我是 CUDA 编程的新手，我正在解决一个需要在一台机器上安装多个 GPU 的问题。我知道为了更好地进行图形编程，需要通过 SLI 组合多个 GPU。但是，对于 CUDA 编程，我是否还需要通过 SLI 组合 GPU？
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不，一般来说，如果您打算使用 GPU 进行计算而不是纯图形应用程序，则您不想使用 SLI。您将能够从 CUDA 程序中将两个 GPU 作为独立设备访问。请注意，您需要在 GPU 之间明确划分工作。
我没有解释为什么 SLI 不适用于计算应用程序，但这是我在 Nvidia 论坛上读到的，并从 IRC 频道的其他人那里听到的。
于 2012-06-20T17:15:46.023 回答
0
您可以在没有 SLI 的多个 GPU 上使用 CUDA，甚至可以在不同的 CUDA 架构之间使用 CUDA，但是您必须编写额外的代码来划分工作并同步划分的子工作。这是一个在 3 个 GPU 上为示例内核 vectorAdd 进行负载平衡的简单程序（GT1030 GPU 一个 Pascal 架构 GPU + 两个 Kepler 架构的 K420 GPU，在同一个任务池中一起工作没有问题）：
/**
* Copyright 1993-2015 NVIDIA Corporation. All rights reserved.
*
* Please refer to the NVIDIA end user license agreement (EULA) associated
* with this source code for terms and conditions that govern your use of
* this software. Any use, reproduction, disclosure, or distribution of
* this software and related documentation outside the terms of the EULA
* is strictly prohibited.
*
*/
/**
* Vector addition: C = A + B.
*
* This sample is a very basic sample that implements element by element
* vector addition. It is the same as the sample illustrating Chapter 2
* of the programming guide with some additions like error checking.
*/
#include <stdio.h>
// For the CUDA runtime routines (prefixed with "cuda_")
#include <cuda_runtime.h>
#include <helper_cuda.h>
// for load balancing between 3 different GPUs
#include "LoadBalancerX.h"
/**
* CUDA Kernel Device code
*
* Computes the vector addition of A and B into C. The 3 vectors have the same
* number of elements numElements.
*/
__global__ void
vectorAdd(const float *A, const float *B, float *C, int numElements)
{
int i = blockDim.x * blockIdx.x + threadIdx.x;
if (i < numElements)
{
C[i] = A[i] + B[i];
}
}
#include<iostream>
#include<map>
int
main(void)
{
int numElements = 1500000;
int numElementsPerGrain = 50000;
size_t size = numElements * sizeof(float);
float *h_A; cudaMallocHost((void**)&h_A,size);
float *h_B; cudaMallocHost((void**)&h_B,size);
float *h_C; cudaMallocHost((void**)&h_C,size);
for (int i = 0; i < numElements; ++i)
{
h_A[i] = rand()/(float)RAND_MAX;
h_B[i] = rand()/(float)RAND_MAX;
}
/*
* default tutorial vecAdd logic
cudaMemcpy(d_A, h_A, size, cudaMemcpyHostToDevice);
cudaMemcpy(d_B, h_B, size, cudaMemcpyHostToDevice);
int threadsPerBlock = 256;
int blocksPerGrid =(numElements + threadsPerBlock - 1) / threadsPerBlock;
vectorAdd<<<blocksPerGrid, threadsPerBlock>>>(d_A, d_B, d_C, numElements);
cudaGetLastError();
cudaMemcpy(h_C, d_C, size, cudaMemcpyDeviceToHost);
*/
/* load-balanced 3-GPU version setup */
class GrainState
{
public:
int offset;
int range;
std::map<int,float *> d_A;
std::map<int,float *> d_B;
std::map<int,float *> d_C;
std::map<int,cudaStream_t> stream;
~GrainState(){
for(auto a:d_A)
cudaFree(a.second);
