1. Subnetting Workshop SARA AKERS SEPTEMBER 2014

2. Disclaimer If you notice any mistakes with any of the slides, please let me know so I can correct. Thank you! Sara Akers sakers@terra.edu

3. SUBNETTING

4. 7.4.63.0 255.255.255.0 Class A (7.4.63.0) – Default subnet mask for a Class A is 255.0.0.0 or /8 The IP address would split up Network vs Host bits like this: NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH (See subnet mask in Title – Since it is a /24 subnet mask, the first 24 bits are on (1) – thus network bits) I only have 8 host bits to work with. I cannot use Network Bits already taken. Therefore, I cannot use 255.255.255 I can only use the host bits when subnetting. In my network, I need at least 8 host addresses per network. (8 is a variable number depending on your network) 7.4.63._ _ _ _ _ _ _ _Count host bits from the right to left; network bits from the left to rt. 7.4.63. 256 128 64 32 16 8 4 2I can get 2 hosts in the first bit from the right (a 0 or a 1) So to get 8 host addresses – my line goes here: 7.4.63. 256 128 64 32 16 | 8 4 2However, we need to add 2 more hosts for Broadcast and Network Address – so I really need 10 hosts - thus my line goes here: 7.4.63. 256 128 64 32 | 16 8 4 2

5. 7.4.63.0 255.255.255.0 7.4.63. _ _ _ _ | _ _ _ _ So everything to the left of the line is now a Network bit and everything to the right of the line is a Host bit. 7.4.63.NNNNHHHH (Giving me approximately 16 Network Address and 16 Host Addresses) I want to solve for Subnet 3. Subnet 0 would look like this: 7.4.63.0000|0000 /28 Subnet 3 would look like this: 7.4.63.00110000 /28 How did I get this? 8 4 2 1 | - Remember I am only working with the network bits to the left of the line. So I use the standard Binary-Decimal conversion of 128, 64, 32, 16, 8, 4, 2, 1 (for a byte) Since I only have 4 network bits to work with, I only use 8, 4, 2, 1. To get Subnet 3 – 0 for the 8 bit, 0 for the 4 bit, 1 for the 2 bit, 1 for the 1 bit.

6. 7.4.63.00110000 /28 What is the IP address now? I need to look at the whole byte. So convert from binary to decimal the following byte – 00110000 (as shown above) Using the Binary to Decimal conversion of 128, 64, 32, 16, 8, 4, 2, 1: 7.4.63.48 (32+16 = 48) This is the network address for subnet 3. The subnet mask would be a count (Binary to Decimal) of all network bits. In our example the network bits are: NNNNNNNN.NNNNNNNN.NNNNNNNN.NNNNHHHH or 11111111.11111111.11111111.11110000 = 255.255.255.240 or /28 (/28 is a count of all network bits turned on) The 1 st available IP Address for our hosts would look like this in binary: 00000111.00000100.00111111.0011 (This is our network address) 0001 (Host address) =7.4.63.49

7. 7.4.63.48 /28 Our last IP address would be all Host bits turned on (except the last one) or: 00000111.00000100.00111111.0011 (This is our network address) 1110 (Host address) =7.4.63.62 The broadcast address would be all Host bits turned on or: 00000111.00000100.00111111.0011 (This is our network address) 1111 (Host address) =7.4.63.63

8. 7.4.63.0 255.255.255.0 Network Address: 7.4.63.48 Subnet Mask: 255.255.255.240 or /28 1 st Available IP Address: 7.4.63.49 Last Available IP Address: 7.4.63.62 Broadcast Address: 7.4.63.63 On to the next problem?

9. Try solving for: 56.2.0.0 255.255.0.0 I need at least 3 subnets. Solve for Subnet 1.

10. 56.2.0.0 255.255.0.0 Class A (56.2.0.0) – Default subnet mask for a Class A is 255.0.0.0 or /8 The IP address would split up Network vs Host bits like this: NNNNNNNN.NNNNNNNN.HHHHHHHH.HHHHHHHH (See subnet mask in Title – Since it is a /16 subnet mask, the first 16 bits are on (1) – thus network bits) I have 16 host bits to work with. I cannot use Network Bits already taken. Therefore, I cannot use 255.255. I can only use the host bits when subnetting. In my network, I need at least 3 subnet addresses per network. (8 is a variable number depending on your network) 56.2._ _ _ _ _ _ _ _._ _ _ _ _ _ _ _Count host bits from the right to left; network bits from the left to rt. 56.2. 2 4 8 16 32 64 128. …I can get 2 network bits in the first bit from the left to right (a 0 or a 1) So to get 3 subnet (network) addresses – my line goes here: 56.2. 2 4 | 8 16 32 64 128. …

11. 56.2.0.0 255.255.0.0 56.2. _ _ | _ _ _ _ _ _. _ _ _ _ _ _ _ _ So everything to the left of the line is now a Network bit and everything to the right of the line is a Host bit. 56.2.NNHHHHHH.HHHHHHHH (Giving me approximately 4 Network Address and 16384 Host Addresses) I want to solve for Subnet 1. Subnet 0 would look like this: 56.2.00 | 000000.00000000 /18 Subnet 1 would look like this: 56.2.01 | 000000.00000000 /18 How did I get this? 2 1 | - Remember I am only working with the network bits to the left of the line. So I use the standard Binary-Decimal conversion of 128, 64, 32, 16, 8, 4, 2, 1 (for a byte) Since I only have 2 network bits to work with, I only use 2, 1. To get Subnet 1 –0 for the 2 bit, 1 for the 1 bit.

