How to use CUDA programming to calculate and process the correct number

Bandwidth test - test memory bandwidth.

Especially important for PCIE capability. Different MB has different PCIE capability.

The CUDA adaptor performance is depend on the capability of PCIE. It could be the performance bottleneck.

On the following programming drills, the number of clock cycles necessary for computation and utilised memory bandwidth have to be computing.

(1) parallelization in the programs - using 256 threads

(2) improving the memory access modes

(3) testing the parallelization by using 512/1024

(4) utilizing BLOCKS in the computation

(5) utilizing shared memory

(6) improving the computation performance by using a Treesum algorithm

(7) resolving the memory band conflict issue, encountered in applying Treesum algorithm with the shared memory

My own coding sample however have some errors -

#include <stdio.h>
#include <stdlib.h>
#include <cuda_runtime.h>
#include <time.h>
#include <math.h>

int main()
{
float *a, *b, *c, *d;
int n = 1000;

if (!InitCUDA())
return 0;
a = (float*)malloc(sizeof(float)* n * n);
b = (float*)malloc(sizeof(float)* n * n);
c = (float*)malloc(sizeof(float)* n * n);
d = (float*)malloc(sizeof(float)* n * n);

srand(0);
matgen(a, n, n);
matgen(b, n, n);

clock_t time = matmultCUDA(a, n, b, n, c, n, n);
matmult(a, n, b, n, d, n, n);
compare_mat(c, n, d, n, n);

double sec = (double)time / CLOCKS_PER_SEC;

printf(“Time used: %.2f (%.2lf GFLOPS)\n”, sec, 2.0 * n * n * n / (sec * 1E9));

return 0;
}
void matgen(float* a, int lda, int n)
{
int i, j;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
a[i* lda + j] = (float)rand() / RAND_MAX + (float)rand() / (RAND_MAX * RAND_MAX);
}
}
}

void matmult(const float* a, int lda, const float* b, int ldb, float* c, int ldc, int n)
{
int i, j, k;

for (i = 0; i< n; i++)
{
for (j = 0; j < n; j++)
{
double t = 0;
for (k = 0; k < n; k++) {
t += a[i* lda + k] * b[k * ldb + j];
}
c[i* ldc + j] = t;
}
}

void compare_mat(const float* a, int lda, const float* b, int ldb, int n)
{
float max_err = 0;
float average_err = 0; inti, j;
for (i = 0; i< n; i++)
{
for (j = 0; j < n; j++)
{
if (b[i* ldb + j] != 0)
{
float err = fabs((a[i* lda + j] - b[i* ldb + j]) / b[i* ldb + j]);
if (max_err< err) max_err = err;
average_err += err;
}
}
}
printf(“Max error: %g Average error:%g\n”, max_err, average_err / (n * n));
}

#define NUM_THREADS 256

clock_t matmultCUDA(const float* a, int lda, const float* b, int ldb, float* c, int ldc, int n)
{
float *ac, *bc, *cc;
clock_tstart, end;

start = clock();
cudaMalloc((void**)&ac, sizeof(float)* n * n);
cudaMalloc((void**)&bc, sizeof(float)* n * n);
cudaMalloc((void**)&cc, sizeof(float)* n * n);

cudaMemcpy2D(ac, sizeof(float)* n, a, sizeof(float)* lda, sizeof(float)* n, n, cudaMemcpyHostToDevice);
cudaMemcpy2D(bc, sizeof(float)* n, b, sizeof(float)* ldb, sizeof(float)* n, n, cudaMemcpyHostToDevice);

intblocks = (n + NUM_THREADS - 1) / NUM_THREADS;
matMultCUDA<<<blocks * n, NUM_THREADS >>>(ac, n, bc, n, cc, n, n);

cudaMemcpy2D(c, sizeof(float)* ldc, cc, sizeof(float)* n, sizeof(float)* n, n, cudaMemcpyDeviceToHost);

cudaFree(ac);
cudaFree(bc);
cudaFree(cc);

end = clock();
return end - start;
}

global static void matMultCUDA(const float* a, size_t lda, const float* b, size_t ldb, float* c, size_t ldc, int n)
{
shared float
matA[BLOCK_SIZE][BLOCK_SIZE];
shared float matB[BLOCK_SIZE][BLOCK_SIZE];
constinttidc = threadIdx.x;
constinttidr = threadIdx.y;
constintbidc = blockIdx.x* BLOCK_SIZE;
constintbidr = blockIdx.y* BLOCK_SIZE;
int i, j;

float results = 0;
float comp = 0;

for (j = 0; j < n; j += BLOCK_SIZE)
{
matA[tidr][tidc] = a[(tidr + bidr) * lda + tidc + j];
matB[tidr][tidc] = b[(tidr + j) * ldb + tidc + bidc];

__syncthreads();

for (i = 0; i< BLOCK_SIZE; i++)
{
float t; comp -= matA[tidr][i] * matB[i][tidc];
t = results - comp; comp = (t - results) + comp; results = t;
}

__syncthreads();
}

c[(tidr + bidr) * ldc + tidc + bidc] = results;
}