p_{\mathbf{s}}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}}^{\mathrm{elec}}) -\mathcal{O}((\mathbf{s}_{ -1},C_{1});p_{\mathbf{s}C}^{\mathrm{elec}}) \wedge D(p_{\mathbf{ s}C}^{\mathrm{elec}}) {}\\ & & +\sum _{j>1,j\notin \mathcal{E}(\underline{p}_{\mathbf{s}})}\left (\mathcal{O}(s_{j};p_{\mathbf{s}C}^{\mathrm{elec}}) -\mathcal{O}(s_{ j};p_{\mathbf{s}}^{\mathrm{elec}})\right ) {}\\ & & +\sum _{j>1,j\in \mathcal{E}(\underline{p}_{\mathbf{s}})}(\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})) {}\\ \end{array}$$

Since $p_{\mathbf{s}C}^{\mathrm{elec}} \leq p_{\mathbf{s}}^{\mathrm{elec}}$ we get

$$\displaystyle{ \varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1}) \leq \sum _{j>1,j\in \mathcal{E}(\underline{p}_{\mathbf{s}})}(\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})). }$$

But for any $j \in \mathcal{E}(\underline{p}_{\mathbf{s}})$, the quantity $\mathcal{O}(s_{j};\underline{p}_{\mathbf{s}}^{-}) \leq \varphi _{j}(\mathbf{s}_{-1},s_{1})$. As $\mathcal{O}(\mathbf{s}_{j};\cdot )$ is non-decreasing a since we have assumed $\underline{p}_{\mathbf{s}C} <\underline{ p}_{\mathbf{s}}$, we get

$$\displaystyle{\mathcal{O}(s_{j};\underline{p}_{\mathbf{s}C}^{-}) \leq \mathcal{O}(s_{ j};\underline{p}_{\mathbf{s}}^{-}) \leq \varphi _{ j}(\mathbf{s}_{-1},s_{1}).}$$

For such j > 1, we thus have

$$\displaystyle\begin{array}{rcl} \varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})& \leq & \varphi _{j}(\mathbf{s}_{-1},C_{1}) -\mathcal{O}(s_{j};p_{\mathbf{s}}^{\mathrm{elec}}{}^{-}) {}\\ & \leq & \varphi _{j}(\mathbf{s}_{-1},C_{1}) -\mathcal{O}(s_{j};p_{\mathbf{s}C}^{\mathrm{elec}}) \leq 0, {}\\ \end{array}$$

from which it follows that $\varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1}) \leq 0.$

Now consider the case where $\underline{p}_{\mathbf{s}} =\underline{ p}_{\mathbf{s}C}:=\underline{ p}$ . Due to the market rule (6), we necessarily have $p_{\mathbf{s}}^{\mathrm{elec}} = p_{\mathbf{s}C}^{\mathrm{elec}}:= p^{\mathrm{elec}}$.

• If $\mathcal{O}((\mathbf{s}_{-1},s_{1});p^{\mathrm{elec}}) \leq \mathcal{O}((\mathbf{s}_{-1},C_{1});p^{\mathrm{elec}}) \leq D(p^{\mathrm{elec}})$, then by the market clearing

  $$\displaystyle{ \varphi _{1}(\mathbf{s}_{-1},s_{1}) = \mathcal{O}(s_{1};p^{\mathrm{elec}}) \leq \mathcal{O}(C_{ 1};p^{\mathrm{elec}}) =\varphi _{ 1}(\mathbf{s}_{-1},C_{1}). }$$

• If $\mathcal{O}((\mathbf{s}_{-1},s_{1});p^{\mathrm{elec}}) \leq D(p^{\mathrm{elec}}) \leq \mathcal{O}((\mathbf{s}_{-1},C_{1});p^{\mathrm{elec}})$, then

  $$\displaystyle\begin{array}{rcl} \varphi _{1}(\mathbf{s}_{-1},s_{1})=\mathcal{O}(s_{1};p^{\mathrm{elec}})& \leq & D(p^{\mathrm{elec}})-\sum _{ j>1}\varphi _{j}(\mathbf{s}_{-1},s_{1})=D(p^{\mathrm{elec}})-\sum _{ j>1}\mathcal{O}(s_{j};p^{\mathrm{elec}}) {}\\ & \leq & D(p^{\mathrm{elec}}) -\sum _{ j>1}\varphi _{j}(\mathbf{s}_{-1},C_{1}) =\varphi _{1}(\mathbf{s}_{-1},C_{1}). {}\\ \end{array}$$

• If $D(p^{\mathrm{elec}}) < \mathcal{O}((\mathbf{s}_{-1},s_{1});p^{\mathrm{elec}}) \leq \mathcal{O}((\mathbf{s}_{-1},C_{1});p^{\mathrm{elec}})$, by the market clearing we get

  $$\displaystyle\begin{array}{rcl} \varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1})& & {}\\ & =& \mathcal{O}((\mathbf{s}_{-1},s_{1}),p^{\mathrm{elec}}) \wedge D(p^{\mathrm{elec}}) -\mathcal{O}((\mathbf{s}_{ -1},C_{1});p^{\mathrm{elec}}) \wedge D(p^{\mathrm{elec}}) {}\\ & & +\sum _{j>1}(\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})) {}\\ & \leq &\sum _{j>1}(\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})) {}\\ & \leq &\sum _{j>1,j\in \mathcal{E}(\underline{p})}(\varphi _{j}(\mathbf{s}_{-1},C_{1}) -\varphi _{j}(\mathbf{s}_{-1},s_{1})). {}\\ \end{array}$$

From (7), we have for $j \in \mathcal{E}(\underline{p})$

$$\displaystyle\begin{array}{rcl} & & \varphi _{j}(\mathbf{s}_{-1},s_{1}) = \mathcal{O}(s_{j},\underline{p}^{-}) +\varDelta ^{-}\mathcal{O}(s_{ j};\underline{p})\dfrac{\left (D(\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},s_{1}),\underline{p}^{-})\right )} {\varDelta ^{-}\mathcal{O}((\mathbf{s}_{ -1},s_{1}),\underline{p})} {}\\ \mbox{ and }\quad \mbox{ }& & \varphi _{j}(\mathbf{s}_{-1},C_{1}) = \mathcal{O}(s_{j};\underline{p}^{-}) +\varDelta ^{-}\mathcal{O}(s_{ j};\underline{p})\dfrac{\left (D(\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-})\right )} {\varDelta ^{-}\mathcal{O}((\mathbf{s}_{ -1},C_{1});\underline{p})}. {}\\ \end{array}$$

