Gnutella Forums - My Suggestion: Making Download Times Better

Whenver I download a file, and the other end is very irradic in its connection speed, the estimated time jumps everywhere from 2 minutes to 2 hours, and it changes a couple of times per second. I find it somewhat difficult to know when I need to look at my computer again if I want to leave it running for a while.

I would suggest making download time estimations stable in the face of connection speed variations by displaying a rolling average of the last few file ETAs which continues to change as new ETAs are computed.

I have read some of the code, and as I understand it, my suggestion could be implemented by changing the DownloadDataLine class in the com.limegroup.gnutella.gui.download package as follows:

Code:

/* Beginning on line 786 */

// If we have a valid rate (can't compute if rate is 0),

// then determine how much time (in seconds) is remaining.

if ( _speed > 0) {

        double kbLeft = ((_size/1024.0) - 

                (_amountRead/1024.0));

        _timeLeft = newTimeRemaining((int)(kbLeft / _speed));

}

And then, add a new private method to do the rolling average, along with some member data:

Code:

// This member data is for the rolling average.

private final int N = 10; // the number of times to save, adjust at leasure

private boolean full = false; // to only use the average when we have N #s

private int index = 0; // the index for the element to replace in times[]

private int times[] = new int[N]; // the list of times

/**

 * This method calculates the new time remaining based on a rolling average.

 * @author bsdman

 */

private int newTimeRemaining(int anotherTime)

{

        times[index++] = anotherTime;

        if(index == N)

        {

                index = 0;

                full = true;

        }

        if(!full)

                return anotherTime; // since we can't average yet

        // otherwise, take the average

        // TODO this average could probably be written more efficeiently

        long total = 0;

        for(int i = 0; i < N; i++)

                total += times[i];

        return (int)(total/N);

}

I hope my code will be useful for getting it done -- if it works correctly. I haven't had an opportunity to test it yet, but since it compiles and I have faith in my programming skills with regard to such a simple problem, I think it will work.

Any comments are welcome.