Originally posted by: drebo
Next you get to learn IPv6 addressing and subnetting! Woo!
FD00:1000:1:1::1/64 is a different network than FD00:1000:1:2::1/64.
How fun is that?
IPv6 was designed by the devil.
That's working for me.Originally posted by: Jeff7181
I was told if you ignore it, it will go away. That's not true?
Originally posted by: RebateMonger
That's working for me.Originally posted by: Jeff7181
I was told if you ignore it, it will go away. That's not true?
Originally posted by: drebo
Next you get to learn IPv6 addressing and subnetting! Woo!
FD00:1000:1:1::1/64 is a different network than FD00:1000:1:2::1/64.
How fun is that?
IPv6 was designed by the devil.
Originally posted by: Jeff7181
A subnet mask tells you which portion of the IP address is the network address and which portion is the client address.
Take the network address 192.168.1.0 for example
It's a class C address, so it can be written 192.168.1.0/24 or 192.168.1.0 and then you specify a subnet mask of 255.255.255.0
The /24 is how many bits (starting from the left) are used for the network portion.
A subnet mask is actually binary...
11111111.11111111.11111111.00000000 = 255.255.255.0 = /24
Lets say you want to subnet that... split it into two networks... you need to borrow bits from the client portion...
11111111.11111111.11111111.10000000 = 255.255.255.128 = /25
Now one of your network addresses is 192.168.1.0/25 and the client range is 192.168.1.1 - 192.168.1.126 (192.168.1.127 becomes the broadcast address for that network)
Your second network address is 192.168.1.128/25 and the client range is 192.168.1.129 - 192.168.1.254 (192.168.1.255 is the broadcast address)
Hope that helps.
*EDIT*
To add a little more info...
You can use a simple formula to calculate subnets...
2^n-2=X where n = the number of bits you borrow from the client portion of the mask and X >= the total number of subnets that will be provided
So if you have a class A address, 10.0.0.0/8 and you want to divide it into at least 10 subnets...
2^n-2 >= 10 (solve for n)
n = 4
So your first network address now becomes 10.0.0.0/12 which can be also written as 10.0.0.0 with a subnet mask of 255.240.0.0
Your second network address will be 10.15.0.0/12
We get that by looking at the second octet (because this is a class A address so we're borrowing bits from the second octet)... 11110000
If we invert that, we get 00001111 which happens to be 15
The third network address will be 10.30.0.0/12
And the forth... 10.45.0.0/12
So...
Network Address * Subnet Mask * Client Range * Broadcast address
10.0.0.0/12 * 255.240.0.0 * 10.0.0.1-10.14.255.254 * 10.14.255.255
10.15.0.0/12 * 255.240.0.0 * 10.15.0.1-10.29.255.254 * 10.29.255.255
10.30.0.0/12 * 255.240.0.0 * 10.30.0.1-10.44.255.254 * 10.44.255.255
etc. etc.
**EDIT**
One more thing... if you want to know how many clients are in each subnet, use the forumula 2^y-2 where y = the total number of bits used for the client portion of the address. So using the above example...
2^20-2 = 1048574
So if I have 10.0.0.0 and i want a subnet mask for 2048 subnets, i do 2^12=4096-2=4094.
Default Subnet mask for a class A address is 11111111.00000000.00000000.00000000 plus the 12 bits I need to borrow, so thats 11111111.11111111.11110000.00000000 which is 255.255.240.0
Can someone confirm?
Your second network address will be 10.15.0.0/12
We get that by looking at the second octet (because this is a class A address so we're borrowing bits from the second octet)... 11110000
If we invert that, we get 00001111 which happens to be 15
The third network address will be 10.30.0.0/12
And the forth... 10.45.0.0/12
So...
Network Address * Subnet Mask * Client Range * Broadcast address
10.0.0.0/12 * 255.240.0.0 * 10.0.0.1-10.14.255.254 * 10.14.255.255
10.15.0.0/12 * 255.240.0.0 * 10.15.0.1-10.29.255.254 * 10.29.255.255
10.30.0.0/12 * 255.240.0.0 * 10.30.0.1-10.44.255.254 * 10.44.255.255
etc. etc.
A subnet mask tells you which portion of the IP address is the network address and which portion is the client address.
Take the network address 192.168.1.0 for example
It's a class C address, so it can be written 192.168.1.0/24 or 192.168.1.0 and then you specify a subnet mask of 255.255.255.0
The /24 is how many bits (starting from the left) are used for the network portion.
A subnet mask is actually binary...
11111111.11111111.11111111.00000000 = 255.255.255.0 = /24
Lets say you want to subnet that... split it into two networks... you need to borrow bits from the client portion...
11111111.11111111.11111111.10000000 = 255.255.255.128 = /25
Now one of your network addresses is 192.168.1.0/25 and the client range is 192.168.1.1 - 192.168.1.126 (192.168.1.127 becomes the broadcast address for that network)
Your second network address is 192.168.1.128/25 and the client range is 192.168.1.129 - 192.168.1.254 (192.168.1.255 is the broadcast address)
Hope that helps.
*EDIT*
To add a little more info...
You can use a simple formula to calculate subnets...
2^n-2=X where n = the number of bits you borrow from the client portion of the mask and X >= the total number of subnets that will be provided
So if you have a class A address, 10.0.0.0/8 and you want to divide it into at least 10 subnets...
2^n-2 >= 10 (solve for n)
n = 4
So your first network address now becomes 10.0.0.0/12 which can be also written as 10.0.0.0 with a subnet mask of 255.240.0.0
Your second network address will be 10.15.0.0/12
We get that by looking at the second octet (because this is a class A address so we're borrowing bits from the second octet)... 11110000
If we invert that, we get 00001111 which happens to be 15
The third network address will be 10.30.0.0/12
And the forth... 10.45.0.0/12
So...
Network Address * Subnet Mask * Client Range * Broadcast address
10.0.0.0/12 * 255.240.0.0 * 10.0.0.1-10.14.255.254 * 10.14.255.255
10.15.0.0/12 * 255.240.0.0 * 10.15.0.1-10.29.255.254 * 10.29.255.255
10.30.0.0/12 * 255.240.0.0 * 10.30.0.1-10.44.255.254 * 10.44.255.255
etc. etc.
**EDIT**
One more thing... if you want to know how many clients are in each subnet, use the forumula 2^y-2 where y = the total number of bits used for the client portion of the address. So using the above example...
2^20-2 = 1048574
when asked which I.P. Addresses are valid for the /27 subnet, I have no clue
what to do?
Thanks in advance.