Ee1034cS10

Ee1034cS10 - Incomplete Gauss Elimination: LU Factorization...

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Incomplete Gauss Elimination: LU Factorization Incomplete Gauss Elimination and LU Factorization: We modify the above CGE method and derive a numerically efficient method. This method is called the LU factorization method and based upon Incomplete Gaus method is called the and based upon Incomplete Gauss Elimination (IGE). LU factorization is important for solving large systems of linear equations, especially those for which there are many “right-hand-side” vectors. That is ,1 , , i A xbi k  where k is quite large. For instance, such problems arise in large linear programming applications. EE103 Lec 4C (SEJ) 1 IGE, Example—No Row Changes 9768 3 0  09 62 5 ,( 1 , 1 , , 1 ) 55 15 4 21 26 3 t Ab x    11 1 10 0 0 9 7 6 8 01 0 0 0 9 6 2 , 5/9 0 1 0 0 80/9 7/3 85/9 EA E A 2/9 0 0 1 0 23/9 2/3 70/9 100 0 9 7 6 8 010 0 0 96 2 E A E E A 22 2 1 2 1 , 0 80 / 81 1 0 0 0 223/ 27 605 / 81 0 23/ 81 0 1 0 0 28 / 27 584 / 81 0 0 9 7 6 8 33 3 2 3 2 1 0 0 0 9 6 2 , 0 0 1 0 0 0 223 / 27 605 / 81 0 0 28 / 223 1 0 0 0 4196 / 669 E AE A E E E AU EE103 Lec 4C (SEJ) 2
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IGE: LU Factorization Note that 3 A is an upper triangular matrix, which we’ll denote by U . We see that 321 UE E E A and that each of the "" E matrices is lower triangular, with “ones” on the main diagonal. We therefore have 1 () EEE U A 3 11 1 1 1 2 3 UA L E E E   A LU 9768 1 0 0 0 L  97 6 8 U 09 62 0 1 0 0 55 15 5 / 9 8 0 / 8 1 1 0 2 1 2 6 2 / 9 23 / 81 28 / 223 1 A    6 2 0 0 223 / 27 605 / 81 0 0 0 4196 / 669 1 10 0 0 01 0 0 5/9 0 1 0 E 2 100 0 010 0 08 0 / 8 110 E 3 0 0 0 0 00 1 0 E EE103 Lec 4C (SEJ) 3 2/9 0 0 1 02 3 / 8 1 0 1 0 0 28 / 223 1 IGE: LU Factorization Ax b LUx b , Ux y Ly b y 1 1 forward substitution b y l n n backward substitution y x u , Ux y x 1,1 22 , 1 1 2 2,2 bly y l , , 1 1, 1 nn n n n yu x x u 33 , 1 13 , 2 2 3 3,3 blyly y l , 1 12 , 2 2, 2 n n n n yuxu x x u EE103 Lec 4C (SEJ) 4
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Forw, Back EE103 Lec 4C (SEJ) 5 Flops: Forw, Back forward substitution backward substitution 1 1 1,1 () b y l bl , n n nn y x u yu x 22 , 1 1 2 2,2 33 , 1 13 , 2 2 y y l blyly y  11 , 1 1, 1 , 1 12 , n n n n n n x u yuxu x x By inspection, we see that there is one flop to determine 1 y ; then there are 3 flops to 3 3,3 l 2 2, 2 n u determine 2 y . The progression of flops is 1, 3, 5, 7, … . That is, the th k term of the sequence is 21 k . Therefore, the total number of flops is, using 1 (1 ) / 2 n kn n  , 2 1 2( 1 ) (2 1) 2 n k n  Th lt i th f l i U h U it i l EE103 Lec 4C (SEJ) 6 The result is the same for solving xy , where is upper triangular.
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IGE: LU Factorization Example of the need to interchange rows ..
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Ee1034cS10 - Incomplete Gauss Elimination: LU Factorization...

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