Name: (1) Kamal Akhtar F.(120450119156)Branch: Mechanical-4C(1)Collage: S.V.M.I.T.
Introductionβ¦
O Iteration method:-
O Why iteration method is need?O When the system of algebraic equation consist of large
number of equation then the direct method becomes cumbersome and time consuming.
O In this case Iteration method provide easy tosolve the system.
O In Iteration method we start from initial approximation to the actual solution and obtain better and better approximation after repeating iteration.
O Iteration method gives the value to the desired accuracy.
Iteration method:-
O(1) Jacobi's methodO(2) Gauss-Seidel method
Jacobi's methodO A set of n equations and n unknowns:
O note: assume all the diagonal elements
are not zero.
11313212111 ... bxaxaxaxa nn
2n2n323222121 bxa...xaxaxa . .
. .
. .
nnnnnnn bxaxaxaxa ...332211
Jacobi's methodO If diagonal elements are zero then we
Rewrite the question asβ¦
O π₯1 =[π1β π11π₯2+π13π₯3+β―+π1ππ₯π ]
π11
O π₯2 =[π2β π21π₯2+π23π₯3+β―+π2ππ₯π ]
π22
O π₯π =[ππβ ππ1π₯2+ππ2π₯2+β―+ππ(πβ1)π₯πβ1 ]
πππ
Jacobi's methodO Suppose that the initial solution isβ¦
O π₯1(0), π₯2
(0), β¦ , π₯π(0) put this all value in the
above equation then we obtain the 1π π‘
approximation to solution as π₯1(1), π₯2
(1), β¦ , π₯π
(1).
O Against the put this value in above equation then we obtain the 2ππ approximation to solution as π₯1
(2), π₯2(2), β¦ , π₯π
(2).
O Continues the process of finding the approximation till get the solution is the desired level of accuracy.
Use the Jacobi method to approximate the solution of the following system of linear equations.
πππ β πππ+πππ = βπβπππ + πππ+ππ = ππππ β ππ β πππ = π
Continue the iterations until two successive approximations are identical when rounded to three significant digits.
O Solution:--O ππ = β
π
π+
π
πππ β
π
πππ
O ππ =π
π+
π
πππ β
π
πππ
O ππ = βπ
π+
π
πππ β
π
πππ
O Because we do not know the actual solution, chooseO ππ = π, ππ = π, ππ = πβ¦Initial approximationO as a convenient initial approximation. So, the first approximation is
O ππ = βπ
π+
π
ππ β
π
ππ β β0.200
O ππ =π
π+
π
π(π) β
π
π(π) β π. πππ
O ππ = βπ
π+
π
ππ β
π
ππ β βπ. πππ
Continuing this procedure, you obtain the sequence of approximations shown in Table
π π₯1 π₯2 π₯3
0 0.000 0.000 0.000
1 -0.200 0.200 -0.429
2 0.146 0.203 -0.517
3 0.192 0.328 -0.416
4 0.181 0.332 -0.421
5 0.185 0.329 -0.424
6 0.186 0.331 -0.423
7 0.186 0.331 -0.423
Because the last two rows in Table are identical, you can conclude that
to three significant digits the solution isβ¦
π₯1=0.186π₯2=0.331π₯3= β 0.423
The Gauss-Seidel MethodO A modification of the Jacobi method called the Gauss-Seidel method.O This modification is no more difficult to use than the Jacobi method,
and it often requires fewer iterations to produce the same degree of accuracy.
O With the Jacobi method, the values of ππ obtained in the nth approximation remain unchanged until the entire(π + π)th approximation has been calculated.
OWith the Gauss- Seidel methodβ¦O On the other hand, you use the new values of each ππ as soon as they
are known. That is, once you have determined ππ from the first equation, its value is then used in the second equation to obtain the new ππ Similarly, the new ππ& ππare used in the third equation to obtain the newππ and so on. This procedure is demonstrated in ππ.
(1)Using the gauss-seidel method to solve the systemβ¦πππ + πππ β ππ = ππ,ππ + πππ + πππ = πππ,ππ + ππ β πππ = ππ,
O Solutionβ¦- rewriting the equation as...
