CEE 618 Scientific Parallel Computing (Lecture 3)

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1 1 / 36 CEE 618 Scientific Parallel Computing (Lecture 3) Linear Algebra Basics using LAPACK Albert S. Kim Department of Civil and Environmental Engineering University of Hawai i at Manoa 2540 Dole Street, Holmes 383, Honolulu, Hawaii 96822

2 2 / 36 Table of Contents 1 Partial Differential Equation: Althernative Method 2 Linear Algebra LU decomposition Numerical Recipes in FORTRAN Linea Algeb PACAKage 3 Eigen Value & Eigen Vector 4 PBS(Portable Batch System

3 Partial Differential Equation: Althernative Method Outline 3 / 36 1 Partial Differential Equation: Althernative Method 2 Linear Algebra LU decomposition Numerical Recipes in FORTRAN Linea Algeb PACAKage 3 Eigen Value & Eigen Vector 4 PBS(Portable Batch System

4 Partial Differential Equation: Althernative Method Convection-Diffusion-Reaction Equation General form C = (D C) (vc) kc (1) t In a steady state without convection and reaction In 2D with a constant diffusion coefficient 0 = (D C) (2) 0 = 2 C x C y 2 (3) Mathematically identical to heat diffusion (C T ) 0 = 2 T x T y 2 (4) Examples? Let s watach some videos in 4 / 36

Partial Differential Equation: Althernative Method Example problem 5 / 36 Solve the following equation using the method of separation of variables: 2 C x 2 + 2 C y 2 = 0

5 Partial Differential Equation: Althernative Method Example problem 5 / 36 Solve the following equation using the method of separation of variables: 2 C x C y 2 = 0 (5) Boundary conditions (0 < x, y < L) 1 C (x = 0, y) = 0 2 C (x, y = 0) = 0 3 C (x, y = L) = 0 4 C (x = L, y) = 10 sin ( ) πy L Figure: How does your solution look like?

6 Partial Differential Equation: Althernative Method Solution 6 / By the method of the separation of variables C (x, y) = X (x) Y (y) (6) Prove. C = 10 πx πy sinh sin sinh π L L (7)

7 Partial Differential Equation: Althernative Method 7 / 36

8 Partial Differential Equation: Althernative Method Solution 8 / By MS Excel: I am nothing but an average of my neighbors. C ij = C i+1,j + C i 1,j + C i,j+1 + C i,j 1 4 (8) Excel setup 1 Open MS Excel 2 Go to File 3 Click Options 4 Go to Formulas 5 Click Enable iterative calculation

9 9 / 36 Outline Linear Algebra 1 Partial Differential Equation: Althernative Method 2 Linear Algebra LU decomposition Numerical Recipes in FORTRAN Linea Algeb PACAKage 3 Eigen Value & Eigen Vector 4 PBS(Portable Batch System

10 10 / 36 Example Linear Algebra Tony is two years odlder than Sam and the sum of their current ages is twenty. How old are Tony and Sam? Use a two by two matrix to solve this problem. T S = 2 (9) T + S = 20 (10)

11 11 / 36 A Linear System Linear Algebra LU decomposition A = a 11 a 12 a 1n a 21 a 22 a 2n a n1 a n2 a nn A x = b (11) x 1 b 1 x 2 b 2, x = and b =.. x n b n n n n 1 n 1 where A is a n n square matrix, and b is a n 1 column vector, of which all elements are known. Then, how can we calculate x?

12 12 / 36 LU decomposition Linear Algebra LU decomposition The square matrix A can be decomposed into A = L U (12) where L and U are lower and upper triangular matrixes, respectively, and calculated as α β 11 β 12 β 1n α 21 α 22 0 L =......, U = 0 β 22 β 2n α n1 α n2 α nn 0 0 β nn Then, A x = (L U) x = L (U x) = b (13) Let s set U x = y, then L y = b (14)

13 Linear Algebra LU decomposition Forward substitution with known L and b to solve for y L y = b α α 21 α α n1 α n2 α nn y 1 y 2. y n = b 1 b 2. b n Then, y 1 = b 1, y 2 = b 2 α 11 y 1, α 11 α (15) Using back substitution, y i = 1 i 1 b i α ij y j α ii (16) j=1 where i = 2, 3,..., n. 13 / 36

14 Linear Algebra LU decomposition Backward substitution with U and y to solve for x Then, β 11 β 12 β 1n β n 1,n 1 β n 1,n 0 0 β nn x n = y n, β nn Using back substitution, x i = 1 β ii where i = n 1, n 2,..., 1. U x = y (17) x 1. x n 1 = x n y 1. y n 1 y n x n 1 = y n 1 β n 1,n x n β n 1,n 1,... (18) y i n j=i+1 β ij x j (19) 14 / 36

