IP Subnet Summarization Simplified

I wrote this article to help simplify the process of combining multiple subnets.
This can be used for route summarization also but there are other better ways to summarize routes,

This article is a result of questions I participate in here at Experts Exchange. This particular question is a practice test question posted at the following link

http://www.9tut.com/ccna-subnetting-questions-4 

I copied it here in case the link breaks years down the line

Question 2


Refer to the exhibit. The Lakeside Company has the internetwork in the exhibit. The Administrator would like to reduce the size of the routing table to the Central Router. Which partial routing table entry in the Central router represents a route summary that represents the LANs in Phoenix but no additional subnets?


A). 10.0.0.0 /22 is subnetted, 1 subnet
        D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

B.) 10.0.0.0 /28 is subnetted, 1 subnet
        D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

C.) 10.0.0.0 /30 is subnetted, 1 subnet
        D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

D.) 10.0.0.0 /22 is subnetted, 1 subnet
        D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

E.) 10.0.0.0 /28 is subnetted, 1 subnet
        D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

F.) 10.0.0.0 /30 is subnetted, 1 subnet
        D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

 

Answer: D

Explanation

All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24¿-> /22 is the best choice -> D is correct.

See the Experts Exchange member's question here
https://www.experts-exchange.com/questions/28114215/summarization.html?anchorAnswerId=39127148#a39127148

10.4.0.0 /24
10.4.1.0 /24
10.4.2.0 /24
10.4.3.0 /24

/24 = 255.255.255.0 = 11111111   11111111   1111111 1   00000000
                                      1st Octet     2nd Octet   3rd Octet              4th Octet
the ones in each octet represents the following
128  64  32  16  8  4  2  1

Adding all digits together 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
All 1s represent network while all 0s represent hosts
A /24 network means you can only have 254 addresses
0 - 255 = gives you 256 addresses Subtract the Network ID and the Broadcast Address
256 -2 = 254

The last 1 (counting from left ie the 24th 1) is the boundary of the network
to understand this, I will replace the 1s with the actual digit they represent for all the octets

128  64  32  16  8  4  2  1      128  64  32  16  8  4  2  1    128  64  32  16  8  4  2   1   00000000

Bit position 24 in reality represents 256
Let me explain with the following

There are 32 possible digits in IPv4

1,2,3,4,5,6,7,8,9,10,11,12 ,13,14,15, 16,17,18,1 9,20,21,22 ,23,24,25, 26,27,28,2 9,30,31,32
That's how the slash notations are calculated

Starting from right to left.
I will 1st add the zero spot then subtract 2

Bit 32 represents 1     # of Hosts = 1              1st address is 0, last address is 0
Bit 31 represents 2     # of hosts = 2               1st address is 0, last address is 1
Bit 30 represents 4     # of hosts = 4               1st address is 0, last address is 3
Bit 29 represents 8     # of hosts = 8               1st address is 0, last address is 7
Bit 28 represents 16   # of hosts = 16             1st address is 0, last address is 15
Bit 27 represents 32   # of hosts = 32             1st address is 0, last address is 31
Bit 26 represents 64   # of hosts = 64             1st address is 0, last address is 63
Bit 25 represents 128 # of hosts = 128           1st address is 0, last address is 127

That is the 4th Octet, now moving to the 3rd Octet
Bit 24 represents 256
Bit 23 represents 512
Bit 22 represents 1024
Bit 21 represents 2048
Bit 20 represents 4096
Bit 19 represents 8192
Bit 18 represents 16384
Bit 17 represents 32768

Moving to the 2nd Octet
Bit 16 represents 65536

I'll stop here.
Let me chip in additional info that may help you in future learning
Take note of 4096, 32768, 65536

You will see those numbers come up in future, this will help you remember them or at least calculate them when dealing with Priority and Bridge ID
You may have come across the question "how many TCP ports or UDP ports are there.
Answer = 0 through 65536 but 1st and last addresses are unusable
              = leaving you with 1 through 65535
             =  65,535 addresses. That was just extra

If you are not bored yet, then let's continue


Going by the example in the question
10.4.0.0 /24 = 254 usable addresses available, 256 total
10.4.1.0 /24 = 254 usable addresses available, 256 total
10.4.2.0 /24 = 254 usable addresses available, 256 total
10.4.3.0 /24 = 254 usable addresses available, 256 total

Breaking it down a little further, the address range will look something like this
10.4.0.0 to  10.4.0.255 = 256 Addresses, remember 1st is Network ID and last is broadcast

There must always be 1 and only 1 Network ID in a network (Identifies the network)
There must also be 1 and only 1 Broadcast Address in a network

We're almost there, just hang on
10.4.0.0 to  10.4.0.255 = 256 Addresses,
10.4.1.0 to  10.4.1.255 = 256 Addresses,
10.4.2.0 to  10.4.2.255 = 256 Addresses,
10.4.3.0 to  10.4.3.255 = 256 Addresses,

If we combine the 1st and 2nd network together, it will look something like this
10.4.0.0 to 10.4.1.255 = 256 + 256 = 512 Addresses, 510 usable

Combine 1st 2nd and 3rd
10.4.0.0 to 10.4.2.255 = 256 + 256 +256 = 768 Addresses, 766 usable

Combine all 4 networks
10.4.0.0 to 10.4.3.255 = 256 + 256 +256 + 256 = 1024 Addresses, 1022 usable

If you're wondering about the other Network IDs and Broadcast IDs of the separate networks before they were combined, just remember that there can on be 1 Network ID and 1 Broadcast ID per network.
When you combine networks, the IDs in the middle disappears. It's like welding 2 metal rods together in a straight line, you will only have 1 rod with 2 ends after joining them.

Now go back to the bit representations above and see which bit accommodates 1024
ie Up to which bit position will be used up to make 1024 addresses available

Correct! (hopefully, if that's what you chose.....lol) Answer is bit 22

Now, after taking you through that whole session, I guess you're probably already considering abandoning networking for something else less tedious.

Well, don't run yet. I have an easy solution for you.
We'll start with the bit positions

11111111   11111111   11111111   00000000  = /24  =  254 usable Addresses, 256 total
11111111   11111111   11111110   00000000  = /23  =  512 usable Addresses, 510 total
11111111   11111111   11111100   00000000  = /22  =  1024 usable Addresses, 1022 total

Notice the pattern, we've just borrowed 2 network bits and converted them to host bits

That is the trick folks!

I hope this helps