在AnyLogic中如何选择通过网络中另一个代理连接的最近代理？

Question

In a model I have connected different agent types in a network through 'link to other agents' objects. I have used a function to create the network:

shopLink.connectTo(this.getNearestAgent(main.shops));
homeLink.connectTo(this.getNearestAgent(main.homes));

So each Factory-agent is connected to the nearest shop and home only. This function is called at the 'at startup' field within the Factory-agent type.

Let's assume the red agent types are factories, yellow represent shops, and green represent homes. Also assume that all Factory-agents contain Person-agents, and I want to send the Person-agents to the most nearby Shop-agent is connected to the Factory-agent. What Java code would I need to use to select the most nearby Shop-agent connected to the Factory-agent?

Answer 1

If your Person agents exist inside a Factory parent agent, and you have connections as shown, then it depends how your connections are set up in Factory:

(a) If Factory's default connections Link to Agents object contains only Shop agent connections, use

getNearestAgent(factory.getConnections())

(b) If Factory's shop connections are in a special Link to Agents object (say shopConnections ) then use

getNearestAgent(factory.shopConnections.getConnections())

(c) If you've mixed connections to all other agent types (shops, factories, etc.) in Factory's default connections Link to Agents you'll have to filter that list first to include only Shop agents so

(Shop) getNearestAgent(filter(factory.getConnections(), f -> f instanceof Shop))

(There are Java subtleties in terms of why case (c) requires the (Shop) bit at the start, which is a Java cast, but (a) doesn't. It's to do with the fact that both getNearestAgent and getConnections are generic methods , and type inference is being used.)