• Nie Znaleziono Wyników

Frontal and multi-frontal solvers:

N/A
N/A
Protected

Academic year: 2021

Share "Frontal and multi-frontal solvers:"

Copied!
35
0
0

Pełen tekst

(1)

Maciej Paszynski

Department of Computer Science

AGH University of Science and Technology, Krakow, Poland maciej.paszynski@agh.edu.pl

http://home.agh.edu.pl/paszynsk

http://www.ki.agh.edu.pl/en/staff/paszynski-maciej http://www.ki.agh.edu.pl/en/research-groups/a2s

Main collaborators Victor Calo (KAUST) Leszek Demkowicz (ICES, UT)

David Pardo (IKERBASQUE)

Frontal and multi-frontal solvers:

Dealing with singularities

(2)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(3)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

Number of operations for partial forward elimination (Schur complement computations)

Computational complexity O(ab2)

(4)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(5)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(6)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(7)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(8)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(9)

COMPUTATIONAL COST OF MULTI-FRONTAL SOLVER

(10)

COMPUTATIONAL COST OF FRONTAL SOLVER

Computational cost of elimination of a single layer O((N0.5)3)=O(N3/2) Number of layers = O(N0.5)

Computational cost of elimination of entire mesh

= computational cost of elimination of a single layer * number of layers O(N0.5N3/2)=O(N2) in 2D

O(N1/3N6/3)=O(N7/3) in 3D

(11)

COMPUTATIONAL COST OF FRONTAL SOLVER

Computational cost of elimination of a single layer O((N0.5)3)=O(N3/2) Number of layers = O(N0.5)

Computational cost of elimination of entire mesh

= computational cost of elimination of a single layer * number of layers O(N0.5N3/2)=O(N2) in 2D

O(N1/3N6/3)=O(N7/3) in 3D

(12)

COMPUTATIONAL COST OF FRONTAL SOLVER

Computational cost of elimination of a single layer O((N0.5)3)=O(N3/2) Number of layers = O(N0.5)

Computational cost of elimination of entire mesh

= computational cost of elimination of a single layer * number of layers O(N0.5N3/2)=O(N2) in 2D

O(N1/3N6/3)=O(N7/3) in 3D

(13)

MODEL ELIPTIC PROBLEM

(14)

COMPUTATIONAL COST OF 3D DIRECT SOLVER

Notation:

N = number of degrees of freedom Ne = number of elements

p = polynomial order of approximation O(N)=O(Ne*p3)

Computational cost of direct solvers =

cost of static condensation + cost of LU factorization Static condensation O(Ne*p9)=O(N*p6)

Cost of LU factorization over regular grid O(N2)

CONCLUSIONS:

For regular grid total cost is O(N*p6+N2) = O(N2)

For other grids it is not always the case (static condensation may dominate)

(15)

UNIFORM REFINEMENTS

Mesh Time of LU factorization Slope = exponent factor = 2 Location = p factor = 1=p0

(16)

UNIFORM REFINEMENTS

MULTI-FRONTAL SOLVER APPROACH

Total cost is O(N*p6+N2) = O(N2) Static condensation

O(N*p6) +

LU factorization O(N2)

(17)

REFINEMENTS TOWARDS POINT SINGULARITY

Mesh Time of LU factorization Slope = exponent factor = 1 Location = p factor = p6

(18)

REFINEMENTS TOWARDS POINT SINGULARITY FRONTAL SOLVER APPROACH

Total cost is O(N*p6+N*p3) = O(N*p6) Static condensation

O(N*p6) +

Schur complent of a single layer O(p6) Number of layers k=O(Ne)=O(N/p3)

…..

Total cost of LU factorization O(p6*k)=O(p6*N/p3)= O(N*p3)

(19)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY

Mesh Time of LU factorization Slope = exponent factor = 1 Location = p factor = p6

(20)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY FRONTAL SOLVER APPROACH

DO NOT USE FRONTAL SOLVER APPROACH Static condensation

O(N*p6) +

Number of dofs in a layer 3*2kp2=O(2kp2)

Number of interfaces dofs in a layer 2*2kp2=O(2kp2) Cost of Schur complement of a single layer O(23kp6)

s=number of layers, N = O = O = O(p32s) Cost of LU factorization O = O(p623s)=O(N3/p3)

…..

k=1 k=2

(21)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY MULTI-FRONTAL SOLVER APPROACH

Static

condensation O(N*p6) +

Number of dofs in a patch O(kp2)

Number of patches in a single layer O(2s-k) Number of interfaces dofs in a patch O(kp2)

Cost of Schur complement of a single layer O(2s-k k3p6) s=number of layers, N = O = O = O(p32s)

Cost of LU factorization O <O(s3p62s)=O(Np3 (log23Ne)) Total cost is < O(N*p6+Np3 ( log23Ne))

k=1 k=2 k=3 k=4 2(s-k) 2 (4-1)=23=8 2(4-2)=22=4 2(4-3)=21=2 2(4-4)=20=1

(22)

ANISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY

Mesh Time of LU factorization Slope = exponent factor = 1 Location = p factor = p6

(23)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY FRONTAL SOLVER APPROACH

Total cost is O(N*p6+N*p3) = O(Np6) Static condensation

O(N*p6) +

Number of dofs in a layer O(p2)

Number of interfaces dofs in a layer O(p2)

…..

