ICND2 – EIGRP Questions

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Question 1

[am4show have=’p2;’]Which EIGRP for IPv6 command can you enter to view the link-local addresses of the neighbors of a device?

A. show ipv6 eigrp 20 interfaces
B. show ipv6 route eigrp
C. show ipv6 eigrp neighbors
D. show ip eigrp traffic

 

Answer: C[/am4show]

Explanation

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

[am4show have=’p2;’]Refer to the exhibit. Based on the exhibited routing table, how will packets from a host within the 192.168.10.192/26 LAN be forwarded to 192.168.10.1?

A. The router will forward packets from R3 to R2 to R1
B. The router will forward packets from R3 to R1
C. The router will forward packets from R3 to R1 to R2
D. The router will forward packets from R3 to R2 to R1 AND from R3 to R1

 

Answer: D[/am4show]

Explanation

From the routing table we learn that network 192.168.10.0/30 is learned via 2 equal-cost paths (192.168.10.9 &192.168.10.5) -> traffic to this network will be load-balancing.

Question 3

[am4show have=’p2;’]Which option describes a difference between EIGRP for IPv4 and IPv6?

A. Only EIGRP for IPv6 advertises all connected networks.
B. Only EIGRP for IPv6 requires a router ID to be configured under the routing process
C. AS numbers are configured in EIGRP but not in EIGRPv3.
D. Only EIGRP for IPv6 is enabled in the global configuration mode.

 

Answer: B[/am4show]

Explanation

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 2.2.2.2 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

[am4show have=’p2;’]Refer to the exhibit. If the router R1 returns the given output and has not had its router ID set manually, what address will EIGRP use as its router ID?

A. 192.168.1.2
B. 172.16.4.1
C. 192.168.10.2
D. 1.1.1.1

 

Answer: D[/am4show]

Question 5

[am4show have=’p2;’]When running EIGRP what is required for RouterA to exchange routing updates with RouterC?

A. AS numbers must be changed to match on all the routers.
B. Loopback interface must be configured so a DR is elected.
C. The no auto-summary command is needed on Router A and Router C.
D. Router B needs to have two network statements, one for each connected network.

 

Answer: A[/am4show]

Explanation

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 10.0.0.0 is not separated by another major network.

Question 6

[am4show have=’p2;’]Refer to the exhibit. A packet with a source IP address of 192.168.2.4 and a destination IP address of 10.1.1.4 arrives at the HokesB router. What action does the router take?

A. forwards the received packet out the Serial0/0 interface
B. forwards a packet containing an EIGRP advertisement out the Serial0/1 interface
C. forwards a packet containing an ICMP message out the FastEthemet0/0 interface
D. forwards a packet containing an ARP request out the FastEthemet0/1 interface

 

Answer: C[/am4show]

Explanation

When a packet with destination IP address of 10.1.1.4 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 192.168.2.0/28 is learned from this interface). So the best answer here should be C – send an ICMP message out of Fa0/0.

Question 7

[am4show have=’p2;’]Refer to the exhibit. Given the output from the show ip eigrp topology command, which router is the feasible successor?

Router# show ip eigrp topology 10.0.0.5 255.255.255.255 IP-EIGRP topology entry for 10.0.0.5/32 State is Passive, Query origin flag is 1, 1 Successor(s), FD is 41152000

A.

10.1.0.3 (Serial0), from 10.1.0.3, Send flag is 0x0
   Composite metric is (46866176/46354176), Route is Internal 
   Vector metric:
     Minimum bandwidth is 56 Kbit
     Total delay is 45000 microseconds
     Reliability is 255/255
     Load is 1/255
     Minimum MTU is 1500
     Hop count is 2 

B.

10.0.0.2 (Serial0.1), from 10.0.0.2, Send flag is 0x0
   Composite metric is (53973248/128256), Route is Internal 
   Vector metric:
     Minimum bandwidth is 48 Kbit
     Total delay is 25000 microseconds
     Reliability is 255/255
     Load is 1/255
     Minimum MTU is 1500
     Hop count is 1

C.

10.1.0.1 (Serial0), from 10.1.0.1, Send flag is 0x0
   Composite metric is (46152000/41640000), Route is Internal Vector metric:
     Minimum bandwidth is 64 Kbit 
     Total delay is 45000 microseconds 
     Reliability is 255/255 
     Load is 1/255 
     Minimum MTU is 1500
     Hop count is 2

D.

10.1.1.1 (SerialO.1), from 10.1.1.1, Send flag is 0x0
   Composite metric is (46763776/46251776), Route is External 
   Vector metric:
     Minimum bandwidth is 56 Kbit
     Total delay is 41000 microseconds
     Reliability is 255/255
     Load is 1/255
     Minimum MTU is 1500
     Hop count is 2

 

Answer: B[/am4show]

Explanation

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

[am4show have=’p2;’]Which type of EIGRP route entry describes a feasible successor?

A. a backup route, stored in the routing table
B. a primary route, stored in the routing table
C. a backup route, stored in the topology table
D. a primary route, stored in the topology table

 

Answer: C[/am4show]

Explanation

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

[am4show have=’p2;’]What are the two default metrics used by EIGRP for route selection? (Choose two)

A. Bandwidth
B. Delay
C. Reliability
D. Load
E. MTU

 

Answer: A B[/am4show]

Explanation

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

[am4show have=’p2;’]Refer to the exhibit.

Which three EIGRP routes will be present in the router R4’s routing table? (Choose three)

A. 172.16.1.0/24
B. 10.1.10.0/30
C. 10.0.0.0/8
D. 10.1.11.0/30
E. 172.16.0.0/16
F. 192.168.1.0/24

 

Answer: C E F[/am4show]