FFTW Vs CUFFT Performance


Can anyone help me with this.

Old Code: Inside fortran

call sfftw_plan_dft_3d(plan,n1,n2,n3,cx,cx,ifset,64)
call sfftw_execute (plan)
call sfftw_destroy_plan (plan)

New Code: Inside Fortran
call tempfft(n1,n2,n3,cx,direction)

#include <stdio.h>
#include <cufft.h>
#include <cutil.h>
#include <cuComplex.h>
#include “cuda.h”

extern “C” void tempfft_(int *n1, int *n2, int *n3,cufftComplex *data, int direction)
int Nx = *n1;
int Ny = *n2;
int Nz = *n3;
cufftComplex *d_data;

    CUDA_SAFE_CALL(cudaMalloc((void**) &d_data, sizeof(cufftComplex)*Nx*Ny*Nz));

    CUDA_SAFE_CALL(cudaMemcpy(d_data, data, Nx*Ny*Nz*sizeof(cufftComplex), cudaMemcpyHostToDevice));

    cufftHandle plan1;
    CUDA_SAFE_CALL(cufftPlan3d(&plan1, Nz, Ny, Nx, CUFFT_C2C));

            CUDA_SAFE_CALL(cufftExecC2C(plan1, (cufftComplex *)d_data, (cufftComplex *)d_data, CUFFT_FORWARD));
            CUDA_SAFE_CALL(cufftExecC2C(plan1, (cufftComplex *)d_data, (cufftComplex *)d_data, CUFFT_INVERSE));

    CUDA_SAFE_CALL(cudaMemcpy(data, d_data, Nx*Ny*Nz*sizeof(cufftComplex), cudaMemcpyDeviceToHost));



When I simulate the above codes inside a big FORTRAN Application

FFTW code takes about 21 minutes for each step while the CUDA code is taking about 66 minutes for each step.

a) Is there any way I can increase the performance ?


It would be better for you to set up the plan outside of this FFT call once and reuse that plan instead of creating a new one every time you want to do an FFT. This assumes of course that you’re doing the same size and type (C2C, C2R, etc.) of FFT everytime.

However, the bigger issue here (which I’m guessing you can’t get away from) is the fact that you’re moving the entire input and output of the FFT across the system bus each time. Even if the FFT is done 10x faster on the card, it might take all that time saved (plus more) just to get the data on and off the card. In a graphics rendering situation, the data is almost always one way (to the card), and lot’s of large data arrays (like textures) are already loaded on the card.