Hence, if $\mathcal{E}(\underline{p})$ is non empty then at least one producer exists, j ≠ 1 such that $\varDelta ^{-}\mathcal{O}(s_{j};\underline{p}) > 0$. and from the desegregation of $\mathcal{O}$ and definition of Δ ⁻ it results that

$$\displaystyle\begin{array}{rcl} & & \varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1}) {}\\ & & =\sum _{j>1,j\in \mathcal{E}(\underline{p})}\varDelta ^{-}\mathcal{O}(s_{ j},\underline{p})\left ( \dfrac{\left (D(\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(C_{1};\underline{p}^{-})\right )} {\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(C_{ 1};\underline{p}^{-})}\right. {}\\ & & \quad \left.- \dfrac{\left (D(\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(s_{1},\underline{p}^{-})\right )} {\mathcal{O}((\mathbf{s}_{-1},s_{1}),\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(s_{ 1};\underline{p}^{-})}\right ). {}\\ \end{array}$$

We note that

$$\displaystyle\begin{array}{rcl} 0& <& \mathcal{O}((\mathbf{s}_{-1},s_{1});\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(C_{ 1};\underline{p}^{-}) {}\\ & <& \mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(C_{ 1};\underline{p}^{-}), {}\\ \end{array}$$

and that $D(\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-}) > 0$ by definition of $\underline{p}$. Then

$$\displaystyle\begin{array}{rcl} & & \varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1}) {}\\ & & \quad \leq \sum _{j>1,j\in \mathcal{E}(\underline{p})}\varDelta ^{-}\mathcal{O}(s_{ j};\underline{p}) \times \left ( \dfrac{\left (D(\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}_{-}) -\mathcal{O}(C_{1};\underline{p}_{-})\right )} {\mathcal{O}((\mathbf{s}_{-1},s_{1});\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(C_{ 1};\underline{p}^{-})}\right. {}\\ & & \qquad \qquad \qquad \qquad \qquad \quad \ \ \left.- \dfrac{\left (D(\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}_{-}) -\mathcal{O}(s_{1};\underline{p}_{-})\right )} {\mathcal{O}((\mathbf{s}_{-1},s_{1});\underline{p}) -\mathcal{O}(\mathbf{s}_{-1},\underline{p}^{-}) -\mathcal{O}(s_{ 1};\underline{p}^{-})}\right ). {}\\ \end{array}$$

Since $D(\underline{p}) \leq \mathcal{O}((\mathbf{s}_{-1},s_{1});\underline{p})$ and $\mathcal{O}(C_{1};\underline{p}^{-}) \geq \mathcal{O}(s_{1};\underline{p}^{-})$, we can deduce that $\varphi _{1}(\mathbf{s}_{-1},s_{1}) -\varphi _{1}(\mathbf{s}_{-1},C_{1}) \leq 0.$ This follows from the fact that when A ≤ B, the map $x\mapsto \dfrac{A - x} {B - x}$ is decreasing on [0, A).

1.1.2 A.0.0.7 B. We Prove the Uniqueness Property (iii)

All Nash equilibria induce the same electricity price and same quantities of electricity bought to each producer.

First, we state the following consequence of the dominance property (i):

Lemma 8.

For any admissible strategy s = (s ₁ ,…,s _J ), such that $\underline{p}(\mathbf{s}) =\underline{ p}(\mathbf{C})$ , if producer j is such that s _j = C _j , then

$$\displaystyle{\varphi _{j}(\mathbf{s}) \geq \varphi _{j}(\mathbf{C}).}$$

Proof.

As arguments are very similar to the proof of (i), we just sketch them. Let s such that $\underline{p}(\mathbf{s}) =\underline{ p}(\mathbf{C}):=\underline{ p}$. Assume that Producer 1 is such that s ₁ = C ₁.

• If $\mathcal{O}(\mathbf{s};\underline{p}) \leq D(\underline{p})$, then by the market clearing

  $$\displaystyle{ \varphi _{1}(\mathbf{s}) = \mathcal{O}(s_{1};\underline{p}) = \mathcal{O}(C_{1};\underline{p}) \geq \varphi _{1}(\mathbf{C}). }$$

• If $D(\underline{p}) < \mathcal{O}(\mathbf{s};\underline{p}) \leq \mathcal{O}(\mathbf{C};\underline{p})$, by the market clearing we get

  $$\displaystyle\begin{array}{rcl} & & \varphi _{1}(\mathbf{s}_{-1},C_{1}) = \mathcal{O}(C_{1};\underline{p}^{-}) +\varDelta ^{-}\mathcal{O}(C_{ 1};\underline{p})\dfrac{\left (D(\underline{p}) -\mathcal{O}((\mathbf{s}_{-1};C_{1});\underline{p}^{-})\right )} {\varDelta ^{-}\mathcal{O}((\mathbf{s}_{ -1};C_{1});\underline{p})} {}\\ \mbox{ and }\quad \mbox{ }& & \varphi _{1}(\mathbf{C}_{-1},C_{1}) = \mathcal{O}(C_{1};\underline{p}^{-}) +\varDelta ^{-}\mathcal{O}(C_{ 1};\underline{p})\dfrac{\left (D(\underline{p}) -\mathcal{O}((\mathbf{C}_{-1};C_{1}),\underline{p}^{-})\right )} {\varDelta ^{-}\mathcal{O}((\mathbf{C}_{ -1};C_{1}),\underline{p})}. {}\\ \end{array}$$

Thus,

$$\displaystyle\begin{array}{rcl} \varphi _{1}(\mathbf{s}_{-1},C_{1}) -\varphi _{1}(\mathbf{C}_{-1},C_{1})&& {}\\ & =& \varDelta ^{-}\mathcal{O}(C_{ 1};\underline{p})\left ( \dfrac{\left (D(\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-})\right )} {\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-})}\right. {}\\ & & \left.- \dfrac{\left (D(\underline{p}) -\mathcal{O}((C_{-1},C_{1});\underline{p}^{-})\right )} {\mathcal{O}((C_{-1},C_{1});\underline{p}) -\mathcal{O}((C_{-1},C_{1});\underline{p}^{-})}\right ). {}\\ \end{array}$$

