I'm trying to process an image using OpenCL 1.1 C++ on my AMD CPU.
The characteristics are:
using CPU: AMD Turion(tm) 64 X2 Mobile Technology TL-60
initCL:CL_DEVICE_IMAGE2D_MAX_WIDTH :8192
initCL:CL_DEVICE_IMAGE2D_MAX_HEIGHT :8192
initCL:timer resolution in ns:1
initCL:CL_DEVICE_GLOBAL_MEM_SIZE in bytes:1975189504
initCL:CL_DEVICE_GLOBAL_MEM_CACHE_SIZE in bytes:65536
initCL:CL_DEVICE_MAX_CONSTANT_BUFFER_SIZE in bytes:65536
initCL:CL_DEVICE_LOCAL_MEM_SIZE in bytes:32768
initCL:CL_DEVICE_MAX_COMPUTE_UNITS:2
initCL:CL_DEVICE_MAX_WORK_GROUP_SIZE:1024
initCL:CL_DEVICE_MAX_WORK_ITEM_DIMENSIONS:3
initCL:CL_DEVICE_MAX_WORK_ITEM_SIZES:dim=0, size 1024
initCL:CL_DEVICE_MAX_WORK_ITEM_SIZES:dim=1, size 1024
initCL:CL_DEVICE_MAX_WORK_ITEM_SIZES:dim=2, size 1024
createCLKernel:mean_value
createCLKernel:CL_KERNEL_WORK_GROUP_SIZE:1024
createCLKernel:CL_KERNEL_LOCAL_MEM_SIZE used by the kernel in bytes:0
createCLKernel:CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE:1
The kernel is for the moment empty:
__kernel void mean_value(image2d_t p_image,
__global ulong4* p_meanValue)
{
}
The execution call is:
cl::NDRange l_globalOffset;
// The global worksize is the entire image
cl::NDRange l_globalWorkSize(l_width, l_height);
// Needs to be determined
cl::NDRange l_localWorkSize;//(2, 2);
// Computes the mean value
cl::Event l_profileEvent;
gQueue.enqueueNDRangeKernel(gKernelMeanValue, l_globalOffset, l_globalWorkSize,
l_localWorkSize, NULL, &l_profileEvent);
If l_width=558 and l_height=328, l_localWorkSize can not be greater than (2, 2) otherwise, I get this error:"Invalid work group size"
Is it because I only have 2 cores ?
Is there a rule to determine l_localWorkSize ?
You can check 2 things using the clGetDeviceInfo function :
CL_DEVICE_MAX_WORK_GROUP_SIZE to check that 4 is not too big for your workgroup and
CL_DEVICE_MAX_WORK_ITEM_SIZES to check that the number of work-items by dimension is not too big.
And the fact the group-size may be limited to the number of cores makes sense : if you have inter work-items communication/synchronization you'll want them to be executed at the same time, otherwise the OpenCL driver would have to emulate this which might be at least hard and probably impossible in the general case.
I read in the OpenCL specs that enqueueNDRangeKernel() succeeds if l_globalWorkSize is evenly divisible byl_localWorkSize. In my case, I can set it up to (2,41).
Related
I'm a bit confused about compute units. I have an nvidia gtx 1650Ti graphics card. When I asked for max_compute_units, it returns 16 units, and max_work_group_size is 1024.
But when I executed the kernel:
int i = get_global_id (0);
result [i] = get_local_id (0);
I get the repeating local id range from 0 to 255. How does this relate to the max_compute_units returned by the graphics card? Is this an error in max_compute_units value and the gpu actually has more compute units than it indicates? Or does OpenCl get_local_id have its own distribution logic not tied to hardware? Thx!
OpenCL ompute units refer to streaming multiprocessors (SMs) on Nvidia GPUs or compute units (CUs) on AMD GPUs. Each SM contains 128 CUDA cores (Pascal and earlier) or 64 CUDA cores (Turing/Volta). For AMD, each CU contains 64 streaming multiprocessors. This refers to the hardware. The more SMs/CUs, the faster the GPU (within the same microarchitecture).
The work group size / local ID refer to how you group threads in software into so-called thread blocks. Thread blocks are useful for matrix multiplications for example, because within a thread block, communication between threads is possible via shared memory. Thread blocks can have different size (sort of an optimization parameter, either 32, 64, 128, 256, 512 or 1024 (max_work_group_size)). Based on your GPU, some intermediate values might also work. On the hardware (at least for Nvidia), the thread blocks are executed as so-called warps (groups of 32 threads) on the SMs. For Turing, one SM can compute 2 warps simultaneously. If you choose the thread block size 16, then each warp only computes 16 threads and the other 16 are idle, so you only get half the performance.