O π₯ =95β11π¦+4π§
83= 1.145 β 0.133π¦ + 0.048π§
O y=104β7π₯β13π§
52= 2 β 0.135 β 0.25π§
O z=71β3π₯+8π¦
29= 2.448 β 0.103π₯ + 0.276π¦
O Assume the initial solution:β¦
O π₯ = π¦ = π§ = 0
Continuing this procedure, you obtain the sequence of approximations shown in Table
n X Y Z
0 0 0 0
1 1.145 1.845 1.821
2 0.987 1.412 1.957
3 1.051 1.369 1.962
4 1.057 1.367 1.962
5 1.057 1.367 1.962
Because the last two rows in Table are identical, you can
conclude that to three significant digits the solution isβ¦
π₯=1.057π¦=1.367π§=1.962
Continuing this procedure, you obtain the sequence of approximations shown in Table
n X Y Z
0 0 0 0
1 1.145 1.845 1.821
2 0.987 1.412 1.957
3 1.051 1.369 1.962
4 1.057 1.367 1.962
5 1.057 1.367 1.962
Because the last two rows in Table are identical, you can
conclude that to three significant digits the solution isβ¦
π₯=1.057π¦=1.367π§=1.962
(2)Using the gauss-seidel method to solve the systemβ¦6π + π + π = πππ,4π + ππ + ππ = πππ,5π + ππ β πππ = ππ,
O Solutionβ¦- rewriting the equation as...
O π₯ =105βπ¦βπ§
6= 17.5 β 0.167π¦ β 0.167π§
O y=155β4π₯β3π§
8= 19.375 β 0.5π₯ β 0.375π§
O z=β65+5π₯+4π¦
10= β6.5 β 0.5π₯ + 0.4π¦
O Assume the initial solution:β¦
O π₯ = π¦ = π§ = 0
Continuing this procedure, you obtain the sequence of approximations shown in Table
n X y z
0 0 0 0
1 17.5 10.625 6.5
2 14.64 9.618 4.667
3 15.114 10.068 5.084
4 14.970 9.984 4.979
5 15.001 10.007 5.003
Because the last two rows in Table are identical, you can
conclude that to three significant digits the solution isβ¦
π₯=15.001π¦=10.007π§=5.003
(3)Using the Gauss-Seidel method to solve the systemβ¦2π β π + ππ = π,π + ππ + ππ = βπ,π + ππ + ππ = π,
start with the initial approximations π = π. π, π = βπ. π, π = π. π
O Solutionβ¦- rewriting the equation as...
O π₯ =3+π¦β2π§
2= 1.5 + 0.5π¦ β π§
O y=β1βπ₯β3π§
3= β0.333 β 0.333π₯ β π§
O z=1βπ₯β2π¦
5= 0.2 β 0.2π₯ β 0.4π¦
O Assume the initial solution:β¦
O π₯ = π¦ = π§ = 0
Continuing this procedure, you obtain the sequence of approximations shown in Table
n X y z
0 0.3 -0.8 0.3
1 0.8 -0.899 0.4
2 0.651 -0.950 0.450
3 0.575 -0.974 0.475
4 0.538 -0.987 0.487
5 0.520 -0.993 0.493
6 0.511 -0.996 0.496
7 0.506 -0.997 0.498
Because the last two rows in Table are identical, you can conclude that to three
significant digits the solution isβ¦
π₯=0.506π¦= β 0.997
π§=0.498
(4)Using the Gauss-Seidel method to solve the systemβ¦3π β π. ππ β π. ππ = π. ππ,0.1π + ππ β π. ππ = βππ. π,0.1π β π. ππ + πππ = ππ. π,
start with the initial approximations π = π. π, π = βπ. π, π = π. π
O Solutionβ¦- rewriting the equation as...
O π₯ =7.85+0.1π¦+0.2π§
3= 2.616 + 0.033π¦ + 0.66z
O π¦ =1β19.3β0.1π₯+0.3π§
7= β2.757 β 0.014π₯ + 0.042π§
O π§ =71.4β0.3π₯+0.2π¦
10= 7.14 β 0.3π₯ β 0.02π¦
O Assume the initial solution:β¦
O π₯ = π¦ = π§ = 0
Continuing this procedure, you obtain the sequence of approximations shown in Table
Because the last two rows in Table are identical, you can conclude that
to three significant digits the solution isβ¦
π₯=3.000π¦= β 2.50000
π§=7.0000
Recommended
![Nonlinear stability of degenerate shock proο¬lesphoward/papers/degnonlinshort.pdfΒ Β· nonlinear Schrodinger equation and the GinzburgβLandau equation [21]. The algebraic (and non-integrable)](https://img.pdfslide.us/doc/110x75/6086050bfe80cf0c283eca7c/nonlinear-stability-of-degenerate-shock-proiles-phowardpapers-nonlinear-schrodinger.jpg)
![11.[36-49]Solution of a Subclass of Lane Emden Differential Equation by Variational Iteration Method](https://img.pdfslide.us/doc/110x75/577d1e5a1a28ab4e1e8e5546/1136-49solution-of-a-subclass-of-lane-emden-differential-equation-by-variational.jpg)
