Subnetting Class B, but My brain is giving me the silent treatment now.

sarahanand

I have completed almost (at least I think), mastering class C. But can't figure out how to determine Class B range.

SysnetNotes

Whats the problem you facing while subnetting class B ?

I will try to explain as i know

Class B Range is from 128 - 191 in the first octet
Class B have 2 network portion and 2 host portion

NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH
[TWO NETWORK PORTION AND TWO HOST PORITION]

NOTE:After each octet consider that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.While subnetting bits will always move to your right side.No change will occur in Octets having 255 in subnets. We can find block size by reducing our CIDR value from next Boundary.

Below are some example

Q :What is the broadcast address of the network 172.27.216.80/28

Our mask is /28.Next Boundary is 32. So 32-28 = 4 Now 2^4 =16 which gives our block size

Given network is 172.27.216.80/28 .We get block size is 16.So next network will be 172.27.216.96.Broadcast address will be the last ip address of 172.27.216.80 network.So it will be 172.27.216.95

Q : What valid host range is the IP address 172.20.232.102 255.255.255.0 a part of?

Here subnet mask is given as 255.255.255.0
Valid host = 2^[Off bits ] -2 means 2^8 -2 =254.we know 2^8 is 256.

Note :Off bits means number of 0’s . 11111111.11111111.1111111.00000000

Q: What is the first valid host on the subnetwork that the node 172.30.119.235 255.255.255.128 belongs to?

Here subnet mask is 255.255.255.128 [11111111.11111111.1111111.10000000 ] .Here CIDR is /25.next boundary is 32.So 32-25 = 7 .So block size is 2^7 = 128

Network address of 172.30.119.235 is 172.30.119.128 and first valid host is 172.30.119.129

jvrlopez

What don't you get?

Every time the last octet "fills" up to 255, the third octet increases by 1.

Here's a sample questions to help you.

Question: What valid host range is the IP address 172.22.69.245/20 a part of?


/20 means that 20 of the possible 32 bits are turned "on" (1s) in each of the 4 octets on your subnet mask, starting from the left.

This reads as 11111111.11111111.11110000.00000000 or 255.255.240.0

Since the first two octets are full, they cannot provide any bits for hosts. Since the 3rd octet has some bits left over to play with, we will start there.

Looking at your mask and your last octet that has some bits left over, take 240 and subtract that from 256. This gives you 16. This is your block size. Every multiple of 16 in that octet will be a valid subnet. Forget about the last octet, we're just "dragging" it along with us.

So with 172.22.0.0, our subnets are as follows:

172.22.0.0
172.22.16.0
172.22.32.0
172.22.48.0
172.22.64.0
172.22.80.0
172.22.96.0
172.22.112.0
172.22.128.0
172.22.144.0
172.22.160.0
172.22.176.0
172.22.192.0
172.22.208.0
172.22.224.0
172.22.240.0

We stop here because another increment of 16 would put us over 255 in the third octet.

So for example, the range of the subnet 172.22.80.0 would be as follows:

172.22.80.0 - Network ID
172.22.80.1 - First usable host IP
172.22.95.254 - Last usable host IP
172.22.95.255 - Broadcast address

Every 1 increase in the 3rd octet means that 255 in the last octet have gone by.

Can you use that to answer the original subnetting question at the start of my post?

instant000

There's a few ways you can do subnetting:

1 - Know the binary math behind it.
2 - Memorize it all.
3 - Learn some tricks.

The mistake some make is that they go to step 3, without doing them in order, starting at 1.

Truth be told, you'd be much better off (long term) to take this approach:
1 - Learn the binary math
2 - Memorize some stuff: Powers of 2, Masks, etc.
3 - Practice on lots of subnetting problems

Once you practice on enough problems, you'll begin to see the patterns in the numbers, and it won't be such a big deal anymore.

For example: If given 3x2, you have memorized the answer as 6.
Of course, you could also collect three objects and add three more then count or (three objects and count them twice) or (two objects and count them three times). Either way, you'd get the answer of 6.

The point is that because you know the actual process of multiplication, when you memorize the multiplication table, you still know what is going on in the background. Then, when you work on larger problems, such as 32 x 4 = 128, you know what is actually going on.

I advise you to go ahead and invest the time in knowing the binary math now. You'll be much further ahead later on, when you start to look at wildcard masks.

You need to get to the point where you can do the subnet math in your head. It's OK to write out a table before a test, when you're trying to minimize errors, but you want to be able to do the problems in your head.

Hope this helps.

sarahanand

Is it between 172.22.64.0 and 172.22.79.255?

jvrlopez

Real close!

You got the concept down, but the question asks for the "valid host range" which means it cannot include the network ID 172.22.64.0 (the first IP in the subnet) or the broadcast address of 172.22.79.255 (the last IP in the subnet). These two addresses cannot be given to a host in any subnet.

So our valid range would be 172.22.64.1 through 172.22.79.254.

Hopefully you can start to see how class B address work in this context!

sarahanand

I so appreciate posting this query. I was almost there, but you're explanations made it more clearer. Thank u very much.

jvrlopez

You're welcome.

Practice makes perfect. It'll come with ease the more you do it. After a while you should be able to visualize it and have it down.