Part 2 of this series begins with one of the harder skills to master in networking; IP Addressing Resolution and Subnetting. Ask anyone who has ever studied to take the CCENT or CCNA exams, and they will most likely agree with me. However, I think that most teachers will over-complicate it by stressing binary conversions, the base2 numbering system that breaks this all down to a series of ones and zeros. I find that it’s easier and faster to work with IP addressing if you use as little binary as possible. Why? Because it’s too slow to work with and it’s an extra step that’s just not necessary. I’ve stressed in my earlier posts that speed is essential when taking the CCENT/CCNA exams. Since IP addressing resolution is prevalent throughout the exams, it only makes sense to learn a few shortcuts.
Please be patient and read the post a couple of times. When you get it (and you will) you're going to wonder why you thought it was so hard in the past.
Without wasting any more time, here is an explanation of the next basic skill in the series:
Resolving an IP Address
– Find the Network, Broadcast, First and Last Host, and Next Network Addresses
A typical question might ask you to find the network address
where a given host address resides, or find the useable host address range for
a given network. The useable address range does not include the network address
(first) or broadcast address (last) for that network. If you are designing an
addressing scheme, you will need to know how many networks you can create, and
where one networks ends, and another begins.
Let’s look at a typical host address and mask, and then find
the network address:
192.168.1.55 /24
or 255.255.255.0 (if you are given the mask in dotted-decimal)
The first thing that you need to know is that the subnet
mask is used to filter out the host information and preserve only the network
information of the address. The mask is used to show you where the network bits
end and the host bits begin, so you can zero out the host bits. When a router
looks at the destination address of a packet, it initially only cares about the
network address in order to make a decision on where to forward the packet
next.
In the example above, /24 means that the first 24 bits of
the address (from left to right) are network bits, which should be left as is,
and the last eight bits should be changed to zeros (0). Recall that there are
32 bits in an IP v4 address. The dotted-decimal mask, 255.255.255.0 means that all of the bits in the first three
octets are set high (or are 1’s), and all of the bits in the last octet are set
low (or are 0’s). When evaluating each bit in the host address you would
compare it to each matching bit in the subnet mask. Just remember the rhyme: If it’s high, let it ride. If it’s low, let
it go. I just made that up (And my English teacher said I would never
amount to anything). Here is how that process would look like in binary.
Notice that only the ones (1’s) are left from the first 24 bits. All of the host bits are changed to zero’s (0’s). In fact, the first three octets from the host address are exactly the same after applying the mask. Two important shortcuts to learn from all of this are:
-
If the mask bits are all 1’s in a given octet, then the matching host octet stay’s the same (like in octets 1-3 in the example above)
- If the mask bits are all 0’s in a given octet, than the matching host octet changes to all zero’s (like in octet 4)
Here are five host
addresses followed by the 3 octets that are easily resolved in red:
166.32.114.184
255.255.224.0 ======>
166.32.114.0
168.168.10.65
255.255.255.224 ======>
168.168.10.65
149.90.165.115 /21 ======> 149.90.165.0
126.5.43.153 /15 ======> 126.5.0.0
150.163.46.151
255.255.128.0 ======>
150.163.46.0
Note: The octet that's left in bold black text would still need to be resolved (coming).
Let’s look at a couple examples where we find the network
address from a given host address.
Finding the Network
Address - Example 1
Find the network address from the following host address and
mask:
172.16.76.125 /18 or 255.255.192.0
Octets 1, 2, and 4 can be resolved with just a glance: 172.16.76.0. Octet 3 still needs resolving.
Here’s the binary
depiction. As you can see, writing it out in binary takes a lot more time than
using the shortcuts.
To resolve the third octet without binary we use the Jump Number method. The Jump Number is the decimal value of the last network bit in the subnet mask (see below). The 18th bit (as in /18) is equal to 64, which means that the network number for that octet will be some increment of 64, starting from zero (0, 64, 128, 192).
Tip: The host number will always be greater than or equal to the network number for the octet that you are resolving with this method
Since the number in the host address for that octet was 76 and it falls between 64 and 128, the network number is 64 and the network
address is 172.16.64.0. The network number for the unresolved
octet will always be an increment of the Jump Number as long as you remember to
start counting at 0.
Finding the Network
Address - Example 2
Find the network address from the following host address and
mask:
188.42.59.45 /21 or 255.255.248.0
Since the network bits end in the third octet again (you can
see this better by looking at the dotted-decimal mask) the unresolved octet
will be octet 3 again. The three octets that can be resolved at a glance are 1, 2
and 4 or 188.42.59.0
This time the last network bit in the subnet mask is the 21st
bit, which is equal to 8. 
Back to our example: The jump number that was found is 8. This means that the network number for that octet will be
some increment of 8, starting from zero (0, 8, 16, 24, 32, 40, 48, 56, 64,
etc.. .). Hint: you can stop
counting once you pass the host number for that octet (59).
Since the number in the host address for that octet was 59,
which falls between 56 and 64, the network number is 56 and the network address
is 172.16.56.0.
I will stop here and we will finish this topic next time. In
the meantime, here are several host addresses for you to resolve to network
addresses:
200.55.116.97 255.255.255.240
88.65.146.77 /13
145.85.117.252
255.255.224.0
196.145.98.177 /27
166.98.225.77
255.255.252.0
I will post the answers tomorrow. Your comments and questions are welcome.
And the answers are:
200.55.116.97 255.255.255.240 ======>
200.55.116.96
(First three octets don’t change; Jump# is 16 – 256-240)
(1st two octets don't change; Octet 4 changes to 0; Jump# is 32 – 256-224)
(1st two octets don't change; Octet 4 changes to 0; Jump# is 4 – 256-252)
Your questions and comments are welcome.
-Jim (rev 7/26/2014)
And the answers are:
(First three octets don’t change; Jump# is 16 – 256-240)
88.65.146.77 /13 ====> 88.64.0.0
(1st octet doesn’t change; Octets 3-4 change to 0; Jump#
is 8 – 13th bit = 8)
145.85.117.252
255.255.224.0 ===> 145.85.96.0 (1st two octets don't change; Octet 4 changes to 0; Jump# is 32 – 256-224)
196.145.98.177 /27 ===>
196.145.98.160
(First
three octets don’t change; Jump# is 32 – 27th
bit = 32)
166.98.225.77
255.255.252.0 ==> 166.98.224.0 (1st two octets don't change; Octet 4 changes to 0; Jump# is 4 – 256-252)
Your questions and comments are welcome.
-Jim (rev 7/26/2014)
Great post Jim, thanks!
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