Understanding Subnetting IP Addresses

[Sh4dow]

I am having a trouble understanding how to subnet ip addresses. I am using the Microsoft Official Academic Course guid 70-293 and their explanations just don't make sense to me.

Can someone give me a few ideas on how to do this in a way that actually makes sense?

For each of the following IP address assignments, specify the # of bits in the subnet identifier, the # of possible Ip addresses in each subnet, the subnet mask for the IP addresses, and the Ip address ranges for the first and last subnet.

10.0.0.0/19

192.168.214.0/29

172.28.0.0/20

yes, I could just use a calculator but I'd like to know how to do this manually!

Thank you!

Edited February 10, 2006 by Sh4dow

[eyeball]

lets take 10.0.0.0/8 its a standard class A address and has a S/N of 255.0.0.0

so if we wanted 8 subnets we do 2 to the power 3 which just happens to be eight (dont want to complicate it yet)

and then we take that 3 and add it to the subnet mask to create an extended Subnet mask like so

1111111.11100000.00000000.00000000

so our extended sn mask is now 11 bits long so we now have 10.0.0.0/11 as opposed to the original 10.0.0.0/8

ok so we are left with 21 zeros's or bits for the host ID so we do 2 to the power 21 and get 2097152.

so in summation we took a 10.0.0.0/8 and subnetted it to 8 subnets each with 2097512 hosts.

the ip range for each subnet is calculated from this below

s/n 1 - 00001010.000 00000.00000000.00000000 = 10.0.0.0/11

s/n 2 - 00001010.001 00000.00000000.00000000 = 10.0.0.32/11

s/n 3 - 00001010.010 00000.00000000.00000000 = 10.0.0.64/11

and so on............

hope this helps you mate

Edited February 10, 2006 by eyeball

[fizban2]

now for the S/N why a 10.0.0.0/27?

[Sh4dow]

I'm actually still lost.

[nmX.Memnoch]

I think something got screwed up in the bottom part of his post...part of the binary data is missing in the middle of each S/N example.

[eyeball]

sorry yes that should have been /11 lol iv edited my post

sorry wasnt thinking straight

[fizban2]

thank goodness,

i thought i was going insane..... couldn't for the life of me figure out where you got 27 from

[nmX.Memnoch]

The binary data is still messed up. Only half of the information for the two middle octets is displayed.

[fizban2]

Lets take your first IP range for example.

10.0.0.0/19

Each 1 or 0 counts as a bit, we have 32 bits in a subnet

we have 19 bits assigned for the network ID giving us

11111111.11111111.11100000.00000000 or a subnet mask of 255.255.224.0

we got the 224 for the 3rd octet by adding the values of each bit, 2^8(128) + 2^7(64) + 2^6(32) = 224

now the remaining 13 bits we have are for the Host ID we use this number to find the number of IPs that can be hosted total. so 2^13 - 2 (can't have a 10.0.0.0 and 10.0.31.255 is for broadcast) = 8190

now we need to find the number of subnets that we have. since the default a class ip range has a subnet of 255.0.0.0 and we have 255.255.224.0 it has changed a bit, lets look at the binary

255.0.0.0 = 11111111.00000000.00000000.00000000

255.255.224.0= 11111111.11111111.11100000.00000000

So to find the number of subnets lets look at out Host ID = 00000.00000000

this is hte number of bits that aren't ones in the subnet mask. to find the number of subnets just start filling in with ones

HostMin: 10.0.0.1 (NetworkID) 00001010.00000000.000 (hostID) 00000.00000001

HostMax: 10.0.31.254 (networkID) 00001010.00000000.000 (HostID) 11111.11111110

so we have 32 subnets that each can have up to 254 clients in each subnet

so based on the questions you asked before

specify the # of bits in the subnet identifier, the # of possible Ip addresses in each subnet, the subnet mask for the IP addresses, and the Ip address ranges for the first and last subnet.

# of bits would be 19

# of Ip address would be 253 ( 0 doesn't count and 255 is reserved for broadcast)

subnet mask would be 255.255.224.0

IP address ranges for the First - 10.0.0.1 - 10.0.0.254 and the Last - 10.0.31.1 - 10.0.31.254 subnets.

let us know which part if holding you up, the more specific the better, i know they are a pain but once you get it, it will seem easly. Feel free to note errors in mine i am a little rusty

[Sprattney]

I happen to have the answers to this, and I would like to know how you get the bits? The actual answer is 11 bits for this problem. Doesn't this have something to do with 8 bits being in the subnet identifier? Which then would be /19-8, giving you eleven? I really don't know, this is all so confusing.

[arctirus]

This website looks like it was made in 1993 but it's very helpful. http://www.learntosubnet.com/

[Stoic Joker]

This website looks like it was made in 1993 but it's very helpful. http://www.learntosubnet.com/

LOL I was about to post the same link but you beat me to it. The site has been there since some time in the late 90s (that I know of...) But it is an excellent tutorial on Subnetting the I recommend frequently.

[chilifrei64]

another good resource to post...

A buddy of mine is a teacher and he post really all of his classes on his website. He does a great job in his classes and on his documents.

Here is the subnetting one

http://www.routeralley.com/ra/docs/ipv4.pdf

The rest of the site is really good too.. Obviously http://www.routeralley.com/

[Sprattney]

Thank-you everyone!

Chilifrei64, your friends sight is very good and very helpfull, thank-you very much!!

[tain]

Welcome back, ChiliFrei!

We've missed you 'round here