Fitness Function - What to do, if value of one Chromosome features is much higher than other ones - scalability

Question

I am trying to write a Fitness Function for my Genetic Algorithm. I have got three features (F1,F2,F3 - All are numeric variables), which are taken into consideration for rating the chromosome. If all of them are equally important I can write such equation:
If their importance would be different, I can multiply features by different constants.
It will all work, if values of all three features would have the same interval.
But what if F1 and F2 could have a value from (0,10) and F3 for example(0,49)?
How could I approximately scale it? Or should I approach it differently?

Answer 1

Scale all your values to [0..1]
Assign a weight / importance to every value
Multiply every scaled value with its weight
Divide the sum of all values from step 3 by the sum of all weights
I don't know which language you are using, so here is some C# code:
Feature Class
public class Feature
{
public double Importance { get; set; }
public double Value { get; set; }
public double LowValue { get; set; }
public double HighValue { get; set; }
public double ScaledValue
{
get
{
// Make sure value is within bounds
double intermediate = Math.Max(Math.Min(LowValue, HighValue),
Math.Min(Math.Max(LowValue, HighValue), Value));
// Scale from 0 to 1 within defined bounds
double scaledValue = (intermediate - LowValue)/(HighValue - LowValue);
return scaledValue;
}
}
}
Example Calculation
// low importance (1), values from [20..-10] (less is better)
var f1 = new Feature
{
Importance = 1,
Value = 0,
LowValue = 20,
HighValue = -10
};
// low importance (1), values from [0..10] (more is better)
var f2 = new Feature
{
Importance = 1,
Value = 1,
LowValue = 0,
HighValue = 10
};
// high importance (5), values from [0..49]
var f3 = new Feature
{
Importance = 5,
Value = 25,
LowValue = 0,
HighValue = 49
};
var features = new[] {f1, f2, f3};
var score = features.Sum(f => f.ScaledValue*f.Importance)/features.Sum(f => f.Importance);
Debug.WriteLine(score);