In your example with the local ID (this is the index in the thread block) betwqeen 0 and 255, your thread block size is 256. You define the thread block size in the kernel call as the "local range". max_work_group_size does not correlate with max_compute_units in any way; both are hardware / driver limitations.
My OpenCL program involves having about 7 billion work-items. In my C++ program, I would set this to my global_item_size:
size_t global_item_size = 7200000000;
If my program is compiled to 64-bit systems (x64), this global size is OK, since SIZE_MAX (the maximum value of size_t) is much larger than 7 billion. However, to ensure backwards compatibility I want to make sure that my program is able to compile to 32-bit systems (x86). On 32-bit systems, SIZE_MAX is about 4 billion, less than my global size, 7 billion. If I would try to set the global size to 7 billion, it would result in an overflow. What can I do in this case?
One of the solutions I was thinking about was to make a multi-dimensional global size and local size. However, this solution requires the kernel to calculate the original global size (because my kernel heavily depends on the global and local size), which would result in a performance loss.
The other solution I considered was to launch multiple kernels. I think this solution would be a little "sloppy" and synchronizing kernels also wouldn't be the best solution.
So my question basically is: How can I (if possible) make the global size larger than the maximum size of size_t? If this is not possible, what are some workarounds?
If you want to avoid batches you can give each kernel more work but effectively wrapping the code in a for loop. E.g.
for (int i = 0; i < WORK_ITEMS_PER_THREAD; ++i)
{
size_t id = WORK_ITEMS_PER_THREAD * get_global_id(0) + i;
...
}
Try to use uint64_t global_item_size = 7200000000ull; to avoid 32-bit integer overflow.
If you are strictly limited to the maximum 32-bit number of work items, you could do the computation in several batches (exchange GPU buffers in between compute steps via PCIe transfer) or you could pack several data items into one GPU thread.
I've shortly started using OpenCL to write programs for GPUs. I'm familiar with basic concepts that are required to write efficient programs in OpenCL, like work-items, work-groups, global-item-size, barriers, etc.
One of my programs involved making about 20 million work-groups with 360 work-items in each work-group. However, for some reason OpenCL couldn't handle that many number of work-groups. All elements of my output array simply remained 0. In addition, OpenCL didn't even start the calculations when I called clEnqueueNDRangeKernel(), since when I viewed the GPU usage stats I didn't see a "spike" that usually happens when I run an OpenCL kernel. I attempted to reduce the number work-groups, to see what is the maximum number of work-groups. It was 5965232 and it is always 5965232. Not more, not less.
I know that the problem is NOT with the number of work-items. It is with the number of work-groups. To prove this, here is my original code, where LIST_SIZE is 360.
global_item_size = 5965232*LIST_SIZE;
local_size = LIST_SIZE;
and a modified version of my code:
global_item_size = 5965232*LIST_SIZE*1.3;
local_size = LIST_SIZE*1.3;
In all the scenarios, the number of work-groups limit was 5965232.
I'm trying to find out what causes this limit and how to check this limit. I understand that there may be a limitation, but what causes this limitation and how can I check check this limit number in OpenCL? I've did a lot of research, but all sites are talking about work-group size limits and not about number of work-group limits.
I'm using the Intel Graphics HD 4000 GPU with an i5-3320M. It has 32 MB of integrated RAM.
5965232*320 = 2147483520 < 2147483647 = 2^31-1 = maximum 32-bit signed integer value
You are dealing with a classical 32-bit integer overflow in the multiplication in line
global_item_size = 5965232*LIST_SIZE;
Try global_item_size = 5965232ull*(uint64_t)LIST_SIZE; instead. Make sure global_item_size is data type uint64_t.
I'm using the Tesla m1060 for GPGPU computation. It has the following specs:
# of Tesla GPUs 1
# of Streaming Processor Cores (XXX per processor) 240
Memory Interface (512-bit per GPU) 512-bit
When I use OpenCL, I can display the following board information:
available platform OpenCL 1.1 CUDA 6.5.14
device Tesla M1060 type:CL_DEVICE_TYPE_GPU
max compute units:30
max work item dimensions:3
max work item sizes (dim:0):512
max work item sizes (dim:1):512
max work item sizes (dim:2):64
global mem size(bytes):4294770688 local mem size:16383
How can I relate the GPU card informations to the OpenCL memory informations ?
For example:
What does "Memory Interace" means ? Is it linked the a Work Item ?
How can I relate the "240 cores" of the GPU to Work Groups/Items ?
How can I map the work-groups to it (what would be the number of Work groups to use) ?
Thanks
EDIT:
After the following answers, there is a thing that is still unclear to me:
The CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE value is 32 for the kernel I use.