Assuming that $\varDelta ^{-}\mathcal{O}(C_{1};\underline{p}) > 0$, we note that

$$\displaystyle\begin{array}{rcl} 0 < \mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-}) \leq \mathcal{O}((C_{ -1},C_{1});\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-}).& & {}\\ \end{array}$$

Since $D(\underline{p}) -\mathcal{O}((C_{-1},C_{1});\underline{p}^{-}) > 0$ by definition of $\underline{p}$,

$$\displaystyle\begin{array}{rcl} \varphi _{1}(\mathbf{s}_{-1},C_{1}) -\varphi _{1}(C_{-1},C_{1})&& {}\\ & \geq & \varDelta ^{-}\mathcal{O}(C_{ 1};\underline{p})\left ( \dfrac{\left (D(\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-})\right )} {\mathcal{O}((C_{-1},C_{1});\underline{p}) -\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-})}\right. {}\\ & & \left.- \dfrac{\left (D(\underline{p}) -\mathcal{O}((C_{-1},C_{1});\underline{p}^{-})\right )} {\mathcal{O}((C_{-1},C_{1});\underline{p}) -\mathcal{O}((C_{-1},C_{1});\underline{p}^{-})}\right ). {}\\ \end{array}$$

As $\mathcal{O}((\mathbf{s}_{-1},C_{1});\underline{p}^{-}) \leq \mathcal{O}((C_{-1},C_{1});\underline{p}^{-})$, we get $\varphi _{1}(\mathbf{s}_{-1},C_{1}) -\varphi _{1}(C_{-1},C_{1}) \geq 0.$ □

We prove that the quantities are the same for all Nash equilibria. Let w an other Nash equilibrium that differs from C. On the global offers we always have $\mathcal{O}(\mathbf{w};\cdot ) \leq \mathcal{O}(\mathbf{C};\cdot )$ that implies $\underline{p}(\mathbf{w}) \geq \underline{ p}(\mathbf{C}).$ Note that when $\underline{p}(\mathbf{C}) = p_{\mbox{ lolc}}$, all admissible strategies s are Nash as $\varphi _{j}(\mathbf{C}) =\varphi _{j}(\mathbf{s}) =\kappa _{j},$ for all j. 

By the offers ordering, it is straightforward to show that

$$\displaystyle{\sum _{j=1}^{J}\varphi _{ j}(\mathbf{w}) \leq \sum _{j=1}^{J}\varphi _{ j}(\mathbf{C}).}$$

Assume that the quantities are not the same, then there exists a producer, say Producer 1, such that $\varphi _{1}(\mathbf{w}) <\varphi _{1}(\mathbf{C}).$ And we also have

$$\displaystyle{ \varphi _{1}(\mathbf{w}) <\varphi _{1}(\mathbf{C}) \leq \mathcal{O}(C_{1};p^{\mathrm{elec}}(\mathbf{C})) \leq \mathcal{O}(C_{ 1};p^{\mathrm{elec}}(\mathbf{w}_{ -1},C_{1}). }$$

If $\underline{p}(\mathbf{C}) =\underline{ p}(\mathbf{w}_{-1},C_{1})$, then by Lemma 8, we have that $\varphi _{1}(\mathbf{w}_{-1},C_{1}) \geq \varphi _{1}(\mathbf{C})$ and hence $\varphi _{1}(\mathbf{w}_{-1},C_{1}) >\varphi _{1}(\mathbf{w})$. In other words, w has a strictly favorable deviation for Producer 1 that contradicts the assumption that w is a Nash equilibrium.

Now if $\underline{p}(\mathbf{C}) <\underline{ p}(\mathbf{w}_{-1},C_{1})$, by (9),

$$\displaystyle{ \varphi _{1}(\mathbf{w}) <\varphi _{1}(\mathbf{C}) \leq \mathcal{O}(C_{1};\underline{p}(\mathbf{C})) \leq \mathcal{O}(C_{1};\underline{p}((\mathbf{w}_{-1},C_{1}))^{-}) \leq \varphi _{ 1}(\mathbf{w}_{-1},C_{1}), }$$

and the same conclusion follows.

We prove that the equilibrium best bid price is unique: $\bar{p}(\mathbf{w}) =\bar{ p}(\mathbf{C})$ , for an other Nash equilibrium w. Assume the contrary, $\bar{p}(\mathbf{w}) >\bar{ p}(\mathbf{C})$. Then by the definition of $\bar{p}(\cdot )$, we have that $D(\underline{p}(\mathbf{w})) < D(\underline{p}(\mathbf{C})).$

From (8) and (9),

$$\displaystyle{\sum _{j=1}^{J}\varphi _{ j}(\mathbf{w}) \leq D(\underline{p}(\mathbf{w})) < D(\underline{p}(\mathbf{C})^{+}) \leq D(\underline{p}(\mathbf{C})) \wedge \mathcal{O}(\mathbf{C};\underline{p}(\mathbf{C})) =\sum _{ j=1}^{J}\varphi _{ j}(\mathbf{C})}$$

that contradicts the fact that Nash equilibria have same clearing quantities.

1.2 Proofs of Lemmas 1 and 2

Proof of Lemma 1.

Although the result of this lemma is intuitive, the proof is rather technical. This is due to our assumptions, in particular regarding demand, that allow the demand function to have discontinuity points and some non-elasticity areas (see Assumption 1).

More precisely, if we define the map $\tau \mapsto \mathcal{O}(\tau;p)$ by

$$\displaystyle{\mathcal{O}(\tau;p) =\sum _{ i=1}^{J}\mathcal{O}(C_{ j}^{\tau }(\cdot );p) =\sum _{ i=1}^{J}\kappa _{ i} \mathbb{1}_{\{p\geq c_{i}+\tau e_{i}\}} =\sum _{ i=1}^{J}\kappa _{ i} \mathbb{1}_{\{\tau \leq \frac{p-c_{i}} {e_{i}} \}},}$$

then we can observe that, for any p > 0 far enough from the c _i , and any τ′ ≥ τ,

$$\displaystyle{\mathcal{O}(\tau ';p) \leq \mathcal{O}(\tau;p)\quad \mbox{ and }\quad \lim _{\epsilon \rightarrow 0^{+}}\mathcal{O}(\tau +\epsilon;p) = \mathcal{O}(\tau;p).}$$

We call $S_{D} =\{ p_{d};\lim _{\epsilon \rightarrow 0^{+}}D(p_{d}+\epsilon ) < D(p_{d})\}$, the set of discontinuity points of the Demand function.

