I'm in the process of practicing subnetting and came across this question, which I cannot find an explanation online.
The question:
Given the address 192.168.20.19/28, which of the following are valid host addresses on this subnet?
Answers are 192.168.20.29 and 192.168.20.17.
Now to my calculation. The fourth binary number below, which is 16, is part of the subnet mask. Which leaves 15 IPs that are available for hosts, and 1 for the broadcast address.
IP = 192.168.20.19/28 /28 = 255.255.255.240
Last octet 192.168.20.19 0 0 0 1 0 0 1 1
4th Subnet Mask octet 1 1 1 1 0 0 0 0
So, my question is, how are 192.168.20.29 and 192.168.20.17 valid host ids?
2 Answers
This seems like a trick question a bit because 192.168.20.19/28 is actually a host address itself.
You were on the right track with /28 using the 4th bit location, so addresses go in blocks of 16. There are actually 14 usable host addresses, and not 15. This is because one is taken up and used to identify the network, the other one is the broadcast address.
If the 4th bit is our deciding factor for address spaces, we know each address block increments by 16.
By writing a few of these network addresses, you get:
192.168.20.0 - .0 (network address), .1-.14 (hosts), .15 (broadcast)
192.168.20.16 - .16 (network address), .17-.30 (hosts), .31 (broadcast)
192.168.20.32 - .32 (network address), .33-.46 (hosts), .47 (broadcast)
After writing it out, you can see that 192.168.20.29 and 192.168.20.17 are valid host addresses forthe 192.168.20.16 network.
For what i know about network subnetting notation,
the subnet 192.168.20.19/28
starts with address 192.168.20.19 and ends with address 192.168.20.31 (last octet 0001111).
Then for me 192.168.20.17 is not a valid host of this subnet, while .29 is a valid host.
This subnet is made of 13 IP address (from 19 to 31).
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