Problem 11

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Contents

  1. Problem 11
    1. Description
    2. Expectation
    3. Code
    4. Profile and Timing
    5. Comments
    6. Contributors

1. Problem 11

1.1. Description

Which of the following will execute faster? 

 increment=xmax/large_number
 do i=1,large_number
   x(i)=i*increment
 enddo

 increment=xmax/large_number
 sum=increment
 do i=1,large_number
   x(i)=sum
   sum=sum+increment
 enddo

1.2. Expectation

The two routines preform basically the same function, however the first performs a multiplication calculation at each iteration whereas the second uses addition. Since multiplication is a more complicated operation, I would expect the first routine to go more slowly, and hopefully a good compiler would figure this out anyway.

1.3. Code

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 
#define XMAX    100
#define BIGNUM  1000000000
 
void case1();
void case2();
 
 
int main( int argc , char* argv[] )
{
        if ( argc < 2 )
        {
                printf("Usage: problem2 [1|2]\n");
                exit(-1);
        }
 
        int caseNum = atoi( *++argv );
 
        if ( caseNum == 1 )
        {
                case1( BIGNUM );
        }
        else if ( caseNum == 2 )
        {
                case2( BIGNUM );
        }
        else
        {
                printf("Usage: problem2 [1|2]\n");
                exit(-1);
        }
 
        exit(0);
}
 
void case1()
{
        // increment=xmax/large_number
        float inc = XMAX/BIGNUM;
 
        // init array
        float *x;
        x = (float *)malloc(sizeof(float)*BIGNUM);
 
        int i;
        // do i=1,large_number
        for(i = 0; i < BIGNUM; i++)
        {
                // x(i)=i*increment
                x[i] = i*inc;
        }
        // enddo
 
        return;
}
 
void case2()
{
        // increment=xmax/large_number
        float inc = XMAX/BIGNUM;
 
        // init array
        float *x;
        x = (float *)malloc(sizeof(float)*BIGNUM);
 
        // sum=increment
        float sum = inc;
 
        int i;
        // do i=1,large_number
        for(i = 0; i < BIGNUM; i++)
        {
                // x(i)=sum
                x[i] = sum;
 
                // sum=sum+increment
                sum = sum + inc;
        }
        // enddo
 
        return;
}

1.4. Profile and Timing

Both cases were run 5 times at each compiler optimization level, using the perl script found here.

The timing results can be found here.


Image:Hw1prob11chart.png

1.5. Comments

Case 1 does not benefit substantially until the O3 optimization level, at which point it is competitive with the performance of case 2. This is likely due to

In case 2, just one level of compiler optimization yields an approximately tenfold performance gain, with little gain from subsequent optimization levels. This is likely due to

1.6. Contributors

Steven Baehr

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