Route Summarization Practice

Subnet Route Summarization Practice Questions

In large internetworks, hundreds, or even thousands of network addresses can exist. It is often problem for routers to maintain these volumes of routes in their routing tables. Route summarization can reduce the number of routes that a router must maintain, because it is a method of representing a series of network numbers in a single summary address. Or in other words, summarization is a method of finding a common network address for all the networks taking part in the summary process.

In Summarization we take multiple small network and logically group them into 1 single big network.

Summarization gives one single network ID to route if some networks are contagious.

A summary route is announced by the summarizing router as long as at least one specific route in its routing table matches the summary route.

When doing summary, forget about the original mask or CIDR notation. Just convert to binary, check where they are separated, count common bits, no matter whether they are 0s or 1s and that is your new summary mask.



For example, in the figure above, router R1 can either send five routing update entries or
summarize of the four addresses into a single network number. If router R1 summarizes the information into a single network number entry, the following things happen:
1. Bandwidth is saved on the link between routers R1 and R2.
2. Router R2 needs to maintain only one route and therefore saves memory.
3. Router R2 also saves CPU resources, because it evaluates packets against fewer entries in its routing table.

Example 1:

Summarize the network given below:
172.16.0.0/24
172.16.1.0/24
172.16.2.0/24
172.16.3.0/24

Solution

The binary equivalents of the shown addresses are as follows:
3rd octet is the significant one, let's convert 3rd octet to binary.
172.16.0.0 = 172.16.000000 00.00000000
172.16.1.0 = 172.16.000000 01.00000000
172.16.2.0 = 172.16.000000 10.00000000
172.16.3.0 = 172.16.000000 11.00000000
Here the first 16 bits (172.16) are fixed means the third portion is varying from 0 to 7.
The last 2 bits of the 3rd octet makes the difference here, first 6 bits of the 3rd octet is identical.
The remaining 6 ones (111111) are added to our fixed 16 bit network portion, 16+6 = 22, or 8+8+6=22, these are the identical bits starting from the leftmost.
Now 22 bits are fixed represents the network portion and the remaining is 12 bits are our host portion.
So the summary address is 172.16.0.0/22.

Example 2:

Summarize the network given below:
172.16.4.0/24
172.16.5.0/24
172.16.6.0/24
172.16.128.0/24

Solution

The binary equivalents of the shown addresses are as follows:
3rd octet is the significant one, let's convert 3rd octet to binary.
172.16.4.0 = 10101100.00010000.00000100.00000000
172.16.5.0 = 10101100.00010000.00000101.00000000
172.16.6.0 = 10101100.00010000.00000110.00000000
172.16.128.0 = 10101100.00010000.10000000.00000000
Here the first 16 bits (172.16) are fixed means the third portion is varying from 0 to 7.
Because of the last subnet (172.16.128.0/24), a good summarization is not possible
with these subnets. Now 16 bits are fixed represents the network portion and the remaining is 16 bits are our host portion.
The summary address is 172.16.0.0/16.
We must drop back to the classful summarization of 172.16.0.0/16. Because discontiguous network addressing can
destroy our network summarization efficiency.

Example 3:

Summarize the network given below:
172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

Solution

172.1.000001 00.00000000
172.1.000001 00.10000000
172.1.000001 01.00000000
172.1.000001 10.00000000
172.1.000001 11.00000000
---------------------------------
172.1.00000100.00000000 - Our summary network.
mask is: 8 bits+8bits+6 bits= /22
or in binary: 11111111.11111111.11111100.00000000=/22
So the summary address is 172.1.4.0/22.

Example 4:

Summarize the network given below:
192.168.1.0/24
192.168.2.0/24
192.168.3.0/24
192.168.4.0/24

Solution

Check for the common bits in all the octets….
192.168.1.0 ---- 11000000.10101000.00000 001.00000000
192.168.2.0 ---- 11000000.10101000.00000 011.00000000
192.168.3.0 ---- 11000000.10101000.00000 011.00000000
192.168.4.0 ---- 11000000.10101000.00000 100.00000000
-----------------------------------------------
192.168.00000000.00000000 - Our summary network.
mask is: 8 bits+8 bits+5 bits(common bits)= /21
or in binary: 11111111.11111111.11111000.00000000=/21
So the summary address is 192.168.0.0/21.
    

Example 5:

Summarize the network given below:
192.168.0.0/24
192.168.1.0/24
192.168.2.0/24
192.168.3.0/24
192.168.4.0/24
192.168.5.0/24
192.168.6.0/24
192.168.7.0/24

Solution

3rd octet is the significant one, lets convert 3rd octet to binary.

192.168.0.0 = 192.168. 00000 000 .0
192.168.1.0 = 192.168. 00000 001 .0
192.168.2.0 = 192.168. 00000 010 .0
192.168.3.0 = 192.168. 00000 011 .0
192.168.4.0 = 192.168. 00000 100 .0
192.168.5.0 = 192.168. 00000 101 .0
192.168.6.0 = 192.168. 00000 110 .0
192.168.7.0 = 192.168. 00000 111 .0
-----------------------------------------------
192.168.00000000.00000000 - Our summary network.
mask is: 8 bits+8bits+5 bits= /21
or in binary: 11111111.11111111.11111000.00000000=/21
So the summary address is 192.168.0.0/21.

Example 6:

Summarize the network given below:
192.168.112.0/24
192.168.113.0/24
192.168.114.0/24
192.168.115.0/24
192.168.116.0/24
192.168.117.0/24
192.168.118.0/24
192.168.119.0/24

Solution

192.168.112.0 = 192.168.01110 000.00000000
192.168.113.0 = 192.168.01110 001.00000000
192.168.114.0 = 192.168.01110 010.00000000
192.168.115.0 = 192.168.01110 011.00000000
192.168.116.0 = 192.168.01110 100.00000000
192.168.117.0 = 192.168.01110 101.00000000
192.168.118.0 = 192.168.01110 110.00000000
192.168.119.0 = 192.168.01110 111.00000000
Based on these binary values, you can see that the first 21 bits on each one of these IP addresses are the same.
Here the first 16 bits (192.168) are fixed means the third portion is varying from 0 to 7.
The last 3 bits of the 3rd octet makes the difference here, first 5 bits of the 3rd octet is identical.
It means we need three bits in third portion that is 11111(000) as we take these 000 for our use.
The remaining 5 ones (11111) are added to our fixed 16 bit network portion, 16+5 = 21. i.e. as each octet is 8 bits, the mask will be 8+8+5=21, these are the identical bits starting from the leftmost.
Now 21 bits are fixed represents the network portion and the remaining is 11 bits are our host portion.
So the summary address is 192.168.112.0/21.

Exercise

Exercise 1:

10.1.1.0/24
10.1.2.0/24
10.1.3.0/24
10.1.4.0/24

Answer:

 Summary: 10.1.0.0/21
 

Exercise 2:

10.10.1.0/27
10.10.1.32/28
10.10.1.48/26
10.10.1.64/25
10.10.1.128/25

Answer:

Summary: 10.10.1.0/24
 

Exercise 3:

10.10.1.0/27
10.10.1.32/28
10.10.1.48/28
10.10.1.64/26
10.10.1.128/29
10.10.1.136/29
10.10.1.144/28

Answer:

Summary: 10.10.1.0/24