VLSM induced paranoia
mattipler
Member Posts: 175
in CCNA & CCENT
Good Evening All,
Just want to double, double check i've got a complete understanding of VLSM.
If I'm given a network lets say 172.16.1.0 /24 and have to provide for a subnet containing 53 hosts, 35 hosts, 20 hosts, 6 hosts and 3 ppp's. The following addressing scheme would work would it not? I'm sure I've got this sorted but had a wave of preexam self doubt! Could someone have a quick look if they get a second and confirm this is alright.
53 hosts 172.20.1.0 255.255.255.192 /26
35 hosts 172.20.1.64 255.255.255.192 /26
20 hosts 172.20.1.128 255.255.255.224 /27
6 hosts 172.20.1.160 255.255.255.248 /29
ppp 1 172.20.1.168 255.255.255.252 /30
ppp 2 172.20.1.172 255.255.255.252 /30
ppp 3 172.20.1.176 255.255.255.252 /30
Nice one.
Manchester Matt
Just want to double, double check i've got a complete understanding of VLSM.
If I'm given a network lets say 172.16.1.0 /24 and have to provide for a subnet containing 53 hosts, 35 hosts, 20 hosts, 6 hosts and 3 ppp's. The following addressing scheme would work would it not? I'm sure I've got this sorted but had a wave of preexam self doubt! Could someone have a quick look if they get a second and confirm this is alright.
53 hosts 172.20.1.0 255.255.255.192 /26
35 hosts 172.20.1.64 255.255.255.192 /26
20 hosts 172.20.1.128 255.255.255.224 /27
6 hosts 172.20.1.160 255.255.255.248 /29
ppp 1 172.20.1.168 255.255.255.252 /30
ppp 2 172.20.1.172 255.255.255.252 /30
ppp 3 172.20.1.176 255.255.255.252 /30
Nice one.
Manchester Matt
Matt of England
Comments

dtlokee Member Posts: 2,378 ■■■■□□□□□□mattipler wrote:Good Evening All,
Just want to double, double check i've got a complete understanding of VLSM.
If I'm given a network lets say 172.16.1.0 /24 and have to provide for a subnet containing 53 hosts, 35 hosts, 20 hosts, 6 hosts and 3 ppp's. The following addressing scheme would work would it not? I'm sure I've got this sorted but had a wave of preexam self doubt! Could someone have a quick look if they get a second and confirm this is alright.
53 hosts 172.20.1.0 255.255.255.192 /26
35 hosts 172.20.1.64 255.255.255.192 /26
20 hosts 172.20.1.128 255.255.255.224 /27
6 hosts 172.20.1.160 255.255.255.248 /29
ppp 1 172.20.1.168 255.255.255.252 /30
ppp 2 172.20.1.172 255.255.255.252 /30
ppp 3 172.20.1.176 255.255.255.252 /30
Nice one.
Manchester Matt
You were given 172.16.1.0/24 and in all your ranges you used 172.20.1.0
Other than that it looks good, I typically would put all the /30's at the upper end, start from .255 and work backwards.The only easy day was yesterday! 
Darthn3ss Member Posts: 1,096i was taught (At school in CCNA 3) to start with your largest networks (and start at .0)........... any real difference in the method?Fantastic. The project manager is inspired.
In Progress: 70640, 70685 
Netstudent Member Posts: 1,693 ■■■□□□□□□□That kinda threw me as well.There is no place like 127.0.0.1 BUT 209.62.5.3 is my 127.0.0.1 away from 127.0.0.1!

NeonNoodle Member Posts: 92 ■■□□□□□□□□Darthn3ss wrote:i was taught (At school in CCNA 3) to start with your largest networks (and start at .0)........... any real difference in the method?
I would do just as you were taught. Subnetting should start at zero and go from the largest to the smallest networks. You can do it differently, but you have to be really careful about choosing a correct subnet address.
For example, let's say you want to create two subnets: Subnet A with four hosts and Subnet B with thirty hosts. Starting with Subnet B, you would have 192.168.1.0/29. Now what would the starting subnet address for Subnet B be? Could it be 192.168.1.8/27? No. It would have to be 192.168.1.32/27.
Now, reversing the procedure, start with Subnet B. Its subnet address will be 192.168.1.0/27. Then Subnet A's address will be 192.168.1.32/29the next freely available address. This helps with figuring out what the next subnet address can be.
If you go from largest subnet to smallest subnet you can keep the addresses together, i.e. the subnet address of a subnet will be number following the broadcast of the previous subnet.
The reason why has to do with how many address bits a subnet needs. Subnet A requires 3 bits for 2^32 host addresses and Subnet B requires 5 bits for 2^52 host addresses. Now the numbers 2^3 and 2^5 are important. For a subnet address to be valid, in this example, it must be divisible by one of those numbers. In the first example, 2^3 divides 0 but 2^5 does not divide 8. In the second example 2^5 divides 0 and 2^3 divides 32.
In general, if your subnet requires X bits then the subnet address must be .2^aX, where a is an integer and 0<=aX<256 because 2^X must divide 2^aX.I recognize the lion by his paw.
Jacob Bernoulli 
Netstudent Member Posts: 1,693 ■■■□□□□□□□As usual a very good explanation neon. It helps to have a strong math backround doesn't it?
I know it did for me.
But yeah, the divisibility rule becomes harder to manage when you start with smaller blocks and you tend to waste IP's. I did a little VLSM writeup a while back that showed this very concept.
http://www.techexams.net/forums/viewtopic.php?t=24119There is no place like 127.0.0.1 BUT 209.62.5.3 is my 127.0.0.1 away from 127.0.0.1! 
mattipler Member Posts: 175Woops! Well spotted dtlokee! School boy error on my part there! Can't afford many of them tomorrow afternoon! At least I've completely understand the concept which is what concerned me a little. Cheers lads.
MattMatt of England