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).
Related
I get the execution time of vector adder with different size of groupsize and I only use one group in this experiment.
groupsize --------execution time
1 ----------------3.6
50 ---------------4.22
100 --------------4.3
200 --------------4.28
300 --------------4.3
400 --------------4.31
500 --------------4.38
600 --------------4.38
700 --------------4.78
800 --------------5.18
900 --------------5.78
1000 -------------6.4
Can I get the conclusion one sm can work about 600 workitems together?
and I have some questions, could anybody can help me?
Why does the execution time increase sharply when groupsize increases from 1 to 50 and from 600 to 1000?
thank you very much
It would be helpful to see some code, both of the kernel and the host enqueueing parameters. The conclusions also depend on what sort of hardware you're running this on - GPU, CPU, accelerator, FPGA, …?
A few ideas:
GPUs typically can run power-of-2 number of threads in parallel in an execution unit. You will likely get better results if you try e.g. 16, 32, 64, 128, etc. CPUs and other accelerators typically have SIMD-widths which are powers of 2 too, for example x86-64 SSE registers can hold 4 floats, AVX 8, AVX512 16, etc. so it most likely will help there, too.
As you can vary group size so freely, I'm going to assume your work-items don't need to coordinate among each other via local memory or barriers. (The problem is embarrassingly parallel.) A group size of 1 in theory allows your compiler, driver, and hardware maximum flexibility for distributing work-items to threads and parallel execution units optimally. So it should not be a surprise that this is the fastest. (Depending on register pressure and memory access patterns it can still sometimes be helpful to manually increase group size for specific types of hardware in the embarrassingly parallel case.)
On GPUs, all items in a work group must run on the same execution unit, in order to be able to coordinate and share local memory. So by increasing the group size, you're limiting the number of execution units the workload can be spread across, and the execution units need to run your work-items serially - you're reducing parallelism. Above 600 you're probably submitting fewer workgroups than your hardware has execution units.
I'm confused by this CL_DEVICE_MAX_COMPUTE_UNITS. For instance my Intel GPU on Mac, this number is 48. Does this mean the max number of parallel tasks run at the same time is 48 or the multiple of 48, maybe 96, 144...? (I know each compute unit is composed of 1 or more processing elements and each processing element is actually in charge of a "thread". What if these each of the 48 compute units is composed of more than 1 processing elements ). In other words, for my Mac, the "ideal" speedup, although impossible in reality, is 48 times faster than a CPU core (we assume the single "core" computation speed of CPU and GPU is the same), or the multiple of 48, maybe 96, 144...?
Summary: Your speedup is a little complicated, but your machine's (Intel GPU, probably GEN8 or GEN9) fp32 throughput 768 FLOPs per (GPU) clock and 1536 for fp16. Let's assume fp32, so something less than 768x (maybe a third of this depending on CPU speed). See below for the reasoning and some very important caveats.
A Quick Aside on CL_DEVICE_MAX_COMPUTE_UNITS:
Intel does something wonky when with CL_DEVICE_MAX_COMPUTE_UNITS with its GPU driver.
From the clGetDeviceInfo (OpenCL 2.0). CL_DEVICE_MAX_COMPUTE_UNITS says
The number of parallel compute units on the OpenCL device. A
work-group executes on a single compute unit. The minimum value is 1.
However, the Intel Graphics driver does not actually follow this definition and instead returns the number of EUs (Execution Units) --- An EU a grouping of the SIMD ALUs and slots for 7 different SIMD threads (registers and what not). Each SIMD thread represents 8, 16, or 32 workitems depending on what the compiler picks (we want higher, but register pressure can force us lower).
A workgroup is actually limited to a "Slice" (see the figure in section 5.5 "Slice Architecture"), which happens to be 24 EUs (in recent HW). Pick the GEN8 or GEN9 documents. Each slice has it's own SLM, barriers, and L3. Given that your apple book is reporting 48 EUs, I'd say that you have two slices.
Maximum Speedup:
Let's ignore this major annoyance and work with the EU number (and from those arch docs above). For "speedup" I'm comparing a single threaded FP32 calculation on the CPU. With good parallelization etc on the CPU, the speedup would be less, of course.
Each of the 48 EUs can issue two SIMD4 operations per clock in ideal circumstances. Assuming those are fused multiply-add's (so really two ops), that gives us:
48 EUs * 2 SIMD4 ops per EU * 2 (if the op is a fused multiply add)
= 192 SIMD4 ops per clock
= 768 FLOPs per clock for single precision floating point
So your ideal speedup is actually ~768. But there's a bunch of things that chip into this ideal number.
Setup and teardown time. Let's ignore this (assume the WL time dominates the runtime).
The GPU clock maxes out around a gigahertz while the CPU runs faster. Factor that ratio in. (crudely 1/3 maybe? 3Ghz on the CPU vs 1Ghz on the GPU).
If the computation is not heavily multiply-adds "mads", divide by 2 since I doubled above. Many important workloads are "mad"-dominated though.
The execution is mostly non-divergent. If a SIMD thread branches into an if-then-else, the entire SIMD thread (8,16,or 32 workitems) has to execute that code.
Register banking collisions delays can reduce EU ALU throughput. Typically the compiler does a great job avoiding this, but it can theoretically chew into your performance a bit (usually a few percent depending on register pressure).
Buffer address calculation can chew off a few percent too (EU must spend time doing integer compute to read and write addresses).
If one uses too much SLM or barriers, the GPU must leave some of the EU's idle so that there's enough SLM for each work item on the machine. (You can tweak your algorithm to fix this.)
We must keep the WL compute bound. If we blow out any cache in the data access hierarchy, we run into scenarios where no thread is ready to run on an EU and must stall. Assume we avoid this.
?. I'm probably forgetting other things that can go wrong.
We call the efficiency the percentage of theoretical perfect. So if our workload runs at ~530 FLOPs per clock, then we are 60% efficient of the theoretical 768. I've seen very carefully tuned workloads exceed 90% efficiency, but it definitely can take some work.
The ideal speedup you can get is the total number of processing elements which in your case corresponds to 48 * number of processing elements per compute unit. I do not know of a way to get the number of processing elements from OpenCL (that does not mean that it is not possible), however you can just google it for your GPU.
Up to my knowledge, a compute unit consists of one or multiple processing elements (for GPUs usually a lot), a register file, and some local memory. The threads of a compute unit are executed in a SIMD (single instruction multiple data) fashion. This means that the threads of a compute unit all execute the same operation but on different data.
Also, the speedup you get depends on how you execute a kernel function. Since a single work-group can not run on multiple compute units you need a sufficient number of work-groups in order to fully utilize all of the compute units. In addition, the work-group size should be a multiple of CL_KERNEL_PREFERRED_WORK_GROUP_SIZE_MULTIPLE.
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)
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.