Tacit collusion games in pool-based electricity markets under transmission constraints

Abstract

Harrington et al. (Math Program Ser B 104:407–435, 2005) introduced a general framework for modeling tacit collusion in which producing firms collectively maximize the Nash bargaining objective function, subject to incentive compatibility constraints. This work extends that collusion model to the setting of a competitive pool-based electricity market operated by an independent system operator. The extension has two features. First, the locationally distinct markets in which firms compete are connected by transmission lines. Capacity limits of the transmission lines, together with the laws of physics that guide the flow of electricity, may alter firms’ strategic behavior. Second, in addition to electricity power producers, other market participants, including system operators and power marketers, play important roles in a competitive electricity market. The new players are included in the model in order to better represent real-world markets, and this inclusion will impact power producers’ strategic behavior as well. The resulting model is a mathematical program with equilibrium constraints (MPEC). Properties of the specific MPEC are discussed and numerical examples illustrating the impacts of transmission congestion in a collusive game are presented.

Appendices

Appendix 1: Static Nash equilibrium models

1.1 Exogenous-ISO model

For each power producer $f \in \mathcal F$, again let $X_f$ denote its generic feasible production region. Then $f$ solves the following optimization problem.

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathop {\text{ maximize }}_{g_f\in X_f}&\pi _f(g_f)=\displaystyle \sum _{i\in \mathcal N }\left[ P_i^0 - \displaystyle \frac{P_i^0}{Q_i^0}\left( \displaystyle \sum _{t\in \mathcal F }g_{ti}+y_{i}\right) \right] g_{fi} - \displaystyle \sum _{i \in \mathcal N }C_{fi}(g_{fi}) \end{array}\qquad \end{aligned}$$

(23)

The ISO solves the following optimization problem.

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathop {\text{ maximize }}_{y} &{} \displaystyle \sum _{i\in \mathcal N }\left[ \int \limits _0^{y_i + G_i}p_i(\tau _i)d\tau _i - \sum _{f\in \mathcal F }C_{fi}(g_{fi})\right] \\ \text{ subject } \, \text{ to }&{} \displaystyle \sum _{i\in \mathcal N }y_i = 0 \\ &{} \displaystyle \sum _{i\in \mathcal N }\mathrm PTDF _{ki}y_i \le T_{k},\quad \forall \; k \in \mathcal A . \end{array} \end{aligned}$$

(24)

By writing out the the KKT conditions of each firm’s and the ISO’s optimization problem, together with the market clearing condition, we obtain a complementarity problem (CP).

1.2 Endogenous-ISO model

For each power producer $f \in \mathcal F$, it solves the following optimization problem.

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathop {\text{ maximize }}_{g_f,\ y} &{} \pi _f(g_f, y)=\displaystyle \sum _{i\in \mathcal N }\left[ P_i^0 - \displaystyle \frac{P_i^0}{Q_i^0}\left( \displaystyle \sum _{t\in \mathcal F }g_{ti} + y_{i}\right) \right] g_{fi} - \displaystyle \sum _{i \in \mathcal N }C_{fi}(g_{fi})\\ \text{ subject } \, \text{ to }&{} g_{f}\in X_f,\\ &{} \mathrm ISO \left\{ \begin{array}{ll} \displaystyle \mathop {\text{ maximize }}_{y} &{} \displaystyle \sum _{i\in \mathcal N }\left[ \int \limits _0^{y_i + G_i}p_i(\tau _i)d\tau _i - \sum _{f\in \mathcal F }C_{fi}(g_{fi})\right] \\ \text{ subject } \, \text{ to }&{} \displaystyle \sum _{i\in \mathcal N }y_i = 0 \\ &{} \displaystyle \sum _{i\in \mathcal N }\mathrm PTDF _{ki}y_i \le T_{k},\quad \forall \; k \in \mathcal A \end{array} \right\} .\\ \end{array} \end{aligned}$$

(25)

As each firm’s problem is an MPEC, by grouping all firms’ problems together, we obtain an equilibrium problem with equilibrium constraints (EPECs).

