Network Expansion to Mitigate Market Power

Abstract

Constrained transmission capacity in electricity networks may give generators the possibility to game the market by specifically causing congestion and thereby appropriating excessive rents. Investment in network capacity can ameliorate such behavior by reducing the potential for strategic behavior. However, modeling Nash equilibria between generators, which explicitly account for their impact on the network, is mathematically and computationally challenging. We propose a three-stage model to describe how network investment can reduce market power exertion: a benevolent planner decides on network upgrades for existing lines anticipating the gaming opportunities by strategic generators. These firms, in turn, anticipate their impact on market-clearing prices and grid congestion. In this respect, we provide the first model endogenizing the trade-off between the costs of grid investment and benefits from reduced market power potential in short-run market clearing. In a numerical example using a three-node network, we illustrate three distinct effects: firstly, by reducing market power exertion, network expansion can yield welfare gains beyond pure efficiency increases. Anticipating gaming possibilities when planning network expansion can push welfare close to a first-best competitive benchmark. Secondly, network upgrades entail a relative shift of rents from producers to consumers when congestion rents were excessive. Thirdly, investment may yield suboptimal or even disequilibrium outcomes when strategic behavior of certain market participants is neglected in network planning.

Appendices

Appendix A: Proof of Proposition 1

First, we derive the Langrangian dual problem for a quadratic problem with a positive definite matrix in a general form:

$$\begin{array}{@{}rcl@{}} \min_{x,y} \quad \tfrac{1}{2} x^{T} D_{x} x + c^{T} \left[\begin{array}{ll} x \\y \end{array}\right] \quad \text{s.t.} \quad A \left[\begin{array}{ll} x \\y \end{array}\right] \leq b \quad (\lambda), \quad B \left[\begin{array}{ll} x \\y \end{array}\right] = e \quad (\nu) \end{array} $$

(10)

where x is the vector of variables that enter the quadratic part of the objective function, and y is the vector of variables that only appear in linear terms. By assumption, D _x is symmetric and strictly positive definite.

The Lagrangian (primal) function is then:

$$\begin{array}{@{}rcl@{}} \mathcal{L}(x,y,\lambda,\nu) \,=\, \tfrac{1}{2} x^{T} D_{x} x \!+ c^{T} \left[\begin{array}{ll} x \\y \end{array}\right] \,+\, \left( A \left[\begin{array}{ll} x \\y \end{array}\right] \!- b\right)^{T} \lambda \,+\, \left( B \left[\begin{array}{ll} x \\y \end{array}\right] \,-\, e\right)^{T} \nu \end{array} $$

(11)

Lemma 1

The Lagrangian dual function for (11) is given by

$$\begin{array}{@{}rcl@{}} \mathcal{\hat{L}}(\lambda,\nu,x,y)\! & = &\!- \tfrac{1}{2} \left( c_{x} + {A_{x}^{T}} \lambda \,+\, {B_{x}^{T}} \nu \right)^{T} (D_{x})^{-1} \left( c_{x} + {A_{x}^{T}} \lambda + {B_{x}^{T}} \nu \right) \,-\, b^{T} \lambda - e^{T} \nu \end{array} $$

(12a)

$$\begin{array}{@{}rcl@{}} \text{where} \quad && D_{x} x + c_{x} + \left( A_{x} \right)^{T} \lambda + \left( B_{x} \right)^{T} \nu = 0, \end{array} $$

(12b)

$$\begin{array}{@{}rcl@{}} && c_{y} + \left( A_{y} \right)^{T} \lambda + \left( B_{y}\right)^{T} \nu = 0, \end{array} $$

(12c)

where subscripts indicate that parts of matrices that are multiplied with the respective variable vector.

