Subnetting -Working out the total available number of Networks and Subnets

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Hello all,

I'm currently working towards the 70-291, and like many on here am finding it quite difficult, to the point that I actually feel quite demotivated.

I've been through the CBT Nuggets (and have learnt quite a bit), and am now beginning the Sybex book. I've just taken the test at the end of the first chapter and came across a question which asked me to work out the total number of networks and subnets.

Eg, if I had a Subnet Mask of 255.255.248.0, I know that there would be 32 different subnets available. What I don't understand, and have been unable to find out (in plain English), is how I would calculate the total number of available networks (including subnets).

Also, could somone please clarify for me what exactly variable length subnet masks are.

Cheers

MaX

 

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The usual 'quick' response to questions about subnetting is http://www.learntosubnet.com/.

<grin>

Variable Length Subnet Mask is just the name when you have different length masks on a system. i.e. - in a company accounts might be given a /28, and sales given a /27.

Harry.

 

What? xD

If you had a subnet of 255.255.248.0 you would have 7 networks, and therefore have 1785 IP addresses.

I'm not sure what your asking.

 

The 70-291 is a hard exam, I used the MS Press Book and asked a few people to get my head around subnetting.

I found the CBT Nugget particularly unhelpful when it came to subnetting.

 

Could you explain how you got this answer?

 

Ok, You say you have a subnetmask of 255.255.248.0. But that is not all you need to know. You also need to have the address of your network. Let me explain. Say your networkaddress is 64.17.43.15. 64 tells you you are working in an A-class network. An A-class network has a standard SNM of 255.0.0.0, in other words, there are 8 bits allready counted for. 255.255.248.0 has 21 bits, so 21 - 8 = 13 bits to play with. 13 bits give you a total amount of 2^13 = roughly 8000 networks. The 11 zeros on your SNM give you 2^11 = 2048 addresses. You need to lose 2, one for the network address and one for the broadcast address, leaving 2046.

If the starting address was however 130.16.25.10, that would give you a B-class to start with, so 16 bits are allready acounted for. leaving 5 bits for your networks (that's 32 networks). The number of hosts stays the same.

 

If he has a subnet mask of 255.255.248.0 (11111111 11111111 11111000 00000000)

He has a total of 7 available network ID's

And there are a total of 255 host ID's available with each network ID, so therefore 7 * 255 = 1785 give or take a few host addresses.

Correct me if I'm wrong.

 

<cough> As Tinus implied - there isn't enough information to say whether it is wrong or right.

There are a *lot* of 'correct' answers depending on assumptions.

Harry.

 

More information please!

 

CCNA Portable command Guide 2nd Edition by Scott Empson actually introduced waht is called The Enhanced Bob Maneuver for Subnetting (or How to Subnet Anything in Under a Minute) it actually very easy to understand, used it myself and still using it to subnet.

VLSM is the more realistic way of subnetting a network to make for the most efficient use of all of the bits.
it also the process of “subnetting a subnet” and using different subnet masks for different networks in your IP plan. What you really have to remember is that you need tomake sure that there is no overlap in any of the addresses. Other books has dealt with VLSM like CCNA: Cisco Certified Network Associate Study Guide(Exam 640-802) 6th edition by Todd Lammle but I find the Bob Enhanced Maneuver esily understandle.

 

How did you work out this 7 network IDs? Please read my other answer on this question.

 

Im getting 7 network segments like this.

Im assuming he has a class C address range so therefore he is using the subnet mask of 255.255.255.0

By using this range he is able to use Classless Interdomain Routing (CIDR) changing the third octet from

11111111

to

11111000

Now, is he is now on the subnet 255.255.248.0 he has an avaliable of 7 network segments:

255 - 248 = 7

So, in the third octet he can create network segments ranging from 1 > 7

If for example he had the subnet mask as follows: 255.255.240.0 (11111111 11111111 11110000 00000000)

Then he would be able to create 15 network segments.

This is going on the principle that he is using the class C address range which uses the subnet mask 255.255.255.0 which ALLOWS him to create 2,097,152 avaliable networks.

His question isnt giving enough information, so everyones answers are just assumptions.

:hhhmmm

 

I think I see where we might be crossing paths xD

Your somehow getting the same answer as me for the hosts.

But your talking about the IP address in its 32 bit form (4 octets)

Im going on the basic subnet mask principles that if your using 255.255.248.0 you are saying that YOU have changed it so you HAVE 7 avaliable network segments and in the last octet you have 255 host addresses (not using the network address and broadcast address).

Can anyone verify that what im saying is true or false? Then I can see what is really going on.

!

 

First: you are not subnetting, you are supernetting.
Second: with 3 bits you get 8 combinations, not 7.
8 * 256 = 2048 as a maximum number of addresses.
Somehow I have the feeling you should review your books on subnetting.

 

I can, It's false. First: every IP-address is 32 bits. The way one makes the calculations does not change nomatter what notation one uses. The dotted decimal notation is just created for the humans, because we are not that good with binairy numbers. You can believe me the computer all recalculates this back to biainry before working with it. The computer doesn't care how you do the calculations, as long as you do them correct.
Second: using 3 bits gives you 8 combinations, not 7.

 

Alright calm down, I was only trying to help, geez.

You seem like the type of person who relishes proving people wrong, great

 

No, I'm not. I am trying to help people with their problems and they are not helped with wrong advise. I have been training students on network technology now for close to 20 years. Helping people is my profession. I also want to help you, because obviously you did not grasp the concept of subnetting. This can cost you points during an exam. If you want I'll give you the full story.

 

I sincerely apologize Tinus1959!

I haven't done what was obviously explained in this topic, I was using what basic knowledge I have from subnet masks TCP/IP etc which I learned from my MS Press 70-270 book!

Right off the mark I was wrong! I should of read the bit that said SUBNETTING

Yet again, my deepest apologies Tinus1959, I'm going to keep my mouth shut from now on !

 

No problem mate. My offer still stands to explain subnetting in detail if you want. Oh, and apologies excepted

 

Since I got a number of PMs asking me to do the explanation on subnetting, I plan to make it a separate thread. I just need some time to do it. Please bear with me.