Appendix A Solving Systems of Nonlinear Equations

Transcription

1 Appendix A Solving Systems of Nonlinear Equations Chapter 4 of this book describes and analyzes the power flow problem. In its ac version, this problem is a system of nonlinear equations. This appendix describes the most common method for solving a system of nonlinear equations, namely, the Newton-Raphson method. This is an iterative method that uses initial values for the unknowns and, then, at each iteration, updates these values until no change occurs in two consecutive iterations. For the sake of clarity, we first describe the working of this method for the case of just one nonlinear equation with one unknown. Then, the general case of n nonlinear equations and n unknowns is considered. We also explain how to directly solve systems of nonlinear equations using appropriate software. A.1 Newton-Raphson Algorithm The Newton-Raphson algorithm is described in this section. A.1.1 One Unknown Consider a nonlinear function f.x/ W R! R. We aim at finding a value of x so that: f.x/ D 0: (A.1) To do so, we first consider a given value of x, e.g., x.0/. In general, we have that f x.0/ 0. Thus, it is necessary to find x.0/ so that f x.0/ C x.0/ D 0. Springer International Publishing AG 2018 A.J. Conejo, L. Baringo, Power System Operations, Power Electronics and Power Systems, 271

2 272 A Solving Systems of Nonlinear Equations Using Taylor series, we can express f x.0/ C x.0/ as: f x.0/ C x.0/ D f x.0/ df.x/.0/ x.0/ 2 C x.0/ C dx 2 d 2.0/ f.x/ C ::: dx 2 (A.2) Considering only the first two terms in Eq. (A.2) and since we seek to find x.0/ so that f x.0/ C x.0/ D 0, we can approximately compute x.0/ as: x.0/ f x.0/ df.x/.0/ : (A.3) dx Next, we can update x as: x.1/ D x.0/ C x.0/ : (A.4) Then, we check if f x.1/ D 0. If so, we have found a value of x that satisfies f.x/ D 0. If not, we repeat the above step to find x.1/ so that f x.1/ C x.1/ D 0 and so on. In general, we can compute x./ as: x.c1/ D x./ f x./ df.x/./ ; (A.5) where is the iteration counter. Considering the above, the Newton-Raphson method consists of the following steps: Step 0: initialize the iteration counter ( D 0) and provide an initial value for x, i.e., x D x./ D x.0/. Step 1: compute x.c1/ using Eq. (A.5). Step 2: check if the difference between the values of x in two consecutive iterations is lower than a prespecified tolerance, i.e., check if ˇˇx.C1/ x./ˇˇ <. If so, the algorithm has converged and the solution is x.c1/. If not, continue at Step 3. Step 3: update the iteration counter C 1 and continue at Step 1. Illustrative Example A.1 Newton-Raphson algorithm for a one-unknown problem We consider the following quadratic function: dx f.x/ D x 2 3x C 2;

3 A Solving Systems of Nonlinear Equations 273 whose first derivative is: df.x/ D 2x 3: dx The Newton-Raphson algorithm proceeds as follows: Step 0: we initialize the iteration counter ( D 0) and provide an initial value for x, e.g., x./ D x.0/ D 0. Step 1: we compute x.1/ using the equation below: x.1/ D x.0/ x.0/ 2 3x.0/ C 2 2x.0/ 3 D C D 0:6667: Step 2: we compute absolute value of the difference between x.1/ and x.0/, i.e., j0:6667 0j D 0:6667. Since this difference is not small enough, we continue at Step 3. Step 3: we update the iteration counter D 0 C 1 D 1 and continue at Step 1. Step 1: we compute x.2/ using the equation below: x.2/ D x.1/ x.1/ 2 3x.1/ C 2 2x.1/ 3 D 0:6667 0: :6667 C 2 2 0: D 0:9333: Step 2: we compute the absolute value of the difference between x.2/ and x.1/, i.e., j0:9333 0:6667j D 0:2666. Since this difference is not small enough, we continue at Step 3. Step 3: we update the iteration counter D 1 C 1 D 2 and continue at Step 1. This iterative algorithm is repeated until the difference between the values of x in two consecutive iterations is small enough. Table A.1 summarizes the results. The algorithm converges in four iterations for a tolerance of Note that the number of iterations needed for convergence by the Newton- Raphson algorithm is small. Table A.1 Illustrative Example A.2: results Iteration x

