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MATH 175: Numerical Analysis II Lecturer: Jomar Fajardo Rabajante 2 nd Sem AY 2012-2013 IMSP, UPLB.

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Presentation on theme: "MATH 175: Numerical Analysis II Lecturer: Jomar Fajardo Rabajante 2 nd Sem AY 2012-2013 IMSP, UPLB."— Presentation transcript:

1 MATH 175: Numerical Analysis II Lecturer: Jomar Fajardo Rabajante 2 nd Sem AY 2012-2013 IMSP, UPLB

2 Numerical Methods for Linear Systems Review (Naïve) Gaussian Elimination Given n equations in n variables. Operation count for elimination step: (multiplications/divisions) Operation count for back substitution:

3 Numerical Methods for Linear Systems Overall (Naïve) Gaussian Elimination takes Take note: we ignored here lower-order terms and we did not include row exchanges and additions/subtractions. WHAT MORE IF WE ADDED THESE STUFFS???!!! KAPOY NA!

4 Numerical Methods for Linear Systems Example: Consider 10 equations in 10 unknowns. The approximate number of operations is If our computations have round-off errors, how would our solution be affected by error magnification? Tsk… Tsk…

5 Numerical Methods for Linear Systems Our goal now is to use methods that will efficiently solve our linear systems with minimized error magnification.

6 1 st Method: Gaussian Elimination with Partial Pivoting When we are processing column i in Gaussian elimination, the (i,i) position is called the pivot position, and the entry in it is called the pivot entry (or simply the pivot). Let [A|b] be an nx(n+1) augmented matrix.

7 1 st Method: Gaussian Elimination with Partial Pivoting STEPS: 1.Begin loop (i = 1 to n–1): 2.Find the largest entry (in absolute value) in column i from row i to row n. If the largest value is zero, signal that a unique solution does not exist and stop. 3.If necessary, perform a row interchange to bring the value from step 2 into the pivot position (i,i).

8 1 st Method: Gaussian Elimination with Partial Pivoting 4. For j = i+1 to n, perform 5. End loop. 6. If the (n,n) entry is zero, signal that a unique solution does not exist and stop. Otherwise, solve for the solution by back substitution.

9 1 st Method: Gaussian Elimination with Partial Pivoting Example: Original matrix (Matrix 0) Matrix 1

10 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 1 1 4 4 0

11 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 2 1 4 4 1

12 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 2 1 4 12 –1

13 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 1 1 4 12 –2

14 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 4 4 4 4 0

15 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 8 4 4 4 4

16 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 12 4 4 0

17 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 1Matrix 2 8 4 4 12 –4

18 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 2Matrix 3

19 1 st Method: Gaussian Elimination with Partial Pivoting Matrix 3 Final Matrix (Matrix 4)

20 1 st Method: Gaussian Elimination with Partial Pivoting Final Matrix Back substitution:

21 1 st Method: Gaussian Elimination with Partial Pivoting a unique solution does not exist

22 1 st Method: Gaussian Elimination with Partial Pivoting There are other pivoting strategies such as the complete (or maximal) pivoting. But complete pivoting is computationally expensive.


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