# EIGRP variance

Sep 10th, 2008

If i need to configure unequal load balancing with ratio for example 4:1 how i should to play with delay to meet that requirement?

first path is R4(bw=1544k,dly=20000)->R1(bw=1544k,dly=20000)->R5(bw=100000k,dly=100)

second path is

R4(bw=100000k,dly=100)->R5(bw=100000k,dly=100)

Hello Mike,

with default K values the EIGRP metric formula is:

256 * (sum of delays in 10s of microseconds)+

256 * 10^7/(Minimum Bandwidth on the path in kbps)

then it is the distance as calculated that has to be up to 4 times the best metric.

To be remembered that in any case the advertised distance must be stricly lower then the Feasible distance: eigrp variance does not allow to bypass the feasibility condition.

you need to increase delay out interface of R4 to R5 so that:

256 * 10^7 / 100000 + 256*( +100) > 256 * 10^7 / /1544 + 256*(20000+100)

feasibility condition

and :

256 * 10^7 / /1544 + 256*(20000+20000+100) < 4 * [256 * 10^7 / 100000 + 256*(x +100)]

I think here the feasibility condition is never satisfied because with no change of delay is R1 more near to destination then the local router R5.

Because R1 is connected with a slower link to R5 and then with another slower link to R4.

R4 has a direct fast connection to R5.

From the point of view of R4 about R5's destination R1 is not nearer to the destination but the opposite and is never seen as a possible alternate path a feasible successor and not usable even by configuring variance.

Hope to help

Giuseppe

Overall Rating: 5 (1 ratings)

## Replies

Giuseppe Larosa Wed, 09/10/2008 - 09:05

Hello Mike,

with default K values the EIGRP metric formula is:

256 * (sum of delays in 10s of microseconds)+

256 * 10^7/(Minimum Bandwidth on the path in kbps)

then it is the distance as calculated that has to be up to 4 times the best metric.

To be remembered that in any case the advertised distance must be stricly lower then the Feasible distance: eigrp variance does not allow to bypass the feasibility condition.

you need to increase delay out interface of R4 to R5 so that:

256 * 10^7 / 100000 + 256*( +100) > 256 * 10^7 / /1544 + 256*(20000+100)

feasibility condition

and :

256 * 10^7 / /1544 + 256*(20000+20000+100) < 4 * [256 * 10^7 / 100000 + 256*(x +100)]

I think here the feasibility condition is never satisfied because with no change of delay is R1 more near to destination then the local router R5.

Because R1 is connected with a slower link to R5 and then with another slower link to R4.

R4 has a direct fast connection to R5.

From the point of view of R4 about R5's destination R1 is not nearer to the destination but the opposite and is never seen as a possible alternate path a feasible successor and not usable even by configuring variance.

Hope to help

Giuseppe

rmv72 Thu, 09/11/2008 - 21:47

Hello!