4 274 A Solving Systems of Nonlinear Equations A.1.2 Many Unknowns The Newton-Raphson method described in the previous section is extended in this section to the general case of a system of n nonlinear equations with n unknowns, as the one described below: 8 f 1.x 1 ; x 2 ;:::;x n / D 0; ˆ< f 2.x 1 ; x 2 ;:::;x n / D 0; : ˆ: f n.x 1 ; x 2 ;:::;x n / D 0; where f i.x 1 ; x 2 ;:::;x n / W R n! R, i D 1;:::;n, are nonlinear functions. The system of equations (A.6) can be rewritten in compact form as: (A.6) f.x/ D 0; (A.7) where: f.x/ D Œf 1.x/ f 2.x/ ::: f n.x/ > D 0: R n! R n, x D Œx 1 x 2 ::: x n >, 0 D Œ00 ::: 0 >, and > denotes the transpose operator. Given an initial value for vector x, i.e., x.0/, we have, in general, that f x.0/ 0. Thus, we need to find x.0/ so that f x.0/ C x.0/ D 0. Using the first-order Taylor series, f x.0/ C x.0/ can be approximately expressed as: f x.0/ C x.0/ f x.0/ C J.0/ x.0/ ; (A.8) where J is the n n Jacobian: 2.x/ J 1 6 : n : :: : : (A.9) :: n.x/ n Since we seek f x.0/ C x.0/ D 0,fromEq.(A.8) we can compute x.0/ as: x.0/ J.0/ 1 f x.0/ : (A.10)

5 A Solving Systems of Nonlinear Equations 275 Then, we can update vector x as: x.1/ D x.0/ C x.0/ : (A.11) In general, we can update vector x as: x.c1/ D x./ J./ 1 f x./ ; (A.12) where is the iteration counter. Considering the above, the Newton-Raphson algorithm consists of the following steps: Step 0: initialize the iteration counter ( D 0) and provide an initial value for vector x, i.e., x D x./ D x.0/. Step 1: compute the Jacobian J using (A.9). Step 2: compute x.c1/ using matrix equation (A.12). Step 3: check every element of the absolute value of the difference between the values of vector x in two consecutive iterations is lower than a prespecified tolerance, i.e., check if ˇˇx.C1/ x./ˇˇ <. If so, the algorithm has converged and the solution is x.c1/. If not, continue at Step 4. Step 4: update the iteration counter C 1 and continue at Step 1. For the sake of clarity, this iterative algorithm is schematically described through the flowchart in Fig. A.1. Illustrative Example A.2 Newton-Raphson algorithm for a two-unknown problem We consider the following system of two equations and two unknowns: ( f 1.x; y/ D x C xy 4; f 2.x; y/ D x C y 3: We aim at finding the values of x and y so that f 1.x; y/ D 0 and f 2.x; y/ D 0. To do so, we use the Newton-Raphson method. First, we compute the partial derivatives: 1.x; y/ D 1 1.x; y/ ˆ< 2.x; y/ 2.x; y/ ˆ: D

6 276 A Solving Systems of Nonlinear Equations n = 0 x = x (0) Compute the Jacobian J ( ) using (A.9) n n Compute x ( +1) using (A.12) x ( n +1) x ( n ) < e? YES END NO n + 1 n Fig. A.1 Algorithm flowchart for the Newton-Raphson method Second, we build the Jacobian matrix: J D 1 C y x : 1 1 Then, we follow the iterative procedure described above: Step 0: we initialize the iteration counter ( D 0) and provide initial values for variables x and y, e.g., x./ D x.0/ D 1:98 and y./ D y.0/ D 1:02, respectively. Step 1: we compute the Jacobian matrix J at iteration D 0: 1 C y J.0/.0/ x D.0/ D C 1:02 1:98 D 1 1 2:02 1:98 : 1 1

