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How Can I build a OD Matrix to distribute pedestrian numbers across the network from an Origin to a Destination getting a value for the number of pedestrains using each link on the network ?

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02-19-2019 10:29 AM
JamesO_Brien
New Contributor

Hello everyone,

I'm new to this. I'm learning by trying out different scenarios.

If I had say 2 origin points A & B, 

A with 2000 people and B with 1000 people.

I want to send them out onto my network to 3 Destination Points

But I want to be able to calculate the number of people using each link on the network.

Is this possible using OD Cost Matrix ?

And if so what would the process be.

Thanks 

James

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2 Replies
JaySandhu
Esri Regular Contributor

The OD in transportation planning has different meaning that the OD solver in Network Analyst. There is no straight forward way to sum trips per OD pair on the links of a network. But it can be done.

To do what you want to do, use the Closest Facility solver. Load your origins A and B as Incidents and load your three destinations as Facilities. Set the CF to solve for three closest facilities and solve the layer. Once the solve is done, Run the GP tool Copy Traversed Source Features with the CF layer as the input network analysis layer. This will now create a new line feature class containing the individual line features for all the paths traversed. Now you will need to do some work to assign the number of trips to each route (OD pair) and then sum up the values per edge.

Jay Sandhu

JamesO_Brien
New Contributor

Hi Jay

Thanks for this. I got it to work with your advise. But I can now see where the difficulty comes in.

NETWORK

table

My problem now is that I don't know the portion of trips coming from each origin for each link. For example from ORG2 to DEST_4 it says 2 trips. I know that one was come from ORG1 and the other from ORG2 but on a larger network this would be difficult to work out. The reason I need to know is I will applied Spectator numbers to the ORG1 and ORG2 and using the number of trips work out the potential number of spectators using each link.

I'll keep working at it. Thanks so much for your help.

James

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