B NET 155 Bootstrap2 .pdf



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B2- Introduction to
Networks
B-NET-155

Bootstrap 2
Optimizing IP Addresses

v1.1

Bootstrap 2
Subnetworks
The creation of a subnetwork allows an important network to be divided into several, equal (or non-equal) sized
subnetworks.
This division enables the broadcast domain of a primary network to be reduced.
Subnetworks are created by using the bits that belong to the Host ID section.
The size of a subnetwork depends on the number of hosts necessary to the subnetwork.

How to optimize an IP address?
In order to optimize an IP address, we’re going to use the bits that belong to the HostID section.
For instance:
For a Class A IP address: xxxx xxxx.xxxx xxxx.xxxx xxxx.xxxx xxxx (24 bits to create subnetworks).
For a Class B IP address: xxxx xxxx.xxxx xxxx.xxxx xxxx.xxxx xxxx (16 bits to create subnetworks).
For a Class C IP address: xxxx xxxx.xxxx xxxx.xxxx xxxx.xxxx xxxx (8 bits to create subnetworks).
1. When the number of hosts is known:
• The address class is identified by our primary address: 192.168.1.0 is a class C address.
• The number of addresses needed is identified in the subnetwork: 11 host addresses + 1 network address + 1
broadcast address.
• The number of bits to be kept in the HostID section must be determined.
bit

8

7

6

5

4

3

2

1

number of available addresses

256

128

64

32

16

8

4

2

For instance:
For an 11-host subnetwork, the following elements are needed:
• 1 network address,
• 11 host addresses,
• 1 broadcast address.
We’re going to borrow 4 bits that will provide us with 16 addresses (3 adresses will not be used).
The subnetwork mask will no longer be a /24 because we will only keep 4 bits in our HostID section (instead of 8, as
before). The new mask will be:
IP address - HostID section -> 32 - 4 = /28

1

2. When the number of subnetworks is known:
• The address class is identified by our primary address: 192.168.1.0 is a class C address.
• The number of subnetworks needed is identified: 11 subnetworks.
• Determine the number of bits to be borrowed from the HostID section.
bit

1

2

3

4

5

6

7

8

number of available subnetworks

2

4

8

16

32

64

128

256

For example:
In order to create 11 subnetworks, we borrow 4 bits that provide us with 16 subnetworks (5 subnetworks will not be
used).
The subnetwork mask will no longer be a /24 because we will only keep 4 bits in our Host ID section (instead of 8, as
before). The new mask will be:
IP address - HostID section -> 32 - 4 = /28

Let’s optimize the 192.168.1.0/24 in order to create the three following subnetworks:
Subnetwork 1
Subnetwork 2
Subnetwork 3
PC1

Fa0

F0/1
Fa0

PC2

F0/2

F0/4

SW1

F0/3

F0/0

F0/1

F0/0

R1

F0/1

R2

F0/0

R3

Fa0

PC3
a) The number of subnetworks and the hosts they/it possess(es) must be identified before beginning the optimization
Subnetwork -1:
1 network address + 4 host addresses (PC1, PC2, PC3 and F0/0 from router 1) + 1 broadcast address = 6 necessary
addresses.

An IP address is not configured on a switch

Subnetwork -2:
1 network address + 2 host addresses (F0/1 from router 1 and F0/0 from router 2) + 1 broadcast address = 4 necessary
addresses.
Subnetwork -3:
1 network address + 2 host addresses (F0/1 from router 2 and F0/0 from router 3) + 1 broadcast address = 4 necessary
addresses.

2

b. The number of bits to be kept for the HostID section of our address (for each subnetwork) must be identified
Subnetwork -1:
6 addresses –> we keep 3 bits that provide 8 addresses (1 network address + 6 host addresses + 1 broadcast address).
Subnetwork -2:
4 addresses –> we keep 2 bits that provide 4 addresses (1 network address + 2 host addresses + 1 broadcast address).
Subnetwork -3:
4 addresses –> we keep 2 bits that provide 4 addresses (1 network address + 2 host addresses + 1 broadcat address).

c. The mask of the subnetwork of our subnetworks must be determined
Subnetwork -1:
We kept 3 bits for the HostID section. Therefore, the remaining bits are in the NetID section: 32 - 3 = /29.
Subnetworks -2/3:
We kept 2 bis for the HostID section. Therefore, the remaining bits are in the NetID section: 32 - 2 = /30.

d. Beginning with the largest, the IP addresses must be attributed to our 3 subnetworks
Subnetwork -1:
• Network address: 192.168.1.0
• Host addresses: 192.168.1.1 to 192.168.1.6
• Broadcast address: 192.168.1.7
• Mask: /29 - 255.255.255.248
Subnetwork -2:
• Network address: 192.168.1.8
• Host addresses: 192.168.1.9 to 192.168.1.10
• Broadcast address: 192.168.1.11
• Mask: /30 - 255.255.255.252
Subnetwork -3:
• Network address: 192.168.1.12
• Host addresses: 192.168.1.13 to 192.168.1.14
• Broadcast address: 192.168.1.15
• Mask: /30 - 255.255.255.252

3

Hands-On Activities
Activity 1
A business has received the following class C IP address: 192.168.1.0
For each of the required subnetworks, you must provide the following:
• The network address
• The first host address
• The last host address
• The broadcast address
• The subnetwork mask
You must cut the 192.168.1.0 address by using the HostID section. Each subnetwork must begin with
192.168.1.X

Subnetworks:
• Subnetwork 1: 12 hosts
• Subnetwork 2: 63 hosts
• Subnetwork 3: 2 hosts
• Subnetwork 4: 7 hosts

Activity 2
Given the following class A address: 10.0.0.0/24,
• How many bits have been borrowed from the HostID section?
• What is this network’s subnetwork mask? (Decimal and binary representation)
• How many subnetworks are there? How many hosts per subnetwork?
• What is the network address and the broadcast address of the 16th subnetwork?
• What is the network address and the broadcast address of the last subnetwork?
• What is the network address and the broadcast address of the primary network?

Activity 3
A business has received the following class B address: 172.16.0.0
This primary network must be subdivided into 430 equal-sized subnetworks.
• How many bits must be borrowed from the HostID section to create 430 subnetworks?
• How many bits remain in the HostID section for each subnetwork?
• How many hosts does each subnetwork have?
• What is the subnetwork mask in CIDR and binary representation ?

4


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