7 A Solving Systems of Nonlinear Equations 277 Table A.2 Illustrative Example A.2: results Iteration x y Step 2: we compute x.1/ and y.1/ using the matrix equation below: x.1/ x.0/ D y.1/ D y.0/ 1:98 1:02 1 C y.0/ x.0/ 1 x.0/ C x.0/ y./ 4 x.0/ C y.0/ D 0 2:02 1: :9900 : 1:0100 Step 3: we compute the difference between x.1/ and x.0/, i.e., j1:9900 1:98j D 0:01, as well as the differences between y.1/ and y.0/, i.e., j1:0100 1:02j D 0:01. Since these differences are not small enough, we continue with Step 4. Step 4: we update the iteration counter D 0 C 1 D 1 and continue with Step 1. This iterative algorithm is repeated until the differences between the values of x and y in two consecutive iterations are small enough. Table A.2 provides the evolution of the values of these unknowns. The algorithm converges in nine iterations for a tolerance of Note that the number of iterations needed by the Newton-Raphson algorithm is rather small. Next, we consider a different initial solution. Table A.3 provides the results. In this case, the algorithm converges in 11 iterations for a tolerance of We conclude that the initial solution does not have an important impact on the number of iterations required for convergence, provided that convergence is attained. However, convergence is not necessarily guaranteed, and the Jacobian may be singular at any iteration. Further details on convergence guarantee and on convergence speed are available in [1].

8 278 A Solving Systems of Nonlinear Equations Table A.3 Illustrative Example A.2: results considering a different initial solution Iteration x y A.2 Direct Solution Generally, the Newton-Raphson method does not need to be implemented. An offthe-self routine (in GNU Octave [2] or MATLAB [3]) embodying the Newton- Raphson algorithm can be used to solve systems of nonlinear equations. Illustrative Examples A.1 and A.2 are solved below using GNU Octave routines. A.2.1 One Unknown The GNU Octave [2] routines below solve Illustrative Example A.1: 1 clc 2 fun 3 x0 = [0]; x = fsolve(fun,x0) 1 function F = NR1(x) 2 % 3 F(1)=x(1)*x(1)-3*x(1)+2; The solution provided by GNU Octave is: 1 x =

9 A Solving Systems of Nonlinear Equations 279 A.2.2 Many Unknowns The GNU Octave routines below solve Illustrative Example A.2: 1 clc 2 fun 3 x0 = [1.98,1.02]; x = fsolve(fun,x0) 1 function F = NR2(x) 2 % 3 F(1)=x(1)+x(1)*x(2)-4; 4 F(2)=x(1)+x(2)-3; 1 x = The solution provided by GNU Octave is: A.3 Summary and Further Reading This appendix describes the Newton-Raphson method, which is the most common method for solving systems of nonlinear equations, as those considered in Chap. 4 of this book. The Newton-Raphson method is based on an iterative procedure that updates the value of the unknowns involved until the changes in their values in two consecutive iterations are small enough. Different illustrative examples are used to show the working of the Newton- Raphson method. Additionally, this appendix explains also how to directly solve a system of nonlinear equations using appropriate software, such as GNU Octave [2]. Additional details can be found in the monograph by Chapra and Canale on numerical methods in engineering [1]. References 1. Chapra, S.C., Canale, R.P.: Numerical Methods for Engineers, 6th edn. McGraw-Hill, New York (2010) 2. GNU Octave (2016): Available at 3. MATLAB (2016): Available at

10 Appendix B Solving Optimization Problems This appendix provides an overview of the general structure of some of the optimization problems considered through the chapters of this book, namely, linear programming, mixed-integer linear programming, and nonlinear programming problems. B.1 Linear Programming Problems The simplest instance of an optimization problem is a linear programming (LP) problem. All variables of an LP problem are continuous and its objective function and constrains are linear. B.1.1 Formulation The general formulation of an LP problem is as follows: min xi ;8i X C i x i subject to i (B.1a) X A ij x i D B j ; j D 1;:::;m; (B.1b) i X D ik x i E k ; k D 1;:::;o; (B.1c) i x i 2 R; i D 1;:::;n; (B.1d) Springer International Publishing AG 2018 A.J. Conejo, L. Baringo, Power System Operations, Power Electronics and Power Systems, 281

