The document summarizes key concepts about network layer addressing in computer networks, including: 1) It discusses IP address classes and issues with classful addressing such as inefficient use of address space and increased router table entries. 2) It introduces subnetting as a way to add hierarchy and flexibility to IP addressing by splitting the host ID portion of an address into subnet ID and host ID bits. 3) It explains how subnet masks are used to indicate which bits of an IP address correspond to the network/subnet ID and which to the host ID. 4) It provides examples of how to determine subnet masks and design subnet addressing schemes based on network and host requirements.

1. Chapter 4
Network Layer
Computer Networking: A
Top Down Approach
4th edition.
Jim Kurose, Keith Ross
Addison-Wesley, July
2007.

2. IP Address Classes

3. Netid and Hostid

4. Issues with Classful Addressing
• Flexibility in Internal addressing:
– Thousands of hosts connected to one large network
• Inefficient Use of Address Space:
– Wastage of limited IP address space.
• Router Table Entries:
– Increase in router table entries.
Example:
• Hosts needed by organization are 5000
• Hosts in B 65,536 and in C 256
– Organization with 5,000 hosts is in a dilemma
– It can only choose to either waste 90% of a Class B address
or use 20 different Class C networks.
– Replacing with 20 C networks will increase entries in routers

5. Subnetting
• Subnetting adds an additional level to the hierarchy
of structures used in IP addressing.
• IP addresses must be broken into three elements
instead of two.
• Network ID is unchanged
• The host ID into a subnet ID and host ID.
• These subnet ID bits are used to identify each
subnet within the network.
• Splitting the host ID into subnet ID and host ID, we
reduce the size of the host ID portion of the address
• Class A networks have 24 bits to split between the
subnet ID and host ID: class B networks have 16, and
class C networks only 8.

6. Subnetting
The more bits we use from the host ID for the subnet ID, the more
subnets we can have but the fewer hosts we can have for each subnet.

7. Subnet Mask
• In classful addressing without subnetting
– Routers use the first octet of the IP address to determine
what the class is of the address
– From class they know which bits are the network ID and
which are the host ID.
• In subnetting, the routers also need to know how that
host ID is divided into subnet ID and host ID.
– This division can be arbitrary for each network.
– There is no way to tell how many bits belong to each simply
by looking at the IP address.
– The additional information about which bits are for the
subnet ID and which for the host ID must be communicated
to devices that interpret IP addresses.
– This information is given in the form of a 32-bit binary
number called a subnet mask.

8. Determining the Subnet Mask
• The subnet mask is a 32-bit binary number
• It is created so that it has a one bit for each corresponding bit of
the IP address that is part of its network ID or subnet ID, and a
zero for each bit of the IP address’s host ID.
• The mask thus tells TCP/IP devices which bits in that IP address
belong to the network ID and subnet ID, and which are part of
the host ID.

9. Determining the Subnet Mask
• Suppose we have the Class B network 154.71.0.0.
• We decide to subnet this using 5 bits for the subnet
ID and 11 bits for the host ID.
• In this case, the subnet mask will have 16 ones for
the network portion (since this is Class B) followed by
5 ones for the subnet ID, and 11 zeroes for the host
ID.
• That's “11111111 11111111 11111000 00000000” in
binary, with the bits corresponding to the subnet ID
highlighted.
• Converting to dotted decimal, the subnet mask would
be 255.255.248.0

10. Determining the Subnet Mask
Example
• A company is granted the site address
210.70.64.0. The company needs six
subnets. What is the subnet Mask?

11. Determining the Subnet Mask
Solution
• Class C address
• 6 subnets means 3 bits in the subnet id
• 3 ones in the subnet mask
• 11111111 11111111 11111111 11100000
• 255.255.255.224 (decimal)

12. Using a Subnet Mask
• Suppose we have a host on a network with an IP of
154.71.150.42 and subnet mask 255.255.248.0
• A router needs to figure out which subnet this
address is on.
154.71.144.0 is the IP address of the subnet to which 154.71.150.42 belongs.

13. Designing a Subnet
•
•
•
•
Analyze the requirements
Determine the subnet mask
Each Subnet address
Host Addresses
Example:
• Network Address: 211.77.20.0
• 8 sub networks
• 32 hosts in each

14. Designing a Subnet

15. Designing a Subnet
211.77.20.0 in binary is
• 11010011 01001101 00010100 00000000
• Subnet 0
– substitute “000” for the subnet ID bits (211.77.20.0)
• Subnet 1
– substitute “001” for the subnet ID bits, to yield the following:
– 11010011 01001101 00010100 00100000 (211.77.20.32 )
• Subnet 2
– substitute 010 for the subnet ID bits to give:
– 11010011 01001101 00010100 01000000 (211.77.20.64)
• Subnet 3
– Substitute 01100000 (211.77.20.96)
• Subnet 7 (last)
– Substitute 11100000 (211.77.20.224)

16. Designing a Subnet

17. Determining Host Address
•
Subnet #0
– First address: 11010011 01001101 00010100 00000000 (211.77.20.0)
– Second address: 11010011 01001101 00010100 00000001
(211.77.20.1)
– Third address: 11010011 01001101 00010100 00000010
(211.77.20.2)
– Last address: 11010011 01001101 00010100 00011111
(211.77.20.31)
•
Subnet #6.
It has “110” for the subnet bits instead of “000”.
– First address: 11010011 01001101 00010100 11000000
(211.77.20.192)
– Second address: 11010011 01001101 00010100 11000001
(211.77.20.193)
– Third address: 11010011 01001101 00010100 11000010
(211.77.20.194)
– Last address: 11010011 01001101 00010100 11011111
– (211.77.20.223)
Similarly for other subnets
•

18. Designing a Subnet
Home Assignment
• A company is granted the site address
172.16.0.0. The company needs 1000
subnets. Design the subnets?

