Galapagos double fitness?

Hi there, It's my first try with Galapagos but I think it's going to be very useful for my particular problem! I have a generic block 100x100. What I did is draw the hypothetical volume aloud by the law. What I need galapagos to do is to manipulate a prism (which is build as a result of an offset of the building site in order to maintain the site proportions) This prism is a hypothetical building with x numbers of floors. Each floor has 2,6m height. So I need galapagos to, on one hand, maximize the number of floors,, and on the other, maximize the area of one floor (every floors has the same area). So again, find the maximum number of floors which are a result of dividing the total available height within the 'legal volume', and finf the maximum floor area.

Sorry If i'm not explaining myself very well.. I'm leaving leaving 2 sketches to clarify, a rhino file and a grasshopper.

Any help would very appreciated!

Thank you in advance.

Andrew

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Permalink Reply by Mirek on February 15, 2015 at 3:45pm

Maybe octopus could help?

http://www.food4rhino.com/project/octopus?etx

Permalink Reply by Andrew on February 15, 2015 at 4:07pm

Thank you for the quick response! I'm trying to understand the Octopus manual, but because I cant understand much... is there a way of finding two maximum fitness values?

Permalink Reply by David Rutten on February 15, 2015 at 4:49pm

Hi Andrew,

I understand your problem, and there is a solution. Galapagos cannot optimize two or more fitnesses. There are ways of doing it (octopus can explore these fitness boundaries I think), but not via Galapagos.

However I suspect you're not really trying to optimize two different fitness dimensions. Floor count and floor area are both different manifestations of a single fitness dimension. You could call it 'maximum total area', or maybe 'maximum total profit', or something else which is dependent on both count and area. Thus you need to combine these two metrics into a single fitness value. If you're looking to maximize the area as a whole, multiply the per floor area by the number of floors. If you're looking to maximize profit, multiply the area of each floor by a per square metre profit for that floor.

Permalink Reply by Andrew on February 15, 2015 at 4:56pm

Hi David, thank you for your response. I understand what you are saying, my problem is that because I'm looking for maximum number of floors (ie.height), I'm not sure how to combine that with square meters..For example a maximum floor are could result in one single large floor instead of a maximum number of floors stacked up

Permalink Reply by David Rutten on February 16, 2015 at 2:08am

Well if maximum number of floors is the overriding goal, then the area per floor is less important. Your total fitness can then be described by the equation

Count*100000 + Area

Of course I picked 100000 out of my behind, you'll have to make sure to pick a value which means a single extra floor is worth more than any growth in area.

Permalink Reply by Andrew on February 16, 2015 at 7:37am

Thank you David, your advice was very useful!

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