11 282 B Solving Optimization Problems where: R is the set of real numbers, C i, 8i, are the cost coefficients of variables x i, 8i, in the objective function (B.1a), A ij, 8i, and B j are the coefficients that define equality constraints (B.1b), 8j, D ik, 8i, and E k are the coefficients that define inequality constraints (B.1c), 8k, n is the number of continuous optimization variables, m is the number of equality constraints, and o is the number of inequality constraints. In compact form, the LP problem (B.1) can be written as: min x C > x (B.2a) subject to Ax D B; Dx E; x 2 R n1 ; (B.2b) (B.2c) (B.2d) where: superscript > denotes the transpose operator, C 2 R n1 is the cost coefficient vector of the variable vector x in the objective function (B.2a), A 2 R mn and B 2 R m1 are the matrix and the vector of coefficients that define equality constraint (B.2b), and D 2 R on and E 2 R o1 are the matrix and the vector of coefficients that define inequality constraint (B.2c). Some examples of LP problems are the dc optimal power flow problem analyzed in Chap. 6 or the economic dispatch problem described in Chap. 7 of this book. B.1.2 Solution One of the most common and efficient methods for solving LP problems is the simplex method [2]. A detailed description of this method can be found, for instance, in [4]. LP problems can be also solved using one of the many commercially available software tools. For example, in this book we use CPLEX [5] under GAMS [3]. Illustrative Example B.1 Linear programming We consider a generating unit with a capacity of 10 MW and a variable cost of $21/MWh. This generating unit has to decide its power output for the following 6 h,

12 B Solving Optimization Problems 283 knowing that the electric energy prices in these hours are $10/MWh, $15/MWh, $22/MWh, $30/MWh, $24/MWh, and $20/MWh, respectively. Considering these data, we formulate the following LP problem: max p1 ;p 2 ;p 3 ;p 4 ;p 5 ;p 6 subject to 10p 1 C 15p 2 C 22p 3 C 30p 4 C 24p 5 C 20p 6 21.p 1 C p 2 C p 3 C p 4 C p 5 C p 6 / 0 p 1 10; 0 p 2 10; 0 p 3 10; 0 p 4 10; 0 p 5 10; 0 p 6 10: The solution of this problem is (note that a superscript in the variables below indicates optimal value): p 1 D 0; p 2 D 0; p 3 p 4 p 5 D 10 MW; D 10 MW; D 10 MW; p 6 D 0: This solution renders an objective function value of $130. A simple input GAMS [3] file to solve Illustrative Example B.1 is provided below: 1 variables z, p1, p2, p3, p4, p5, p6; 3 equations fobj, eq1a, eq1b, eq2a, eq2b, eq3a, eq3b, eq4a, eq4b, eq5a, eq5b, eq6a, eq6b; 5 fobj.. z=e=10*p1+15*p2+22*p3+30*p4+24*p5+20*p6-21*(p1+p2+p3+p4+ p5+p6);

13 284 B Solving Optimization Problems 7 eq1a.. 0=l=p1; 8 eq1b.. p1=l=10; 9 eq2a.. 0=l=p2; 10 eq2b.. p2=l=10; 11 eq3a.. 0=l=p3; 12 eq3b.. p3=l=10; 13 eq4a.. 0=l=p4; 14 eq4b.. p4=l=10; 15 eq5a.. 0=l=p5; 16 eq5b.. p5=l=10; 17 eq6a.. 0=l=p6; 18 eq6b.. p6=l=10; 20 model example_lp /all/; 21 solve example_lp using lp maximizing z; 23 display z.l, p1.l, p2.l, p3.l, p4.l, p5.l, p6.l; The part of the GAMS output file that provides the optimal solution is given below: variable z.l = variable p1.l = variable p2.l = variable p3.l = variable p4.l = variable p5.l = variable p6.l = B.2 Mixed-Integer Linear Programming Problems A mixed-integer linear programming (MILP) problem is an LP problem in which some of the optimization variables are not continuous but integer. B.2.1 Formulation The general formulation of a MILP problem is as follows: min xi ;8iIy`;8` X C i x i C X` i R`y` (B.3a)