15 15 / 36 Linear Algebra LU decomposition Combined matrix of α s and β s with less memory Using α ii = 1 where i = 1, 2,..., n L U C = β 11 β 12 β 13 β 1,n 1 β 1n α 21 β 22 β 23 β 2,n 1 β 2n α 31 α 32 β 33 β 3,n 1 β 3n α n 1,1 α n 1,2 α n 1,3 β n 1,n 1 β n 1,n α n1 α n2 α n3 α n,n 1 β nn

16 16 / 36 Example: Linear Algebra LU decomposition A = A x = b (20), b = 1 0 0, x = x 1 x 2 x 3 =? (21) C = , x = (22) However, C does not directrly represent L and U of matrix A because pivoting exchanges row index during the LU decomposition.

17 Makefile Linear Algebra LU decomposition 1 Use files in /opt/cee618s13/class03/ to solve this problem using ludcmp and lubksb subroutines from NRF Use LAPACK routines 2 of DGETRF and DGETRS. 3 Check how to link LAPACK in Makefile. 1 Section 2.3 of Numerical Recipes in FORTRAN 77, available at 2 LAPACK user s guide at 17 / 36

18 18 / 36 Linear Algebra Numerical Recipes in FORTRAN Using subroutines in NRF: ludcmp & lubksb 1 program LU i m p l i c i t none 3 i n t e g e r : : i, j, indx ( 3 ) r e a l : : a ( 3, 3 ) = ( / 1., 1., 2., 3., 1., 3., 1., 2., 4. / ) 5 r e a l : : d, b ( 3 ) = ( / 1., 0., 0. / ) 7 open (11, f i l e = l u. dat )! Display the given matrix, A and b 9 w r i t e ( 1 1, * ) do i =1,3 1 w r i t e (11, " (4(2 x, F12. 6 ) ) " ) ( a ( i, j ), j =1,3), b ( i ) enddo 3! Decomposition of the given m a trix A c a l l ludcmp ( a, 3, 3, indx, d ) 5! Display the decomposed matrix, A and b w r i t e ( 1 1, * ) 7 do i =1,3 w r i t e (11, " (4(2 x, F12. 6 ) ) " ) ( a ( i, j ), j =1,3) 9 enddo! Solving f o r x w ith the decomposed matrix using b a c k s u b s i t u t i o n 1 c a l l lubksb ( a, 3, 3, indx, b )! Display the decomposed matrix, A and the solution x 3 w r i t e ( 1 1, * ) do i =1,3 5 w r i t e (11, " (4(2 x, F12. 6 ) ) " ) ( a ( i, j ), j =1,3), b ( i ) enddo 7 stop end./codes/lu/lu3.f90

19 19 / 36 Linear Algebra Results using ludcmp & lubksb Numerical Recipes in FORTRAN = (23) /codes/LU/lu.dat

20 Linear Algebra Numerical Recipes in FORTRAN SUBROUTINE ludcmp ( a, n, np, indx, d ) 2 INTEGER n, np, indx ( n ),NMAX REAL d, a ( np, np ), TINY 4 PARAMETER (NMAX=500,TINY=1.0e 20) INTEGER i, imax, j, k 6 REAL aamax, dum, sum, vv (NMAX) d=1. 8 do 12 i =1,n aamax=0. 0 do 11 j =1,n i f ( abs ( a ( i, j ) ). gt. aamax ) aamax=abs ( a ( i, j ) ) 2 11 continue i f ( aamax. eq. 0. ) pause s i n g u l a r m a t r ix i n ludcmp 4 vv ( i ) = 1. / aamax 12 continue 6 do 19 j =1,n do 14 i =1, j 1 8 sum=a ( i, j ) do 13 k =1, i 1 0 sum=sum a ( i, k ) * a ( k, j ) 13 continue 2 a ( i, j ) =sum 14 continue 20 / 36

21 Linear Algebra Numerical Recipes in FORTRAN 1 SUBROUTINE lubksb ( a, n, np, indx, b ) INTEGER n, np, indx ( n ) 3 REAL a ( np, np ), b ( n ) INTEGER i, i i, j, l l 5 REAL sum i i =0 7 do 12 i =1,n l l =indx ( i ) 9 sum=b ( l l ) b ( l l ) =b ( i ) 1 i f ( i i. ne. 0 ) then do 11 j = i i, i 1 3 sum=sum a ( i, j ) * b ( j ) 11 continue 5 else i f (sum. ne. 0. ) then i i = i 7 e n d i f b ( i ) =sum 9 12 continue do 14 i =n,1, 1 1 sum=b ( i ) do 13 j = i +1,n 3 sum=sum a ( i, j ) * b ( j ) 21 / 36

