使用OpenMP和PThreads的並行程序比順序的要慢

Question

對於矩陣乘法的以下程序的並行化，我遇到了問題。 優化版本比順序版本慢或快幾個。 我已經准備好尋找錯誤，但是找不到它……我也在另一台機器上進行了測試，但是得到了相同的結果……

感謝您的幫助

主要：

int main(int argc, char** argv){

    if((matrixA).size != (matrixB).size){
     fprintf(ResultFile,"\tError for %s and %s - Matrix A and B are not of the same size ...\n", argv[1], argv[2]);
    }
    else{
     allocateResultMatrix(&resultMatrix, matrixA.size, 0);

     if(*argv[5] == '1'){ /* Sequentielle Ausfuehrung */
      begin = clock();
      matrixMultSeq(&matrixA, &matrixB, &resultMatrix);
      end = clock();
     };

     if(*argv[5] == '2'){ /* Ausfuehrung mit OpenMP */
      printf("Max number of threads: %i \n",omp_get_max_threads());
      begin = clock();
      matrixMultOmp(&matrixA, &matrixB, &resultMatrix);
      end = clock();
     };

     if(*argv[5] == '3'){ /* Ausführung mittels PThreads */
      pthread_t  threads[NUMTHREADS];
      pthread_attr_t attr;
      int i;
      struct parameter arg[NUMTHREADS];

      pthread_attr_init(&attr); /* Attribut initialisieren */

      begin = clock();

      for(i=0; i<NUMTHREADS; i++){ /* Initialisierung der einzelnen Threads */
       arg[i].id = i;
       arg[i].num_threads = NUMTHREADS;
       arg[i].dimension = matrixA.size;
       arg[i].matrixA = &matrixA;
       arg[i].matrixB = &matrixB;
       arg[i].resultMatrix = &resultMatrix;
       pthread_create(&threads[i], &attr, worker, (void *)(&arg[i]));
      }

      pthread_attr_destroy(&attr);

      for(i=0; i<NUMTHREADS; i++){ /* Warten auf Rückkehr der Threads */
       pthread_join(threads[i], NULL);
      }

      end = clock();
    }

    t=end - begin;
    t/=CLOCKS_PER_SEC;
    if(*argv[5] == '1')
      fprintf(ResultFile, "\tTime for sequential multiplication: %0.10f seconds\n\n", t);
    if(*argv[5] == '2')
      fprintf(ResultFile, "\tTime for OpenMP multiplication: %0.10f seconds\n\n", t);
    if(*argv[5] == '3')
      fprintf(ResultFile, "\tTime for PThread multiplication: %0.10f seconds\n\n", t);
    }
  }
}

void matrixMultOmp(struct matrix * matrixA, struct matrix * matrixB, struct matrix * resultMatrix){
  int i, j, k, l;
  double sum = 0;

  l = (*matrixA).size;
#pragma omp parallel for private(j,k) firstprivate (sum)
  for(i=0; i<=l; i++){
   for(j=0; j<=l; j++){
      sum = 0;
      for(k=0; k<=l; k++){
         sum = sum + (*matrixA).matrixPointer[i][k]*(*matrixB).matrixPointer[k][j];
      }
      (*resultMatrix).matrixPointer[i][j] = sum;
    }
  }
}

void mm(int thread_id, int numthreads, int dimension, struct matrix* a, struct matrix* b, struct matrix* c){
  int i,j,k;
  double sum;
  i = thread_id;
  while (i <= dimension) {
    for (j = 0; j <= dimension; j++) {
      sum = 0;
      for (k = 0; k <= dimension; k++) {
    sum = sum + (*a).matrixPointer[i][k] * (*b).matrixPointer[k][j];
      }
      (*c).matrixPointer[i][j] = sum;
    }
    i+=numthreads;
 }
}

void * worker(void * arg){
  struct parameter * p = (struct parameter *) arg;
  mm((*p).id, (*p).numthreads, (*p).dimension, (*p).matrixA, (*p).matrixB, (*p).resultMatrix);
  pthread_exit((void *) 0);
}