We call S _κ  = { p _c ; D(p _c ) = ∑ κ _i }, the set of prices that make demand coincide with some accumulation of production capacities.

We observe that $p^{\mathrm{elec}}(\tau ) \in \{ c_{i} +\tau e_{i},i = 1,\ldots,j\} \cup S_{D} \cup S_{\kappa }$. In particular, from Definition 1, $\underline{p}(\tau ) =\inf \{ p > 0;\mathcal{O}(\tau;p) > D(p)\},$ and we obtain that $D(\underline{p}(\tau +\epsilon )) \leq \mathcal{O}(\tau +\epsilon;\underline{p}(\tau +\epsilon )) \leq \mathcal{O}(\tau;\underline{p}(\tau +\epsilon ))$ from which we conclude that $\underline{p}(\tau +\epsilon ) \geq \underline{ p}(\tau )$.

Now we prove the right continuity of $\tau \mapsto \underline{p}(\tau )$. Let us fix a τ.

• (i) We first consider the case $D(\underline{p}(\tau )) < \mathcal{O}(\tau;\underline{p}(\tau ))$.

This means that $\underline{p}(\tau )$ is of the form c _ℓ +τ ℓ, for a given ℓ. Then when ε > 0 is small enough, we also have $\underline{p}(\tau +\epsilon ) = c_{\ell} + (\tau +\epsilon )e_{\ell}$. Indeed, D(c _ℓ + (τ +ε)e _ℓ ) ≤ D(c _ℓ +τ e _ℓ ) and for a small enough ε,

$$\displaystyle{\mathcal{O}(\tau;c_{\ell} +\tau e_{\ell}) =\kappa _{\ell} +\sum _{i\neq \ell}\kappa _{i} \mathbb{1}_{\{\tau \leq \frac{c_{\ell}-c_{i}} {1-e_{i}/e_{\ell}} }\} = \mathcal{O}(\tau +\epsilon;c_{\ell} + (\tau +\epsilon )e_{\ell}).}$$

Thus, $D(c_{\ell} + (\tau +\epsilon )e_{\ell}) < \mathcal{O}(\tau +\epsilon;c_{\ell} + (\tau +\epsilon )e_{\ell})$ which implies that $\underline{p}(\tau ) + e_{\ell}\epsilon = c_{\ell} + (\tau +\epsilon )e_{\ell} \geq \underline{ p}(\tau +\epsilon )$ and hence $e_{\ell}\epsilon \geq \underline{ p}(\tau +\epsilon ) -\underline{ p}(\tau ).$

(ii) We consider next the case $D(\underline{p}(\tau )) > \mathcal{O}(\tau;\underline{p}(\tau ))$.

This means that $\underline{p}(\tau ) \in S_{D}$ is at a discontinuity point, say p _d of the demand, $\underline{p}(\tau ) = p_{d}$. Then, for any δ > 0,

$$\displaystyle{D(\underline{p}(\tau )+\delta ) < \mathcal{O}(\tau;\underline{p}(\tau )+\delta ).}$$

But,

$$\displaystyle{\mathcal{O}(\tau;p_{d}+\delta ) =\sum _{ i=1}^{J}\kappa _{ i} \mathbb{1}_{\{\tau \leq \frac{p_{d}+\delta -c_{i}} {e_{i}} \}},}$$

and we can choose δ to be small enough so that $\tau \neq \frac{p_{d}+\delta -c_{i}} {e_{i}}$. Then, for a small enough ε,

$$\displaystyle{D(\underline{p}(\tau )+\delta ) < \mathcal{O}(\tau;\underline{p}(\tau )+\delta ) = \mathcal{O}(\tau +\epsilon;\underline{p}(\tau )+\delta ),}$$

which implies that $\underline{p}(\tau )+\delta \geq \bar{ p}(\tau +\epsilon )$, so we obtain $\delta \geq \underline{ p}(\tau +\epsilon ) -\underline{ p}(\tau ) \geq 0.$ (iii) We consider now the case $D(\underline{p}(\tau )) = \mathcal{O}(\tau;\underline{p}(\tau ))$.

This means that $\underline{p}(\tau ) \in S_{\kappa }$, say $\underline{p}(\tau ) = p_{c}$ Then, for any δ > 0,

$$\displaystyle{D(\underline{p}(\tau )+\delta ) < \mathcal{O}(\tau;\underline{p}(\tau )+\delta ).}$$

But,

$$\displaystyle{\mathcal{O}(\tau;p_{c}+\delta ) =\sum _{ i=1}^{J}\kappa _{ i} \mathbb{1}_{\{\tau \leq \frac{p_{c}+\delta -c_{i}} {e_{i}} \}}}$$

and we can choose δ small enough such that $\tau \neq \frac{p_{c}+\delta -c_{i}} {e_{i}}$. Then, for ε small enough,

$$\displaystyle{D(\underline{p}(\tau )+\delta ) < \mathcal{O}(\tau;\underline{p}(\tau )+\delta ) = \mathcal{O}(\tau +\epsilon;\underline{p}(\tau )+\delta )}$$

which implies that $\underline{p}(\tau )+\delta \geq \underline{ p}(\tau +\epsilon )$, so we get $\delta \geq \underline{ p}(\tau +\epsilon ) -\underline{ p}(\tau ) \geq 0.$ The right-continuity of $\tau \mapsto \bar{p}(\tau )$ follows, by definition as $\bar{p}(\tau )$ is a continuous transformation of $\underline{p}(\tau )$. □

Proof of Lemma 2.

The proof consists in a complete analysis of the entire combination of situations, but each situation is elementary.

