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Network Layer

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Network Layer Topics to Cover

Logical Addressing

I nternet Protocol

Address M apping Delivery, F orwarding, Routing

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IPv4 Addresses

An IPv4 address is a 32-bit address thatuniquely and universally defines theconnection of a device (for example, acomputer or a router) to the Internet.

The address space of IPv4 is 2 32 or

4,294,967,296.

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Dotted-decimal Notation and Binary notation for an IPv4 address

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Change the foll owing I Pv4 addresses f rom binary notation to dotted-decimal notation.

Example

Solution We replace each group of 8 bits with i ts equi valent decimal number (see Appendix B) and add dots for separation.

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Change the following I Pv4 addresses f rom dotted-decimal notation to binary notation.

Example

Solution We replace each decimal number with its binary equi valent (see Appendix B).

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F ind the er ror, i f any, in the foll owing I Pv4 addresses.

Example

Solution a. There must be no leading zero (045).b. There can be no c. Each number needs to be less than or equal to 255.d. A mixture of binary notation and dotted-decimal

notation i s not al lowed.

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In classful addressing, the addressspace is divided into five classes:A, B, C, D, and E.

Note

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Finding the classes in binary and dotted-decimal notation

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F ind the class of each address.a. 0 0000001 00001011 00001011 11101111b. 110 00001 10000011 00011011 11111111c. 14 .23.120.8

d. 252 .5.15.111

Example

Solution a. The f ir st bi t i s 0. This is a class A address.

b. The f ir st 2 bi ts are 1; the third bi t is 0. This is a class C address.

c. The f ir st byte is 14; the class is A.d. The f ir st byte is 252; the class is E.

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Number of blocks and block size in classful IPv4 addressing

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In classful addressing, a large part of theavailable addresses were wasted.

Note

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Defaul t masks for classful addressing

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Classful addressing, which is almostobsolete, is replaced with classlessaddressing.

Note

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Subnet Addressing

Subnet addressing introduces another hierarchical levelTransparent to remote networksSimplifies management of multiplicity of LANsMasking used to find subnet number

Originaladdress

Subnettedaddress

Net ID Host ID1 0

Net ID Host ID1 0 Subnet ID

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Subnets

IP address: subnet part (high order bits)host part (low order

bits)Whats a subnet ? device interfaces withsame subnet part of IPaddresscan physically reacheach other withoutintervening router

223.1.1.1

223.1.1.2

223.1.1.3

223.1.1.4 223.1.2.9

223.1.2.2

223.1.2.1

223.1.3.2 223.1.3.1

223.1.3.27

network consisting of 3 subnets

LAN

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Subnets

How many? 223.1.1.1

223.1.1.3

223.1.1.4

223.1.2.2 223.1.2.1

223.1.2.6

223.1.3.2 223.1.3.1

223.1.3.27

223.1.1.2

223.1.7.0

223.1.7.1 223.1.8.0 223.1.8.1

223.1.9.1

223.1.9.2

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Supernetting (Address Aggregation)

Allows several networks to be combined tocreate a larger range of addresses.For example this allows an organisation tocombine several Class C blocks together.It aggregates a contiguous block of addresses to form a single larger network.

Identified by less bits of 1s (i.e. shorter network part) in the mask than default one.

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Supernetting

Summarize a contiguous group of class C addressesusing variable-length maskExample: 200.158.16.0/20IP Address (200.158.16.0) & mask length (20)IP add = 11001000 10011110 00010000 00000000Mask = 11111111 11111111 1111 0000 00000000Contains 16 Class C blocks:From 11001000 10011110 0001 0000 00000000i.e. 200.158.16.0Up to 11001000 10011110 0001 1111 00000000i.e. 200.158.31.0

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Supernet : an example

Internet part

Host ID

Local part

11000000 11100110 000101 00 00000000

11000000 11100110 000101 01 0000000011000000 11100110 000101 10 0000000011000000 11100110 000101 11 00000000

Network ID

4 Class C IP network addresses

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Supernet : an example

Four class C networks were put together to form thesupernet:

192.230.20. 0

192.230.21. 0192.230.22. 0

192.230.23. 0

The address of lowest network together with thesupernet mask is required in the global routers for routing : 192.230.20.0 / 22

11000000 11100110 010100 00 00000000

11000000 11100110 010100 01 00000000

11000000 11100110 010100 10 00000000

11000000 11100110 010100 11 00000000

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Classless Addressing

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F igur e below shows a block of addresses, in both binary and dotted-decimal notation, granted to a small business that needs 16 addr esses.

