solution04.tex 13 KB
Newer Older
sp2668's avatar
sp2668 committed
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
\documentclass[11pt,a4paper,fleqn]{scrartcl}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[colorlinks=true, citecolor=blue, linkcolor=blue, filecolor=blue,urlcolor=blue]{hyperref}
\hypersetup{
     colorlinks   = true,
     citecolor    = gray
}
\usepackage{wrapfig}

\usepackage{caption}
\captionsetup{format=plain, indent=5pt, font=footnotesize, labelfont=bf}

\setkomafont{disposition}{\scshape\bfseries}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{bbm}
\usepackage{mathtools}
% \usepackage{epsfig}
% \usepackage{grffile}
%\usepackage{times}
%\usepackage{babel}
\usepackage{tikz}
\usepackage{paralist}
\usepackage{color}
\usepackage[top=3cm, bottom=2.5cm, left=2.5cm, right=3cm]{geometry}
%\setlength{\mathindent}{1ex}

% PGF
\usepackage{pgfplots}
\usepackage{pgf}
\usepackage{siunitx}
\usepackage{xfrac}
\usepackage{calculator}
\usepackage{calculus}
\usepackage{eurosym}
\usepackage{booktabs}
%\sisetup{per-mode=fraction,%
%	fraction-function=\sfrac}

%\newcommand{\eur}[1]{\EUR{#1}\si{\per\kilo\meter}}
\pgfplotsset{
  compat=newest,
  every axis/.append style={small, minor tick num=3}
}

%\usepackage[backend=biber,style=alphabetic,url=false,doi=false]{biblatex}
%\addbibresource{sheet01_biber.bib}
% \addbibresource{/home/coroa/papers/refs.bib}

\newcommand{\id}{\mathbbm{1}}
\newcommand{\NN}{{\mathbbm{N}}}
\newcommand{\ZZ}{{\mathbbm{Z}}}
\newcommand{\RR}{{\mathbbm{R}}}
\newcommand{\CC}{{\mathbbm{C}}}
\renewcommand{\vec}[1]{{\boldsymbol{#1}}}

\renewcommand{\i}{\mathrm{i}}

\newcommand{\expect}[1]{\langle\,#1\,\rangle}
\newcommand{\e}[1]{\ensuremath{\,\mathrm{#1}}}

\renewcommand{\O}{\mc{O}}
\newcommand{\veps}{\varepsilon}
\newcommand{\ud}[1]{\textup{d}#1\,}

\newcommand{\unclear}[1]{\color{green}#1}
\newcommand{\problem}[1]{\color{red}#1}
\newcommand{\rd}[1]{\num[round-mode=places,round-precision=1]{#1}}

%\DeclareSIUnit{\euro}{\EUR}
\DeclareSIUnit{\dollar}{\$}
\newcommand{\eur}{\text{\EUR{}}}

\usepackage{palatino}
\usepackage{mathpazo}
\setlength\parindent{0pt}
\usepackage{xcolor}
\usepackage{framed}
\definecolor{shadecolor}{rgb}{.9,.9,.9}

\def\cap{\text{Cap}}
\def\floor{\text{Floor}}
\def\l{\lambda}
\def\m{\mu}
\def\d{\partial}
\def\cL{\mathcal{L}}
\def\co2{CO${}_2$}

\def\mw{\text{ MW}}
\def\mwh{\text{ MWh}}
\def\gw{\text{ GW}}
\def\gwh{\text{ GWh}}
\def\emwh{\text{ \euro/MWh}}
\def\bemwh{\text{ [\euro/MWh]}}
\newcommand{\ubar}[1]{\text{\b{$#1$}}}

%=====================================================================
%=====================================================================
\begin{document}

\begin{center}
sp2668's avatar
sp2668 committed
106
107
 \textbf{\Large Energy System Modelling }\\
 {SS 2018, Karlsruhe Institute of Technology}\\
sp2668's avatar
sp2668 committed
108
 {Institute for Automation and Applied Informatics}\\ [1em]
sp2668's avatar
sp2668 committed
109
110
 \textbf{\textsc{\Large Solution IV: Electricity Markets}}\\
 \small Will be worked on in the exercise session on Tuesday, 17 July 2018.\\[1.5em]
sp2668's avatar
sp2668 committed
111
112
113
114
115
116
117
118
119
120
\end{center}

\vspace{1em}

%=============== ======================================================
\paragraph{Solution IV.1 \normalsize (Shadow prices of limits on consumption).}~\\
%=====================================================================