14 B Solving Optimization Problems 285 subject to X A ij x i C X` G`j u` D B j ; 8j; (B.3b) i X D ik x i C X` H`k u` E k ; 8k; (B.3c) i x i 2 R; 8i; (B.3d) y` 2 I; ` D 1;:::;p; (B.3e) where: I is the set of integer variables, C i, 8i, and R`, 8`, are the cost coefficients of variables x i, 8i, and y`, 8`, respectively, in the objective function (B.3a), A ij, 8i; G`j, 8`; and B j are the coefficients that define equality constraints (B.3b), 8j, D ik, 8i; H`k, 8`; and E k are the coefficients that define inequality constraints (B.3c), 8k, and p is the number of integer optimization variables, In compact form, MILP problem (B.3) can be written as: min x;y C > x C R > y (B.4a) subject to Ax C Gy D B; Dx C Hy E; x 2 R n1 ; y 2 I p1 ; (B.4b) (B.4c) (B.4d) (B.4e) where: C 2 R n1 and R 2 R p1 are the cost coefficient vectors of the variable vectors x and y, respectively, in objective function (B.2a), A 2 R mn, G 2 R mp, and B 2 R m1 are the matrices and vector of coefficients that define equality constraint (B.2b), D 2 R on, H 2 R op, and E 2 R o1 are the matrices and vector of coefficients that define inequality constraint (B.2c), Some examples of MILP problems are the unit commitment problem described in Chap. 7 or the self-scheduling problem analyzed in Chap. 8 of this book.

15 286 B Solving Optimization Problems B.2.2 Solution MILP problems can be solved using branch-and-cut methods. A detailed description of these methods can be found, for instance, in [4]. MILP problems can also be solved using one of the many commercially available software tools. For example, in this book we use CPLEX [5] under GAMS [3]. Illustrative Example B.2 Mixed-integer linear programming We consider again the data of Illustrative Example B.1. However, in this case, we assume that the generating unit has a minimum power output of 2 MW and a fixed cost of $25. Considering these data, we formulate the following MILP problem: min p1 ;p 2 ;p 3 ;p 4 ;p 5 ;p 6 ;u 1 ;u 2 ;u 3 ;u 4 ;u 5 ;u 6 subject to 10p 1 C 15p 2 C 22p 3 C 30p 4 C 24p 5 C 20p 6 21.p 1 C p 2 C p 3 C p 4 C p 5 C p 6 / 25.u 1 C u 2 C u 3 C u 4 C u 5 C u 6 / 2u 1 p 1 10u 1 ; 2u 2 p 2 10u 2 ; 2u 3 p 3 10u 3 ; 2u 4 p 4 10u 4 ; 2u 5 p 5 10u 5 ; 2u 6 p 6 10u 6 ; u 1 ; u 2 ; u 3 ; u 4 ; u 5 ; u 6 2f0; 1g: In this example it is necessary to include binary variables to represent the on/off status of the generating unit at each time period. The solution of this problem is (note that a superscript in the variables below indicates optimal value): p 1 D 0; p 2 D 0; p 3 D 0; p 4 p 5 D 10 MW; D 10 MW;

16 B Solving Optimization Problems 287 p 6 D 0; u 1 D 0; u 2 D 0; u 3 D 0; u 4 D 1; u 5 D 1; u 6 D 0: Contrary to the solution of Illustrative Example B.1, we can see that in this example it is not optimal to turn on the generating unit at the third time period as a result of its fixed cost. This solution renders an objective function value of $70. A simple input GAMS [3] file to solve Illustrative Example B.2 is provided below: 1 variables z, p1, p2, p3, p4, p5, p6; 3 binary variables u1, u2, u3, u4, u5, u6; 5 equations fobj, eq1a, eq1b, eq2a, eq2b, eq3a, eq3b, eq4a, eq4b, eq5a, eq5b, eq6a, eq6b; 7 fobj.. z=e=10*p1+15*p2+22*p3+30*p4+24*p5+20*p6-21*(p1+p2+p3+p4+ p5+p6)-5*(u1+u2+u3+u4+u5+u6); 9 eq1a.. 2*u1=l=p1; 10 eq1b.. p1=l=10*u1; 11 eq2a.. 2*u2=l=p2; 12 eq2b.. p2=l=10*u2; 13 eq3a.. 2*u3=l=p3; 14 eq3b.. p3=l=10*u3; 15 eq4a.. 2*u4=l=p4; 16 eq4b.. p4=l=10*u4; 17 eq5a.. 2*u5=l=p5; 18 eq5b.. p5=l=10*u5; 19 eq6a.. 2*u6=l=p6; 20 eq6b.. p6=l=10*u6; 22 model example_milp /all/; 23 solve example_milp using mip maximizing z; 25 display z.l, p1.l, p2.l, p3.l, p4.l, p5.l, p6.l, u1.l, u2.l, u3. l, u4.l, u5.l, u6.l;

