EIGRP

[aliy3507]

Can someone help me finding appropriate answer to the question in attached snapshot.

which one would be correct option

1) network 172.16.0.0 0.15.255.255

2) network 128.0.0.0 127.255.255.255

[MHM Cisco World]

no attachment 

[Flavio Miranda]

Hi

1) network 172.16.0.0 0.15.255.255

you can use online calculator for this

https://asecuritysite.com/ip/routesum

[dovega]

1) network 172.16.0.0 0.15.255.255

[MHM Cisco World]

172.20.0.0 0.0.127.255 <- this right for me' why it include all IP for both interface.

[Elliot Dierksen]

The inverse mask in the "network" statement doesn't need to match exactly the physical interface that you are trying to include. It needs to be wide enough to INCLUDE the physical interface you are trying to match. The route advertised from that that network will match the subnet of the physical interface.

[Valerio Passeri]

A) network 128.0.0.0 127.255.255.255

Just convert the wildcard mask to the corresponding subnet mask:

127.255.255.255 -> 128.0.0.0

Then, does network 128.0.0.0/1 include both interfaces on R1? Yes, but it's too broad a statement. Let's look a little further.

B) network 172.16.0.0 0.0.255.255

This equals network 172.16.0.0/255.255.0.0 = 172.16.0.0/16

This seems to be the correct answer, but I would take a look to the other ones, to see if there is a more specific network statement.

c) network 172.20.0.0 0.0.127.255

It's network 172.20.0.0/255.255.128.0

Then,

172.20.20.17 AND 255.255.128.0 = 172.20.0.0

172.26.20.12 AND 255.255.128.0 = 172.26.0.0

To me, wrong answer.

D) network 172.20.0.0 0.3.255.255

It's 172.20.0.0/255.252.0.0

Then,

172.20.20.17 AND 255.252.0.0 =172.20.0.0

172.26.20.12 AND 255.252.0.0 = 172.20.0.0

To me, correct answer.