HNDE Lecture note..Labuduwa Galle.. upload by sika

1. Information Technology III
– EE3104
Lecture 03 – Addressing in a Network

2. Addressing with TCP/IP
Address
Port
Address
Logical
Address
Physical
Address
MAC
Address
IP
Address
Port Address
Identify a
Computer
Identify a
Process

3. Physical Address
 Stored in the Network Interface Card (NIC)
 A hardware setting set by the manufacturer of NIC .
 Unchangeable
Ex :- MAC address
 For Ethernet, the MAC address is a 48 bit or 12 Hex
number
Ex : 5A:B3:87:F1:93:7C
5A-B3-87-F1-93-7C
 MAC address operates in the Data Link Layer (Layer 2)
CRC/FCS DATA DA SA SOF PreambleEthernet Frame

4. Logical Address
 Address scheme depends on the used protocol
 Widely used protocol is TCP/IP
Ex :- IP Address
192.168.16.53
10.39.40.3
 Logical address operates at the Network Layer
(Layer 3)

5. IP Address
 The worldwide IP Address (high level) is decided by Internet
Assigned Numbers Authority (IANA)
 Within Sri Lanka Internet Address authority is Asia Pacific Network
Information Center (APNIC )
 IP Addresses
IP Version 6 (IPV6) (IPng)
128 bit scheme
IP Version 4 (IPV4)
32 bit scheme

6. IP Version 4 (IPV4)
 The 32 bits are represented in following manner.
Byte 1. Byte 2. Byte 3. Byte 4
( 1 byte = 8 bits)

7. IP Version 4 (IPV4) cont.

8. IP Version 4 (IPV4) cont.
 The minimum value of a byte
00000000 = 0
 The maximum value of a byte
11111111 = 255
 The minimum IP Address
0.0.0.0
 The maximum IP Address
255.255.255.255

9. Example
 Change the following IPV4 addresses from binary to dotted decimal notation
 10000001 00001011 00001011 11101111
 11000001 10000011 00011011 11111111
 11100111 11011011 10001011 01101111
 11111001 10011011 11111011 00001111
 Change the following IPV4 addresses from dotted decimal notation to binary
format
 111.56.45.78
 221.34.7.82
 241.8.56.12
 75.45.34.78

10. Network ID and Host ID
 IP Addresses - Network ID + Host ID
 Part of the IP Address is allocated to Network ID
 Remaining part is allocated to Host ID (Computer ID)

11. Classes of IP addresses
Class Net ID Host ID
A 1 Byte 3 Bytes
B 2 Bytes 2 Bytes
C 3 Bytes 1 Byte

12. Five IP Address Classes
IP
Address
Class
Format Purpose Higher
Order
Bit(s)
Address range No Bits
Network
/ Host
Max. Host
A N.H.H.H Large organizations 0 1.0.0.0 to
126.0.0.0
7/24 16777214
B N.N.H.H Medium size
organizations
1,0 128.1.0.0 to
191.254.0.0
14/16 65634
C N.N.N.H Small organizations 1,1,0 192.0.1.0 to
223.255.254.0
21/8 254
D N/A Multicast groups 1,1,1,0 224.0.0.0 to
239.255.255.255
N/A N/A
E N/A Experimental 1,1,1,1 240.0.0.0 to
254.255.255.255
N/A N/A

13. Classes of IP addresses cont.
 Class D is introduced for Multicasting
 Class E is reserved

14. Class A
10 . 0 . 0 . 0
0000 1010 . 0000 0000 . 0000 0000 . 0000 0000
 1 byte (8 bits) allocated for Network ID and remaining 3 bytes
(24 bits) allocated to Host ID
The maximum number of Networks
28 = 256
The maximum number of Hosts per each network
224
 Both in Network ID and Host ID all 0s and all 1s are reserved for special
purposes.
The actual maximum no. Networks = 28 - 2 = 254
The actual maximum no. of Hosts per Network =224 - 2
Network ID Host ID

15. Class
Theoretical Maximum
number of Networks
Theoretical Maximum
number of Hosts per
Networks
A 28 =256 224=16777216
B 216= 65536 216= 65536
C 224=16777216 28 =256

16. 0
1 0
1 1 0
Class A byte 1
Class B byte 1
Class C byte 1
Class Minimum Network ID Maximum Networks ID
A
00000000
0
01111111
127
B
10000000.00000000
128.0
10111111.11111111
191.255
C
11000000.00000000.00000000
192.0.0
11011111.11111111.11111111
223.255.255

17. Network Address and Broadcast Address
 For the Network Address,
 the Host ID part of the IP Address will be considered as All 0s
 For the Broadcast Address,
 the Host ID part of the IP Address will be considered as All 1s
 Ex : 103.58.35.1
This is a Class A address
Net ID is = 103
Host ID is = 58.35.1
 Network Address  103.0.0.0
 Broadcast Address  103.255.255.255

18. 198 . 8 . 0 . 1
1100 0110 . 0000 1000 . 0000 0000 . 0000 0001
 The actual maximum no. Networks
= 224 - 2
 The actual maximum no. of Hosts per Network
=28 - 2 = 254
 Network Address :
1100 0110 . 0000 1000 . 0000 0000 . 0000 0000 (198.8.0.0)
 Broadcast Address :
1100 0110 . 0000 1000 . 0000 0000 . 1111 1111 (198.8.0.255)
Network ID : 3 bytes (24 bits) Host ID : 1 byte (8 bits)
Class C

