Block LU Factorization Lecture 24

Block LU FactorizationLecture 24

MA471 Fall 2003

Example Case

1) Suppose we are faced with the solution of a linear system Ax=b

2) Further suppose:1) A is large (dim(A)>10,000) 2) A is dense3) A is full4) We have a sequence of different b vectors.

Problems• Suppose we are able to compute the

matrix –– It costs N2 doubles to store the matrix– E.g. for N=100,000 we require 76.3 gigabytes

of storage for the matrix alone.– 32 bit processors are limited to 4 gigabytes of

memory– Most desktops (even 64 bit) do not have 76.3

gigabytes

– What to do?

Divide and Conquer

P0 P1 P2 P3

P4 P5 P6 P7

P8 P9 P10 P11

P12 P13 P14 P15

One approach is to assume we have a square number of processors.We then divide the matrix into blocks – storing one block per processor.

Back to the Linear System

• We are now faced with LU factorization of a distributed matrix.

• This calls for a modified LU routine which acts on blocks of the matrix.

• We will demonstrate this algorithm for one level.

• i.e. we need to construct matrices L,U such that A=LU and we only store single blocks of A,L,U on any processor.

Constructing the Block LU Factorization

A00 A01 A02

A10 A11 A12

A20 A21 A22

=

L00 0 0

L10 1 0

L20 0 1

*

U00 U01 U02

0 ?11 ?12

0 ?21 ?22

First we LU factorize A00 and look for the above block factorization. However, we need to figure out what each of the entries are:

A00 = L00*U00 (compute by L00, U00 by LU factorization)

A01 = L00*U01 => U01 = L00\A01A02 = L00*U02 => U02 = L00\A02

A10 = L10*U00 => L10 = A10/U00A20 = L20*U00 => L20 = A20/U00

A11 = L10*U01 + ?11 => ?11 = A11 – L10*U01..

contA00 = L00*U00 (compute by L00, U00 by LU factorization)

A01 = L00*U01 => U01 = L00\A01A02 = L00*U02 => U02 = L00\A02

A10 = L10*U00 => L10 = A10/U00A20 = L20*U00 => L20 = A20/U00

A11 = L10*U01 + ?11 => ?11 = A11 – L10*U01A12 = L10*U02 + ?12 => ?12 = A12 – L10*U02A21 = L20*U01 + ?21 => ?21 = A21 – L20*U01A22 = L20*U02 + ?22 => ?22 = A22 – L20*U02

In the general case:Anm = Ln0*U0m + ?nm => ?nm = Anm – Ln0*U0m

Summary First Stage

A00 A01 A02

A10 A11 A12

A20 A21 A22

=

L00 0 0

L10 1 0

L20 0 1

*

U00 U01 U02

0 ?11 ?12

0 ?21 ?22

First step: LU factorize uppermost block diagonal

Second step: a) compute U0n = L00\A0n n>0 b) compute Ln0 = An0/U00 n>0

Third step: compute ?nm = Anm – Ln0*U0m, (n,m>0)

Now Factorize Lower SE Block

?11 ?12

?21 ?22=

L11 0

L21 1*

U11 U12

0 ??22

We repeat the previous algorithm this time on the two by two SE block.

End Result

A00 A01 A02

A10 A11 A12

A20 A21 A22

=

L00 0 0

L10 L11 0

L20 L21 L22

*

U00 U01 U02

0 U11 U12

0 0 U22

Matlab Version

Parallel AlgorithmP0 P1 P2

P3 P4 P5

P6 P7 P8

P0: A00 = L00*U00 (compute by L00, U00 by LU factorization)

P1: U01 = L00\A01P2: U02 = L00\A02

P3: L10 = A10/U00P6: L20 = A20/U00

P4: A11 <- A11 – L10*U01P5: A12 <- A12 – L10*U02P7: A21 <- A21 – L20*U01P8: A22 <- A22 – L20*U02

In the general case:Anm = Ln0*U0m + ?nm => ?nm = Anm – Ln0*U0m

Parallel Communication

L00U00 U01 U02

L10 A11 A12

L20 A21 A22

P0: L00,U00 =lu(A)

P1: U01 = L00\A01P2: U02 = L00\A02

P3: L10 = A10/U00P6: L20 = A20/U00

P4: A11 <- A11 – L10*U01P5: A12 <- A12 – L10*U02P7: A21 <- A21 – L20*U01P8: A22 <- A22 – L20*U02

In the general case:Anm = Ln0*U0m + ?nm => ?nm = Anm – Ln0*U0m

Communication Summary

P0: L00,U00 =lu(A)

P1: U01 = L00\A01P2: U02 = L00\A02

P3: L10 = A10/U00P6: L20 = A20/U00

P4: A11 <- A11 – L10*U01P5: A12 <- A12 – L10*U02P7: A21 <- A21 – L20*U01P8: A22 <- A22 – L20*U02

P0: sends L00 to P1,P2 sends U00 to P3,P6

P1: sends U01 to P4,P7P2: sends U02 to P5,P8

P3: sends L10 to P4,P5P4: sends L20 to P7,P8

P0 P1 P2

P3 P4 P5

P6 P7 P8

L00U00 U01 U02

L10 A11 A12

L20 A21 A22

Upshot

Notes:1) I added an MPI_Barrier purely to separate the LU factorization and the backsolve.2) In terms of efficiency we can see that quite a bit of time is spent in MPI_Wait

compared to compute time.3) The compute part of this code can be optimized much more – making the parallel

efficiency even worse.

a b

(a) P0: sends L00 to P1,P2 sends U00 to P3,P6

(b) P1: sends U01 to P4,P7(c) P2: sends U02 to P5,P8

(d) P3: sends L10 to P4,P5(e) P4: sends L20 to P7,P8

cde

(f) P4: sends L11 to P5 sends U11 to P7

(g) P1: sends U12 to P8

(h) P3: sends L21 to P8

f

1st stage: 1st stage:

g

h

Block Back Solve

• After factorization we are left with the task of using the distributed L and U to compute the backsolve:

U00L00 U01 U02

L10 U11L11 U12

L20 L21 U22L22

Block distribution of L and U

P0 P1 P2

P3 P4 P5

P6 P7 P8

Recall

• Given an LU factorization of A namely, L,U such that A=LU

• Then we can solve Ax=b by• y=L\b• x=U\y

Distributed Back Solve

L00 0 0

L10 L11 0

L20 L21 L22

=

y0

y1

y2

b0

b1

b2

P0: solve L00*y0 = b0 send: y0 to P3,P6P3: send: L10*y0 to P4P4: solve L11*y1 = b1-L10*y0 send: y1 to P7P6: send: L20*y0 to P8\P7: send: L21*y1 to P8P8: solve L22*y2 = b2-L20*y0-L21*y1Results: y0 on P0, y1 on P4, y2 on P8

P0 P1 P2

P3 P4 P5

P6 P7 P8

Matlab Code

Back Solve

After the factorization we computed a solution to Ax=b

This consists of two distributed block triangular systems to solve

Barrier Between Back Solves

This time I inserted an MPI_Barrier call between the backsolves. This highlights the serial nature of the backsolves..

Example Codehttp://www.math.unm.edu/~timwar/MA471F03/blocklu.m

http://www.math.unm.edu/~timwar/MA471F03/parlufact2.c