這是帶有時間的輸出：開始計算矩陣/SimpleMatrixA.txt和矩陣/SimpleMatrixB.txt的結果矩陣...矩陣A的大小：6個元素矩陣B的大小：6個元素連續乘法的時間：0.0000030000秒

Starting calculating resultMatrix for matrices/SimpleMatrixA.txt and matrices/SimpleMatrixB.txt ...
    Size of matrixA: 6 elements
    Size of matrixB: 6 elements
    Time for OpenMP multiplication: 0.0002440000 seconds

Starting calculating resultMatrix for matrices/SimpleMatrixA.txt and matrices/SimpleMatrixB.txt ...
    Size of matrixA: 6 elements
    Size of matrixB: 6 elements
    Time for PThread multiplication: 0.0006680000 seconds

Starting calculating resultMatrix for matrices/ShortMatrixA.txt and matrices/ShortMatrixB.txt ...
    Size of matrixA: 100 elements
    Size of matrixB: 100 elements
    Time for sequential multiplication: 0.0075190002 seconds

Starting calculating resultMatrix for matrices/ShortMatrixA.txt and matrices/ShortMatrixB.txt ...
    Size of matrixA: 100 elements
    Size of matrixB: 100 elements
    Time for OpenMP multiplication: 0.0076710000 seconds

Starting calculating resultMatrix for matrices/ShortMatrixA.txt and matrices/ShortMatrixB.txt ...
    Size of matrixA: 100 elements
    Size of matrixB: 100 elements
    Time for PThread multiplication: 0.0068080002 seconds

Starting calculating resultMatrix for matrices/LargeMatrixA.txt and matrices/LargeMatrixB.txt ...
    Size of matrixA: 1000 elements
    Size of matrixB: 1000 elements
    Time for sequential multiplication: 9.6421155930 seconds

Starting calculating resultMatrix for matrices/LargeMatrixA.txt and matrices/LargeMatrixB.txt ...
    Size of matrixA: 1000 elements
    Size of matrixB: 1000 elements
    Time for OpenMP multiplication: 10.5361270905 seconds

Starting calculating resultMatrix for matrices/LargeMatrixA.txt and matrices/LargeMatrixB.txt ...
    Size of matrixA: 1000 elements
    Size of matrixB: 1000 elements
    Time for PThread multiplication: 9.8952226639 seconds

Starting calculating resultMatrix for matrices/HugeMatrixA.txt and matrices/HugeMatrixB.txt ...
    Size of matrixA: 5000 elements
    Size of matrixB: 5000 elements
    Time for sequential multiplication: 1981.1383056641 seconds

Starting calculating resultMatrix for matrices/HugeMatrixA.txt and matrices/HugeMatrixB.txt ...
    Size of matrixA: 5000 elements
    Size of matrixB: 5000 elements
    Time for OpenMP multiplication: 2137.8527832031 seconds

Starting calculating resultMatrix for matrices/HugeMatrixA.txt and matrices/HugeMatrixB.txt ...
    Size of matrixA: 5000 elements
    Size of matrixB: 5000 elements
    Time for PThread multiplication: 1977.5153808594 seconds

Answer 1

正如評論中已經提到的那樣，您的第一個也是主要的問題是使用clock() 。 它返回程序執行的處理器時間。 你所尋找的是在程序執行的掛鍾時間。 在順序代碼中，它們是相同的，但是具有多個核心，這是完全不正確的。 幸運的是，OpenMP已經涵蓋了您：使用函數omp_get_wtime()代替。

最后，您需要更大的矩陣才能看到多線程帶來的任何好處。 如果創建/管理線程的開銷比線程正在執行的實際工作更昂貴，那么您將永遠不會從並行中看到任何好處。 因此，計時6x6矩陣乘法毫無意義。 我將從1000x1000開始，至少檢查2000x2000和8000x8000。