Proof

The Lagrange dual function is defined as the infimum of the Lagrange primal function over the primal decision variables⁸. Differentiating (11) with respect to x and y yields

$$\begin{array}{@{}rcl@{}} \frac{\partial \mathcal{L}}{\partial x} &=& D_{x} x + c_{x} + \left( A_{x} \right)^{T} \lambda + \left( B_{x} \right)^{T} \nu = 0 \end{array} $$

(13a)

$$\begin{array}{@{}rcl@{}} & \Rightarrow& x^{*} = - \left( D_{x}\right)^{-1} \left( c_{x} + \left( A_{x} \right)^{T} \lambda + {B_{x}^{T}} \nu \right) \end{array} $$

(13b)

$$\begin{array}{@{}rcl@{}} \frac{\partial \mathcal{L}}{\partial y} &= &c_{y} + \left( A_{y} \right)^{T} \lambda + {B_{y}^{T}} \nu = 0 \end{array} $$

(13c)

Note that $\mathcal {L}$ is strictly convex in x and y such that the first order equality conditions are necessary and sufficient to ensure a global minimum. Plugging optimality condition (13b) into the (primal) Lagrangian (11)

$$\begin{array}{@{}rcl@{}} \mathcal{L}&=&\tfrac{1}{2} \left[\left[(-D_{x}^{-1})(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right]^{T}D_{x}\left[(-D_{x}^{-1})(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right]\right] \\ &&+{c_{x}^{T}}\left[(-D_{x}^{-1})(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right]+{c_{y}^{T}}y \\ &&+\left[A_{x}\left[(-D_{x}^{-1})(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right]+A_{y}y-b\right]^{T}\lambda \\ &&+\left[B_{x}\left[(-D_{x}^{-1})(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right]+B_{y}y-e\right]^{T}\nu \end{array} $$

(13d)

$$\begin{array}{@{}rcl@{}} &=&\tfrac{1}{2}\left[(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)^{T}D^{-1}_{x}(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right] \\ &&-\left[c_{x}^{T}D_{x}^{-1}(c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)+(c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)^{T}D_{x}^{-1}{A_{x}^{T}}\lambda \,+\, (c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)^{T}D_{x}^{-1}{B_{x}^{T}}\nu \right] \\ &&+c_{y}^{T}y+y^{T}{A_{y}^{T}}\lambda + +y^{T}{B_{y}^{T}}\nu \,-\, b^{T}\lambda - e^{T}\nu \end{array} $$

(13e)

where we have made use of $(D_{x}^{-1})^{T}=D_{x}^{-1}$ due to symmetry of D _x . Now observe that from (13c) ${A^{T}_{y}}\lambda + {B^{T}_{y}}\nu =-c_{y}$ and obtain

$$\begin{array}{@{}rcl@{}} \mathcal{L}&=&\tfrac{1}{2}\left[(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)^{T}D^{-1}_{x}(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right] \\ &&\,-\,\left[c_{x}^{T}D_{x}^{-1}(c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)\,+\,(c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)^{T}D_{x}^{-1}{A_{x}^{T}}\lambda \,+\, (c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)^{T}D_{x}^{-1}{B_{x}^{T}}\nu \right] \\ &&+{c_{y}^{T}}y-\!y^{T}c_{y} - b^{T}\lambda - e^{T}\nu \end{array} $$

(13f)

The inner products ${c_{y}^{T}}y$ and y ^T c _y in the last row of (13f) are of dimension (1×1), thus necessarily symmetric and cancel out. Moreover, straightforward algebra yields that the two terms in brackets in the first and second line of (13f) are identical. Therefore,

$$\begin{array}{@{}rcl@{}} \mathcal{\hat{L}}(\lambda,\nu)\,=\,-\tfrac{1}{2}\left[(c_{x}\,+\,{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)^{T}D_{x}^{-1}(c_{x}\,+\,{A_{x}^{T}}\lambda + {B_{x}^{T}} \nu)\right]\!\,-\,b^{T}\lambda \!- e^{T}\nu \end{array} $$

(13g)

