Usable Subnets Explanation

cardinalphin

Good morning,

I'm having a hard time understanding the answer to the following question:

You have an IP of 156.233.42.56 with a subnet mask of 7 bits. How many hosts and subnets are possible assuming that subnet 0 is not used?

126 subnets and 510 hosts

128 subnets and 512 hosts

510 subnets and 126 hosts

I know the correct answer is 126 subnets and 510 hosts. I believe I understand the 510 hosts. I don't fully understand the 126 subnets. Why do we subtract 2 when we're not enabling subnet zero instead of just 1? I feel like everything I've learned previously indicates we calculate subnets as 2^^n where n indicates the number of bits borrowed for the subnet. I get what the answer is, just don't understand the why.

Thanks in advance!

gespenstern

510 hosts is the right answer, that's for sure. There are 512 IPs for 32 (overall IPv4 bits) - 16 (Class B by definition) - 7 = 9 bits for hosts, which is 2^9=512 IPs, out of which the first one is the network ID address and the last one is the broadcast address, therefore 512-2=510.

Regarding subnets... It should be 2^7 which is 128, subtract the subnet 0 then we get 127.

The only explanation I have is they meant both "subnet zero" and "subnet all-ones", but forgot to mention the second in the question. Subnet zero is not all-ones subnet, but at some point the internet treated them the same -- it was not allowed/discouraged to use BOTH.

Overall, just a little rant, Cisco is so tiring with their hanging on this stupidity which is classful addressing and routing that is not in use for >20 years. Who cares? We are about to forget IPv4 altogether but rest assured long after IPv6 gets implemented in >99% networks Cisco will be teaching a full-on classful IPv4 addressing. My ass...

cardinalphin

Thanks for the clarification. That is exactly the information I was hoping to get. I'm just getting started in my CCNA prepwork and I don't recall coming across the subnet all 1's yet.

If anyone strolls across this thread later, Cisco has a page explaining this:
https://www.cisco.com/c/en/us/support/docs/ip/dynamic-address-allocation-resolution/13711-40.html

Thanks again for your help!

gespenstern

Good luck with your CCNA! As usual, I recommend to use the ICND1 + ICND2 route to the goal because of too much info. Wendell Odom, for example, has 2 insanely thick books of ~600 pages each and it's really hard, considering how much one has to memorize, to retain everything from them by the exam time.

Seadgs

cardinalphin wrote: »

Thanks for the clarification. That is exactly the information I was hoping to get. I'm just getting started in my CCNA prepwork and I don't recall coming across the subnet all 1's yet.

If anyone strolls across this thread later, Cisco has a page explaining this:
https://www.cisco.com/c/en/us/support/docs/ip/dynamic-address-allocation-resolution/13711-40.html

Thanks again for your help!

Here's a tip to get faster as well...

simply write out the binary conversion table for the last octet when figuring out your subnet mask etc... : 128 64 32 16 8 4 2 1

Cover up the 1 and boom there is your 2^(bits) table. Continue on as needed.

AvgITGeek

I always remember:
128 192 224 240 248 252 254 255 as subnets
128 64 32 16 8 4 2 1 as block size
You will need to adjust for number of hosts and subnets based on the class of network unless VLSM is in play.
I highly recommend you see what is happening in binary. Sometimes that is the quickest way for me to find the network address based on a host address and subnet mask.

willieb

Great replies...


Here's my logical thinking when getting a question like this on an exam. I've written it out for explanation but a lot of it can be done in your head after plenty of practice, or by writing down only the 2^n line:


This is a class B so the default subnet mask is /16 or 255.255.0.0 or 11111111.11111111.00000000.00000000


It's stating you have an additional 7 bits used for the network. This leaves 9 bits for hosts.


So this is /23 or 255.255.252.0 or 11111111.11111111.11111110.00000000


Looking at the 3rd and 4th octet of the SM in binary, which is where the subnetting is taking place:





Total Host bits = 9 = 2^9 = 512-2 not usable = 510 usable Hosts
Total Net bits = 7 = 2^7 = 128-1 subnet 0 = 127 Nets


Thinking logically to eliminate wrong answers, the last answer is obviously incorrect based on our answer above.


The question doesn't mention to remove the 2 non usable hosts, so hosts could be 512 or 510


The 2 choices for total subnets are 126 and 128. It said to assume subnet 0 is not used. For 128 it is used so that eliminates that answer.


This leaves one answer which is the one I would choose.


126 subnets and 510 hosts