Question

For this assignment, write a parallel program in C++ using OpenMP for vector addition. Assume A, B, C are three vectors of equal length. The program will add the corresponding elements of vectors A and B and will store the sum in the corresponding elements in vector C (in other words C[i] = A[i] + B[i]). Every thread should execute an approximately equal number of loop iterations.

As an input vector A, initialize its size to 10,000 and elements from 1 to 10,000.

So, A[0] = 1, A[1] = 2, A[2] = 3, … , A[9999] = 10000.

Input vector B will be initialized to the same size with opposite inputs.

So, B[0] = 10000, B[1] = 9999, B[2] = 9998, … , B[9999] = 1

Using the above input vectors A and B, create output Vector C which will be computed as

C[ i ] = A[ i ] + B[ i ];

You should check whether your output vector value is 10001 in every C[ i ].

First, start with 2 threads (each thread adding 5,000 vectors), and then do with 4, and 8 threads. Remember sometimes your vector size can not be divided equally by the number of threads. You need to slightly modify the pseudo code to handle the situation accordingly. (Hint: If you have p threads, first (p - 1) threads should have the same input size and the last thread will take care of whatever the remainder portion.) Check the running time from each experiment and compare the result. Report your findings from this project in a separate paragraph.

Your output should show a team of treads does evenly distributed work, but a big vector size might cause an issue in output. You can create mini version of original vector in much smaller size of 100 (A[0] = 1, A[1] = 2, A[2] = 3, … , A[99] = 100) and run with 6 threads once and take a snapshot of your output. And run with original size with 2, 4, and 8 threads to compare running times.

Pseudocode for Assignment

mystart = myid*n/p; // starting index for the individual thread

myend = mystart+n/p; // ending index for the individual thread

for (i = mystart; i < myend; i++) // each thread computes local sum

do vector addition // and later all local sums combined

Answer 1

#include <stdlib.h> //malloc and free
#include <stdio.h> //printf
#include <omp.h> //OpenMP

// Very small values for this simple illustrative example
#define ARRAY_SIZE 8 //Size of arrays whose elements will be added together.
#define NUM_THREADS 4 //Number of threads to use for vector addition.

/*
* Classic vector addition using openMP default data decomposition.
*
* Compile using gcc like this:
*     gcc -o va-omp-simple VA-OMP-simple.c -fopenmp
*
* Execute:
*     ./va-omp-simple
*/
int main (int argc, char *argv[])
{
   // elements of arrays a and b will be added
   // and placed in array c
   int * a;
   int * b;
   int * c;
  
int n = ARRAY_SIZE; // number of array elements
   int n_per_thread; // elements per thread
   int total_threads = NUM_THREADS; // number of threads to use
   int i; // loop index
  
// allocate spce for the arrays
a = (int *) malloc(sizeof(int)*n);
   b = (int *) malloc(sizeof(int)*n);
   c = (int *) malloc(sizeof(int)*n);

// initialize arrays a and b with consecutive integer values
   // as a simple example
for(i=0; i<n; i++) {
a[i] = i;
}
for(i=0; i<n; i++) {
b[i] = i;
}   
  
   // Additional work to set the number of threads.
   // We hard-code to 4 for illustration purposes only.
   omp_set_num_threads(total_threads);
  
   // determine how many elements each process will work on
   n_per_thread = n/total_threads;
  
// Compute the vector addition
   // Here is where the 4 threads are specifically 'forked' to
   // execute in parallel. This is directed by the pragma and
   // thread forking is compiled into the resulting exacutable.
   // Here we use a 'static schedule' so each thread works on
   // a 2-element chunk of the original 8-element arrays.
   #pragma omp parallel for shared(a, b, c) private(i) schedule(static, n_per_thread)
for(i=0; i<n; i++) {
       c[i] = a[i]+b[i];
       // Which thread am I? Show who works on what for this samll example
       printf("Thread %d works on element%d\n", omp_get_thread_num(), i);
}
  
   // Check for correctness (only plausible for small vector size)
   // A test we would eventually leave out
   printf("i\ta[i]\t+\tb[i]\t=\tc[i]\n");
for(i=0; i<n; i++) {
       printf("%d\t%d\t\t%d\t\t%d\n", i, a[i], b[i], c[i]);
}
  
// clean up memory
free(a); free(b); free(c);
  
   return 0;
}