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Page 1: Iterative methods

ESCUELA DE INGENIERÍA DE PETROLEOS

RUBEN DARIO ARISMENDI RUEDA

Page 2: Iterative methods

ESCUELA DE INGENIERÍA DE PETROLEOS

CHAPTER 4: ‘Iterative Methods to solve lineal ecuation systems’

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ESCUELA DE INGENIERÍA DE PETROLEOS

INTRODUCTIONIn the next presentetion, there will be some examples that are already solved in excel document but the interesting part is to see how each different method behaves.

To use Jacobi method, the matrix has to has a prevailing diagonal. ( the addition of the other terms are less than the term of the diagonal)

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ESCUELA DE INGENIERÍA DE PETROLEOS

1. JACOBI.

Jacobi method consist to solve each lineal ecuation for Xi. You can asume the initial vector.

10 1 3 191 8 2 29

-1 -1 6 8

3x3 MATRIX

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ESCUELA DE INGENIERÍA DE PETROLEOS

n 0 1 2 3 4 5 6 7 8

x1 0 1,9 1,1375 0,91833333 0,99848958 1,00789757 0,99926888 0,99935904 1,00014015

x2 0 3,625 3,05416667 2,91927083 3,00222222 3,0069553 2,99898315 2,99947252 3,00015295

x3 0 1,33333333 2,25416667 2,03194444 1,97293403 2,00011863 2,00247548 1,99970867 1,99980526

Tol

x1 1,9 -0,7625 -0,21916667 0,08015625 0,00940799 -0,00862869 9,0162E-05 0,0007811

x2 3,625 -0,57083333 -0,13489583 0,08295139 0,00473307 -0,00797215 0,00048938 0,00068043

x3 1,33333333 0,92083333 -0,22222222 -0,05901042 0,02718461 0,00235684 -0,00276681 9,659E-05

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ESCUELA DE INGENIERÍA DE PETROLEOS

n 9 10 11 12 13 14

x1 1,00004313 0,99998223 0,99999805 1,00000186 0,99999997 0,99999983

x2 3,00003117 2,9999824 2,99999913 3,00000172 2,99999989 2,99999985

x3 2,00004885 2,00001238 1,9999941 1,99999953 2,0000006 1,99999998

Tol

x1 -9,702E-05 -6,0898E-05 1,5817E-05 3,8105E-06 -1,8866E-06 -1,369E-07

x2 -0,00012179 -4,877E-05 1,6729E-05 2,5923E-06 -1,8324E-06 -3,0965E-08

x3 0,00024359 -3,6468E-05 -1,8278E-05 5,4244E-06 1,0671E-06 -6,1983E-07

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ESCUELA DE INGENIERÍA DE PETROLEOS

n 15 16 17x1 1,00000002 1,00000001 1x2 3,00000003 3,00000001 3x3 1,99999995 2,00000001 2Tol x1 1,8904E-07 -8,8137E-09 -1,6392E-08x2 1,7207E-07 -1,6636E-08 -1,3945E-08x3 -2,7978E-08 6,0186E-08 -4,2416E-09

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ESCUELA DE INGENIERÍA DE PETROLEOS

2. GAUSS-SEIDEL.

The algoritm is the same for the Jacobi method. But it makes an exception with the next expression that improve the Jacobi’s method.

nipara

a

bxaxa

xii

i

n

ij

kjij

i

j

kjij

ki ....,,2,11

1

1

1

1

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ESCUELA DE INGENIERÍA DE PETROLEOS

10 1 3 19

1 8 2 29

-1 -1 6 8

3x3 MATRIX

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ESCUELA DE INGENIERÍA DE PETROLEOS

n 0 1 2 3 4 5

x1 0 1,9 0,896875 1,01126953 0,99874858 1,00013821

x2 0 3,3875 2,95924479 3,00458632 2,99949577 3,00005588

x3 0 2,21458333 1,97601997 2,00264264 1,99970739 2,00003235

Tolerancia

x1 1,9 -1,003125 0,11439453 -0,01252096 0,00138963

x2 3,3875 -0,42825521 0,04534153 -0,00509055 0,00056011

x3 2,21458333 -0,23856337 0,02662268 -0,00293525 0,00032496

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n 6 7 8 9 10

x1 0,99998471 1,00000169 0,99999981 1,00000002 1

x2 2,99999382 3,00000068 2,99999992 3,00000001 3

x3 1,99999642 2,0000004 1,99999996 2 2

Tolerancia

x1 -0,0001535 1,6983E-05 -1,8779E-06 2,0769E-07 -2,2969E-08

x2 -6,2052E-05 6,8584E-06 -7,5864E-07 8,3894E-08 -9,2783E-09

x3 -3,5925E-05 3,9735E-06 -4,3942E-07 4,8598E-08 -5,3745E-09

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With the two last examples, we can see that the Gauss-Seidel method is more efficient than the Jacobi method. Because it gets to the answer in a less number of iterations.


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