Subnetting Subnetting is not in the CIS221-3 Syllabus It was covered in CIS110 last year We are revising it because: We need to study supernetting The technique is similar to subnetting Some students find it difficult 16/11/10 06-Subnetting

Revision of Classful IP Addressing © Tanenbaum, Prentice Hall International 16/11/10 06-Subnetting

Revision of Classful IP Addressing Class ID 1st Octet Networks Hosts Purpose A 01 1-126 126 16,777,214 Large Networks 01111111 127 Loopback B 10 128-191 16,382 65,534 Medium Networks C 110 192-223 2,097,152 254 Small Networks D 1110 224-239 Multicast E 1111 240-255 Experimental 16/11/10 06-Subnetting

Revision of Classful IP Addressing Note that for every range of network Ids and host IDs, two cannot be used. Network Ids of all 0s means this network Network Ids of all 1s means all networks Host Ids of all 0s are used to define an IP address for the network itself Host Ids of all 1s are used for a broadcast address for all hosts on the network Therefore for both host and network Ids which are N bits long, the number of valid addresses is 2N – 2 16/11/10 06-Subnetting

Net Masks Address Class Net Mask Class A 255.0.0.0 11111111000000000000000000000000 Class B 255.255.0.0 11111111111111110000000000000000 Class C 255.255.255.0 11111111111111111111111100000000 16/11/10 06-Subnetting

Classful Routing When a router receives a packet, it examines the destination address in the IP header It checks the first byte of the address and determines the address class and the net mask for this class. It then applies does a bitwise AND between the destination address and the net mask to determine the destination network address It then looks up the destination network address in its routing table and finds the IP address of the router it should forward the packet to and which interface this address can be reached by Finally, it routes the packet 16/11/10 06-Subnetting

Example of Classful Routing Suppose the destination address is 164.36.4.6 The first byte is between 128 and 191, so it is a Class B address The net mask for a Class B address is 255.255.0.0 Bitwise AND the destination address and the default net mask to give 164.36.0.0 as the destination network address Look up 164.36.0.0 in the routing table and obtain the IP address of the next router and the interface by which it can be reached (say 132.15.234.1 by serial 0) Route the packet with destination address 164.36.4.6 out of the serial 0 interface. 16/11/10 06-Subnetting

Revision of Subnetting With subnetting the host ID part of the IP address is further sub-divided into a subnetwork ID and a host ID This allows organisations to structure the Class A, B or C addresses into a further hierarchy where each subnetwork can be identified by a different subnetwork address Only routers within the subnetwork need to have entries for the subnetwork addresses and knowledge of the subnet mask used All other routers on the Internet just have one entry for the network address, thus reducing routing table size 16/11/10 06-Subnetting

Example 1: 204.33.180.0 – Class C We have a Class C network address (with an 8-bit host ID) given above and we want to use it to create six subnetworks each having up to 30 hosts We need 3 bits to define the subnet and 5 to define the hosts to give 23 – 2 = 6 subnetwork addresses each with 25 – 2 = 30 hosts With subnetting there is always just one byte that we need to work with. It is the byte in which the boundary between the subnet ID and the host ID is found. This is known as the interesting byte. It is the last byte in the network address that is not all 0s or the last byte in the subnet mask that is not all 1s 16/11/10 06-Subnetting

Example 1: 204.33.180.0 – Class C In the case of a Class C address, the interesting byte is the last byte. The subnet mask will be the all 1s in the first 3 bytes and the final byte will be 111|000002 = 22410 The ‘|’ represents the boundary between the subnet ID and the host ID Subnet mask is therefore 255.255.255.224 16/11/10 06-Subnetting

Example 1: 204.33.180.0 – Class C So, if we are given a host address 204.33.180.106 we need to bitwise AND it with the subnet mask 255.255.255.224 to get the address of its subnet. Again we only need to look at the interesting (last) byte: 22410 = 111|000002 10610 = 011|010102 ANDed 011|000002 = 9610 16/11/10 06-Subnetting

Example 1: 204.33.180.96 – Class C So the subnetwork address is: 204.33.180.96 9610 = 011|000002 The first valid host address is: 204.33.180.97 9710 = 011|000012 The broadcast address is: 204.33.180.127 12710 = 011|111112 The last valid host address is: 204.33.180.126 12610 = 011|111102 16/11/10 06-Subnetting