Appendix 2: Algebraic reformulation of the collusion model

This appendix provides the derivation of the model (18) from (17). The key idea is that certain variables in the optimality conditions of the ISO’s optimization problem are redundant. For the ease of argument, we re-provide the optimality conditions in the following.

$$\begin{aligned}&y\ \mathrm free , \qquad P^0 - B(G + y) - \mu \mathbf 1 - \mathrm PTDF ^T\lambda = 0 \end{aligned}$$

(26)

$$\begin{aligned}&\mu \ \mathrm free , \qquad \mathbf 1 ^T y = 0 \end{aligned}$$

(27)

$$\begin{aligned}&0\le \lambda \perp \mathrm PTDF ^T\lambda - T \le 0, \end{aligned}$$

(28)

which is derived under the affine inverse demand function assumption: $p(d) = P^0 - Bd$.

The purpose of the following derivation is to derive explicit expressions of $y$ and $\mu$ with respect to $G$ and $\lambda$, hence eliminating the need to keep the (redundant) variables $y$ and $\mu$ in the model. Without loss of generality, we assume that node $N$ is designated as the hub node, and introduce the following notations. Let $\breve{P}^0, \breve{G}$ and $\breve{y}$ denote the $\mathfrak R ^{N-1}$ vectors excluding the $N$th component of $P^0, G$ and $y$, respectively. Also let $P_{N}^0, Q_{N}^0$ and $G_{N}$ denote the $N$th component of the vectors $P^0, Q^0$, and $G$, respectively. Further use $\breve{B}$ to denote the $(N-1)\times (N-1)$ matrix resulting from deleting the $N$th row and column of $B$, and $\breve{\mathbf{1 }}$ to represent an $(N-1)$ vector of 1’s. Notice that by two equations in (26) and (27) we can have the following linear system of $\mu$ and $\breve{y}$ with $G$ and $\lambda$ as parameters:

$$\begin{aligned} \breve{P}^0-\breve{B}(\breve{G}+\breve{y})&= \mu \breve{\mathbf{1 }} + \mathrm PTDF ^T\lambda \end{aligned}$$

(29)

$$\begin{aligned} P_N^0-\displaystyle \frac{P_N^0}{Q_N^0}(G_N-\breve{\mathbf{1 }}^T\breve{y})&= \mu . \end{aligned}$$

(30)

where Eq. (29) is the same as in (26) for all the nodes other than the hub node; while Eq. (30) is derived using (26) at the hub node (where $\mathrm PTDF _{kN}$ for each $k\in \mathcal A$ is 0) and Eq. (27). Rewriting the linear system (29) and (30) into a matrix form yields the following.

$$\begin{aligned} \left[ \begin{array}{cc} \breve{B} &{} \breve{\mathbf{1 }} \\ \displaystyle - \breve{\mathbf{1 }}^T &{} \frac{Q_N^0}{P_N^0} \\ \end{array} \right] \left[ \begin{array}{c} \breve{y} \\ \mu \\ \end{array} \right] = \left[ \begin{array}{c} \breve{P}^0 - \breve{B}\breve{G} - \mathrm PTDF ^T\lambda \\ Q^0_N - G_N \\ \end{array} \right] . \end{aligned}$$

(31)

Recall that matrix $\breve{B} = \mathrm Diag (b_i)$ is an $(N - 1) \times (N - 1)$ diagonal matrix, with $b_i = P_i^0/Q_i^0,\ i = 1, \ldots , N - 1$. Under the assumption that $P_i^0 > 0$ and $Q_i^0 > 0$ for each $i = 1, \ldots , N,\, \breve{B}$ is positive definite, and so is the skew-symmetric coefficient matrix in (31)

$$\begin{aligned} \left[ \begin{array}{ll} \breve{B} &{} \breve{\mathbf{1 }} \\ \displaystyle - \breve{\mathbf{1 }}^T &{} \frac{Q_N^0}{P_N^0} \\ \end{array} \right] . \end{aligned}$$

Hence, the coefficient matrix is nonsingular, and the linear system (31) has a unique solution. Let $H\in \mathfrak R ^{2L \times N}$ denote the $\mathrm PTDF$ matrix, and $\breve{H}\in \mathfrak R ^{2L \times (N-1)}$ be the $H$ matrix with its $N$th column removed. Then the solution of (31) can be expressed as follows.

$$\begin{aligned} \mu&= \kappa - \omega \mathbf 1 ^TG - \breve{\rho }^T\breve{H}^T\lambda \end{aligned}$$

