I am going through the sample code of NVIDIA provided at link
In the sample kernels code (file oclReduction_kernel.c) reduce4 uses the technique of
1) unrolling and removing synchronization barrier for thread id < 32.
2) Apart from this the code uses the blockSize checks to sum the data in local memory. I think there in OpenCL we have get_local_size(0/1) to know the work group size. Block Size is confusing me.
I am not able to understand both the points mentioned above. Why and how these things helping out in optimization? Any explanation on reduce5 and reduce6 will be helpful as well.
You have that pretty much explained in slide 21 and 22 of https://docs.nvidia.com/cuda/samples/6_Advanced/reduction/doc/reduction.pdf which #Marco13 linked in comments.
As reduction proceeds, # “active” threads decreases
When s <= 32, we have only one warp left
Instructions are SIMD synchronous within a warp.
That means when s <= 32:
We don’t need to __syncthreads()
We don’t need “if (tid < s)” because it doesn’t save any work
Without unrolling, all warps execute every iteration of the for loop
and if statement
And by https://www.pgroup.com/lit/articles/insider/v2n1a5.htm:
The code is actually executed in groups of 32 threads, what NVIDIA
calls a warp.
Each core can execute a sequential thread, but the cores execute in
what NVIDIA calls SIMT (Single Instruction, Multiple Thread) fashion;
all cores in the same group execute the same instruction at the same
time, much like classical SIMD processors.
Re 2) blockSize there looks to be size of the work group.
Related
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 optimization guide of Beignet, an open source implementation of OpenCL targeting Intel GPUs
Work group Size should be larger than 16 and be multiple of 16.
As two possible SIMD lanes on Gen are 8 or 16. To not waste SIMD
lanes, we need to follow this rule.
Also mentioned in the Compute Architecture of Intel Processor Graphics Gen7.5:
For Gen7.5 based products, each EU has seven threads for a total of 28 Kbytes of general purpose register file (GRF).
...
On Gen7.5 compute architecture, most SPMD programming models employ
this style code generation and EU processor execution. Effectively,
each SPMD kernel instance appears to execute serially and independently within its own SIMD lane.
In actuality, each thread executes a SIMD-Width number of kernel instances >concurrently. Thus for a SIMD-16 compile of a compute
kernel, it is possible for SIMD-16 x 7 threads = 112 kernel instances
to be executing concurrently on a single EU. Similarly, for SIMD-32 x
7 threads = 224 kernel instances executing concurrently on a single
EU.
If I understand it correctly, using the SIMD-16 x 7 threads = 112 kernel instances as a example, in order to run 224 threads on one EU, the work group size need to be 16. Then the OpenCL compiler will fold 16 kernel instances into a 16 lane SIMD thread, and do this 7 times on 7 work groups, and run them on a single EU?
Question 1: am I correct until here?
However OpenCL spec also provide vector data types. So it's feasible to make full use of the SIMD-16 computing resources in a EU by conventional SIMD programming(as in NEON and SSE).
Question 2: If this is the case, using vector-16 data type already makes explicit use of the SIMD-16 resources, hence removes the at-least-16-item-per-work-group restrictions. Is this the case?
Question 3: If all above are true, then how does the two approach compare with each other: 1) 112 threads fold into 7 SIMD-16 threads by OpenCL compiler; 2) 7 native threads coded to explicitly use vector-16 data types and SIMD-16 operations?
Almost. You are making the assumptions that there is one thread per workgroup (N.B. thread in this context is what CUDA calls a "wave". In Intel GPU speak a work item is a SIMD channel of a GPU thread). Without subgroups, there is no way to force a workgroup size to be exactly a thread. For instance, if you choose a WG size of 16, the compiler is still free to compile SIMD8 and spread it amongst two SIMD8 threads. Keep in mind that the compiler chooses the SIMD width before the WG size is known to it (clCompileProgram precedes clEnqueueNDRange). The subgroups extension might allow you to force the SIMD width, but is definitely not implemented on GEN7.5.
OpenCL vector types are an optional explicit vectorization step on top of the implicit vectorization that already happens automatically. Were you to use float16 for example. Each of the work items would be processing 16 floats each, but the compiler would still compile at least SIMD8. Hence each GPU thread would be processing (8 * 16) floats (in parallel though). That might be a bit overkill. Ideally we don't want to have to explicitly vectorize our CL by using explicit OpenCL vector types. But it can be helpful sometimes if the kernel is not doing enough work (kernels that are too short can be bad). Somewhere it says float4 is a good rule of thumb.
I think you meant 112 work items? By native thread do you mean CPU threads or GPU threads?
If you meant CPU threads, the usual arguments about GPUs apply. GPUs are good when your program doesn't diverge much (all instances take similar paths) and you use the data enough times to mitigate the cost transferring it to and from the GPU (arithmetic density).
