VLSM - what do you do with the unused IPs ? are we able to group / reuse them ? - brainstorm

[SJ K]

Hi all,

At the end of all the VLSM subnetting over at the thread below.

http://computernetworkingnotes.com/subnetting-supernetting-and-vlsm/vlsm.html

The user is left with 20 ips unused; on his very last VLSM is 192.168.1.224/30 it gaves him another 8 more subnets of 2 usable hosts each.

In which he has only utilize 3 of those, leaving behind 5 subnets left (20 ips)

3 of the subnets used below

Subnet	Subnet 57	Subnet 58	Subnet 59
Network ID	224	228	232
First host	225	229	233
Last host	226	230	234
Broadcast ID	227	231	235

q1) Is there anyway whereby instead of having left with the choice to use the 5 subnets each with 2 usable hosts for future usages, is there any way i can consolidate the remaining 5 subnets into a larger subnet with more hosts ?

if i set a new subnet with 192.168.1.240/29, will there be any conflict ? as some of the other subnet  192.168.1.232/29 will be in conflict with the subnet 59 as shown above, but i will not be using them. I will only use 192.168.1.240/29.

So far from VLSM, i am seeing it as taking a subnet and break it further to smaller subnet , but can we also like group the remaining smaller subnets at the end into a bigger one ?

Regards,
Noob

[Jon Marshall · Accepted solution]

q1) yes because VLSM means you can use different subnet masks but you cannot overlap so you need to work out where the subnet boubndaries are.

So 192.168.1.232 255.255.255.252 is your last used subnet.

You could then have -

192.168.1.236 255.255.255.252

and then

192.168.1.240 255.255.255.240

or a different combination eg instead of the last subnet you could have -

192.168.1.240 255.255.255.248
192.168.1.248 255.255.255.248

You just have to know the bit boundaries.

Jon

[SJ K]

Hi Jon, Kumaresan,

Thanks for replying

Is there any quick tip / formula to quickly work out the bit boundaries ?

I am thinking of using the remaining ips (E.g. 20), closest = 2^4=16 which means I have to reserved 4 hosts bits without causing overlapping of the previous subnets and hosts, which also indirectly means that we will have additional 4 network bits = 24+4= /28

Regards,
Noob
 

[Jon Marshall]

Is there any quick tip / formula to quickly work out the bit boundaries ?

Easiest way is to subtract from 256

ie. 192.168.10.0 255.255.255.192

so take the last octet and 256 - 192 = 64

so your subnets go up in 64's ie.

192.168.10.0
192.168.10.64
192.168.10.128
192.168.10.192

you can do this with any subnet mask and it will show you the bit boundaries.

Jon

[SJ K]

Hi Jon,

Using my example above, even though I have already found the boundaries

192.168.1.232 255.255.255.252

(256-252=4) hence

192.168.1.232

192.168.1.236

192.168.1.240

192.168.1.244

192.168.1.248

192.168.1.252

===================================================

So how do I know from looking at the list above what could be the next subnet that can be use to be merge with the remaining together ?

How do we see that it is .240 onwards and not .236 ?

Regards,

Noob

[Jon Marshall]

What subnet mask do you want to use ?

Jon

[SJ K]

Hi Jon,
 

Assuming that the subnet mask I want to use will be the one that allows me to have the largest congregation of the remaining available ips.

How do i gauge from the boundaries subnets of /30 above, to start at which one ?

So far, i am writing down the binaries to see what will be the next available subnet to congregate which is .240 onwards, but how to do it without writing the binaries calculation ?

 

Regards,
Noob

[Jon Marshall]

So the starting point is 192.168.1.236.

Try applying the maths I suggested to the above eg.

255.255.255.192 = 256 - 192 = 64

so you know the last subnet = 192.168.1.192 which doesn't work.

255.255.255.224 = 256 - 192 = 32 ie. your subnets go up in 32s.

Here your subnets go up in 32s and if you counted from the start the last available subnet would be 192.168.1.224 which is no good.

255.255.255.240 = 256 - 240 = 16 so your subnets go up in 16s. Again if you counted from the start the last useable subnet is 192.168.1.240 which is okay because it starts after 192.168.1.236.

255.255.255.248 = 256 - 248 = 8 which will also work but you are looking for the largest subnet.

So you now know the largest subnet you can get from your remaining IPs is -

192.168.1.240 255.255.255.240

if you wanted to use the other IPs you would then need -

192.168.1.236 255.255.255.252 

ie. 256 - 252 = 4 and if you counted up from the start 192.168.1.236 is a valid subnet and the next one is 192.168.1.240 so it works.

If you entered 192.168.1.236 255.255.255.240 what that actually works out as is 192.168.1.224 255.255.255.240 which is not what you want because you have already used some of those IPs.

That is what I meant about bit boundaries.

You get used to the maths very quickly so I have been saying count up from the start but you can just as easily multiply the number in your head to work out available subnets and it does become second nature with practice.

Jon

[SJ K]

Hi Jon,

Thanks for the insight, using your theory, I hope i am right, i have applied it on

http://www.cisco.com/c/en/us/support/docs/ip/routing-information-protocol-rip/13788-3.html#cidr  on figure 5

Can you tell me what would be the best assignments to utilize all the available IPs left ? with the maximum congregation of ips available (no of subnet not a concern)

My calculation works out to be

202.15.5.112/28

then

202.15.5.128/25

----------------------------------

I will still be losing the IPs from 100 onwards to 111 though ( can only be use for point to point)

Let me know if i am right ? Thank you!

Regards,
Noob
 

[Jon Marshall]

That looks good to me.

Jon

[SJ K]

Thank you Jon.

My best mentor.

Last try Jon if you dont mind, before i officially declare i am VLSM savvy.

http://www.tutorialspoint.com/ipv4/ipv4_vlsm.htm

Congregation of available IPs  =

192.168.1.232/29 then

192.168.1.240/28

Am i right ?

Regards,
Noob

[Jon Marshall]

Again, looks good to me.

Think you've got a good understanding of it now both using binary and with the shortcut.

I'll be waiting for your next subject :-)

Jon

[SJ K]

Hi Jon,

Yea! Thanks to you.

Earlier I have posted a thread on DCE/DTE, let me know if you have any opinions

https://supportforums.cisco.com/discussion/12479536/dte-and-dce-equipments-modem

Thanks!

[Kumaresan Chinnasamy · Accepted solution]

Hi Noob,

We can use networks like below,

192.168.1.236/30
192.168.1.240/28

0r

192.168.1.236/30

192.168.1.240/29

192.168.1.248/29

We can group remaining networks but the number of ip's should be 2 to the power of something like above.