View
217
Download
0
Category
Tags:
Preview:
Citation preview
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).
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
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|<.
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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.
© Copyright 1999 Daniel Tylavsky
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.
© Copyright 1999 Daniel Tylavsky
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)|<
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
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.
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
4. Check for convergence:
)( 1xf
5. If converged, end, otherwise perform next iteration with i=2:
)()(11 kkkkkk xfxJxxxx
© Copyright 1999 Daniel Tylavsky
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
q
© Copyright 1999 Daniel Tylavsky
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
15.0
)sin(10
15.0
)cos(10
22
222
2
q
q
V
VV
Break equation into real and imaginary parts:
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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
© Copyright 1999 Daniel Tylavsky
Nonlinear Problem Solvers
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
The End
Recommended