12. 56.2.01000000.00000000 /18 What is the IP address now? I need to look at the whole byte. So convert from binary to decimal the following byte – 01000000 (as shown above) Using the Binary to Decimal conversion of 128, 64, 32, 16, 8, 4, 2, 1: 56.2.64.0 This is the network address for subnet 1. The subnet mask would be a count (Binary to Decimal) of all network bits. In our example the network bits are: NNNNNNNN.NNNNNNNN.NNHHHHHHHH.HHHHHHHH or 11111111.11111111.11000000.00000000 = 255.255.192.0 or /18 (/18 is a count of all network bits turned on) The 1 st available IP Address for our hosts would look like this in binary: 00111000.00000010.01(This is our network address) 000000. 00000001 (Host address) =56.2.64.1

13. 56.2.64.0 /18 Our last IP address would be all Host bits turned on (except the last one) or: 00111000.00000010.01(This is our network address) 111111.11111110 (Host address) =56.2.127.254 The broadcast address would be all Host bits turned on or: 00111000.00000010.01(This is our network address) 111111.11111110 (Host address) =56.2.127.255

14. 56.2.64.0 255.255.192.0 Network Address: 56.2.64.0 Subnet Mask: 255.255.192.0 or /18 1 st Available IP Address: 56.2.64.1 Last Available IP Address: 56.2.127.254 Broadcast Address: 56.2.127.255

15. ANDing ANDing is the process a router uses to determine the network address for a packet. Which interface should this packet go out? It takes the destination address and adds it to the subnet mask to determine the number address destination address + subnet mask = network address For example, let’s and a packet going to 10.25.6.1 /24. Which network does this belong to? To determine this I need to convert from decimal to binary.

16. 10.25.6.1 /24 10.25.6.1 00001010.00011001.00000110.00000001Destination Address 255.255.255.011111111.11111111.11111111.00000000Subnet Mask -------------------------------------------------------------------------------- 10.25.6.0 00001010.00011001.00000110.00000000Network Address How did I get this? 0+0=0 0+1=0 1+0=0 1+1=1 Then convert the Network address back to decimal to get – 10.25.6.0 Now the router “knows” which network this packet belongs and can send it out the appropriate interface.

17. 155.64.95.7 /16 155.64.95.710011011.01000000.01011111.00000111Destination Address 255.255.0.011111111.11111111.00000000.00000000Subnet Mask -------------------------------------------------------------------------------- 155.64.0.010011011.01000000.00000000.00000000Network Address How did I get this? 0+0=0 0+1=0 1+0=0 1+1=1 Then convert the Network address back to decimal to get – 155.64.0.0

18. 25.9.173.46 /9 25.9.173.4600011001.00001001.10101101.00101110Destination Address 255.128.0.011111111.10000000.00000000.00000000Subnet Mask -------------------------------------------------------------------------------- 25.0.0.000011001.00000000.00000000.00000000Network Address How did I get this? 0+0=0 0+1=0 1+0=0 1+1=1 Then convert the Network address back to decimal to get – 25.0.0.0

19. VLSM VLSM or Variable Length Subnet Mask is a more efficient way to subnet a network. You do not waste as many IP addresses. Let’s look at this picture of our network: Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts

20. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts So in our picture above: LAN - Sydney needs 30 hosts LAN – Melbourne needs 30 hosts LAN – Perth needs 30 hosts LAN – Brisbane needs 60 hosts WAN – Connection between Sydney and Melbourne (2 host addresses – one for each router connection) WAN – Connection between Melbourne and Perth (2 host addresses) WAN – Connection between Perth and Brisbane (2 host addresses) WAN – Connection between Brisbane and Sydney (2 host addresses)

21. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts In VLSM – always solve for the most networks down to the least networks. Let’s say we were given the IP address 192.168.187.0 255.255.255.0 How would we use this address to subnet for all the networks shown in the picture? (4 LAN and 4 WAN) Starting with the “most networks” we would first solve for the LAN Brisbane network which needs 60 host addresses Remember we can only work with host bits to subnet; we cannot use network bits already used. So all we have to work with are the last 8 host bits (See Subnet Mask – which determines this)

22. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts LAN Brisbane needs 60 hosts from IP Address 192.168.187.0 /24 192.168.187._ _ _ _ _ _ _ _Count host bits from the right; network bits from the left 192.168.187. 256 128 64 32 16 8 4 2I can get 2 hosts in the first bit from the right (0 or 1) So to get 60 host addresses – my line goes here: 192.168.187. 256 128 | 64 32 16 8 4 2 192.168.187.NN | HHHHHH In VLSM – start with subnet 0

23. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts LAN - BRISBANE 60 hosts 192.168.187.NN | HHHHHH In VLSM – start with subnet 0 So the network address is: 192.168.187.00 | 000000or 192.168.187.0 The subnet mask is: 11111111.11111111.11111111.11000000or 255.255.255.192 or /26 The first IP host address to assign is: 192.168.187.00 | 000001or 192.168.187.1 The last IP host address to assign is: 192.168.187.00 | 111110or 192.168.187.62 The broadcast address for this network is: 192.168.187.00 | 111111or 192.168.187.63 Since our next largest network is either LAN – Sydney, LAN-Melbourne, or LAN-Perth, we can do these in any order.

24. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts LAN – SYDNEY 30 hosts 192.168.187.NNN | HHHHH In VLSM – start with subnet 0; Already used; Now let’s use Subnet 1: So the network address is: 192.168.187.010 | 00000or 192.168.187.64 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.224 or /27 The first IP host address to assign is: 192.168.187.010 | 00001or 192.168.187.65 The last IP host address to assign is: 192.168.187.010 | 11110or 192.168.187.94 The broadcast address for this network is: 192.168.187.010 | 11111or 192.168.187.95

25. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts LAN – MELBOURNE 30 hosts 192.168.187.NNN | HHHHH So the network address is: 192.168.187.011 | 00000or 192.168.187.96 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.224 or /27 The first IP host address to assign is: 192.168.187.011 | 00001or 192.168.187.97 The last IP host address to assign is: 192.168.187.011 | 11110or 192.168.187.126 The broadcast address for this network is: 192.168.187.011 | 11111or 192.168.187.127

26. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts LAN – PERTH 30 hosts 192.168.187.NNN | HHHHH So the network address is: 192.168.187.100 | 00000or 192.168.187.128 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.224 or /27 The first IP host address to assign is: 192.168.187.100 | 00001or 192.168.187.129 The last IP host address to assign is: 192.168.187.100 | 11110or 192.168.187.158 The broadcast address for this network is: 192.168.187.100 | 11111or 192.168.187.159

27. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts WAN – Sydney to Melbourne - 2 hosts needed (one for each router on each end) 192.168.187.NNNNNN | HH So the network address is: 192.168.187.101000 | 00or 192.168.187.160 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.230 or /30 The first IP host address to assign is: 192.168.187.101000 | 01or 192.168.187.161 The last IP host address to assign is: 192.168.187.101000 | 11or 192.168.187.162 The broadcast address for this network is: 192.168.187.101000 | 11or 192.168.187.163

28. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts WAN – Melbourne to Perth- 2 hosts needed (one for each router on each end) 192.168.187.NNNNNN | HH So the network address is: 192.168.187.101001 | 00or 192.168.187.164 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.230 or /30 The first IP host address to assign is: 192.168.187.101001 | 01or 192.168.187.165 The last IP host address to assign is: 192.168.187.101001 | 10or 192.168.187.166 The broadcast address for this network is: 192.168.187.101001 | 11or 192.168.187.167

29. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts WAN –Perth to Brisbane- 2 hosts needed (one for each router on each end) 192.168.187.NNNNNN | HH So the network address is: 192.168.187.101010 | 00or 192.168.187.168 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.230 or /30 The first IP host address to assign is: 192.168.187.101010 | 01or 192.168.187.169 The last IP host address to assign is: 192.168.187.101010 | 10or 192.168.187.170 The broadcast address for this network is: 192.168.187.101010 | 11or 192.168.187.171

30. Brisbane Sydney Melbourne Perth 30 hosts 60 hosts 30 hosts WAN –Brisbane to Sydney - 2 hosts needed (one for each router on each end) 192.168.187.NNNNNN | HH So the network address is: 192.168.187.101011 | 00or 192.168.187.172 The subnet mask is: 11111111.11111111.11111111.11100000or 255.255.255.230 or /30 The first IP host address to assign is: 192.168.187.101011 | 01or 192.168.187.173 The last IP host address to assign is: 192.168.187.101011 | 10or 192.168.187.174 The broadcast address for this network is: 192.168.187.101011 | 11or 192.168.187.175

31. VLSM Network for Australia ConnectionNetwork Address LAN Brisbane192.168.187.0 /26 LAN Sydney192.168.187.64 /27 LAN Melbourne192.168.187.96 /27 LAN Perth192.168.187.128 /27 WAN Sydney to Melbourne192.168.187.160 /30 WAN Melbourne to Perth192.168.187.164 /30 WAN Perth to Brisbane192.168.187.168 /30 WAN Brisbane to Sydney192.168.187.172 /30 Notice the subnet mask varies thus VLSM (Variable Length Subnet Mask)