I am a complete beginner to gpgpu and opencl. I am unable to answer the following two questions about GPGPU in general,
a) Suppose I have a piece of code suitable to be run on a gpu (executes the exact same set of instructions on multiple data). Assume I already have my data on the gpu. Is there any way to look at the specifications of the cpu and gpu, and estimate the potential speed gains? For example, how can I estimate the speed gains (excluding time taken to transfer data to the gpu) if I ran the piece of code (running exact same set of instructions on multiple data) on AMDs R9 295X2 gpu (http://www.amd.com/en-us/products/graphics/desktop/r9/2...) instead of intel i7-4770K processor (http://ark.intel.com/products/75123)
b) Is there any way to estimate the amount of time it would take to transfer data to the gpu?
Thank you!
Thank you for the responses! Given the large number of factors influencing speed gains, trying and testing is certainly a good idea. However, I do have a question on the GFLOPS approach mentioned some responses; GFLOPS metric was what I was looking at before posting the question.
I would think that GFLOPS would be a good way to estimate potential performance gains for SIMD type operations, given that it takes into account difference in clock speeds, cores, and floating point operations per cycle. However, when I crunch numbers using GFLOPS specifications something does not seem correct.
The Good:
GFLOPS based estimate seems to match the observed speed gains for the toy kernel below. The kernel for input integer "n" computes the sum (1+2+3+...+n) in a brute force way. I feel, the kernel below for large integers has a lot of computation operations. I ran the kernel for all ints from 1000 to 60000 on gpu and cpu (sequentially on cpu, without threading), and measured the timings.
__kernel void calculate(__global int* input,__global int* output){
size_t id=get_global_id(0);
int inp_num=input[id];
int si;
int sum;
sum=0;
for(int i=0;i<=inp_num;++i)
sum+=i;
output[id]=sum;
}
GPU on my laptop:
NVS 5400M (www.nvidia.com/object/nvs_techspecs.html)
GFLOPS, single precision: 253.44 (en.wikipedia.org/wiki/List_of_Nvidia_graphics_processing_units)
CPU on my Laptop:
intel i7-3720QM, 2.6 GHz
GFLOPS (assuming single precision): 83.2 (download.intel.com/support/processors/corei7/sb/core_i7-3700_m.pdf). Intel document does not specify if it is single or double
CPU Time: 3.295 sec
GPU Time: 0.184 sec
Speed gains per core: 3.295/0.184 ~18
Theoretical Estimate of Speed gains with Using all 4 cores: 18/4 ~ 4.5
Speed Gains based on FLOPS: (GPU FLOPS)/(CPU FLOPS) = (253.44/83.2) = 3.0
For the above example GLOPS based estimate seems to be consistent with those obtained from experimentation, if the intel documentation indeed specifies FLOPS for single and not double precision. I did try to search for more links for flops specification for the intel processor on my laptop. The observed speed gain also seems good, given that I have a modest GPU
The Problem:
The FLOPS based approach seems to give a much lower than expected speed gains, after factoring gpu price, when comparing AMDs R9 295X2 gpu (www.amd.com/en-us/products/graphics/desktop/r9/295x2#) with intels i7-4770K (ark.intel.com/products/75123):
AMDs FLOPS, single precision: 11.5 TFLOPS (from above mentioned link)
Intels FLOPS, single precision: (num. of cores) x (FLOPS per cycle per core) x (clock speed) = (4) x (32 (peak) (www.pcmag.com/article2/0,2817,2419798,00.asp)) x (3.5) = 448 GFLOPS
Speed Gains Based on FLOPS = (11.5 TFLOPS)/(448) ~ 26
AMD GPUs price: $1500
Intel CPUs price: $300
For every AMD R9 295X2 gpu, I can buy 5 intel i7-4770K cpus, which reduces the effective speed gains to (26/5) ~ 5. However, this estimate is not at all consistent with the 100-200x, increase in speed one would expect. The low estimate in speed gains by the GFLOPS approach makes my think that something is incorrect with my analysis, but I am not sure what?
You need to examine the kernel(s). I myself am learning CUDA, so I couldn't tell you exactly what you'd do with OpenCL.
But I would figure out roughly how many floating point operations one single instance of the kernel will perform. Then find the number of floating point operations per second each device can handle.
number of kernels to be launched * (n floating-point operations of kernel / throughput of device (FLOPS)) = time to execute
The number of kernels launched will depend on your data.
A) Normally this question is never answered. Since we are not speaking at 1.05x speed gains. When the problem is suitable, the problem is BIG enough to hide any overheads (100k WI), and the data is already in the GPU, then we are speaking of speeds of 100-300x. Normally nobody cares if it is 250x or 251x.