Number of layers k=O(Ne)=O(N/p3)

Total cost of LU factorization O(p6*k)=O(p6*N/p3)= O(N*p3)

(24)

ISOTROPIC REFINEMENTS TOWARDS FACE SINGULARITY

Mesh Time of LU factorization

Slope = exponent factor = 1.5 Location = p factor = p1.5

(25)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY FRONTAL SOLVER APPROACH

Static condensation O(N*p6) +

Number of dofs in a layer 22kp2=O(22kp2)

Number of interfaces dofs in a layer 22kp2=O(22kp2) Cost of Schur complement of a single layer O(26kp6) s=number of layers, N = O = O(p322s)

Cost of LU factorization O = O(p626s)=O(N3/p3)

…..

k=1 k=2

DO NOT USE FRONTAL SOLVER APPROACH

(26)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY MULTI-FRONTAL SOLVER APPROACH

Static condensation O(N*p6) +

Number of dofs in a patch O(2kp2) Numbers of patches in a layer O(22(s-k))

Number of interfaces dofs in a patch O(2kp2)

Cost of Schur complement of a single layer O(22(s-k)23kp6) s=number of layers, N = O = O(p322s)

Cost of LU factorization O = O(p623s)=O(N1.5*p1.5) Total cost is O(N*p6+N1.5*p1.5)

k=1 k=2 k=3 22(s-k) 22(4-1)=22*3=64 22(4-2)=22*2=16 22(4-3)=22=4

(27)

ANISOTROPIC REFINEMENTS TOWARDS FACE SINGULARITY

Mesh Time of LU factorization Slope = exponent factor = 1 Location = p factor = p6

(28)

ISOTROPIC REFINEMENTS TOWARDS EDGE SINGULARITY FRONTAL SOLVER APPROACH

Total cost is O(N*p6+N*p3) = O(Np6) Static condensation

O(N*p6) +

Number of dofs in a layer O(p2)

Number of interfaces dofs in a layer O(p2)

…..

Number of layers k=O(Ne)=O(N/p3)

Total cost of LU factorization O(p6*k)=O(p6*N/p3)= O(N*p3)

(29)

COMPARISON OF NUMERICAL AND THEORETICAL SCALABILITY EXPONENT FACTORS

FOR REFINEMENTS TOWARDS A SINGLE ENTITY

(30)

POINT + ANISOTROPIC EDGE SINGULARITY

Mesh Time of LU factorization

(31)

POINT + ANISOTROPIC FACE SINGULARITY

Mesh Time of LU factorization

(32)

ANISOTROPIC EDGE + ANISOTROPIC FACE SINGULARITY

Mesh Time of LU factorization

(33)

POINT + ANISOTROPIC EDGE + ANISOTROPIC FACE SINGULARITY

Mesh Time of LU factorization

(34)

NUMERICAL SCALABILITY EXPONENT FACTORS

FOR REFINEMENTS TOWARDS MULTIPLE SINGULARITIES

(35)

PAPERS

Maciej Paszyński, David Pardo, Damian Goik

PERFORMANCE OF DIRECT SOLVERS ON H-ADAPTED GRIDS submitted to SIAM Journal of Numerical Analysis, 2013

Cytaty

Powiązane dokumenty

The elimination tree created for the initial mesh is updated when the mesh is refined (elimination tree is constructed dynamically, during mesh refinements).. • Local matrices

Computational cost for a leaf = cost of elimination of interior degrees of freedom Computational cost for a node = cost for son1 + cost for son2 +.. cost of elimination of

(A1)-(A) 4 -(AN) (generation of frontal matrices at leaves of the elimination trees) (A2) 3 (merging contributions at father nodes). (E2) 3 (elimination of fully

Generation of frontal matrices at leaves of the eliminaton tree expressed as the execution of graph grammar productions (A1)-(A) 4 -(AN)... PROCESS OF

Lecture 1: Frontal and multi-frontal solvers: orderings, elimination trees, refinement trees The lecture introduces the frontal and multi-frontal solver algorithms on the example of

COMPUTATIONAL COST ESTIMATES FOR PARALLEL SHARED MEMORY ISOGEOMETRIC MULTI-FRONTAL SOLVERS,. Computers and Mathematics with Applications, 67(10)

Figure: Convergence of tumor volumes for quescient cells, for different submodels sim1, sim2, sim3, for the averaged model (sim1+sim2+sim3)/3, for the supermodel, with respect to

rIGA with optimal size of macro elements (16 in this case) cubic B-splines is one order of magnitude faster than FEM and IGA-FEM. Daniel Garcia, David Pardo, Lisandro Dalcin,