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EIGRP Questions

March 23rd, 2017 in ICND2 200-105 Go to comments

Question 1


The “show ipv6 eigrp neighbors” command displays the neighbors discovered by the EIGRPv6. Notice that the neighbors are displayed by their link-local addresses.


Question 2


From the routing table we learn that network is learned via 2 equal-cost paths ( & -> traffic to this network will be load-balancing.

Question 3


To configure EIGRP for IPv6 we must explicitly specify a router ID before it can start running. For example:

ipv6 router eigrp 1
eigrp router-id
no shutdown

Notice that EIGRP for IPv6 router-id must be an IPv4 address. EIGRP for IPv4 can automatically pick-up an IPv4 to use as its EIGRP router-id with this rule:
+ The highest IP address assigned to a loopback interface is selected as the router ID.
+ If there are not any loopback addresses configured, the highest IP address assigned to any other active interface is chosen as the router ID

EIGRPv3 also uses the AS number (for example: ipv6 eigrp 1 under interface mode).

Question 4

Question 5


We notice that 3 routers are using different AS numbers so they do not become neighbors and cannot exchange their routing updates. We need to choose only one AS number and use it on all 3 routers to make them exchange routing updates.

In this case we don’t need to use the “no auto-summary” command because network is not separated by another major network.

Question 6


When a packet with destination IP address of arrives at HokesB, it will look up in the routing table to find the most specific path. In this case no path is found so HokesB must inform to the source host that the destination is unreachable on the interface it has received this packet (it is Fa0/0 because the network is learned from this interface). So the best answer here should be C – send an ICMP message out of Fa0/0.

Question 7


First we must notice that all the 4 answers are parts of the “show ip eigrp topology” output. As you can see, there are 2 parameters in the form of [FD/AD] in each answer. For example answer C has [46152000/41640000], it means that the FD of that route is 46152000 while the AD is 41640000.

To become a feasible successor, a router must meet the feasibility condition:

To qualify as a feasible successor, a router must have an AD less than the FD of the current successor route

In four answer above, only answer B has an AD of 128256 and it is smaller than the FD of the current successor route (41152000) so it is the feasible successor -> B is correct.

Question 8


Feasible successor is a route whose Advertised Distance is less than the Feasible Distance of the current best path. A feasible successor is a backup route, which is not stored in the routing table but stored in the topology table.

Question 9


The formula to caculate EIGRP metric is:

metric = [K1 * bandwidth + (K2 * bandwidth)/(256 – load) + K3 * delay] * [K5/(reliability + K4)]

By default, K1 = 1, K2 = 0, K3 = 1, K4 = 0, K5 = 0 which means that the default values use only bandwidth & delay parameters while others are ignored. The metric formula is now reduced to:

metric = bandwidth + delay

Note: But remember the bandwidth here is defined as the slowest bandwidth in the route to the destination & delay is the sum of the delays of each link.

Question 10

Comments (5) Comments
  1. Anonymous
    March 30th, 2017

    Why F is correct? R3 is not advertised that network right?

  2. Anonymous
    March 30th, 2017

    ^ its on Q10….

  3. gotthatpma
    April 17th, 2017

    Yeah, can anyone clarify question 10?

  4. yup yup
    April 21st, 2017

    Yes, please provide explaination to question 10.

  5. Kumo
    April 24th, 2017

    Q10. – class A – class B – class C

    Because auto-summary is enabled, eigrp will behave like a classful protocol.
    Therefore only route with default mask will be shown in the routing table.