CUDA表现质疑

Question

Since i didnt got a response from the CUDA forum, ill try it here:

After doing a few programs in CUDA ive now started to obtain their effective bandwidth. However i have some strange results, for example in the following code, where i can sum all the elements in a vector(regardless of dimension), the bandwidth with the Unroll Code and the "normal" code seems to have the same median result(around 3000 Gb/s) I dont know if im doing something wrong(AFAIK the program works fine) but from what ive read so far, the Unroll code should have a higher bandwidth.

#include <stdio.h>
#include <limits.h>
#include <stdlib.h>
#include <math.h>
#define elements 1000
#define blocksize 16    


__global__ void vecsumkernel(float*input, float*output,int nelements){



    __shared__ float psum[blocksize];
    int tid=threadIdx.x;

    if(tid + blockDim.x * blockIdx.x < nelements)
    psum[tid]=input[tid+blockDim.x*blockIdx.x];
    else
    psum[tid]=0.0f;
    __syncthreads();

    //WITHOUT UNROLL

    int stride;     
    for(stride=blockDim.x/2;stride>0;stride>>=1){
            if(tid<stride)
                    psum[tid]+=psum[tid+stride];
    __syncthreads();
    }
    if(tid==0)
            output[blockIdx.x]=psum[0];


    //WITH UNROLL
 /*
    if(blocksize>=512 && tid<256) psum[tid]+=psum[tid+256];__syncthreads();
    if(blocksize>=256 && tid<128) psum[tid]+=psum[tid+128];__syncthreads();
    if(blocksize>=128 && tid<64) psum[tid]+=psum[tid+64];__syncthreads();


    if (tid < 32) {
            if (blocksize >= 64) psum[tid] += psum[tid + 32];
            if (blocksize >= 32) psum[tid] += psum[tid + 16];
            if (blocksize >= 16) psum[tid] += psum[tid + 8];
            if (blocksize >=  8) psum[tid] += psum[tid + 4];
            if (blocksize >=  4) psum[tid] += psum[tid + 2];
            if (blocksize >=  2) psum[tid] += psum[tid + 1];
    }*/

    if(tid==0)
            output[blockIdx.x]=psum[0];



}

void vecsumv2(float*input, float*output, int nelements){
    dim3 dimBlock(blocksize,1,1);
    int i;

    for(i=((int)ceil((double)(nelements)/(double)blocksize))*blocksize;i>1;i(int)ceil((double)i/(double)blocksize)){
            dim3 dimGrid((int)ceil((double)i/(double)blocksize),1,1);
            printf("\ni=%d\ndimgrid=%u\n ",i,dimGrid.x);

            vecsumkernel<<<dimGrid,dimBlock>>>(i==((int)ceil((double)(nelements)/(double)blocksize))*blocksize ?input:output,output,i==((int)ceil((double)(nelements)/(double)blocksize))*blocksize ? elements:i);
    }

 }

 void printVec(float*vec,int dim){
    printf("\n{");
    for(int i=0;i<dim;i++)
            printf("%f ",vec[i]);
    printf("}\n");
 }

 int main(){
    cudaEvent_t evstart, evstop;
    cudaEventCreate(&evstart);
    cudaEventCreate(&evstop);


    float*input=(float*)malloc(sizeof(float)*(elements));
    for(int i=0;i<elements;i++)
            input[i]=(float) i;


    float*output=(float*)malloc(sizeof(float)*elements);



    float *input_d,*output_d;

    cudaMalloc((void**)&input_d,elements*sizeof(float));

    cudaMalloc((void**)&output_d,elements*sizeof(float));



    cudaMemcpy(input_d,input,elements*sizeof(float),cudaMemcpyHostToDevice);


    cudaEventRecord(evstart,0);

    vecsumv2(input_d,output_d,elements);

    cudaEventRecord(evstop,0);
    cudaEventSynchronize(evstop);
    float time;
    cudaEventElapsedTime(&time,evstart,evstop);
    printf("\ntempo gasto:%f\n",time);
    float Bandwidth=((1000*4*2)/10^9)/time;
    printf("\n Bandwidth:%f Gb/s\n",Bandwidth);


    cudaMemcpy(output,output_d,elements*sizeof(float),cudaMemcpyDeviceToHost);


    cudaFree(input_d);
    cudaFree(output_d);
    printf("soma do vector");
    printVec(output,4);



   }

Answer 1

Your unrolled code has a lot of branching in it. I count ten additional branches. Typically branching within a warp on a GPU is expensive because all threads in the warp end up waiting on the branch (divergence).

See here for more info on warp divergence:

http://forums.nvidia.com/index.php?showtopic=74842

Have you tried using a profiler to see what's going on?

Answer 2

3000 Gb/s Does not make sense. The max bus speed of PCIe is 8Gb/s on each direction.

Take a look at this paper Parallel Prefix Sum to gain insight on how to speed up your implementation. Also consider that the thrust library have this already implemented in the Reductions module

Answer 3

your not-unrolled code is invalid. For stride<32 some threads of the same warp enter the for-loop, while the others do not. Therefore, some (but not all) threads of the warp hit the __syncthreads() . CUDA specification says that when that happens, the behaviour is undefined.

It can happen that warp gets out of sync and some threads already begin loading next chunk of data, halting on next instances of __syncthreads() while previous threads are still stuck in your previous loop.

I am not sure though if that is what you are going to face in this particular case.

Answer 4

I see you're doing Reduction Sum in kernel. Here's a good presentation by NVIDIA for optimizing reduction on GPUs. You'll notice that the same code that was giving a throughput of 2 GB/s is optimized to 63 GB/s in this guide.