\newcommand{\thistitle}{Lecture 9: Complex Markets} \input{title.tex} \section{Optimisation Revision} \begin{frame} \frametitle{Optimisation problem} We have an \alert{objective function} $f: \R^k \to \R$ \begin{equation*} \max_{x} f(x) \end{equation*} [$x = (x_1, \dots x_k)$] subject to some \alert{constraints} within $\R^k$: \begin{align*} g_i(x) & = c_i \hspace{1cm}\leftrightarrow\hspace{1cm} \l_i \hspace{1cm} i = 1,\dots n \\ h_j(x) & \leq d_j \hspace{1cm}\leftrightarrow\hspace{1cm} \m_j \hspace{1cm} j = 1,\dots m \end{align*} $\l_i$ and $\m_j$ are the \alert{KKT multipliers} we introduce for each constraint equation; they measure the change in the objective value of the optimal solution obtained by relaxing the constraints (for this reason they are also called \alert{shadow prices}). \end{frame} \begin{frame} \frametitle{KKT conditions} The \alert{Karush-Kuhn-Tucker (KKT) conditions} are necessary conditions that an optimal solution $x^*,\m^*,\l^*$ always satisfies (up to some regularity conditions): \begin{enumerate} \item \alert{Stationarity}: For $l = 1,\dots k$ \begin{equation*} \frac{\d \cL}{\d x_l} = \frac{\d f}{\d x_l} - \sum_i \l_i^* \frac{\d g_i}{\d x_l} - \sum_j \m_j^* \frac{\d h_j}{\d x_l} = 0 \end{equation*} \item \alert{Primal feasibility}: \begin{align*} g_i(x^*) & = c_i \\ h_j(x^*) &\leq d_j \end{align*} \item \alert{Dual feasibility}: $\m_j^* \geq 0$ \item \alert{Complementary slackness}: $\m_j^* (h_j(x^*) - d_j) = 0$ \end{enumerate} \end{frame} \begin{frame} \frametitle{min/max and signs} If the problem is a \alert{maximisation} problem (e.g. welfare maximisation), then \alert{$\m_j^* \geq 0$} since $\m_j = \frac{\d \cL}{\d d_j}$ and if we increase $d_j$ in the constraint $h_j(x) \leq d_j$, then the feasible space can only get bigger. Since if $X \subseteq X'$ \begin{equation*} \max_{x\in X} f(x ) \leq \max_{x\in X'} f(x) \end{equation*} then the objective value at the optimum point can only get bigger, and thus $\m_j^* \geq 0$. (If $d_j \to \infty$ then the constraint is no longer binding, if $d_j \to -\infty$ then the feasible space vanishes.) If however the problem is a \alert{minimisation} problem (e.g. cost minimisation) then we can use \begin{equation*} \min_{x\in X} f(x) = - \max_{x\in X} \left[ -f(x)\right] \end{equation*} We can keep our definition of the Lagrangian and almost all the KKT conditions, but we have a change of sign \alert{$\m_j^* \leq 0$}, since \begin{equation*} \min_{x\in X} f(x ) \geq \min_{x\in X'} f(x) \end{equation*} The $\l_i^*$ also change sign. \end{frame} \section{Welfare maximisation revision} \begin{frame}{KKT and Welfare Maximisation 1/2} Apply KKT now to maximisation of total economic welfare: \begin{align*} \max_{\{d_b\}, \{g_s\}} f(\{d_b\}, \{g_s\}) = \left[ \sum_b U_b (d_b) - \sum_s C_s (g_s) \right] \end{align*} subject to the balance constraint: \begin{align*} g(\{d_b\}, \{g_s\}) = \sum_b d_b - \sum_s g_s = 0 \hspace{1cm} \leftrightarrow \hspace{1cm} \l \end{align*} and any other constraints (e.g. limits on generator capacity, etc.). Our optimisation variables are $\{x\} = \{d_b\} \cup \{g_s\}$. We get from KKT stationarity at the optimal point: \begin{align*} 0 & = \frac{\d f}{\d d_b} - \l^* \frac{\d g}{\d d_b} = U_b'(d_b^*) - \l^* = 0 \\ 0 & = \frac{\d f}{\d g_s} - \l^* \frac{\d g}{\d g_s} = -C_s'(g_s^*) + \l^* = 0 \end{align*} \end{frame} \begin{frame}{KKT and Welfare Maximisation 2/2} So at the optimal point of maximal total economic welfare we get the same result as if everyone maximises their own welfare separately based on the price $\l^*$: \begin{align*} U_b'(d_b^*) = \l^* \\ C_s'(g_s^*) = \l^* \end{align*} This is the CENTRAL result of microeconomics. If we have further inequality constraints that are binding (e.g. capacity constraints), then these equations will receive additions with $\m_i^* > 0$. \end{frame} \section{Optimise Single Node with Linear Generation Costs and Demand Utility} \begin{frame}[fragile] \frametitle{Simplified world: linear generation costs, linear demand utility} We will now turn to a simpler world: all the generator cost functions are linear \begin{align*} C_s(g_s) = o_s g_s \end{align*} and each generator has limited output $0 \leq g_s \leq G_s$. The marginal cost function is a constant $C_s'(g_s) = o_s$. The quantity $G_s$ and marginal cost $o_s$ define a \alert{supply offer}. All the consumer utility functions are also linear \begin{align*} U_b(d_b) = v_b d_b \end{align*} and each consumer has limited consumption $0 \leq d_b \leq D_b$. The marginal utility function is a constant $U_b'(d_b) = v_b$. The quantity $D_b$ and marginal utility $v_b$ define a \alert{demand bid}. \end{frame} \begin{frame}{Supply-demand linear example: generator offers} Example from Kirschen and Strbac pages 56-58. The following generators offer into the market for the hour between 0900 and 1000 on 20th April 2016: \ra{1.1} \begin{table}[!t] \begin{tabular}{lrr} \toprule Company & Quantity [MW] & Marginal cost [\$/MWh]\\ \midrule Red & 200 & 12 \\ Red & 50 & 15 \\ Red & 150 & 20 \\ Green & 150 & 16 \\ Green & 50 & 17 \\ Blue & 100 & 13 \\ Blue & 50 & 18 \\ \bottomrule \end{tabular} \end{table} \end{frame} \begin{frame}{Supply-demand linear example: consumer bids} The following consumers make bids for the same period: \ra{1.1} \begin{table}[!t] \begin{tabular}{lrr} \toprule Company & Quantity [MW] & Marginal utility [\$/MWh]\\ \midrule Yellow & 50 & 13 \\ Yellow & 100 & 23 \\ Purple & 50 & 11 \\ Purple & 150 & 22 \\ Orange & 50 & 10 \\ Orange & 200 & 25 \\ \bottomrule \end{tabular} \end{table} \end{frame} \begin{frame}{Supply-demand example: Curve} If the bids and offers are stacked up in order, the supply and demand curves meet with a demand of 450~MW at a system marginal price of $\l^* = 16$~\$/MWh. \centering \includegraphics[width=10cm]{supply-bid-example} \source{Kirschen \& Strbac} \end{frame} \begin{frame}{Supply-demand example: Revenue and Expenses} Dispatch and revenue/expense of each company: \ra{1.1} \begin{table}[!t] \begin{tabular}{l|rr|rr} \toprule Company & Production& Consumption & Revenue& Expense\\ & [MWh] & [MWh] & [\$] & [\$] \\ \midrule Red & 250 & & 4000 & \\ Blue & 100 & & 1600 & \\ Green & 100 & & 1600 & \\ Orange && 200 && 3200 \\ Yellow && 100 && 1600 \\ Purple && 150 && 2400 \\ \midrule Total & 450 & 450 & 7200 & 7200 \\ \bottomrule \end{tabular} \end{table} \end{frame} \begin{frame}[fragile] \frametitle{Simplified world: linear generation costs, perfectly inelastic demand} For the analysis of the KKT equations, we will simplify even further. We consider a single demand bid of volume $D$ so that the demand does not respond to price changes (i.e. the demand is \alert{perfectly inelastic}) up to a very high marginal utility $v >> o_s\,\, \forall s$, i.e. \begin{align*} U(d) = vd \end{align*} for $d\leq D$. $v$ is sometimes called the \alert{Value Of Lost Load (VOLL)}. \end{frame} \begin{frame}[fragile] \frametitle{Simplify representation of consumers and generators} In this case we get for our welfare maximisation: \begin{align*} \max_{d, \{g_s\} }\left[ vd - \sum_s o_s g_s \right] \end{align*} subject to: \begin{align*} d - \sum_s g_s = 0 \hspace{1cm} & \leftrightarrow \hspace{1cm} \l \\ d \leq D \hspace{1cm} & \leftrightarrow \hspace{1cm} \m \\ g_s \leq G_s \hspace{1cm}& \leftrightarrow\hspace{1cm} \bar{\m}_s \\ - g_s \leq 0 \hspace{1cm}& \leftrightarrow\hspace{1cm} \ubar{\m}_s \end{align*} \end{frame} \begin{frame}{Simplest example: one generator type, perfectly inelastic demand} Suppose all generators have the same marginal cost $o$ and we represent their total dispatch by $g$ and total capacity by $G$ \begin{align*} \max_{d, g }\left[ vd - og\right] \end{align*} such that: \begin{align*} d - g & = 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \l \\ d & \leq D \hspace{1cm} \leftrightarrow \hspace{1cm} \m \\ g & \leq G \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m} \\ - g & \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m} \end{align*} \end{frame} \begin{frame}{Simplest example: one generator type, perfectly inelastic demand} If $D < G$ then since $v >> o$, it will be always be welfare-maximising to dispatch to satisfy the load, i.e. \begin{equation*} g^* = d^* = D \end{equation*} If the demand is non-zero then since $g^* > 0$ by complementarity we have $\ubar{\m}^* = 0$. Since $D < G$ then $g^* < G$ and by complementarity we have $\bar{\m}^* = 0$. To compute $\l^*$ we use stationarity: \begin{align*} 0 & = \frac{\d \cL}{\d g } = \frac{\d f}{\d g} - \sum_i \l_i^* \frac{\d g_i}{\d g} - \sum_j \m_j^* \frac{\d h_j}{\d g} = - o + \l^* - \bar{\m}^* + \ubar{\m}^* \end{align*} Thus $\l^* = o$, which is the cost per unit of supplying extra demand. The \alert{generator sets the price}. There is no generator short-term profit and a large consumer surplus. For the load $\m^*$ can be non-zero because $d^*=D$: \begin{align*} 0 & = \frac{\d \cL}{\d d} = \frac{\d f}{\d d} - \sum_i \l_i^* \frac{\d g_i}{\d d} - \sum_j \m_j^* \frac{\d h_j}{\d d} = v - \l^* - \m^* \end{align*} $\m^* = v - \l^*$ is the marginal benefit of each increase in demand. \end{frame} \begin{frame}{Simplest example: one generator type, perfectly inelastic demand} For the case $D < G$: \includegraphics[width=12cm]{supply-bigger-demand-single.pdf} \end{frame} \begin{frame}{Simplest example: one generator type, perfectly inelastic demand} If $D > G$ then the generator will dispatch up to its maximum capacity \begin{equation*} g^* = d^* = G \end{equation*} For its lower limit we have $\ubar{\m}^* = 0$. From stationarity: \begin{align*} 0 & = \frac{\d \cL}{\d g } = \frac{\d f}{\d g} - \sum_i \l_i^* \frac{\d g_i}{\d g} - \sum_j \m_j^* \frac{\d h_j}{\d g} = - o + \l^* - \bar{\m}^* + \ubar{\m}^* \end{align*} Thus $\l^* = o + \bar{\m}^*$. To find $\l^*$ we have to look at the demand: \begin{align*} 0 & = \frac{\d \cL}{\d d} = \frac{\d f}{\d d} - \sum_i \l_i^* \frac{\d g_i}{\d d} - \sum_j \m_j^* \frac{\d h_j}{\d d} = v - \l^* - \m^* \end{align*} Since $d^* < D$, $\m^* = 0$, $\l^* = v$ and thus $\bar{\m}^* = v-o$, which is the marginal benefit of increasing the generator capacity $G$. The \alert{demand sets the price}. There is no consumer surplus and the generator makes a large profit. $\bar{\m}^*$ is the \alert{inframarginal rent}, i.e. the difference between the market price and the generator's marginal cost. It is also know as the \alert{contribution margin} towards paying the fixed costs of the generator. \end{frame} \begin{frame}{Simplest example: one generator type, perfectly inelastic demand} For the case $D > G$: \includegraphics[width=12cm]{demand-bigger-supply.pdf} \end{frame} \begin{frame}{Next simplest example: several generators, fixed demand} Suppose we have several generators with dispatch $g_s$ and strictly ordered operating costs $o_s$ such that $o_s < o_{s+1}$. We now maximise \begin{align*} \max_{\{d, g_s\}} \left[ vd - \sum_s o_s g_s \right] \end{align*} such that \begin{align*} d - \sum_s g_s & = 0 \hspace{1cm} \leftrightarrow \hspace{1cm} \l \\ d & \leq D \hspace{1cm} \leftrightarrow \hspace{1cm} \m \\ g_s & \leq G_s \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_s \\ - g_s & \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_s \end{align*} \end{frame} \begin{frame}{Next simplest example: several generators, fixed demand} Stationarity gives us for each generator $g_s$: \begin{align*} 0 = \frac{\d \cL}{\d g_s} = -o_s + \l^* - \bar{\m}^*_s + \ubar{\m}^*_s \end{align*} and from complementarity we get \begin{align*} \bar{\m}_s(g_s^* - G_s) = 0 \hspace{2cm} \ubar{\m}_sg_s^* = 0 \end{align*} We can see by inspection that we will dispatch the cheapest generation first. Suppose that we have enough generation for the demand, i.e. $D < \sum_s G_s$. [If $D > \sum_s G_s$ we have the same situation as for a single generator, i.e. $\l^* = v$, so that the demand sets the price.] Find the generator $m$ on the margin where the supply curve intersects with the demand $D$, i.e. the $m$ where $\sum_{s=1}^{m-1} G_s < D < \sum_{s=1}^{m} G_s$. For $s \leq m-1$ we have $g_s^* = G_s$, $\ubar{\m}^*_s = 0$, $\bar{\m}^*_s = \l^* - o_s$. $\bar{\m}^*_s$ are the \alert{inframarginal rents}. For $s = m$ we have $g_m^* = D - \sum_{s=1}^{m-1} G_s$ to cover what's left of the demand. Since $0 < g_m^* < G_m$ we have $\ubar{\m}^*_m = \bar{\m}^*_m = 0$ and thus $\l^* = o_m$. \end{frame} \begin{frame}{Next simplest example: several generators, fixed demand} Specific example of two generators with $G_1 = 300$~MW, $G_2 = 400$~MW, $o_1 = 10$~\euro/MWh, $o_2 = 30$~\euro/MWh and $D = 500$~MW. In