# How to find the subnets?

Posts: 2Registered Users ■■■□□□□□□□

Here is the question and answer I can get the hosts no problem 2^10 -2 1022 not to bad . Now to find the subnets I don't remember how to do it. I tried to read the guides on here, but for some reason I still just don't get it. If some one could just do a quick run through of how to find them that would be greatly appreciated.
Question: How many subnets and hosts per subnet can you get from the network 172.23.0.0/22?
Answer: 64 subnets and 1022 hosts

Also I have a second question.

I saw this chart in a different post (chart below) how could I use it to help me with this question. I have my own method of using this chart, but I think there is a more efficient way. Could someone could explain how they would use this chart? I memorized this char awhile age in high school and it stuck with me I dont know why, but I feel it can only help on the exam

Chart:

128 192 224 240 248 252 254 255
25 26 27 28 29 30 31 32
17 18 19 20 21 22 23 24
128 64 32 16 8 4 2 1
127 63 31 15 7 3 1 0

Chart is supposed to be formatted. I am sure you guys can figure it out.

Thanks,
Jake

• Posts: 111Member
/22 makes the 3rd octet the most interesting octet( /8+/8+/6 )with the networks being as followed:

172.23.0.0 - 172.23.3.255
172.23.4.0 - 172.23.7.255
172.23.8.0 - 172.23.11.255

So for example if you had an ip address of 172.23.5.250,it would be in the 172.23.4.0 subnetwork with a broadcast of 172.23.7.255.......
• Posts: 44Member ■■□□□□□□□□
first you have to look at the first octet to determine what is the class of the network.
172 means its a class B network, with netmask /16
the network in question subnet this class B network by a /22 mask, so extra network bits
generated is 22-16 = 6, thats why 64 subnets.
all the remaining bit will be the host bits, which means host part contain 10 bits,
thats why 1024 hosts(minus 2, the all 0s and all 1s)=1022 hosts

for the second question, i just burn into my mind the following
128 192 224 240 248 252 254 255
128 64 32 16 8 4 2 1(and the fact that all the bits value adding up to the right of any number X = X-1.eg. 8-1=4+2+1)

although memorize all yours suggestion wont hurt. but its too much for me to memorise. ^-^
• Posts: 879Member
The question is confusing because it should be saying something like "if you have a /16 network, how many /22 subnets does it contain?". With the current wording of the question you just have to assume that you have a "class B" mask of 255.255.0.0 (/16) and splitting that into /22s. If you don't make these assumptions, the question doesn't have 1 single answer because you could subnet 172.23.0.0/22 into /23s, and then subnet those further into smaller subnets, which gives you the answer 2 to a lot (1024?).

It would be helpful if they could finally give up on classes - it's been what, 20 years or so since CIDR was introduced?
• Posts: 10Member ■□□□□□□□□□
I'd recommend looking at "Routing TCP/IP, Volume 1, Second Edition" Chapter 1 section IPv4 Addresses. This gives a very good explanation on IP addresses, subnetting and how to find them and troubleshoot them. I found the best way for me to learn subnetting is to break it down to the binary level and figure it out that way. After that learn the tricks to get through it quickly in a way you understand.
• Posts: 160Member ■■■□□□□□□□
to find the subnets u have to count how many 1's you have after the default.. in this example the default would be /16 (class since you subnet to /22, you have 6 more bits, than is the same as the hosts except for that -2 part. so it would be 2^(6) = 64.