| Author |
Did I subnet this correctly?
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| Hi all.. I used freak's subnettng guide for the first time, and I just gotta say... It's freaking great! are the only works that come to my mind right now..
This whole subnetting mess just started making sence to me.. when I finally put the time into learning it from start to end ( I know this isn't the end  )
But, can someone tell me if I did this correctly please? (I tried to come up with some weird numbers to see where I would screw up)
Original IP: 205.10.100.x
Mask: 255.255.255.0
Amount of subnets needed: 8
New subnet mask: 255.255.255.240
IP Ranges are:
205.10.100.17 - 205.10.100.30
205.10.100.33 - 205.10.100.46
205.10.100.49 - 205.10.100.62
205.10.100.65 - 205.10.100.78
205.10.100.81 - 205.10.100.94
205.10.100.97 - 205.10.100.110
205.10.100.113 - 205.10.100.126
etc..
etc..
etc..
Thanks! | |
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| And how about this scenario:
Original IP: 150.100.x.x
Subnest mask: 255.255.0.0
Amout of subnets required: 24
New Subnet mask: 255.255.248.0
IP Ranges:
150.100.9.1 - 150.100.17.254
150.100.18.1 - 150.100.26.254
150.100.27.1 - 150.100.35.254
150.100.36.1 - 150.100.44.254
150.100.45.1 - 150.100.53.254
150.100.54.1 - 150.100.62.254
150.100.63.1 - 150.100.71.254
150.100.72.1 - 150.100.80.254
etc..
etc..
Thanks again.. | |
| cahillrobert 2002-06-27, 7:49 pm |
| Oops... The subnetd you were listing already had 7 network ranges and were only half way done the list.
Easy way to rememeber
Quick 1
if you need n number of subnets divide into 256 to find the total number of ip addresses given
256/8 = 32
Now you will need to figure out:
1. Will I get the right number of IP addresses
2. Can the number be fit into a subnet.
If you needed 5 subnets.
256 / 5 = 51.2 well that won't work.
need to break the subnets into something smaller.
Bit Address Subnet
25 = 128 255.255.255.128
26 = 64 255.255.255.192
27 = 32 255.255.255.224
---- or -----
Quick 2
One other quick way once you know the number of ip addresses (32)
256 - 32 = 224
Don't fret. This process will become a no-brainer after a while. After the nth time you've done this process in your head in a flash of a nano-second, just smile. | |
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| cahillrobert,
Thanks for the reply, but I'm sorry i'm not sure what you mean..
are you saying I did this right but didn't finishing listing the possible ip ranges?
Or did I mess up somewhere?
And I don't really understand your chart:
Bit Address Subnet
25 = 128 255.255.255.128
26 = 64 255.255.255.192
27 = 32 255.255.255.224
Any hints? | |
| cahillrobert 2002-06-27, 8:15 pm |
| quote:
Bit Address Subnet
25 = 128 255.255.255.128
26 = 64 255.255.255.192
27 = 32 255.255.255.224
Bit...IP Address..Subnet Mask
25 =..128 ........255.255.255.128
26 =...64 ........255.255.255.192
27 =...32 ........255.255.255.224
Very simply if you need a specific number of subnets divide that number into 256
will give you the number of ip addresses that network range will carry.
A. The number of hosts address will be two less.
B. One address will be required for the router interface.
As Above If I wanted 2 networks. 256/2
= This network will carry 128 ip addresses.
As Above I want 4 networks 256 /4 = 64
The network will be divided 4 times each carrying 64 addresses. Now if you do the following 256-64 = 192. Magically the octect (subnet address) you need is releaved 255.255.255.192
Continuing on:
8 networks 256/8 = 32 (addresses)
256-32=224 (the subnet) 255.255.255.224
Clearer?! | |
| cahillrobert 2002-06-27, 8:21 pm |
| Why 256?
When everything is 255.255.255..
Because the 0 is a real number beinging that it's all 1 0's .
So there are really 256 numbers available within the address spaces we use.
To check whether you did it right. If you want 8 subnets there should ONLY be 8 line items for subnet ranges. More or less - Oops. | |
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| imho i think he should just stick with good old power of 2 method for now...
any other way would prolly just add to his confusion as I think it is happening now.
when he master how to do it, and understand how and why it is done/needed etc, the rest become obvious.
plus, i am pretty sure microsoft do not account for subnet 0, not sure on that part... but if it does not, he is gonna die with your numbers ... | |
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| cahillrobert,
Yes, it's more clear now.. except one thing that i'm still not getting..
when you say 64 addresses and 32 addresses, what exactly are you talking about? It's probably something very simple, but I'd like you to show me exactly how these "address" would look..
And also, did I get my subnetting right? or was it wrong?
Thanks again.. | |
| cahillrobert 2002-06-27, 9:09 pm |
| Wat I am referring to when I indicate:
32 addresses
64 addresses
are the number of IP Address you will have on each network.
Each of the following 8 network ranges support - 32 IP addresses
192.168.1.0 - 192.168.1.31
192.168.1.32 - 192.168.1.63
192.168.1.64 - 192.168.1.95
192.168.1.96 - 192.168.1.127
...
192.168.1.224 - 192.168.1.255
each of these address will use the subnet mask of 255.255.255.224
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If 4 subnets then ...
192.168.1.0 - 192.168.1.63
192.168.1.64 - 192.168.1.127
192.168.1.128 - 192.168.1.191
192.168.1.192 - 192.168.1.255
64 IP Addresses for each of these networks & will use the mask of 255.255.255.192
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Mike has a valid point and concern. Microsoft is not in the router business and has a tendency to not use what is called subnet-zero so MS's take on things are somewhat scewed. I have been using the above mechanism for many years without failing an exam, installation (router, switch, or server) yet. | |
| BlackEvoVII 2002-08-11, 2:14 pm |
| Even though this is late, very late.
THe reason why he got 240 for the network portion is mostely because he was using a classful addressing scheme that probably supported only RIP I(being 00000000 is reserved as well as 11111111 is resevered) or in his case the 4 bits allocated to the network portion would be 0000 and 1111. Thus a 224 network portion would only give him 6 useable hosts. |
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