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Answer for:

ip/subnets

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jedi_k9master

Answer is C: 192.168.1.40/29

.248 mask uses 5 bits (1111 1000).
.42 IP in binary is (0010 1010)

The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP ...
1111 1000 AND
0010 1010 equals
0010 1000 - which is .40

/24 is standard class C mask.
adding the 5 bits from the .248 mask gives /29
thus answer C

As an aside: the broadcast address would be the HIGHEST binary value without changing the output of an AND operation of the subnet mask and IP ...

0010 1000 - Base subnet
0010 1111 - (.47) Last three 1's is the highest value without changing the AND operation .. as in .. making it .48 would result in 0011 0000 which changes the outcome of the AND operation against the subnet mask)
Thus broadcast address is: .47
usable addresses: .47 - .40 = 7 .. less 2 = 5 usable addresses in that range
(take away 2 because the base subnet and broadcast address can't be used as end device IPs)

Hope that makes sense.
I'm admittedly quite rusty on my IP theory so if I've got anything wrong, please excuse me .. 9and others please feel free to correct me.