Iterative methods
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Transcript of Iterative methods
ESCUELA DE INGENIERÍA DE PETROLEOS
RUBEN DARIO ARISMENDI RUEDA
ESCUELA DE INGENIERÍA DE PETROLEOS
CHAPTER 4: ‘Iterative Methods to solve lineal ecuation systems’
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)
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
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
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
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
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
ESCUELA DE INGENIERÍA DE PETROLEOS
10 1 3 19
1 8 2 29
-1 -1 6 8
3x3 MATRIX
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
ESCUELA DE INGENIERÍA DE PETROLEOS
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
ESCUELA DE INGENIERÍA DE PETROLEOS
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.