Let us suppose the opposite, that is there exists $0 \leq t < t' \leq \mathfrak{p}$ such that the emission levels are $\mathcal{W}(t') > \mathcal{W}(t)$.

We define the function $\tau \mapsto I(\tau )$ valued in the subsets of {1, …, J} that lists the producers in the electricity market producing at tax level τ:

$$\displaystyle{i \in I(\tau )\quad \mbox{ if}\quad \varphi _{i}(\tau ) > 0.}$$

In particular we have for all $\tau \in [0,\mathfrak{p}]$,

$$\displaystyle{\mathcal{W}(\tau ) =\sum _{i\in I(\tau )}e_{i}\varphi _{i}(\tau ).}$$

• (i) We first examine the situation I(t′) = I(t).

  To shorten the expressions, we adopt the following notation

  $$\displaystyle{I(t) = I\quad \mbox{ and }\quad I(t') = I'.}$$

  (i-a) :

  If $\sum _{i\in I}\varphi _{i}(t) = D(t)$ then, from the demand constraint (DC) and the emission levels hypothesis (EH), we have

  $$\displaystyle{ \sum _{i\in I}\varphi _{i}(t) = D(t) \geq D(t') \geq \sum _{i\in I'}\varphi _{i}(t') }$$

  (DC)

  $$\displaystyle{ \sum _{i\in I}\varphi _{i}(t)e_{i} <\sum _{i\in I'}\varphi _{i}(t')e_{i}. }$$

  (EH)

  We denote by $\hat{I}$ the subset of I of index such that $c_{i} + \mathit{te}_{i} =\underline{ p}(t)$. In particular, when $j \in I\setminus \hat{I}$, then $\varphi _{j}(t) =\kappa _{j}$.

  Note that there exists at most one index (say ℓ) in the set $\hat{I} \cap \widehat{ I'}$. If $j \in \hat{ I}\setminus \widehat{I'}$ and $k \in \widehat{ I'}\setminus \hat{I}$, then, by the definition of the sets

  $$\displaystyle{ \begin{array}{ll} c_{j} + e_{j}t = c_{\ell} + e_{\ell}t, &\quad c_{k} + e_{k}t < c_{j} + e_{j}t, \\ c_{j} + e_{j}t' < c_{\ell} + e_{\ell}t', &\quad c_{k} + e_{k}t' = c_{\ell} + e_{\ell}t', \\ c_{j} + e_{j}t' < c_{k} + e_{k}t,&\quad c_{k} + e_{k}t < c_{\ell} + e_{\ell}t,\end{array} }$$

  from which we easily deduce that

  $$\displaystyle{ \max \{e_{j},j \in \hat{ I}\setminus \widehat{I'}\} < e_{\ell} <\min \{ e_{k},k \in \widehat{ I'}\setminus \hat{I}\}. }$$

  (A.2)

  Now we decompose the sets I and I′ in the demand constraint (DC) and the emission levels hypothesis (EH) as follows:

  $$\displaystyle\begin{array}{rcl} & & \sum _{n\in I\setminus \hat{I}\cup \widehat{I'}}\kappa _{n} +\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}}\varphi _{i}(t) +\sum _{k\in \widehat{I'}\setminus \hat{I}}\kappa _{k} \\ & & \qquad \geq \sum _{n\in I\setminus \hat{I}\cup \widehat{I'}}\kappa _{n} +\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}}\kappa _{i} +\sum _{k\in \widehat{I'}\setminus \hat{I}}\varphi _{k}(t'), {}\end{array}$$

  (DC)
  $$\displaystyle\begin{array}{rcl} & & \sum _{n\in I\setminus \hat{I}\cup \widehat{I'}}e_{n}\kappa _{n} + e_{\ell}\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}}e_{i}\varphi _{i}(t) +\sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}\kappa _{k} \\ & & \qquad <\sum _{n\in I\setminus \hat{I}\cup \widehat{I'}}e_{n}\kappa _{n} + e_{\ell}\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}}e_{i}\kappa _{i} +\sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}\varphi _{k}(t'). {}\end{array}$$

  (EH)

  After simplification, we obtain

  $$\displaystyle{ \varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}}\varphi _{i}(t) +\sum _{k\in \widehat{I'}\setminus \hat{I}}\kappa _{k} \geq \varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}}\kappa _{i} +\sum _{k\in \widehat{I'}\setminus \hat{I}}\varphi _{k}(t'), }$$

  (DC)
  $$\displaystyle{ e_{\ell}\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}}e_{i}\varphi _{i}(t) +\sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}\kappa _{k} < e_{\ell}\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}}e_{i}\kappa _{i} +\sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}\varphi _{k}(t'). }$$

  (EH)

  Assume first that $\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}}\varphi _{i}(t) \geq \varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}}\kappa _{i}$. Equivalently, we have

  $$\displaystyle{\varphi _{\ell}(t) -\varphi _{\ell}(t') \geq \sum _{i\in \hat{I}\setminus \widehat{I'}}(\kappa _{i} -\varphi _{i}(t))}$$

  and from (A.2),

  $$\displaystyle{e_{\ell}\left (\varphi _{\ell}(t) -\varphi _{\ell}(t')\right ) \geq \sum _{i\in \hat{I}\setminus \widehat{I'}}e_{i}(\kappa _{i} -\varphi _{i}(t)).}$$

  By combining the above with the emission levels hypothesis (EH), we obtain the following contradiction: $\sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}\kappa _{k} <\sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}\varphi _{k}(t')$.

  Assume now that $\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}}\varphi _{i}(t) <\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}}\kappa _{i}$. Multiplying the demand constraint (DC) by $\hat{e}:=\min \{ e_{k},k \in \widehat{ I'}\setminus \hat{I}\}$, we get

  $$\displaystyle{ \sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}(\kappa _{k} -\varphi _{k}(t')) \geq \hat{ e}\left (\varphi _{\ell}(t) -\varphi _{\ell}(t')\right ) +\hat{ e}\sum _{i\in \hat{I}\setminus \widehat{I'}}(\kappa _{i} -\varphi _{i}(t)). }$$

  But from (EH) and (A.2), we also have

  $$\displaystyle{ \sum _{k\in \widehat{I'}\setminus \hat{I}}e_{k}(\kappa _{k} -\varphi _{k}(t')) < e_{\ell}\left (\varphi _{\ell}(t) -\varphi _{\ell}(t')\right ) + e_{\ell}\sum _{i\in \hat{I}\setminus \widehat{I'}}(\kappa _{i} -\varphi _{i}(t)), }$$

  then

  $$\displaystyle{ 0 \geq (\hat{e} - e_{\ell})\left (\varphi _{\ell}(t) -\varphi _{\ell}(t')\right ) + (\hat{e} - e_{\ell})\sum _{i\in \hat{I}\setminus \widehat{I'}}(\kappa _{i} -\varphi _{i}(t)), }$$

  which contradicts our assumption.

