How to find the Range to divide the ip addresses ?

tahakirmani

Hi,

I have 7 Networks and the ip address is 184.54.0.0 with sub net of 255.255.255.0
I have 200,100,100,50,2,2,2 Hosts in these 7 Networks.

I have understand how to perform VLSM, but unable to understand how to divide this IP address for 7 Networks.

If i take 2^7= 64 Hosts per network then it will not fullfil the requirements of 200,100 hosts.

Kindly guide me how to find the range for subnetting?

Thanks
Taha

neno2014

I am still very much a student so I am interested in how this will be solved.

2^7=128-2=126 hosts per subnet by the way.

This is just a stab in the dark but if you are required to have that many hosts in one network then you will need to use a 255.255.254.0 subnet so you can get a total of 510 hosts.

Don't be to hard on me I am still a student.

esr0159

Hi,
is your subnet mask really 255.255.255.0 (Class C) or 255.255.0.0 (Class ? The max number of host based on your given is 254 (2^8-2)

tahakirmani

Thank You so Much for the Response


So it means my sub net will be something like this..
184.54.0.0
255.255.255.0
7 Networks
Maximum Numbers of Host = 200;


2^8-2= 256-2=254


Network 1:
Network :184.54.0.0IP: 184.54.0.1-184.54.0.254
Brodcast: 184.54.0.255



Network 2:
Network :184.54.1.0IP: 184.54.1.1-184.54.1.254
Brodcast: 184.54.1.255



Network 3:
Network :184.54.2.0IP: 184.54.2.1-184.54.2.254
Brodcast: 184.54.2.255

.
.
.
.

tahakirmani

Why 2^8 when i have 7 networks?

davenull

Like esr0159 said, the way the problem is presented there is no solution. The subnet mask of 255.255.255.0 gives you only the last octet to play with, which is not enough to accommodate a total of 456 users that you need.

The lowest number of bits we need to have to fit 456 users is 9 (2^9=512), so your mask would have to be at least /23 or 255.255.254.0 before we can even start dividing the pie.

tahakirmani

Sorry that was my writing fault, Its /23, but still i am unable to understand why 2^9 when i have 7 Networks, as per my knowledge it should be 2^3=8 for 7 Networks?
Kindly help me on this i am very confused in sloving Fixed Length Sub netting problems.

neno2014

Ahh, that is more doable but as I am terrible at vlsm on any other class other than a class C. I am trying this on paper but I am confusing myself, I am looking forward to seeing this solved by someone. I am still trying this and will test with packet tracer.



just a quick update, it does not seem that this is possible. The first three required networks take up all of the available ip addresses. The networks I get are 184.54.0.0, 184.54.1.0, 184.54.1.128 after which point we are out of available addresses.

Can someone confirm?

tahakirmani

Can You solve this using Fixed Length Sub netting ?

davenull

yep, the /23 still doesn't solve it according to my napkin math:

184.54.0.0 /23 - borrowing a bit to fit 200 users to get 2 subnets:
184.54.0.0/24 ---a slice for 200 users
184.54.1.0/24 --- divide in 2:
184.54.1.0/25 --- first 100
184.54.1.128/25 --- second 100
... and we can't fit 50 anywhere

neno2014

Is vlsm a requirement for this situation? Is this from an assignment, can we get more information?

I have done something similar to this on a class C, with fewer needed hosts, but this one seems to be lacking some essential information.

tahakirmani

Yes, i need to solve this as a fixed length not as VLSM, I am unable to understand how to solve this using subnetting.

tahakirmani

davenull wrote: »

yep, the /23 still doesn't solve it according to my napkin math:

184.54.0.0 /23 - borrowing a bit to fit 200 users to get 2 subnets:
184.54.0.0/24 ---a slice for 200 users
184.54.1.0/24 --- divide in 2:
184.54.1.0/25 --- first 100
184.54.1.128/25 --- second 100
... and we can't fit 50 anywhere

Is this the VLSM method?

esr0159

tahakirmani can you post the updated problem?

tahakirmani

I have 7 Networks and the
IP Address = 184.54.0.0 with
SUB NET = 255.255.254.0

I have 200,100,100,50,2,2,2 Hosts in these 7 Networks.

How to solve the above Problem using Simple Sub-net Method?

esr0159

255.255.254.0 (/23) still doesn't fit the host requirements. (check davenull's post above),

Taha, maybe you should try some basic subnetting first

If the subnet mask for the network is 255.255.252.0 (/22)
200 hosts 184.54.0.0/24 (2^8=256)
100 hosts 184.54.1.0/25 (2^7=12)
100 hosts 184.54.1.128/25 (2^7=12)
50 hosts 184.54.2.0/26 (2^6=64)
2 hosts 184.54.2.64/30 (2^2=4)
2 hosts 184.54.2.68/30 (2^2=4)
2 hosts 184.54.2.72/30 (2^2=4)

tahakirmani

Thank you so Much esr0159 for the Reply, You did it using VLSM? Can yiu tell me how to do this using Normal Subnetting ?

Kai123

If you are using a given IP address, can you can chop and change the host bits but you can never go above the network IP you were given?

For example, 171.128.0.0/17

171.128 is locked, the rest you can chop and change as needed? Its actually hard to find a VLSM guide that uses "Class B" or A subnets, its all Class C, which is easy to understand and does not help with the more confusing subnets below /24. Even decent ANDing guides A and B subnetting is hard to come by. If anyone knows of one lets us know! I'm sure Tahakirmani will appreciate it as well

davenull

I think to really understand vlsm, you have to practice doing it in binary, or at least seeing what's happening on the binary level. Then it doesn't matter whether it's Class C or B or A, the principles remain the same. The same applies to summarization.

When in doubt, I literally imagine what it looks like in binary and that helps a lot.