VLSM Subnets

ukman2003

Hi everyone.

I'm working on a problem for a buddy of mine, who is having problems with VLSM. I thought sure no problem, but it seems I'm having a little trouble also.

The requirements for the network are as follows:

Starting network address: 172.16.0.0/16
Must use VLSM so that address requirements can be met using the 172.16.0.0/16 network.

East Network Section

The N-East (Northeast) LAN 1 will require 4000 host IP addresses.
The N-East (Northeast) LAN 2 will require 4000 host IP addresses.
The SE-BR1 (Southeast Branch 1) LAN 1 will require 1000 host IP addresses.
The SE-BR1 (Southeast Branch 1) LAN 2 will require 1000 host IP addresses.
The SE-BR2 (Southeast Branch 2) LAN 1 will require 500 host IP addresses.
The SE-BR2 (Southeast Branch 2) LAN 2 will require 500 host IP addresses.
The SE-ST1 (Southeast Satellite 1) LAN 1 will require 250 host IP addresses.
The SE-ST1 (Southeast Satellite 1) LAN 2 will require 250 host IP addresses.
The SE-ST2 (Southeast Satellite 2) LAN 1 will require 125 host IP addresses.
The SE-ST2 (Southeast Satellite 2) LAN 2 will require 125 host IP addresses.

West Network Section

The S-West (Southwest) LAN 1 will require 4000 host IP addresses.
The S-West (Southwest) LAN 2 will require 4000 host IP addresses.
The NW-BR1 (Northwest Branch 1) LAN 1 will require 2000 host IP addresses
The NW-BR1 (Northwest Branch 1) LAN 2 will require 2000 host IP addresses
The NW-BR2 (Northwest Branch 2) LAN 1 will require 1000 host IP addresses
The NW-BR2 (Northwest Branch 2) LAN 2 will require 1000 host IP addresses

Central Network Section

The Central LAN 1 will require 8000 host IP addresses.
The Central LAN 2 will require 4000 host IP addresses.

The WAN links between each of the routerswill require an IP address for each end of the link. Which I believe there are 13 WAN links each as a /30.

So far I have as follows:

172.16.0.0/16 -- Starting Network Address.
Class B Address -- 1 Subnet 65,534 Hosts

East Network --- Need 11, 750 host IP addresses.

NE-LAN 1 - 172.16.0.0/20
.1 -.14 .15-Broadcast address

NE-LAN 2 - 172.16.16.0/20
.17-.30 .31-B
255.255.240.0

SE-BR1-LAN 1 - 172.16.32.0/22
.33-34 .35-B
255.255.252.0

SE-BR1- LAN 2 - 172.16.36.0/22
.37-.38 .39-B
255.255.252.0

SE-BR2- LAN 1 - 172.16.40.0/25
.41-166 .167-B
255.255.255.128

SE-BR2- LAN 2 - 172.16.168.0/25
.169-.254 .255-B
255.255.255.128

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Comments

elphrank0

Write out your powers of 2 if you do not know them, subtract 2 from them and thats the valid hosts you get for class b. Start with the 32k and thats a /17 working backwards.16k is a /18 etc

Forsaken_GA

It's a good idea to leave yourself some room for growth as well if you're able to. Those /25's will cut it awful close. I don't know if that's part of the parameters of the exercise, but it is a consideration for real world design

Of course, I'd just **** and use IPPlan

mella060

A good idea is to start with your largest networks (host requirements) first and work your way down. So start with the 8000 host network, then the 4000 host networks and the 2000 host networks and so on.

8000 hosts...2 ^ 13 - 2 = 8190...so you would need 13 host bits for the 8000 host networks.

172.16.0.0 /19 would be the network address of the 8000 host network with an increment of 32

172.16.0.1 - 172.16.31.255 would be the range of addresses.

For the next network...4000 hosts...2 ^ 12 - 2 = 4096...so you would need 12 host bits for the 4000 host networks.

172.16.32.0 /20 would be the network address for the 4000 host network in central office. Increment is 16

172.16.32.1 - 172.16.47.255 would be the range of addresses.

The N-East (Northeast) LAN 1 - 4000 addresses - 172.16.48.0 /20
The N-East (Northeast) LAN 2 - 4000 addresses - 172.16.64.0 /20
The S-West (Southwest) LAN 1 - 4000 addresses - 172.16.80.0 /20
The S-West (Southwest) LAN 2 - 4000 addresses - 172.16.96.0 /20

The NW-BR1 (Northwest Branch 1) LAN 1 - 2000 addresses - 172.16.112.0 /21
The NW-BR1 (Northwest Branch 1) LAN 2 - 2000 addresses - 172.16.120.0 /21
The NW-BR2 (Northwest Branch 2) LAN 1 - 1000 addresses - 172.16.128.0 /22
The NW-BR2 (Northwest Branch 2) LAN 2 - 1000 addresses - 172.16.132.0 /22

