Conceptual problem

[siouxchief]

Hows it going everyone?

I just have a question about networks.
I was doing a bit of footprinting on the local businesses (a WAN i suppose)
in my area and i found that a lot of them are on the one isdn network
(we dont have broadband).

from 1.2.3.0 --1.2.3.255 as an example this is the size not the actual i.ps

Using a tool call networkview i mapped them out.

Just have 2 questions really.

1:Networkview identified some of them as servers and others as not a "snmp device".
What does this mean?
Does this mean some have a network behind the 1 i.p. and the others are just one
computer maybe?

2:More importantly,i would be interested if anyone knows how people discover the i.ps
of the many computers in these businesses from knowing the 1 i.p. address of the server.
Some of these business have maybe 20 computers with maybe open shares but how do
you know the location of these machines from knowing their 1 i.p. number on the isdn network.

This could be a piece of cake for some of ye.
looking forward to hear what ye think.

[SDK]

I'm gonna give a try to answer your question..

1:Networkview identified some of them as servers and others as not a "snmp device".

SNMP is Simple Network Management Protocol. This is a procotol to manage router, switch, server from another location on the network and not directly siting infront of the computer. Some of you server might have configure to not allow Management by the SNMP witch have a BIG flaw right now.

2:More importantly,i would be interested if anyone knows how people discover the i.ps of the many computers in these businesses from knowing the 1 i.p. address of the server.
Some of these business have maybe 20 computers with maybe open shares but how do
you know the location of these machines from knowing their 1 i.p. number on the isdn network

Their program that map a entire network by pinging very IP from a certain range. If you know the first IP, you can ping the entire Network.

Hope it help.

[Custy_J]

it can be much more complex, but I think thats a pretty good answer SDK..