The estimation is difficult to make, since the platforms are completely different. Not only on clock speeds, but memory latency and caches, as well as bus speeds and processing elements.
I cannot give you a clear answer on this, other than try it and measure.
B) The time to copy the memory is completely dependent on the GPU-CPU bus speed (PCI bus). And that is the HW limit, in practice you will always have less speed than that on copying. Generally you can apply the rule of three to solve the time needed, but there is always a small driver overhead that depends on the platform and device. So, copying 100 bytes is usually very slow, but copying some MB is as fast as the bus speed.
The memory copying speed is usually not a design constrain when creating a GPGPU app. Since it can be hided in many ways (pinned memory, etc..), that nodoby will notice any speed decrease due to memory operations.
You should not make any decisions on whether the problem is suitable or not for GPU, just by looking at the time lost at memory copy. Better measures are, if the problem is suitable, and if you have enough data to make the GPU busy (otherwise it is faster to do it in CPU directly).
Potential speed gain highly depends on algorithm implementation. It's difficult to forecast performance level unless you're developing come very simple applications (like simplest image filter). In some cases, estimations can be done, using memory system performance as basis, as many algorithms are bandwidth-bound.
You can calculate transmission time by dividing data amount on GPU memory bandwidth for Device-internal operations. Look at hardware characteristics to get it, or calculate if you know memory frequency & bus width. For Host-Device operations, PCI-E bus speed is the limit usually.
If code is easy(is what lightweight- cores of gpu need) and is not memory dependent then you can approximate to :
Sample kernel:
Read two 32-bit floats from memory and
do calcs on them for 20-30 times at least.
Then write to memory once.
New: GPU
Old: CPU
Gain ratio = ((New/Old) - 1 ) *100 (%)
New= 5000 cores * 2 ALU-FPU per core * 1.0 GHz frequency = 10000 gflops
Old = 10 cores * 8 ALU-FPU per core * 4.0GHz frequency = 320 gflops
((New/Old) - 1 ) *100 ===> 3000% speed gain.
This is when code uses registers and local memory mostly. Rarely hitting global mem.
If code is hard( heavy branching + fake recursivity + non-uniformity ) only 3-5 times speed gain. it can be equal or less than CPU performance for linear code ofcourse.
When code is memory dependant, it will be 1TB/s(GPU) divided by 40GB/s(CPU).
If each iteration needs to upload data to gpu, there will be pci-e bandwidth bottlenect too.
loads are usually classified into 2 categories
bandwidth bound - more time is spent on fetches from global-memory. Even increasing cpu clock freq doesn't help. problems like sorting. bandwidth capacity is measured using GBPS
compute bound - directly proportional to cpu horse-power. problems like matrix multiplication. compute capacity is measured using GFLOPS
there is a tool clpeak which tries to programmatically measure these
its very important to classify your problem to measure its performance & choose the right device(knowing their limits)
say if you compare intel-HD-4000 & i7-3630(both on same chip) in https://github.com/krrishnarraj/clpeak/tree/master/results/Intel%28R%29_OpenCL
i7 is comparatively better at bandwidth(plus no transfer overheads)
in terms of compute, gpu is 4-5 times faster than i7
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.
I want to calculate the theoretical peak performance of graphics hardware. Well, actually I want to understand the calculation.
Example with a AMD Radeon HD 6670:
The AMD Accelerated Parallel Processing Programming Guide (http://developer.amd.com/download/AMD_Accelerated_Parallel_Processing_OpenCL_Programming_Guide.pdf) tells me in the middle of page 6-42 to take the number of Stream Cores (96), multiply it by the number of operations per cycle for each Stream Core (let's take an ADD with Single Precision, which would be 5) and multiply that by the core clock (800 MHz). That results to:
96 * 5 FLOPS * 800MHz = 384,000 MFLOPS = 384 GFLOPS
The very same document tells me on page D-4 that this particular device has a peak throughput of 768 GFLOPS, which is twice of what I just calculated. Wikipedia and the AMD homepage state the same.
So my question is: Where am I missing the factor of two?
I am not sure about AMD hardware, but I remember that NVIDIA counted MAD (multiply-add) operation as two flops. Since MADs are performed in one cycle, the theoretical performance was multiplied by two.
480 processing elements * 2 operations per cycle(single addition pipeline + single multiplication pipeline per element) * 800MHz = 768 GFLOPS
When the code has too many levels of branching, it drops to 1-4 shader per compute unit which means 6-24 of them and this translates to as low as 10-40 GFlops which is even slower than a single cpu core.
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'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.