Re: [rrg] IPv4 & IPv6 routing scaling problems

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Paul,
While I am still waiting for you to convince me, that DV is superior to  
Dijkstra, let me expand on the 10x10 chessboard grid example where you have  
184758 shortest path from the left upper corner A to the right lower corner  
B.
Grab one of them! Then envision that each of its 20 best next hops  might 
be replaced by a 3-hops-detour: 
1 hop to some more remote node, followed by 2 (explicitly enforced) best  
next hops, where each of which must not go back to where the packet come  
from.So there will be "20 over 1" = 20 paths with just one such detour.  
Additionally there will be "20 over 2" paths with 2 such detours, etc.  etc.
Altogether there will be 2^20 = 1048576 paths which include 0,1,2,...19 or  
20 such detours.
 
184758 x 1048576  =193,730,707,456. I.e.by enabling such little  obviously 
smallest possible detours
there are 193,730,707,456 paths from A to B.
No comes the worst: With DV you won't get to know that any such detouring  
path exists !!
 
You have no rearview mirror. DV restricts the number of routes  enormously 
and  prevents traffic engineering which deserves its  name.
 
Heiner  
 
 
 
 
In einer eMail vom 14.02.2010 21:11:21 Westeuropäische Normalzeit schreibt  
[e-mail]:

 
In einer eMail vom 14.02.2010 17:00:37 Westeuropäische Normalzeit  schreibt 
[e-mail]:

Hi,

I've only just stumbled across your TARA stuff, and am  reading 
through your mails. Re "Forget DV!", and double-plus to  that!


Well, then try to convince me while viewing the following:
I was told that an average route  contains about 20 hops. Imagine a  
chessboard like grid however with 10x10 rows and columns instead of 8x8.  Imagine 
each DFZ-router has only 4 neighboring DFZ-routers (which is  certainly less 
than in reality).Imagine the ingress at the upper left corner  and the 
egress at the lower right corner. Then there are (20 over 10) =  20x19x...x11 / 
(1x2x3....x10) = 184 758 shortest paths in-between.
 
Consider that there are more than 10,000 egress DFZ-routers ! 184 756  
times 10 000 = 1 847 560 000 shortest routes. However there are  multitudes 
thereof, if you envisioned (loopfree of course) detours as well!  And if you 
envisioned that DFZ routers have more than 4 neighbor nodes  (BTW, can anyone 
provide useful valid figures so that we can discuss density  issues and 
aspects?)
 
What a nonsense of algorithm !!! DV is a holy cow from the early  90's!
 
Please convince me, why provisioning and managing all these routes would  
be reasonable? 
Particularly, when you can do it without!  
 
Heiner