GT540M 上的低性能 CUDA 代码

Question

在 GeForce GT540M 上执行以下代码示例大约需要 750 毫秒，而在 GT330M 上执行相同代码大约需要 250 毫秒。

将 dev_a 和 dev_b 复制到 CUDA 设备内存需要大约 350 毫秒 GT540M 和大约 250 毫秒。 “addCuda”的执行和复制回主机在 GT540M 上需要大约 400 毫秒，在 GT330M 上需要大约 0 毫秒。

这不是我所期望的，所以我检查了设备的属性，发现 GT540M 设备除了多处理器数量之外，其他方面都超过或等于 GT330M - GT540M 有 2 个，GT330M 有 6 个。这真的是真的吗?如果是这样，它真的会对执行时间产生如此大的影响吗？

#include "cuda_runtime.h"
#include "device_launch_parameters.h"

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

#define T 512
#define N 60000*T

__global__ void addCuda(double *a, double *b, double *c) {
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if(tid < N) {
        c[tid] = sqrt(fabs(a[tid] * b[tid] / 12.34567)) * cos(a[tid]);
    }
}

int main() {
    double *dev_a, *dev_b, *dev_c;

    double* a = (double*)malloc(N*sizeof(double));
    double* b = (double*)malloc(N*sizeof(double));
    double* c = (double*)malloc(N*sizeof(double));

    printf("Filling arrays (CPU)...\n\n");
    int i;
    for(i = 0; i < N; i++) {
        a[i] = (double)-i;
        b[i] = (double)i;
    }

    int timer = clock();
    cudaMalloc((void**) &dev_a, N*sizeof(double));
    cudaMalloc((void**) &dev_b, N*sizeof(double));
    cudaMalloc((void**) &dev_c, N*sizeof(double));
    cudaMemcpy(dev_a, a, N*sizeof(double), cudaMemcpyHostToDevice);
    cudaMemcpy(dev_b, b, N*sizeof(double), cudaMemcpyHostToDevice);

    printf("Memcpy time: %d\n", clock() - timer);
    addCuda<<<(N+T-1)/T,T>>>(dev_a, dev_b, dev_c);
    cudaMemcpy(c, dev_c, N*sizeof(double), cudaMemcpyDeviceToHost);

    printf("Time elapsed: %d\n", clock() - timer);

cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
free(a);
free(b);
free(c);

return 0;
}

设备的设备属性：

GT540M：

Major revision number:         2
Minor revision number:         1
Name:                          GeForce GT 540M
Total global memory:           1073741824
Total shared memory per block: 49152
Total registers per block:     32768
Warp size:                     32
Maximum memory pitch:          2147483647
Maximum threads per block:     1024
Maximum dimension 0 of block:  1024
Maximum dimension 1 of block:  1024
Maximum dimension 2 of block:  64
Maximum dimension 0 of grid:   65535
Maximum dimension 1 of grid:   65535
Maximum dimension 2 of grid:   65535
Clock rate:                    1344000
Total constant memory:         65536
Texture alignment:             512
Concurrent copy and execution: Yes
Number of multiprocessors:     2
Kernel execution timeout:      Yes

GT330M

Major revision number:         1
Minor revision number:         2
Name:                          GeForce GT 330M
Total global memory:           268435456
Total shared memory per block: 16384
Total registers per block:     16384
Warp size:                     32
Maximum memory pitch:          2147483647
Maximum threads per block:     512
Maximum dimension 0 of block:  512
Maximum dimension 1 of block:  512
Maximum dimension 2 of block:  64
Maximum dimension 0 of grid:   65535
Maximum dimension 1 of grid:   65535
Maximum dimension 2 of grid:   1
Clock rate:                    1100000
Total constant memory:         65536
Texture alignment:             256
Concurrent copy and execution: Yes
Number of multiprocessors:     6
Kernel execution timeout:      Yes

Answer 1

我认为从设备到主机的复制不可能是 ~0 毫秒。我建议检查该副本是否存在stg错误

Answer 2

查看多处理器的数量。