subnetting workshop sara akers september 2014. disclaimer if you notice any mistakes with any of the...

Subnetting WorkshopSARA AKERS

SEPTEMBER 2014

Disclaimer If you notice any mistakes with any of the slides, please let me know so I can correct.

Thank you!

Sara [email protected]

SUBNETTING

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 2 I 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 2 However, 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

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.

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 1st 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

7.4.63.48 /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

7.4.63.0 255.255.255.0 Network Address: 7.4.63.48

Subnet Mask: 255.255.255.240 or /28

1st 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?

Try solving for: 56.2.0.0 255.255.0.0

I need at least 3 subnets. Solve for Subnet 1.

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 . …

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.

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 1st 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

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

56.2.64.0 255.255.192.0 Network Address: 56.2.64.0

Subnet Mask: 255.255.192.0 or /18

1st Available IP Address: 56.2.64.1

Last Available IP Address: 56.2.127.254

Broadcast Address: 56.2.127.255

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.

10.25.6.1 /24 10.25.6.1 00001010.00011001.00000110.00000001 Destination Address

255.255.255.0 11111111.11111111.11111111.00000000 Subnet Mask--------------------------------------------------------------------------------

10.25.6.0 00001010.00011001.00000110.00000000 Network Address

How did I get this?0+0=00+1=01+0=01+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.


255.255.0.0 11111111.11111111.00000000.00000000 Subnet Mask--------------------------------------------------------------------------------

155.64.0.0 10011011.01000000.00000000.00000000 Network Address

How did I get this?0+0=00+1=01+0=01+1=1



255.128.0.0 11111111.10000000.00000000.00000000 Subnet Mask--------------------------------------------------------------------------------

25.0.0.0 00011001.00000000.00000000.00000000 Network Address

How did I get this?0+0=00+1=01+0=01+1=1


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

30 hosts

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

So in our picture above:LAN - Sydney needs 30 hostsLAN – Melbourne needs 30 hostsLAN – Perth needs 30 hostsLAN – Brisbane needs 60 hostsWAN – 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)

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 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 192.168.187.0 255.255.255.0How 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)

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

LAN Brisbane needs 60 hosts from IP Address 192.168.187.0 /24192.168.187._ _ _ _ _ _ _ _ Count host bits from the right; network bits from the left192.168.187. 256 128 64 32 16 8 4 2 I 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 2192.168.187.NN | HHHHHHIn VLSM – start with subnet 0

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

LAN - BRISBANE 60 hosts192.168.187.NN | HHHHHHIn VLSM – start with subnet 0So the network address is: 192.168.187.00 | 000000 or 192.168.187.0The subnet mask is: 11111111.11111111.11111111.11000000 or 255.255.255.192 or /26The first IP host address to assign is: 192.168.187.00 | 000001 or 192.168.187.1The last IP host address to assign is: 192.168.187.00 | 111110 or 192.168.187.62The broadcast address for this network is: 192.168.187.00 | 111111 or 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.

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

LAN – SYDNEY 30 hosts192.168.187.NNN | HHHHHIn VLSM – start with subnet 0; Already used; Now let’s use Subnet 1:So the network address is: 192.168.187.010 | 00000 or 192.168.187.64The subnet mask is: 11111111.11111111.11111111.11100000 or 255.255.255.224 or /27The first IP host address to assign is: 192.168.187.010 | 00001 or 192.168.187.65The last IP host address to assign is: 192.168.187.010 | 11110 or 192.168.187.94The broadcast address for this network is: 192.168.187.010 | 11111 or 192.168.187.95

Akers, Sara

Notice our line moved. For 30 hosts we ONLY really need the last 5 bits (See Subnetting Section for how to determine How many hosts needed?)

Akers, Sara

Even though our line moved, we already used the IP address: 00100000 (or 32 - see last slide). So we have to stick with the line we used in the last example (see last slide) and while network vs host bits looks like this: NNNHHHHH; Our bits for subnet 1 would look like this: 010 instead of 000 on the nework side. How can i double check this? The network address should be the next number after the broadcast address in the last slide.

Akers, Sara

Remember line changed - so subnet mask (network bits) changed

Akers, Sara

Do you notice a pattern developing? Once you figure out Nework address (which is easy in VLSM - next number after last Broadcast address) +1 for 1st IP. Figure out Last IP +1 for Broadcast.

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

LAN – MELBOURNE 30 hosts192.168.187.NNN | HHHHHSo the network address is: 192.168.187.011 | 00000 or 192.168.187.96The subnet mask is: 11111111.11111111.11111111.11100000 or 255.255.255.224 or /27The first IP host address to assign is: 192.168.187.011 | 00001 or 192.168.187.97The last IP host address to assign is: 192.168.187.011 | 11110 or 192.168.187.126The broadcast address for this network is: 192.168.187.011 | 11111 or 192.168.187.127

Akers, Sara

We can leave our line where it is. Same # of hosts needed

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

LAN – PERTH 30 hosts192.168.187.NNN | HHHHHSo the network address is: 192.168.187.100 | 00000 or 192.168.187.128The subnet mask is: 11111111.11111111.11111111.11100000 or 255.255.255.224 or /27The first IP host address to assign is: 192.168.187.100 | 00001 or 192.168.187.129The last IP host address to assign is: 192.168.187.100 | 11110 or 192.168.187.158The broadcast address for this network is: 192.168.187.100 | 11111 or 192.168.187.159

Akers, Sara

We can leave our line where it is. Same # of hosts needed

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

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

Akers, Sara

Notice our line is on the move again - 2 host addresses needed + 1 network address + 1 broadcast address = 4 addresses total. Thus the last 2 bits are host addresses.

Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

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

Akers, Sara


Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

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

Akers, Sara


Brisbane

Sydney Melbourne

Perth

30 hosts

60 hosts

30 hosts

30 hosts

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

Akers, Sara


VLSM Network for AustraliaConnection Network Address

LAN Brisbane 192.168.187.0 /26

LAN Sydney 192.168.187.64 /27

LAN Melbourne 192.168.187.96 /27

LAN Perth 192.168.187.128 /27

WAN Sydney to Melbourne 192.168.187.160 /30

WAN Melbourne to Perth 192.168.187.164 /30

WAN Perth to Brisbane 192.168.187.168 /30

WAN Brisbane to Sydney 192.168.187.172 /30

Notice the subnet mask varies thus VLSM (Variable Length Subnet Mask)

subnetting workshop sara akers september 2014. disclaimer if you notice any mistakes with any of the...

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