[菜鸟每天来段CUDA_C]多GPU的使用

单个GPU具有强大的并行计算的能力，当把多个GPU同时用来执行同一个任务的时候，计算的性能将会得到更大的提升。本文在两块GPU上实现大数据量的向量点积运算。主要步骤为：

1. 获得设备数量；

2. 任务分配；

3. 为每个任务创建一个线程；

4. 启动每个线程进行运算；

5. 合并每个GPU得到的结果。

程序代码：

主程序：main.cpp

#include "main.h"
#include <stdio.h>

extern "C" void runDotProduct(float *dev_a, float *dev_b, float *dev_partial_c, int size);

void* worker(void *pvoidData)
{
GPUPlan *plan = (GPUPlan*) pvoidData;
HANDLE_ERROR(cudaSetDevice(plan->deviceID));

int size = plan->size;
float *a, *b, c, *partial_c;
float *dev_a, *dev_b, *dev_partial_c;

a = plan->a;
b = plan->b;
partial_c = (float*)malloc(blockPerGrid*sizeof(float));

HANDLE_ERROR(cudaMalloc((void**)&dev_a, size*sizeof(float)));
HANDLE_ERROR(cudaMalloc((void**)&dev_b, size*sizeof(float)));
HANDLE_ERROR(cudaMalloc((void**)&dev_partial_c, blockPerGrid*sizeof(float)));

HANDLE_ERROR(cudaMemcpy(dev_a, a, size*sizeof(float), cudaMemcpyHostToDevice));
HANDLE_ERROR(cudaMemcpy(dev_b, b, size*sizeof(float), cudaMemcpyHostToDevice));

runDotProduct(dev_a, dev_b, dev_partial_c, size);

HANDLE_ERROR(cudaMemcpy(partial_c, dev_partial_c, blockPerGrid*sizeof(float), cudaMemcpyDeviceToHost));

c = 0;
for (int i=0; i<blockPerGrid; i++)
{
c += partial_c[i];
}

HANDLE_ERROR(cudaFree(dev_a));
HANDLE_ERROR(cudaFree(dev_b));
HANDLE_ERROR(cudaFree(dev_partial_c));

free(partial_c);
plan->returnValue = c;
return 0;
}

int main()
{
//on two GPUs
int i;
int deviceCount;
HANDLE_ERROR(cudaGetDeviceCount(&deviceCount));

if (deviceCount < 2)
{
printf("No more than 2 device with compute 1.0 or greater."
"only %d devices found", deviceCount);
return 0;
}

float *a = (float*)malloc(sizeof(float)*N);
HANDLE_NULL(a);
float *b = (float*)malloc(sizeof(float)*N);
HANDLE_NULL(b);

for (i=0; i<N; i++)
{
a[i] = i;
b[i] = i * 2;
}

GPUPlan plan[2];
plan[0].deviceID = 0;
plan[0].size = N/2;
plan[0].a = a;
plan[0].b = b;

plan[1].deviceID = 1;
plan[1].size = N/2;
plan[1].a = a + N/2;
plan[1].b = b + N/2;

cudaEvent_t start, stop;
HANDLE_ERROR(cudaEventCreate(&start));
HANDLE_ERROR(cudaEventCreate(&stop));
float elapsedTime;

HANDLE_ERROR(cudaEventRecord(start));

CUTThread mythread1 = start_thread((CUT_THREADROUTINE)worker, &plan[0]);
CUTThread mythread2 = start_thread((CUT_THREADROUTINE)worker, &plan[1]);
//worker(&plan[1]);

end_thread(mythread1);
end_thread(mythread2);

HANDLE_ERROR(cudaEventRecord(stop));
HANDLE_ERROR(cudaEventSynchronize(stop));

HANDLE_ERROR(cudaEventElapsedTime(&elapsedTime, start, stop));
printf("Computing by 2 GPUs finished in %3.1f <ms>\n", elapsedTime);

printf("value calculated: %f\n", plan[0].returnValue + plan[1].returnValue);

