Hi EEpeeps,

I want to learn tough VLSM but iam confuse on these a friend on cisco forum asked me some question so that i can practice for it but iam only able to answer easy ones.

Pls tell me what can be answer of "Not so easy ones" and "MUST" describe/explain how you solved it yourself this is what real question is :)

You are given an address of 76.222.112.64/22

First easy ones.

1. What are the network and broadcast addresses of this network?

2. How many hosts can be in the network?

Now not so easy.

3. Suppose you need to subnet that network. You need 6 subnets with at least 30 addresses each, 4 subnets with at least 100 addresses each, and 1 subnet with no more than 6 and no less than 4 host addresses. Is that possible? If so, how would you do it?

4. You are asked to summarize your network with a 76.216.232.111/17 network. What will the summary address be like provided you need the smallest summary possible?

But NOW I see that I am wrong. Forget everything I wrote above regarding #4. I see that the difference is in the 2:nd octet, not the third!

We have these networks:

* 76.222.112.64/22, which fools us to believe that that is the network address, but it isn´t. With a /22, the network address is 76.222.112.0/22.

* 76.216.232.111/17, which also fools us. The network address with that mask is 76.216.128.0/17.

so:

76.222.112.64/22

76.216.232.111/17

The first difference fro left to right is in the 2:nd octet. So the network/netmask-boundary should be in that 2:nd octet, right?

222 and 216. How do we define them as narrow as possible? If you cant see it by eyes I suggest that you write the numbers down in binaries:

128 64 32 16 8 4 2 1

222 = 1 1 0 1 1 1 1 0

216 = 1 1 0 1 1 0 0 0

So, how many bits are common for those numbers, from left to right? 11011, right which means 5 bits. So the network number is 11011 binary = 216.

Where is the boundary? First in 1:st octet we have 8 bits, then we have 5 bits in 2:nd octet, which give us 13 bits in the netmask. Conclusion: /13.

So the solution should be 76.216.0.0/13 which includes 76.[216-223].x.x

Sorry for bringing this confusion. :-)

/Kvistofta