Subnet Masks

[Allen]

I have recently completed a course called Network Management which covered some basics on installing and administering a network, mainly in a client-server setup. I have pretty good all round knowledge on the subject and I am know trying to increase my potential and start studying towards getting an MCSE qualification.

I have started by getting a CBT program for Administrering Windows XP Professional (exam 70-270, the core client OS exam I going to do as part of the MCSE on Server 2003 certification) as I believe it would be a good starting point, being the OS I currently have the best knowledge of. I have started the course and I am doing pretty well, however, the course really doesn't explain subnet masks very well (and tbh I wouldn't have thought it would, I would expect to cover them more in a Server 2003 orientated course) but some of the test questions for the Networking module actually ask in detail about subnet masks.

I covered subnet masks very brielfy on my Network Management course, but I don't have a good enough knowledge of how to work them out in my head as I will have to in an exam situation. I have tried looking on the net for further help and I only ever seem to end up staring at pages and pages of text and pointless diagrams which really do not help me at all.

Does anyone out there know of any websites which just cover an easy way to work out valid IPs on a subnet using a simple process that can be done in your head (i.e. not cheating and using the AND calculation on a scientific calculator as I have seen some do) and that does not cover the reasoning why you need to do it, as I kind of understand why already.

I know how to work out small amounts of binary, probably enough to (albeit slowly) convert an IP address into binary for part of the calculation.

Any help with this would be greatly appreciated.

[kidzer]

I shall give a brief overview of Subnetting, see if this helps at all;

- The purpose of a subnet mask, when combined with an IP address is to split that address into two portions, the Network ID, and Host ID.

- Each class of IP address has a different default subnet mask, Class A: 255.0.0.0, Class B 255.255.0.0 and so on and so on.

- In short, when the subnet mask is converted to binary (e.g. Class A would be 11111111.00000000.00000000.00000000), the 1's represent the bits used for the Network ID, and the 0's the Host ID.

- This 'mask' can be changed though, to create sub-networks, commonly used to divide a network further into more logical partitions and the like.

- Now, given that the 1's = Network, and 0's = Host, once we have the number of bits for each, we can work out the number of hosts/networks avaliable, like so...

Network: 192.168.0.0
S/N Mask: 255.255.255.224 (Class C IP Default: 255.255.255.0) - 3 bits of the Host portion have been "borrowed", and it is only the last octet we work with in this case - because it is all you are changing. Were you subnetting a Class A address you would use the last 3 octets in your calculations, the last 2 for Class B etc.

Number of Networks = 2^n = 2^3 = 8
Number of usable Networks = 2^n - 2 = 2^3 - 2 = 6 Usable networks

Number of Hosts = 2^n = 2^5 = 32
Number of usable Hosts = 2^n - 2 = 30 Usable hosts per Network


where N = the number of bits for the network/host ID.

The usable Networks/Hosts comes about because 2 IP addresses for each subnet are 'reserved', one for the Network ID, and the other for the Broadcast Address.


You must also remember that the first and last subnets you work out are unusable, because the first one will have a Network ID of the 'Major' Network, being the IP Range before it was sub-netted, and the final one will have the Broadcast Address of the 'Major' Network.

Now, I think I've got that all correct, but should a Moby-Dick of the like come along and say i'm wrong, trust them and not me!

[Moby-Dick]

sounds about right to me chap - I'll have a look in the morning when I'm less tired.

classles subnets are quite handy too.

Allen can you post an example question ?

Oddly enough I'd take 290/291 before doing the client OS exam.

[kidzer]

sounds about right to me chap - I'll have a look in the morning when I'm less tired.

As will I

It might look different tomorrow morning!

[TiG]

The problem is that taking subnets as a subject on its own is hard to take, it makes no sense and unless looked at with the boundaries of why you would do things like this its hard to understand.

If you take this in conjunction with understanding Network Address Translation (NAT), and why these things are important then it will make a lot more sense when you really try and get your head round it all.

And finally once you want to play here is the ultimate play tool

http://www2.rad.com/networks/2005/ipkit/main.htm

TiG

[kidzer]

Ooh, that is funky...its like a web based version of Packet Tracer (also a funky little app)

[Allen]

Thanks for the replies so far.

Allen can you post an example question ?

OK, the first question I came across was basically this:

You administer your company's network shown in the exhibit. You have just added a new host to the network (Host A). You need to manually configure the TCP/IP properties for Host A. At SRV1, you run ipconfig and discover that the subnet mask is 255.255.255.224.

Assuming there are no other hosts on the network othat than those shown in the exhibit, which of the following are valid IP addresses for Host A? (Choose all that apply.)

And there's a diagram attached showing the following:



Sorry for the paint skills, but I didn't wanna copy the image directly, just in case.

