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# Jeff's Subnetting CHart
Below is a chart my course instructor, Jeff, sent me to study with for the CCNA.
I'm typing this out by hand in the hopes of making it stick in my head better.
Notes:
Each IPv4 address has 8 bits and each bit has a place, each place has a value.
This is positional notation.
If we explode the third octate, x.x._ _ _ _ _ _ _ _.x
The values for each place in an octet are 128 64 32 16 8 4 2 1
Adding these values together will get us: 128 192 224 240 248 252 254 255 which is the decimal mask based upon position.
Pretending we are in the third octet for example, if we were to list out masks in slash notation we would have:
/17 /18 /19 /20 /21 /22/ 23 /24
Putting those lines together get us a lot of information such as, a mask in decimal, block size, and decimal mask in slash.
128 192 224 240 248 252 254 255 (mask in decimal)
128 64 32 16 8 4 2 1 (positional notation and block size)
/17 /18 /19 /20 /21 /22 /23 /24 (mask in slash notation)
That is an excelent chart to use, we can add the slash notation depending upon which octet we are working in.
For example:
Given Class B 172.25.0.0 we need 0 hosts over 6 networks
Figuring they gave you the network ID an dsince it is class B the default mask is /16 or 255.255.0.0
We'd start from 172.25.0.0/16, the first two octets are network bits but based on the mask the 3rd and 4th octets are completly available. We need to subnet to meet their
requirements of 80 hosts per network and at least 6 networks. Since the last two octets are completely avilable we have 16 bits available to do that.
Using the the `2^h-2` and `2^n` formulas, where h=number of hosts bits left over and n=number of hosts bits borrowed.
2^6-2=62 but 2^7-2=126, so we need to ensure however many bits we borrow we leave at least 7 to meet the hosts requirements.
They want over 6 new networks so we need to borrow all of the rest of the bits from the 16 we had available, which is 9. We already set aside 7 for the host requirements
(16-7=9). We add the 9 bits (locking bits in place from left to right) to the bits we already have representing network bits (/16, the first two octets), which is 16+9 (16+9=25)
and that gives us a new mask of /25.
We can double check by subtracting the total number of bits from our 25 to see if we have the amount of bits needed for the required hosts, 32-25=7.
Another forumla we can use, 2^n, can be used to find the number of new networks. 2^9=512. We have lots of networks and should definietely be over 6.
Our subnets are going to start at the address we were given, 172.25.0.0, and go up from there by increments of 128.
New subnets:
1. 172.25.0.0/25
2. 172.25.0.128/25
3. 172.25.1.0/25
4. 172.25.1.128/25
5. 172.25.2.0/25
6. 172.25.2.128/25
7. 172.25.3.0/25
8. 172.25.3.128/25
# Finding the Network Address
Given 192.168.23.12/29
We're in the fourth octet so adding slash notation as well for the fourth octet
128 192 224 240 248 252 254 255
128 64 32 16 8 4 2 1
/25 /26 /27 /28 /29 /30 /31 /32
/29 gives a block size of 8.
Subnet ID: 192.168.23.8/29
First Usable: 192.168.23.9
Last Usable: 192.168.23.14
Broadcast: 192.168.23.15 (The final address before the next block)
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