Suppose that the utility for the electricity consumption of an industrial company is given by
\[
sp2668's avatar
sp2668 committed
121
 U(q) = 70q - 3q^2 [\textrm{\euro}/h] \quad , \quad q_{min}=2\leq q \leq q_{max}=10,
sp2668's avatar
sp2668 committed
122
123
124
125
\]
where $q$ is the demand in MW and $q_{min}, q_{max}$ are the minimum and maximum demand. \\
[1em]
Assume that the company is maximising its net surplus for a given electricity price $\pi$, i.e. it maximises $\max_{q} \left[U(q) -
sp2668's avatar
sp2668 committed
126
  \pi q\right]$.
sp2668's avatar
sp2668 committed
127
\begin{enumerate}[(a)]
sp2668's avatar
sp2668 committed
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
 \begin{shaded} \item If the price is $\pi = 5$~\euro/MWh, what is the optimal
  demand $q^*$?  What is the value of the KKT multiplier $\mu_{max}$
  for the constraint $q \leq q_{max}=10$ at this optimal solution?
  What is the value of $\mu_{min}$ for $q \geq q_{min} = 2$?\end{shaded}
 We convert the exercise to an optimisation problem with objective
 \begin{equation}
  \max_q U(q) - \pi q
 \end{equation}
 with constraints
 \begin{align}
  q  & \leq q_{max} \hspace{1cm}\leftrightarrow\hspace{1cm} \m_{max}  \\
  -q & \leq -q_{min} \hspace{1cm}\leftrightarrow\hspace{1cm} \m_{min}
 \end{align}

 From stationarity we get:
 \begin{align}
  0 & =   \frac{\d}{\d q} \left(U(q) - \pi q\right) - \m_{max} \frac{\d}{\d q} (q-q_{max})- \m_{min} \frac{\d}{\d q} (-q+q_{min}) \\
    & =  U'(q) - \pi - \m_{max} + \m_{min} \label{eq:2stat}
 \end{align}

 The marginal utility curve is $U'(q) = 70 - 6q$ [\euro/MWh]. At
 $\pi = 5$, the demand would be determined by $5=70-6q$, i.e. $q =
  65/6 = 10.8333$, which is above the consumption limit
 $q_{max} = 10$. Therefore the optimal demand is $q^* = 10$, the upper limit is binding $\mu_{max}
  \geq 0$ and the lower limit is non-binding $\mu_{min} = 0$.

 To determine the value of $\mu_{max}$ we use \eqref{eq:2stat} to get
 $\m_{max} = U'(q^*) - \pi = U'(10) - 5 = 5$.

 \begin{shaded}
  \item Suppose now the electricity price is $\pi = 60$~\euro/MWh. What are
  the optimal demand $q^*$, $\mu_{max}$ and $\mu_{min}$ now?
 \end{shaded}

 At $\pi = 60$, the demand would be determined by $60=70-6q$, i.e. $q = 10/6 = 1.667$, which is below the consumption limit $q_{min} = 2$. Therefore the optimal demand is $q^* = 2$, the upper limit is non-binding $\mu_{max}
  = 0$ and the lower limit is binding $\mu_{min} \geq 0$.

 To determine the value of $\mu_{min}$ we use \eqref{eq:2stat} to get
 $\m_{min} =  \pi - U'(q^*) = 60 - U'(2) = 2$.
sp2668's avatar
sp2668 committed
167
168
169
\end{enumerate}

%=============== ======================================================
sp2668's avatar
sp2668 committed
170
\paragraph{Solution IV.2 \normalsize (Economic dispatch in a single bidding zone).}~\\
sp2668's avatar
sp2668 committed
171
172
173
174
%=====================================================================

Consider an electricity market with two generator types, one with variable cost $c = 20\emwh$, capacity $K = 300\mw$ and a dispatch rate of $Q_1$~[MW] and another with variable cost $c=50\emwh$, capacity $K=400\mw$ and a dispatch rate of $Q_2$~[MW]. The demand has utility function $U(Q) = 8000Q - 5Q^2$~[\euro/h] for a consumption rate of $Q$~[MW].
\begin{enumerate}[(a)]
sp2668's avatar
sp2668 committed
175
176
177
178
 \begin{shaded}\item What are the objective function and constraints required for an optimisation problem to maximise short-run social welfare in this market?\end{shaded}

