计算CUFFT的性能 [英] Calculating performance of CUFFT

问题描述

我运行CUFFT上的块（N * N / p）分为多个GPU，我有一个问题，计算的性能。首先，了解我的操作方式：

I am running CUFFT on chunks (N*N/p) divided in multiple GPUs, and I have a question regarding calculating the performance. First, a bit about how I am doing it:

1. 向每个GPU发送N * N / p个块


2. 对p个GPU中的每一行进行成批的1-D FFT


3. 获取N * N / p个块返回主机 - 对整个数据集执行转置


4. 同上第1步


5. 同上第2步

1. Send N*N/p chunks to each GPU
2. Batched 1-D FFT for each row in p GPUs
3. Get N*N/p chunks back to host - perform transpose on the entire dataset
4. Ditto Step 1
5. Ditto Step 2

Gflops =（1e-9 * 5 * N * N * lg（N * N））/执行时间

as：

执行时间= Sum（每个GPU的行和列FFT的memcpyHtoD + kernel + memcpyDtoH次）

这是在多个GPU上评估CUFFT性能的正确方法吗？是否有其他方法可以表示FFT的性能？

Is this the correct way to evaluate CUFFT performance on multiple GPUs? Is there any other way I could represent the performance of FFT?

谢谢。

推荐答案

如果要进行复杂变换，操作计数是正确的（对于实值变换，它应该为2.5 N log2（N）），但GFLOP公式不正确。在并行多处理器操作中，通常的吞吐量计算是

If you are doing a complex transform, the operation count is correct (it should be 2.5 N log2(N) for a real valued transform), but the GFLOP formula is incorrect. In a parallel, multiprocessor operation the usual calculation of throughput is

operation count / wall clock time

在你的情况下，假设GPU并行运行，测量挂钟时间（即整个操作花费多长时间）执行时间或使用：

In your case, presuming the GPUs are operating in parallel, either measure the wall clock time (ie. how long the whole operation took) for the execution time, or use this:

execution time = max(memcpyHtoD + kernel + memcpyDtoH times for row and col FFT for each GPU)

因为它是，你的计算代表串行执行时间。考虑到multigpu方案的开销，我预计您获得的计算性能数值会比单个GPU上的等效变换低。

As it stands, your calculation represents the serial execution time. Allowing for the overheads from the multigpu scheme, I would expect that the calculated performance numbers you are getting will be lower than the equivalent transform done on a single GPU.

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