Confused on a subnetting question....

Class B

9 subnet bits
7 host bits (-2)

Yes, but I though you could only use the last octet for the two networks 0 and 128...which is where I got 2 subnets...

I'm confused now!

No, the last octet is split 1-7

172.26.0.0 255.255.255.128

xxxxxxxx.xxxxxxxx.12345678.9xxxxxxx = 9 subnet bits
xxxxxxxx.xxxxxxxx.xxxxxxxx.x1234567= 7 host bits (-2)

HAMP wrote: »

I'm confused now!

Class B - default subnet mask ... so begin in 3rd octet

HAMP wrote: »

I'm confused now!

Agreed, here was my math...
172.26.0.0
255.255.255.128

256-128 = 128 which would be 0 and 128...or 2 subnets...right?! Unless there's something special about the 3rd and 4th octet being 0's that I just don't realize...Still don't see where 512 subnets is coming from

172.26.0.0 and 172.26.0.128

Xyro wrote: »

Class B - default subnet mask ... so begin in 3rd octet

Wait, I think I see it....hang on..

prdemon

the third octet in your network address is 0 and your subnetmask in the third octet is 255.
that means all 8 bits are for the network and you borrowed one bit from the last octet , thats where your 128 is comibg from

no!all! wrote: »

Agreed, here was my math...
172.26.0.0
255.255.255.128

256-128 = 128 which would be 0 and 128...or 2 subnets...right?! Unless there's something special about the 3rd and 4th octet being 0's that I just don't realize...Still don't see where 512 subnets is coming from

172.26.0.0 and 172.26.0.128

That is how I have known it to work. It would seem I am missing something also.

FloOz

You guys that are confused are looking at the network as class C. Remember the default mask for 172.16.0.0/16 is Class B.

xnx

no!all! wrote: »

So, I've been using subnettingquestions.com and I came across one that I'm stuck on...

Q: How many subnets and hosts per subnet can you get from the network 172.26.0.0 255.255.255.128?
A: 512 Subnets and 126 Hosts

I got 2 subnets and 126 hosts...where does 512 come from?!

Class B Address: /16 Default (255.255.0.0)
3rd octet = 255 = 8 bits
4th octet = 128 = 1 bit

9 bits total for number of subnets
2^9 = 512

7 bits left from the 16 for hosts
(2^7) -2 = 126 hosts per subnet

prdemon wrote: »

the third octet in your network address is 0 and your subnetmask in the third octet is 255.
that means all 8 bits are for the network and you borrowed one bit from the last octet , thats where your 128 is comibg from

I believe I need to revisit subnetting. I don’t remember a formula when the network is ‘0’, while the mask is ‘255’. I didn’t know that made a difference. I am so glad I found this site!!!!!

I was using the same formula that “no!all!” is using

xnx

HAMP wrote: »

I believe I need to revisit subnetting. I don’t remember a formula when the network is ‘0’, while the mask is ‘255’. I didn’t know that made a difference. I am so glad I found this site!!!!!

I was using the same formula that “no!all!” is using

No worries, subnetting can be tricky when you start learning, I can only do it easily because I learnt binary and hex many years ago

Yea, HAMP, you and I were thinking the same thing....thanks xnx for the info!

I went to the subnetting site to do a few questions, and I came across this one:

What valid host range is the IP address 172.24.89.220 255.255.255.224 a part of?

I gotten the correct answer, and I know how I did it.
172.24.89.193 through to 172.24.89.222

To me, I see having 8 subnets with 30 usable host each subnet (240 host)

I don't see 2048 subnets with, which I gotten from the formula you guys just gave.

omi2123

even though they used a class C subnet mask, but by default it's a class B network & the subnet counts from the third octet......512 subnet is correct...

HAMP wrote: »

What valid host range is the IP address 172.24.89.220 255.255.255.224 a part of?

To me, I see having 8 subnets with 30 usable host each subnet (240 host)

I don't see 2048 subnets with, which I gotten from the formula you guys just gave.

No, there are 11 subnets bits here. Again, it is a Class B you are working with.

xxxxxxxx.xxxxxxxx.12345678.123xxxxxx = 11 Subnet bits
xxxxxxxx.xxxxxxxx.xxxxxxxx.xxx12345 = 5 Host bits

11 Subnet bits = 2048
5 host bits = 32 (-2)

OfWolfAndMan

Just remember these numbers:
0-126 (First octet)= Class A. 255.0.0.0 Subnet mask
127 is a loopbavk address
128-191=Class B (255.255.0.0 default subnet mask)
192-223=Class C (255.255.255.0 subnet)
224 and up is generally multicast
After assessing class, look at how many bits they added. 172 is class B, therefore having 255 in third octet is 8 added net bits, along with the other one in the fourth octet. That leaves seven host bits. Subnets = 2^9 Hosts total = 2^7-2 (Which excluded network and broadcast)

I see what my problem is, but it’s still confusing. As what FloOz and omi2123 mentioned, it is a class B and not a class C. I’ve been looking at the mask and using a formula based on the mask classes and not the IP addy class. (did I say that right? lol). That seems to be a different formula that I not able to do In my head just yet.

OfWolfAndMan

The ones that aren't already subnetted are pretty simple once you get used to the formula. It's subnetting the IP addresses that are already subnetted that make it a bit more fun

You'll really want to know your subnetting as well when you start summarizing addresses in OSPF/EIGRP for your ABRs/ASBRs.