Now, we can set up the dual Lagrangian problem, consisting in maximizing $\hat {L}$ over λ and ν, while optimality conditions concerning x and y are added as constraints. By construction, further constraints comprise weak positivity of decision variables attached to inequality constraints

$$\begin{array}{@{}rcl@{}} \max_{\lambda,\nu} \quad && \!\!-\tfrac{1}{2}\left[(c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)^{T}D_{x}^{-1}(c_{x}\,+\,{A_{x}^{T}}\lambda \,+\, {B_{x}^{T}} \nu)\right]\!\,-\,b^{T}\lambda \,-\, e^{T}\nu \end{array} $$

(14a)

$$\begin{array}{@{}rcl@{}} \text{s.t.} \qquad && D_{x} x + c_{x} + \left( A_{x} \right)^{T} \lambda + \left( B_{x} \right)^{T} \nu = 0, \end{array} $$

(14b)

$$\begin{array}{@{}rcl@{}} && c_{y} + \left( A_{y} \right)^{T} \lambda + \left( B_{y}\right)^{T} \nu = 0 \end{array} $$

(14c)

$$\begin{array}{@{}rcl@{}} && \lambda \geq 0, \, \nu \in \mathbb{R} \end{array} $$

(14d)

Next, we apply this result to our problem. Rewriting the primal problem (1a) - (1h) in matrix notation in the fashion of (10) yields for the objective function:

$$\begin{array}{@{}rcl@{}} \min_{g^{F},d,\delta,\mathfrak{D}} - \sum\limits_{n \in \mathfrak{N}} \left[\left( a_{n}-\tfrac{1}{2} b_{n} d_{n}\right) d_{n}\right] - \sum\limits_{s \in \mathfrak{F}} {c^{G}_{s}}{g^{F}_{s}} {\hat{=}} \min_{g^{F},d,\delta,\mathfrak{D}} \, \tfrac{1}{2}d^{T}{\Delta} d + \gamma^{T} \left[\begin{array}{lll} d \\ g^{F} \\ \delta \end{array}\right] \\ \end{array} $$

(15a)

where the variable vectors are composed as follows:

$$\begin{array}{@{}rcl@{}} \underset{(\tilde{N}\times1)}{d}= \left[\begin{array}{lllll} d_{1} \\ {\vdots} \\ d_{\tilde{N}} \end{array}\right], \underset{(\tilde{F}\times1)}{g^{F}}= \left[\begin{array}{lllll} {g^{F}_{1}} \\ {\vdots} \\ g^{F}_{\tilde{F}} \end{array}\right], \underset{(N\times1)}{\delta} = \left[\begin{array}{lllll} \delta_{1} \\ {\vdots} \\ \delta_{N} \end{array}\right] \end{array} $$

(15b)

and the parameter matrices as⁹:

$$\begin{array}{@{}rcl@{}} \underset{(\tilde{N} \times \tilde{N})}{\Delta} \,=\, \left[\begin{array}{ccc} b_{1} & {\ldots} & 0 \\ {\vdots} & {\ddots} & {\vdots} \\ 0 & {\ldots} & b_{\tilde{N}} \end{array} \right],~~\underset{((\tilde{N}+\tilde{F}+N)\times 1)}{\gamma}\,=\,\!\left[-a_{1} {\ldots} \,-\,a_{\tilde{N}} \quad {c_{1}^{G}} {\ldots} c_{\tilde{F}}^{G} \quad 0 {\ldots} 0 \right]^{T} \\ \end{array} $$

(15c)

For convenience, let $\left |\mathfrak {N}\right | = \tilde {N}$, and $\left |\mathfrak {F}\right | = \tilde {F}$ render the number of nodes at which demand is located, and number of fringe plants respectively. Moreover, observe that Δ fulfills our assumptions of symmetry and strict positive definiteness. Expressing constraints (1b) - (1h) in matrix notation:

$$\begin{array}{@{}rcl@{}} A \left[\begin{array}{lll} d \\ g^{F} \\ \delta \end{array}\right] \leq b \quad (\lambda) \quad &{\hat{=}} \quad \left[A_{d} : A_{g^{F}\delta}\right]\left[\begin{array}{lll} d \\ g^{F} \\ \delta \end{array}\right] \leq b \quad (\lambda), \end{array} $$

