并行计算学习之用MPI实现梯形积分法

Posted 2022-03-29 cola-1998

• 梯形积分法

基本思想是，将x轴上区间划分成n个等长的子区间。估计介于函数图像以及每个子区间内梯形区域的面积。

　　设子区间端点为x_i和x_i+1 ，长度h=x_i+1 - x_i， 同样的两条垂直线的长度为f(x_i)和f(x_i+1)

　　　　那么面积为：h/2[ f(x_i) + f(x_i+1) ]

n个区间是等分的，如果两条垂直线包围区域的边界分别为a和b，那么

　　　　　　h = (b-a) / n

• 设计并行程序的四个步骤

1. 将问题的解决方案划分成多个任务
2. 在任务间识别出需要的通信信道（识别出联系，是否需要交流）
3. 将任务聚合成复合任务 --->减少通信
4. 在核上分配复合任务

• 串行代码

/*  Input  a ,b , n*/
h = (b-a)/n
approx  = (f(a) + f(b))/2.0

for(i =1; i <= n-1; i++)
    x_i = a + i * h;
    approx = h * approx


approx = h*approx

• 　并行程序

     伪代码：

Get a,b,n;
h = (b-a)/n;
local_n = n/comm_sz;
local_a = a + my_rank * local_n * h;
local_b = local_a + local_n*h;
local_integral = Trap(local_a, local_b, local_n, h)  /*Trap函数作用*/

if( my_rank != 0 )
     Send local_integral to process 0;

else /*my_rank == 0*/
      total_integral = local_integral;
      for(proc = 1;proc < comm_sz; proc++)
           Receive local_integral from proc;
           total_integral += local_integral;
     

if(my_rank == 0)
      print result;

　　

int main(void)
     int my_rank , comm_sz, n = 1024, local_n;
     double a = 0.0, b=3.0, h, local_a,local_b;
     double local_int,total_int;
     int source;    

     MPI_Init(NULL,NULL);
     MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
     MPI_Comm_size(MPI_COMM_WORLD, &comm_sz);
   
     h = (b-a)/n;
     local_n = n/comm_sz;
     
     local_a = a + my_rank * local_n * h;
     local_b = local_a +local_n * h;
     local_int = Trap(local_a , local_b , local_n, h )

    if (my_rank != 0)
            MPI_Send(&local_int , 1 , MPI_DOUBLE , 0, 0 , MPI_COMM_WORLD);
    else
             total_int = local_int;
             for(source = 1 ; source < comm_sz ; source++)
                    MPI_Recv(&local_int, 1, MPI_DOUBLE, source, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
                    total_int += local_int;
             
    

   if (my_rank == 0 )
           printf("With n = %d trapezoids, our estimate\\n",n);
           printf("of the integral from %f to %f = %.15e\\n",a,b,total_int);
   

   MPI_Finalize();
   return 0;