Optimal workgroup size for sum reduction in OpenCL - opencl

Question

I am using the following kernel for sum reduciton.
__kernel void reduce(__global float* input, __global float* output, __local float* sdata)
{
// load shared mem
unsigned int tid = get_local_id(0);
unsigned int bid = get_group_id(0);
unsigned int gid = get_global_id(0);
unsigned int localSize = get_local_size(0);
unsigned int stride = gid * 2;
sdata[tid] = input[stride] + input[stride + 1];
barrier(CLK_LOCAL_MEM_FENCE);
// do reduction in shared mem
for(unsigned int s = localSize >> 2; s > 0; s >>= 1)
{
if(tid < s)
{
sdata[tid] += sdata[tid + s];
}
barrier(CLK_LOCAL_MEM_FENCE);
}
// write result for this block to global mem
if(tid == 0) output[bid] = sdata[0];
}
It works fine, but I don't know how to choose the optimal workgroup size or number of workgroups if I need more than one workgroup (for example if I want to calculate the sum of 1048576 elements). As far as I understand, the more workgroups I use, the more subresults I will get, which also means that I will need more global reductions at the end.
I've seen the answers to the general workgroup size question here. Are there any recommendations that concern reduction operations specifically?

Answer 1

This question is a possible duplicate of one I answered a while back:
What is the algorithm to determine optimal work group size and number of workgroup.
Experimentation will be the best way to know for sure for any given device.
Update:
I think you can safely stick to 1-dimensional work groups, as you have done in your sample code. On the host, you can try out the best values.
For each device:
1) query for CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE.
2) loop over a few multiples and run the kernel with that group size. save the execution time for each test.
3) when you think you have an optimal value, hard code it into a new kernel for use with that specific device. This will give a further boost to performance. You can also eliminate your sdata parameter in the device-specific kernel.
//define your own context, kernel, queue here
int err;
size_t global_size; //set this somewhere to match your test data size
size_t preferred_size;
size_t max_group_size;
err = clGetKernelWorkGroupInfo(kernel, device_id, CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE, sizeof(size_t), preferred_size, NULL);
//check err
err = clGetKernelWorkGroupInfo(kernel, device_id, CL_KERNEL_WORK_GROUP_SIZE, sizeof(size_t), max_group_size, NULL);
//check err
size_t test_size;
//your vars for hi-res timer go here
for (unsigned int i=preferred_size ; i<=max_group_size ; i+=preferred_size){
//reset timer
test_size = (size_t)i;
err = clEnqueueNDRangeKernel(queue, kernel, 1, NULL, &global_size, &test_size, 0, NULL, NULL);
if(err){
fail("Unable to enqueue kernel"); //implement your own fail function somewhere..
}else{
clfinish(queue);
//stop timer, save value
//output timer value and test_size
}
}
The device-specific kernel can look like this, except the first line should have your optimal value substituted:
#define LOCAL_SIZE 32
__kernel void reduce(__global float* input, __global float* output)
{
unsigned int tid = get_local_id(0);
unsigned int stride = get_global_id(0) * 2;
__local float sdata[LOCAL_SIZE];
sdata[tid] = input[stride] + input[stride + 1];
barrier(CLK_LOCAL_MEM_FENCE);
for(unsigned int s = LOCAL_SIZE >> 2; s > 0; s >>= 1){
if(tid < s){
sdata[tid] += sdata[tid + s];
}
barrier(CLK_LOCAL_MEM_FENCE);
}
if(tid == 0) output[get_group_id(0)] = sdata[0];
}