[VLSM] Overlapping Subnets?

Simrid

Hi Guys,

I'm just revising for my ICND1, going back over some VLSM. I have a question in my book which is:

"R1 has configured interface FA0/0 with the ip address 10.5.48.1 255.255.240.0 command. Which of the following subnets, when configured on another interface on R1, would not be considered an overlapping VLSM subnbet?

A. 10.5.0.0 255.255.240.0
B. 10.4.0.0 255.254.0.0
C. 10.5.32.0 255.255.224.0
D. 10.5.0.0 255.255.128.0"


The answer, is A, but why? I mean, I guessed this questions correctly purely based on the fact it's a 10.5.0.0 address and the share the same subnet.

Could someone perhaps explain what an overlapping VLSM subnet is and how I can spot them?

Thanks everyone & have a great day!

Simrid

Ltat42a

Easiest way to figure questions like this out is to determine the range of the network in the question. Determine what network is 10.5.48.1/20 in, and what the broadcast address is. You'll be able to determine which of the four answers is correct.

Simrid

Ltat42a wrote: »

Easiest way to figure questions like this out is to determine the range of the network in the question. Determine what network is 10.5.48.1/20 in, and what the broadcast address is. You'll be able to determine which of the four answers is correct.

Thanks for your time and reply. So to double check, is it a matter of finding out the network ranges, looking at the host and broadcast addresses and seeing which are viable?

Priston

An overlapping VLSM subnet is a subnet that shares some of the same addresses

192.168.125.0 255.255.255.0
192.168.125.1 - 192.168.125.254

192.168.96.0 255.255.224.0
192.168.96.1 - 10.5.127.254

192.168.125.1 - 192.168.125.254 is within the range of 192.168.96.1 - 10.5.127.254 so they overlap

satishtech

Overlapping example * :
192.168.255.192/26 Subnet 1
192.168.255.224/27 Subnet 2

Subnet 1 : ( /26 = 64)
Network number : 192.168.255.192
First Host : 192.168.255.193
Last Host : 192.168.255.254

Range = 193 to 254

Subnet 2 : (/27 = 32)
Network Number : 192.168.255.224
First Host : 192.168.255.225
Last Host : 192.168.255.254

Range: 225 to 254

There is over lapping as we see the same
Host IP addresses in both the Subnets.

Another Example *:
Subnet1:
192.168.5.15 / 24
network 192.168.5.0 (range from 192.168.5.0 - 192.168.5.255)

Subnet2:
192.168.6.240 / 22
network 192.168.4.0 (range form 192.168.4.0 - 192.168.7.255

Again Host IP addresses from subnet 1 show up in subnet 2.

*Source : Yahoo answers.

Jon_Cisco

The trick is to be able to determine the range quickly in your head.
For me I look for block size and see what fits in it. Everyone subnets a little different.

What this question is trying to get you to recognize is not that the mask is the same but that the .0.0 subnet is different then the .48.0 subnet.
What you can't have is two network ranges that include the same addresses. Perhaps the one meant to trick you is answer B because if you look at it quickly you might not realize the range of 10.4.0.0 -10.5.255.255

A great site to practice is Subnetting Practice Questions

satishtech

The reason why A 10.5.0.0 255.255.240.0 is correct .
10.5.0.0 / 20
/20 = 16
Network number = 10.5.0.0
First Host = 10.5.0.1
Last Host = 10.5.15.254

The Hosts lie between 10.5.0.1 to 10.5.15.254

And in the reference subnet 10.5.48.1 255.255.240.0
The hosts lies between 10.5.48.1 to 10.5.63.254.

No identical Host IP's in both subnets.