hi i am sorry but i am not too sure... most of the qns pertaining to ccna...should be asked here?
i have one qn...i am a little confused...
Router A interfaces with addresses 220.127.116.11 and 10.1.1.1.
Router B is connected to router A over a serial link and has addresses 10.1.1.2 and 18.104.22.168. the following are commands that would be part of a complete RIP configuration on router B with which router B advertises out all interfaces and about all route.
the options 1 to 3 are the answers.....i cant even figure out how the connection looks like....can someone plz explain ...also why the answers r such....million thanx in advance....
Since Router A and B are connected between each other via serial link, this requires them to be in the same network. If you look at the info given, the two address common to both routers are the 10.1.1.1 [Router A] and 10.1.1.2 [Router B].
Now, for Router B to advertise the networks it knows about out all interfaces, we must enter the following commands:
Router rip - enable rip
Network 10.0.0.0 - advertise to this network
Network 22.214.171.124 - advertise to this network
As you can see, all 3 commands are required in order to satisfy your question.
p.s I've renamed your thread to a more appropriate topic.
hi guys...i have one more qn...
i read somewhere that to aid in converting to hex tt its a good practice to write down the 16x table even before the exam starts...
does anyone noe how this helps?
thanx a mill in advance;)