VLSM so difficult

kobebrian

forbesl

Here's a couple of links that will help you with completing this exercise:

http://www.geocities.com/msaqibilyas/rs/ipadd.zip
http://www.geocities.com/msaqibilyas/rs.html

On this next page, select "VLSM tutorial" from the top left drop down menu:
http://sky.fit.qut.edu.au/~salleh/tutorial.htm

ccna_wannaB

You're Right VLSM is a HOG!

Starting address: 172.16.160.0 /21

Anywho start with the highest number of host addresses required and subnet.

623 Hosts

Step 1)
First find out how many host bits you will need using the formula:
2^n - 2 = # of hosts

2^n - 2 = 623
2^n = 625
n = 10

So 10 bits is going to make up the host portion.

If you plug 10 back into the formula you will have:

2^10 - 2= 1022

399 Host addresses will be wasted but that is the best we can do.

Also, this formula tells us that there will be a total of 1024 addresses per subnet.


Step 2)
What is the mask going to be?
Since 10 bits make up the host portion, the mask will look like this:

11111111.11111111.11111 1 00.00000000 or /22

First portion being the Network, next being Subnet, and lastly the host portion

Step 3)
What are the subnets for this mask?
The starting address being 172.16.160.0, this will be your begining point.

Since each subnet will contain at least 1024 addresses as indicated by Step 1, add 1024 to the starting address and you will end up with 2 avalible subnets.

172.16.160.0 + 1024 = 172.16.164.0

Subnet 1 = 172.16.160.0 /22
Subnet 2 = 172.16.164.0 /22

Notice in Step 2 there is only 1 Subnet bit and therefore only two avalible subnets for the /22 mask.

Step 4)
Identify the Interface Address and Broadcast Address
From Subnet 1, we can identify the first host address as 172.16.160.1 /22 which will be the R3 fa0 interface and the broadcast address as 172.16.163.255 /22. You have now found the first valid address for the segment requiring at least 623 hosts.

Step 5)
Choose the next highest host requirement.
The next highest host requirement is 510 hosts.

2^n - 2 = 510
n = 9

Therefore,

2^9 - 2 = 510

This time we have used up every single host address avalible for this subnet.

Each subnet will contain 512 addresses.

11111111.11111111.11111 11 0.00000000 or /23

Two subnet bits tell us there will be a total of 4 subnets for the /23 network.

Since we have already used up addresses 172.16.160.0 - 172.16.163.255, our next starting address will be 172.16.164.0 which is the start of Subnet 2 in Step 3.

Subnet 1 = 172.16.164.0 /23
Subnet 2 = 172.16.166.0 /23
Subnet 3 = 172.16.168.0 /23
Subnet 4 = 172.16.168.0 /23

172.16.164.1 /23 can be used as R1's fa0 interface.

Continue this process for all the addresses