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How does EIGRP really calculate it's metric?

nikhil.kulkarni
Level 1
Level 1

Hi,

I was reading through the CCNP chapters and got confused on how EIGRP really calculates it metric.

Considering all the values as default the formula says:

Metric =   ( 10^7 / least-bw + cumalative delay ) * 256

What does it mean when you say cumalative delay is 10 of microsecs?

Now say there is a subnet 10.10.10.0 /24 connected on router A. Router A is connected to Router B, both on LAN on Gigabit ethrenet interfaces and ping from router A to Router A takes 1 ms. How will Router B really calculate the metric?

Regards,

Nikhil

7 Replies 7

cadet alain
VIP Alumni
VIP Alumni

Hi,

The delay that EIGRP uses is not a calculated delay but the default one or the configured one( with the delay command) that appear in the outputof sh interface command.the same goes for the bandwidth( configured with bandwidth command).

Regards.

Alain

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Ok, I still have some problem understanding the FD and the AD from any router to a given network:

For explaining I am taking the below example:

The network 137.72.128.31 /32 is the loopback 0 address of the RouterA interface:

I do a show topology table for this IP and get the below information

The router has calculated a FD to it's own IP and given the value which is perfectly correct

RouterA#sh ip eigrp topo | b 137.72.128.31/32

P 137.72.128.31/32, 1 successors, FD is 128256

Here is how the calculation was done:

RouterA#sh int lo 0 | in BW

  MTU 1514 bytes, BW 8000000 Kbit, DLY 5000 usec,

RouterA#

This gives us the formula:

(10^7 / 8000000 + delay ) * 256

(1.25 + 500) * 256

(1 + 500) * 256

501 * 256

128256

The above calculation works perfect and gives the FD as 128256. So RouterA has calculated a metric to it's own directly connected network.

Till this part I am clear but here is where my confusion starts.

Consider RouterA is connected to RouterB with a link of bandwidth 2 Gig. Now the FD from RouterB to the loopback 0 interface of RouterA should be the AD that RouterA advertises(that means it's own FD) plus the Metric that RouterB calculates on it's own.

Now this is how RouterB calculates it metric:

RouterB#sh int po 1 | i BW

  MTU 1500 bytes, BW 2000000 Kbit, DLY 10 usec,

RouterB#

The least BW is 2Gig(2000000 Kbit) and the delay would be 5000 + 10=5010

This gives us the formula:

(10^7 / 2000000 + delay) * 256

(5 + 501) * 256

506 * 256

129536

As per my understanding, from the above the FD calculated by RouterB to loopback interface of RouterA should actually be 129536 + 128256 = 257792

But the output shows this as 129536 which is the calculated metric of RouterB to the loopback interface of RouterA.

RouterB#sh ip eigrp topo | be 137.72.128.31

P 137.72.128.31/32, 1 successors, FD is 129536

So, now my question is where am I going wrong or is my understanding wrong on the FD part?

Hi,

your understanding is wrong, each router calculates its FD to the destination with the formula you used( with least bandwidth and cumulative delays) and advertise it as RD to its downstream neighbour, the routers don't sum up the RD with the FD to the neigbour to get the FD.

Regards.

Alain

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ok Alain..I get that now. Thanks for the clarification.

Regards,

Nikhil

Alain,

Close. Routers don't actually send the calculated FD value to peers to use as RD. instead, each router sends the metric vectors to peers, who calculate the RD from the values, then factor in the outbound interface metric info toward that peer to determine the local computed metric. Clear?

Sent from Cisco Technical Support iPad App

Hi Donald,

Yes I realized i had given a wrong information and striked the error but thanks for pointing out.

Regards.

Alain

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Don't forget to rate helpful posts.

nikhil.kulkarni
Level 1
Level 1

thanks folks.!! things are a lot clear to me now!!

Regards,

Nikhil

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