# EIGRP variance

May 5th, 2007

Do you find this answer correct? cause I am thinking that best answer is D and not B.

=============

The TestKing EIGRP network is displayed in the following topology diagram:

|------20------ R1 -----10------------|

|

R5------10------ R3 ------10------ R2

|

|------20------ R4 ------25-----------|

You work as a network technician at TestKing.com. Study the exhibits carefully. If the command "variance 3" was added to the EIGRP configuration of TestKing5, which path or paths would be chosen to route traffic from TestKing5 to network X?

A. TestKing5-TestKing2-TestKing1

B. TestKing5-TestKing2-TestKing1 and TestKing5-TestKing3-TestKing1.

C. TestKing5-TestKing3-TestKing1 and TestKing5-TestKing4-TestKing1.

D.TestKing5-TestKing2-TestKing1,TestKing5-TestKing3-TestKing1, and TestKing5-TestKing4-TestKing1.

Explanation:

In this question the variance 3 command is used. Since the lowest metric is 10. 10*3 = 30

meaning we can use all routes with a metric of 30 and under which is answer B(5-2-1 and

===========

Friend,

You need to always remember that the route should satisfy the FC first i.e the Advertised distance should be less than FD of the current successor

When the variance is set to 3, the metric is changed to 20*3=60.

Now the path through R1 which satisfies the FC, has a metric of 30 and hence is entered into the routing table.

But if you consider the path through R4, the metric may be 45 which is less than 60 but it does not satisfy the FC at all.

The advertised distance is still 25 which is greater than the FD (20) via R3 and hence not entered in the routing table

Actually this example is taken from the cisco site itself

http://www.cisco.com/warp/public/103/19.html#topic1

HTH, rate if it does

Narayan

Overall Rating: 4.5 (4 ratings)

## Replies

royalblues Mon, 05/07/2007 - 09:12

Friend,

You need to always remember that the route should satisfy the FC first i.e the Advertised distance should be less than FD of the current successor

When the variance is set to 3, the metric is changed to 20*3=60.

Now the path through R1 which satisfies the FC, has a metric of 30 and hence is entered into the routing table.

But if you consider the path through R4, the metric may be 45 which is less than 60 but it does not satisfy the FC at all.

The advertised distance is still 25 which is greater than the FD (20) via R3 and hence not entered in the routing table

Actually this example is taken from the cisco site itself

http://www.cisco.com/warp/public/103/19.html#topic1

HTH, rate if it does

Narayan

mounir.mohamed Mon, 05/07/2007 - 11:39

Dear Mohamed,

the variance command specify the average between the metrics

of the routes that will selected to form unequal-load balance,

personally i'm using this formula to calalcue

the average between both routes and use it In the variance

command Lower-metric/high-metric=variance 10003712/2503680=3.99 So 4 is the magic number in such case.

Just take a look in Jeff Doyle V1 2nd-ed, Chapter 7.

Best Regards,

Mounir Mohamed