22 22 / 36 Linear Algebra Linea Algeb PACAKage Using subroutines in LAPACK: dgetrf & dgetrs program LUlapack 2 i m p l i c i t none i n t e g e r : : i, j, i p i v ( 3 ), i n f o 4 double p r e c i s i o n : : a ( 3, 3 ) = ( / 1., 1., 2., 3., 1., 3., 1., 2., 4. / ) double p r e c i s i o n : : b ( 3 ) = ( / 1., 0., 0. / ) 6 open (11, f i l e = lu l a p a c k. dat ) 8! Display the given matrix, A and b w r i t e ( 1 1, * ) 0 do i =1,3 w r i t e (11, " (4(2 x, F12. 6 ) ) " ) ( a ( i, j ), j =1,3), b ( i ) 2 end do! Decomposition of the given matrix A 4 c a l l d g e t r f (3,3, a, 3, i p i v, i n f o )! Display the decomposed matrix, A and b 6 w r i t e ( 1 1, * ) do i =1,3 8 w r i t e (11, " (4(2 x, F12. 6 ) ) " ) ( a ( i, j ), j =1,3), b ( i ) end do 0! Solving f o r x with the decomposed matrix using b a c k s u b s i t u t i o n c a l l dgetrs ( N,3,1, a, 3, i p i v, b, 3, i n f o ) 2! Display the decomposed matrix, A and the solution x w r i t e ( 1 1, * ) 4 do i =1,3 w r i t e (11, " (4(2 x, F12. 6 ) ) " ) ( a ( i, j ), j =1,3), b ( i ) 6 end do stop 8 end./codes/lulapack/lu3dlapack.f90

23 23 / 36 Archives Linear Algebra Linea Algeb PACAKage Linear Equations at Individual at Single, REAL at Double, REAL at dgetrf at dgetrs at dgetri at

24 24 / 36 Specifically Linear Algebra Linea Algeb PACAKage call dgetrf ( 3, 3, a, 3, ipiv, info ) call dgetrs( N, 3, 1, a, 3, ipiv, b, 3, info )

25 25 / 36 Outline Eigen Value & Eigen Vector 1 Partial Differential Equation: Althernative Method 2 Linear Algebra LU decomposition Numerical Recipes in FORTRAN Linea Algeb PACAKage 3 Eigen Value & Eigen Vector 4 PBS(Portable Batch System

26 Eigen Value & Eigen Vector Eigen Value & Eigen Vector Example: Rotate to principal axes the quadratic surface In matrix form this equation is or x 2 + 6xy 2y 2 2yz + z 2 = 24 (24) ( x y ) z x y z = 24 (25) X T MX = 24 (26) The characteristic equation of this matrix is 1 µ µ µ = µ3 +13µ 12 = (µ 1) (µ + 4) (µ 3) = 0 (27) The characteristic values are µ = 1, 4, / 36

27 Eigen Value & Eigen Vector 27 / 36 From ( x y ) z x y z = 24 (28) relative to the principal axes (x, y, z ), the quadratic sufrace equation becomes ( x y z ) = 24 (29) or or where x 2 + ( 4) y z 2 = 24 (30) x y z X T M X = 24 (31) M = (32)

28 Eigen Value & Eigen Vector 28 / 36 Eigen vectors are ( 1 10, ( 3 5,, ( 3, 2, ) 0 3, ) 1 35 ) 1 14 for µ = 1 (33) for µ = 4 (34) for µ = 3 (35) or x y z C X = X X T C T = X T = x y z (36)

29 Eigen Value & Eigen Vector 29 / 36 In other words, = (37) In the eigen vector matrix, the columns can be exchanged and signs can be reverted. It is a matter of using right-handed or left-haded coordinates. Using transformed coordinates makes the problem mathematically so convenient. In quantum mechanics, eigen values are energy and eigen vectors are quantum states.

