1. Unit 24 Network Design IP Addressing (Part 3)

2. Objectives to date… Classes of IP addresses (A, B and C) Using Subnetting and Subnet Masks The Limitations of Classful IP addressing Classless Inter-Domain Routing (CIDR)

3. Objectives to date… CIDR – classless inter domain routing Having IP addresses with a ‘prefix’ 156.46.16.6 /20 Identifying the Network ID, Host ID 10011100.00010001.00100000.00000000 Calculating the number of hosts available on a subnet (2^12) = 4096 (2^12)-2 = 4094 Why? Calculating the Subnet Mask 255.255.240.0

4. Objectives for today… To determine from the CIDR IP Address the IP address range we can use (the first and last IP addresses, and therefore all combinations in between) To use the CIDR address for subnetting To determine the IP addresses that can be allocated for each subnet To use the Classful and Classless IP addressing for Task 2 Logical topology diagrams

5. CIDR – Valid Host Addresses Once you have found the network ID, subnet mask and have calculated the number of hosts/machines, according to the CIDR ‘prefix’, you now need to identify all IP addresses valid for this network. Look again at the table identifying the binary network ID, host ID and subnet mask…

6. CIDR: Network ID, Host ID, Subnet Mask Decimal15617426 /20 Binary10011100000100010010101000000110 Network ID10011100000100010010000000000000 Host ID 000000000000 Subnet Mask11111111 255 11111111 255 11110000 240 00000000 0 20 bits 12 bits

7. CIDR – Valid Host Addresses From the binary IP address (network ID) how do we identify the valid IP address range, i.e. the actual IP addresses that can be allocated to hosts given the IP address of 156.17.42.6 /20?

8. CIDR – Valid Host Addresses Decimal15617466 /20 Network ID10011100 156 00010001 17 00100000 32 00000000 0 First Valid IP address 10011100000100010010000000000001 Decimal IP address 15617321 Last Valid IP address 10011100000100010010111111111110 Decimal IP address 1561747254 The binary address cannot contain all zeros – the last bit is therefore a 1 The binary address cannot contain all ones – the last bit is therefore a 0

9. Exercises… Log onto the shared area Emerson A\Unit 24\IP Addressing\Week 3\CIDR Exercises part 2 Complete question 5a

10. Subnetting using CIDR When subnetting using CIDR you need to consider a few factors How many subnets do you need? How many devices (hosts) on each subnet? Do you need to plan for expansion in the future?

11. Subnetting using CIDR Here’s our Network ID in binary 10011100. 00010001. 00100000. 00000000 How many subnets? 4 in this example How many bits will I need to use if I need 4 subnets? 2^2=4 (There are only 4 possible combinations if I use 2 bits) 00100000 00100100 00101000 00101100 12 bits

12. Subnetting: Subnet 1 11111110 254 00100011 35 00010001 17 10011100 156 Last IP Address 00000001 1 00100000 32 00010001 17 10011100 156 First IP Address 00000000001000000001000110011100Network ID 20 bits for the network 2 Subnet bits 10 Host bits

13. Subnetting: Subnet 2 Network ID10011100000100010010010000000000 First IP Address 10011100 156 00010001 17 00100100 36 00000001 1 Last IP Address 10011100 156 00010001 17 00100111 39 11111110 254 20 bits for the network 2 Subnet bits 10 Host bits

14. Subnetting: Subnet 3 Network ID10011100000100010010100000000000 First IP Address 10011100 156 00010001 17 00101000 32 00000001 1 Last IP Address 10011100 156 00010001 17 00101011 35 11111110 254 20 bits for the network 2 Subnet bits 10 Host bits

15. Subnetting: Subnet 4 Network ID10011100000100010010110000000000 First IP Address 10011100 156 00010001 17 00101100 44 00000001 1 Last IP Address 10011100 156 00010001 17 00101111 47 11111110 254 20 bits for the network 2 Subnet bits 10 Host bits

16. Exercises… Log onto the shared area Emerson A\Unit 24\IP Addressing\Week 3\CIDR Exercises part 2 Complete question 5b and question 6 Show all binary working out