  (i-b) :

  If $\sum _{i\in I}\varphi _{i}(t) < D(t)$ then, for all i ∈ I, $\varphi _{i}(t) =\kappa _{i}$ and (EH) is necessarily false.

• (ii) We examine the situation I(t′) ≠ I(t).

We add the following shortened notation: $I(t) \cap I'(t) = II'$.

We break down I and I′ into the sets II′, $I\setminus I'$ and $I'\setminus I$. We denote by $\hat{I}$ the set of index i ∈ I such that $c_{i} + te_{i} =\underline{ p}(t)$. In particular, when $j \in I\setminus \hat{I}$, then $\varphi _{j}(t) =\kappa _{j}$.

We first derive some generic relations between the emission rates for these.

Among the indexes in the set II′, we observe that at most one index exists (say ℓ) in the set $\hat{I} \cap \widehat{ I'}$. If $j \in \hat{ I}\setminus \widehat{I'}$, if $k \in \widehat{ I'}\setminus \hat{I}$, then, by the definition of the sets

$$\displaystyle{ \begin{array}{ll} c_{j} + e_{j}t = c_{\ell} + e_{\ell}t, &\quad c_{k} + e_{k}t < c_{j} + e_{j}t, \\ c_{j} + e_{j}t' < c_{\ell} + e_{\ell}t', &\quad c_{k} + e_{k}t' = c_{\ell} + e_{\ell}t', \\ c_{j} + e_{j}t' < c_{k} + e_{k}t,&\quad c_{k} + e_{k}t < c_{\ell} + e_{\ell}t,\\ \end{array} }$$

from which, we easily deduce that

$$\displaystyle{ \begin{array}{rl} \hat{e}:=\max \left \{e_{j},j \in II' \cap \left (\hat{I}\setminus \widehat{I'}\right )\right \} < e_{\ell} <\min \left \{e_{k},k \in II' \cap \left (\widehat{I'}\setminus \hat{I}\right )\right \}:=\hat{ e}'. \end{array} }$$

(A.3)

For $j \in I\setminus I'$ and $k \in I'\setminus I$, we have

$$\displaystyle{ c_{j} + e_{j}t < c_{k} + e_{k}t\quad \mbox{ and }\quad c_{j} + e_{j}t' > c_{k} + e_{k}t' }$$

from which, we also easily deduce that

$$\displaystyle{ \max \{e_{k},k \in I'\setminus I\} <\min \{ e_{j},j \in I\setminus I'\}. }$$

(A.4)

For the same j and k, for $(\hat{c},\hat{e})$ representative of index in $II' \cap \hat{ I}\setminus \widehat{I'}$, and $(\hat{c}',\hat{e}')$ representative of index in $II' \cap \widehat{ I'}\setminus \hat{I}$, we also have

$$\displaystyle{ \begin{array}{ll} c_{j} + e_{j}t& \leq \hat{ c} +\hat{ e}t \\ c_{j} + e_{j}t'& >\hat{ c} +\hat{ e}t' \end{array} \quad \mbox{ and }\quad \begin{array}{ll} c_{k} + e_{k}t& >\hat{ c}' +\hat{ e}'t \\ c_{k} + e_{k}t'& \leq \hat{ c}' +\hat{ e}'t' \end{array} }$$

from which, we deduce that

$$\displaystyle{ \begin{array}{rl} \min \{e_{j},j \in I\setminus I'\}& > (e_{\ell},\hat{e}) \vee \max \{ e_{k},k \in I'\setminus I\} \\ \max \{e_{k},k \in I'\setminus I\}& < (e_{\ell},\hat{e}') \wedge \min \{ e_{j},j \in I\setminus I'\}. \end{array} }$$

(A.5)

We divide the analysis in cases. In the first one the demand is fully satisfied for the price p ^elec(t).

(ii-a) :

If $\sum _{i\in I}\varphi _{I}(t) = D(p^{\mathrm{elec}}(t))$,

$$\displaystyle{ \sum _{i\in I\setminus I'}\varphi _{I} +\sum _{i\in II'}\varphi _{i}(t) = D(p^{\mathrm{elec}}(t)) \geq D(p^{\mathrm{elec}}(t')) \geq \sum _{ i\in II'}\varphi _{i}(t') +\sum _{i\in I'\setminus I}\varphi _{i}(t'), }$$

(DC)

$$\displaystyle{ \sum _{i\in I\setminus I'}\varphi _{i}(t)e_{i} +\sum _{i\in II'}\varphi _{i}(t)e_{i} <\sum _{i\in II'}\varphi _{i}(t')e_{i} +\sum _{i\in I'\setminus I}\varphi _{i}(t)e_{i}. }$$

(EH)

We must then examine the following two subcases, relative to the situations where the demand is satisfied or not at the price p ^elec(t′).

