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Home > Archive > CCIE > March 2004 > rip v1 versus rip v2
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rip v1 versus rip v2
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| bkhanh 2004-03-04, 4:46 pm |
| Hi all guru outhere:
can someone kindly show me the right answer to this question, which I found very confusing:
Below is the question:
Which four features does RIP version 2 provide, which were impossible with RIP version 1? (Choose four)
A. Poison reverse and classless routing.
B. Classless routing and split-horizon.
C. Poison reverse and updates to multicast address.
D. Classless routing and updates to multicast address.
E. Route tag and classless routing.
The correct answer given was A, B, C, E with
this explanation:
A, B: Class routing is provided by RIP version 2, but was impossible with RIP v2.
C, E: Multicast updates (224.0.0.9) were impossible with RIP v1
Both Classless Routing and Multicast updates (224.0.0.9) were impossible with RIP v1 (all other answers have one thing that RIP v1 had.).
In my opinion, Rip v2 provide classless routing and update to the multicast address that Rip V1 doesn't. So the only answer I think correct is D. Then why this question asked to choose 4 to begin with and why the correct answer given was A,B,C,E? Also why the explanation given above to the answer A,B,C,E is so contradiction with its own answer!!!
HELP!!!!!!!!!!!!!
thanks a bunch
ckb | |
| locaso 2004-03-04, 10:37 pm |
| Ripv2 is a classless routing protocol, so all the answers that have this (A,B,D,E)are chosen, now Split horizon, Poison Reverse and Triggered updates are Distance Vectors methods(Rip1 and Rip2) to avoid routing loops.
I think the same as you in that D would be the only two options that Rip1 could not do. In every answer but D, there is at least one thing that Rip1 would do. So I think this question is a MISTAKE or have some typo. Where did you get it? many books have mistakes. Do not worry about it.
Locaso.  | |
| bkhanh 2004-03-05, 12:41 am |
| thanks locaso.
The question appears in one of the sponsored site for ccie. It is "question of the day".
Most of my colleagues can't tell what wrong with this question. At last, I found someone who shares the same view.
kcb |
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