Which three descriptions are correct based on the exhibited output? (Choose three)
A. R1 is configured with the variance command.
B. The route to 10.2.0.0/16 was redistributed into EIGRP.
C. A default route has been redistributed into the EIGRP autonomous system.
D. R1 is configured with the ip summary-address command.
Answer: A C D
From the routing table above, we see that network 172.16.1. can be reached via 2 unequal paths (with FD of 23072000 & 20640000) so surely R1 has been configured with the “variance” command -> A is correct.
By configuring a default route and redistribute it into EIGRP you will get the line “D *EX 0.0.0.0/0 …” line in the routing table of that router -> C is correct.
From the line “10.2.0.0/16 is a summary, 00:16:18, Null0″ we know that this network has been summarized with the “ip summaray-address” command (notice that 10.2.0.0 is not the major network of net-> D is correct.
Refer to the exhibit. Which two statements are true? (Choose two)
A. The eigrp stub command prevents queries from being sent from R2 to R1.
B. The eigrp stub command will automatically enable summarization of routes on R2.
C. The eigrp stub command prevents all routes except a default route from being advertised to R1.
D. Router R1 will advertise connected and summary routes only.
E. Router R1 will advertise connected and static routes. The sending of summary routes will not be permitted.
F. Router R1 is configured as a receive-only neighbor and will not send any connected, static or summary routes.
Answer: A D
The command “eigrp stub” turns R1 into a stub router so R2 will never send any query to R1 because R2 knows that a stub router will only route packets for networks it has explicitly advertised -> A is correct.
The command “eigrp stub” is same as “eigrp stub connected summary” command because connected and summarized routes are advertised by default -> D is correct.
Note: Because the network 192.168.50.0 is not advertised by “network” statement, it is necessary to redistribute connected route with the “redistribute connected” command.
Refer to the exhibits. Router B should advertise the network connected to the E0/0/0 interface to router A and block all other network advertisements. The IP routing table on router A indicates that it is not receiving this prefix from router B.
What is the probable cause of the problem?
A. An access list on router B is causing the 192.168.3.16/28 network to be denied.
B. An access list on router B is causing the 192.168.3.32/28 network to be denied.
C. The distribute list on router B is referencing a numbered access list that does not exist on router B.
D. The distribute list on router B is referencing the wrong interface.
This is an unclear question. The question says “Router B should advertise the network connected to the E0/0/0 interface to router A and block all other network advertisements. The IP routing table on router A indicates that it is not receiving this prefix from router B.” That means the network 192.168.3.16/28 (including the IP 192.168.3.21/28) is not received on router A -> A is the most suitable answer.
Note: Distribute list are used to filter routing updates and they are based on access lists.
Study the exhibit carefully. What must be done on router A in order to make EIGRP work effectively in a Frame Relay multipoint environment?
A. Issue the command bandwidth 56 on the physical interface.
B. Issue the command bandwidth 56 on each subinterface.
C. Issue the command bandwidth 224 on each subinterface.
D. Issue the command bandwidth 224 on the physical interface.
In Frame Relay, all neighbors share the same bandwidth, regardless of the actual CIR of each individual PVC. In this case the CIR of each PVC is the same so we can find the bandwidth of the main interface (multipoint connection interface) by 56 x 4 = 224.
Notice that if the bandwidth on each PVC is not equal then we get the lowest bandwidth to multiply.
Refer to the exhibit. ROUTE Enterprises has many stub networks in their enterprise network, such as router B and its associated network. EIGRP is to be implemented on router A so that neither the prefix for the S/0/0/0 interface nor the prefixes from router B appear in the routing tables for the router in the enterprise network.
Which action will accomplish this goal?
A. Declare router B a stub router using the eigrp stub command.
B. Use the passive-interface command for interface Serial0/0/0.
C. Use a mask with the network command to exclude interface Serial0/0/0.
D. Implement a distribute list to exclude the link prefix from the routing updates.
If we declare router B a stub router then the routers in Enterprise Network still learn about the network for S0/0/0 interface and the network behind router B -> A is not correct.
If we use the passive-interface command on s0/0/0 interface then router A & B can not become neighbor because they don’t exchange hello messages -> A can not send traffic to the network behind B -> B is not correct.
Theoretically, we can use a distribute list to exclude both the link prefix and the prefix from router B but it is not efficient because:
+ We have many stub networks so we will need a “long” distribute list.
+ We declare networks in stub routers (like router B) while filter them out at router A -> it is a waste.
I am not totally sure about answer C because if we “use a mask with the network command to exclude interface Serial0/0/0″ then router A and B can not become neighbors and the situation is same as answer B. But from many discussions about this question, maybe C is the best answer.
*Mar 20 12:12:06: %DUAL-5-NBRCHANGE: IP-EIGRP 1: Neighbor 10.1.4.3 (Serial0) is down: stuck in active
*Mar 20 12:15:23: %DUAL-3-SIA: Route 10.1.1.0/24 stuck-in-active state in IP-EIGRP 1.