The geographical location of the item on the network map generally starts from the node where the mapping software is running. The other objects on the network should then be mapped based on whether there is a direct link from the starting node to it or goes thorough another node.
eg : from my pc, there is no direct connection to node 4, it goes through a hub. So the my pc and node 4 will be centered round the hub. Similarly, if the connection goes through the same hub and then through a switch to go to node 105, the switch then is the center for all hubs, nodes connected through it. and so on
Regards Saswata
----- Original Message ----- From: "q u a s i" quasar@vsnl.net To: linuxers@mm.ilug-bom.org.in Sent: Saturday, November 16, 2002 12:18 AM Subject: Re: [ILUG-BOM] OT - Topology plotting
At 02:02 morn 11/15/02 -0800, Kishor wrote:
hello guys, I'm facing a vexing problem...one of my colleagues is writing a network mgmt system thats queries our WAN devices, collects info on their 'neighbours' and is supposed to plot the topology of the network. so basically...the input to the plotting program would be a list of: nodeID,{set of its neighbours}
From this, it has to graphically plot the links. I'll give an example of the issues...how do we know which node should be at the center? how about redundant links between the same two locations?
But does a network necessarily have a center? In the sense that I did not understand what you mean by a center. Anyway the plot will be relative. So suppose one WAN device is the center of some type of star topology then it will have multiple links with different devices with the help of which you can determine its 'centerness'. Some redundant links will overlap and others could be mostly weeded out by a little external knowledge I think. But I am not sure I understand you fully.
Does anyone know of any mathematical algorithms for solving such a problem? well..would be glad if anyone can help!
Graph theory deals with this. But I have not understood your problem enough to see any need for any such algorithm... Looks like an interesting problem though, and we can continue it off list if it so interests you.
cheers, quasi