this case $m=2$, $g_1^* = G_1 = 300$~MW, $g_2^* = d - G_1 = 200$~MW, $\l^* = o_2$, $\ubar{\m}_i = 0$, $\bar{\m}_2 = 0$ and $\bar{\m}_1 = o_2 - o_1$. \centering \includegraphics[width=8cm]{supply-demand-two.pdf} \end{frame} \begin{frame}{From welfare maximisation to cost minimisation} For the case $D > \sum_s G_s$ we can instead imagine that the demand is rigidly fixed to $D$ and that instead we have a dummy generator with dispatch $g_d = D-\sum_s G_s$ that represents \alert{load shedding}. In this case we can substitute $d = D - g_d$ to get \begin{align*} \max_{\{g_d, g_s\}} \left[ vD - vg_d - \sum_s o_s g_s \right] \end{align*} such that \begin{align*} D - g_d - \sum_s g_s & = 0 \hspace{1cm} \leftrightarrow \hspace{1cm} \l \\ g_s & \leq G_s \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_s \\ - g_s & \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_s \end{align*} Since $vD$ is a constant, we can use $\max_{x\in X} \left[ -f(x)\right] = - \min_{x\in X} f(x)$ to recast this as a minimisation of the total generator costs, absorbing $g_d$ into the set $\{g_s\}$. The constant $vD$ is dropped. \end{frame} \begin{frame}{From welfare maximisation to cost minimisation} We have turned the maximisation of total welfare into \alert{cost minimisation}: \begin{align*} \min_{\{g_s\}} \sum_s o_s g_s \end{align*} such that: \begin{align*} \sum_s g_s - d & = 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \l \\ g_s & \leq G_s \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_s \\ - g_s & \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_s \end{align*} The most expensive generator has $o_s = v$ and $G_s = \infty$ and represents \alert{load shedding}. We've replaced the symbol $D$ with $d$ for simplicity going forward ($d$ is now a constant). NB: Because the signs of the KKT multipliers change when we go from maximisation to minimisation, we've also changed the sign of the balance constraint to keep the marginal price $\l$ positive. \end{frame} \section{Optimise nodes in a network} \begin{frame}[fragile] \frametitle{Welfare optimisation for several nodes in a network} Now let's suppose we have several nodes $i$ with different loads and different generators, with flows $f_\ell$ in the network lines $\ell$. Now we have additional optimisation variables $f_\ell$ AND additional constraints for welfare maximisation: \begin{align*} \max_{\{d_{i,b}\},\{g_{i,s}\},\{f_\ell\}}\left[\sum_{i,b} U_{i,b}(d_{i,b}) - \sum_{i,s} C_{i,s} (g_{i,s}) \right] \end{align*} such that demand is met either by generation or by the network at each node $i$ \begin{align*} \sum_{b} d_{i,b} - \sum_{s} g_{i,s} + \sum_\ell K_{i\ell}f_\ell = 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \l_i \end{align*} Note there is now a \alert{market price for each node}. As before, generator constraints are satisified \begin{align*} g_{i,s} & \leq G_{i,s} \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_{i,s} \\ - g_{i,s} & \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_{i,s} \end{align*} \end{frame} \begin{frame}[fragile] \frametitle{Linear cost minimisation at several nodes in a network} For cost minimisation we have a fixed load $d_i$ at each node, and absorb load-shedding above a value $v$ into a dummy generator. Now we minimise over $f_\ell$ and $g_{i,s}$ for the case of linear cost functions: \begin{align*} \min_{\{g_{i,s}\},\{f_\ell\}} \sum_{i,s} o_{i,s} g_{i,s} \end{align*} such that demand is met either by generation or by the network at each node $i$ \begin{align*} \sum_{s} g_{i,s} - d_i = \sum_\ell K_{i\ell}f_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \l_i \end{align*} and generator constraints are satisified \begin{align*} g_{i,s} & \leq G_{i,s} \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_{i,s} \\ - g_{i,s} & \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_{i,s} \end{align*} \end{frame} \begin{frame}[fragile] \frametitle{Several generators at different nodes in a network} In addition we have constraints on the line flows. First, they have to satisfy Kirchoff's Voltage Law around each closed cycle $c$: \begin{align*} \sum_{c} C_{\ell c} x_\ell f_\ell = 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \l_c \end{align*} and in addition the flows cannot overload the thermal limits, $|f_\ell| \leq F_\ell$ \begin{align*} f_\ell \leq F_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_\ell \\ - f_\ell \leq F_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_\ell \end{align*} \end{frame} \begin{frame}[fragile] \frametitle{Simplest example: two nodes connected by a single line} At node 1 we have demand of $d_1 = 100$~MW and a generator with costs $o_1 = 10$~\euro/MWh and a capacity of $G_1 = 300$~MW. At node 2 we have demand of $d_2 = 100$~MW and a generator with costs $o_2 = 20$~\euro/MWh and a capacity of $G_2 = 300$~MW. What happens if the capacity of the line connecting them is $F = 0$? What about $F = 50$~MW? What about $F = \infty$? \end{frame} \begin{frame}[fragile] \frametitle{Simplest example: two nodes connected by a single line} \centering \begin{circuitikz} \draw [ultra thick] (1,13) node[anchor=south]{1} -- (4,13); \draw (2.5,13) -- +(0,0.5) to [short,i^=$|f| \leq F$] +(5,0.5) -- +(5,0); \draw [ultra thick] (6,13) node[anchor=south]{2} -- (9,13); \draw (1.5,13) -- +(0,-0.5) node[sground]{}; \draw (3,12) node[vsourcesinshape, rotate=270](V1){} (V1.left) -- +(0,0.6); \draw (1.5,11) node{$d_1$}; \draw (3,11) node{$g_{1}$}; \draw (6.5,13) -- +(0,-0.5) node[sground]{}; \draw (8,12) node[vsourcesinshape, rotate=270](V2){} (V2.left) -- +(0,0.6); \draw (6.5,11) node{$d_2$}; \draw (8,11) node{$g_{2}$}; \end{circuitikz} \centering \begin{columns}[T] \begin{column}{5cm} \centering $g_1 - d_1 = f \quad \leftrightarrow \quad \l_1$ \hspace{1cm} $d_1 =$ 100~MW $G_1 =$ 300~MW $o_1 = 10$~\euro/MWh \end{column} \begin{column}{5cm} \centering $g_2 - d_2 = -f \quad \leftrightarrow \quad \l_2$ \hspace{1cm} $d_2 =$ 100~MW $G_2 =$ 300~MW $o_2 = 20$~\euro/MWh \end{column} \end{columns} \end{frame} \begin{frame}[fragile] \frametitle{Simplest example: two nodes connected by a single line} Out optimisation problem has objective function: \begin{equation*} \min_{g_1,g_2,f} \left[o_1 g_1 + o_2 g_2 \right] \end{equation*} subject to the following constraints: \begin{align*} g_1 - d_1 = f \hspace{1cm}\leftrightarrow\hspace{1cm} \l_1 \\ g_2 - d_2 = -f \hspace{1cm}\leftrightarrow\hspace{1cm} \l_2 \\ g_1 \leq G_1 \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_1 \\ - g_1 \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_1 \\ g_2 \leq G_2 \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_2 \\ - g_2 \leq 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_2 \\ f \leq F \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m} \\ - f \leq F \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m} \end{align*} \end{frame} \begin{frame}[fragile] \frametitle{Two nodes: Case $F = 0$ } For the case $F = 0$ the nodes are like two separated islands, $f^* = 0$. The generator on each island provides the demand separately, so: \begin{equation*} g_1^* = d_1 \hspace{2cm} \textrm{and} \hspace{2cm} g_2^* = d_2 \end{equation*} Neither generator has any binding constraints, since in each case the demand (100~MW) is less than the generator capacity (300~MW), so \begin{equation*} \bar{\m}_1^* = \ubar{\m}_1^* = \bar{\m}_2^* = \ubar{\m}_2^* = 0 \end{equation*} From stationarity for each site we get \begin{equation*} 0 = \frac{\d \cL}{\d g_{i}} = o_i - \l_i^* - \bar{\m}_i^* + \ubar{\m}_i^* \end{equation*} Thus we have at each site $\l_i^* = o_i$, as if we had optimised the nodes separately. \end{frame} \begin{frame}[fragile] \frametitle{Two nodes: Case $F =$ 50~MW } For the case $F = $50~MW the cheaper node 1 will export to the more expensive node 2 as much as the restricted capacity $F$ allows: \begin{equation*} f^* = F = 50\textrm{ MW} \end{equation*} Generator 1 covers 50~MW of the demand from node 2: \begin{equation*} g_1^* = d_1+f^* = 150\textrm{ MW} \hspace{2cm} \textrm{and} \hspace{2cm} g_2^* = d_2 - f^* = 50\textrm{ MW} \end{equation*} Neither generator has any binding constraints, so \begin{equation*} \bar{\m}_1^* = \ubar{\m}_1^* = \bar{\m}_2^* = \ubar{\m}_2^* = 0 \end{equation*} and thus we have again different prices at each $\l_i^* = o_i$. For the flow: \begin{equation*} 0 = \frac{\d \cL}{\d f} = 0 + \l_1^* - \l_2^* - \bar{\m}^* + \ubar{\m}^* \end{equation*} Only the upper limit is binding, so we get $\ubar{\m}^* = 0$ and $\bar{\m}^* = \l_1^* - \l_2^* = o_1 - o_2 = -10 $ \euro/MWh. $\bar{\m}^*$ is the cost reduction if we expand the transmission capacity $F$ by $\varepsilon$, allowing us to substitute some of the expensive generation at node 2 with cheap generation from node 1. \end{frame} \begin{frame}[fragile] \frametitle{Two nodes: Case $F = \infty$ } For the case $F = \infty$ we have unrestricted capacity, so it is like merging the two nodes to one node. Now all the demand is covered by the cheapest node: \begin{equation*} f^* = d_2 = 100\textrm{ MW} \end{equation*} Generator 1 covers all the demand: \begin{equation*} g_1^* = d_1+d_2 = 200\textrm{ MW} \hspace{2cm} \textrm{and} \hspace{2cm} g_2^* = 0 \end{equation*} Only generator 2 has a non-zero KKT multiplier, so at node 1 we have $\l_1^* =o_1$ and at node 2 we have: \begin{equation*} \ubar{\m}_2^* = \l_2^* - o_2 \end{equation*} From KKT for the flow $f$ we have no constraints so $\bar{\m}^* = \ubar{\m}^* = 0$ and from stationarity \begin{equation*} 0 = \frac{\d \cL}{\d f} = 0 + \l_1^* - \l_2^* - \bar{\m}^* + \ubar{\m}^* \end{equation*} i.e. $\l_1^* = \l_2^*$. We have price equalisation, as if it were a single node. \end{frame} \begin{frame}{Two node: demand payments versus generation revenue} Now let's compare for our examples what each demand pays $\l_i^*d_i$ and what each generator receives as revenue $\l_i^*g_i^*$ from each market. \ra{1.1} \begin{table}[!t] \begin{tabular}{rrrrrrrrr} \toprule Case & $\l_1^*$ & $\l_2^*$ & $\l_1^*d_1$ & $\l_2^*d_2$ & $\sum_i \l_i^*d_i$ & $\l_1^*g_1^*$ & $\l_2^*g_2^*$ & $\sum_i \l_i^*g_i^*$ \\ & [\euro/MWh] & [\euro/MWh] & [\euro/h] & [\euro/h] & [\euro/h] & [\euro/h] & [\euro/h] & [\euro/h] \\ \midrule $F = 0$ & 10 & 20 & 1000 & 2000 & 3000 & 1000 & 2000 & 3000 \\ $F = 50$ & 10 & 20 & 1000 & 2000 & 3000 & 1500 & 1000 & 2500 \\ $F = \infty$ & 10 & 10 & 1000 & 1000 & 2000 & 2000 & 0 & 2000 \\ \bottomrule \end{tabular} \end{table} NB: In the case with $F=50$, total demand payments are 3000~\euro/h, whereas the generators are only receiving 2500~\euro/h. Where is the missing money (500~\euro/h) going? Answer: to the network operator for service of doing arbitrage, buying low and selling high. \end{frame} \begin{frame}{Congestion rent} Due to the congestion of the transmission line, the marginal cost of producing electricity can be different at node 1 and node 2. The competitive price at node 2 is higher than at node 1 -- this corresponds to \alert{locational marginal pricing}, or \alert{nodal pricing}. Since consumers pay and generators get paid the price in their local market, in case of congestion there is a difference between the total payment of consumers and the total revenue of producers -- this is the \alert{merchandising surplus} or \alert{congestion rent}, collected by the network operator. For each line it is given by the price difference in both regions times the amount of power flow between them: \begin{align*} \text{Congestion rent =}\:\Delta \lambda\times f \end{align*} \end{frame} \begin{frame}{Congestion rent: Two node example} Returning to our two node example: \ra{1.1} \begin{table}[!t] \begin{tabular}{rrrrrr} \toprule Case & Demand pays & Generator gets & $\l_2^* - \l_1^*$ & flow $f$ & Cong. rent \\ & [\euro/h] & [\euro/h] & [\euro/MWh] & [MW] & [\euro/h] \\ \midrule $F = 0$ & 3000 & 3000 & 10 & 0 & 0 \\ $F = 50$ & 3000 & 2500 & 10 & 50 & 500 \\ $F = \infty$ & 2000 & 2000 & 0 & 100 & 0 \\ \bottomrule \end{tabular} \end{table} To get a congestion rent, we need congestion somewhere in the network to cause a price difference between the nodes, as well as a non-zero flow between the nodes. \end{frame} \begin{frame}[fragile] \frametitle{Congestion rent} In this example we saw that the sum of what consumers pay does not always equal the sum of generator revenue. In fact if we take the balance constraint and sum it weighted by the market price at each node we find \begin{align*} \sum_i \l_i^* d_i - \sum_i \l_i^* \sum_{s} g^*_{i,s} = -\sum_i \l_i^* \sum_\ell K_{i\ell}f^*_\ell \end{align*} The quantity for each $\ell$ \begin{equation*} -f_\ell^*\sum_i