17 288 B Solving Optimization Problems The part of the GAMS output file that provides the optimal solution is given below: variable z.l = variable p1.l = variable p2.l = variable p3.l = variable p4.l = variable p5.l = variable p6.l = variable u1.l = variable u2.l = variable u3.l = variable u4.l = variable u5.l = variable u6.l = B.3 Nonlinear Programming Problems A nonlinear programming (NLP) problem is an optimization problem in which the objective function and/or some of the constraints are nonlinear. B.3.1 Formulation The general formulation of an NLP problem is as follows: min xi ;8i f.x 1 ;:::;x n / (B.5a) subject to where: A j.x 1 ;:::;x n / D 0; 8j; (B.5b) D k.x 1 ;:::;x n / 0; 8k; (B.5c) x i 2 R; 8i; (B.5d) f.x 1 ;:::;x n /: R n! R is the nonlinear objective function (B.5a), A j.x 1 ;:::;x n /: R n! R are the nonlinear functions that define equality constraints (B.5b), 8j, and D k.x 1 ;:::;x n /: R n! R are the nonlinear functions that define inequality constraints (B.5c), 8k.

18 B Solving Optimization Problems 289 Problem (B.5) can be rewritten in compact form as: min x f.x/ (B.6a) subject to A.x/ D 0; D.x/ 0; x 2 R n ; (B.6b) (B.6c) (B.6d) where: f.x/: R n! R is the nonlinear objective function (B.6a), A.x/: R n! R m is the nonlinear function that defines equality constraint (B.6b), and D.x/: R n! R o is the nonlinear function that defines inequality constraint (B.6c). Some examples of NLP problems are the state estimation problem described in Chap. 5 or the ac optimal power flow problem analyzed in Chap. 6 of this book. B.3.2 Solution Solving NLP problems is generally more complicated than solving LP problems or MILP problems. NLP problems can be solved using one of the many commercially available software tools. For example, in this book we use CONOPT [1] under GAMS [3]. Further information about NLP problems can be found, for instance, in [4]. Illustrative Example B.3 Nonlinear programming We consider again the data of Illustrative Example B.1. However, in this case, we assume that the generating unit has a quadratic cost function so that the cost is: c t D 15p t C 2p 2 t ; t D 1;:::;6: Considering these data, we formulate the following NLP problem: max p1 ;p 2 ;p 3 ;p 4 ;p 5 ;p 6 10p 1 C 15p 2 C 22p 3 C 30p 4 C 24p 5 C 20p 6 15.p 1 C p 2 C p 3 C p 4 C p 5 C p 6 / 2 p 2 1 C p2 2 C p2 3 C p2 4 C p2 5 C p2 6

19 290 B Solving Optimization Problems subject to 0 p 1 10; 0 p 2 10; 0 p 3 10; 0 p 4 10; 0 p 5 10; 0 p 6 10: The solution of this problem is (note that a superscript in the variables below indicates optimal value): p 1 D 0; p 2 D 0; p 3 p 4 p 5 p 6 D 1:75 MW; D 3:75 MW; D 2:25 MW; D 1:25 MW: This solution renders an objective function value of $47.5. A simple input GAMS [3] file to solve Illustrative Example B.3 is provided below: 1 variables z, p1, p2, p3, p4, p5, p6; 3 equations fobj, eq1a, eq1b, eq2a, eq2b, eq3a, eq3b, eq4a, eq4b, eq5a, eq5b, eq6a, eq6b; 5 fobj.. z=e=10*p1+15*p2+22*p3+30*p4+24*p5+20*p6-15*(p1+p2+p3+p4+ p5+p6)-2*(p1*p1+p2*p2+p3*p3+p4*p4+p5*p5+p6*p6); 7 eq1a.. 0=l=p1; 8 eq1b.. p1=l=10; 9 eq2a.. 0=l=p2; 10 eq2b.. p2=l=10; 11 eq3a.. 0=l=p3; 12 eq3b.. p3=l=10; 13 eq4a.. 0=l=p4; 14 eq4b.. p4=l=10; 15 eq5a.. 0=l=p5; 16 eq5b.. p5=l=10;

