A company has a forest with 4 sites. Subnets are as follows:
MainOffice 172.16.1.0 Subnet: 255.255.255.0 Gateway 172.16.1.254
Site1 192.168.12.0 Subnet: 255.255.255.0
Site 2 192.168.13.0 Subnet: 255.255.255.0
Site 3 192.168.14.0 Subnet: 255.255.255.0
Site 4 192.168.15.0 Subnet: 255.255.255.0
You add a new server to the MainOffice and it needs to be able to communicate to all sites.
Which route command would you run?
A.
route add -p 192.168.8.0 netmask 255.255.252.0 172.16.1.254
B.
route add -p 192.168.0.0 netmask 255.255.248.0 172.16.1.254
C.
route add -p 192.168.12.0 netmask 255.255.252.0 172.16.1.254
D.
route add -p 192.168.12.0 netmask 255.255.240.0 172.16.1.254
What I understand is that if you will have 6 subnets, then you need 6 bits for the network ID. What means you would have a mask 255.255.252.0
||||||||.||||||||.||||||00.00000000
255.255.252.0
252 = 128 + 64 + 32 + 16 + 8 + 4 (6 bits)
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192.168.12.0: 11000000 10101000 00001100 00000000
192.168.15.255: 11000000 10101000 00001111 11111111
^^^^^^^^ ^^^^^^^^ ^^^^^^– ——– 22 bits
Count the matching bits.
192.168.12.0/24: 11000000 10101000 00001100 xxxxxxxx
192.168.13.0/24: 11000000 10101000 00001101 xxxxxxxx
192.168.14.0/24: 11000000 10101000 00001110 xxxxxxxx
192.168.15.0/24: 11000000 10101000 00001111 xxxxxxxx
^^^^^^^^ ^^^^^^^^ ^^^^^^– 22 bits
/22- The prefix length for the route, all identical bits amount to 192.168.12.0/22 –
convert it to a mask, 255.255.252.0.
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Alignment go messed up.
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