19. Classless Addressing
•To overcome address depletion
•No classes, but the address are still granted in blocks.
•The size of the block( the number of addresses) varies based
on the nature and size of the entity.
• Household: 2 addresses
•Large organization: thousands of addresses
• ISP: thousands or hundreds of thousands based on the
number of customers it may serve.
•The number of addresses in a block must be power of two
(2,4,8…..).
•Classless Inter Domain Routing (CIDR)

20. Mask:Slash Notation
• Each mask is made of some ones from
the left and followed by some 0s.
• Instead of 255.255.255.224
– Mask has 27 1s
– Attach this number to a classless address
–x.y.z.t/n
• n defines the number of bits that are same in
every block.

21. Classless Addressing: Mask
• The n leftmost bits are 1s
• 32-n rightmost bits are 0s
• In CIDR, n has a value from 0 to 32 and
used to show the mask in the form /n.
• A block of addresses can be defined as
x.y.z.t/n in which x.y.z.t defines one of
the addresses and the /n defines the
mask.
Example: 172.31.1.2/25

22. Classless Addressing: Mask

23. Address allocation with Classful
Addressing
Need 10 addresses
Class C. Give them 256.
Need 200 addresses
Class C. Give them 256.
Need 500 addresses
Class B. Give them 65,536.
Need 1000 addresses
Class B. Give them 65,536.
Need 4000 addresses
Class B. Give them 65,536.
Fig courtesy: Ward Abingdon and Witney College CCNA Exploration

24. Address allocation with
Classless Addressing
Need 10 addresses
/28. Give them 16.
Need 200 addresses
/24. Give them 256.
Need 500 addresses
/23. Give them 512.
Need 1000 addresses
/22. Give them 1024.
Need 4000 addresses
/20. Give them 4096.
Fig courtesy: Ward Abingdon and Witney College CCNA Exploration

25. Classless Addressing
• A small organization is given a block with a
beginning address 205.16.37.24/29.What is
the range of the block?
• Number of bits for hosts =32-29=3
• Total hosts=8
• The first address is 205.16.37.24
• The last address is 205.16.37.31

26. Classless Addressing
• A block of addresses is granted to a small
organization. We know that one of the
addresses is 205.16.37.39/28.What is the
first address in the block?
The given address in binary is
• 11001101 00010000 00100101 00100111
• If we set right most 4 bits to 0 we get
• 11001101 00010000 00100101 00100000
– 205.16.37.32 (decimal)

27. Classless Addressing
• A block of addresses is granted to a small
organization. We know that one of the
addresses is 167.199.170.82/27.What is
the first address in the block?

28. Classless Addressing
Solution
• The last byte is 01010010
• Changing the last 5 bits to 0s we get
• 01000000 or 64
• The first address in the block is
167.199.170.64/27

29. Subnetting
• An organization is granted the block
17.12.40.0/26. The organization has
three offices and needs to divide the
addresses into three sub-blocks of
32,16 and 16 addresses.What are the
subnet addresses and the range of
addresses for each subnet?

30. Subnetting
Solution:
• For subnet 1 that requires 32 addresses
– The mask is 32-5=27
• For subnet 2 that requires 16 addresses
– The mask is 32-4=28
• For subnet 3 that requires 16 addresses
– The mask is 32-4=28
• This means we have the masks 27,28,28
with the organization mask being 26

31. Subnetting
Solution:
• For subnet 1
– The first address is 17.12.40.0/27
– The last address is 17.12.40.31/27
• For subnet 2
– The first address is 17.12.40.32/28
– The last address is 17.12.40.47/28
• For subnet 3
– The first address is 17.12.40.48/28
– The last address is 17.12.40.63/28

32. Subnetting
• An organization is granted the block
130.34.12.64/26. The organization
needs to have four subnets with 16
hosts. What are the subnet addresses
and the range of addresses for each
subnet?

33. Subnetting
• For all subnets
– The mask is 32-4=28
• For subnet 1
– The first address is 130.34.12.64/28
– The last address is 130.34.12.79/28
• For subnet 2
– The first address is 130.34.12.80/28
– The last address is 130.34.12.95/28
• For subnet 3
– The first address is 130.34.12.96/28
– The last address is 130.34.12.111/28
• For subnet 4
– The first address is 130.34.12.112/28
– The last address is 130.34.12.127/28

34. Subnetting
Home Assignment
An organization is granted a block of addresses with the
beginning address 14.24.74.0/24. There are 232−24= 256
addresses in this block. The organization needs to have
11 subnets as shown below:
a. two subnets, each with 64 addresses.
b. two subnets, each with 32 addresses.
c. three subnets, each with 16 addresses.
d. four subnets, each with 4 addresses.
Design the subnets.