As for the multiple choice answers, it just gives a few IP addresses all starting 192.168.1 and ending 5, 11, 60, 15, 124 and 31. Obviously I know that 11 is being used by the server so that's off the list. However, without sitting there and working out (in a rather complicated way) the complete ranges for the whole subnet I wouldn't have a clue which to choose. Now I think I've just worked out the table for the valid IP addresses, but I'm not 100% sure. This is what I got:

192.168.1.33-62
192.168.1.65-94
192.168.1.97-126
192.168.1.128-158
192.168.1.193-221

So that would mean that 60 and 124 are the only valid IP addresses from the list that I can assign to Host A. However, upon checking the answer it says that the only valid IPs are 5 and 15 (11 is valid and being used by the server). I was taught that the first and last range of the IP addresses in the table you create are unavailable (in this case 1-30 and 225-254), and I even tested this on a setup in class by setting the subnet to 255.255.255.224 and the first IP address it would let me assign to the Host was 33.

Oddly enough I'd take 290/291 before doing the client OS exam.

Well, since I have no experience in IT yet and I only have an NVQ Level 4 in Network Management (and hopefully soon my first MCP) I thought the best way for me getting into what I want to do was to get a job in first line support/helpdesk and work on my qualifications whilst getting experience. Do you think it would be easier doing the 4 cores first then? Or would it depend what I wanted to do?

[Allen]

Now my head hurts...

[TiG]

You've got half right you've worked out correctly that 255.255.255.224 is a /27 subnet meaning there are 30 hosts in each segment. However it starts at 192.168.1.1

Which means you get values from 192.168.1.1 to 192.168.1.31

So ticking all items that are valid would be for me
192.168.1.5
192.168.1.15

192.168.1.31 is however the broadcast address for that subnet. Therefore i believe it shouldn't be used for a host.

TiG

[b0redom]

Yeah - that's what I make it too.

Never heard of not being able to use the first or last subnets. That would make the netmask 255.255.255.128 pretty useless!

[Allen]

OK, thanks TiG.

b0redom, it may have just been me misinterperating the bit about not being able to use the first and last, as in the network ID and the broadcast address that kidzer mentioned above.

Now I just need a quick way to work out the ranges without having to draw up a table and work out the binary in my head. If there is one...

[b0redom]

Well I may be wrong, but can't you just do:

256-subnet-2 = useable hosts?

so 192.168.0.1/255.255.255.192

(It's 256 because you go 0-255 not 1-255)

256-192= 64
64-2 = 62 usable hosts per subnet

192.168.0.1/255.255.255.128

256-128 = 128
128-2 = 126 usable hosts per subnet

If you go up a class, you work out the subnet size in the same way, but then multiply it by 256.

It's unlikely you'll get that high though as it would quicly become unmanageable.

[Moby-Dick]

You can't use the first and last address in a given subnet as the first is the network address and the last the broadcast.

I don't think you'll have to work out all the possible ranges as there are enough clues in the question ( ie you know the subnet has to contain the addresses of the other hosts on that subnet )

I found that the core exams made the Client os exam more relvent as its looking at the client from the config , not the day to day usage side.

try and get hold of a book from the old MSCE NT core exam called Networking Essentials - its got some very good overviews in it.

[kidzer]

I may have been wrong with the first/last subnet business, that's just what i've been taught (or atleast, how I interpreted it, I know i'm not confusing it with the Network ID/Broadcast Address for each subnet, because we were taught that too) - so I assumed it was correct!

I have always thought it a bit odd though, because the addresses were being split up and so (for example) 192.168.0.0 was only the network address of the first sub-network because the 'Major' (or Un-Subnetted) network (being 192.168.0.0/24) no longer existed.

Hopefully my inaccurate ramblings haven't put you too far off track!

[Allen]

You can't use the first and last address in a given subnet as the first is the network address and the last the broadcast.

I don't think you'll have to work out all the possible ranges as there are enough clues in the question ( ie you know the subnet has to contain the addresses of the other hosts on that subnet )

Hmm, OK, so am I correct in saying then that a subnet is one range of IPs, not a group of ranges using the same mask? I think the terminology is getting me confused.

In a 255.255.255.224 subnet there can be 32 IPs in each segment, however only 30 are available to be used as the first is the network address, the last is the broadcast IP. So this would give us the following:

Net  Valid    Broadcast
0    1-30     31
32   33-62    63
64   65-94    95
96   97-126   127
128  129-158  159
160  161-190  191
192  193-222  223
224  225-254  255

However, out of the available answers (which were 5, 11, 60, 15, 124 and 31) only 31 is invalid according to the above, and 11 is being used by the server. So, since the server is 11 and the correct answers were 5 and 15 it leads me to think that the subnet is only 1-30 and not all of the above ranges, which is what I thought it was...

[TiG]

192.168.1.0 /27 is a subnet (the first one in your code list)
192.168.1.32/27 is the second

A subnet is an IP address and a subnet mask. You are correct in your statement that your subnet in your question is just the top subnet.

So tell me what IP addresses can be used in this range

88.97.191.200 with subnet mask of 255.255.255.248

TiG