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Question 28/48
You have a server named Server1 that runs Windows Server 2016. Server1 has the following routing table.
What will occur when Server1 attempts to connect to a host that has an IP address of 172.20.10.50?
What will occur when Server1 attempts to connect to a host that has an IP address of 172.20.10.50?
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I would say the answer is A!
IP: 172.20.10.50 applies to route: 172.16.0.0 / 255.240.0.0
IP: 172.16.0.1
= 1010 1100.0001 0000.0000 0000.0000 0001
NM: 255.240.0.0
= 1111 1111.1111 0000.0000 0000.0000 0000
Here you see the first 12 bits determine the Network ID
NID: 172.16.0.0
IP: 172.20.10.50
= 1010 1100.0001 0100.0000 1010.0011 0010
It has the same first 12 bits as IP: 172.16.0.1
So IP: 172.16.0.1/12 is in the same Network as IP: 172.20.10.50/12
Now I hope to have proven to you that 172.20.10.50 applies to the route 172.16.0.0 / 255.240.0.0
(The last IP in NID:172.16.0.0 with 12 bits: 172.31.255.255/12
IP: 172.31.255.255
= 1010 1100.0001 1111.1111 1111.1111 1111
It has the same first 12 bits as IP: 172.16.0.1
Address range: 172.16.0.1 - 172.31.255.254
)