A router received three destination prefixes: 10.0.0.0/8, 10.0.0.0/16, and 10.0.0.0/24. When the show ip route command is executed, which output does it return?
A.
Gateway of last resort is 172.16.1.1 to network 0.0.0.0 o E2 10.0.0.0/8 [110/5] via 192.168.1.1, 0:01:00, Ethernet0 o E2 10.0.0.0/16[110/5] via 192.168.2.1, 0:01:00, Ethernet1 o E2 10.0.0.0/24[110/5] via 192.168.3.1, 0:01:00, Ethernet2
B.
Gateway of last resort is 172.16.1.1 to network 0.0.0.0 o E2 10.0.0.0/8 [110/5] via 192.168.1.1, 0:01:00, Ethernet0
C.
Gateway of last resort is 172.16.1.1 to network 0.0.0.0 o E2 10.0.0.0/24[110/5] via 192.168.3.1, 0:01:00, Ethernet2
D.
Gateway of last resort is 172.16.1.1 to network 0.0.0.0 o E2 10.0.0.0/16[110/5] via 192.168.2.1, 0:01:00, Ethernet1 o E2 10.0.0.0/24[110/5] via 192.168.3.1, 0:01:00, Ethernet2
A new route will replace an old route only if it has a better AD. In our case the three routes have the same AD with different prefix lengths, this means all 3 routes will be installed in the routing table.
Explanation
For the purpose of installing routes in the routing table, the router considers different prefix lengths as different destinations.
That is why multiple routes from the same and/or different routing protocols are installed in the routing table.
Tie breaker is longest match rule, that selects the route with the longest subnet mask (prefix length) from among routes already in the routing table.
Reference: https://www.cisconetsolutions.com/introduction-to-routing-protocols/
so A is correct
Can someone explain. At first i thougt answer A is right because every präfix have his own next hop interface but also answer B could be possible, because the /16 and /24 subnets are included in the /8 subnet?
right, it could be B, i take what i said earlier back lol je ne connais pas la reponse maintenant lol
upvoted 2 times
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