Question

在我目前的项目中，我试图用cuBLAS计算大（n> 2000）矩阵的逆矩阵。执行逆计算，但由于某些原因，计算时间明显慢于在MATLAB中完成的计算时间。

我已经使用我的实现语言和性能结果附加了对随机矩阵执行的样本计算。

对于可能导致这种减速的任何帮助或建议将不胜感激。

提前谢谢。

比较

cuBLAS与MATLAB

N = 500：cuBLAS~0.130秒，MATLAB~0.066秒 - > 〜1.97x较慢

N = 1000：cuBLAS~0.898秒，MATLAB~0.311秒 - > 〜2.89x慢

N = 2000：cuBLAS~6.667秒，MATLAB~0.659秒 - > ~10.12x慢

N = 4000：cuBLAS~51.860秒，MATLAB~4.296秒 - ＆gt; ~12.07x慢

C ++代码

#include <string>
#include <cuda_runtime.h>
#include <cublas_v2.h>
#include <conio.h>

#define CUDA_CALL(res, str) { if (res != cudaSuccess) { printf("CUDA Error : %s : %s %d : ERR %s\n", str, __FILE__, __LINE__, cudaGetErrorName(res)); } }
#define CUBLAS_CALL(res, str) { if (res != CUBLAS_STATUS_SUCCESS) { printf("CUBLAS Error : %s : %s %d : ERR %d\n", str, __FILE__, __LINE__, int(res)); } }

static cudaEvent_t cu_TimerStart;
static cudaEvent_t cu_TimerStop;

void d_CUDATimerStart(void)
{
    CUDA_CALL(cudaEventCreate(&cu_TimerStart), "Failed to create start event!");
    CUDA_CALL(cudaEventCreate(&cu_TimerStop), "Failed to create stop event!");

    CUDA_CALL(cudaEventRecord(cu_TimerStart), "Failed to record start event!");
}

float d_CUDATimerStop(void)
{
    CUDA_CALL(cudaEventRecord(cu_TimerStop), "Failed to record stop event!");

    CUDA_CALL(cudaEventSynchronize(cu_TimerStop), "Failed to synch stop event!");

    float ms;

    CUDA_CALL(cudaEventElapsedTime(&ms, cu_TimerStart, cu_TimerStop), "Failed to elapse events!");

    CUDA_CALL(cudaEventDestroy(cu_TimerStart), "Failed to destroy start event!");
    CUDA_CALL(cudaEventDestroy(cu_TimerStop), "Failed to destroy stop event!");

    return ms;
}

float* d_GetInv(float* L, int n)
{
    cublasHandle_t cu_cublasHandle;
    CUBLAS_CALL(cublasCreate(&cu_cublasHandle), "Failed to initialize cuBLAS!");

    float** adL;
    float** adC;
    float* dL;
    float* dC;
    int* dLUPivots;
    int* dLUInfo;

    size_t szA = n * n * sizeof(float);

    CUDA_CALL(cudaMalloc(&adL, sizeof(float*)), "Failed to allocate adL!");
    CUDA_CALL(cudaMalloc(&adC, sizeof(float*)), "Failed to allocate adC!");
    CUDA_CALL(cudaMalloc(&dL, szA), "Failed to allocate dL!");
    CUDA_CALL(cudaMalloc(&dC, szA), "Failed to allocate dC!");
    CUDA_CALL(cudaMalloc(&dLUPivots, n * sizeof(int)), "Failed to allocate dLUPivots!");
    CUDA_CALL(cudaMalloc(&dLUInfo, sizeof(int)), "Failed to allocate dLUInfo!");

    CUDA_CALL(cudaMemcpy(dL, L, szA, cudaMemcpyHostToDevice), "Failed to copy to dL!");
    CUDA_CALL(cudaMemcpy(adL, &dL, sizeof(float*), cudaMemcpyHostToDevice), "Failed to copy to adL!");
    CUDA_CALL(cudaMemcpy(adC, &dC, sizeof(float*), cudaMemcpyHostToDevice), "Failed to copy to adC!");

    d_CUDATimerStart();

    CUBLAS_CALL(cublasSgetrfBatched(cu_cublasHandle, n, adL, n, dLUPivots, dLUInfo, 1), "Failed to perform LU decomp operation!");
    CUDA_CALL(cudaDeviceSynchronize(), "Failed to synchronize after kernel call!");

    CUBLAS_CALL(cublasSgetriBatched(cu_cublasHandle, n, (const float **)adL, n, dLUPivots, adC, n, dLUInfo, 1), "Failed to perform Inverse operation!");
    CUDA_CALL(cudaDeviceSynchronize(), "Failed to synchronize after kernel call!");

    float timed = d_CUDATimerStop();

    printf("cublas inverse in: %.5f ms.\n", timed);

    float* res = (float*)malloc(szA);

    CUDA_CALL(cudaMemcpy(res, dC, szA, cudaMemcpyDeviceToHost), "Failed to copy to res!");

    CUDA_CALL(cudaFree(adL), "Failed to free adL!");
    CUDA_CALL(cudaFree(adC), "Failed to free adC!");
    CUDA_CALL(cudaFree(dL), "Failed to free dL!");
    CUDA_CALL(cudaFree(dC), "Failed to free dC!");
    CUDA_CALL(cudaFree(dLUPivots), "Failed to free dLUPivots!");
    CUDA_CALL(cudaFree(dLUInfo), "Failed to free dLUInfo!");

    CUBLAS_CALL(cublasDestroy(cu_cublasHandle), "Failed to destroy cuBLAS!");

    return res;
}

int main()
{
    int n = 1000;
    float* L = (float*)malloc(n * n * sizeof(float));
    for(int i = 0; i < n * n; i++)
        L[i] = ((float)rand()/(float)(RAND_MAX));

    float* inv = d_GetInv(L, n);

    printf("done.");
    _getch();

    return 0;
}

MATLAB代码

A = rand(1000);
tic
X = inv(A);
toc

系统信息：

GPU：GTX 780 3gb

CPU：i7-4790S @ 3.20 GHz

Answer 1

正如@RobertCrovella所说，你不应该使用批量小矩阵API进行单个大矩阵反演。

基本上，您可以使用与代码中相同的方法，但使用非批处理版本的getrf()和getri()来最大化大矩阵的性能。

getrf()你可以在这里找到它。

http://docs.nvidia.com/cuda/cusolver/index.html#cuds-lt-t-gt-getrf

对于getri()，尽管CUDA工具包未提供getri()来解决AX=I，其中A由getrf()进行LU设置，但它确实提供了getrs()一个AX=B来解决B=I。您需要做的就是在致电getrs()之前设置appc ti build -p ios。

http://docs.nvidia.com/cuda/cusolver/index.html#cuds-lt-t-gt-getrs

cuBLAS矩阵逆矩阵比MATLAB慢得多

1 个答案: