I dont have an HD5850 but how can I know maximum workgroup size of it for opencl ? What is the preferred floating point vector width for HD5850? I suspected it was 5 but did not work on a friends computer who has 5850. Tried width 4 but did not work fast enough now I suspect work group size is not optimal. Doing NBody for 25k 50k and 100k particles consists of float8 variables for x,y,z, vx,vy,vz.
Thanks.
If you need the OpenCL specifics at development time but don't have access to the hardware, try http://clbenchmark.com. For example, the HD 5850 page is here: http://clbenchmark.com/device-environment.jsp?config=11975982. It shows CL_DEVICE_PREFERRED_VECTOR_WIDTH_FLOAT=4.
Use clGetDeviceInfo to poll for CL_DEVICE_MAX_WORK_GROUP_SIZE. I think the 5850 will have this at 256, but that may not be optimal for your kernel.
Use the same technique to poll for CL_DEVICE_PREFERRED_VECTOR_WIDTH_FLOAT, which I think is 4 on your card.
clGetDeviceInfo
Related
I am using two graphics card for opencl code
using profiling, my GTX 630 kepler is running faster than GTX650 Ti for each method request.
after profiling i found out some differences for both graphics card. But i am not able to understand what occupancy, l1_global_load_hit, l1_global_load_miss, active_warps and active_cycles are less for GTX650 Ti. Can any one please help me understand these terms in a more better way.
Decrease local work group size from 1024 down to 512 or 256 maybe even 64, then try again. This will leave more local memory per wave of threads. So more will execute simultaneously henceforth occupying more ALUs.
Don't forget to make the total number of threads a multiple of 768(number of cores of your faster card) to actually fill it evenly through all cores.(not just multiple of 384 like 1k-is which is not good for your faster card)
My pyopencl kernel program is started with global size of (512,512), I assume it will run 512x512=262,144 times. I want to find the minimum value of a function in my 512x512 image but I don't want to return 262,144 floats to my CPU to calculate the min. I want to run another kernel (possibly waiting in the queue ) to find the min value of all 262,144 pixels then just send that one float to the CPU. I think this would be faster. Should my waiting kernel's global size be (1,1), ? I hope the large 262,144 Buffer of floats that I created using mf.COPY_HOST_PTR will not cross the GPU/CPU bus before I call the next kernel.
Thanks
Tim
Andreas is right: reduction is the solution. Here is a nice article from AMD explaining how to implement simple reduction. It discusses different approaches and the gain in terms of performance they bring. The example in the article is about summing all the elements and not to find the minimum, but it's fairly trivial to modify the given codes.
BTW, maybe I don't understand well you first sentence, but a kernel launched with a global size of (512, 512) will not run 262,144 times but only one time with 262,144 threads scheduled.
Use a reduction kernel to find the minimum.
I am wondering how to chose optimal local and global work sizes for different devices in OpenCL?
Is it any universal rule for AMD, NVIDIA, INTEL GPUs?
Should I analyze physical build of the devices (number of multiprocessors, number of streaming processors in multiprocessor, etc)?
Does it depends on the algorithm/implementation? Because I saw that some libraries (like ViennaCL) to assess correct values just tests many combination of local/global work sizes and chose best combination.
NVIDIA recommends that your (local)workgroup-size is a multiple of 32 (equal to one warp, which is their atomic unit of execution, meaning that 32 threads/work-items are scheduled atomically together). AMD on the other hand recommends a multiple of 64(equal to one wavefront). Unsure about Intel, but you can find this type of information in their documentation.
So when you are doing some computation and let say you have 2300 work-items (the global size), 2300 is not dividable by 64 nor 32. If you don't specify the local size, OpenCL will choose a bad local size for you. What happens when you don't have a local size which is a multiple of the atomic unit of execution is that you will get idle threads which leads to bad device utilization. Thus, it can be benificial to add some "dummy" threads so that you get a global size which is a multiple of 32/64 and then use a local size of 32/64 (the global size has to be dividable by the local size). For 2300 you can add 4 dummy threads/work-items, because 2304 is dividable by 32. In the actual kernel, you can write something like:
int globalID = get_global_id(0);
if(globalID >= realNumberOfThreads)
globalID = 0;
This will make the four extra threads do the same as thread 0. (it is often faster to do some extra work then to have many idle threads).
Hope that answered your question. GL HF!
If you're essentially making processing using little memory (e.g. to store kernel private state) you can choose the most intuitive global size for your problem and let OpenCL choose the local size for you.
See my answer here : https://stackoverflow.com/a/13762847/145757
If memory management is a central part of your algorithm and will have a great impact on performance you should indeed go a little further and first check the maximum local size (which depends on the local/private memory usage of your kernel) using clGetKernelWorkGroupInfo, which itself will decide of your global size.
I can not understand what work_dim is for in clEnqueueNDRangeKernel()?
So, what is the difference between work_dim=1 and work_dim=2?
And why work items are grouped into work groups?
A work item or a work group is a thread running on the device (or neither)?
Thanks ahead!
work_dim is the number of dimensions for the clEnqueueNDRangeKernel() execution.
If you specify work_dim = 1, then the global and local work sizes are unidimensional. Thus, inside the kernels you can only access info in the first dimension, e.g. get_global_id(0), etc.
If you specify work_dim = 2 or 3, then you must also specify 2 or 3 dimensional global and local worksizes; in such case, you can access info inside the kernels in 2 or 3 dimensions, e.g. get_global_id(1), or get_group_id(2).
In practice you can do everything in 1D, but for dealing with 2D or 3D data, it maybe simpler to directly use 2/3 dimensional kernels; for example, in the case of 2D data, such as an image, if each thread/work-item is to deal with a single pixel, each thread/work-item could deal with the pixel at coordinates (x,y), with x = get_global_id(0) and y = get_global_id(1).
A work-item is a thread, while work-groups are groups of work-items/threads.
I believe the division work-groups / work-items is related with the hardware architecture of GPUs and other accelerators (e.g. Cell/BE); you can map the execution of work-groups to GPU Stream Multiprocessors (in NVIDIA talk) or SPUs (in IBM/Cell talk), while the corresponding work-itens would run inside the execution units of the Stream MultiProcessors and/or SPUs. It's not uncommon to have work group size = 1 if you are executing kernels in a CPU (e.g. for a quad-core, you would have 4 work groups, each one with one work item - though in my experience it's usually better to have more workgroups than CPU cores).
Check the OpenCL reference manual, as well as the OpenCl manual for whichever device your are programming. The quick reference card is also very helpful.
Searching through the NVIDIA forums I found these questions, which are also of interest to me, but nobody had answered them in the last four days or so. Can you help?
Original Forum Post
Digging into OpenCL reading tutorials some things stayed unclear for me. Here is a collection of my questions regarding local and global work sizes.
Must the global_work_size be smaller than CL_DEVICE_MAX_WORK_ITEM_SIZES?
On my machine CL_DEVICE_MAX_WORK_ITEM_SIZES = 512, 512, 64.
Is CL_KERNEL_WORK_GROUP_SIZE the recommended work_group_size for the used kernel?
Or is this the only work_group_size the GPU allows?
On my machine CL_KERNEL_WORK_GROUP_SIZE = 512
Do I need to divide into work groups or can I have only one, but not specifying local_work_size?
To what do I have to pay attention, when I only have one work group?
What does CL_DEVICE_MAX_WORK_GROUP_SIZE mean?
On my machine CL_DEVICE_MAX_WORK_GROUP_SIZE = 512, 512, 64
Does this mean, I can have one work group which is as large as the CL_DEVICE_MAX_WORK_ITEM_SIZES?
Has global_work_size to be a divisor of CL_DEVICE_MAX_WORK_ITEM_SIZES?
In my code global_work_size = 20.
In general you can choose global_work_size as big as you want, while local_work_size is constraint by the underlying device/hardware, so all query results will tell you the possible dimensions for local_work_size instead of the global_work_size. the only constraint for the global_work_size is that it must be a multiple of the local_work_size (for each dimension).
The work group sizes specify the sizes of the workgroups so if CL_DEVICE_MAX_WORK_ITEM_SIZES is 512, 512, 64 that means your local_work_size can't be bigger then 512 for the x and y dimension and 64 for the z dimension.
However there is also a constraint on the local group size depending on the kernel. This is expressed through CL_KERNEL_WORK_GROUP_SIZE. Your cumulative workgoupsize (as in the product of all dimensions, e.g. 256 if you have a localsize of 16, 16, 1) must not be greater then that number. This is due to the limited hardware resources to be divided between the threads (from your query results I assume you are programming on a NVIDIA GPU, so the amount of local memory and registers used by a thread will limit the number of threads which can be executed in parallel).
CL_DEVICE_MAX_WORK_GROUP_SIZE defines the maximum size of a work group in the same manner as CL_KERNEL_WORK_GROUP_SIZE, but specific to the device instead the kernel (and it should be a a scalar value aka 512).
You can choose not to specify local_work_group_size, in which case the OpenCL implementation will choose a local work group size for you (so its not a guarantee that it uses only one workgroup). However it's generally not advisiable, since you don't know how your work is divided into workgroups and furthermore it's not guaranteed that the workgroupsize chosen will be optimal.
However, you should note that using only one workgroup is generally not a good idea performancewise (and why use OpenCL if performance is not a concern). In general a workgroup has to execute on one compute unit, while most devices will have more then one (modern CPUs have 2 or more, one for each core, while modern GPUs can have 20 or more). Furthermore even the one Compute Unit on which your workgroup executes might not be fully used, since several workgroup can execute on one compute unit in an SMT style. To use NVIDIA GPUs optimally you need 768/1024/1536 threads (depending on the generation, meaning G80/GT200/GF100) executing on one compute unit, and while I don't know the numbers for amd right now, they are in the same magnitude, so it's good to have more then one workgroup. Furthermore, for GPUs, it's typically advisable to have workgroups which at least 64 threads (and a number of threads divisible by 32/64 (nvidia/amd) per workgroup), because otherwise you will again have reduced performance (32/64 is the minimum granuaty for execution on gpus, so if you have less items in a workgroup, it will still execute as 32/64 threads, but discard the results from unused threads).