 The optimisation problem has objective function:
 \begin{equation*}
sp2668's avatar
sp2668 committed
179
  \max_{Q,Q_1,Q_2}\left[ U(Q) - C_1(Q_1) - C_2(Q_2) \right] =     \max_{Q,Q_1,Q_2} \left[8000Q-5Q^2 - c_1Q_1 - c_2Q_2 \right]
sp2668's avatar
sp2668 committed
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
 \end{equation*}
 with constraints:
 \begin{align*}
  Q - Q_1 - Q_2 & = 0 \leftrightarrow \l              \\
  Q_1           & \leq K_1 \leftrightarrow \bar{\m}_1 \\
  Q_2           & \leq K_2 \leftrightarrow \bar{\m}_2 \\
  -Q_1          & \leq 0 \leftrightarrow \ubar{\m}_1  \\
  -Q_2          & \leq 0 \leftrightarrow \ubar{\m}_2
 \end{align*}

 \begin{shaded}\item Write down the Karush-Kuhn-Tucker (KKT) conditions for this problem.\end{shaded}

 Stationarity gives for $Q$:
 \begin{equation*}
  \frac{\d U}{\d Q} - \l  = 8000 - 10Q - \l = 0
 \end{equation*}
 Stationarity for $Q_1$ gives:
 \begin{equation*}
  -\frac{\d C_1}{\d Q_1} + \l - \m_1  =  -c_1+ \l - \bar{\m}_1 + \ubar{\m_1} = 0
 \end{equation*}
 Stationarity for $Q_2$ gives:
 \begin{equation*}
  -\frac{\d C_2}{\d Q_2} + \l - \m_2  =  -c_2+ \l - \bar{\m}_2 + \ubar{\m_2} = 0
 \end{equation*}
 Primal feasibility is just the constraints above. Dual feasibility is $\bar{\m}_i,\ubar{\m}_i \geq 0$ and complementary slackness is $\bar{\m}_i(Q_i-K) = 0$ and $\ubar{\m}_i Q_i = 0$ for $i=1,2$.

 \begin{shaded}\item Determine the optimal rate of production of the generators and the value of all KKT multipliers. What is the interpretation of the respective KKT multipliers?\end{shaded}
 The marginal utility at the full output of the generators, $K_1
  + K_2 = $ 700~MW is $U'(700) = 8000 - 10\cdot700 = 1000$ \euro/MWh,
 which is higher than the costs $c_i$, so we'll find optimal rates
 $Q_1^* = K_1$ and $Q_2^* = K_2$ and $Q^* = K_1+K_2$. This means $\l
  = U'(K_1+K_2) = 1000$ \euro/MWh, which is the market price. Because
 the lower constraints on the generator output are not binding, from
 complementary slackness we have $\ubar{\m}_i = 0$. The upper
 constraints are binding, so $\bar{\m}_i \geq 0$.
 From stationarity $\bar{\m}_i =
  \l - c_i$, which is the increase in social welfare if Generator $i$
 could increase its capacity by a marginal amount.
sp2668's avatar
sp2668 committed
218
219
220
\end{enumerate}

%=============== ======================================================
sp2668's avatar
sp2668 committed
221
\paragraph{Solution IV.3 \normalsize (efficient dispatch in a two-bus power system).}~\\
sp2668's avatar
sp2668 committed
222
223
224
%=====================================================================

\begin{figure}[h]
sp2668's avatar
sp2668 committed
225
226
227
228
 \centering
 \includegraphics[width=14cm]{two-bus}
 \caption{A simple two-bus power system.}
 \label{twobus}
sp2668's avatar
sp2668 committed
229
230
\end{figure}

sp2668's avatar
sp2668 committed
231
Consider the two-bus power system shown in Figure \ref{twobus}, where the two nodes represent two markets, each with different total demand, and one generator at each node. At node A the demand is $D_A = 2000 \si{\mega\watt}$, whereas at node B the demand is $D_B = 1000 \si{\mega\watt}$. Furthermore, there is a transmission line with a capacity denoted by $F_{AB}$. The marginal cost of production of the generators connected to buses A and B are given respectively by the following expressions:
sp2668's avatar
sp2668 committed
232
\begin{align*}
sp2668's avatar
sp2668 committed
233
234
 MC_A & = 20 + 0.03 P_A \hspace{1cm}\eur/\si{\mega\watt\hour}  \\
 MC_B & = 15 + 0.02 P_B \hspace{1cm} \eur/\si{\mega\watt\hour}
sp2668's avatar
sp2668 committed
235
236
237
238
239
\end{align*}

Assume that the demand $D_*$ is constant and insensitive to price, that energy is sold at its marginal cost of production and that there are no limits on the output of the generators.

