a basic subnetting question

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192 is a C-class. It starts (binairy) with 110 and that nominates a C-Class. But, you could join all 256 networks into one supernetted B-class.

 

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O hell, why don't I read the complete thread before I answer.
Sorry folks.

 

Yes but that is subnetting Freddy which the question specifically prohibited

 

Yep, i'll go with that, even though I think even this is an example of subnetting as the mask is not the natural mask for the IP class. I suppose a pretty strict definition of the word "subnet" is what is needed here.

 

Well maybe my understandings wrong but a standard C class subnet mask would mean 24 bits for network address and 8 bits for host. So potentially we have 2 power 24 networks, obviously we need to take off a few as not all combinations of address are valid depending on the standard and the hardware. Whether we call each network a subnet or not I'm not sure, or whether a subnet is a subset of a standard network.

I thought routers routed to a network ? So surely there would be two networks in the example not three ? The hosts and devices would have host addresses on the networks, the wire between the routers would be part of neither network ?

 

No. It takes it's own network block to accomplish that. Each network interface on a router has to have a separate network address and subnet mask, otherwise the router would not know where to send specific packets.

 

Am I gonna have to enter this thread, or do y'all have it sorted out?

 

Well, you're already here running up your post count....

 

As if I needed this thread to increase my already monstrous post count. heehee!

...though, it does seem to have helped some of y'all's! ;)

 

Ok well by network interface i guess you mean the router to router links ? As the standard LAN ports can be on the same network ? Since theres only two routers and one link, that would be two interfaces no and hance two networks ?

 

???

255.255.255.0 is the natural mask for a 192.168.0.0 type (class C, if you will) network.

That's why it solves the problem without subnetting.

 

The connection between the routers is its own network.

Using a class C network for such a network is a bit wasteful, and it is usual to use a /30 subnet mask. But it does have to be a separate network to the two LANs

 

No.

Say one LAN is 192.168.1.0
Second LAN is 192.168.2.0

The third LAN for just the two routers could be 192.168.0.0 - one router would have the interface on that LAN (ie the connection to the other router) as, say,
192.168.0.1 255.255.255.0

and the other router would have its interface at the other end of the connection as, say,
192.168.0.2 255.255.255.0

The routers would be set up to route between the three networks (so that, for example, you could route from a host machine on LAN A to the router which is connected to LAN B, for management purposes)

If you think of the two routers as connecting over a WAN link form different ends of the country, the need for a specific network for just that connection might make more sense.

(And yes, I know in practice no-one would set up a WAN link between just two routers using up a whole Class C address space - although if they were using really old equipment and RIP v1 as the routing protocol, who knows....)

 

It occurs to me that the original requirements could be satisfied in the following manner:

Use a serial link between the two routers and run PPP over it.
Configure the routers to be an Ethernet bridge.

Now you can put the PCs on just one network.

I know this isn't do-able on cheap domestic routers, but the Lucent unit I was playing with some time ago mentioned this as a possibility.

Harry.

 

We have to take into account that the original requirements as described in the question may be b@ll@cks and we may be chasing an answer that doesn't exist...

 

My bad. This thread has prompted me to re-read ALL my networking books from scratch. Its amazing how much you forget when not actually using it every day.

 

On the possibility of showing how little i actually know abtou subnetting ( im still trying to get my head around the concepts. for my 70-291 exam)

i would like to contradict you.

and address of 192.168.1.1 /24 you would have 254 hosts on the network numbered from 1-254. but if you were to have a network of say 192.168.1.1 /16 you would have the possibility of 65534 hosts on the network as the subnet mask splits the address into 2 parts.

im my first example of 192.168.1.1/24 you get this

Address in decimal
192.168.1.1

Address in Binary
11000000.10101000.00000001.00000001

subnet mask of 255.255.255.0
in binary 11111111.11111111.11111111.00000000

so now if you AND the subnet mask and the IP address together you get.

11000000.10101000.00000001.00000001 Ip Address
11111111.11111111.11111111.00000000 Snet mask
00000000.00000000.00000001.00000001 Host Identifyer

you have now split the network into 2 parts

192.168.1 = the network ID
0.0.0.1 = the host ID.

this gives a posible of 254 hosts on the network like you mentioned



so now lets do the same for the 192.168.1.1/16 subnet.

Address in Binary
11000000.10101000.00000001.00000001

subnet mask of 255.255.0.0
in binary 11111111.11111111.00000000.00000000

so now if you AND the subnet mask and the IP address together you get.

11000000.10101000.00000001.00000001 Ip Address
11111111.11111111.00000000.00000000 Snet mask
00000000.00000000.00000001.00000001 Host Identifyer

you have now split the network into 2 parts

192.168 = the network ID
0.0.1.1 = the host ID.

this gives a posible of 65534 hosts on the network.

or thats how i understand subnetting. altho i need to brush up on how to split the network up for example if a company wanted to support 300 clients on the 192.168.1.0 network what subnet mask would you apply well its not /24 and its not /16 how i work it out takes a while and i need to get my head round.


Post Edited to correct simple mathmatical mistake as pointed out by Harry in the post below.

 

Er - no.

192.168.1.1/16 is a host adress, not a network address. The network address for that mask would be 192.168.0.0.

A /16 network would normaly have 65534 possible hosts.

And you could get 300 hosts on 192.168.0.0/23

Harry.

 

Have i missed something crucial here i know that 192.168.1.1 is a host on the 192.168.1.0 network.

what i was trying to explain is that the subnet mask 255.255.0.0 or /16 are the same, and it splits the ip address into 2 chunks the first part would be the netowrk ID in the example its the first 16 octets so 192.168. then the next 16 octects would be host ID.


and i think i have just noticed my slight miscalculation. i did 254*2 not 254*254. im gonna edit my post to make it correct thanx for pointing my mistake out.

 

I think you meant bits, not octets here! Octets are bytes!

Harry.