Subnetting question

jekyll

Given the 194.168.14.0 IP address, we borrow 2 bits to create subnetworks. How many hosts can we have in each subnetwork?

Answer: 256

Why? Why 256 and not something else?

Jon_Cisco

That answer does not look right to me. Where are you taking the question and answer from?

stryder144

Since there is no slash notation used in the IP address, I will assume it is a class C network.  If you borrow 2 of the 8 host bits that are available, you get 6 bits for the hosts.  2 to the 6th power is 64, subtract 2 (one address for the network and one for the broadcast), that leaves you with 62 possible hosts...unless my math and/or logic is wrong.

kaiju

Where did you find this poorly worded trick question?

AvgITGeek

Probably one of those sites that will be blocked here.

kaiju

Based off of normal nth power math coupled to subnetting where you subtract the network and broadcast addresses, a possible answer would be 254 and not 256. To support 256 addresses you would need a /23 which would provide 510 addresses (512 - 2). 

Since 194.168.14.0 falls within the class C range we automatically know there are 24 network bits. If you borrow 2 bits from the remaining hosts bits you end up with a /26 So the answer that @stryder144 provided is correct. A /26 would support 62 hosts in each subnetwork.

stryder144

@kaiju...thank you for validating what I wrote.  My public math skills leave something to be desired.

jekyll

kaiju said:

Based off of normal nth power math coupled to subnetting where you subtract the network and broadcast addresses, a possible answer would be 254 and not 256. To support 256 addresses you would need a /23 which would provide 510 addresses (512 - 2). 

Since 194.168.14.0 falls within the class C range we automatically know there are 24 network bits. If you borrow 2 bits from the remaining hosts bits you end up with a /26 So the answer that @stryder144 provided is correct. A /26 would support 62 hosts in each subnetwork.

stryder144 said:

Since there is no slash notation used in the IP address, I will assume it is a class C network.  If you borrow 2 of the 8 host bits that are available, you get 6 bits for the hosts.  2 to the 6th power is 64, subtract 2 (one address for the network and one for the broadcast), that leaves you with 62 possible hosts...unless my math and/or logic is wrong.

One last question. Is it always 2 to the (8 minus the bits borrowed)th power minus 2, or does it vary depending on A, B or C class network types? An example, assuming a class B network, would be:

Given the 129.168.14.0 IP address, we borrow 2 bits to create subnetworks. How many hosts can we have in each subnetwork?

Would that also leave me with 62 possible hosts? I want to be able to calculate this for any network class. Perhaps there are mathematical steps involved that I am missing.

Jon_Cisco

Try this practice site. It helped me a lot to see the question and answer right away and then try another one.

https://www.subnetting.net/Subnetting.aspx?mode=practice

AvgITGeek

None of your questions are telling us what the subnet mask/newtork address is.

I would highly suggest learning the binary way to do this. Learning that made everything make sense to me when I was learning subnetting.

First step is to figure out what class network you have. A, B or C. Class A networks will always begin with a binary 0. Class B with a binary 1 (128)  and Class C with a binary 11 (192). Your network starts with 129 which means it is a class B network. The default subnet mask for class B is 255.255.0.0. If you use two bits for subnets you will have a 255.255.192.0 subnet mask which allows for 4 subnets and 16382 hosts.

In binary. Bold is default subnet mask for a class B network. Normal font is your 2 bits you are borrowing  for hosts which will give you four subnets (block size of 64) and the Italic are your host bits - 2.

11111111.11111111.11000000.00000000

Since I'm crappy at math, I always remember the following:

For network block size (Left to right): 128, 64, 32, 16, 8, 4, 2, 1 Number of networks: 2, 4, 8, 16, 32, 64, 128

Hosts are counted from right to left: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384 (subtract 2 from each to account for network and broadcast addresses)

stryder144

As @AvgITGeek stated above, binary helps us better visualize the network and the hosts.  The number of bits left after you have determined the network bits (subnet mask) is the number of bits left for the hosts.  For instance, a Class A network, such as 10.0.0.0, has 8 of the 32 bits, leaving 24 bits for hosts.  2 to the 24th power is 16,777,216, subtract the network address and the broadcast address and you get 16,777,214 assignable host addresses.  Using the 10.0.0.0 network, the network address is 10.0.0.0, the first assignable address for a host is 10.0.0.1, the last assignable address for a host is 10.255.255.254, and the broadcast address is 10.255.255.255.   

paul78

The OP did say 194.168.14.0. That IP falls into a /17 range allocated by RIPE to Virgin Media and the /24 is assigned to Futureview Internet Services. 

I love random questions.