for(auto b:d_B)
cudaFree(b.second);
for(auto c:d_C)
cudaFree(c.second);
for(auto s:stream)
cudaStreamDestroy(s.second);
}
};
class DeviceState
{
public:
int gpuId;
int amIgpu;
};
LoadBalanceLib::LoadBalancerX<DeviceState,GrainState> lb;
lb.addDevice(LoadBalanceLib::ComputeDevice<DeviceState>({0,1})); // 1st cuda gpu in computer
lb.addDevice(LoadBalanceLib::ComputeDevice<DeviceState>({1,1})); // 2nd cuda gpu in computer
lb.addDevice(LoadBalanceLib::ComputeDevice<DeviceState>({2,1})); // 3rd cuda gpu in computer
//lb.addDevice(LoadBalanceLib::ComputeDevice<DeviceState>({3,0})); // CPU single core
for(int i=0;i<numElements;i+=numElementsPerGrain)
{
lb.addWork(LoadBalanceLib::GrainOfWork<DeviceState,GrainState>(
[&,i](DeviceState gpu, GrainState& grain){
if(gpu.amIgpu)
{
cudaSetDevice(gpu.gpuId);
cudaStreamCreate(&grain.stream[gpu.gpuId]);
cudaMalloc((void **)&grain.d_A[gpu.gpuId], numElementsPerGrain*sizeof(float));
cudaMalloc((void **)&grain.d_B[gpu.gpuId], numElementsPerGrain*sizeof(float));
cudaMalloc((void **)&grain.d_C[gpu.gpuId], numElementsPerGrain*sizeof(float));
}
},
[&,i](DeviceState gpu, GrainState& grain){
if(gpu.amIgpu)
{
cudaSetDevice(gpu.gpuId);
cudaMemcpyAsync(grain.d_A[gpu.gpuId], h_A+i, numElementsPerGrain*sizeof(float), cudaMemcpyHostToDevice,grain.stream[gpu.gpuId]);
cudaMemcpyAsync(grain.d_B[gpu.gpuId], h_B+i, numElementsPerGrain*sizeof(float), cudaMemcpyHostToDevice,grain.stream[gpu.gpuId]);
}
},
[&,i](DeviceState gpu, GrainState& grain){
if(gpu.amIgpu)
{
int threadsPerBlock = 1000;
int blocksPerGrid =numElementsPerGrain/1000;
vectorAdd<<<blocksPerGrid, threadsPerBlock, 0, grain.stream[gpu.gpuId]>>>(grain.d_A[gpu.gpuId], grain.d_B[gpu.gpuId], grain.d_C[gpu.gpuId], numElements-i);
}
else
{
for(int j=0;j<numElementsPerGrain;j++)
{
const int index = j+i;
h_C[index]=h_A[index]+h_B[index];
}
}
},
[&,i](DeviceState gpu, GrainState& grain){
if(gpu.amIgpu)
{
cudaMemcpyAsync(h_C+i, grain.d_C[gpu.gpuId], numElementsPerGrain*sizeof(float), cudaMemcpyDeviceToHost,grain.stream[gpu.gpuId]);
}
},
[&,i](DeviceState gpu, GrainState& grain){
if(gpu.amIgpu)
{
cudaStreamSynchronize(grain.stream[gpu.gpuId]);
}
}
));
}
/* load-balance setup end*/
// run 100 times
size_t nanoseconds=0;
for(int i=0;i<100;i++)
{
nanoseconds += lb.run();
}
std::cout<<nanoseconds/100.0<<"ns ("<<((numElements*12.0/(nanoseconds/100.0)))<<"GB/s)"<<std::endl;
std::cout<<"??"<<std::endl;
for (int i = 0; i < numElements; i+=numElementsPerGrain)
{
std::cout<<h_A[i]<<" + "<<h_B[i]<<" = "<<h_C[i]<<std::endl;
}
auto z = lb.getRelativePerformancesOfDevices();
std::cout<<"work distribution to devices:"<<std::endl;
for(auto zz:z)
{
std::cout<<zz<<"% ";
}
std::cout<<std::endl;
cudaFreeHost(h_A);
cudaFreeHost(h_B);
cudaFreeHost(h_C);
return 0;
}
于 2022-02-25T20:38:24.743 回答