On Mon, May 14, 2001 at 05:59:35PM -0400, Thomas Gainer wrote:

Kind of a newbie questions, but I would like to know what the consensus is.

Hmm, until it breaks the application? Latency is what matters. That is unless your TTL expires prematurely; in this case defined as before the applications endpoints can reach each other (which in turn could theoretically be a function of latency as well). -- Christian Kuhtz <ck@arch.bellsouth.net> -wk, <ck@gnu.org> -hm Sr. Architect, Engineering & Architecture, BellSouth.net, Atlanta, GA, U.S. "I speak for myself only.""

Subject: Re: How many routed hops would be considered too many to leave an AS Douglas Adams, had he done routing, would say 42. Sadly he's no longer with us. You won't get a concensus or a summary, but if I post you're sure to get many opinions with people correcting or disagreeing, or ranting about nonsequiter things in a pedantic way, so I'll try to help, or at least incite a flurry of vindicative responses. I believe the answer is A/ It Depends, and B/ Usually 6 to 8 routed hops for each AS. In general, a network operator must be considerate of the lowest-common-denominator traffic on one's network. With regards to hop-count, many would say that Windows 95 (98?) is the LCD, as it emits IP packets (by default) with a TTL of 32 (Thirty-Two). I believe Win2k (maybe 98) has a default TTL of 128. I assume that WinME and WinXP are at 128 or more as well. It is fairly reasonable to consider that most (90+%) traffic goes through at most 4 ASes: (no AS) Client -> AS 1. Upstream AS of "client" AS 2. "Tier 1" Upstream to #1 AS 3. "Tier 1" Peer and Upstream to #4 AS 4. Upstream AS of "server" -> Server (no AS) Since there are 4 ASes, and 32 hops, each AS should get something close to 8 hops. Add a few at each end for aggregation, and you can get give each AS 6 hops, with some leftover for end-network traffic. MIN_HOPS_PER_AS = ( LCD_TTL - (END_TTL * 2) ) / MAX_REG_AS_HOPS) or, MIN_HOPS_PER_AS = ( ( 32 - (2 * 2)) / 4 ) MIN_HOPS_PER_AS = ( ( 32 - 4) / 4 ) MIN_HOPS_PER_AS = ( 28 / 4 ) MIN_HOPS_PER_AS = 7 It is interesting to note that more IP elements, without more meshing, will create more hops. For example, a SONET network without significant meshing will have more IP hops than a fully meshed IP/ATM network. There are many many things of more significant importance than how many IP/L3 hops a network has, however. Certain ATM and MPLS networks create logical meshes which decrease (IP L3) hop counts in certain circumstances. Pros/Cons all the way around. -alan Thus spake Thomas Gainer (TGainer@e-xpedient.com) on or about Mon, May 14, 2001 at 05:59:35PM -0400:

Kind of a newbie questions, but I would like to know what the consensus is.

Thanks,

Thomas Gainer