20 B Solving Optimization Problems eq6a.. 0=l=p6; 18 eq6b.. p6=l=10; 20 model example_nlp /all/; 21 solve example_nlp using nlp maximizing z; 23 display z.l, p1.l, p2.l, p3.l, p4.l, p5.l, p6.l; The part of the GAMS output file that provides the optimal solution is given below: variable z.l = variable p1.l = variable p2.l = variable p3.l = variable p4.l = variable p5.l = variable p6.l = B.4 Summary and Further Reading This appendix provides brief formal descriptions of the three types of optimization problems considered in this book, namely, LP problems, MILP problems, and NLP problems. A detailed description of these problems can be found in the monograph by Sioshansi and Conejo [4]. References 1. CONOPT (2016). Available at 2. Dantzig, G.B.: Linear Programming and Extensions. Princeton University Press, Princeton, NJ (1963) 3. GAMS (2016). Available at 4. Sioshansi, R., Conejo, A.J.: Optimization in Engineering. Models and Algorithms. Springer, New York (2017) 5. The ILOG CPLEX (2016). Available at

21 Index A ac source Angular frequency, 18 Ordinary frequency, 18 Period, 18 Phasorial representation, 18 Root mean square, 18 Sinusoidal representation, 18 Active and reactive power decoupling, 81 Admittance matrix, 101 Alternating current (ac), 17 B Balanced three-phase circuits, 17 Active power, 43 Apparent power, 44 Balanced three-phase sequence, 18 Common star connection, 38 Currents, 23 Delta currents, 25 Equivalence wye-delta, 28 Exercises, 52 How to measure power?, 44 Instantaneous power, 43 Line currents, 23 Line voltages, 22 Magnitudes, 21 Negative sequence, 20 Phase voltages, 22 Positive sequence, 19 Power, 42 Reactive power, 44 Voltages, 21 E Economic dispatch, 197, 209 Capacity limits of transmission lines, 212 Cost function, 213 Description, 211 Example, 210, 214 Example: impact of transmission capacity limits, 215 Example: locational marginal prices, 216 Example: marginal prices, 211 Formulation, 213 GAMS code, 225 Locational marginal prices, 216 Marginal prices, 211 Power balance, 213 Power bounds, 212 Power flows through transmission lines, 211 Reference node, 212 Electrical line, 71 Capacity, 79 Efficiency, 80 Geometric mean radius, 78 Inductance, 76 Model, 71 Parameters, 75 Reactance, 79 Regulation, 80 Resistance, 75 Resistivity, 75 Springer International Publishing AG 2018 A.J. Conejo, L. Baringo, Power System Operations, Power Electronics and Power Systems, 293

22 294 Index G Generator and motor, 56 Efficiency, 58 Three-phase generator, 56 Three-phase motor, 57 I Introduction, 1 K Kirchhoff s laws, 99 L Load, 68 Constant impedance, 69 Constant power, 71 Constant voltage, 71 Induction motor, 69 Induction motor efficiency, 70 Model, M Magnetic constant, 76 Market clearing auction, 242 Bids, 233 Consumer surplus, 244 Consumption bid curve, 244 Example, 248, 252, 257 Formulation, 246 Formulation: multi period, 251 Formulation: single period, 247 Formulation: transmission-constrained multi-period, 256 GAMS code, 264 Introduction, 234 Locational marginal prices, 257 Market clearing price, 250 Market operator, 234 Offers, 233 Participants, 242 Producer surplus, 244 Production offer curve, 243 Profit of generating units, 250 Social welfare, 244 N Network-constrained unit commitment, 197 Example, 218 Formulation, 217 GAMS code, 226 Newton-Raphson method, 111 O Optimal power flow, 165 Active power limits, 168 Concluding remarks, 185 dc example, 178 dc formulation, 177 dc GAMS code, 189 Description, 166 Example, 171 Formulation, 170 GAMS code, 185 Introduction, 165 Objective function, 169 Power balance, 166 Power flows through transmission lines, 167 Reactive power limits, 168 Security, 179 Solution, 171 Transmission line limits, 168 Voltage angle limits, 169 Voltage magnitude limits, 169 Optimization problems, 281 Linear programming, 281 Linear programming: example, 282 Linear programming: formulation, 281 Linear programming: GAMS code, 283 Linear programming: simplex method, 282 Linear programming: solution, 282 Mixed-integer linear programming, 284 Mixed-integer linear programming: branch-and-bound methods, 286 Mixed-integer linear programming: example, 286 Mixed-integer linear programming: formulation, 284 Mixed-integer linear programming: GAMS code, 287 Mixed-integer linear programming: solution, 286 Non-linear programming, 288 Non-linear programming: example, 289 Non-linear programming: formulation, 288 Non-linear programming: GAMS code, 290 Non-linear programming: solution, 289

