MATH 175: Numerical Analysis II Lecturer: Jomar Fajardo Rabajante 2 nd Sem AY 2012-2013 IMSP, UPLB.
MATH 175: Numerical Analysis II
description
Transcript of MATH 175: Numerical Analysis II
![Page 1: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/1.jpg)
MATH 175: Numerical Analysis II
Lecturer: Jomar Fajardo Rabajante2nd Sem AY 2012-2013
IMSP, UPLB
![Page 2: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/2.jpg)
Numerical Methods for Linear Systems
Review (Naïve) Gaussian EliminationGiven n equations in n variables.
• Operation count for elimination step:(multiplications/divisions)
• Operation count for back substitution:
3
3nO
2
2nO
![Page 3: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/3.jpg)
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!
323
323 nOnOnO
![Page 4: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/4.jpg)
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…
3343103
![Page 5: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/5.jpg)
Numerical Methods for Linear Systems
Our goal now is to use methods that will efficiently solve our linear systems with minimized error
magnification.
![Page 6: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/6.jpg)
1st 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.
![Page 7: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/7.jpg)
1st 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).
![Page 8: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/8.jpg)
1st 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.
jii,i
j,ij RRaa
R
![Page 9: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/9.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
Example:
8121
12841244221
8112
12842211244
Original matrix (Matrix 0) Matrix 1
![Page 10: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/10.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
120
1244Matrix 1 Matrix 2
11
44 0
![Page 11: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/11.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
12101244
Matrix 1 Matrix 2
21
44 1
![Page 12: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/12.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
121101244
Matrix 1 Matrix 2
21
412 –1
![Page 13: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/13.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
2121101244
Matrix 1 Matrix 2
11
412 –2
![Page 14: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/14.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
212
01101244
Matrix 1 Matrix 2
44
44 0
![Page 15: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/15.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
212
401101244
Matrix 1 Matrix 2
84
44 4
![Page 16: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/16.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
212
0401101244
Matrix 1 Matrix 2
124
412 0
![Page 17: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/17.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
8112
12842211244
4212
0401101244
Matrix 1 Matrix 2
84
412 –4
![Page 18: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/18.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
2412
1100401244
4212
0401101244
Matrix 2 Matrix 3
![Page 19: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/19.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
2412
1100401244
1412
1000401244
Matrix 3 Final Matrix (Matrix 4)
![Page 20: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/20.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
1412
1000401244
Final Matrix Back substitution:
1x 121244x
1212z4y4x1y44y
1z1z
![Page 21: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/21.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
112
7204101290
8212
000104012191
a unique solution does not exist
![Page 22: MATH 175: Numerical Analysis II](https://reader035.fdocuments.in/reader035/viewer/2022062218/568164eb550346895dd75719/html5/thumbnails/22.jpg)
1st Method: Gaussian Elimination with Partial Pivoting
• There are other pivoting strategies such as the complete (or maximal) pivoting. But complete pivoting is computationally expensive.