subnetting question help

ipcipher

I am not sure how the answer to the questions below are correct. Can someone break it down and explain how they arrived at the answer?

Question: What is the first valid host on the subnetwork that the node 172.21.67.191/23 belongs to?
Answer: 172.21.66.1

Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0 255.255.254.0?
Answer: 128 subnets and 510 hosts

RynoR

Hi ipcipher, have a look at this article below i think the explanations are quite good. Also there is allot of resources on subnetting available lots of videos on youtube to check out.

http://www.networkworld.com/article/2224957/cisco-subnet/ccna-prep--learning-how-to-subnet-properly.html

davenull

It seems like everyone has their own method of subnetting.

For the first question, the mask is /23, the next boundary is /24. 24 - 23 = 1. You raise 2 to the power of 1 to find the 'magic number', which is 2. Your networks are going to be 172.21.0.0, 172.21.2.0, 172.21.4.0 etc. You see that you got 67 there, so the network that it belongs to is 66, where the first host is 172.21.66.1

For the second question, it looks like 7 bits were borrowed for subnets, so you have 2**7 = 128 subnets. You still have the whole 4th octet and 1 more bit from the 3rd octet for your hosts, so the total number of bits for your hosts is 8+1=9. Then 2**9 = 512, subtract 2 for network and broadcast address to get 510.

spacenoxx

ipcipher wrote: »

I am not sure how the answer to the questions below are correct. Can someone break it down and explain how they arrived at the answer?

Question: What is the first valid host on the subnetwork that the node 172.21.67.191/23 belongs to?
Answer: 172.21.66.1

Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0 255.255.254.0?
Answer: 128 subnets and 510 hosts

I always use the logic of doing
1. 2^ (/32 - /X) = Hosts (where X is more than 24)
2. 2^ (/24 - /X) = C Class Networks (as in 256) (Where X is between 16-24)
3. 2^ (/16 - /X) = B Class Networks (as in 256) or C Class Networks (65536) (where X is between 8-16)

It looks more complicated but it isnt. For instance /21 = 2^3 (24-21 =3) = 8 C Class Subnets

which helps when breaking a 172.16.0.0 network into possible subnets such as 172.16.0.0, 172.16.8.0, 172.16.16.0 so on

Similarly /13 in a 10.0.0.0 network means 2^3 (16-13=3) = 8 B Class networks such as 10.0.0.0, 10.8.0.0, 10.16.0.0 and so on..

To Answer your question
1: 172.21.67.191/23 => 24-23 => 1 = 2^1 = 2 C Class Subnets = 172.21.0.0, 172.21.2.0 ....172.21.66.0, 172.21.68.0 (stop here)
So 172.21.66.0 as 172.21.68.0 is past the given IP
Valid Host = 172.21.66.1 - 172.21.67.254

172.30.0.0 255.255.254.0?, 254 and /23 are the same. In this case when mask is given in DDN you use 255 instead of /32 or /24 etc. Which means

256-254 = 2X (where X is IPs if the "interesting octet" is 4th)
256-254 = 2X (where X is C Class Subnets if the "interesting octet" is 3rd
256-254 = 2X (where X is B Class Subnets if the "interesting octet" is 2nd

In the above question its in 3rd octet so 2 C Class Subnets

The Subnet size got Doubled ( as in 2 C Class = 512 Hosts = 510 valid Hosts)
so Number of Subnets gets Halved = 256/2 = 128 Subnets

Atleast thats how I think in my head.

JockVSJock

You can always check your answers here:

Online IP Subnet Calculator