Talk:Enhanced Interior Gateway Routing Protocol

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Talk page archive

Archive 1

Reworking

Hello to anyone. I have partially reworked the article on December 26th (from IP address 158.193.152.134, just forgot to log in). I would like to hear anyone's suggestions to article contents, completeness and clarity. There are still crucial parts missing:

description of diffusing computations as proposed by Dijkstra and Scholten
description of the DUAL finite state machine itself
description of the overall EIGRP algorithm
Authentication of routing peers
What else?

I am planning to add those in due time. Paluchpeter 16:39, 28 December 2006 (UTC)[reply]

More things to add: discuss autonomous systems and AS numbers, Reliable Transport Protocol (how EIGRP uses multicast to send updates to the multicast address 224.0.0.10, and if no acknowledgment is received from known neighbors, it switches to unicast), and classless wildcard host mask (subnet mask complement). Dejaphoenix (talk) 10:03, 30 March 2008 (UTC)[reply]

Maximum EIGRP Hop Count

I've seen a lot of conflicting arguments regarding what the actual maximum hop count for EIGRP is, to the point where I was watching a snippet of a video from Train Signal on EIGRP and the instructor, Chris Bryant CCIE #12933, mentioned that he's seen arguments on the net about it (even discrepancies in Cisco's own documentation). In the video, he shows that This was the the router itself claims that the maximum hop count is 100, not 220; he runs the "show ip protocols" command on a Cisco 2500 I believe:

R1#show ip protocols

Routing Protocol is "eigrp 100"
 Outgoing update filter list for all interfaces is not set
 Incoming update filter list for all interfaces is not set
 Default networks flagged in outgoing updates
 Default networks accepted from incoming updates
 EIGRP metric weight K1=1, K2=0, K3=1, K4=0, K5=0
 EIGRP maximum hopcount 100
 EIGRP maximum metric variance 1

However, I was reading elsewhere that the default is 100, and that it could be adjusted. If the user runs 'router eigrp <AS#>' then 'metric maximum-hops', the maximum hop count can be changed from 1-255, according to IOS help:

R1#conf t

Enter configuration commands, one per line. End with CNTL/Z.

R1(config)#router eigrp 100

R1(config-router)#metric ?

 holddown      Enable EIGRP holddown
 maximum-hops  Advertise EIGRP routes greater than <hops> as unreachable
 weights       Modify EIGRP metric coefficients

R1(config-router)#metric maximum-hops ?

 <1-255>  Hop count

R1(config-router)#metric maximum-hops 255

R1(config-router)#

I suppose that this is something that should probably be checked. Daedalus01 17:11, 1 June 2007 (UTC)[reply]

Division vs. multiplication error in "multiple metrics" section

Currently, in the "Multiple metrics" section of this article, the following sentences appear:

Analogously, the interface delay is a configurable static parameter expressed in units of tens of microseconds. Dividing this interface delay value by 10 yields a delay in units of microseconds that is used in the weighted formula.

I believe that some part of these sentences is incorrect. If one has a measurement in tenths of microseconds, he should multiply by 10 to obtain the measurement in microseconds. I considered performing a simple edit to change "dividing" to "multiplying," but I figured I'd post something here first to make sure that the true error did not lie somewhere else. If there is no further comment on this issue within the next few days, I will go ahead and make the edit.

Douglyuckling 21:58, 17 June 2007 (UTC)[reply]

A good point! You're absolutely right. I will make the change. Thank you for pointing this out!

Paluchpeter 17:12, 12 July 2007 (UTC) The above is incorrect. The interfaces have a delay in microseconds. For example a serial link has a delay of 20,000 microseconds. Eigrp uses 10s of microseconds in its metric. Therefore to find the delay metric you would add all of the delay for a path and then divide by 10. Maybe this isn't clear, but if you had 20,000 dollars and wanted to find out how many 10s of dollars you had, you wouldn't multiply by 10, you would divide. You would have 2000 10 dollar bills not 200,000. —Preceding unsigned comment added by 155.45.81.25 (talk) 20:24, 17 October 2007 (UTC)[reply]