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ITERATIVE TECHNIQUES ITERATIVE TECHNIQUES FOR SOLVINGFOR SOLVING
NON-LINEAR SYSTEMS NON-LINEAR SYSTEMS (AND LINEAR SYSTEMS)(AND LINEAR SYSTEMS)
Jacobi Iterative Technique
Consider the following set of equations.
15
11
25
6
83
102
311
210
432
4321
4321
321
xxx
xxxx
xxxx
xxx
Convert the set Ax = b in the form of x = Tx + c.
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
324
4213
4312
321
xxx
xxxx
xxxx
xxx
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
03
02
14
04
02
01
13
04
03
01
12
03
02
11
)()()(
)()()()(
)()()()(
)()(
xxx
xxxx
xxxx
xxx )(
Start with an initial approximation of:
.and,, )()()()( 0000 04
03
02
01 xxxx
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
14
13
12
11
(0)(0)
(0)(0)(0)
(0)(0)(0)
(0)(0)
)(
)(
)(
x
x
x
x )(
8750110001
27272600001
41
3
12
11
.and.
,.,.)()(
)()(
xx
xx
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
13
12
24
14
12
11
23
14
13
11
22
13
12
21
)()()(
)()()()(
)()()()(
)()(
xxx
xxxx
xxxx
xxx )(
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
13
124
14
12
113
14
13
112
13
121
)()()(
)()()()(
)()()()(
)()(
kkk
kkkk
kkkk
kk(k)
xxx
xxxx
xxxx
xxx
k 0 1 2 3
0.0000 0.6000 1.0473 0.9326
0.0000 2.2727 1.7159 2.0530
0.0000 -1.1000 -0.8052 -1.0493
0.0000 1.8750 0.8852 1.1309
)( kx1
)( kx2
)( kx3
)( kx4
Results of Jacobi Iteration:
Gauss-Seidel Iterative Technique
Consider the following set of equations.
15
11
25
6
83
102
311
210
432
4321
4321
321
xxx
xxxx
xxxx
xxx
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
13
124
14
12
113
14
13
112
13
121
)()()(
)()()()(
)()()()(
)()(
kkk
kkkk
kkkk
kk(k)
xxx
xxxx
xxxx
xxx
8
15
8
1
8
3
10
11
10
1
10
1
5
1
11
25
11
3
11
1
11
1
5
3
5
1
10
1
324
14213
14
1312
13
121
)()()(
)()()()(
)()()()(
)()(
kkk
kkkk
kkkk
kk(k)
xxx
xxxx
xxxx
xxx
k 0 1 2 3
0.0000 0.6000
0.6000
1.0300
1.0473
1.0065
0.9326
0.0000 2.3272
2.2727
2.0370
1.7159
2.0036
2.0530
0.0000 -0.9873
-1.1000
-1.0140
-0.8052
-1.0025
-1.0493
0.0000 0.8789
1.8750
0.9844
0.8852
0.9983
1.1309
)( kx1
)( kx2
)( kx3
)( kx4
Results of Gauss-Seidel Iteration:(Blue numbers are for Jacobi iterations.)
It required 15 iterations for Jacobi method and 7 iterations for Gauss-Seidel method to arrive at the solution with a tolerance of 0.00001.
The solution is: x1= 1, x2 = 2, x3 = -1, x4 = 1
Newton’s Iterative TechniqueNewton’s Iterative Technique
)()()()( )( 11
11
kkkk XFXJXX
Given:
0
0
0
0
321
3213
3212
3211
nn
n
n
n
xxxxf
xxxxf
xxxxf
xxxxf
,......,,,
....
,......,,,
,......,,,
,......,,,
nn
n
n
n
xxxxf
xxxxf
xxxxf
xxxxf
XF
,......,,,
....
,......,,,
,......,,,
,......,,,
)(
321
3213
3212
3211
n
nnn
n
n
x
f
x
f
x
f
x
f
x
f
x
fx
f
x
f
x
f
XJ
...
............
...
...
)(
21
2
2
2
1
2
1
2
1
1
1
Jacobian Matrix:
Consider the following set of non-linear equation:
04
4
1
01
322
21
23
21
23
22
21
xxx
xx
xxx
322
213213
23
213212
23
22
213211
4
4
1
1
xxxxxxf
xxxxxf
xxxxxxf
,,
,,
,,
422
202
222
21
31
321
3
3
2
3
1
3
3
2
2
2
1
2
3
1
2
1
1
1
xx
xx
xxx
x
f
x
f
x
fx
f
x
f
x
fx
f
x
f
x
f
XJ )(
Make an initial guess:
1
1
10 )(X
2
751
20 .)( )(XF
422
202
2220 )( )(XJ
)()()()( )( 01
001 XFXJXX
2
751
2
422
202
222
1
1
11
1 .)(X
333330
875000
7916701
.
.
.)(X
)()()()( )( 11
112 XFXJXX
059050
487850
5034801
.
.
.
)( )(XF
4750001583341
6666600583341
6666607500015833411
..
..
...
)( )(XJ
059050
487850
503480
4750001583341
6666600583341
666660750001583341
333330
875000
7916701
2
.
.
.
..
..
...
.
.
.)(X
)()()()( )( 11
112 XFXJXX
236070
866030
4473303
.
.
.)(X
236070
866030
4408104
.
.
.)(X
238100
866070
5236502
.
.
.)(X
Example of Gauss-Seidel Iteration
42
0
06
3122
3
31221
31
21
xxx
xee
xxxxxxx
x 1
3x 1
2 x 2 x 1 x 3 6
x 2 ln x 3 ex 1
x 3
x 22 4
2 x 1
x 1 1 x 2 1 x 3 1
x 1 1.81712 x 2 0.17733 x 3 1.09199
x 1 1.94859 x 2 0.0518 x 3 1.0257x 1 1.98336 x 2 0.11868 x 3 1.00484
x 1 1.9597 x 2 0.14625 x 3 1.01511
x 1 1.95112 x 2 0.13583 x 3 1.02033
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