Manually subnetting

Hi!!

I'm busy studying for my 70-291 exam and actually I think and it goes well except manually determine subnets without a calculator….. I just can’t find a way to do it without a scientific calculator or some kind of table – neither allowed to the exam! Can anyone help me with a step-by-step guide of how to do this.

Example:

Networks needed (subnets): 20
Max Hosts on each network: 600
IP Address to start from: 130.131.0.0 / 16 (B class)

How do I calculate the custom subnet for this + the start and end IP address of each network?

Thank you in advance,
Kristian
Kristian_MoritsenAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

bmquintasCommented:
I used this site for intro, http://www.ralphb.net/IPSubnet/  and this one is ok too: http://www.learntosubnet.com/

0
jeopboyCommented:
If you can multiply and divide by 2, you can do this on paper or in your head.

I find it's easiest to start from the basic fact of a Class C subnet:

Class C = /24 = 256-2 = 254 hosts  

For more than 254 hosts, you will supernet, for fewer than 254, you will subnet.  
You always use 256 as your base to make the calcs easier.

For supernetting, multiply by 2, then subtract 2 (network and broadcast addresses).  Each time you double the number of class C subnets, you decrease the subnet mask in / notation by 1.
For subnetting, divide by 2, then subtract 2.  Each time you divide by 2, increase the subnet mask in / notation by 1.

Subnet: 256/2 = 128-2 = 126 hosts with a /25 mask (1/2 of a Class C)
Supernet: 2*256 - 2 hosts = 510 hosts with a /23 mask.(2 Class C's)

Keep multiplying or dividing by 2 until you meet your hosts needs.

I'll change your example so that I start at the same place but I need 1100 hosts and 5 networks.  Based on this, /23 isn't enough.  Therefore we repeat the multiplication process:

        2*2*256 = 4*256 = 1024-2 = 1022 hosts with a /22 mask.
        This is 4 Class C's supernetted together.
Still not enough, repeat:
       2*2*2*256 = 8*256 = 2048-2 = 2046 hosts with a /21 mask.
        This is 8 Class C's supernetted together and covers our needs.

So now you need to know where each range starts and stops.

You know your start:  130.131.0.0 is the first subnet so the first usable address is 130.131.0.1

You know your mask is /21 and you need 8 class-C's.  This means you simply count up 8 subnets from the start, (remembering to include the first 130.131.0 subnet), then put .255 in the last octet for broadcast.  So in my example, I get:
        Subnet 1: 130.131.0.0/21   Broadcast: 130.131.7.255

For the next and subsequent subnets, add the number of class C's to the 3rd octet and repeat:
        Subnet 2: 130.131.8.0/21  Broadcast: 130.131.15.255

Repeat this for however many networks you need.  The pattern becomes pretty clear so you can simply add the number of Class C's to the 3rd Octet of each previous subnet and broadcast

I'm not sure what you mean by first and last IP.  I assumed you meant network and broadcast addresses.

Hope this helps.
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Windows Networking

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.