(15d)

$$\begin{array}{@{}rcl@{}} B \left[\begin{array}{lll} d \\ g^{F} \\ \delta \end{array}\right] = e \quad (\nu) \quad &{\hat{=}} \quad \left[B_{d} : B_{g^{F}\delta}\right]\left[\begin{array}{lll} d \\ g^{F} \\ \delta \end{array}\right] = e \quad (\nu) \end{array} $$

(15e)

with sub-matrices $A_{d},A_{g^{F}\delta },B_{d},B_{g^{F}\delta }$ capturing the relevant entries for the respective constraints, and accordingly composed vectors b,e,λ, and ν. We now just have to plug the matrices and vectors into the formulation of the dual Lagrangian problem (14a) - (14d). Straightforward algebra yields for the expression $(c_{x}+{A_{x}^{T}}\lambda + {B_{x}^{T}}\nu )$:

$$\begin{array}{@{}rcl@{}} \left[c_{x}\,+\,{A_{x}^{T}}\lambda + {B_{x}^{T}}\nu\right] \!{\hat{=}}\! \left[\left[\begin{array}{c}-a_{1} \\ {\vdots} \\ -a_{\tilde{N}} \end{array} \right] \,+\, {A_{d}^{T}}\lambda + {B_{d}^{T}}\nu \right] \,=\, \left[\begin{array}{c}-a_{1} + p_{1} - \phi_{1} \\ {\vdots} \\ -a_{\tilde{N}} + p_{\tilde{N}} - \phi_{\tilde{N}} \end{array} \right] \end{array} $$

(16a)

obviously rendering the first term of the dual objective:

$$\begin{array}{@{}rcl@{}} & -&\tfrac{1}{2}\left[(c_{x}+{A^{T}_{x}}\lambda + {B^{T}_{x}}\nu)^{T}D_{x}^{-1}(c_{x}+{A^{T}_{x}}\lambda+ {B^{T}_{x}}\nu)\right] \\ &&{\hat{=}} -\tfrac{1}{2}\left[\left[\begin{array}{c}-a_{1} + p_{1} - \phi_{1} \\ {\vdots} \\ -a_{\tilde{N}} + p_{\tilde{N}} - \phi_{\tilde{N}} \end{array} \right] \right]^{T}{\Delta}^{-1}\left[\left[\begin{array}{c}-a_{1} + p_{1} - \phi_{1} \\ {\vdots} \\ -a_{\tilde{N}} + p_{\tilde{N}} - \phi_{\tilde{N}} \end{array} \right]\right] \\ &&{\hat{=}} -\tfrac{1}{2}\sum\limits_{n \in \mathfrak{N}}\tfrac{1}{b_{n}}\left( a_{n}-p_{n}+\phi_{n}\right)^{2} \end{array} $$

(16b)

For the second term of (14a) it immediately follows:

$$\begin{array}{@{}rcl@{}} b^{T}\lambda + e^{T}\nu =\sum\limits_{l}(\bar{f}_{l}+e_{l})(\bar{\mu}_{l}+\underline{\mu}_{l}) + \sum\limits_{n} p_{n}\left( \sum\limits_{s \in \mathfrak{S}_{n}}{g^{S}_{s}}\right) + \sum\limits_{s \in \mathfrak{F}}\beta_{s}\bar{g}^{F}_{s} \end{array} $$

(16c)

Bringing (16b) and (16c) together thus yields the dual objective. The constraint set of the dual is abstractly given by:

$$\begin{array}{@{}rcl@{}} D_{x} x &+& c_{x} + \left( A_{x} \right)^{T} \lambda + \left( B_{x} \right)^{T} \nu {\hat{=}} {\Delta} d - \left[a_{1} \; {\ldots} \; a_{\tilde{N}}\right] + \left( A_{d}\right)^{T} \lambda + \left( B_{d}\right)^{T} \nu = 0, \end{array} $$

