Which one of the following IP addresses is the last valid host in the subnet using mask
255.255.255.224?
A.
192.168.2.63
B.
192.168.2.62
C.
192.168.2.61
D.
192.168.2.60
E.
192.168.2.32
Explanation:
With the 224 there are 8 networks with increments of 32
One of these is 32 33 62 63 where 63 is broadcast so 62 is last valid host out of given choices.
I don’t get it! If .244 is the last octet in the mask, then the binary equivalent is 11100000. 128+64+32= 224. The last five zeros in the host portion would leave you with 16, 8, 4, 2 , and 1. Added up, they equal 31. 31 (00011111) would be a broadcast. This would leave you with 1 – 30 reserved for hosts, and 0 reserved for the network ID digit.
.62 would be represented as 00111110, which crosses over into the network portion of the mask.
Please enlighten me, if I’m wrong!
You are assuming that the first subnet is being used, where all three network bits in the last octet are 0. However, there are 8 subnets available with this subnet mask. So the final octet in this case would be 001|00000, or the 192.168.2.32 network. The addresses in this network are 192.168.2.32 – 192.168.2.63, where 192.168.2.32 is the network address, and 192.168.2.63 is the broadcast address, leaving 192.168.2.62 as the last valid host.
To answer such question you have to convert all options in binary. let’s say
.63 = 00111111
.62 = 00111110
.61 = 00111101
.60 = 00111100
.32 = 00100000
now to qualify for that last valid host, that address’s last bit should be 0 and rest of the host bits should be 1. so there is just one address which qualify and that is .62
hope that explains