If you meant GPU threads (the GEN SIMD8 or SIMD16 critters). There is no (publicly visible) way to program the GPU threads explicitly at the moment (EDIT see the subgroups extension (not available on GEN7.5)). If you were able to, it'd be a similar trade off to assembly language. The job is harder, and the compiler sometimes just does a better job than we can, but when you are solving a specific problem and have better domain knowledge, you can generally do better with enough programming effort (until hardware changes and your clever program's assumptions becomes invalidated.)
My current question is exetension of previous
SIMD-8,SIMD-16 or SIMD-32 in opencl on gpgpu question.
I understand the concept of SIMD programming on GPU. It says all the scalar instructions on different work items are executed together in a warp/SIMD width group/Wavefront. My understanding here is that if we write a packed vector instruction in kernel code, compiler converts that instruction into scalars. And while execution all the work items in a simd width group execute the same instruction.
1) Now if we use a builtin like mad provided by opencl how this will be executed on the gpu ? Will all the work-items execute this as mad or this will be turned into scalar first?
2) If mad is being executed on the all workitems will the SIMD width get reduce from 32 to 16 or 16 - 8 ?
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'm taking my first steps in OpenCL (and CUDA) for my internship. All nice and well, I now have working OpenCL code, but the computation times are way too high, I think. My guess is that I'm doing too much I/O, but I don't know where that could be.
The code is for the main: http://pastebin.com/i4A6kPfn, and for the kernel: http://pastebin.com/Wefrqifh I'm starting to measure time after segmentPunten(segmentArray, begin, eind); has returned, and I end measuring time after the last clEnqueueReadBuffer.
Computation time on a Nvidia GT440 is 38.6 seconds, on a GT555M 35.5, on a Athlon II X4 5.6 seconds, and on a Intel P8600 6 seconds.
Can someone explain this to me? Why are the computation times are so high, and what solutions are there for this?
What is it supposed to do: (short version) to calculate how much noiseload there is made by an airplane that is passing by.
long version: there are several Observer Points (OP) wich are the points in wich sound is measured from an airplane thas is passing by. The flightpath is being segmented in 10.000 segments, this is done at the function segmentPunten. The double for loop in the main gives OPs a coordinate. There are two kernels. The first one calculates the distance from a single OP to a single segment. This is then saved in the array "afstanden". The second kernel calculates the sound load in an OP, from all the segments.
Just eyeballing your kernel, I see this:
kernel void SEL(global const float *afstanden, global double *totaalSEL,
const int aantalSegmenten)
{
// ...
for(i = 0; i < aantalSegmenten; i++) {
double distance = afstanden[threadID * aantalSegmenten + i];
// ...
}
// ...
}
It looks like aantalSegmenten is being set to 1000. You have a loop in each
kernel that accesses global memory 1000 times. Without crawling though the code,
I'm guessing that many of these accesses overlap when considering your
computation as a whole. It this the case? Will two work items access the same
global memory? If this is the case, you will see a potentially huge win on the
GPU from rewriting your algorithm to partition the work such that you can read
from a specific global memory only once, saving it in local memory. After that,
each work item in the work group that needs that location can read it quickly.
As an aside, the CL specification allows you to omit the leading __ from CL
keywords like global and kernel. I don't think many newcomers to CL realize
that.
Before optimizing further, you should first get an understanding of what is taking all that time. Is it the kernel compiles, data transfer, or actual kernel execution?
As mentioned above, you can get rid of the kernel compiles by caching the results. I believe some OpenCL implementations (the Apple one at least) already do this automatically. With other, you may need to do the caching manually. Here's instructions for the caching.
If the performance bottle neck is the kernel itself, you can probably get a major speed-up by organizing the 'afstanden' array lookups differently. Currently when a block of threads performs a read from the memory, the addresses are spread out through the memory, which is a real killer for GPU performance. You'd ideally want to index array with something like afstanden[ndx*NUM_THREADS + threadID], which would make accesses from a work group to load a contiguous block of memory. This is much faster than the current, essentially random, memory lookup.
First of all you are measuring not the computation time but the whole kernel read-in/compile/execute mumbo-jumbo. To do a fair comparison measure the computation time from the first "non-static" part of your program. (For example from between the first clSetKernelArgs to the last clEnqueueReadBuffer.)
If the execution time is still too high, then you can use some kind of profiler (such as VisualProfiler from NVidia), and read the OpenCL Best Practices guid which is included in the CUDA Toolkit documentation.
To the raw kernel execution time: Consider (and measure) that do you really need the double precision for your calculation, because the double precision calculations are artificially slowed down on the consumer grade NVidia cards.