(16d)

$$\begin{array}{@{}rcl@{}} c_{y}&+&{A_{y}^{T}}\lambda + {B_{y}^{T}}\nu {\hat{=}} \left[c_{1}^{G} \; {\ldots} \; c_{\tilde{F}}^{G} \quad 0 \; {\ldots} \; 0 \right]^{T} + A_{g^{F}\delta}^{T}\lambda + B_{g^{F}\delta}^{T}\nu =0, \end{array} $$

(16e)

cf. (14b) - (14c). Plugging in the relevant matrices reproduces KKT conditions (2a) - (2c) of the original problem:

$$\begin{array}{@{}rcl@{}} {c^{G}_{s}} - p_{n, s \in \mathfrak{S}_{n}} + \beta_{s}-\psi_{s} &=& 0 \quad \forall s \in \mathfrak{F} \end{array} $$

(16f)

$$\begin{array}{@{}rcl@{}} - a_{n} + b_{n} d_{n} + p_{n} - \phi_{n} &=& 0 \quad \forall n \in \mathfrak{N} \end{array} $$

(16g)

$$\begin{array}{@{}rcl@{}} \sum\limits_{k} B_{kn}p_{n} + \sum\limits_{l} H_{ln} \left( \overline{\mu}_{l} - \underline{\mu}_{l}\right) + \genfrac\{\}{0pt}{0}{\gamma \quad \text{if }n = \hat{n}}{0 \quad \text{else} \hspace{0.55 cm}} &= &0 \quad \forall n \end{array} $$

(16h)

Therefore, the equivalent Lagrangian dual problem to (1a) - (1h) is given by

$$\begin{array}{@{}rcl@{}} \max_{p,\overline{\mu},\underline{\mu},\beta, \psi, \phi, \gamma} &-& \tfrac{1}{2} \sum\limits_{n \in \mathfrak{N}} \frac{1}{b_{n}} \left[a_{n}- p_{n} + \phi_{n}\right]^{2} - \sum\limits_{l}(\bar{f}_{l}+e_{l})(\bar{\mu}_{l}+\underline{\mu}_{l}) \\ &-& \sum\limits_{n} p_{n}\left( \sum\limits_{s \in \mathfrak{S}_{n}}{g^{S}_{s}}\right)- \sum\limits_{s \in \mathfrak{F}}\beta_{s}\bar{g}^{F}_{s} \end{array} $$

(17a)

$$\begin{array}{@{}rcl@{}} \text{s.t.}&& (16\mathrm{f}), (16\mathrm{g}), (16\mathrm{h}) \end{array} $$

(17b)

$$\begin{array}{@{}rcl@{}} && \bar{\mu},\underline{\mu},\beta,\psi,\phi \geq 0, \, p,\gamma \in \mathbb{R} \end{array} $$

(17c)

□

Appendix B: KKT conditions of the strategic players

The following equations and inequalities constitute the set of KKT conditions for all strategic players. Each vector satisfying Eqs. (18a–19l) constitutes a stationary point of the spot market equilibrium problem. First all KKT conditions without complementarity requirements.

$$\begin{array}{@{}rcl@{}} -p_{n, s \in \mathfrak{S}_{n}} + {c^{G}_{s}} + {\beta^{S}_{s}} - {\psi^{S}_{s}} - \iota^{S}_{ni, s \in \mathfrak{S}_{n}, s \in \mathfrak{S}_{i}}\\ + \xi^{S}_{i, s \in \mathfrak{S}_{i}}p_{n, s \in \mathfrak{S}_{n}} &= 0 & \quad \forall \ s \notin \mathfrak{F} \end{array} $$

