c语言mpi程序的output进程数不同

Question

I need to write down a parallel code for this formula

I wrote a mpi program in c language for parallel programming

#include <stdio.h>
#include <math.h>
#include<mpi.h>
double sum(int n);

int main(void){

    int my_rank,comm_sz, n=1;
    double local_sum, total_sum;
    int source;
    int local_n;

MPI_Init(NULL,NULL);
MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
MPI_Comm_size(MPI_COMM_WORLD, &comm_sz);


local_n= n/comm_sz;
local_sum=sum(local_n);



if ( my_rank != 0) {
     MPI_Send (&local_sum , 1, MPI_DOUBLE , 0, 0, MPI_COMM_WORLD ) ;
 }

 else{
     total_sum = local_sum;
     for(source=1;source<comm_sz;source++){
     MPI_Recv (&local_sum , 1, MPI_DOUBLE , source , 0, MPI_COMM_WORLD , MPI_STATUS_IGNORE ) ;
     total_sum+=local_sum;
    }
  }
if(my_rank==0){
     printf("Sum is=%lf",total_sum);
     printf("\n");
}
  MPI_Finalize();
  return 0;

}


double sum(int n){
int i;
double cal_sum=0;
for (i =0;i <= n;i++) {
cal_sum = cal_sum + 4*(pow (-1, i))/((2*i)+1);
}
return cal_sum;
}

The output should be same whatever the number of process is. But as per my code when I run the program for 1 process the result is different than the number of process 8.

For example, if n= 1, p= 1 the summation is 2.667

 Whereas, if n= 1, p= 8 the summation is 32.00

But as per my understanding even if p= 8 the result of summation should be 2.667

Answer 1

There is logical error in your code.

Scenario 1: When comm_sz = 1;

For the iteration i = 0 : n=1, so loop condition 0<=1 is true
Value of cal_sum will be 4

For the iteration i = 0 : n=1, so loop condition 1<=1 is true
Value of cal_sum will be 4 + - (4/3) =  2.667

Scenario 2: When comm_sz = 2;

In my_rank = 0 ;

For the iteration i = 0 : n=0.5, so loop condition 0<=0 is true
So Value of cal_sum will be 4;
For the iteration i = 1 : n=0.5, so loop condition is false
exits the loop with cal_sum as 4;

In my_rank = 1 ;

For the iteration i = 0 : n=0.5, so loop condition 0<=0 is true
So Value of cal_sum will be 4;

For the iteration i = 1 : n=0.5, so loop condition is false
exits the loop with cal_sum as 4;

Hence the resulting value is comm_sz * 4 = 8 .

Basically, the same thing happens, even if you have 8 processes, the sum will be 32.

The result from your code will always be comm_sz*4 if comm_sz > 1 .

I hope this will assist you to understand the issue in your code.

You should also consider how to distribute the tasks across different processes.

Hint : Say if you have comm_size=4 and n=8 , the local_sum for each process should calculate n/comm_size iterations but with different starting and ending indexes . ie rank 0 should calculate sum for i=0 to 1 , rank 1 for i=2 to i=3 ... rank 4 should calculate i=6 to i=8 .