12d.1 two example parallel programs using mpi unc-wilmington, c. ferner, 2007 mar 209, 2007

12d.1

Two Example Parallel Programs using MPI

UNC-Wilmington, C. Ferner, 2007 Mar 209, 2007

12d.2

Matrix Multiplication• Matrices are multiplied together using the

dot product of each row of the first matrix with each column of the second matrix

* =

A B C

12d.3

Matrix Multiplication

• For each value at row i and column j, the result is the dot product of the ith row from A and the jth column from B:

1

0,,, *

N

kjkkiji BAC

12d.4

Matrix Multiplication• For each row i from [0..N-1] and each

column j from [0..N-1] the value for position [i][j] of the resulting matrix is computed:

for (i = 0; i < N; i++) for (j = 0; j < N; j++) { C[i][j] = 0; for (k = 0; k < N; j++) C[i][j] += A[i][k] * B[k][j]; }

12d.5

Matrix Multiplication• This can be implemented

on multiple processors where each processor is responsible for computing a different set of rows in the final matrix

• As long as each processor has the parts of the A and B matrix, they can do this without communication

C

12d.6

Matrix Multiplication• If there are N rows and P processors,

then each processor is responsible for N/P rows.

• Each processor is responsible for the rows from my_rank * N/P up to (but excluding) (my_rank + 1) * N/P

my_rank = 0

my_rank = 1

my_rank = 2

0 * N/P

1 * N/P

2 * N/P

3 * N/P

{{{

12d.7

Matrix Multiplication• This is coded as:

for (i = 0 + my_rank * N/P; i < 0 + (my_rank + 1) * N/P; i++) for (j = 0; j < N; j++) { C[i][j] = 0; for (k = 0; k < N; j++) C[i][j] += A[i][k] * B[k][j]; }

12d.8

Matrix Multiplication• One Problem: What if N/P is not an

integer? • The last processor has fewer than N/P

rows for which it is responsible.• The code on the previous slide will cause

the last processors (or last couple of processors) to compute beyond the last row of the matrix

12d.9

Matrix Multiplication• This is dealt with as follows:

blksz = (int) ceil((float) N / P);

for (i = 0 + my_rank * blksz; i < min(N, 0 + (my_rank + 1) * blksz); i++) for (j = 0; j < N; j++) { C[i][j] = 0; for (k = 0; k < N; j++) C[i][j] += A[i][k] * B[k][j]; }

12d.10

Matrix Multiplication• For example suppose N=13 and P=4.

Then:

blksz = ceiling(13/4) = 4

Processor 0 : i = [0*4..1*4) = [0..4)Processor 1 : i = [1*4..2*4) = [4..8)Processor 2 : i = [2*4..3*4) = [8..12)Processor 3 : i = [3*4..min(13,4*4))=[12..13)

12d.11

Matrix Multiplication• The assignment deals with the parallel

execution of matrix multiplication

12d.12

Numerical Integration

• Suppose we have a non-negative, continuous function f and we want to compute the integral of f from a to b:

a b

y

x

xf

dxxfb

a

12d.13


• We can approximate the integral by dividing the area into trapezoids and summing the area of the trapezoids

a b

y

x

xf

12d.14


• If we use equal width partitions, then each partition is h=(a+b)/n

a b

y

x

xf

12d.15


• The area of the ith trapezoid is:

a b

y

x

xf

h

1ixf

ixf

ii xf+xfh 12

1

12d.16


• The area for all trapezoids is: ...

2

1

2

12110 +xf+xfh+xf+xfh

nn xf+xfh+ 12

1...

nxf++xf+xf+xfh

= ...222 210

hxf++xf+xf+xf+xf= nn 1210 ...2/2/

12d.17

Numerical Integration Sequential program

double f(double x);

main (int argc, char *argv[])

{

int N, i;

double a, b, h, x, integral;

char *usage = "Usage: %s a b N \n";

double elapsed_time;

struct timeval tv1, tv2;

12d.18


if (argc < 4) {

fprintf (stderr, usage, argv[0]);

return -1;

}

a = atof(argv[1]);

b = atof(argv[2]);

N = atoi(argv[3]);

12d.19


gettimeofday(&tv1, NULL);

h = (b - a) / N;

integral = (f(a) + f(b))/2.0;

x = a + h;

for (i = 1; i < N; i++) {

integral += f(x);

x += h;

}

integral = integral*h;

gettimeofday(&tv2, NULL);

12d.20


elapsed_time = (tv2.tv_sec - tv1.tv_sec) +

((tv2.tv_usec - tv1.tv_usec) / 1000000.0);

printf ("elapsed_time=\t%lf seconds\n",

elapsed_time);

printf ("With N = %d trapezoids, \n", N);

printf ("estimate of integral from %f to %f = %f\n", N, a, b, integral);

}

12d.21


double f(double x)

{

return 6*x*x - 5*x;

}

12d.22


$ ./integ 1 3 10000

a = 1.000000, b = 3.000000, N = 10000

elapsed_time= 0.000567 seconds

With N = 10000 trapezoids,

estimate of integral from 1.000000 to 3.000000 = 32.000000

12d.23

Numerical Integration Parallel program

• Each processor will be responsible for computing the area of a subset of trapezoids

a b

y

x

xf{

P0

{

P1

{P

2

12d.24


double f (double x);

int main(int argc, char *argv[])

{

int N, P, mypid, blksz, i;

double a, b, h, x, integral, localA, localB,

total;

char *usage = "Usage: %s a b N \n";

double elapsed_time;

struct timeval tv1, tv2;

int abort = 0;

12d.25


a = atof(argv[1]);

b = atof(argv[2]);

N = atoi(argv[3]);

MPI_Bcast (&a, 1, MPI_DOUBLE, 0,

MPI_COMM_WORLD);

MPI_Bcast (&b, 1, MPI_DOUBLE, 0,

MPI_COMM_WORLD);

MPI_Bcast (&N, 1, MPI_INT, 0, MPI_COMM_WORLD);

h = (b - a) / N;

12d.26


blksz = (int) ceil ( ((float) N) / P);

localA = a + mypid * blksz * h;

localB = min(b, a + (mypid + 1) * blksz * h);

integral = (f(localA) + f(localB))/2.0;

x = localA + h;

for (i = 1; i < blksz && x <= localB; i++) {

integral += f(x);

x += h;

}

integral = integral*h;

12d.27


MPI_Reduce (&integral, &total, 1, MPI_DOUBLE, MPI_SUM, 0, MPI_COMM_WORLD);

if (mypid == 0) printf ("integral = %f\n", total);}

float f(float x)

{

return 6*x*x - 5*x;

}

12d.28


$ mpicc mpiInteg.c -o mpiInteg -lm$ mpirun -nolocal -np 4 mpiInteg 1 3 10000

elapsed_time= 0.001416 secondsintegral = 32.000000

12d.1 two example parallel programs using mpi unc-wilmington, c. ferner, 2007 mar 209, 2007

Documents

matrix slide

b matrix

n j cij

b c slide

n rows

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