Finding minimum in GPU slower than CPU - cuda

Question

I have implemented this code: http://www.cuvilib.com/Reduction.pdf in order to calculate the sum of the elements of a matrix.
However in GPU it runs much slower than in CPU.
I got i7 processor and NVIDIA GT 540M graphics card.
Is it supposed to be that way or something else?
EDIT: I use version 3 of the above code in Ubuntu 13.04 and I compile it using Eclipse Nsight. The size of the matrix is 2097152 elements. It executes in 3.6 ms whereas the CPU version in around 1.0 ms. Below is the whole code:
#include <stdio.h>
#include <stdlib.h>
#include <thrust/sort.h>
#include <sys/time.h>
#include <omp.h>
#include <iostream>
#include <algorithm>
#define MIN(a,b) (((a)<(b))?(a):(b))
static const int WORK_SIZE = 2097152;
int find_min(int *a,int length){
int min = a[0];
for (int i=1;i<length;i++)
if (a[i]<min)
min=a[i];
return min;
}
__global__ static void red_min(int *g_idata,int *g_odata) {
extern __shared__ int sdata[];
unsigned int tid = threadIdx.x;
unsigned int i = blockIdx.x * blockDim.x + threadIdx.x;
sdata[tid]= g_idata[i];
__syncthreads();
for(unsigned int s=blockDim.x/2; s > 0; s >>= 1) {
if (tid<s) {
sdata[tid] = MIN(sdata[tid],sdata[tid + s]);
}
__syncthreads();
}
if (tid == 0)
g_odata[blockIdx.x] = sdata[0];
}
int main(void) {
int *d1,*d2;
int i,*result;
int *idata,*fdata;
srand ( time(NULL) );
result = (int *)malloc(sizeof(int));
idata = (int *)malloc(WORK_SIZE*sizeof(int));
fdata = (int *)malloc(WORK_SIZE*sizeof(int));
cudaMalloc((int**)&d1,WORK_SIZE*sizeof(int));
cudaMalloc((int**)&d2,WORK_SIZE*sizeof(int));
for (i = 0; i < WORK_SIZE; i++){
idata[i] = rand();
fdata[i] = i;
}
struct timeval begin, end;
gettimeofday(&begin, NULL);
*result = find_min(idata,WORK_SIZE);
printf( "Minimum Element CPU: %d \n", *result);
gettimeofday(&end, NULL);
int time = (end.tv_sec * (unsigned int)1e6 + end.tv_usec) - (begin.tv_sec * (unsigned int)1e6 + begin.tv_usec);
printf("Microseconds elapsed CPU: %d\n", time);
cudaMemcpy(d1,idata,WORK_SIZE*sizeof(int),cudaMemcpyHostToDevice);
cudaEvent_t start, stop;
cudaEventCreate( &start);
cudaEventCreate( &stop);
cudaEventRecord(start,0);
int num_blocks = 16384;
bool flag = true;
while (num_blocks>0){
if (flag) {
red_min<<<num_blocks,128,128*sizeof(int)>>>(d1,d2);
}
else {
red_min<<<num_blocks,128,128*sizeof(int)>>>(d2,d1);
}
num_blocks /= 128;
flag = !flag;
}

