ITERATIVE METHODS
Basic Procedure:
Ensure that the system is arranged diagonally dominant.
Algebraically solve each linear equation for xi using an initial guess (if given, otherwise, use 0)
Solve for each xi and repeat
Use absolute relative approximate error after each iteration to check if error is within a pre-specified tolerance; or until 1 of variables’ answer repeats.
Algorithm A set of n equations and n unknowns:
a11 x1 + a12 x2 + a13 x3 + ... + a1n xn = b1 a21 x1 + a22 x2 + a23 x3 + ... + a2n xn = b2 . . .
If: the diagonal elements are non-zero
. . .
an1 x1 + an 2 x2 + an 3 x3 + ... + ann xn = bn
Rewrite each equation solving for the corresponding unknown
Rewriting each equation
x1 =
c1 − a12 x 2 − a13 x3 − a1n x n a11
From Equation 1
c2 − a21 x1 − a23 x3 − a2 n xn a22
From equation 2
x2 =
xn −1 = xn =
cn −1 − an −1,1 x1 − an −1, 2 x2 − an −1,n − 2 xn − 2 − an −1,n xn
From equation n-1
an −1,n −1
cn − an1 x1 − an 2 x2 − − an,n −1 xn −1 ann
From equation n
Calculate the Absolute Relative Approximate Error
∈a i =
−x
new i
old i
x
new i
x
×100
So when has the answer been found? The iterations are stopped when the absolute relative approximate error is less than a prespecified tolerance for all unknowns.
5
GAUSS JACOBI METHOD
Consider: 2 x1 −5 x2 +x3 =15 4 x1 −x2 +x3 =7 x1 +2 x2 −10 x3 =36
Re-arrange:
4 x1 −x2 +x3 =7 2 x1 −5 x2 +x3 =15 x1 +2 x2 −10 x3 =36
Working Equations: 7 +x2 −x3 x1 = 4 −15 +2 x1 +x3 x2 = 5 −36 +x1 +2 x2 x3 = 10
Results in tabular form: i
X1
X2
x3
0
0
0
0
1
1.7500
-3.0000
-3.6000
2
1.9000
-3.0200
-4.0250
3
2.0013
-3.0450
-4.0140
4
1.9923
-3.0023
-4.0089
5
2.0017
-3.0049
-4.0012
6
1.9991
-2.9996
-4.0008
7
2.0003
-3.0005
-4.0000
8
1.9999
-2.9999
-4.0001
9
2.0001
-3.0001
-4.0001
Îľ a -
Results in tabular form:
Îľ a
i
X1
X2
x3
0
0
0
0
1
1.7500
-3.0000
-3.6000
-
2
1.9000
-3.0200
-4.0250
-
3
2.0013
-3.0450
-4.0140
-
4
1.9923
-3.0023
-4.0089
-
5
2.0016
-3.0049
-4.0012
-
6
1.9991
-2.9996
-4.0008
-
7
2.0003
-3.0005
-4.0000
-
8
1.9999
-2.9999
-4.0001
-
9
2.0000
-3.0001
-4.0000
-
10
2.0000
-3.0000
-4.0000
-
-
Gauss Siedel Method Uses every current values
•
Results in tabular form:
(x1,x2,x3=0)
i
X1
X2
x3
1
1.7500
-2.3000
-3.8850
2
2.1463
-2.9185
-3.9691
3
2.0127
-2.9887
-3.9965
4
2.0020
-2.9985
-3.9995
5
2.0003
-2.9998
-3.9999
6
2.0000
-3.0000
-4.0000
7
2.0000
-3.0000
-4.0000
Îľ a
Results in tabular form:
(x1,x2,x3=0)
Îľ a
i
X1
X2
x3
1
1.7500
-2.3000
-3.8850
-
2
2.1463
-2.9185
-3.9691
-
3
2.0126
-2.9888
-3.9965
-
4
2.0019
-2.9985
-3.9995
-
5
2.0002
-29998
-3.9999
-
6
2.0000
-3.0000
-4.0000
-
7
2.0000
-3.0000
-4.0000
-
Relaxation Method
Working Equation:
Rewrite the system in Residual form •
Divide the equation by the negative of the highest coefficient)
For the iterations:
Select the highest |Rn| Xnew= highest Rn + Xn
Consider: 3 x1 −2 x2 −9 x3 =5 5 x1 +x2 −3 x3 =2 3 x1 +7 x2 −3 x3 =−13
Re-arrange: 5 x1 +x2 −3 x3 =2 3 x1 +7 x2 −3 x3 =−13 3 x1 −2 x2 −9 x3 =5
Working Equations: −x1 −0.2 x2 +0.6 x3 +0.4 =R1 −0.4286 x1 −x2 +0.4286 x3 −1.8571 =R2 0.3333 x1 −0.2222 x2 −x3 −0.5556 =R3
Results in tabular form:
(x1,x2,x3=0)
X1
R1
X2
R2
x3
R3
0
0.4000
0
-1.8571
0
-0.5556
0
0.7714
-1.8571
0
0
-0.1430
0.7714
0
-1.8571
-0.3306
0
0.1142
0.7714
0.0661
-2.1877
0
0
0.1876
0.7714
0.1787
-2.1877
0.0804
0.1876
0
Îľ a
Results in tabular form:
(x1,x2,x3=0)
Îľ a
X1
R1
X2
R2
x3
R3
0
0.4000
0
-1.8571
0
-0.5556
0.7714
-1.8571
0
-0.1430
-
-0.3306
0.1142
-
0
0.1876
-
0
-
0.0596
-
0
-
0.7714
0 0.0661
0.9501
0.9859
-2.1877
0.1787
0.0804
0
0.0038
0.0358
0.0293
0
0.0140
0.0119
-
0
0.0088
-
0
-
0
-0.0008
-
-0.0008
-0.0003
-
0
-0.0001
-
0
-
-0.0028
-2.1737
0.0024 0.0017 0.9876
-2.1699
0 0.0001
0.9877
0.0038
0
-2.1707
0
0.1876
-
0.2472
0.2560
0.2559
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