K_{i\ell} \l_i^* = f_\ell (\l_{\textrm{end}}^* - \l_{\textrm{start}}^*) \end{equation*} is called the \alert{congestion rent} and is the money the network operator receives for transferring power from a low price node (start) to a high price node (end), `buy it low, sell it high'. It is zero if: a) the flow is zero or b) the price difference is zero. \end{frame} \begin{frame}[fragile] \frametitle{Two nodes, quadratic cost function} \begin{equation*} \min_{g_1,g_2,f} \left[C_1(g_1) + C_2(g_2) \right] \end{equation*} \centering \begin{circuitikz} \draw [ultra thick] (1,13) node[anchor=south]{1} -- (4,13); \draw(2.5,13) -- +(0,0.5) to [short,i^=$f \leq F \leftrightarrow \bar{\mu}\hspace{1cm} -f \leq F \leftrightarrow \ubar{\mu}$] +(5,0.5) -- +(5,0); \draw [ultra thick] (6,13) node[anchor=south]{2} -- (9,13); \draw (1.5,13) -- +(0,-0.5) node[sground]{}; \draw (3,12) node[vsourcesinshape, rotate=270](V2){} (V2.left) -- +(0,0.6); \draw (1,11) node{$d_1 = 500$}; \draw (3,11) node{$g_{1}$}; \draw (2.5,10) node{$C'_1(g_1) = 10 + 0.01 g_{1}$}; \draw (2.5,9) node{$g_1 - d_1 = f \leftrightarrow \l_1$}; \draw (6.5,13) -- +(0,-0.5) node[sground]{}; \draw (8,12) node[vsourcesinshape, rotate=270](V2){} (V2.left) -- +(0,0.6); \draw (6,11) node{$d_2 = 1500$}; \draw (8,11) node{$g_{2}$}; \draw (7.5,10) node{$C'_2(g_2) = 13 + 0.02 g_{2}$}; \draw (7.5,9) node{$g_2 - d_2 = -f \leftrightarrow \l_2$}; \end{circuitikz} \end{frame} \begin{frame}[fragile] \frametitle{Two nodes, quadratic cost function} From stationarity \begin{align*} 0 & = \frac{\d \cL}{\d g_1} = C'_1(g_1^*) - \l_1^* \\ 0 & = \frac{\d \cL}{\d g_2} = C'_2(g_2^*) - \l_2^* \\ 0 & = \frac{\d \cL}{\d f} = 0 + \l_1^* - \l_2^* - \bar{\m}^* + \ubar{\m}^* \end{align*} \end{frame} \begin{frame}{Outcome with no transmission capacity $F = 0$} \centering $F=0,f^*=0,g_1^* = 500, g_2^* = 1500, \l_1^* = 15, \l_2^* = 43, \ubar{\m}^* = 0, \bar{\m}^* = \l_1^* - \l_2^* = -28$ \begin{tikzpicture} % left bottom right top \node[anchor=south west,inner sep=0] (image) at (0,0) { \includegraphics[trim=0 0.45cm 0 0cm,clip=true,width=10cm]{strbac_separate_markets.pdf}}; \draw (3,-0.2) node{$d_1 = 500,$ $d_2 = 1500$}; \draw (3,-0.5) node{$g_1^* = 500, g_2^* = 1500$}; \end{tikzpicture} \end{frame} \begin{frame}{Outcome with unlimited transmission capacity $F = \infty$} \centering $F=\infty,f^*=933,g_1^* = 1433, g_2^* = 567, \l_1^* = \l_2^* = 24.33, \ubar{\m}^* = \bar{\m}^* = 0$ \begin{tikzpicture} % left bottom right top \node[anchor=south west,inner sep=0] (image) at (0,0) { \includegraphics[trim=0 0.45cm 0 0cm,clip=true,width=10cm]{strbac_single_market.pdf}}; \draw (3,-0.2) node{$d_1 = 500, d_2 = 1500$}; \draw (7.5,-0.2) node{$g_1^* = 1433, g_2^* = 567$}; \end{tikzpicture} \end{frame} \begin{frame}{Outcome with constrained transmission capacity $F = 400$} \centering $F=400,f^*=400,g_1^* = 900, g_2^* = 1100, \l_1^* = 19, \l_2^* = 35, \ubar{\m}^* = 0, \bar{\m}^* = \l_1^* - \l_2^* = -16$ \begin{tikzpicture} % left bottom right top \node[anchor=south west,inner sep=0] (image) at (0,0) { \includegraphics[trim=0 0.45cm 0 0cm,clip=true,width=10cm]{strbac_constrained_market.pdf}}; \draw (3.3,-0.2) node{$d_1 = 500,$}; \draw (3.3,-0.5) node{$d_2 = 1500$}; \draw (5.2,-0.2) node{$g_1^* = 900,$}; \draw (5.2,-0.5) node{$g_2^* = 1100$}; \end{tikzpicture} \end{frame} \begin{frame}{Comparison} \ra{1.1} \begin{table}[!t] \begin{tabular}{lrrr} \toprule & Separate markets & Single market & Constrained market \\ \midrule $d_1$ [MW] & 500 & 500 & 500 \\ $g_1^*$ [MW] & 500 & 1433 & 900 \\ $\l_1^*$ [\euro/MWh] & 15 & 24.33 & 19 \\ $d_2$ [MW] & 1500 & 1500 & 1500 \\ $g_2^*$ [MW] & 1500 & 567 & 1100\\ $\l_2^*$ [\euro/MWh] & 43 & 24.33 & 35 \\ $f^*$ [MW] & 0 & 933 & 400\\ $\ubar{\mu}^*$ [\euro/MWh] & 0 & 0 & 0 \\ $\bar{\mu}^*$ [\euro/MWh] & -28 & 0 & -16 \\ $\sum_{s}\lambda_{s}\times g_s$ [\euro] & 72000 & 48660 & 55600 \\ $\sum_{s}\lambda_{s}\times d_s$ [\euro] & 72000 & 48660 & 62000 \\ congestion rent & 0 & 0 & 6400 \\ \bottomrule \end{tabular} \end{table} \end{frame} \section{Advanced: 3-node example} \begin{frame}[fragile] \frametitle{PTDF formulation of linearised optimal power flow} For the 3-node example it is easier to switch to the \alert{Power Transfer Distribution Factors} (PTDF) formulation of the power flow $f_\ell = \sum_k \textrm{PTDF}_{\ell k} p_k$. Keep our objective \begin{align*} \min_{\{g_{i,s}\},\{f_\ell\}} \sum_{i,s} o_{i,s} g_{i,s} \end{align*} and the same generator constraints and line constraints: \begin{align*} f_\ell \leq F_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_\ell \\ - f_\ell \leq F_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_\ell \end{align*} but instead of energy conservation at each node and cycle constraints we have: \begin{equation*} f_\ell = \sum_k \textrm{PTDF}_{\ell k} p_k = \sum_k \textrm{PTDF}_{\ell k} \left( \sum_{s} g_{k,s} - d_k \right) \end{equation*} and one overall conservation constraint: \begin{equation*} \sum_i p_i = \sum_i\left( \sum_{s} g_{i,s} - d_i \right) = 0 \end{equation*} \end{frame} \begin{frame}{An example 3-node system} The cheapest generators are are at node 1, most demand is at node 3. \begin{columns}[c] \begin{column}{7cm} \includegraphics[width=7cm,angle=0,origin=c]{kirschen_figure} $\sum_i d_i = 410$~MW \end{column} \begin{column}{7cm} \begin{table}[!t] \begin{tabular}{ccc} \toprule Generator & Capacity & Marginal cost\\ & (MW) & (\euro/MWh)\\ \midrule A & 140 & 7.5\\ B & 285 & 6\\ C & 90 & 14\\ D & 85 & 10\\ %\midrule\\ \bottomrule Line & Reactance & Capacity\\ & (p.u.) & (MW)\\ \midrule $1\to 2$ & 0.2 & 126\\ $1\to 3$ & 0.2 & 250\\ $2\to 3$ & 0.1 & 130\\ \bottomrule \end{tabular} \end{table} \end{column} \end{columns} \source{Kirschen and Strbac} \end{frame} \begin{frame}{Power flows and feasible injections} \begin{columns}[c] \column{.4\textwidth} \includegraphics[width=3cm]{3nodes4} \begin{align*} \textbf{PTDF} = \begin{matrix} 1\to 2 \\ 1\to 3 \\ 2\to 3 \end{matrix} \begin{pmatrix} 2/5 & -1/5 & 0 \\ 3/5 & 1/5 & 0 \\ 2/5 & 4/5 & 0 \end{pmatrix} \end{align*} Power flows: \begin{align*} f_{1\to 2} &= \frac{1}{5}\left(2p_1 - p_2\right)\\ f_{1\to 3} &= \frac{1}{5}\left(3p_1 +p_2\right)\\ f_{2\to 3} &= \frac{2}{5}\left(p_1 + 2p_2\right) \end{align*} \column{.5\textwidth} Implications of six