(18a)

$$\begin{array}{@{}rcl@{}} -\iota^{S}_{ni, s \in \mathfrak{S}_{n}} + \beta^{SF}_{si} - \psi^{SF}_{si} + {\xi_{i}^{S}}\left( {c^{G}_{s}} + \beta_{s}\right) &= 0 & \quad \forall \ s \in \mathfrak{F},i \end{array} $$

(18b)

$$\begin{array}{@{}rcl@{}} -\sum\limits_{s \in \left( \mathfrak{S}_{n} \cap \mathfrak{S}_{i}\right)}{g^{S}_{s}} - \sum\limits_{s \in \left( \mathfrak{S}_{n} \cap \mathfrak{F}\right)}\zeta^{S}_{si} + \eta^{S}_{ni} + \sum\limits_{k}\theta^{S}_{ki}B_{nk} && \\ + {\xi^{S}_{i}}\tfrac{1}{b_{n}}\left( p_{n} - a_{n} - \phi_{n}\right) + {\xi^{S}_{i}}\sum\limits_{s \in \mathfrak{S}_{n}}{g^{S}_{s}} &= 0& \quad \forall \ n,i \end{array} $$

(18c)

$$\begin{array}{@{}rcl@{}} b_{n}\eta^{S}_{ni} + \iota^{S}_{ni} - \phi^{S}_{ni} - {\xi^{S}_{i}}\left( a_{n} - b_{n}d_{n}\right) &= 0 & \quad \forall \ n,i \end{array} $$

(18d)

$$\begin{array}{@{}rcl@{}} \sum\limits_{k}B_{kn}\iota_{ki}^{S} + \sum\limits_{l}H_{ln}\left( \overline{\mu}_{li}^{S} - \underline{\mu}_{li}^{S}\right) + \genfrac\{\}{0pt}{0}{{\gamma^{S}_{i}} \quad \text{if }n = \hat{n}}{0 \quad \text{else} \hspace{0.55 cm}} &= 0 & \quad \forall \ n,i \end{array} $$

(18e)

$$\begin{array}{@{}rcl@{}} \zeta^{S}_{si} + {\xi^{S}_{i}}\overline{g}^{F}_{s} &\geq 0 & \quad \forall \ s \in \mathfrak{F}, i \end{array} $$

(18f)

$$\begin{array}{@{}rcl@{}} -\zeta^{S}_{si} &\geq 0 & \quad \forall \ s \in \mathfrak{F}, i \end{array} $$

(18g)

$$\begin{array}{@{}rcl@{}} -\eta^{S}_{ni} + {\xi^{S}_{i}}\left[\tfrac{1}{b_{n}}\left( \phi_{n} + a_{n} - p_{n}\right)\right] &\geq 0 & \quad \forall \ n,i \end{array} $$

(18h)

$$\begin{array}{@{}rcl@{}} \sum\limits_{n}\theta_{ni}^{S} H_{ln} + {\xi^{S}_{i}}\left( \bar{f}_{l}+e_{l}\right) &\geq 0 & \quad \forall \ l,i \end{array} $$

(18i)

$$\begin{array}{@{}rcl@{}} -\sum\limits_{n}\theta_{ni}^{S}H_{ln} + {\xi^{S}_{i}}\left( \bar{f}_{l}+e_{l}\right) &\geq 0 & \quad \forall \ l,i \end{array} $$

(18j)

$$\begin{array}{@{}rcl@{}} \theta_{\hat{n}i}^{S} &= 0 & \quad \forall \ i \end{array} $$

(18k)

$$\begin{array}{@{}rcl@{}} \overline{g}^{S}_{s} - &\geq 0 & \quad \forall \ s \notin \mathfrak{F} \end{array} $$

(18l)

$$\begin{array}{@{}rcl@{}} {g^{S}_{s}} &\geq 0 & \quad \forall \ s \notin \mathfrak{F} \end{array} $$

(18m)

$$\begin{array}{@{}rcl@{}} {c^{G}_{s}} - p_{n, s \in \mathfrak{S}_{n}} + \beta_{s} - \psi_{s} &= 0 & \quad \forall \ s \in \mathfrak{F}, i \end{array} $$