Example

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The first address in the block can befound by setting the rightmost32 n bits to 0s.

Note

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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 f irst address in the block?

Solution The binary representation of the given address is 11001101 00010000 00100101 00100111

If we set 3228 rightmost bits to 0, we get

11001101 00010000 00100101 0010000 or

205.16.37.32 .

Example

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The last address in the block can befound by setting the rightmost32 n bits to 1s.

Note

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F ind the last address for the block in last Example.

Solution The binary representation of the given address is

11001101 00010000 00100101 00100111 If we set 32 28 rightmost bits to 1, we get 11001101 00010000 00100101 00101111

or

205.16.37.47

Example

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The number of addresses in the blockcan be found by using the formula2 32n .

Note

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F ind the number of addresses in last Example.

Example

Solution The value of n is 28, which means that number

of addr esses is 2 32 28

or 16.

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Solutiona. The first address can be found by ANDing the givenaddresses with the mask. ANDing here is done bit by

bit. The result of ANDing 2 bits is 1 if both bits are 1s;

the result is 0 otherwise.

Example (continued)

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b. The last address can be found by ORing the givenaddresses with the complement of the mask. ORinghere is done bit by bit. The result of ORing 2 bits is 0 if

both bits are 0s; the result is 1 otherwise. The

complement of a number is found by changing each 1to 0 and each 0 to 1.

Example (continued)

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c. The number of addresses can be found bycomplementing the mask, interpreting it as a decimalnumber, and adding 1 to it.

Example (Contd.)

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A network configuration for the block 205.16.37.32/28

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The first address in a block isnormally not assigned to any device;it is used as the network address that

represents the organizationto the rest of the world.

Note

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Configuration and Addresses in a Subnetted Network

32

16

16

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Three-level hierarchy in an IPv4 address

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An ISP is granted a block of addresses starting with 190.100.0.0/16(65,536 addresses). The ISP needs to distribute these addresses tothree groups of customers as follows:a. The first group has 64 customers; each needs 256

addresses.

b. The second group has 128 customers; each needs 128addresses.c. The third group has 128 customers; each needs 64

addresses.Design the subblocks and find out how many addresses are stillavailable after these allocations.

Example

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Solution

Example (continued)

Group 1 F or this group, each customer needs 256 addresses. This

means that 8 (log2 256) bits are needed to def ine each host. The prefix length is then 32 8 = 24. The addr esses are

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Example (continued)

Group 2 F or this group, each customer needs 128 addresses. This means that 7 (log2 128) bits are needed to def ine each host. The pref ix length i s then 32 7 = 25. The addr esses

are

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Example (continued) Group 3

F or this group, each customer needs 64 addresses. This means that 6 (log 2 64) bi ts are needed to each host. The pref ix l ength is then 32 6 = 26. The addresses are

Number of granted addresses to the I SP: 65,536 Number of allocated addresses by the I SP: 40,960 Number of avai lable addresses: 24,576

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An example of address allocation and distribution by an ISP

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Example

An organization has a class C network: 200.1.1.0,and it wants to form subnets for 4 departmentswith the number of hosts as follows:Subnet A: 72 hosts

Subnet B: 35 hostsSubnet C: 20 hostsSubnet D: 18 hostsThere are 145 hosts in total.

Provide a possible arrangement of the networkaddress space, together with the respective rangeof IP addresses for each subnet. Explain your work!