将功能转移到Intel Xeon Phi所需的时间

Question

Is there a predefined time that is required for offload call to transfer the data(parameters) of a function from host to Intel MIC(Xeon Phi coprocessor 3120 series)?

Specifically I do offload call ("#pragma offload target(mic)") for a function that I want to be executed on MIC. The function has 15 parameters(pointers and variables) and I have already confirmed the right passing of the parameters on MIC. However I have simplified the code with purpose to check the time for the passing of the parameters and so it contains just one simple "printf()" function. I use the "gettimeofday()" of "sys/time.h" header file for measuring time as it seems in the code below:

Some hardware informations for the host: Intel(R) Core(TM) i7-3770 CPU @ 3.40GHz / CentOS release 6.8 / PCI Express Revision 2.0

main.c

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/time.h>
#include <string.h>

__attribute__ (( target (mic))) unsigned long long ForSolution = 0;
__attribute__ (( target (mic))) unsigned long long sufficientSol = 1;
__attribute__ (( target (mic))) float timer = 0.0;

__attribute__ (( target (mic))) void function(float *grid, float *displ, unsigned long long *li, unsigned long long *repet, float *solution, unsigned long long dim, unsigned long long numOfa, unsigned long long numLoops, unsigned long long numBlock, unsigned long long thread, unsigned long long blockGrid, unsigned long long station, unsigned long long bytesSol, unsigned long long totalSol, volatile unsigned long long *prog);

   float    *grid, *displ, *solution;
   unsigned long long   *li,repet;
   volatile unsigned long long  *prog;
   unsigned long long dim = 10, grid_a = 3, numLoops = 2, numBlock = 0;
   unsigned long long thread = 220, blockGrid = 0, station = 12;
   unsigned long long station_at = 8, bytesSol, totalSol;

   bytesSol = dim*sizeof(float);
   totalSol = ((1024 * 1024 * 1024) / bytesSol) * bytesSol;



   /******** Some memcpy() functions here for the pointers*********/                   



gettimeofday(&start, NULL);

   #pragma offload target(mic) \
        in(grid:length(dim * grid_a * sizeof(float))) \
        in(displ:length(station * station_at * sizeof(float))) \
        in(li:length(dim * sizeof(unsigned long long))) \
        in(repet:length(dim * sizeof(unsigned long long))) \
        out(solution:length(totalSol/sizeof(float))) \
        in(dim,grid_a,numLoops,numBlock,thread,blockGrid,station,bytesSol,totalSol) \
        in(prog:length(sizeof(volatile unsigned long long))) \
        inout(ForSolution,sufficientSol,timer)
   {
        function(grid, displ, li, repet, solution, dim, grid_a, numLoops, numBlock, thread, blockGrid, station, bytesSol, totalSol, prog);
   }

    gettimeofday(&end, NULL);  

    printf("Time to tranfer data on Intel Xeon Phi: %f sec\n", (((end.tv_sec - start.tv_sec) * 1000000.0 + (end.tv_usec - start.tv_usec)) / 1000000.0) - timer);
    printf("Time for calculations: %f sec\n", timer);

function.c

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/time.h>
#include <string.h>
#include <omp.h>

void function(float *grid, float *displ, unsigned long long *li, unsigned long long *repet, float *solution, unsigned long long dim, unsigned long long numOfa, unsigned long long numLoops, unsigned long long numBlock, unsigned long long thread, unsigned long long blockGrid, unsigned long long station, unsigned long long bytesSol, unsigned long long totalSol, volatile unsigned long long *prog)
{
    struct timeval      timer_start, timer_end;

    gettimeofday(&timer_start, NULL);

printf("Hello World!!!\n");


    gettimeofday(&timer_end, NULL);

    timer = ((timer_end.tv_sec - timer_start.tv_sec) * 1000000.0 + (timer_end.tv_usec - timer_start.tv_usec)) / 1000000.0 ;  
}

Results of terminal:

Time to tranfer data on Intel Xeon Phi: 3.512706 sec
Time for calculations: 0.000002 sec
Hello World!!!

The code require 3.5 seconds to complete the "offload target". Is the above result normal? Is there any way to reduce that significant time delay of offload call?

Answer 1

Let's look at the steps here:

a) For the very first #pragma offload the MIC is initialised; which probably includes resetting it, booting a stripped down Linux (and waiting for it to start all the CPUs, initialise its memory management, start a psuedo-NIC driver, etc), and uploading your code to the device. This probably takes multiple seconds alone.

b) All the input data is uploaded to the MIC.

c) The function is executed.

d) All the output data is downloaded from the MIC.

For raw data transfers over PCI Express Revision 2.0 (x16) the max. bandwidth is 8 GB/s; however you're not going to get max. bandwidth. From what I remember communication with the Phi involves shared ring buffers and "doorbell" IRQs with "pseudo NIC" drivers on both sides (on the host, and on the coprocessor's OS); and with all the handshaking and overhead I'd be surprised if you get half the max. bandwidth.

I think that the total amount of code uploaded, data uploaded and data downloaded is well over 1 GiB (eg the out(solution:length(totalSol/sizeof(float))) is 1 GiB all by itself). If we're assuming "about 4 GiB/s" that's at least another ~250 ms.

My suggestion is to do everything twice; and measure the difference between the first time (which includes initialising everything) and the second time (when everything is already initialised) to determine how long it takes to initialise the coprocessor. The second measurement (minus the time to execute the function) will tell you how long the data transfers took.