constraints $|f_\ell | \leq F_\ell$ on $p_1$ and $p_2$ ($p_3$ is not independent since $\sum_i p_i = 0$; graphic $Z_i$ is $p_i$): feasible space is rectangle. \includegraphics[width=4.7cm]{feasible_example1}\\ \includegraphics[width=4.7cm]{feasible_example2} \end{columns} \end{frame} \begin{frame}{Economic dispatch (ignore grid)} If we ignore the grid constraints, we dispatch the cheapest generators first, market price set by marginal generator $A$: $\lambda= o_A = 7.5\emwh$. \begin{columns}[c] \column{.5\textwidth} \begin{align*} g_A &= 125\mw\\ g_B &= 285\mw\\ g_C &= g_D = 0 \end{align*} \includegraphics[width=5.2cm]{feasible_example3}\\ $p_1 = 360\mw, p_2=-60\mw$. \column{.5\textwidth} \begin{align*} F_{1\to 2} &= 126\mw\\ F_{1\to 3} &= 250\mw\\ F_{2\to 3} &= 130\mw \end{align*} \begin{align*} f_{1\to 2} &= \frac{1}{5}\left(2p_1 - p_2\right)=156\mw\\ f_{1\to 3} &= \frac{1}{5}\left(3p_1 +p_2\right)=204\mw\\ f_{2\to 3} &= \frac{2}{5}\left(p_1 + 2p_2\right)=96\mw \end{align*} First line limit is violated! Optimal point (black dot) is outside the red feasible space. \end{columns} \end{frame} \begin{frame}{Economic redispatch: problem} What can we do? Transfer $x$~MW from generator $A$ at node 1 to generator $D$ at node 3 to relieve the line. \begin{columns}[c] \column{.5\textwidth} \begin{align*} g_A &= 125 \to g_A = 125 - x \\ g_D &= 0 \to g_D = x \end{align*} \includegraphics[width=5.2cm]{feasible_example4}\\ $p_1 = 360 - x, p_2=-60$. \column{.5\textwidth} \begin{align*} F_{1\to 2} &= 126\\ F_{1\to 3} &= 250\\ F_{2\to 3} &= 130 \end{align*} \begin{align*} f_{1\to 2} &= \frac{1}{5}\left(2(p_1-x) - p_2\right)=156 - \frac{2}{5}x = 126\\ f_{1\to 3} &= \frac{1}{5}\left(3p_1 +p_2\right)=159\\ f_{2\to 3} &= \frac{2}{5}\left(p_1 + 2p_2\right)=66 \end{align*} Solution? $\frac{2}{5}x = 30$, i.e. $x = 75$. \end{columns} \end{frame} \begin{frame}{Economic redispatch: solution} Solution with $x=75$~MW transfer from generator $A$ at node 1 to generator $D$ at node 3 to relieve the line $f_{1\to 2}$. \begin{columns}[c] \column{.5\textwidth} \begin{align*} g_A &= 125 \to g_A = 125 - x = 50 \\ g_D &= 0 \to g_D = x = 75 \end{align*} Redispatch cost: $187.5$ \euro/h \includegraphics[width=5.2cm]{feasible_example4}\\ $p_1 = 360 - x = 285, p_2=-60$. \column{.5\textwidth} \begin{align*} F_{1\to 2} &= 126\\ F_{1\to 3} &= 250\\ F_{2\to 3} &= 130 \end{align*} \begin{align*} f_{1\to 2} &= \frac{1}{5}\left(2p_1 - p_2\right)=126\\ f_{1\to 3} &= \frac{1}{5}\left(3p_1 +p_2\right)=159\\ f_{2\to 3} &= \frac{2}{5}\left(p_1 + 2p_2\right)=66 \end{align*} \end{columns} \end{frame} \begin{frame}{Nodal prices} The \alert{nodal marginal price} is equal to the minimal system cost of supplying an additional megawatt of load at this node. For node 1, there is a cheap generator $A$ to locally supply the demand, so $\lambda_1 = o_A = 7.5$. For node 3, it cannot be supplied from node 1 because of the transmission constraint on $f_{1\to 2}$, so it must be supplied locally $\lambda_3 = o_D = 10$. What about node 2? We cannot supply from node 1, node 2's generator $C$ is expensive, so what about from node 3? Suppose we increase demand by $\varepsilon$, i.e. $p_2 \to p_2 - \varepsilon$. Because of the binding transmission constraint $f_{1\to 2} = \frac{1}{5}\left(2p_1 - p_2\right)=126$ we need to compensate by decreasing $p_1 \to p_1 -\frac{1}{2}\varepsilon$. This means we need to increase generator $D$ and $p_3 \to p_3 + \frac{3}{2} \varepsilon$ so we get \begin{align*} \lambda_2 = 1.5 o_D - 0.5 o_A = 11.25 \end{align*} NB: This price is a composite, not the marginal price of any generator! \end{frame} \begin{frame}{Nodal prices} Economic operation of the three-node system using nodal pricing. \begin{table}[!t] \begin{tabular}{lrrrr} \toprule & Node 1 & Node 2 & Node 3 & System\\ \midrule Consumption (MW) & 50 & 60 & 300 & 410\\ Production (MW) & 335 & 0 & 75 & 410\\ Nodal marginal price (\euro/MWh) & 7.5 & 11.25 & 10 & -\\ Consumer payments (\euro/h) & 375 & 675 & 3000 & 4050 \\ Generator revenue (\euro/h) & 2512.5 & 0 & 750 & 3262.5\\ Congestion rent (\euro/h) & & & & 787.5\\ \bottomrule \end{tabular} \end{table} \end{frame} \begin{frame}{Nodal prices} Congestion rent: \begin{table}[!t] \begin{tabular}{lrrrr} \toprule Connection & Flow & 'From' price & 'To' price & Surplus\\ & (MW) & (\euro/MWh) & (\euro/MWh) & (\euro/h)\\ \midrule $1\to 2$ & 126 & 7.5 & 11.25 & 427.5\\ $1\to 3$ & 159 & 7.5 & 10 & 397.5\\ $2\to 3$ & 66 & 11.25 & 10 & -82.5\\ Total & & & & 787.5\\ \bottomrule \end{tabular} \end{table} Note the counter-intuitive flow from node $2$ (higher price) to node $3$ (lower price)! This is because of grid constraints from KVL. Note also that the line itself does not have to be congested to have a congestion rent (see e.g. line $1\to 3$). It suffices that somewhere in the network there is congestion (here: $1\to 2$). \end{frame} \begin{frame}{Example slightly changed} The \alert{nodal marginal price} is equal to the minimal system cost of supplying an additional megawatt of load at this node. \begin{columns}[c] \column{.5\textwidth} \begin{align*} g_A &= 47.5\mw \\ g_B &= 285\mw\\ g_C &= 0\mw\\ g_D &=77.5\mw \end{align*} Nodal prices: \begin{align*} \lambda_1 &= o_A = 7.5\emwh\\ \lambda_3 &= o_D = 10 \emwh\\ \lambda_2 &= 2\times o_A - 1\times o_D = 5\emwh \end{align*} \column{.5\textwidth} \begin{align*} F_{1\to 2} &= 126\mw\\ F_{1\to 3} &= 250\mw\\ F_{2\to 3} &= \mathbf{65}\mw \end{align*} \begin{align*} f_{1\to 2} &= \frac{1}{5}\left(2p_1 - p_2\right)=125\mw\\ f_{1\to 3} &= \frac{1}{5}\left(3p_1 +p_2\right)=157.5\mw\\ f_{2\to 3} &= \frac{2}{5}\left(p_1 + 2p_2\right)=65\mw \end{align*} \end{columns} The nodal marginal price at node $2$ is lower than the marginal cost of any generator! \end{frame} \begin{frame}{Negative nodal prices} %{\centering \begin{center} \includegraphics[width=8cm,angle=0,origin=c]{harris} \source{Harris} \end{center} Generator A has marginal costs 60\euro/MWh, generator B has marginal costs 30\euro/MWh. The line between E and D is constrained to $25$ MW. The additional load of 10 MW at node E allows cheap generator B to substitute some of expensive generator A and thus \alert{reduces} the system cost by 300 \euro/h, so $\lambda_E=-30\emwh$! \end{frame} \begin{frame}{Nodal prices for Germany} Spot