cuBLAS矩陣逆運算比MATLAB慢得多

Question

在我當前的項目中，我正在嘗試使用cuBLAS計算大型（n> 2000）矩陣的逆。 進行了逆計算，但是由於某些原因，與在MATLAB中進行計算相比，計算時間明顯慢。

我已經附上了使用兩種語言以及性能結果對我的隨機矩陣執行的示例計算。

對於可能導致此速度下降的任何幫助或建議，將不勝感激。

先感謝您。

對照

cuBLAS與MATLAB

N = 500：cuBLAS〜0.130秒，MATLAB〜0.066秒->〜1.97倍慢

N = 1000：cuBLAS〜0.898秒，MATLAB〜0.311秒->〜2.89倍慢

N = 2000：cuBLAS〜6.​​667秒，MATLAB〜0.659秒->〜10.12x慢

N = 4000：cuBLAS〜51.860秒，MATLAB〜4.296秒->〜12.07x慢

C ++代碼

#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代碼

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

系統信息：

GPU：GTX 780 3GB

CPU：i7-4790S @ 3.20 GHz

Answer 1

正如@RobertCrovella所說，您不應將批處理的小矩陣API用於單個大矩陣反轉。

基本上，您可以使用與代碼中相同的方法，但是使用非批處理版本的getrf()和getri()來最大化大型矩陣的性能。

對於getrf()您可以在這里找到它。

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

對於getri() ，盡管CUDA工具包不提供getri()來解決AX=I ，其中A由getrf() LU支持，但確實提供了getrs()來解決AX=B 您需要做的就是在調用getrs()之前設置B=I

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