However, my device has a CL_DEVICE_MAX_COMPUTE_UNITS value of 30.
In the OpenCL 1.1 Api, it is written (p. 15):
Compute Unit: An OpenCL device has one or more compute units. A work-group executes on a single compute unit
It seems that either something is incoherent here, or that I didn't fully understand the difference between Work-Groups and Compute Units.
As previously stated, when I set the number of Work Groups to 32, the programs fails with the following error:
Entry function uses too much shared data (0x4020 bytes, 0x4000 max).
The value 16 works.
Addendum
Here is my Kernel signature:
// enable double precision (not enabled by default)
#ifdef cl_khr_fp64
#pragma OPENCL EXTENSION cl_khr_fp64 : enable
#else
#error "IEEE-754 double precision not supported by OpenCL implementation."
#endif
#define BLOCK_SIZE 16 // --> this is what defines the WG size to me
__kernel __attribute__((reqd_work_group_size(BLOCK_SIZE, BLOCK_SIZE, 1)))
void mmult(__global double * A, __global double * B, __global double * C, const unsigned int q)
{
__local double A_sub[BLOCK_SIZE][BLOCK_SIZE];
__local double B_sub[BLOCK_SIZE][BLOCK_SIZE];
// stuff that does matrix multiplication with __local
}
In the host code part:
#define BLOCK_SIZE 16
...
const size_t local_work_size[2] = {BLOCK_SIZE, BLOCK_SIZE};
...
status = clEnqueueNDRangeKernel(command_queue, kernel, 2, NULL, global_work_size, local_work_size, 0, NULL, NULL);
The memory interface doesn't mean anything to an opencl application. It is the number of bits the memory controller has for reading/writing to the memory (the ddr5 part in modern gpus). The formula for maximum global memory speed is approximately: pipelineWidth * memoryClockSpeed, but since opencl is meant to be cross-platform, you won't really need to know this value unless you are trying to figure out an upper bound for memory performance. Knowing about the 512-bit interface is somewhat useful when you're dealing with memory coalescing. wiki: Coalescing (computer science)
The max work item sizes have to do with 1) how the hardware schedules computations, and 2) the amount of low-level memory on the device -- eg. private memory and local memory.
The 240 figure doesn't matter to opencl very much either. You can determine that each of the 30 compute units is made up of 8 streaming processor cores for this gpu architecture (because 240/30 = 8). If you query for CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE, it will very likey be a multiple of 8 for this device. see: clGetKernelWorkGroupInfo
I have answered a similar questions about work group sizing. see here, and here
Ultimately, you need to tune your application and kernels based on your own bench-marking results. I find it worth the time to write many tests with various work group sizes and eventually hard-code the optimal size.
Adding another answer to address your local memory issue.
Entry function uses too much shared data (0x4020 bytes, 0x4000 max)
Since you are allocating A_sub and B_sub, each having 32*32*sizeof(double), you run out of local memory. The device should be allowing you to allocate 16kb, or 0x4000 bytes of local memory without an issue.
0x4020 is 32 bytes or 4 doubles more than what your device allows. There are only two things I can think of that may cause the error: 1) there could be a bug with your device or drivers preventing you from allocating the full 16kb, or 2) you are allocating the memory somewhere else in your kernel.
You will have to use a BLOCK_SIZE value less than 32 to work around this for now.
There's good news though. If you only want to hit a multiple of CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE as a work group size, BLOCK_SIZE=16 already does this for you. (16*16 = 256 = 32*8). To better take advantage of local memory, try BLOCK_SIZE=24. (576=32*18)
In the clEnqueueNDRangeKernel should global_work_size parameter be a power of 2?
If not and it is not power of two which error (if at all) is returned?
UPD
Based on the answers : global and local work sizes should not be power of two.
What aabout relation between workgroup size and wavefront size?:
if wavefront size is 64 and local_work_size < 64 - in each lock-step 64 work-item will execute,while (64 - local_work_size) will be work_items which "do nothing".
if 128 > local_work_size > 64 - how will the execution be? In even lock-step entire wavefront will be executed (64 work-items) and in one one local_work_size % 64
Its not necessary that that global work size is a power of 2, it can be any positive integer and less than the maximum number of work items allowed by the device.
The values doesn't need to be a power of 2 but it has to be a number divisible by the work group size.
As it is already said, it does not have to be a power of 2. But in order to have a good performance, you have to choose a local work size that is a multiple of 32 (see this related question: Questions about global and local work size)
Therefore, as your local work size must be a divider of your global work size, you will likely have a power of two (one source of optimization is to choose a global work size bigger than necessary; in order to choose a good local work size, you have to try some)