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EIGRP Variance & Traffic distribution count

Vivek Ganapathi
Enthusiast
Enthusiast

Hello All,

Have a question on the EIGRP traffic distribution count. Is there a possibility to perform the traffic distribution in 5:2 ?  Below is the calculation to perform a 5:1 ratio distribution  (Diagram is attached for the setup).

Note : Using only delay to perform the composite metric calculation

 

Path via R2 from R1 to reach Loopback on R4

Metric = 256*[sum of delay/10]

=256*[10+10+500]

=256*520

=133120

Path via R3 from R1 to reach Loopback on R4

=256*2510=642560

So. Path via R2 would be successor & path via R3

As Feasible successor.

Variance = FD of FS/FD of Successor

So, 642560/133120 = 4.82

Variance = 5 (whole number)

Lowest metric (FD of the successor)* 5 = 256 * [delay/10]

So, looking at my metric figures

133120 x 5 = 256 * [delay/10]

665600/256 = delay/10

26000 = delay

Now, to determine the local interface delay (LD), we need to subtract the relative distance delay in order to get the LD.

R3 - R4 has a cumulative delay of 5100.

So, 26000 - 5100 = 20900

Now, by setting the delay value of 2090 on the R1 interface connecting to R3, i would achieve 5:1. What can be done to get 5:2 ?

Thanks

Vivek       

1 Accepted Solution

Accepted Solutions

Giuseppe Larosa
Hall of Fame Master Hall of Fame Master
Hall of Fame Master

Hello Vivek,

instead of using the equation

133120 x 5 = 256 * [delay/10]

you would use the following:

133120 x 5/2 = 256 * [delay/10]

you get total delay = 13000

you subtract the known delay components and you get

dx = 13000 - 5100 = 7900

so by setting interface delay to 790 tens of microseconds you should get a 5:2 ratio in the metric values on the two paths

Let's verify:

790+10+500 = 1300

on the other path you have

10+10+500  = 520

1300 /520 = 2,5 = 5/2

Hope to help

Giuseppe

View solution in original post

4 Replies 4

bava_ccna
Beginner
Beginner

Interesting,I am also looking for this solution..

-bava

Giuseppe Larosa
Hall of Fame Master Hall of Fame Master
Hall of Fame Master

Hello Vivek,

instead of using the equation

133120 x 5 = 256 * [delay/10]

you would use the following:

133120 x 5/2 = 256 * [delay/10]

you get total delay = 13000

you subtract the known delay components and you get

dx = 13000 - 5100 = 7900

so by setting interface delay to 790 tens of microseconds you should get a 5:2 ratio in the metric values on the two paths

Let's verify:

790+10+500 = 1300

on the other path you have

10+10+500  = 520

1300 /520 = 2,5 = 5/2

Hope to help

Giuseppe

Hi Giuseppe

This was very helpful. I have been struggling on the math for quite a long time on this. I will test this in my lab tonight & update the post.

Again, a great help !!

Regards

Vivek

Execellent Giuseppe...!! Thanks for the post Vivek

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