23 Index 295 P Per-unit system, 46 Base value, 48 Definition, 46 Example, 47, 50 Procedure, 51 Permittivity, 79 Phasor, 18 Power, 42 Active power, 43 Apparent power, 44 How to measure power?, 44 Instantaneous power, 43 Reactive power, 44 Power flow, 97 Applications, 97 Concluding remarks, 130 dc formulation, 121 Decoupled, 119 Distributed slack, 120 Equations, 104 Example, 117 Example in Octave, 130 Exercises, 132 Introduction, 97 Nodal equations, 98 Outcome, 114 Slack, PV, and PQ nodes, 109 Solution, 110 Power markets, 10 Day-ahead market, 12 Futures market, 11 Intra-day markets, 12 Pool, 11 Real-time market, 12 Power system Fundamentals, 17 Model components, 55 Power system components Examples, 83 Power system operations, 9 Day-ahead operation, 9 Hours before power delivery, 10 Minutes before power delivery, 10 Power system structure, 1 Centralized operation, 7 Distribution, 4 Economic layer, 5 Generation, 2 Market operation, 8 Physical layer, 1 Regulatory layer, 7 Supply, 4 Transmission, 3 Power transformer, 58 Connections, 61 Denomination, 66 Model, 67 Per-unit analysis, 66 Transformation ratio, 60 S Scope of the book, 12 What we do, 13 What we do not do, 13 Security-constrained optimal power flow, 179 n 1 security, 180 n k security, 180 Concluding remarks, 185 Corrective approach, 180 Description, 180 Example, 182 Formulation, 180 GAMS code, 190 Introduction, 179 Preventive approach, 180 Self-scheduling, 234 Description, 234 Example, 237, 240 Formulation, 236 GAMS code, 263 Introduction, 233 Self-scheduling and market clearing auction, 233 Final remarks, 262 Introduction, 233 Sinusoidal ac source, 18 State estimation, 137 Cumulative distribution function, 155 Erroneous measurement detection, 152 Erroneous measurement detection: 2 test, 152 Erroneous measurement detection: Example, 153 Erroneous measurement identification, 154 Erroneous measurement identification: Example, 155 Erroneous measurement identification: Normalized residual test, 154 Estimation, 140 Estimation: Example, 143 Exercises, 160 Measurements, 138 Non-observable, 148 Observability, 145 Observability: Example, 148 Observable, 148

24 296 Index State estimation (cont.) Residuals, 154 System state, 137 Systems of nonlinear equations, 110, 271 Direct solution, 278 Direct solution: many unknowns, 279 Direct solution: one unknown, 278 Jacobian, 274 Newton-Raphson algorithm, 271 Newton-Raphson algorithm: example, 272, 275 Newton-Raphson algorithm: many unknowns, 274 Newton-Raphson algorithm: one unknown, 271 Taylor series, 272, 274 U Unit commitment, 197, 198 Costs of generating units, 199 Costs of generating units: Fixed costs, 199 Costs of generating units: Shut-down costs, 200 Costs of generating units: Start-up costs, 200 Costs of generating units: Variable costs, 200 Example, 205 Formulation, 204 GAMS code, 223 Generating units, 199 Logical expressions, 201 Logical expressions: Example, 201 Network-constrained unit commitment, 216 Planning horizon, 199 Power balance, 204 Power bounds, 202 Ramping limits, 202 Security constraints, 204 Unit commitment and economic dispatch, 197 Exercises, 228 Final remarks, 222 GAMS codes, 222 Introduction, 197