Lecture #18 EEE 574 Dr. Dan Tylavsky Nonlinear Problem Solvers

Lecture #18

EEE 574

Dr. Dan Tylavsky

Nonlinear Problem Solvers

© Copyright 1999 Daniel Tylavsky

Nonlinear Problem Solvers Two traditional methods for solving simultaneous nonlinear eqautions:

– Guass-Seidel Method (Overview - available in many text books and easy to understand.)

– Newton Raphson Method (We’ll work through in detail).


Nonlinear Problem Solvers Gauss- Seidel Method

– Suppose we wish to find solutions to equation:

0)( xf– Let’s write this equation in the following fixed-point form:

)(xgx – We can always get this form. If by no other means we can ways write:

)()( xgxfxx


Nonlinear Problem Solvers– Then iterate as follows:

)(1 kk xgx

– Where we start with an initial estimate, x0.

– Depending on the characteristics of the problem, the result may be fast or slow to converge, or the solution process may diverge.– With power flow problems that have been examined, the process usually converges.

– Converged when, |xk+1-xk|<.


Nonlinear Problem Solvers• Think-Pair-Square: Find a solution to the following problem starting with an initial estimate of x0=0.4. If x is

a bus voltage, determine a reasonable convergence criteria and be able to justify it.

02.2 xx


Nonlinear Problem Solvers– Asymptotic Rate of Convergence: Largest integer r such that:

.arg,0*

*1

kelforxx

xxrk

k

– Gauss-Seidel method has linear convergence, i.e., r=1.

– Define error as:*xxkk

i

0.5710.1310.0280.01

3.37610 3

1.16410 3

3.97210 4

1.34910 4

4.70310 5

1.5110 5

6.12710 60.57082

6.127099 106

i

100 i

0 5 101 10

6

1 105

1 104

1 103

0.01

0.1

1Plot of Error v. Iteration Number

Interation Index



For linear convergence, asymptotic rate of convergence:

.arg'1*

*1

kelforxx

xxC

k

k

**1

1*

*1

loglog

log)'log(

xxxxC

xx

xxC

kk

k

k

Incremental improvement in estimate is constant on a log scale.


Nonlinear Problem Solvers Gauss- Seidel Method

– Advantages• Simple conceptually• Simple to program• No LDU factorization.• Sparsity is used simply• Iterations take little execution time.• Often more robust than Newton-Raphson

– Disadvantages• Sometimes not as robust.• Slow linear convergence rate.


Nonlinear Problem Solvers Newton’s Method

– Single Equation: Looking for the roots of

0)( xf

f(x)

Approach

1. Guess at solution x=x0

x0

2. Represent f(x) as a linear approximation about x=x0 using a Taylor series expansion:

0)()(

)()(0

0

o

x

xxx

xfxfxf3. Solve for a better estimate of

x*:

x

xf

xfxxx

)(

)()(

0

001

x

xf

xfxx

)(

)(0

001

x

xf

xfxx

k

kkk

)(

)(1

x1x2x*

4. Converged when:|f(xk)|<


Nonlinear Problem Solvers• Think-Pair-Square: Using Newton’s method, find a solution to the following problem starting with an initial estimate of x0=0.4. If f(x) is sum of

real/reactive power flow into a bus, and x represents bus voltage, determine a reasonable convergence criteria and be able to justify it.

02.2 xx


Nonlinear Problem Solvers Compare iterates from G-S and Newton.

xi

0.41.80.74782610.30422660.18188520.17091020.17082040.1708204

xi

0.40.040.19840.160637

0.1741960.1696560.171217

0.1706850.1708670.1708050.170826

G-S Newton

Newton starts off worse because it is more sensitive to:• An accurate initial guess.• The “smoothness” of f(x) near x0.



1.8

0.45

x2

x .2

21 x1 0.5 0 0.5 1 1.5 2

0.5

0

0.5

1

1.5

2

x1 x2

x0

-1.8

x3x4

xi

0.41.80.74782610.30422660.18188520.17091020.17082040.1708204

Newton

x0

x1

x2

x3

x4



1.62918

7.210865 107

Ni

GSi

100 i0 5 10

1 107

1 106

1 105

1 104

1 103

0.01

0.1

1

10

– Asymptotic Rate of Convergence: Largest integer r such that:

rk

k

xx

xx

*

*1

0

– Newton’s method has quadratic convergence, i.e., r=2.

