VLSM on Class B

mguy

I can do VLSM on Class C, but Class B got me :

Say you need

Network 1: 200 users
Network 2: 14 users
Network 3: 2 users

Given: 172.16.0.0 /23


I would think I'd give Network 1
172.16.0.0 /23
Range[172.16.0.0 - 172.16.3.255]

So network 2 (14 users) has to start with
172.16.0.0/28
So network 3 (2 users) will be
172.16.4.0/30

Help! Ionno how..

mguy

Now that I think about it. We are assigned 172.16.0.0/23 which means we have Range[172.16.0.0 - 172.16.3.255]

N1: 172.16.0.0 /24
N2:172.16.1.0/28
N3:172.16.1.16/30

correct?

elderkai

You can't have your networks overlap, though. You say the range for N1 is 172.16.0.0 - 172.16.3.255, but then you have N2 start at 172.16.1.0 which belongs to N1's range. You'd have to start N2 at 172.16.4.0/28.

mguy

mguy wrote: »

Now that I think about it. We are assigned 172.16.0.0/23 which means we have Range[172.16.0.0 - 172.16.3.255]

N1: 172.16.0.0 /24
N2:172.16.1.0/28
N3:172.16.1.16/30

correct?

is there overlap in this?

oli356

Yes, because if network 1 is 172.16.0.0 - 172.16.3.255 and network 2 starts at 172.16.1.0, well it can't start at 1, has to start at 172.16.4.0 like elderkai said.

I'm good at subnetting, hopeless at VLSM, need to do some studying

elderkai

Yeah, that's what I said in my post. You just need to change the third ocet to start at 4 for N2. It takes some practice and getting used to, but I'm sure you'll get the hang of it. You pretty much have it.

xbuzz

You're right in your second post, if I understand the question correctly mguy.

You have the 172.16.0.0/23 subnet at your disposal. That subnet isn't used at all, which is where the other posters are getting confused I think. The question requires that you split that subnet into further small subnets to satisfy the criteria of 200 users in 1 subnet, 14 in the next, and 4 users in the final subnet, in which case you're right that:

N1: 172.16.0.0 /24
N2:172.16.1.0/28
N3:172.16.1.16/30

...is the answer.

EDIT: You are however wrong in thinking that the network 172.16.0.0/23 gives you the range 172.16.0.0 -> 172.16.3.255. It actually gives you the range 172.16.0.0 -> 172.16.1.255, but that doesn't effect the answer to the question in this case.

mguy

xbuzz wrote: »

You're right in your second post, if I understand the question correctly mguy.

You have the 172.16.0.0/23 subnet at your disposal. That subnet isn't used at all, which is where the other posters are getting confused I think. The question requires that you split that subnet into further small subnets to satisfy the criteria of 200 users in 1 subnet, 14 in the next, and 4 users in the final subnet, in which case you're right that:

N1: 172.16.0.0 /24
N2:172.16.1.0/28
N3:172.16.1.16/30

...is the answer.

EDIT: You are however wrong in thinking that the network 172.16.0.0/23 gives you the range 172.16.0.0 -> 172.16.3.255. It actually gives you the range 172.16.0.0 -> 172.16.1.255, but that doesn't effect the answer to the question in this case.

Yup that range is correct. VLSM is starting to lock in to place.

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