(18n)

$$\begin{array}{@{}rcl@{}} - a_{n} + b_{n} d_{n} + p_{n} - \phi_{n} &= 0 & \quad \forall \ n \in \mathfrak{N},i \end{array} $$

(18o)

$$\begin{array}{@{}rcl@{}} \sum\limits_{k} B_{kn} p_{k} + \sum\limits_{l} H_{ln} \left( \overline{\mu}_{l} - \underline{\mu}_{l}\right) + \genfrac\{\}{0pt}{0}{\gamma \quad \text{if }n = \hat{n}}{0 \quad \text{else} \hspace{0.55 cm}} &= 0 & \quad \forall \ n,i \end{array} $$

(18p)

$$\begin{array}{@{}rcl@{}} -\sum\limits_{s \in \mathfrak{S}_{n}} \left( {g^{S}_{s}} + {g^{F}_{s}}\right) + \sum\limits_{k} B_{nk} \delta_{k} + d_{n} &=0 & \quad \forall \ n,i \end{array} $$

(18q)

$$\begin{array}{@{}rcl@{}} \overline{f}_{l} + e_{l} - \sum\limits_{k} H_{lk} \delta_{k} &\geq 0 & \quad \forall \ l,i \end{array} $$

(18r)

$$\begin{array}{@{}rcl@{}} \overline{f}_{l} + e_{l} + \sum\limits_{k} H_{lk} \delta_{k} &\geq 0 & \quad \forall \ l,i \end{array} $$

(18s)

$$\begin{array}{@{}rcl@{}} \overline{g}^{F}_{s} - {g^{F}_{s}} &\geq 0 & \quad \forall \ s \in \mathfrak{F}, i \end{array} $$

(18t)

$$\begin{array}{@{}rcl@{}} {g^{F}_{S}} &\geq 0 & \quad \forall \ s \in \mathfrak{F}, i \end{array} $$

(18u)

$$\begin{array}{@{}rcl@{}} d_{n} &\geq 0 & \quad \forall \ n,i \end{array} $$

(18v)

$$\begin{array}{@{}rcl@{}} \delta_{\hat{n}} &= 0 & \quad \forall \ i \end{array} $$

(18w)

Below, we list all complementarity requirements which are replaced by disjunctive constraints.

$$\begin{array}{@{}rcl@{}} \sum\limits_{i} \left( \zeta^{S}_{si} + {\xi^{S}_{i}}\overline{g}^{F}_{s}\right) \!&\leq& r^{\beta}_{s}K^{\beta}_{s}, \quad \beta_{s} \leq \left( 1\,-\,r^{\beta}_{s}\right)K^{\beta}_{s} \quad \forall \ s \in \mathfrak{F} \end{array} $$

(19a)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i}\left( -\zeta^{S}_{si}\right) \!&\leq& r^{\psi}_{s}K^{\psi}_{s}, \quad \psi_{s} \leq \left( 1-r^{\psi}_{s}\right)K^{\psi}_{s} \quad \forall \ s \in \mathfrak{F} \end{array} $$

(19b)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i}\left( -\eta^{S}_{ni} + {\xi^{S}_{i}}\left[\tfrac{1}{b_{n}}\left( \phi_{n} + a_{n} - p_{n}\right)\right]\right) \!&\leq& r^{\phi}_{n}K^{\phi}_{n}, \quad \phi_{n} \leq \left( 1-r^{\phi}_{n}\right)K^{\phi}_{n} \quad \forall \ n \end{array} $$

(19c)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i}\left( \sum\limits_{n}\theta_{ni}^{S} H_{ln} \!+ {\xi^{S}_{i}}\left( \bar{f}_{l}+e_{l}\right)\right) \!&\leq& r^{\overline{\mu}}_{l}K^{\overline{\mu}}_{l}, \quad \overline{\mu}_{l} \leq \left( 1-r^{\overline{\mu}}_{l}\right)K^{\overline{\mu}}_{l} \quad \forall \ l \end{array} $$