(32)

$$\begin{aligned} \breve{y}&= \breve{R}^0 + \breve{\varUpsilon }G - \breve{\varXi }\breve{H}^T\lambda , \end{aligned}$$

(33)

where

$$\begin{aligned}&\omega =\displaystyle \frac{1}{\sum _{i = 1}^N\frac{Q_i^0}{P_i^0}},\ \kappa =\omega \sum _{i=1}^N Q_i^0,\\&\breve{\rho }=\left[ \begin{array}{l}\rho _1\\ \vdots \\ \rho _{N-1}\end{array}\right] \in \mathfrak R ^{N-1}\ \mathrm with \\&\rho _i=\omega \displaystyle \frac{Q_i^0}{P_i^0},\ \breve{R}^0=\left[ \begin{array}{c} R^0_1\\ \vdots \\ R^0_{N-1}\end{array}\right] \in \mathfrak R ^{N-1} \ \mathrm with \ R_i^0=Q_i^0-\displaystyle \frac{Q_i^0}{P_i^0}\kappa ,\\&\breve{\varUpsilon }\in \mathfrak R ^{(N-1)\times N} \ \mathrm with \ \breve{\varUpsilon }_{ij}=\left\{ \begin{array}{cc} \omega \displaystyle \frac{Q_i^0}{P_i^0}-1,&{} i=j\\ \omega \displaystyle \frac{Q_i^0}{P_i^0},&{}i\ne j,\\ \end{array} \right. \\&\breve{\varXi }\in \mathfrak R ^{(N-1)\times (N-1)} \ \mathrm with \ \breve{\varXi }_{ij}=\left\{ \begin{array}{cc} \displaystyle \frac{Q_i^0}{P_i^0}(1-\rho _i),&{} i=j\\ -\omega \displaystyle \frac{Q_i^0}{P_i^0}\displaystyle \frac{Q_j^0}{P_j^0},&{}i\ne j.\\ \end{array} \right. \end{aligned}$$

With (32) and (33), the three conditions in the ISO’s optimality conditions (26)–(28) can be condensed into one single complementarity constraint:

$$\begin{aligned} 0\le \lambda \perp (T-H\breve{R}^0)-H\breve{\varUpsilon }G+H\breve{\varXi }H^T\lambda \ge 0. \end{aligned}$$

Let $r := T - H\breve{R}^0 \in \mathfrak R ^{2L}, \varDelta := -H\breve{\varUpsilon }\in \mathfrak R ^{2L \times N}$ and $M := H\breve{\varXi }H^T \in \mathfrak R ^{2L\times 2L}$. The above complementarity system can then be written as

$$\begin{aligned} 0 \le \lambda \perp r + \varDelta G + M\lambda \ge 0. \end{aligned}$$

(34)

Furthermore, with (33) and Eq. (27), each firm’s payoff function $\pi _f(g_f, y) = p(G + y)^Tg_f - C_f(g_f)$ can be re-written as a function of $(G, \lambda )$ as follows

$$\begin{aligned} \pi _f(g, \lambda ) = \kappa \mathbf 1 ^T g_f-\omega G^T E g_f+\lambda ^T \varDelta g_f - C_f(g_f),\quad \forall \; f\in \mathcal F , \end{aligned}$$

(35)

which is exactly Eq. (19). Hence, we have completed the derivation of (18) from (17).

Proof of Lemma 1

Given the assumptions in Lemma 1, the matrix $\breve{\varXi }$ is strongly diagonally dominant by the fact that for each $i = 1, \ldots , N - 1$,

$$\begin{aligned} \varXi _{ii} = \displaystyle \frac{Q_i^0}{P_i^0}(1 - \rho _i) = \frac{Q_i^0}{P_i^0}\left( 1 - \omega \frac{Q_i^0}{P_i^0}\right) > \frac{Q_i^0}{P_i^0}\left( 1 - \omega \frac{Q_i^0}{P_i^0} - \omega \frac{Q_N^0}{P_N^0}\right) = \sum _{i \ne j = 1}^{N - 1} |\varXi _{ij}|. \end{aligned}$$

Since $\breve{\varXi }$ is also symmetric and has positive diagonal entries, it is then a positive definite matrix. Consequently, $M \!=\! H \breve{\varXi } H^T$ is a symmetric, positive semi-definite matrix.$\quad \square$