\begin{enumerate}[(a)]
sp2668's avatar
sp2668 committed
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
 \begin{shaded}\item Calculate the price of electricity at each bus, the production
  of each generator, the flow on the line, and the value of any KKT
  multipliers for the following cases:\end{shaded}
 Use the following nomenclature: price $\lambda_i$, production $Q^{S}_i$, flow $F$.
 \begin{enumerate}[(i)]
  \begin{shaded}\item The line between buses A and B is disconnected.\end{shaded}
  $\l_A= 80\emwh$, $\l_B=35\emwh$,

  $Q_{A}^S=2000$ MW, $Q_B^S=1000$ MW, $F=0$

  \begin{shaded}\item The line between buses A and B is in service and has an unlimited capacity.\end{shaded}
  $\l_A= 53\emwh$, $\l_B=53\emwh$,

  $Q_{A}^S=1100\mw$, $Q_B^S=1900$ MW, $F=-900\mw$

  \begin{shaded}\item The line between buses A and B is in service and has an unlimited capacity, but the maximum output of Generator B is 1500~MW.\end{shaded}
  $\l_A= 65\emwh$, $\l_B=65\emwh$,

  $Q_{A}^S=1500\mw$, $Q_B^S=1500$ MW, $F=-500\mw$

  \begin{shaded}\item The   line between buses A and B is in service and has an unlimited capacity, but the maximum output of Generator A is 900~MW. The output of Generator B is unlimited.\end{shaded}
  $\l_A= 57\emwh$, $\l_B=57\emwh$,

  $Q_{A}^S=900\mw$, $Q_B^S=2100$ MW, $F=-1100\mw$

  \begin{shaded}\item The line between buses A and B is in service but its capacity is limited to 600~MW. The output of the generators is unlimited.\end{shaded}
  $\l_A= 62\emwh$, $\l_B=47\emwh$,

  $Q_{A}^S=1400\mw$, $Q_B^S=1600$ MW, $F=-600\mw$
 \end{enumerate}
 \begin{shaded}\item Calculate the generator revenues, generator profits, consumer payments and consumer net surplus for all the cases considered in the above problem. Who benefits from the line connecting these two buses?\end{shaded}
 Generator revenues $R_{i}$, generator costs $C_{i}$, generator profits $P_{i}$, consumer payments $E_{i}$. Find the generator profits by substracting the costs from the revenue. Costs are given by integrating the marginal cost, i.e. $C_A = 20Q + 0.015Q^2$ and $C_B = 15Q + 0.01Q^2$. The generator at $B$ and the consumers at $A$ benefit from the line (price increases at $B$, decreases at $B$).
 \begin{table}[!h]
  \centering
  \begin{tabular}{lccccc}
   \toprule
   Case          & i      & ii     & iii    & iv     & v      \\
   \midrule
   $E_A$ (\euro) & 160000 & 106000 & 130000 & 114000 & 124000 \\
   \midrule
   $E_B$ (\euro) & 35000  & 53000  & 65000  & 57000  & 47000  \\
   \midrule
   $R_A$ (\euro) & 160000 & 58300  & 97500  & 51300  & 86800  \\
   \midrule
   $R_B$ (\euro) & 35000  & 100700 & 97500  & 119700 & 75200  \\
   \midrule
   $C_A$ (\euro) & 100000 & 40150  & 63750  & 30150  & 57400  \\
   \midrule
   $C_B$ (\euro) & 25000  & 64600  & 45000  & 75600  & 49600  \\
   \midrule
   $P_A$ (\euro) & 60000  & 18150  & 33750  & 21150  & 29400  \\
   \midrule
   $P_B$ (\euro) & 10000  & 36100  & 52500  & 44100  & 25600  \\
   \bottomrule
  \end{tabular}
 \end{table}

 \begin{shaded}\item Calculate the congestion surplus for case (v). For what values of the flow on the line between buses A and B is the congestion surplus equal to zero?\end{shaded}
 Congestion surplus is 9000 \euro:
 \begin{equation*}
  \left(E_A + E_B\right) - (R_A + R_B) = |F|\times (\l_A - \l_B)
 \end{equation*}
 Congestion surplus is equal to zero when the flow $F=0$, or when it is equal to the unconstrained value $F=-900\mw$ (then $\l_A = \l_B$).
sp2668's avatar
sp2668 committed
303
304
305
\end{enumerate}

%=============== ======================================================
sp2668's avatar
sp2668 committed
306
%\paragraph{Solution IV.4 \normalsize (bidding in africa with pypsa).}~\\
sp2668's avatar
sp2668 committed
307
308
309
310
311
%=====================================================================



\end{document}