(19d)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i}\left( \sum\limits_{n}\theta_{ni}^{S} H_{ln} + {\xi^{S}_{i}}\left( \bar{f}_{l}+e_{l}\right)\right) \!&\leq& r^{\underline{\mu}}_{l}K^{\underline{\mu}}_{l}, \quad \underline{\mu}_{l} \leq \left( 1-r^{\underline{\mu}}_{l}\right)K^{\underline{\mu}}_{l} \quad \forall \ l \end{array} $$

(19e)

$$\begin{array}{@{}rcl@{}} \overline{g}^{S}_{s} - {g^{S}_{s}} \!&\leq &r^{\beta^{S}}_{s}K^{\beta^{S}}_{s}, \quad {\beta^{S}_{s}} \leq \left( 1-r^{\beta^{S}}_{s}\right)K^{\beta^{S}}_{s} \quad \forall \ s \notin \mathfrak{F} \end{array} $$

(19f)

$$\begin{array}{@{}rcl@{}} {g^{S}_{s}} \!&\leq& r^{\psi^{S}}_{s}K^{\psi^{S}}_{s}, \quad {\psi^{S}_{s}} \leq \left( 1- r^{\psi^{S}}_{s}\right)K^{\psi^{S}}_{s} \quad \forall \ s \notin \mathfrak{F} \end{array} $$

(19g)

$$\begin{array}{@{}rcl@{}} \overline{f}_{l} + e_{l} - \sum\limits_{k} H_{lk} \delta_{k} \!&\leq& r^{\overline{\mu}^{S}}_{l}K^{\overline{\mu}^{S}}_{l}, \quad \sum\limits_{i}\overline{\mu}^{S}_{li} + \overline{\mu}_{l} \!\leq\! \left( 1\!-r^{\overline{\mu}^{S}}_{l}\right)K^{\overline{\mu}^{S}}_{l} \quad \forall \ l \end{array} $$

(19h)

$$\begin{array}{@{}rcl@{}} \overline{f}_{l} + e_{l} - \sum\limits_{k} H_{lk} \delta_{k} \!&\leq&\! r^{\underline{\mu}^{S}}_{l}K^{\underline{\mu}^{S}}_{l}, \quad \sum\limits_{i}\underline{\mu}^{S}_{li} + \underline{\mu}_{l} \leq \left( 1\!-r^{\underline{\mu}^{S}}_{l}\right)K^{\underline{\mu}^{S}}_{l} \quad \forall \ l \end{array} $$

(19i)

$$\begin{array}{@{}rcl@{}} \overline{g}^{F}_{s} - {g^{F}_{s}} &\leq& r^{\beta^{SF}}_{s}K^{\beta^{SF}}_{s}, \quad \sum\limits_{i}\beta^{SF}_{si} + \beta_{s} \leq \left( 1-r^{\beta^{SF}}_{s}\right)K^{\beta^{SF}}_{s} \quad \forall \ s \in \mathfrak{F} \end{array} $$

(19j)

$$\begin{array}{@{}rcl@{}} {g^{F}_{s}} &\!\leq\!& r^{\psi^{SF}}_{s}K^{\psi^{SF}}_{s}, \quad \sum\limits_{i} \psi^{SF}_{si} \,+\, \psi_{s} \!\leq\! \left( 1\!-r^{\psi^{SF}}_{s}\right)K^{\psi^{SF}}_{s} \quad \forall \ s \in \mathfrak{F} \end{array} $$

(19k)

$$\begin{array}{@{}rcl@{}} d_{n} &\leq& r^{\phi^{S}}_{n}K^{\phi^{S}}_{n}, \qquad \sum\limits_{i}\phi^{S}_{ni} + \phi_{n} \leq \left( 1-r^{\phi^{S}}_{n}\right)K^{\phi^{S}}_{n} \quad \forall \ n \end{array} $$

(19l)