Just to add something to what Joe has written, your auther is rite when he says use PRECISE routes. For instance, currenlty your router needs to advertise

1.1.1.0

1.1.1.1

1.1.1.2

1.1.1.3

and you know that 1.1.1.0 0.0.0.3 will let you advertise 4 IPs from 1.1.1.0 to 1.1.1.3, but what if you also need to 1.1.1.4, then this 0.0.0.3 will not work, as you would have to additionaly specify the route for 1.1.1.4 0.0.0.0 as adding the bit in the wild mask will make it 0.0.0.7, which equates to 8 IPs, meaning from 1.1.1.0 to 1.1.1.7, whereas you only need 1.1.1.0 to 1.1.1.4 to be advertised, hence, you would adv 1.1.1.0 0.0.0.3, and 1.1.1.4 0.0.0.0 ...

However there is another way on advanced level, where you can even go further for bit by bit subnetting, but i guess that does not come under the scope of this discussion :)

HTH

Mohsin

Hello All,

I am most grateful for your responses and my apologies for the late reply.

To ascertain that i now i understand this combined wild-casting, would i be correct if the addreses are now written as follows?.

GIVEN ADDRESSES

1.1.1.0/32

1.1.1.1/32

1.1.1.2/32

1.1.1.3/32

ADVERTISED AS

1.1.1.0 0.0.0.0

1.1.1.1 0.0.0.0

1.1.1.2 0.0.0.0

1.1.1.3 0.0.0.0

Would this mean the same as the

1.1.1.0 0.0.0.3?, and if so where does the 3 come from knowing well that the given addreses are /32

Resolving the misery 3 in the wild cast mask, would make things much clearer

Many thanks in advance.

Okay, do you understand subnet masks and how they are derived ie.

1.1.1.0 255.255.255.252 gives you

1.1.1.0

1.1.1.1

1.1.1.2

1.1.1.3

If you understand the above then the 0.0.0.3 is really just a different

way to write it and is called an inverse mask. Easiest way to remember it is

256 - 252 - 1 = 3 so

255.255.255.252 = 0.0.0.3

So 255.255.255.240 would be

256 - 240 = 16 - 1 = 15 so

255.255.255.240 = 0.0.0.15

255.255.255.192 would be

256 - 192 = 64 - 1 = 63 so

255.255.255.192 = 0.0.0.63

Hope this makes sense.

Jon

Hi Giuslar,

I don't really agree with you on that. Network commands are maybe processed sequentially, but they do no work like ACLs. In ACLs, the first matching command is worked on and rest are skipped. However network commands search for the "More specific prefix of subnet" and hence uses the most precise route for the routing.

Addtionaly, if you don't care for the subnets based routing, you can never understand the VLSM.

Hello Mohsin,

I was trying to provide an analogy for someone like the thread's originator that is studying EIGRP.

My experience with OSPF for example is that network area statements are treated as ACL statements: you cannot move a subnet to a new area by adding a new network area X command if the subnet already matches a present network area command, from this kind of scenarios comes the need to be as precise as possible to be ready for future developements.

VLSM:

the ability to advertise subnets with a variable subnet mask is a routing protocol's property that for EIGRP and RIPv2 is influenced by the auto-summary command : they both by default summarize at major network boundary (Class A,B,C )when sending updates out an interface that is in another major network (another Class A,B,C)

This VLSM capability is not influenced by the network commands inside the router process that actually act as placeholders for one or more connected subnets.

In fact, EIGRP accepts network commands with wildcard masks only from release 12.0(5)T and I used routers that only allowed to provide the network command. Even those routers were capable of VLSM.

VLSM capable = carry subnet mask in updates = classless routing protocol

Hope to help

Giuseppe

Hello All,

A very big thanks to you all for your willingnes to assist me.

To be sure i now understood this wildcard cast, let say i am asked to advertise the following on OSPF or EIGRP, since they both require wildcard masking.

2.2.2.64 - 2.2.2.75/32

would i be correct to advertise the network as 2.2.2.64 0.0.0.127?.

Because all i could come up with from your explanations was that,we summarise given addresses and then minus it from 255.

Again, does that mean that if the same question was asked but now with 64 sitting in the third octet i.e 2.2.64.0 - 2.2.75.0/24

Would i then be correct to advertise the network as 2.2.64.0 0.0.127.0?

Kindly accept my apologies in advance if at all my question is becoming a pain to anyone.

Regards

not quite there yet...

with a 2.2.2.64 0.0.0.127 you would probably get an error message saying incosistant network/mask (something like that)

I wildcard mask of 0.0.0.127 translates to 255.255.255.128 mask (or /25) with that mask you can only break it down to two available networks

ex:

You have the private class C of 192.168.1.0 /24. You want to break it down to two subnets, they would then become

192.168.1.0/25 (.1-.126 are usable)

192.168.1.128 /25 (.129 - .254 are usable)

so for the statement you included, it would become 2.2.2.0 0.0.0.127 or you could get alot more specific & do something like 2.2.2.64 0.0.0.15 (which would cover .64 - .79)

As for you example with the 64 in the third octect, that isn't quite correct either. You cannot have the 64 in the third octect with a wildcard mask of 127 in the 3rd octet...Also, once you put a number other than 0 in the third octet, then the 4th octet in the wildcard mask automatically becomes 255.

With an example of 192.168.1.0 255.255.255.0 (/24)

When you have the wildcard mask for this, it would be 0.0.0.255. The reason for that is everything is reversed

192.168.1.0 = network number

192.168.1.1 - .254 = usable hosts

192.168.1.255 = broadcast address

You figure out this info from the combination of the address & the subnet mask

With the wildcard, you still figure out the same info

subnet 192.168.1.0

wilcard 0.0.0.255

What the 0 in the first octet represents is that your address has to match exactly to the 192

The 0 in the 2nd octet means you have to match exactly to the 168

The 0 in the third octet means it has to match exactly to the 1

The 255 in the last octet, means it can be any number within the range.

This logic is the complete opposite of a normal subnet mask. In a normal subnet mask. 255.255.255.0...The 255's must match & the 0 can be anything.

I know it seems confusing but once you get the hang of it, it is pretty easy. You really need to start out with understanding subnetting before going on to wildcard masks & supernetting (you attempted supernetting in your 2nd example)

To try and help with the subnetting....

Take the binary number 11111111. In decimal form it is 255.

Each of these ones have a numerical weight to it

128 64 32 16 8 4 2 1 (number weight)

1 1 1 1 1 1 1 1 (binary number)

Add up the number weights is how you get to 255

Take the subnet 192.168.1.0 255.255.255.192 (62 usable hosts) You can figure out the usable hosts by calculating the binary value

192 = 11000000

Look at the number weight above your last one which is 64 then subtract two (network & bcast). The ones represent your network portion & the 0's represent your hosts portion. When all ones, it represents a bcast address.

So if you were to break down the 192.168.1.0 255.255.255.192 (/26) into usable subnets it would look like this...

192.168.1.0 /26

192.168.1.64 /26

192.168.1.128 /26

192.168.1.192 /26

so it is broken down into 4 usable subnets. Each subnet with its own network, hosts, bcast.

After you understand this, subnetting & supernetting will fall into place.