30 Eigen Value & Eigen Vector./codes/eigen/eigvv.f90 30 / 36 PROGRAM EIGENVV 2 IMPLICIT NONE INTEGER : : I, INFO, J, N, LWORK 4 DOUBLE PRECISION : : DUMMY(1,1) DOUBLE PRECISION, ALLOCATABLE, DIMENSION ( :, : ) : : A, B, VR 6 DOUBLE PRECISION, ALLOCATABLE, DIMENSION ( : ) : : ALPHAR, ALPHAI, BETA, WORK open (11, f i l e = mat. i n, s t a t u s = old ) 8 read ( 1 1, * ) N LWORK = 8 * N 0 a l l o c a t e (A(N,N),B(N,N),ALPHAR(N),ALPHAI (N),BETA(N),VR(N,N),WORK(LWORK) ) B = 0. 0 ; do i = 1, N; B( i, i ) = 1. 0 ; end do 2 READ ( 1 1, * ) ( ( A( I, J ), J =1,N), I =1,N) CALL DGGEV( N, V,N,A,N,B,N, ALPHAR, ALPHAI,BETA,DUMMY,1,VR,N,WORK,LWORK, INFO ) 4 w r i t e ( *, * ) Eigen values are ( diagonal ) : w r i t e ( *, " ( 3(2X, F12. 8 ) ) " ) ( ( A( i, j ), J =1,N), I =1,N) 6 w r i t e ( *, * ) c a l l eigvec_norm (N,VR) 8 w r i t e ( *, * ) Eigen vectors are : w r i t e ( *, " ( 3(2X, F12. 8 ) ) " ) ( (VR( i, j ), J =1,N), I =1,N) 0 w r i t e ( *, * ) deallocate (A,B,ALPHAR, ALPHAI,BETA,VR,WORK) 2 contains 4 subroutine eigvec_norm (N,VR) 6 DOUBLE PRECISION : : VR(N,N) i n t e g e r : : N, i, j 8 double p r e c i s i o n : : norm do i = 1, 3 0 norm = DOT_PRODUCT (VR( :, i ), VR( :, i ) ) VR( :, i ) = VR( :, i ) / s q r t ( norm ) 2 enddo end subroutine eigvec_norm 4 end program

31 31 / 36 Makefile Eigen Value & Eigen Vector s r c r o o t =eigvv 2 s r c f i l e =$ ( s r c r o o t ). f90 e x e f i l e =$ ( s r c r o o t ). x 4 a l l : 6 i f o r t $ ( s r c f i l e ) o $ ( e x e f i l e ) l l a p a c k 8 run :. / $ ( e x e f i l e ) 0 2 e d i t : vim $ ( s r c f i l e ) 4 clean : 6 rm f *. x *. o./codes/eigen/makefile

32 32 / 36 Output Eigen Value & Eigen Vector. / eigvv. x 2 Eigen values are ( diagonal ) : Eigen vectors are : /codes/eigen/output.dat

33 33 / 36 Outline PBS(Portable Batch System 1 Partial Differential Equation: Althernative Method 2 Linear Algebra LU decomposition Numerical Recipes in FORTRAN Linea Algeb PACAKage 3 Eigen Value & Eigen Vector 4 PBS(Portable Batch System

34 PBS(Portable Batch System sample0.pbs & sample1.pbs 34 / 36 1 #PBS S / bin / bash #PBS V 3 uname n echo $PBS_O_JOBID./codes/PBS/sample0.pbs #! / bin / bash 2 #PBS l w a l l t i m e =12:00:00 #PBS N MyJob 4 #PBS V uname n 6 echo $PBS_O_JOBID cd $PBS_O_WORKDIR 8 pwd./codes/pbs/sample1.pbs

35 PBS(Portable Batch System sample2.pbs 35 / 36 1 #! / bin / bash #PBS l host= f r a c t a l 3 #PBS l w a l l t i m e =12:00:00 #PBS l s e l e c t =1: mpiprocs =4: ncpus=4 5 #PBS N Sample #PBS V 7 #PBS j oe cd $PBS_O_WORKDIR 9 ### put your s p e c i f i c job here a f t e r time command ### time l s laf 1 ####################################################### q s t a t f $PBS_JOBID./codes/PBS/sample2.pbs

36 36 / 36 Commands PBS(Portable Batch System 1 $ qsub < sample0.pbs 2 $ qstat The first comman is to submit a job described in "sample0.pbs" to a queueing system, i.e. "torque". The second comman is to monitor a status of the job, of which job number was assigned automatically by the first command. Observe the directory since each command of "qsub" will generate two files with the job number. Look at contents of newly generated files.

CEE 618 Scientific Parallel Computing (Lecture 7): OpenMP (con td) and Matrix Multiplication

1 / 26 CEE 618 Scientific Parallel Computing (Lecture 7): OpenMP (con td) and Matrix Multiplication Albert S. Kim Department of Civil and Environmental Engineering University of Hawai i at Manoa 2540 Dole