1. 30/11/1008-Supernetting1 Revision of Classful IP Addressing © Tanenbaum, Prentice Hall International

2. 30/11/1008-Supernetting2 Revision of Classful IP Addressing ClassClass ID1 st OctetNetworksHostsPurpose A011-12612616,777,214Large Networks 01111111127Loopback B10128-19116,38265,534Medium Networks C110192-2232,097,152254Small Networks D1110224-239Multicast E1111240-255Experimental

3. 30/11/1008-Supernetting3 Classless Inter-Domain Routing (CIDR) PROBLEM 1 Hierarchical IP address space classes are not the right size to meet demand Not enough organisations want Class C addresses with 254 hosts Too many organisations want Class B addresses with 65534 hosts, but they typically have nothing like as many hosts as can be supported by Class B What is needed is something in between Class B and Class C

4. 30/11/1008-Supernetting4 Classless Inter-Domain Routing (CIDR) PROBLEM 2 ISPs typically support hundreds or thousands of different Class B and Class C networks Many of the ISPs were too late to be allocated a Class A address and hence apply for blocks of Class B and C addresses Without CIDR, every router on the Internet needs to have all of these addresses in its routing tables even though they all route to the same ISP Large routing tables wastes router memory and also network bandwidth when routing information is transmitted Looking up addresses in large routing tables wastes router processing time

5. 30/11/1008-Supernetting5 Classless Inter-Domain Routing (CIDR) SOLUTION CIDR where: –Classful addressing is abandoned –Network Ids no longer have to to be an integral number of bytes long –Routers can no longer determine the net mask from the first few bits of the address –Routers have to include the length of the network ID within their routing tables

6. 30/11/1008-Supernetting6 Classless Inter-Domain Routing (CIDR) With CIDR networks are represented in routing tables and elsewhere by an IP address followed by a “/” and then a prefix length that defines the length of the network ID. EXAMPLE:192.10.64.0/19

7. 30/11/1008-Supernetting7 Example 1:192.10.64.0/19 An ISP has been allocated the network address 192.10.64.0/19. What is the network mask and what are the first and last host addresses in the range? How many host addresses can the ISP allocate?

8. 30/11/1008-Supernetting8 Example 1:192.10.64.0/19 The net mask for this network can be found by setting the network ID to all 1s Network address: 11000000 00001010 010|00000 00000000 = 192.10.64.0/19 Net mask: 11111111 11111111 111|00000 00000000 = 255.255.224.0

9. 30/11/1008-Supernetting9 Example 1:192.10.64.0/19 The network address is: 010|00000 00000000 = 192.10.64.0/19 The first network host address is: 010|00000 00000001 = 192.10.64.1 The broadcast address for the network is: 010|11111 11111111 = 192.10.95.255 The last host address for the network is: 010|11111 11111110 = 192.10.95.254 There are 2 13 host addresses (32 - 19 = 13) in this range. so the ISP can allocate 8192 - 2 = 8190 host addresses

10. 30/11/1008-Supernetting10 Example 2: 194.160.0.0 - 194.175.255.255 A company has been allocated the address range 194.160.0.0 to 194.175.255.255. What network address should it use and what prefix length? It’s hard to see what the interesting byte is here. It cannot be bytes 1 though, because it is the same at both ends of the range. It is byte 2 (the first byte that is not the same in both ends of the range).

11. 30/11/1008-Supernetting11 Example 2: 194.160.0.0 - 194.175.255.255 The interesting (2 nd ) byte of the first address is: 160 10 = 10100000 2 The interesting byte of the last network address is: 175 10 = 10101111 2 Find the boundary between the network and the host parts of the address. It is just before the first bit that is different between the first and last addresses i.e. after the 4 th bit. So prefix length is 12 (8 bits from 1 st byte + 4 bits from 2 nd byte)

12. 30/11/1008-Supernetting12 Example 2: 194.160.0.0 - 194.175.255.255 Network address is:194.160.0.0/12 Network mask is:255.240.0.0 First host address is:194.160.0.1 Last host address is:194.175.255.254 Broadcast address is:194.175.255.255

13. 30/11/1008-Supernetting13 Example 2: 194.160.0.0 - 194.175.255.255 There are 12 bits used for the network ID and 20 bits can be used for host addresses (ie 2 20 – 2 = 1,048,574 addresses are supported)

14. 30/11/1008-Supernetting14 Exercise 1:210.83.224.0/22 A company has been allocated the network address 210.83.224.0/22. What is the network mask and what are the first and last host addresses in the range? How many host addresses can be allocated?

