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Subject: Re: [OMPI users] Stream interactions in CUDA
From: Rolf vandeVaart (rvandevaart_at_[hidden])
Date: 2012-12-13 09:18:02


Hi Justin:

I assume you are running on a single node. In that case, Open MPI is supposed to take advantage of the CUDA IPC support. This will be used only when messages are larger than 4K, which yours are. In that case, I would have expected that the library would exchange some messages and then do a DMA copy of all the GPU data. For messages smaller than that, the library does synchronous copies through host memory. (cuMemcpy).

Let me try out your application and see what I see.

Rolf

>-----Original Message-----
>From: users-bounces_at_[hidden] [mailto:users-bounces_at_[hidden]]
>On Behalf Of Jens Glaser
>Sent: Wednesday, December 12, 2012 8:12 PM
>To: Open MPI Users
>Subject: Re: [OMPI users] Stream interactions in CUDA
>
>Hi Justin
>
>from looking at your code it seems you are receiving more bytes from the
>processors then you send (I assume MAX_RECV_SIZE_PER_PE >
>send_sizes[p]).
>I don't think this is valid. Your transfers should have matched sizes on the
>sending and receiving side. To achieve this, either communicate the message
>size before exchanging the actual data (a simple MPI_Isend/MPI_Irecv pair
>with one MPI_INT will do), or use a mechanism provided by the MPI library for
>this. I believe MPI_Probe is made for this purpose.
>
>As to why the transfers occur, my wild guess would be: you have set
>MAX_RECV_SIZE_PER_PE to something large, which would explain the size
>and number of the H2D transfers.
>I am just guessing, but maybe OMPI divides the data into chunks. Unless you
>are using intra-node Peer2Peer (smcuda), all MPI traffic has to go through the
>host, therefore the copies.
>I don't know what causes the D2H transfers to be of the same size, the library
>might be doing something strange here, given that you have potentially asked
>it to receive more data then you send - don't do that. Your third loop actually
>does not exchange the data, as you wrote, it just does an extra copying of
>data which in principle you could avoid by sending the message sizes first.
>
>Concerning your question about asynchronous copying. If you are using
>device buffers (and it seems you do) for MPI, then you will have to rely on the
>library to do asynchronous copying of the buffers (cudaMemcpyAsync) for
>you. I don't know if OpenMPI does this, you could check the source. I think
>MVAPICH2 does. If you really want control over the streams, you have to the
>D2H/H2D copying yourself, which is fine unless you are relying on peer-to-
>peer capability - but it seems you don't. If you are manually copying the data
>you can give any stream parameter to the cudaMemcpyAsync calls you prefer.
>
>My general experiences can be summarized as: achieving true async MPI
>computation is hard if using the CUDA support of the library, but very easy if
>you are using only the host routines of MPI. Since your kernel calls are async
>with respect to host already, all you have to do is asynchronously copy the
>data between host and device.
>
>Jens
>
>On Dec 12, 2012, at 6:30 PM, Justin Luitjens wrote:
>
>> Hello,
>>
>> I'm working on an application using OpenMPI with CUDA and GPUDirect. I
>would like to get the MPI transfers to overlap with computation on the CUDA
>device. To do this I need to ensure that all memory transfers do not go to
>stream 0. In this application I have one step that performs an MPI_Alltoall
>operation. Ideally I would like this Alltoall operation to be asynchronous. Thus
>I have implemented my own Alltoall using Isend and Irecv. Which can be
>found at the bottom of this email.
>>
>> The profiler shows that this operation has some very odd PCI-E traffic that I
>was hoping someone could explain and help me eliminate. In this example
>NPES=2 and each process has its own M2090 GPU. I am using cuda 5.0 and
>OpenMPI-1.7rc5. The behavior I am seeing is the following. Once the Isend
>loop occurs there is a sequence of DtoH followed by HtoD transfers. These
>transfers are 256K in size and there are 28 of them that occur. Each of these
>transfers are placed in stream0. After this there are a few more small
>transfers also placed in stream0. Finally when the 3rd loop occurs there are 2
>DtoD transfers (this is the actual data being exchanged).
>>
>> Can anyone explain what all of the traffic ping-ponging back and forth
>between the host and device is? Is this traffic necessary?
>>
>> Thanks,
>> Justin
>>
>>
>> uint64_t scatter_gather( uint128 * input_buffer, uint128
>> *output_buffer, uint128 *recv_buckets, int* send_sizes, int
>> MAX_RECV_SIZE_PER_PE) {
>>
>> std::vector<MPI_Request> srequest(NPES), rrequest(NPES);
>>
>> //Start receives
>> for(int p=0;p<NPES;p++) {
>>
>>
>MPI_Irecv(recv_buckets+MAX_RECV_SIZE_PER_PE*p,MAX_RECV_SIZE_PER
>_PE,MPI
>> _INT_128,p,0,MPI_COMM_WORLD,&rrequest[p]);
>> }
>>
>> //Start sends
>> int send_count=0;
>> for(int p=0;p<NPES;p++) {
>>
>MPI_Isend(input_buffer+send_count,send_sizes[p],MPI_INT_128,p,0,MPI_C
>OMM_WORLD,&srequest[p]);
>> send_count+=send_sizes[p];
>> }
>>
>> //Process outstanding receives
>> int recv_count=0;
>> for(int p=0;p<NPES;p++) {
>> MPI_Status status;
>> MPI_Wait(&rrequest[p],&status);
>> int count;
>> MPI_Get_count(&status,MPI_INT_128,&count);
>> assert(count<MAX_RECV_SIZE_PER_PE);
>>
>cudaMemcpy(output_buffer+recv_count,recv_buckets+MAX_RECV_SIZE_PE
>R_PE*p,count*sizeof(uint128),cudaMemcpyDeviceToDevice);
>> recv_count+=count;
>> }
>>
>> //Wait for outstanding sends
>> for(int p=0;p<NPES;p++) {
>> MPI_Status status;
>> MPI_Wait(&srequest[p],&status);
>> }
>> return recv_count;
>> }
>>
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