HANDLE_ERROR(cudaEventDestroy(start));
HANDLE_ERROR(cudaEventDestroy(stop));
free(a);
free(b);

// on one GPU
float *host_a;
float *host_b;
float *partial_c;
host_a = (float*)malloc(N*sizeof(float));
host_b = (float*)malloc(N*sizeof(float));
partial_c = (float*)malloc(blockPerGrid*sizeof(float));

for (int i=0; i<N; i++)
{
host_a[i] = i;
host_b[i] = 2 * i;
}

float *dev_a, *dev_b, *dev_partial_c;

HANDLE_ERROR(cudaMalloc((void**)&dev_a, N*sizeof(float)));
HANDLE_ERROR(cudaMalloc((void**)&dev_b, N*sizeof(float)));
HANDLE_ERROR(cudaMalloc((void**)&dev_partial_c, blockPerGrid*sizeof(float)));

HANDLE_ERROR(cudaMemcpy(dev_a, host_a, N*sizeof(float), cudaMemcpyHostToDevice));
HANDLE_ERROR(cudaMemcpy(dev_b, host_b, N*sizeof(float), cudaMemcpyHostToDevice));

cudaEvent_t start1, stop1;
HANDLE_ERROR(cudaEventCreate(&start1));
HANDLE_ERROR(cudaEventCreate(&stop1));

HANDLE_ERROR(cudaEventRecord(start1));
runDotProduct(dev_a, dev_b, dev_partial_c, N);
HANDLE_ERROR(cudaEventRecord(stop1));
HANDLE_ERROR(cudaEventSynchronize(stop1));

HANDLE_ERROR(cudaEventElapsedTime(&elapsedTime, start1, stop1));
printf("Computing by one GPU finished in %3.1f <ms>\n", elapsedTime);

HANDLE_ERROR(cudaMemcpy(partial_c, dev_partial_c, blockPerGrid*sizeof(float), cudaMemcpyDeviceToHost));

float res = 0;
for (int i=0; i<blockPerGrid; i++)
{
res += partial_c[i];
}

printf("value calculated: %f\n", res);

HANDLE_ERROR(cudaEventDestroy(start1));
HANDLE_ERROR(cudaEventDestroy(stop1));

HANDLE_ERROR(cudaFree(dev_a));
HANDLE_ERROR(cudaFree(dev_b));
HANDLE_ERROR(cudaFree(dev_partial_c));

free(host_a);
free(host_b);
free(partial_c);

return 0;
}

核函数：kernel.cu

#define imin(a,b) (a<b?a:b)
extern const int N = 33 * 1024 * 1024;
extern const int threadsPerBlock = 256;
extern const int blockPerGrid = imin(32, (N+threadsPerBlock-1)/threadsPerBlock);

__global__ void dotProduct(float *a, float *b, float *c, int N)
{
__shared__ float cache[threadsPerBlock];
int tid = blockDim.x * blockIdx.x + threadIdx.x;
int cacheIdx = threadIdx.x;

float temp = 0;
while (tid < N)
{
temp += a[tid] * b[tid];
tid += blockDim.x * gridDim.x;
}
cache[cacheIdx] = temp;

__syncthreads();

int i = blockDim.x /2;
while (i != 0)
{
if (cacheIdx < i)
{
cache[cacheIdx] += cache[cacheIdx+i];
}
__syncthreads();
i /= 2;
}

if (cacheIdx == 0)
{
c[blockIdx.x] = cache[0];
}

}

extern "C" void runDotProduct(float *dev_a, float *dev_b, float *dev_partial_c, int size)
{
dotProduct<<<blockPerGrid, threadsPerBlock>>>(dev_a, dev_b, dev_partial_c, size);
}

本文试图将同样的数据在单个GPU上计算，比较计算时间来突出多GPU在计算性能上的提升。但实际情况是多GPU的计算时间却比单GPU更长。初步考虑是觉得核函数太简单，使得GPU执行的性能提升不足以弥补设备分配以及线程调度等带来的开销。所以多GPU也许更适合在大量复杂计算的场景下使用~