15. 30/11/1008-Supernetting15 Exercise 1 : 210.83.224.0/22 Interesting byte is the 3 rd byte as the boundary is after the 22 nd bit 224 10 = 111000|00 2 To find the netmask set all the network bits to 1 and all the host bits to 0 111111|00 2 =252 10 Net mask is therefore 255.255.252.0

16. 30/11/1008-Supernetting16 Exercise 1 : 210.83.224.0/22 The network address is: 111000|00 00000000 : 210.83.224.0/22 The first network host address is: 111000|00 00000001 : 210.83.224.1 The broadcast address for the network is: 111000|11 11111111 : 210.83.227.255 The last host address for the network is: 111000|11 11111110 : 210.83.227.254

17. 30/11/1008-Supernetting17 Exercise 1 : 210.83.224.0/22 There are 2 10 = 1024 host addresses within this range (32 – 22 = 10) Therefore 1024 – 2 = 1022 host addresses can be allocated

18. 30/11/1008-Supernetting18 Exercise 2: 192.16.128.0 - 192.16.135.255 A company has been allocated the address range 192.16.128.0 to 192.16.135.255. What network address should it use and what prefix length?

19. 30/11/1008-Supernetting19 Exercise 2: 192.16.128.0 - 192.16.135.255 The interesting byte of the first network address is: 128 10 = 10000|000 2 The interesting byte of the last network address is: 135 10 = 10000|111 2 Find the boundary between the network and the host parts of the address. It is just before the first bit that is different between the bytes from the first and last network addresses. Which will give us 21 1s in the net mask. So prefix length is 21

20. 30/11/1008-Supernetting20 Exercise 2: 192.16.128.0 - 192.16.135.255 Network address is: 192.16.128.0/21 Network mask is:255.255.248.0 First host address is :192.16.128.1 Last host address is:192.16.135.254 Broadcast address is:192.16.135.255

21. 30/11/1008-Supernetting21 Exercise 2: 192.16.128.0 - 192.16.135.255 There are 21 bits used for the network ID and 11 (32 - 21) bits can be used for host addresses (ie 2 11 – 2 = 2046 addresses are supported)

22. 30/11/1008-Supernetting22 Shortcut with Calculations For both subnetting and supernetting, there is a shortcut which allows the calculations to be done in decimal rather than in binary. The shortcut is to find the interesting byte in the network mask and subtract it from 256. The result will give you a power of two that is the size of each range of network/subnet addresses in the interesting byte. This is how engineers can subnet in their heads! Not recommended for examinations, other than to check answers as it does not demonstrate understanding.

23. 30/11/1008-Supernetting23 Subnetting Example 1 with Shortcut Host Address = 204.33.180.106 Subnet Mask= 255.255.255.224. Interesting Byte is 4 th Byte Magic No = 256 - 224 = 32 This is the size of each subnet address range in this byte Find multiples of magic no either side of Interesting Byte value 106 is >= 96 (3x32) and < 128 (4x32) Network address = 204.33.180.96 Broadcast address for subnet = 204.33.180.127 First host address in subnet = 204.33.180.97 Last host address in subnet = 204.33.180.126

24. 30/11/1008-Supernetting24 Supernetting Exercise 2 with Shortcut Network Address= 192.16.128.0/21 Network Mask = 255.255.248.0 Interesting Byte is 3 rd Byte Magic No = 256 - 248 = 8 This is the size of the address ranges in this byte Find multiples of magic no either side of Interesting Byte value 128 is >=128 (16x8) and < 136 (17x8) Network address = 192.16.128.0/21 Broadcast address = 192.16.135.255 First host address in range = 196.16.128.1 Last host address in range = 196.16.135.254