(ii-a-1) :

If $\sum _{i\in I'}\varphi _{i}(t') < D(p^{\mathrm{elec}}(t'))$, then $\varphi _{i}(t') =\kappa _{i}$ for all i ∈ I′ and

$$\displaystyle{ \sum _{j\in I\setminus I'}\varphi _{j}(t) +\sum _{i\in II'}\varphi _{i}(t) >\sum _{i\in II'}\kappa _{i} +\sum _{k\in I'\setminus I}\kappa _{k}, }$$

(DC)

$$\displaystyle{ \sum _{j\in I\setminus I'}\varphi _{j}(t)e_{j} +\sum _{i\in II'}\varphi _{i}(t)e_{i} <\sum _{i\in II'}\kappa _{i}e_{i} +\sum _{k\in I'\setminus I}\kappa _{k}e_{k}. }$$

(EH)

As $\varphi _{i}(t) =\kappa _{i}$ when $i \in (I\setminus \hat{I}) \cap II'$, we can simplify the two sides of (DC) and (EH) by the sum over $(I\setminus \hat{I}) \cap II'$. The remaining part of II′ is $\{\ell\}\cup \left (\hat{I}\setminus \widehat{I'} \cap II'\right )$:

$$\displaystyle{ \sum _{j\in I\setminus I'}\varphi _{j}(t) +\varphi _{\ell} +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\varphi _{i}(t) >\kappa _{\ell} +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\kappa _{i} +\sum _{k\in I'\setminus I}\kappa _{k}, }$$

(DC)

$$\displaystyle{ \sum _{j\in I\setminus I'}e_{j}\varphi _{j}(t) + e_{\ell}\varphi _{\ell} +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}e_{i}\varphi _{i}(t) < e_{\ell}\kappa _{\ell} +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}e_{i}\kappa _{i} +\sum _{k\in I'\setminus I}e_{k}\kappa _{k}. }$$

(EH)

Then we multiply (DC) by $\bar{e}:= (e_{\ell},\hat{e}) \vee \max \{ e_{k},k \in I'\setminus I\}$, and we obtain by (A.5)

$$\displaystyle{ \sum _{j\in I\setminus I'}e_{j}\varphi _{j}(t) +\bar{ e}\varphi _{\ell} +\bar{ e}\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\varphi _{i}(t) >\bar{ e}\kappa _{\ell} +\bar{ e}\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\kappa _{i} +\sum _{k\in I'\setminus I}e_{k}\kappa _{k}. }$$

We subtract with (EH):

$$\displaystyle{ (\bar{e} - e_{\ell})\varphi _{\ell} +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\varphi _{i}(t) > (\bar{e} - e_{\ell})\kappa _{\ell} +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\kappa _{i}. }$$

But $\bar{e} \geq e_{\ell}$ when ℓ exists, and $\bar{e} \geq \hat{ e} \geq e_{i}$ for $i \in \hat{ I}\setminus \widehat{I'} \cap II'$. So we obtain our contradiction.

(ii-a-2) :

If $\sum _{i\in I'}\varphi _{i}(t') = D(p^{\mathrm{elec}}(t'))$, then

$$\displaystyle{ \sum _{j\in I\setminus I'}\varphi _{j}(t) +\sum _{i\in II'}\varphi _{i}(t) >\sum _{i\in II'}\varphi _{i}(t') +\sum _{k\in I'\setminus I}\varphi _{k}(t'), }$$

(DC)

$$\displaystyle{ \sum _{j\in I\setminus I'}\varphi _{j}(t)e_{j} +\sum _{i\in II'}\varphi _{i}(t)e_{i} <\sum _{i\in II'}\varphi _{i}(t')e_{i} +\sum _{k\in I'\setminus I}\varphi _{k}(t')e_{k}. }$$

(EH)

We decompose $I\setminus I' = \left (I\setminus (I' \cup \hat{ I})\right ) \cup \hat{ I}\setminus I'$ and $I'\setminus I = \left (I'\setminus (I \cup \widehat{ I'})\right ) \cup \widehat{ I'}\setminus I$:

$$\displaystyle{ \sum _{j\in I\setminus (I'\cup \hat{I})}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}\varphi _{j}(t) +\sum _{i\in II'}\varphi _{i}(t) >\sum _{i\in II'}\varphi _{i}(t') +\sum _{k\in \widehat{I'}\setminus I}\varphi _{k}(t') +\sum _{k\in I'\setminus (I\cup \widehat{I'})}\kappa _{k}, }$$

(DC)

$$\displaystyle\begin{array}{rcl} & & \sum _{j\in I\setminus (I'\cup \hat{I})}e_{j}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}e_{j}\varphi _{j}(t) +\sum _{i\in II'}e_{i}\varphi _{i}(t) \\ & & \qquad \qquad <\sum _{i\in II'}e_{i}\varphi _{i}(t') +\sum _{k\in \widehat{I'}\setminus I}e_{k}\varphi _{k}(t') +\sum _{k\in I'\setminus (I\cup \widehat{I'})}e_{k}\kappa _{k}. {}\end{array}$$

(EH)

We also break down the set $II' = (I \cap I')$:

$$\displaystyle{ II' = \left (II'\cap \{\ell\}\right ) \cup \left (II' \cap \hat{ I}\setminus \widehat{I'}\right ) \cup \left (II' \cap \widehat{ I'}\setminus \hat{I}\right ) \cup \left (I\setminus \hat{I} \cap I'\setminus \widehat{I'})\right ). }$$

$$\displaystyle\begin{array}{rcl} & & \sum _{j\in I\setminus (I'\cup \hat{I}))}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}\varphi _{j}(t) +\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\varphi _{i}(t) +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}\varphi _{i}(t) \\ & & >\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\varphi _{i}(t') +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}\varphi _{i}(t') +\sum _{k\in \widehat{I'}\setminus I}\varphi _{k}(t') +\sum _{k\in I'\setminus (I\cup \widehat{I'})}\kappa _{k}, {}\end{array}$$

(DC)

$$\displaystyle\begin{array}{rcl} & & \sum _{j\in I\setminus (I'\cup \hat{I})}e_{j}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}e_{j}\varphi _{j}(t) + e_{\ell}\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}e_{i}\varphi _{i}(t) +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}e_{i}\varphi _{i}(t) \\ & & <e_{\ell}\varphi _{\ell}(t')+\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}e_{i}\varphi _{i}(t')+\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}e_{i}\varphi _{i}(t')+\sum _{k\in \widehat{I'}\setminus I}e_{k}\varphi _{k}(t')+\sum _{k\in I'\setminus (I\cup \widehat{I'})}e_{k}\kappa _{k}. {}\end{array}$$

(EH)

