Subnet Masks

[IAntDemo]

ok, let me just say this.

Ok, I need to know what and How does subnetting work?

O look, I googled it:

http://www.dialogic.com/support/helpweb/safepipe/SP/courses/ip/

Jeez so much info

Ok maybe I was abit harsh to put google up as what's google but come on, If I want to know how to burn a dvd, I google it as a fast response, now if I couldn't find out on google I simply goto a forums just like this.

 

[Kage]

Thats one thing that annoys me about this bloody forum, and thats that even though someone has replied with an answer that answers the question;

Root (A fantastic reply),

an argument carries on, taking away the point of the thread, and takes away from the fact that it has been replied to in a positive manner...

 

[lhuser]

root has it more elaborate. I'll just give you a crash course to make things simple, but if you want something well done, root's got a better solution than mine, hands down.

Class range net IDs
Here's a list of common private IP addresses:
Class A= 10.0.0.1 to 10.255.255.254 // SM = 255.0.0.0 default // Bcast= 10.255.255.255 // Ranges from 1-127
Class B= 172.16.0.1 to 172.16.255.254 // SM 255.0.0.0 default // Bcast= 172.16.255.255 // Ranges from 128-191
Class C= 192.168.0.1 to 192.168.0.254 // SM 255.255.255.0 default // Bcast=192.168.0.255 // Ranges from 192-224
Class D= 224-248
Class E=249-255?

Class D and E, I don't know them by heart, but they are experimental, so you don't worry about them, Of course, you can always use them for you home network, just not on the internet.

SUBNET
A subnet basically divides the NetID and the HostID. Take a very common Class C network.

192.168.1.1
255.255.255.0

The long version looks like this:
11000000.10101000.00000001.00000001
11111111.11111111.11111111.00000000

All the 1s hide the network portion, remaining with the .1 at the end for the hosts. You can have 254 hosts in an IP address (255 is the broadcast address. Common in DHCP)


So, the NetID is 192.168.1 and the HostID is .1

On a class B, like 172.16.0.1, and a SM (Subnet Mask) 255.255.0.0, the NetID is 172.16 and the HostID is .0.1, so there's a lot more hosts available. A Class A network, with IP 10.0.0.1 with a SM of 255.0.0.0 gives the most hosts. 10 s the NetID, and the .0.0.1 is the HostID.

Also, you can divide the hostID so that a portion of it may be ussed for subnetting. The best solution to save you the long way and giving you headaches, you can use a subnet calculator. Here's one for you to practice on:
http://www.subnet-calculator.com/

Oh Kage, if you see the argument going on, why not split the thread and remove these nonsense?

 

[root]

- What is the host address of the following IP addresses (assuming default subnet mask is used) : 125.125.250.45, 120.125.250.75, and 194.194.150.20?

what do you mean by host address?
I'll go with what I assume you mean host address, by saying that I'm assuming that the host address is the first available (or last available) address in the network space.

to answer this question you first need to appreciate network classes, IP ranges are defined as being of 3 classes, (A, B or C) these have address ranges as such.
class A 0.0.0.0 - 127.255.255.255.255 (default mask 255.0.0.0)
class B 128.0.0.0 - 191.255.255.255 (default mask 255.255.0.0)
Class C 192.0.0.0 - 223.255.255.255 (default mask 255.255.255.0)

so...
125.125.250.45, class A (mask 255.0.0.0 / network address x.0.0.0)
first address in range 125.0.0.1 (last address in range 125.255.255.255 -broadcast 125.255.255.255)
120.125.250.75, class A (mask 255.0.0.0 / network address x.0.0.0)
first address in range 120.0.0.1 (last address in range 120.255.255.254 -broadcast 120.255.255.255)
194.194.150.20 class C (mask 255.255.255.0 / network address x.x.x.0)
first address in range 194.194.150.1 (last address in range 194.194.150.254 -broadcast 194.194.150.255)

- Is the following subnet mask valid: 255.255.255.16

No.
valid subnets are ones where the binary looks like this
11111111 . 11111111 . 11111111 . 11111111 (255.255.255.255 (/32))
11111111 . 11111111 . 11111111 . 11111000 (255.255.255.248 (/29)
11111111 . 11111111 . 11111111 . 00000000 (255.255.255.0 (/24))
the subnet 255.255.255.16 looks like
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
you can't have a 1 appearing in the middle of 0's cause then when you do the bitwise and operation you get results like this