Answer 1

GT540M is a mobile GPU, so I assume you're running on a laptop, and furthermore you may be hosting the X display on the 540M GPU.
I built a complete version of your code:
#include <stdio.h>
#include <stdlib.h>
#include <thrust/sort.h>
#include <sys/time.h>
#include <omp.h>
#include <iostream>
#include <algorithm>
#define MIN(a,b) (((a)<(b))?(a):(b))
static const int WORK_SIZE = 2097152;
int find_min(int *a,int length){
int min = a[0];
for (int i=1;i<length;i++)
if (a[i]<min)
min=a[i];
return min;
}
__global__ static void red_min(int *g_idata,int *g_odata) {
extern __shared__ int sdata[];
unsigned int tid = threadIdx.x;
unsigned int i = blockIdx.x * blockDim.x + threadIdx.x;
sdata[tid]= g_idata[i];
__syncthreads();
for(unsigned int s=blockDim.x/2; s > 0; s >>= 1) {
if (tid<s) {
sdata[tid] = MIN(sdata[tid],sdata[tid + s]);
}
__syncthreads();
}
if (tid == 0)
g_odata[blockIdx.x] = sdata[0];
}
int main(void) {
int *d1,*d2;
int i,*result;
int *idata,*fdata;
srand ( time(NULL) );
result = (int *)malloc(sizeof(int));
idata = (int *)malloc(WORK_SIZE*sizeof(int));
fdata = (int *)malloc(WORK_SIZE*sizeof(int));
cudaMalloc((int**)&d1,WORK_SIZE*sizeof(int));
cudaMalloc((int**)&d2,WORK_SIZE*sizeof(int));
for (i = 0; i < WORK_SIZE; i++){
idata[i] = rand();
fdata[i] = i;
}
struct timeval begin, end;
gettimeofday(&begin, NULL);
*result = find_min(idata,WORK_SIZE);
printf( "Minimum Element CPU: %d \n", *result);
gettimeofday(&end, NULL);
int time = (end.tv_sec * (unsigned int)1e6 + end.tv_usec) - (begin.tv_sec * (unsigned int)1e6 + begin.tv_usec);
printf("Microseconds elapsed CPU: %d\n", time);
cudaMemcpy(d1,idata,WORK_SIZE*sizeof(int),cudaMemcpyHostToDevice);
cudaEvent_t start, stop;
cudaEventCreate( &start);
cudaEventCreate( &stop);
cudaEventRecord(start,0);
int num_blocks = 16384;
bool flag = true;
int loops = 0;
while (num_blocks>0){
if (flag) {
red_min<<<num_blocks,128,128*sizeof(int)>>>(d1,d2);
}
else {
red_min<<<num_blocks,128,128*sizeof(int)>>>(d2,d1);
}
num_blocks /= 128;
flag = !flag;
loops++;
}
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
float et = 0.0f;
cudaEventElapsedTime(&et, start, stop);
printf("GPU time: %fms, in %d loops\n", et, loops);
int gpuresult;
if (flag)
cudaMemcpy(&gpuresult, d1, sizeof(int), cudaMemcpyDeviceToHost);
else
cudaMemcpy(&gpuresult, d2, sizeof(int), cudaMemcpyDeviceToHost);
printf("GPU min: %d\n", gpuresult);
return 0;
}
compiled it:
$ nvcc -O3 -arch=sm_20 -o t264 t264.cu
and ran it on a M2050 GPU, RHEL 5.5, CUDA 5.5, Xeon X5650 CPU
$ ./t264
Minimum Element CPU: 288
Microseconds elapsed CPU: 1217
GPU time: 0.621408ms, in 3 loops
GPU min: 288
$
So my CPU results were pretty close to yours, but my GPU results were about 5-6x faster. If we compare M2050 to your GT540M, we see that the M2050 has 14 SMs whereas the GT540M has 2. More importantly, the M2050 has about 5x the memory bandwidth of your GT540M GPU (28.8GB/s peak theoretical for GT540M vs. ~150GB/s peak theoretical for M2050)
Since a well written parallel reduction is a memory bandwidth constrained code on GPUs, the speed difference between your GPU and my GPU makes sense.
So I would say your results are probably about what is expected, and to get better results you will probably need a faster GPU.
Also, if your GT540M is also hosting an X display, it's possible that the GPU timing is corrupted by display activity. If we are timing a single kernel, this is not normally an issue - the kernel execution interrupts the display processing briefly. But when we are timing a sequence of kernels in succession, it's possible for the display tasks to jump in and execute in-between kernel calls (the GPU is multi-tasking when it is asked to both support a display and also process CUDA code). Therefore, this may be a possible performance impact in your case as well.