For index i in the last subset $(I\setminus \hat{I} \cap I'\setminus \widehat{I'})$, we have $\varphi _{i}(t) =\kappa _{i}$ and $\varphi _{i}(t') =\kappa _{i}$, so we simplify (DC) and (EH) from this last subset. Thus,

$$\displaystyle\begin{array}{rcl} & & \sum _{j\in I\setminus (I'\cup \hat{I})}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}\varphi _{j}(t) +\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\varphi _{i}(t) +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}\kappa _{i} \\ & & \qquad >\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\kappa _{i} +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}\varphi _{i}(t') +\sum _{k\in \widehat{I'}\setminus I}\varphi _{k}(t') +\sum _{k\in I'\setminus (I\cup \widehat{I'})}\kappa _{k}, {}\end{array}$$

(DC)

$$\displaystyle\begin{array}{rcl} & & \sum _{j\in I\setminus (I'\cup \hat{I})}e_{j}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}e_{j}\varphi _{j}(t) + e_{\ell}\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}e_{i}\varphi _{i}(t) +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}e_{i}\kappa _{i} \\ & & \qquad <e_{\ell}\varphi _{\ell}(t')+\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}e_{i}\kappa _{i}+\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}e_{i}\varphi _{i}(t')+\sum _{k\in \widehat{I'}\setminus I}e_{k}\varphi _{k}(t')+\sum _{k\in I'\setminus (I\cup \widehat{I'}}e_{k}\kappa _{k}. {}\end{array}$$

(EH)

We multiply (DC) by $\bar{e}:= (e_{\ell},\hat{e}) \vee \max \{ e_{k},k \in I'\setminus I\}$, we get by (A.5)

$$\displaystyle\begin{array}{rcl} & & \sum _{j\in I\setminus (I'\cup \hat{I})}e_{j}\kappa _{j} +\sum _{j\in \hat{I}\setminus I'}e_{j}\varphi _{j}(t) +\bar{ e}\varphi _{\ell}(t) +\bar{ e}\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\varphi _{i}(t) +\bar{ e}\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}\kappa _{i} {}\\ & & >\bar{ e}\varphi _{\ell}(t') +\bar{ e}\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}\kappa _{i} +\bar{ e}\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}\varphi _{i}(t') +\sum _{k\in \widehat{I'}\setminus I}e_{k}\varphi _{k}(t') +\sum _{k\in I'\setminus (I\cup \widehat{I'}}e_{k}\kappa _{k}. {}\\ \end{array}$$

We subtract (EH)

$$\displaystyle\begin{array}{rcl} & & (\bar{e} - e_{\ell})\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\varphi _{i}(t) +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(\bar{e} - e_{i})\kappa _{i} {}\\ & & > (\bar{e} - e_{\ell})\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\kappa _{i} +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(\bar{e} - e_{i})\varphi _{i}(t'). {}\\ \end{array}$$

We arrange the terms

$$\displaystyle\begin{array}{rcl} & & (\bar{e} - e_{\ell})\varphi _{\ell}(t) +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\varphi _{i}(t) +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(\bar{e} - e_{i})\kappa _{i} {}\\ & & > (\bar{e} - e_{\ell})\varphi _{\ell}(t') +\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\kappa _{i} +\sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(\bar{e} - e_{i})\varphi _{i}(t'). {}\\ \end{array}$$

If ℓ exists, then $\bar{e} = e_{\ell}$ and

$$\displaystyle{ \begin{array}{rl} \sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(e_{\ell} - e_{i})\left (\kappa _{i} -\varphi _{i}(t')\right )& >\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(e_{\ell} - e_{i})\left (\kappa _{i} -\varphi _{i}(t)\right ), \\ \sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(e_{\ell} -\hat{ e}')\left (\kappa _{i} -\varphi _{i}(t')\right )& >\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(e_{\ell} -\hat{ e})\left (\kappa _{i} -\varphi _{i}(t)\right ).\end{array} }$$

(A.6)

But $\hat{e} < e_{\ell} <\hat{ e}'$, and the contradiction follows. If ℓ does not exist, then $\bar{e} =\hat{ e} \vee \max \{ e_{k},k \in I'\setminus I\}$

$$\displaystyle{ \begin{array}{rl} \sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(\bar{e} - e_{i})\left (\kappa _{i} -\varphi _{i}(t')\right )& >\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} - e_{i})\left (\kappa _{i} -\varphi _{i}(t)\right ), \\ \sum _{i\in \widehat{I'}\setminus \hat{I}\cap II'}(\bar{e} -\hat{ e}')\left (\kappa _{i} -\varphi _{i}(t')\right )& >\sum _{i\in \hat{I}\setminus \widehat{I'}\cap II'}(\bar{e} -\hat{ e})\left (\kappa _{i} -\varphi _{i}(t)\right ).\end{array} }$$

(A.7)

But $\max \{e_{k},k \in I'\setminus I\} <\hat{ e}'$, and the contradiction follows.

(ii-b) :

If $\sum _{i\in I}\varphi _{i}(t) < D(p^{\mathrm{elec}}(t))$ then for all i ∈ I, $\varphi _{i}(t) =\kappa _{i}$.

(ii-b-1) :

If $\sum _{i\in I'}\varphi _{i}(t') < D(p^{\mathrm{elec}}(t'))$, then $\varphi _{i}(t') =\kappa _{i}$ for all i ∈ I′. Moreover, we have that $\mathcal{O}(t,\underline{p}(t)) \geq D(\underline{p}(t))+\varepsilon ) \geq D(\underline{p}(t')) > \mathcal{O}(t',\underline{p}(t'))$ and (DC)–(EH) becomes

$$\displaystyle{ \sum _{j\in I\setminus I'}\kappa _{j} >\sum _{k\in I'\setminus I}\kappa _{k}, }$$

(DC)

$$\displaystyle{ \sum _{j\in I\setminus I'}e_{j}\kappa _{j} <\sum _{k\in I'\setminus I}e_{k}\kappa _{k}. }$$

(EH)

Then, we multiply (DC) by $\min \{e_{j};j \in I\setminus I'\} \geq \max \{ e_{k};k \in I'\setminus I\}$, and we obtain a contradiction with (EH).

(ii-b-2) :

If $\sum _{i\in I'}\varphi _{i}(t') = D(p^{\mathrm{elec}}(t'))$, we go back to the analysis of the case (ii-a-2), with the main difference that all quantities $\varphi _{i}(t)$ are now equal to κ _i . We go to inequalities (A.6) and (A.7) which are simplified as the right-had sides are now zero. The contradiction follows with the same arguments

⊓ ⊔