19. Example
 Find the class of each address
 00000001 00001011 00001011 11101111
 11000001 10000011 00011011 11111111
 10100111 11011011 10000101 01101111
 11110011 10011011 11111011 00001111
 Find the class of each address
 227.12.14.87
 193.14.56.22
 14.23.120.8
 252.5.15.111

20. Example
 Find the number of addresses in the block, the first address and the last
address for following addresses
 73.22.17.25
 180.8.17.9
 200.11.8.45

21. Classful Addressing

22. Classful Addressing - Subnet Mask
Net ID part : All 1’s
Host ID part : All 0’s
• A network mask or a default mask in classful addressing is a 32-bit number with
n leftmost bits all set to 1s and (32 − n) rightmost bits all set to 0s.
• Since n is different for each class in classful addressing, we have three default
masks in classful addressing as shown in Figure

23. Masking Concept

24. AND Operation

25. Example
 Find the beginning address (network address) for
following network addresses.
 23.56.7.91
 132.6.17.85
 201.180.56.5

26. Subnetting/Supernetting
and
Classless Addressing

27. A network with two levels of hierarchy (not
subnetted)

28. A network with three levels of hierarchy
(subnetted)

29. Addresses in a network with and without
subnetting

30. Default mask and subnet mask

31. Example 1
What is the subnetwork address if the
destination address is 200.45.34.56 and the
subnet mask is 255.255.240.0?

32. Solution
11001000 00101101 00100010 00111000
11111111 11111111 11110000 00000000
11001000 00101101 00100000 00000000
The subnetwork address is 200.45.32.0.

33. Example 2
What is the subnetwork address if the
destination address is 19.30.80.5 and the
mask is 255.255.192.0?
Solution
See Figure 5.6

34. Comparison of a default mask and a subnet
mask

35. Example 3
A company is granted the site address
201.70.64.0 (class C). The company needs
six subnets. Design the subnets.

36. Solution (Continued)
The company needs six subnets. This
number 6 is not a power of 2. The next
number that is a power of 2 is 8 (23). We
need 3 more 1s in the subnet mask. The
total number of 1s in the subnet mask is 27
(24 + 3).
The total number of 0s is 5 (32 - 27). The
mask is

37. Solution (Continued)
11111111 11111111 11111111 11100000
or
255.255.255.224
The number of subnets is 8.
The number of addresses in each subnet
is 25 (5 is the number of 0s) or 32.

38. Example 3

39. Example 4
A company is granted the site address
181.56.0.0 (class B). The company needs
1000 subnets. Design the subnets.

40. Solution (Continued)
The mask is
11111111 11111111 11111111 11000000
or
255.255.255.192.
The number of subnets is 1024.
The number of addresses in each subnet is 26
(6 is the number of 0s) or 64.

41. Example 4

42. SUPERNETTING

43. The problem
 Most Class A or B network addresses have already
been assigned.
 The problem is compounded by the fact that Class
C networks are limited to a maximum of 254 hosts.

44. Solution
 To create a supernetwork, or supernet, an
organization uses a block of IP addresses assigned
to several Class C networks to create one large
network.
 With supernetting you can combine small networks
into one larger network.

45. A supernetwork

46. Rules:
** The number of blocks must be a power of 2 (1,
2, 4, 8, 16, . . .).
** The blocks must be contiguous in the address
space (no gaps between the blocks).
** The third byte of the first address in the
superblock must be evenly divisible by the number
of blocks. In other words, if the number of blocks is
N, the third byte must be divisible by N.

47. Example 5
A company needs 600 addresses. Which of the following set of class C
blocks can be used to form a supernet for this company?
198.47.32.0 198.47.33.0 198.47.34.0
198.47.32.0 198.47.42.0 198.47.52.0 198.47.62.0
198.47.31.0 198.47.32.0 198.47.33.0 198.47.52.0
198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0

48. Solution
1: No, there are only three blocks.
2: No, the blocks are not contiguous.
3: No, 31 in the first block is not divisible by 4.
4: Yes, all three requirements are fulfilled.

49. In subnetting,
we need the first address of the
subnet and the subnet mask to
define the range of addresses.

50. In supernetting,
we need the first address of
the supernet
and the supernet mask to
define the range of addresses.

51. Comparison of subnet, default,
and supernet masks

52. Example 6
We need to make a supernetwork out of 16 class C
blocks. What is the supernet mask?
Solution
We need 16 blocks. For 16 blocks we need to change
four 1s to 0s in the default mask. So the mask is
11111111 11111111 11110000 00000000
or
255.255.240.0

53. Example 7
A supernet has a first address of 205.16.32.0
and a supernet mask of 255.255.248.0. A
router receives three packets with the
following destination addresses:
205.16.37.44
205.16.42.56
205.17.33.76
Which packet belongs to the supernet?

54. Solution
We apply the supernet mask to see if we can find the
beginning address.
205.16.37.44 AND 255.255.248.0  205.16.32.0
205.16.42.56 AND 255.255.248.0  205.16.40.0
205.17.33.76 AND 255.255.248.0  205.17.32.0
Only the first address belongs to this supernet.

55. Example 8
A supernet has a first address of 205.16.32.0 and a
supernet mask of 255.255.248.0. How many blocks
are in this supernet and what is the range of
addresses?
Solution
The supernet has 21 1s. The default mask has 24 1s.
Since the difference is 3, there are 23 or 8 blocks in
this supernet. The blocks are 205.16.32.0 to
205.16.39.0. The first address is 205.16.32.0. The
last address is 205.16.39.255.