Need help with Class B VLSM

[sabregod]

Hello to all,

I've been practicing subnetting for hosts requirement and I can't seem to figure it out. I've uploaded the question with instructor answers and my answer..

I'm at a dead end since i can't seem to know how he got those numbers. Am I doing something wrong? Here's my calculation:

https://i.imgur.com/D9DuZYz.jpg

Also, I don't know how the 3 reserved addresses he gave out relevant to this question at all. I'm a complete beginner so any help would be greatly appreciated.

[Julio E. Moisa]

Hi

VLSM is not complicated like it looks. If you are subnetting for hosts, you need to use this formula:

(2^n)-2 >= #hosts

and verify the available bits to get the required bits for "n"

Let me check the requirements and provide that to you with explanation

:-)

[Julio E. Moisa]

Hi

Please check the attached file and apologies for my English.

Hope it is useful

:-) 

[sabregod]

My issue is actually at host 100 and onwards. I'm getting different answers from my instructor. Does it have something to do 3 reserved addresses?

[Georg Pauwen]

Hello,

in addition to Julio's posts, there are a few useful online calculators such as the one linked below.

http://www.subnet-calculator.com/

This particular one shows you the host ranges available with different masks.

[Julio E. Moisa]

Hi

Attached you will the completed VLSM, the teacher has some restrictions so you cannot use some subnets and take in consideration the teacher is playing with the mind in some requirements the teacher takes the last subnet to complete the requirement and use the first to subnetting for the next requirement which is valid. 

Good exercise :-)

Please dont forget to rate the comment if it answered your question.

:-)

[sabregod]

On the 3rd requirement, can we use 172.128.51.0/25? Why did my teacher use .128/25? This is where I don't understand, I thought we could just pick any subnets

[Julio E. Moisa]

Hi

Yes, you can use the 172.128.51.0/25 and actually that should be used to keep an order during the VLSM process.

I am assuming your teacher used the subnet 128/25 to play with the minds, but it's valid although he is not keeping an order. 

If you dont have a order it is an easy way to make mistakes in the subnetting process so you need to be focused on the procedure.

:-)

[sabregod]

Thanks for answering my questions. So basically the addresses 172.128.56.0/21, 172.128.51.76/30 and 172.128.52.0/30 are all pointless?

I also asked another person about my problem and he says:

"In your addressing scheme, you assigned the 100 host space 172.128.51.0/25. This space goes from 172.128.51.0-172.128.51.127.

This overlaps with the reserved /30 above.

Your professor assigned the 100 host space 172.128.51.128/25, and then split 172.128.51.0/25 into smaller subnets, being careful not to overlap with the reserved /30. "

What do you think about his statement?

[chrihussey]

The other person you asked had a response similar to mine. The fact that the your professor said that the 172.128.51.76 /30 was to be reserved eliminated the 172.128.51.0 /25 from being assigned as a single block. It's purpose is to make sure you have an understanding of how to appropriately assign subnets and to figure out VLSM.

If you can understand why the answer the professor provided differed from yours and why your answer was basically incorrect you should get the point.

[sabregod]

Since 172.128.51.76 /30 is from a different subnet mask than 172.128.51.0 /25. How did 172.128.51.0 /25 got eliminated? This is the most confusing part. I just repeated what the other person said but I honestly don't know the logic behind it. Hope someone can explain it thoroughly :)

[chrihussey]

172.128.51.76 /30 uses the .76 thru .79 IPs. 172.128.51.0 /25 uses 172.128.51.0 - 127 IPs. So If you have reserved the 172.128.51.76 /30 for say a point to point link, or for whatever else, how could you possibly assign the 172.128.51.0 /25 to a LAN segment or anything else. There is an overlap of assigned IP space and a duplication of assigned IPs. If one was to do this traffic to anything in the 172.128.51.76 /30 range would have two possible destinations. Not good.

To put it another way, once you assign a smaller subnet of a /25 range, you cannot assign that /25 range somewhere else.

Ask away, once you understand it make perfect sense.

Regards

[sabregod]

Can you explain how 172.128.51.76 /30 uses the .76 thru .79 IPs?

[Julio E. Moisa]

Hi

It can be checked with the subnet mask 

172.128.51.76 /30

/30 = 255.255.255.252

Formula:

256 - 252 = 4

So the subnets will be increased by 4 in the 4th octet.

172.128.51.76 /30
172.128.51.80 /30

You have:

Subnet 172.128.51.76/30
Valid IP addresses:  172.128.51.77   <-- it can be used under an interface or any host
                                172.128.51.78   <-- it can be used under an interface or any host

Broadcast IP: 172.128.51.79  <--- it cannot be used under an interface, host either.

A subnet mask /30 can provide just 2 valid IP addresses. Often used to create a point to point to connect two layer 3 devices.

:-)

[sabregod]

If 172.128.51.76 /30 ranges from 172.128.51.77 to 172.128.51.79.

And 172.128.51.0/25 ranges from 172.128.0.1 to 172.128.127.254. Does that mean we couldn't use 172.128.51.0/25 for the 3rd host requirement?