lecture06.B

lecture06.B - INTRODUCTION TO NUMERICAL SIMULATION LECTURE...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 6.B. QR Factorization 1 T ODAY S O UTLINE : ± Singular Problems ± Projection Formulas ± Modified Gram-Schmidt Algorithm QR F ACTORIZATION Singular Problems – LU Factorization Fails ± Struts Example The resulting nodal matrix is SINGULAR: ² LU decomposition fails ² But a solution exists! Actually, many…. load force joint strut ( ) 3 , 0 = L f r 0 0 * * = + = + y x L y L x f f f f 0 0 y y u x x u y x = = + = ε 3 0 0 0 0 y x u u 2 M r b r 1 M r 2 M r b r {} solutions infinite , 2 1 M M span b r r r
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 6.B. QR Factorization 2 2 M r 1 M r 3 M r { } 3 2 1 , , M M M span b r r r r b r projection of onto { } 3 2 1 , , M M M span r r r + + + 1 0 0 0 0 1 0 1 1 1 0 0 0 1 1 0 1 1 LU will fail No pivots available i = 1 i = 1 v 2 v 1 v 3 + = + + + 1 1 1 1 0 0 0 1 1 0 1 1 3 2 1 v v v 1 M r 2 M r 3 M r b r 1 M r 2 M r 3 M r b r { } exists solution , , 3 2 1 M M M span b r r r r plane solutions infinite L.D. are and 2 1 M M r r { {{ 3 2 1 3 2 1 1 1 0 M M M b x x x r r r r + = e.g. { { { 3 2 1 3 2 1 1 0 1 M M M b x x x r r r r + =
Background image of page 2
I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 6.B. QR Factorization 3 Recall weighted sum of columns view of systems of equations b M x M x M x b b b x x x M M M N N N N N r r L r r M M L = + + + = 2 2 1 1 2 1 2 1 2 1 M is singular but b is in the span of the columns of M so there is a solution, actually lots of them. How do we find them? ± Systems of Linear Equations – Summary Table { } M range b { } M range b L.I. columns Solution exists and is unique Use LU No solutions Find the “closest” L.D. columns Infinite solutions exist Find one … or all No solutions Find the “closest” v 4 i = 1 i = 1 v 2 v 1 v 3 = 0 1 1 1 2 1 1 1 1 1 1 1 4 3 2 1 v v v v = 0 1 1 1 2 1 1 1 0 0 1 1 4 3 2 1 step LU 1 after v v v v R R R = 0 1 1 1 2 4 3 2 1 v v v v G G G G G G G G R G 1 = Let R = 1
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 6.B.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 12

lecture06.B - INTRODUCTION TO NUMERICAL SIMULATION LECTURE...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online