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appendixa solving systems of nonlinear equations chapter 4 of this book describes and analyzes the power ow problem in its ac version this problem is a system of nonlinear equations ...

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                    AppendixA
                    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.
                        Forthesakeofclarity,wefirstdescribetheworkingofthismethodforthecaseof
                    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-RaphsonAlgorithm
                    TheNewton-Raphsonalgorithm is described in this section.
                    A.1.1         OneUnknown
                                                                  . /
                    Consider a nonlinear function f x W R ! R. We aim at finding a value of x so that:
                                                                           . /
                                                                         f  x D0:                                                   (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                                                                          271
                    A.J. Conejo, L. Baringo, Power System Operations, Power Electronics and Power
                    Systems, https://doi.org/10.1007/978-3-319-69407-8
              272                                            A Solving Systems of Nonlinear Equations
                 Using Taylor series, we can express f x.0/ C x.0/ as:
                                                                                 
                                                    . /  .0/    x.0/ 2     2 . / .0/
                f  x.0/ C x.0/ D f x.0/ Cx.0/       df x      C             d f x       C:::
                                                       dx              2        dx2
                                                                                             (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:
                                                            .0/
                                            x.0/  f x           :                         (A.3)
                                                              
                                                            . /  .0/
                                                         df x
                                                           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      ;                      (A.5)
                                                                 
                                                              . /  ./
                                                            df x
                                                             dx
              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
                                                                                    ˇ .C1/    ./ˇ
                                                                                    ˇ            ˇ
                 iterations is lower than a prespecified tolerance , i.e., check if x      x
                 <.Ifso,thealgorithmhasconvergedandthesolutionisx.C1/.Ifnot,continue
                 at Step 3.
              •  Step 3: update the iteration counter     C 1 and continue at Step 1.
              Illustrative Example A.1 Newton-Raphsonalgorithmforaone-unknownproblem
                 Weconsider the following quadratic function:
                                               . /     2
                                              f x Dx 3xC2;
                A Solving Systems of Nonlinear Equations                                                    273
                whosefirst derivative is:
                                                      df.x/ D 2x3:
                                                        dx
                   TheNewton-Raphsonalgorithm proceeds as follows:
                •  Step0:weinitialize 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:
                                            . /2       . /
                                            x 0    3x0 C2                 2
                             . /     . /                                  0 30C2
                            x 1 D x 0                            D0                      D0:6667:
                                                    . /
                                                 2x 0 3                     203
                                                                                             . /       . /
                •  Step 2: we compute absolute value of the difference between x 1 and x 0 , i.e.,
                    0:66670 D0:6667.Sincethisdifferenceisnotsmallenough,wecontinueat
                   j             j
                   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:
                                      
                                    . / 2      . /
                                   x 1    3x1 C2                            2
                     . /    . /                                      0:6667 30:6667C2
                   x 2 D x 1                           D0:6667                                   D0:9333:
                                          . /
                                       2x 1 3                              20:66673
                                                                                                  . /       . /
                •  Step 2: we compute the absolute value of the difference between x 2 and x 1 ,
                   i.e., 0:9333  0:6667 D 0:2666. Since this difference is not small enough, we
                        j                   j
                   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
                                                                                      4
                algorithm converges in four iterations for a tolerance of 1  10         .
                   Note that the number of iterations needed for convergence by the Newton-
                Raphson algorithm is small.
                                                                                                             
                Table A.1 Illustrative                                                      Iteration  x
                ExampleA.2:results                                                          0          0
                                                                                            1          0.6667
                                                                                            2          0.9333
                                                                                            3          0.9961
                                                                                            4          1.0000
               274                                             A Solving Systems of Nonlinear Equations
               A.1.2     ManyUnknowns
               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
                                            ˆ .                /
                                               f  x ;x ;:::;x    D0;
                                            ˆ1 1 2            n
                                            ˆ
                                            ˆ
                                            ˆ
                                            < .                /
                                               f  x ;x ;:::;x    D0;
                                               2   1  2       n                                  (A.6)
                                            ˆ :
                                            ˆ :
                                            ˆ :
                                            ˆ
                                            ˆ
                                            : .                /
                                               f  x ;x ;:::;x    D0;
                                               n   1  2       n
                        .             /    n
               where f x ;x ;:::;x      W R !R,iD1;:::;n,arenonlinearfunctions.
                       i  1  2       n
                  Thesystemofequations (A.6) can be rewritten in compact form as:
                                                       . /
                                                     f x D0;                                     (A.7)
               where:
                                                   >         n      n
                   . /    Œ  . /    . /        . /
               •  f x D f x f x :::f x               D0:R !R ,
                            1     2 >        n
                       Œ            
               •  x D x x :::x         ,
                         1  2      n
                       Œ         >
               •  0 D 00:::0 ,and
               •  >denotesthetranspose operator.                                               
                                                            .0/                              .0/
                  Givenaninitial value for vector x, i.e., x  , we have, in general, that f x    ¤0.
                                         .0/          .0/      .0/
               Thus,weneedtofindx           sothatf x     Cx       D0.Usingthefirst-orderTaylor
                        .0/       .0/
               series, f x   Cx       can be approximately expressed as:
                                        .0/       .0/     .0/     .0/   .0/
                                      f x    Cx       f x       CJ x ;                        (A.8)
               where J is the n  n Jacobian:
                                            2@f . /        . /         . / 3
                                                   x @f x          @f   x
                                                 1       1        1
                                            6 @x         @x          @x    7
                                            6      1       2           n   7
                                                  . /      . /         . /
                                               @f  x @f x          @f   x
                                            6 2          2        2     7
                                            6                              7
                                            6 @x         @x          @x    7
                                       J D 6       1       2           n   7:                    (A.9)
                                            6 :           :    :      :    7
                                                  :       :     ::    :
                                            6 :           :           :    7
                                            4     . /      . /         . / 5
                                               @f  x @f x          @f   x
                                                 n       n        n
                                                 @x      @x          @x
                                                   1       2           n
                                    .0/      .0/                                            .0/
                  Since we seek f x     Cx        D0,fromEq.(A.8)wecancomputex                 as:
                                               .0/       .0/1  .0/
                                            x J               f x    :                      (A.10)
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...Appendixa solving systems of nonlinear equations chapter this book describes and analyzes the power ow problem in its ac version is a system appendix most common method for namely newton raphson an iterative that uses initial values unknowns then at each iteration updates these until no change occurs two consecutive iterations forthesakeofclarity werstdescribetheworkingofthismethodforthecaseof just one equation with unknown general case n considered we also explain how to directly solve using appropriate software raphsonalgorithm thenewton described section oneunknown consider function f x w r aim nding value so d do rst given e g have thus it necessary nd c springer international publishing ag j conejo l baringo operations electronics https doi org taylor series can express as df dx considering only terms eq since seek approximately compute next update check if found satises not repeat above step on where counter consists following steps initialize provide i difference between lower t...

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