Based on the exhibited output, which three statements are true? (Choose three.)
R1#show ip eigrp topology
IP-EIGRP Topology Table for process 200
Codes: P – Passive, A – Active, U – Update, Q – Query, R -Reply, r – Reply Status
P 192.168.1.64/28 1 successors, FD is 281600 via Connected,
Ethernet0
P 192.168.1.32/28 2 successors, FD is 40512000 via Connected,
Serial1
P 192.168.1.48/28 1 successors, FD is 40537600
via 192.168.1.66, (40537600/40512000), Ethernet0
via 192.168.1.17, (41024000/40512000), Serial0
via 192.168.1.33, (41024000/40512000), Serial1
P 192.168.1.16/28 1 successors, FD is 40512000 via Connected,
Serial0
A.
R1 is in AS 200.
B.
R1 will load balance between three paths to reach the 192.168.1.48/28 prefix because all
three paths have the same advertised distance (AD) of 40512000.
C.
The best path for R1 to reach the 192.168.1.48/28 prefix is via 192.168.1.66.
D.
40512000 is the advertised distance (AD) via 192.168.1.66 to reach the 192.168.1.48/28
prefix.
E.
All the routes are in the passive mode because these routes are in the hold-down state.
F.
All the routes are in the passive mode because R1 is in the query process for those
routes.
Explanation:
In the statement “IP-EIGRP Topology Table for process 200″, process 200 here means AS
200 -> A is correct.
There are 3 paths to reach network 192.168.1.48/28 but there is only 1 path in the routing
table (because there is only 1 successor) so the path with least FD will be chosen -> path
via 192.168.1.66 with a FD of 40537600 will be chosen -> C is correct.
The other parameter, 40512000, is the AD of that route -> D is correct.