the node with negative price. If we increase the load here, it will relieve a line somewhere that allows more flow from a cheap node to an expensive node, thus reducing the system cost. \centering \includegraphics[width=9cm]{lmp.png} \source{PyPSA \\(Python for Power System Analysis)} \end{frame} \section{The European Market} \begin{frame} \frametitle{Existing bidding zones} \begin{columns}[T] \begin{column}{5cm} \includegraphics[width=5.8cm]{european_bidding_zones.png} \end{column} \begin{column}{8cm} \begin{itemize} \item Bids for German electricity take place in a \alert{giant bidding zone} encompassing both Germany and Luxembourg (Austria was separated from the German bidding zone in October 2018) \item This means that transmission constraints are only visible to the market at the \alert{borders} to the other national zones \item Internal transmission constraints are \alert{ignored} - market bids are handled as if they do not exist \item Only KCL enforced on most borders - KVL much harder \end{itemize} \end{column} \end{columns} \end{frame} \begin{frame} \frametitle{The Problem} Renewables are not always located near demand centres, as in this example from Germany. \begin{columns}[T] \begin{column}{6cm} \includegraphics[width=6cm]{scigrid-load} \end{column} \begin{column}{6cm} % left bottom right top \includegraphics[trim=0 0cm 0 1cm,width=6.3cm,clip=true]{scigrid-wind} \end{column} \end{columns} \end{frame} \begin{frame} \frametitle{The Problem} \begin{columns}[T] \begin{column}{5.5cm} \vspace{.5cm} \includegraphics[width=6cm]{scigrid-loading} \end{column} \begin{column}{6cm} \begin{itemize} \item This leads to \alert{overloaded lines} in the middle of Germany, which cannot transport all the wind energy from North Germany to the load in South Germany \item It also overloads lines in neighbouring countries due to \alert{loop flows} (unplanned physical flows `according to least resistance' which do not correspond to traded flows) \item It also \alert{blocks imports and exports} with neighbouring countries, e.g. Denmark \end{itemize} \end{column} \end{columns} \end{frame} \begin{frame} \frametitle{Solution 1: Redispatch after energy market clearing} These problems are \alert{not visible} in the day-ahead electricity market, which treats the whole of Germany and Austria as a single bidding zone. It dispatches wind in North Germany as if there was no internal congestion... To ensure that the physical limits of transmission are not exceeded, the network operator must \alert{`re-dispatch'} power stations and \alert{curtail} (Einspeisemanagement) renewables to restore order. This is \alert{costly} (0.8 redispatch + 0.6 RE-compensation = 1.4 billion EUR in 2017 - although exceptional circumstances in 1st quarter) and results in \alert{lost CO$_2$-free generation} (5.5 TWh curtailment of RE and CHP in 2017). \alert{International redispatch} is sometimes also required (Multilateral Remedial Actions = MRA). Furthermore, there are \alert{no market incentives} to reinforce the North-South grid, to locate more power stations in South Germany or to build storage / P2X in North Germany. \end{frame} \begin{frame} \frametitle{Redispatch in Germany: Trend} \centering \includegraphics[width=11cm]{Volume-and-expenditures-for-redispatch-and-curtailment-from-2015-to-2019-in-Germany_W640.jpg} \source{\hrefc{https://doi.org/10.1016/j.apenergy.2020.116201}{Xiong et al (2021)}} \end{frame} \begin{frame} \frametitle{Redispatch in Europe: Future} The Joint Research Centre (JRC) of the European Commission calculates rising redispatch if the existing bidding zone configuration is kept. \centering \includegraphics[height=6.5cm]{jrc-redispatch.png} \source{\hrefc{https://publications.jrc.ec.europa.eu/repository/handle/JRC137685}{Thomassen et al, 2024}} \end{frame} \begin{frame} \frametitle{Solution 2: Smaller bidding zones to ``see'' congested boundaries} \begin{columns}[T] \begin{column}{5cm} \includegraphics[width=6cm]{nordel.png} \end{column} \begin{column}{5cm} \begin{itemize} \item In Scandinavia they have solved this by introducing \alert{smaller bidding zones} \item Now congestion at the boundaries between zones is taken into account in the \alert{implicit auctions} of the market \item This is also done in Italy (again, a long country), where prices for small consumers are \alert{uniformised} for fairness \end{itemize} \end{column} \end{columns} \end{frame} \begin{frame} \frametitle{Solution 3: Nodal pricing} \begin{columns}[T] \begin{column}{6cm} \vspace{.5cm} \includegraphics[width=7cm]{lmp.png} \end{column} \begin{column}{5cm} \begin{itemize} \item The ultimate solution, as used in the US and other markets, is \alert{nodal pricing}, which exposes all transmission congestion \item Considered too complex and subject to market power to be used in Europe, but this is questionable... \item Here we see clearly why many argue for a North-South German split \end{itemize} \end{column} \end{columns} \end{frame} \begin{frame} \frametitle{First step: Split Germany North-South} \begin{columns}[T] \begin{column}{5cm} \vspace{.5cm} \includegraphics[width=5.2cm]{DE_split.png} \end{column} \begin{column}{6cm} \vspace{.7cm} \begin{itemize} \item Initial price difference could average up to 12 EUR/MWh \item Prices would converge with more network expansion \item Redispatch costs reduced by 39\% in 2025, 58\% in 2035 (assuming NEP 2030 transmission projects get built) \item Politically difficult, may require, like Italy, uniformised price on consumer side \end{itemize} \end{column} \end{columns} \source{\hrefc{https://www.strommarkttreffen.org/2018-02_Fraunholz&Hladik_Zwei_Preiszonen_in_D.pdf}{Fraunholz \& Hladik, 2018}} \end{frame} \begin{frame} \frametitle{Solution 1.5: Flow-based market coupling} Flow-based market coupling can be used in zonal markets to see precise individual line constraints via PTDF, instead of ``boxing'' the feasible space like ATC/NTC schemes do. \centering \includegraphics[width=13cm]{fbmc} \source{Van den Bergh, Boury, Delarue} \end{frame} \begin{frame} \frametitle{Solution 1.5: Flow-based market coupling} \begin{columns}[T] \begin{column}{6cm} \vspace{.5cm} \includegraphics[height=7cm]{core-fbmc.png} \end{column} \begin{column}{7cm} \vspace{.8cm} \begin{itemize} \item From May 2015 to