We are have roughly 12,000 polygons which represent residential subdivisions. Each polygon has an the total homes in a field named “Count_”within the attribute table. The “ We are trying to merge the polygons into groups of between 700-800 homes while using another layer containing the major roads (lines) as a boundaries. We have tried aggregate and dissolve tools with minimal success. Any suggestions or help would be greatly appreciated.
grouping with a numeric constraint is going to be complicated by the fact that the numeric optimum may aggregate polygons that are separated by some distance.
What rule set is in place to consider adjacency ?
Is adjacency as important than the numeric combined sum?
Do you have a 'seed' location to begin from? every start location can conceivably produce a different aggregation and aggregation pattern?
Just a few thoughts. Perhaps an image showing the existing distribution etc would facilitate a more comprehensive solution
Dan, thanks for the quick reply. We do need them to be adjacent polygons. If you can imagine a city with hundreds of subdivions. Each with say 50 homes. We need to group the subdivisions on larger groups consisting of say 250 homes, but use the major roads as boundaries as we wouldn’t want to combine neighborhoods that are located across a major freeway from each other. I also looked at the ESRI business analyst tool to possibly create territories, but couldn’t figure out how to make it work.
Since that is a major constraint... begin their and subdivide first by the major highways manually. From that point you could put a coarse 'Fishnet' over each division and get the house counts that fall within each grid square (use the centroid of each of your polygons or an intersect).
Have a look at what you have then and see if any other rules emerge
I ended up creating a polygon layer using the roads as boundaries. I then did a join the sum the count column from the smaller polygons within each larger one. Then I merged and split the larger polygons to get them as close as I could to an equal number. It was time intensive, but it worked. Thanks for your help.