The SE-BR1 (Southeast Branch 1) LAN 1 - 1000 addresses - 172.16.136.0 /22
The SE-BR1 (Southeast Branch 1) LAN 2 - 1000 addresses - 172.16.140.0 /22

The SE-BR2 (Southeast Branch 2) LAN 1 - 500 addresses - 172.16.144.0 /23
The SE-BR2 (Southeast Branch 2) LAN 2 - 500 addresses - 172.16.146.0 /23
The SE-ST1 (Southeast Satellite 1) LAN 1 - 250 addresses - 172.16.148.0 /24
The SE-ST1 (Southeast Satellite 1) LAN 2 - 250 addresses - 172.16.149.0 /24
The SE-ST2 (Southeast Satellite 2) LAN 1 - 125 addresses - 172.16.150.0 /25
The SE-ST2 (Southeast Satellite 2) LAN 2 - 125 addresses - 172.16.150.128 /25

The WAN links could be...

172.16.151.0 /30
172.16.151.4 /30
172.16.151.8 /30 and so on

This could be all wrong but i believe it is way to do it...

ConstantlyLearning

mella060 wrote: »

A good idea is to start with your largest networks (host requirements) first and work your way down. So start with the 8000 host network, then the 4000 host networks and the 2000 host networks and so on.

8000 hosts...2 ^ 13 - 2 = 8190...so you would need 13 host bits for the 8000 host networks.

172.16.0.0 /19 would be the network address of the 8000 host network with an increment of 32

172.16.0.1 - 172.16.31.255 would be the range of addresses.

For the next network...4000 hosts...2 ^ 12 - 2 = 4096...so you would need 12 host bits for the 4000 host networks.

172.16.32.0 /20 would be the network address for the 4000 host network in central office. Increment is 16

172.16.32.1 - 172.16.47.255 would be the range of addresses.

The N-East (Northeast) LAN 1 - 4000 addresses - 172.16.48.0 /20
The N-East (Northeast) LAN 2 - 4000 addresses - 172.16.64.0 /20
The S-West (Southwest) LAN 1 - 4000 addresses - 172.16.80.0 /20
The S-West (Southwest) LAN 2 - 4000 addresses - 172.16.96.0 /20

The NW-BR1 (Northwest Branch 1) LAN 1 - 2000 addresses - 172.16.112.0 /21
The NW-BR1 (Northwest Branch 1) LAN 2 - 2000 addresses - 172.16.120.0 /21
The NW-BR2 (Northwest Branch 2) LAN 1 - 1000 addresses - 172.16.128.0 /22
The NW-BR2 (Northwest Branch 2) LAN 2 - 1000 addresses - 172.16.132.0 /22

The SE-BR1 (Southeast Branch 1) LAN 1 - 1000 addresses - 172.16.136.0 /22
The SE-BR1 (Southeast Branch 1) LAN 2 - 1000 addresses - 172.16.140.0 /22

The SE-BR2 (Southeast Branch 2) LAN 1 - 500 addresses - 172.16.144.0 /23
The SE-BR2 (Southeast Branch 2) LAN 2 - 500 addresses - 172.16.146.0 /23
The SE-ST1 (Southeast Satellite 1) LAN 1 - 250 addresses - 172.16.148.0 /24
The SE-ST1 (Southeast Satellite 1) LAN 2 - 250 addresses - 172.16.149.0 /24
The SE-ST2 (Southeast Satellite 2) LAN 1 - 125 addresses - 172.16.150.0 /25
The SE-ST2 (Southeast Satellite 2) LAN 2 - 125 addresses - 172.16.150.128 /25

The WAN links could be...

172.16.151.0 /30
172.16.151.4 /30
172.16.151.8 /30 and so on

This could be all wrong but i believe it is way to do it...

I was going to do up a reply but yours is exactly what mine would have been.

ukman2003

I forgot to note that there is no need to calculate for future growth.

How I understood it, was that HQ a single router connected by three WAN links to the EAST, WEST, and CENTRAL hubs would receive the first subnet, Central would receive the second subnet, EAST would receive the third subnet, and WEST would receive the fourth subnet. Pretty much divided into four equal subnetworks.

For example the subnet for HQ which should be done first is given as
172.16.0.0 to 172.16.63.255

The last 256 addresses should start with 172.16.63.0
So the WAN Links would be assigned to the first subnet
HQ to Central
HQ to East
HQ to West

Then to Central I have a Router connected to two LANs.
Central LAN 1
Central LAN 2
No WAN subnets are needed for Central.

Then the East Section
(see original post)

6 WAN Link subnets for EAST section using the last 256 addresses of the EAST subnet.

Then the West Section
(see original post)

4 WAN Link subnets for WEST section using the last 256 addresses of the WEST subnet.