Eigrp unequal load balance

[acbenny]

Hi Expert,

With attach is two eigrp digram, I have few question on it. Hope you can help

From the eigrp(1) diagram, if I am in the view of Router E to destination Router A, then

(Q1) does the reported distance is that I high light in the diagram ?

(Q2) Since the Feasible distance is 20, the if use "variance 2" can acheive unequal load balance

which only path (RouterE->RouterB->RouterA) achieve two crateria

crateria 1 --(RouterE->RouterB->RouterA) cost is 30 < FD*2=40

crateria 2---reported distance of (RouterB->RouterA) is less than FD (20)

From the eigrp(2) diagram, if I am in the view of Router A to destination Router F, the

(Q1) does the reported distance is that I high light in the diagram ?

(Q2) Since Feasible distance is 20, the if use "variance 2" can acheive unequal load balance

it is total 3 path to Router F ( Router A-C-F ) (Router A-D-G-F ) ( RouterA-D-G-H-F )?

[Jon Marshall]

Hi Jack

The first one is correct. The route from E to A via B will be a feasible successor and with a variance of 2 it will load balance 2:1 across both path.

The second example i'm not so sure about. Where are you applying the variance command ?

If your'e doing it on Router A then there will only be two paths to F. I don't believe you would see the A-D-G-H-F path. Even if you did when the packet got to Router G that router would always forward the packet straight to F rather than via H. The path taken is relevant locally to each router.

HTH

Jon

[acbenny]

Hi Jon,

Thanks for your help, since I am a new guy in routing, can I have one more example.

With the attach new diagram, If route from RouterA to Router F, then Feasible Distance is 20

Q1 Does the reported distance are RouterD-G-F and RouterB-E-H-F ?

Q2 Router B and Router D become feasible successor ?

Q3 if set variance 2 in Router A, then totally 3 path to Router F

First one, Router A-C-F

Second one, Router A-B-E-H-F

Third one, Router A-D-G-F

[Jon Marshall]

Hi Jack

No problem with more examples.

This one is a bit more complicated. Bear in mind also that the examples you are using are based purely on link costs which in the real world is not how EIGRP does things but there's nothing wrong with this approach. In fact as you may know Jeff Doyle uses the same cost examples in his CCIE book for simplicity. So based on using costs alone.

A very important point to note and one which i overlooked in your previous example (apologies) is that you are not measuring FD to a router but to a network. So if we assume that the network we are working on is the network between F -> H which we will refer to as Net1.

The feasible distance from A -> Net 1 with the path A -> C -> F -> Net1 = 10 + 10 + 5. Note that you must include the cost of the Net1 link.

As you can probably see all your paths have an FD of 25 to get to Net1 so they would all be seen as equal cost paths to Net1.

Notice also that in contrast to what i said before Router G will also have 2 equal cost paths to Net1 via Router F and Router H.

Hope this makes sense.

Jon