June 2022 flow-based market coupling was only applied in Germany, France, Netherlands, Belgium and Luxembourg \item From June 2022 it has extended across northern `Core' countries in continental Europe \end{itemize} \end{column} \end{columns} \source{\hrefc{https://www.jao.eu/core-fb-mc}{JAO Core FBMC}} \end{frame} \section{Storage Optimisation} \begin{frame}[fragile] \frametitle{Storage equations} Now, like the network case where we add different nodes $i$ with different loads, for storage we have to consider different time periods $t$. This is called \alert{multi-period optimisation}. Label conventional generators by $s$, storage by $r$ and now minimise \begin{align*} & \min_{\{g_{i,s,t}\},\{g_{i,r,t,\textrm{charge}}\},\{g_{i,r,t,\textrm{discharge}}\},\{f_{\ell,t}\}}\\ &\left[ \sum_{i,s,t} o_{i,s} g_{i,s,t} + \sum_{i,r,t} o_{i,r,\textrm{charge} }\, g_{i,r,t,\textrm{charge}} + \sum_{i,r,t} o_{i,r,\textrm{discharge}}\, g_{i,r,t,\textrm{discharge}} \right] \end{align*} The power balance constraints are now (cf. Lecture 5) for each node $i$ and time $t$ that the demand is met either by generation, storage or network flows: \begin{align*} \sum_{s} g_{i,s,t} + \sum_{r}(g_{i,r,t,\textrm{discharge}} - g_{i,r,t,\textrm{charge}}) - d_{i,t} = \sum_\ell K_{i\ell}f_{\ell,t} \hspace{0.5cm}\leftrightarrow\hspace{0.5cm} \l_{i,t} \end{align*} Now we have a market price $\l_{i,t}$ for each node $i$ and time $t$. \end{frame} \begin{frame}[fragile] \frametitle{Storage equations} We have constraints on normal generators \begin{align*} 0 \leq g_{i,s,t} \leq G_{i,s} \hspace{0.5cm}\leftrightarrow\hspace{0.5cm} \m_{i,s,t} \end{align*} and on the storage \begin{align*} 0 & \leq g_{i,r,t,\textrm{discharge}} \leq G_{i,r,\textrm{discharge}} \hspace{0.5cm}\leftrightarrow\hspace{0.5cm} \m_{i,r,t,\textrm{discharge}} \\ 0 & \leq g_{i,r,t,\textrm{charge}} \leq G_{i,r,\textrm{charge}} \hspace{0.5cm}\leftrightarrow\hspace{0.5cm} \m_{i,r,t,\textrm{charge}} \end{align*} The energy level of the storage (or `state of charge') is given by \begin{align*} e_{i,r,t} = \eta_0e_{i,r,t-1} + \eta_1g_{i,r,t,\textrm{charge}} - \eta_2^{-1} g_{i,r,t,\textrm{discharge}} \hspace{0.5cm}\leftrightarrow\hspace{0.5cm} \tilde \l_{i,r,t} \end{align*} The KKT multiplier $\tilde \l_{i,r,t}$ is the \alert{value of the storage medium} at this time. The storage state of charge is limited by its energy capacity $E_{i,r}$ \begin{align*} 0 \leq e_{i,r,t} \leq E_{i,r} \hspace{0.5cm}\leftrightarrow\hspace{0.5cm} \m_{i,r,t} \end{align*} \end{frame} \begin{frame}[fragile] \frametitle{Idea of storage} Storage does `buy it low, sell it high' \alert{arbitrage}, like network, but in time rather than space, i.e. between cheap times (e.g. with lots of zero-marginal-cost renewables) and expensive times (e.g. with high demand, low renewables and expensive conventional generators). \end{frame} \begin{frame}[fragile] \frametitle{Storage charges at low prices, discharges at high prices} Simplified example from \urlc{https://model.energy} For Germany with only wind and hydrogen storage to meet a flat 100~MW demand. Average charging price (with electrolyser): 43 \euro/MWh Average discharging price (with turbine): 144 \euro/MWh %single_node_storage_bids.ipynb \centering \includegraphics[width=8cm]{storage_bids.pdf} \end{frame} \begin{frame} \frametitle{Relation of storage bidding to storage medium value} The KKT stationarity for the discharge variable $g_{t,\textrm{discharge}}$ (ignoring $i,r$ indices for now) is \begin{equation*} 0 = \frac{\d \cL}{\d g_{t,\textrm{discharge}}} = \eta_2^{-1} \tilde \lambda_{t}^* -\l_t^* + \ubar{\m}^*_{t,\textrm{discharge}} - \bar{\m}^*_{t,\textrm{discharge}} \quad \forall t \end{equation*} Note that this has exactly the same structure as a conventional generator with marginal cost $o_s = \eta_2^{-1} \tilde \lambda_{t}^*$ based on a fuel cost $\tilde \lambda_t^*$. So the storage medium value (sometimes called \alert{Belman value} or \alert{marginal storage value}) sets how the storage bids into the electricity market. Doing the same for the the charge variable $g_{t,\textrm{charge}}$ we get from stationarity \begin{equation*} 0 = \frac{\d \cL}{\d g_{t,\textrm{charge}}} = - \eta_1 \tilde \lambda_{t}^* + \l_t^* + \ubar{\m}^*_{t,\textrm{charge}} - \bar{\m}^*_{t,\textrm{charge}} \quad \forall t \end{equation*} Note that this has exactly the same structure as a flexible demand bidding with a willingness to pay of $\eta_1 \tilde \lambda_{t}^*$. The charger is willing to pay up to $\eta_1 \tilde \lambda_{t}^*$ for electricity because if it wants to produce 1~MWh of storage, it needs $1/\eta_1$~MWh of electricity. If it pays $\eta_1 \tilde \lambda_{t}^*$~\euro/MWh or less for $1/\eta_1$~MWh it will pay up to $\lambda_{t}^*$ and still can make a profit charging the storage medium. In the model.energy example the storage value $\tilde \lambda_{t}^*$ for hydrogen storage is the hydrogen price. \end{frame} \begin{frame} \frametitle{Relation of storage bidding to storage medium value} For a hydro dam the marginal storage value $\tilde \lambda_{t}^*$ is the value of the water: the opportunity cost of using up water that could be used to generate at later times with high electricity prices. \centering \includegraphics[height=6cm]{hydro_dam.jpg} \source{Wikipedia} \end{frame} \begin{frame}[fragile] \frametitle{Storage plus network equations} Finally for the flows we repeat the constraints for each time $t$. We have KVL for each cycle $c$ and time $t$ \begin{align*} \sum_{\ell} C_{\ell c} x_\ell f_{\ell,t} = 0 \hspace{1cm}\leftrightarrow\hspace{1cm} \l_{c,t} \end{align*} and in addition the flows cannot overload the thermal limits, $|f_{\ell,t}| \leq F_\ell$ \begin{align*} f_{\ell,t} \leq F_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \bar{\m}_{\ell,t} \\ - f_{\ell,t} \leq F_\ell \hspace{1cm}\leftrightarrow\hspace{1cm} \ubar{\m}_{\ell,t} \end{align*} \end{frame} \begin{frame}[fragile] \frametitle{Preview: Investment optimisation} Preview for next time: Next time we will also optimise \alert{investment} in the \alert{capacities} of generators, storage and network lines, to maximise \alert{long-run efficiency}. We will promote the capacities $G_{i,s}$, $G_{i,r,*}$, $E_{i,r}$ and $F_{\ell}$ to optimisation variables. \end{frame} \end{document}