1 HAMPIRAN NUMERIK SOLUSI PERSAMAAN LANJAR Pertemuan 5 Matakuliah: K0342 / Metode Numerik I Tahun: 2006 TIK:Mahasiswa dapat meghitung nilai hampiran numerik.

Presentasi berjudul: "1 HAMPIRAN NUMERIK SOLUSI PERSAMAAN LANJAR Pertemuan 5 Matakuliah: K0342 / Metode Numerik I Tahun: 2006 TIK:Mahasiswa dapat meghitung nilai hampiran numerik."— Transcript presentasi:

1 1 HAMPIRAN NUMERIK SOLUSI PERSAMAAN LANJAR Pertemuan 5 Matakuliah: K0342 / Metode Numerik I Tahun: 2006 TIK:Mahasiswa dapat meghitung nilai hampiran numerik solusi persamaan lanjar

2 2 PERTEMUAN 5 HAMPIRAN NUMERIK SOLUSI PERSAMAAN LANJAR TIK:Mahasiswa dapat meghitung nilai hampiran numerik solusi persamaan lanjar

3 3 Bentuk umum persamaan lanjar (linier) bujur sangkar: Untuk i = 1,2,3,…,n

4 4 Solusi persamaan lanjar Eliminasi Gauss Eliminasi Gauss- Jordan Matriks balikan Dekomposisi LU Iterasi Jacobi Iterasi Gauss- Seidel

5 5 Gauss Elimination Consider n equations in n unknowns The solution vector, x, for this system of equations remains unchanged if either of the following fundamental row operations are performed: –Multiply or divide any equation by a constant. –Replace any equation by the sum of that equation and any other equation.

6 6 Forward Elimination b’ k = b k (a 11 ) – b 1 (a k1 ) k = 2,3,4,…,n

7 7 Forward Elimination b’ k = b k (a ’ 22 ) – b 2 (a k2 ) k = 3,4,…,n

8 8 Forward Elimination

9 9 Back Substitution Finally

10 10 Gauss Elimination

11 11 Gauss Elimination

12 12 Gauss Elimination

13 13 Example Forward Elimination

14 14 Example Back Substitution

15 15 GAUSS-JORDAN ELIMINATON

16 16 GAUSS-JORDAN ELIMINATON

17 17 Contoh: Selesaikan SPL berikut dengan eliminasi Gass-Jordan Jawaban: Matriks perluasan(augmented matrix) dari koefisien SPL: b ’ 1 = b 1 /3

18 18 b ’ 2 = b 2 – 0.1 b 1 b ’ 3 = b 3 – 0.3 b 1 b ” 2 = b ’ 2 /7.033333

19 19 b ” 1 = b ’ 1 + 0.033333 b ” 2 b ” 3 = b ’ 3 + 0.190000 b” 2 b ”’ 3 = b ” 3 /10.01200

20 20 b ”’ 1 = b ” 1 + 0.068063 b ”’ 3 b ”’ 2 = b ’’ 2 + 0.190000 b”’ 3

21 21 Iterative Methods Consider the linear system

22 22 Solution Methods Gauss Elimination – Subject to roundoff errors and ill conditioning Iterative methods -- Alternative to elimination method Take initial guess of solution and then iterate to obtain improved estimates of the solution Jacobi and Gauss-Seidel methods Work well for large sets of equations

23 23 Iterative Methods Rearrange the equations so that an unknown is on the left-hand side of each equation: Initial guess

24 24 Jacobi Method Initial guess Next approximation of the solution

25 25 Jacobi Method After k iterations of this process

26 26 Example – Jacobi Method System of 3 equations in 3 unknowns Rearrange

27 27 Example – Jacobi Method Initial guess After 20 iterations

28 28 Errors and Stopping Criteria How do we know when to stop? Two major sources of error in numerical methods: –Roundoff error: Computers represent quantities with a finite number of digits –Truncation error: Numerical methods employ approximations to represent exact mathematical operations and quantities Consider the error between the numerical and analytical solutions –True value –Approximate value

29 29 Error Measures True value = Approximate value + Error  = Error = True value - Approximate value  r = relative error d = significant digits

30 30 Example Pi ~ 3.1416 Better approximation x = 3.1415927. Find the error, relative error and the number of significant digits in the approximation.

31 31 Error Perkiraan  A is the approximate error between the current approximate value and our previous approximate value

32 32 Example Estimate exp(x) for x = 0.5 by adding more and more terms to the sequence and computing the errors after adding each new term. Add terms until the estimate is valid to three significant digits. From a calculators x = 1.648721271

33 33 Example Using only the first term of the series, Using the first and second terms of the series,and so on … # of TermsResult  11 0.393 21.5 0.09 31.625 0.014 41.645833333 0.0017 51.648437500 0.00017 61.648697917 0.000014

34 34 Jacobi Methode We stop when

35 35 Gauss Seidel Methode The current approximation of one of the unknowns is available for use after each step. This information could be used immediately in the calculation of the next unknown

36 36 Contoh metode Gauss-Seidel after 1 iteration of the method convenient initial guess

37 37 after 2 nd iterations after 9 th iterations after 1 st iteration

38 38 Terima kasih