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!
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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.
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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
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