I want to do a least cost path analysis and the least cost path has to go through several different points instead of having just a start and end destination. Is this possible?
I am using ArcGIS 10.2.2
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1if the points are ordered you might use a series of cost-path analyses to create a path from origin to destination 1, destination 1 to 2, etc... – dof1985 Mar 17 '15 at 17:28
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Yes, I have thought of that. Is there no way of doing it all at once? – Stefanie Mar 17 '15 at 22:01
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It depands. Scripting is the best option, other than manual use of tools, since model builder might be problematic. I guess that if you need to the it several time with many points it might be beter to either write a script or build a model. A workflow might work in phases - within each the "BEST_SINGLE" option is computed from a destination to a set of sources, thus giving the best way. Than the selected source should be removed from the source/origin layer and set as the new destination, and so on... – dof1985 Mar 18 '15 at 11:16
In Customize make sure that Network Analyst is turned on. Then, under Network Analyst Tools, under Analysis, open Make Route Layer. Input the set of points you would like to have your route travel through, including your starting point and your ending point. If the start and ending points are the same (i.e. finishing the route by returning to the start) I would just copy the start point so that points #1 and #100 (or whatever your last record is) are at the same location.
So in order to find the least cost path make sure you check the box that says "Reorder Stops to Find Optimal Route" and then update "Preserve Ordering of Stops" to whichever option suits your needs. If you are starting and ending at the same place choose Preserve_Both -- that's where the copied start/end point comes in. Press OK.
Because this is a Network Analysis tool it is important that your input points have been Geocoded and there is a network already set. Once you create the Route layer you will right click the layer and press Solve. Good luck.
