Sign in with
Sign up | Sign in
Your question
Solved

Probably far beyond the scope of this forum

Last response: in Networking
Share
June 2, 2012 8:13:11 AM

sorry if it is, but what the hell, i'll ask just in case someone out there might know:

if you understand the concepts of network addressing, such as classful vs clasless ...read on, if not, don't bother.

if you are given a network to assign addresses ok.....using VLSM CIDR....

for one thing you are given the address 172.16.1.0 /28 to assign for WAN serial links and you need to subnet it ok?

your first subnet is going to be prefix /29, you have a binary address of 10101100.00010000.00000001.00000000

your mask is 11111111.11111111.11111111.11111000

if you had to assign 3 disparate wan network links....would you take that first /29 with the 255.255.255.248 mask and use that for one of the networks, that is all you can do right? then goto /30 for the next network, leaving you with /31 for the last network which only gives you a point to point connection (2 addresses total).....

my confusion is, it seems there is more that you can do here, but i don't see it. i can just arbitrarily change network bits to suit my needs as long as i don't assign hosts outside of the last 4 bits? something i'm missing because i'm looking at one topology
that is using the /30 mask with a 192.168.10.0 network ..and assigned 192.168.10.4 as one network on one serial wan link, and then goes and assigns 192.168.10.8 /30 to another serial wan link connected to the first...where is this leeway coming from for addressing when using /30, it is only 2 bits...

More about : scope forum

June 2, 2012 6:25:13 PM

Serials links between two routers only require 2 addresses /30 for a single link. Any more addresses would just be wasted. Don't use /31 for this application, that is just a single address.

For serial links between routers, a /28 network can be subnetted into 8x /30 subnets = 2 host addresses per subnet (1 host address per router for each side of the network). This gives you enough subnets for 8 links.

For example:

NETWORK 1
Network Address: 172.16.1.0
Router 1: s0/0/0 - 172.16.1.1
Router 2: s0/0/0 - 172.16.1.2
Broadcast Address: 172.16.1.3

----

NETWORK 2
Network Address: 172.16.1.4
Router 3: s0/0/0 - 172.16.1.5
Router 4: s0/0/0 - 172.16.1.6
Broadcast Address: 172.16.1.7

and so on...

Refer to this chart to better visualize subnet proportions:
http://wb2bwu.org/drupal617/sites/default/files/picture...
m
0
l
June 3, 2012 4:21:40 AM

i think you have helped me to finally narrow down where my brain is refusing to compute...because the funny thing is, i just did this same technique with a /21 address to /24 giving it a bunch of 254 hosts.....i realize why in my mind there is a difference, it has to do with the bits in the subnet mask...the fact that doing this same thing with a /24 gives the ability to keep flipping that 3rd octet number, where as in this situation you cant do that. lemme show you what i mean, and this is the exact point of my confusion:


subnet mask /30 = 11111111.11111111.11111111.11111100

so you take that 5th and 6th bit for your network and keep switching those hosts so that you can assign the 2 addresses right? that is what you said in essence. here is where i'm confused, look at the very first subnet that you listed 172.16.1.0 ok? to get the first addresses in that network 1.1 1.2 etc you need to change the very final bits in the 4th octet, the question is, how can you do that? they are not masked under the subnet mask? THAT is where i cannot understand this. thank you for the reply.
m
0
l
Related resources
June 3, 2012 6:10:07 AM

subnet mask /30 = 11111111.11111111.11111111.11111100 only represents the number of possible host addresses allowed.

The 1's are network bits, they are lock in and don't change.
The 0's can change.

Given there are only 2 0's for a /30 that gives only 4 possible combinations:

11111111.11111111.11111111.111111-------00
11111111.11111111.11111111.111111-------01
11111111.11111111.11111111.111111-------10
11111111.11111111.11111111.111111-------11

Remember the 1's already in place cannot change therefor only the last 2 0's can be altered thus only 4 addresses allowed for /30.
Also note that for ANY /30 address those same 4 binary masks apply.

As you lower you subnet masks, for example go from /30 to /29, the last network bit into a host bit so it would equal:

11111111.11111111.11111111.11111000

Now you have 3 0's, giving you more combinations, thus more addresses.

The number of 1's is always = to the /30 so:
/30 = 30 1's
/29 = 29 1's
/28 = 28 1's
etc

Each time you lower the slash notation number, you get more host bits thus more address.
m
0
l
June 3, 2012 6:42:34 AM

i understand what you said in your first post, where you put "network 1" "network 2", what are the masks? i think that will help me clear this up. i had a 10.1.16.0 /21 network, and i assigned it a /24 mask, and then just started flipping the number in the 3rd octet like 10.1.17.1-254, then 10.1.18.1-255 all under the same mask, that may have been wrong, but it seems like something has to do be done differently then the standard subnet process which will just frontload so many hosts and then you wind up with one subnet that has 1000 hosts, 2 subnets with 500, 3 with 250, 4 with 125 approx, etc. so to even it out, i did that, maybe it was wrong i dunno. i'm definitely missing something. you see what im sayin? i know how to do your standard subnetting....but taking a subnet that is /21 that has 2k hosts or whatever, then going to /22 then /23 , it just doesn't distribute hosts well when you have 8 networks that all need like 200 hosts ....
m
0
l

Best solution

June 3, 2012 8:06:34 AM

1. The first step is to figure out how many IP addresses you need per subnet. You have done this and the number is 200.

2. The next step is to figure out which subnet can accommodate for 200 IP addresses.

3. /25 gives a total of 128 addresses, 126 of which are usable host addresses. so that is not enough if you need 200 hosts per network.

4. /24 gives a total of 256 addresses, 254 of which are usable host addresses, this can accommodate the 200 IP addresses specified, per subnet, with some left over addresses.

The little chart below demonstrates how 8 /24 subnets can fix into a /21 subnet.
------------------------------------------------------------
10.1.16.0 to 10.1.16.255 /24 |
10.1.17.0 to 10.1.17.255 /24 |
10.1.18.0 to 10.1.18.255 /24 |
10.1.19.0 to 10.1.19.255 /24 |
10.1.20.0 to 10.1.20.255 /24 |---------/21
10.1.21.0 to 10.1.21.255 /24 |
10.1.22.0 to 10.1.22.255 /24 |
10.1.23.0 to 10.1.23.255 /24 |
------------------------------------------------------------

Note that with VLSM you can manipulate the subnets within the /21 above to make them bigger or smaller by dividing or adding subnets together by at even number intervals.

The bottom line is you can have any combination of subnets within a particular network mask (/21 in example) as long as they fit and don't overlap your good to go.
Share
June 3, 2012 8:32:08 AM

Best answer selected by Frizzo.
m
0
l
!