Precision Limitation


Nonlinear Problem Solvers Newton-Raphson Method

– Apply Newton’s Method to Simultaneous Equations: Looking for the roots of:

0

0

0

0

),,(

),,(

),,(

0)(

1

12

11

nn

n

n

xxf

xxf

xxf

xf


Nonlinear Problem Solvers Repeat same Newton-type steps:

1. Make ‘good’ guess at solution:

Tnxxxxx 002

01

0 ,, 2. Approximate f(x) using a Taylor series expansion about point x0:

0)()(

)()(

0)()(

)()(

00

00

00

0

xxJxf

JacobianxJx

xfwhere

xxx

xfxfxf



0

02

01

21

2

2

2

1

2

1

2

1

1

1

00

0

))((

n

xn

nnn

n

n

x

x

x

x

f

x

f

x

f

x

f

x

f

x

fx

f

x

f

x

f

xxJ

3. Solve linearized equations for a better estimate of x*:

001

001000 )()(0)()(

xxx

xfxJxxxJxf



4. Check for convergence:

)( 1xf

5. If converged, end, otherwise perform next iteration with i=2:

)()(11 kkkkkk xfxJxxxx


Nonlinear Problem Solvers Ex: Find the solution to the following 2-bus power flow problem:

E1=1@0o E2=V2@q2

P2+jQ2=1+j1

j0.15

15.0

)sin()cos(11

)@(,15.0

111

15.0

222222

222

222

2

*

2122

j

VjVVj

VEj

VEj

Ej

EEjQP

qq

q


Nonlinear Problem Solvers Bringing all terms to one side and multiplying

numerator and denominator by (-j) gives:

15.0

)sin()cos(110

15.0

)sin()cos(11

222222

222222

jVVjVj

j

VjVVj

qq

qq

15.0

)sin(10

15.0

)cos(10

22

222

2

q

q

V

VV

Break equation into real and imaginary parts:


Nonlinear Problem Solvers Setting x1=q2,x2=V2 gives:

magnitudevoltagebusx

anglevoltagebusxwhere

xxV

xxxVV

2

1

1222

122222

22

15.0

)sin(10

15.0

)sin(10

15.0

)cos(10

15.0

)cos(10

q

q

1. Make a ‘good’ guess at solution:

0.1

0.0x

2a. Evaluate f(x) at x0:

1

1

15.0

)0cos(111

15.0

)0sin(0.11

)( 20xf



15.0

)cos(*215.0

)cos(1

15.0

)sin(15.0

)cos(1

15.0

)sin(15.0

)sin(1

15.0

)cos(*15.0

)sin(1

)(

12

2

1222

2

212

1

1222

1

2

1

2

12

2

112

1

12

1

1

xx

x

xxx

x

fxx

x

xxx

x

f

x

x

xx

x

fxx

x

xx

x

f

xJ

2b. Calculate analytical form of the Jacobian


Nonlinear Problem Solvers2c. Evaluate the Jacobian at best estimate of solution:

667.60

0667.6

15.0

)0cos(1*2

15.0

)0sin(*115.0

)0sin(

15.0

)0cos(*1

15.0

)cos(*2

15.0

)sin(15.0

)sin(

15.0

)cos(

)(

0

0

1212

112

x

x xxxx

xxx

xJ


Nonlinear Problem Solvers3. Solve the linear approximation for a better estimate of x*.

15.0

15.0

1

1

667.60

0667.6

)()(

1

0

02

12

01

11

0 xfxJxx

xxx

x

85.0

15.0

15.01

15.00

02

02

01

01

12

11

xx

xx

x

xx

4. Check for convergence.

213.0

153.0

15.0

)15.0cos(85.85.1

15.0

)15.0sin(85.1

)( 21xf



qk

00.150.18650990.19032740.19037230.1903723

Vk

10.850.79842430.79277620.79270940.7927094

Iteration Results

012345

Norm_fk

10.2136

0.0194-42.2385·10-83.1157·10

0



0 1 2 3 4 51 10

161 10

151 1014

1 10131 1012

1 10111 1010

1 1091 108

1 1071 106

1 1051 104

1 1030.010.1

1Newton Power Flow Convergence

Iteration Index

Nor

m o

f M

ism

atch

es

1

0

Norm_f k

50 k

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Lecture #18 EEE 574 Dr. Dan Tylavsky Nonlinear Problem Solvers