Algorithm Timing - Tilted Forum Project Discussion Community

RAGEAngel9 · 04-14-2005, 05:33 PM

Hey guys quick question that I can't seem to get.
I'm trying to show how a program I wrote improves (or worsens) as the number of processors increases. However, I can't for the life of me determine a way to represent the time it take for one of the funcitons to run.

Code:

    double term = start;
    double newTerm = 0;
    double precision=1e-6;
    double sum = start;
    int i =0;
    // Calculate each new term and add to total 
    //Stop when the previous term is less than the precision
    for( i = p;term > precision ; )
    {
        double newFac = 1;
        // Calculate the nessesary jump given i and the (n)umber of processors
        for(int j = 1;j <= n ; j++ )
        {
            newFac = newFac * ( x / ( i+ j ));
            
        }
        newTerm = newFac;
        //Get new term
        term = term * newTerm;
        // Increment sum by new term
        sum += term;
        i=i+n;
    }

I know the inner loop runs n times each time the main loop runs. But I can't figure out how to determine the number of times the main loop runs in any sort of formulaic way.

Yes, it's homework, but 90% of the assignment was developing and writing the algorithm.
Thanks

vinaur · 04-14-2005, 08:04 PM

Is this it, because I don't see any declarations for start, p and x. It would be nice to know what those are.

theFez · 04-14-2005, 08:21 PM

If you can use perl, there are some good benchmarking features in the benchmark module.

I've used them a few times to test performance of different algorithms.

http://www.perldoc.com/perl5.8.0/lib/Benchmark.html

SiNai · 04-15-2005, 01:15 PM

Are we talking theoretical big-oh stuff or empirical timing, or both? You can use System.currentTimeMillis() for actual timings in milliseconds.

Theoretical.. You're incrementing your outer loop by a constant? 0(n^2) then.. (Can't tell what n is..)

04-14-2005, 05:33 PM	#1 (permalink)
RAGEAngel9 Insane Location: Vermont	Algorithm Timing Hey guys quick question that I can't seem to get. I'm trying to show how a program I wrote improves (or worsens) as the number of processors increases. However, I can't for the life of me determine a way to represent the time it take for one of the funcitons to run. Code: double term = start; double newTerm = 0; double precision=1e-6; double sum = start; int i =0; // Calculate each new term and add to total //Stop when the previous term is less than the precision for( i = p;term > precision ; ) { double newFac = 1; // Calculate the nessesary jump given i and the (n)umber of processors for(int j = 1;j <= n ; j++ ) { newFac = newFac * ( x / ( i+ j )); } newTerm = newFac; //Get new term term = term * newTerm; // Increment sum by new term sum += term; i=i+n; } I know the inner loop runs n times each time the main loop runs. But I can't figure out how to determine the number of times the main loop runs in any sort of formulaic way. Yes, it's homework, but 90% of the assignment was developing and writing the algorithm. Thanks

04-14-2005, 08:04 PM	#2 (permalink)
vinaur Insane	Is this it, because I don't see any declarations for start, p and x. It would be nice to know what those are. __________________ If you multiply that by infinity and take it to the depths of forever, you will, perhaps, get just a glimpse of what I am talking about. --Meet Joe Black--

04-14-2005, 08:21 PM	#3 (permalink)
theFez Crazy Location: here and there	If you can use perl, there are some good benchmarking features in the benchmark module. I've used them a few times to test performance of different algorithms. http://www.perldoc.com/perl5.8.0/lib/Benchmark.html __________________ # chmod 111 /bin/Laden Last edited by theFez; 04-14-2005 at 08:22 PM.. Reason: link

04-15-2005, 01:15 PM	#4 (permalink)
SiNai Dreams In Digital Location: Iowa	Are we talking theoretical big-oh stuff or empirical timing, or both? You can use System.currentTimeMillis() for actual timings in milliseconds. Theoretical.. You're incrementing your outer loop by a constant? 0(n^2) then.. (Can't tell what n is..) __________________ I can't seem to remember now What it was like- to live life, before you.. symbiont