Example 2: 132.40.17.200 – Class B Suppose the subnet mask is 255.255.255.128 The interesting byte is the last byte in the subnet mask which is not all 1s (i.e. the last byte) 12810 = 1|00000002 20010 = 1|10010002 ANDed 1|00000002 = 12810 There will be 29- 2 valid subnetwork addresses and 27 - 2 valid host addresses 16/11/10 06-Subnetting

Example 2: 132.40.17.200 – Class B So the subnetwork address is: 132.40.17.128 12810 = 1|00000002 The first valid host address is: 132.40.17.129 12910 = 1|00000012 The broadcast address is: 132.40.17.255 25510 = 1|11111112 The last valid host address is: 132.40.17.254 25410 = 1|11111102 16/11/10 06-Subnetting

Example 3: 120.19.51.6 – Class A Suppose the subnet mask is 255.252.0.0 The interesting byte is the last byte in the subnet mask which is not all 1s (i.e. the 2nd byte) 25210 = 111111|002 1910 = 000100|112 ANDed 000100|002 = 1610 There will be 26- 2 valid subnetwork addresses and 218- 2 valid host addresses 16/11/10 06-Subnetting

Example 3: 120.19.51.6 – Class A So the subnetwork address is: 120.16.0.0 1610 = 000100|002 The first valid host address is: 120.16.0.1 The broadcast address is: 120.19.255.255 1910 = 000100|112 The last valid host address is: 120.19.255.254 16/11/10 06-Subnetting

Exercise 1: 86.20.38.0 – Class A How many valid subnet addresses and host addresses are supported by the subnetwork address 86.20.38.0 with subnet mask 255.255.224.0? Find the first valid host address, last valid host address and broadcast address for the subnetwork 16/11/10 06-Subnetting

Exercise 1: 86.20.38.0 – Class A Interesting byte is the 3rd byte 22410 = 111|000002 3810 = 001|001102 ANDed 001|000002 = 3210 There will be 211- 2 valid subnetwork addresses and 213- 2 valid host addresses 16/11/10 06-Subnetting

Exercise 1: 86.20.38.0 – Class A So the subnetwork address is: 86.20.32.0 3210 = 001|000002 The first valid host address is: 86.20.32.1 The broadcast address is: 86.20.63.255 6310 = 001|111112 The last valid host address is: 86.20.63.254 16/11/10 06-Subnetting

Exercise 2: 160.16.18.99 - Class B How many valid host address belong to the same subnet as 160.16.18.99 if the subnet mask is 255.255.255.192? What are the first and last valid host addresses in this subnet? 16/11/10 06-Subnetting

Exercise 2: 160.16.18.99 - Class B Interesting byte is the last byte 19210 = 11|0000002 9910 = 01|1000112 ANDed 01|0000002 = 6410 There will be 210- 2 valid subnetwork addresses and 26- 2 valid host addresses 16/11/10 06-Subnetting

Exercise 2: 160.16.18.99 - Class B So the subnetwork address is: 160.16.18.64 6410 = 01|0000002 The first valid host address is: 160.16.18.65 6510 = 01|0000012 The broadcast address is: 160.16.18.127 12710 = 01|1111112 The last valid host address is: 160.16.18.126 12610 = 01|1111102 16/11/10 06-Subnetting

Exercise 3: 193.25.18.0 – Class C A customer has been allocated the Class C network address 193.25.18.0 and wishes to create up to 30 subnets each of which having a maximum of 6 hosts. What subnet mask should he use? What will be the subnet address of the first subnet? What will be the broadcast address of this subnet? 16/11/10 06-Subnetting

Exercise 3: 193.25.18.0 – Class C For Class C addresses the interesting byte is the last byte 30 subnets require 25 bits 6 hosts requires 23 bits Last byte of subnet mask is 11111|0002 = 24810 Full Subnet Mask is 255.255.255.248 First theoretical subnetwork address is the network address given (193.25.18.0), but this would have a subnetwork address of all 0s 00000|0002, which is used to identify the subnetwork and this subnetwork is not normally used for hosts 16/11/10 06-Subnetting

Exercise 3: 193.25.18.0 – Class C So the first subnetwork address is: 193.25.18.8 810 = 00001|0002 The first valid host address in first subnet is: 193.25.18.9 910 = 00001|0012 The broadcast address of first subnet is: 193.25.18.15 1510 = 000011112 The last valid host address is: 193.25.18.14 1410 = 000011102 16/11/10 06-Subnetting