11000000 . 10101000 . 00000000 . 00000001 (192.168.0.1)
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
11000000 . 10101000 . 00000000 . 00000000 (resultant)

and that's not in the same range as
11000000 . 10101000 . 00000000 . 10010110 (192.168.0.150)
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
11000000 . 10101000 . 00000000 . 00010000 (resultant)

and yet at the same time is in the same network as
11000000 . 10101000 . 00000000 . 11001000 (192.168.0.200)
11111111 . 11111111 . 11111111 . 00010000 (255.255.255.16)
11000000 . 10101000 . 00000000 . 00000000 (resultant)
so your range is what, 192.168.0.1 - 192.168.0.255 BUT excluding certain addresses?

- How many subnets are possible with the subnet maks below, and how many hosts are available per subnet: 255.255.255.128?

what's with the vague questions!! that would very much depend on the class of the network!
for example if you took a class A network with that subnet then there are 131072 available subnets with 126 hosts persubnet

a class C networ there are 2 subnets available each 126 hosts per subnet

(i'll assume it's meant to be a class C)

one way to see this is to write out the subnet as binary
11111111.11111111.11111111.10000000
then do a logical NOT opperation, (invert the bits)
00000000.00000000.00000000.01111111
then convert back to decimal.
1111111 = 127
255.255.255.128 = 127 addresses available. then subtract 1 as it's the broadcast address and can't be a host address.

the maximum amount of subnets is the amount of times you can fit this number into the maximum amount of addresses

maximum amount of addresses = 255 (well 254 really cause there is the broadcast)
so you have one network of 127 addresses (including broadcast)
then you can have another network of 127 addresses.
so now you can see there are 254 addresses in the 255 address range used, you can only fit 2 subnets in this range.

255.255.255.128 give 2 subnets, with 126 host in each

 

[BamaFan29891]

Well the arguement can end here, cause I hate it too.

If I had wanted a response from Google, I would have Google'd it Antdemo.

But, I wanted a response from someone that I could maybe ask more questions regarding the subject to. So I got my response, I can now read it, and if I have more questions, I will go on with it.

Thanks for the reply Root, +1.

I am sorry for the vagueness of some of the questions, but you can tell why I was having some problems with them, .

I will read it again some more, to try and understand it better.

 

[root]

- Same question: 255.255.192.0?

same again
11111111.11111111.11000000.00000000
invert
00000000.00000000.00111111.11111111 =
11111111111111 (in binary) = 16,383 (in decimal)
take one away for the broadcast
=16282 host addresses available available on each network

in a non sub netted (mask = 255.255.0.0) network there are:
11111111.11111111.00000000.00000000
(invert)
00000000.00000000.11111111.11111111
1111111111111111 (in binary) = 65,535 (in decimal)
16,383 (one subnet)
+16,383 = 32,766 (two subnets)
+16,383 = 49,149 (three subnets)
+16,383 = 65,532 (four subnets)

giving 4 subnets available

- Given the following IP address and subnet mask, what ist he host and network address of each: 192.168.1.25 using subnet mask of 255.255.255.0?

basically the subnet mask is indicating the amount of bits that can change.
as you'll have seen above 255.255.255.0 means that nothing in the first three octets can change, any number in the last octet will satisfy the bitwise and operation.

so the 192.168.1.25 / 255.255.255.0
has ranges 192.168.1.0 -> 192.168.1.254 with a broadcast address of 192.168.1.255

 

[IAntDemo]

Well the arguement can end here, cause I hate it too.

If I had wanted a response from Google, I would have Google'd it Antdemo.

But, I wanted a response from someone that I could maybe ask more questions regarding the subject to. So I got my response, I can now read it, and if I have more questions, I will go on with it.

Thanks for the reply Root, +1.

I am sorry for the vagueness of some of the questions, but you can tell why I was having some problems with them, .

I will read it again some more, to try and understand it better.

Just keep revising it until you think you have it.

give it about an hour maybe and you will have learnt something

 

[root]

I am sorry for the vagueness of some of the questions, but you can tell why I was having some problems with them, .

I will read it again some more, to try and understand it better.

some of those questions are really terribly vague! like the one that asks how many subnets you can have with a given mask, that question really does require more detail before you can give an answer...

I think that if there are going to be questions like that then you should be asking in the test for them to clarify the questions.