Can you predict the runtime of a CUDA kernel?
To what degree can one predict / calculate the performanc开发者_开发百科e of a CUDA kernel?
Having worked a bit with CUDA, this seems non trivial.
But a colleage of mine, who is not working on CUDA, told me, that it cant be hard if you have the memory bandwidth, the number of processors and their speed?
What he said seems not to be consistent with what I read. This is what I could imagine could work. What do you think?
Memory processed
------------------ = runtime for memory bound kernels ?
Memory bandwidth
or
Flops
------------ = runtime for computation bound kernels?
Max GFlops
Such calculation will barely give good prediction. There are many factors that hurt the performance. And those factors interact with each other in a extremely complicated way. So your calculation will give the upper bound of the performance, which is far away from the actual performance (in most cases).
For example, for memory bound kernels, those with a lot cache misses will be different with those with hits. Or those with divergences, those with barriers...
I suggest you to read this paper, which might give you more ideas on the problem: "An Analytical Model for a GPU Architecture with Memory-level and Thread-level Parallelism Awareness".
Hope it helps.
I think you can predict a best-case with a bit of work. Like you said, with instruction counts, memory bandwidth, input size, etc.
However, predicting the actual or worst-case is much trickier.
First off, there are factors like memory access patterns. Eg: with older CUDA capable cards, you had to pay attention to distribute your global memory accesses so that they wouldn't all contend for a single memory bank. (The newer CUDA cards use a hash between logical and physical addresses to resolve this).
Secondly, there are non-deterministic factors like: how busy is the PCI bus? How busy is the host kernel? Etc.
I suspect the easiest way to get close to actual run-times is basically to run the kernel on subsets of the input and see how long it actually takes.
精彩评论