Refer to the exhibit. ROUTE.com has just implemented this EIGRP network. A network
administrator came to you for advice while trying to implement load balancing across part of
their EIGRP network. If the variance value is configured as 2 on all routers and all other
metric and K values are configured to their defaultvalues, traffic from the Internet to the
data center will be load balanced across how many paths?
A.
1
B.
2
C.
3
D.
4
E.
5
Explanation:
First we should list all the paths from the Internet to the data center:
+ A-B-C-H with a metric of 70 (40 + 15 + 15)
+ A-B-E-H with a metric of 60 (40+10+10)
+ A-D-E-H with a metric of 30(10+10+10)
+ A-D-E-B-C-H with a metric of 60 (10+10+10+15+15)
+ A-D-E-F-G-H with a metric of 70 (10+10+10+20+20)
+ A-F-G-H with a metric of 60 (20+20+20)
+ A-F-E-H with a metric of 40 (20+10+10)
So the path A-D-E-H will be chosen because it has the best metric. But EIGRP can support
unequal cost path load balancing. By configuring the variance value of 2, the minimum
metric is increased to 60 (30 * 2) and all the routes that have a metric of less than or equal
to 60 and satisfy the feasibility condition will beused to send traffic.
Besides the main path A-D-E-H we have 4 more paths that have the metric of less than or
equal to 60 (we also include the Advertised Distances of these routes for later comparison):
+ A-B-E-H with an AD of 20
+ A-D-E-B-C-H with an AD of 50
+ A-F-G-H with an AD of 40
+ A-F-E-H with an AD of 20
Now the last thing we need to consider is the feasible condition. The feasible condition
states:
“To qualify as a feasible successor, a router must have an AD less than the FD of the
current successor route”
The FD of the current successor route here is 30 (notice that the variance number is not
calculated here). Therefore there are only 2 paths that can satisfy this conditions: the path
A-B-E-H& A-F-E-H.
In conclusion, traffic from the Internet to the data center will be load balanced across 3
paths, including the main path (successor path) -> C is correct.