multithreading - 并行执行比顺序执行慢，即使代码是 "heavy"

Question

有很多问题会问“为什么我的并行循环比顺序循环慢”，答案是“循环内所需的工作很少；尝试进行更多迭代”，例如 this one .

我有一个循环，每次迭代大约需要 0.5 分钟。我认为这个循环足够“重”，与线程相关的任何开销都无关紧要。事实证明，并行执行速度较慢。我的代码是 C 语言，我使用 OpenMP 进行并行化。我的代码结构如下:

int main() {
  malloc and calculate group G1 of arrays;
  #pragma omp parallel
  {
    printf("avaible threads: %i\n", omp_get_num_threads());
    malloc group G2 of arrays;
    #pragma omp for
    for (int i = 0; i < N; i++) {
      calculate G2 arrays' elements through numerical integration;
      do some linear algebra with G1 and G2 to assemble the system Ax=b;
      solve Ax = b;
    }
  }
  return 0;
}

一些澄清:

数组 G1 组在循环内没有被修改，它们被用作辅助变量

执行迭代的顺序无关紧要

G1 组需要大约 0.6 GiB 的内存

G2 组每个线程需要大约 0.8 GiB 的内存

所有线性代数都是使用 Intel MKL 完成的(带有顺序线程层，请参阅其 link advisor)

对于 N=6，串行执行需要 3 分钟，而并行(3 个线程)需要 4 分钟。对于 N=30，串行为 15 分钟，并行(3 个线程)为 17 分钟。

我试图就为什么会发生这种情况提出一个假设。也许它与 CPU 缓存和我的阵列的大小有关？

这是我正在使用的计算机的一些信息:

Linux 4.15.0-74-generic #84-Ubuntu
Architecture:        x86_64
CPU op-mode(s):      32-bit, 64-bit
Byte Order:          Little Endian
CPU(s):              4
On-line CPU(s) list: 0-3
Thread(s) per core:  1
Core(s) per socket:  4
Socket(s):           1
NUMA node(s):        1
Vendor ID:           GenuineIntel
CPU family:          6
Model:               58
Model name:          Intel(R) Core(TM) i7-3770 CPU @ 3.40GHz
Stepping:            9
CPU MHz:             3392.363
CPU max MHz:         3400,0000
CPU min MHz:         1600,0000
L1d cache:           32K
L1i cache:           32K
L2 cache:            256K
L3 cache:            8192K
NUMA node0 CPU(s):   0-3

一个实际例子

下面的代码演示了我在做什么。我试图让它尽可能简单和小巧。这取决于 Netlib 的 LAPACK 和 OpenBLAS(使用 USE_OPENMP=1 构建)。我的实际程序有数千行。

编译:

  gcc -O2 -m64 -fopenmp test.c -lopenblas -llapack -lpthread -lm

运行:

  export OMP_NUM_THREADS=3
  ./a.out 30

    // test.c
    #include <complex.h>
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    #include <time.h>
    #include <omp.h>

    // LAPACK routines
    extern void
    zgesv_ (int *n, int *nrhs, _Complex double *a, int *lda, int* ipiv,
            _Complex double *b, int *ldb, int* info);

    extern void
    zgemv_ (char *trans, int *m, int *n, _Complex double *alpha, _Complex double *a,
            int *lda, _Complex double *x, int *incx, _Complex double *beta,
            _Complex double *y, int *incy);

    int
    main (int argc, char **argv)
    {
        srand(300);
        if (argc != 2) {
            printf("must have 1 argument: number of iterations!\n");
            exit(1);
        }
        const int nf = atoi(argv[1]);
        printf("loop will have %i iterations\n", nf);
        clock_t begin, end;
        double time_spent;
        begin = clock();
        int ne = 2296;
        int ne2 = ne * ne;
        _Complex double* restrict pot = malloc(ne2 * sizeof(_Complex double));
        for (int i = 0; i < ne; i++) {
            for (int k = i; k < ne; k++) {
                pot[i * ne + k] = (double) rand() / RAND_MAX;
                pot[i * ne + k] *= I;
                pot[i * ne + k] += (double) rand() / RAND_MAX;
                pot[k * ne + i] = pot[i * ne + k];
            }
        }
        char trans = 'N';
        _Complex double one = 1.0;
        _Complex double zero = 0.0;
        #pragma omp parallel
        {
            int n = ne;
            int ipiv[n]; //pivot indices
            int info;
            int nrhs = 1;
            #pragma omp single
            {
                printf("avaible threads: %i\n", omp_get_num_threads());
            }
            _Complex double* restrict zl = malloc(ne2 * sizeof(_Complex double));
            _Complex double* restrict ie = malloc(ne2 * sizeof(_Complex double));
            _Complex double gpr;
            #pragma omp for
            for (int i = 0; i < nf; i++) {
                printf("i = %i from thread %d\n", i, omp_get_thread_num());
                for (int m = 0; m < ne; m++) {
                    for (int k = m; k < ne; k++) {
                        gpr = cexp(k - m);
                        zl[m * ne + k] = gpr * pot[m * ne + k];
                        zl[k * ne + m] = zl[m * ne + k];
                    }
                }
                ie[0] = 1.0;
                for (int m = 1; m < ne; m++) {
                    ie[m] = 0.0;
                }
                zgesv_(&n, &nrhs, zl, &n, ipiv, ie, &n, &info);
                // Check for the exact singularity
                if (info > 0) {
                    printf("The diagonal element of the triangular factor of ZL,\n");
                    printf("U(%i,%i) is zero, so that ZL is singular;\n", info, info);
                    printf("the solution could not be computed.\n");
                }
                zgemv_(&trans, &ne, &ne, &one, zl, &ne, ie, &nrhs, &zero, ie, &nrhs);
                for (int p = 0; p < ne2; p++) {
                    gpr = 0.0;
                    for (int m = 0; m < ne; m++) {
                        gpr += ie[m] * cexp(-m * 5.4) / 4.1;
                    }
                }
            }
            free(zl);
            free(ie);
        }
        free(pot);
        end = clock();
        time_spent = (double) (end - begin) / CLOCKS_PER_SEC;
        printf("Success. Elapsed time: %.2f minutes\n", time_spent / 60.0);
        return 0;
    }

Answer 1

我有个教授说

If the experimental results did not agree with the theoretical ones, a question you must ask is How are you measuring?

正如@Gilles 在对问题的评论中指出的那样:使用 omp_get_wtime()而不是 clock() .那是因为 clock()返回 所有线程的累积 CPU 时间 , 而 omp_get_wtime()返回挂钟时间(见 OpenMP Forums)。