Real Networkers don’t use Decimal! Part 2 Planning Subnets October 18, 2004

Steps to planning a subnet Obtain starting block of addresses Determine how many subnets and how many hosts on each subnet you will need Determine how many host bits to “borrow” Create a subnet mask Create a subnet table listing, for each subnet: Subnet Address and Mask Host Address Range Broadcast address

Starting Block of addresses Determined by a network address and a mask At first, every starting blocks will be a simple Class A, B or C network. Variable Length Subnetting allows a subnet to be further subnetted

Determining # of Subnets & Hosts Some information will be provided Assume if only one number is provided (e.g. # of subnets), the goal is to maximize the other (e.g. # of hosts) In the real world be sure to discuss growth plans with the client!

# of Usable Subnets = # of Possible Subnets - 2 How many bits to borrow Remember the formula’s! # of Possible Subnets = 2Number of subnet bits # of Possible Hosts = 2Number of host bits # of Usable Hosts = (2Number of host bits) – 2 Note: Networks running RIP protocol version 1 can’t use the first (Subnet 0) and the last (all 1’s) subnet . Older texts will show the formula # of Usable Subnets = # of Possible Subnets - 2

Number of host bits The number of host bits will be determined by your subnet mask. It is a good idea to confirm that there will be enough host bits to meet the specifications.

Class C Sample Problem Subnet the 222.33.4.0 network to create at least 10 networks while maximizing the number of hosts. Step 1. Determine # of bits to borrow The formula 2 N ≥ 10 Look at your Powers of 2 table to find a value ≥ 10 You will need to borrow 4 bits.

Class C Problem Step 2 Step 2: Determine your mask. Start with the Class C default mask: 11111111.11111111.11111111.00000000 Change the 4 (from previous step) leftmost host bits into subnet bits 11111111.11111111.11111111.11110000 Write the mask: 255.255.255.240 or /28

Class C Problem Step 3. Identify your subnets The formula determines that there are 16 possible subnets. Subnet # Bits 0000 8 1000 1 0001 9 1001 2 0010 10 1010 3 0011 11 1011 4 0100 12 1100 5 0101 13 1101 6 0110 14 1110 7 0111 15 1111

Class C Problem Binary subnets 11011110 00100001 00000100 00000000 00000100 00010000 00000100 10010000 00000100 00100000 00000100 10100000 00000100 00110000 00000100 10110000 00000100 01000000 00000100 11000000 00000100 01010000 00000100 11010000 00000100 01100000 00000100 11100000 00000100 01110000 00000100 11110000 For example subnet 11 is 11011110 00100001 00000100 10110000

Class C Subnets Dotted Decimal notation Dotted Decimal notation ignores subnetting and converts the 32 bit address into 8 bit chunks. The 11th subnet: 11011110.00100001.00000100.10110000 becomes 222.33.4.176 What will be the 12th subnet in dotted decimal?

Class C Subnets Dotted Decimal notation # Bits Dotted Decimal 11011110.00100001 00000100.00000000 222.33.4.0 8 00000100.10000000 222.33.4.128 1 .00010000 222.33.4.16 9 .10010000 222.33.4.144 2 .00100000 222.33.4.32 10 .10100000 222.33.4.160 3 .00110000 222.33.4.48 11 .10110000 222.33.4.176 4 .01000000 222.33.4.64 12 .11000000 222.33.4.192 5 .01010000 222.33.4.80 13 .11010000 222.33.4.208 6 .01100000 222.33.4.96 14 .11100000 222.33.4.224 7 .01110000 222.33.4.112 15 .11110000 222.33.4.240

# of Host Addresses = 2Number of host bits Since number of possible host addresses depends on the number of host bits, use the same formula used to determine the number of subnets. # of Host Addresses = 2Number of host bits 4 host bits means 16 possible host addresses

Possible Hosts on Subnet 11 # Bits Dotted Deceimal 11011110.00100001 00000100.11010000 222.33.4.176 8 00000100.11011000 222.33.4.184 1 .11010001 222.33.4.177 9 .11011001 222.33.4.185 2 .11010010 222.33.4.178 10 .11011010 222.33.4.186 3 .11010011 222.33.4.179 11 .11011011 222.33.4.187 4 .11010100 222.33.4.180 12 .11011100 222.33.4.188 5 .11010101 222.33.4.181 13 .11011101 222.33.4.189 6 .11010110 222.33.4.182 14 .11011110 222.33.4.190 7 .11010111 222.33.4.183 15 .11011111 222.33.4.191 Subnet and host bits are combined to create the dotted decimal

Special host addresses A host address of all 0 bits is the network identifier or network address. 11011110.00100001.00000100.00000000 222.33.4.0/24 11011110.00100001.00000100.10110000 222.33.4.176/28 Subnet 11 from our sample A host address of all 1 bits is the layer 3 broadcast address for that network. 11011110.00100001.00000100.11111111 222.33.4.255/24 11011110.00100001.00000100.10111111 222.33.4.191/28 The broadcast address for subnet 11 from our sample These two addresses can not be assigned to a host. # of Usable Host Addresses = (2Number of host bits) - 2

Unusable Subnets The all 0’s and all 1’s subnets should also not be used when using classful (RIP version 1) routing. The all 0’s subnet could be confused with the original network. 11011110.00100001.00000100.00000000 222.33.4.0/24 11011110.00100001.00000100.00000000 222.33.4.0/28 A broadcast to the all 1’s subnet could not be distinguished from a broadcast to all hosts of the original network 11011110.00100001.00000100.11111111 222.33.4.255/24 11011110.00100001.00000100.11111111 222.33.4.255/28 Broadcast to subnet 15

Formulas # of Possible Subnets = 2Number of subnet bits # of Possible Hosts = 2Number of host bits # of Usable Hosts = (2Number of host bits) - 2 # of Usable Subnets with RIP version 1 = (2Number of subnet bits) – 2

Class B Sample Problem Subnet the 172.123.0.0 network to create at least 5 networks while maximizing the number of hosts. Step 1. Determine # of bits to borrow The formula 2 N ≥ 5 Look at the Powers of 2 table to find a value ≥ 5 You will need to borrow 3 bits.

Class B Problem Step 2 Step 2: Determine your mask. Start with the Class B default mask: 11111111.11111111.00000000.00000000 Change the 3 leftmost host bits (from previous step) into subnet bits 11111111.11111111.11100000.00000000 Write the mask: 255.255.224.0 or /19

Class B Problem Step 3. Identify your subnets: The formula determines that there are 8 possible Subnets. # Bits Dotted Decimal 10101100.01111011 00000000.00000000 172.123.0.0 4 10000000.00000000 172.123.128.0 1 00100000.00000000 172.123.32.0 5 10100000.00000000 172.123.160.0 2 01000000.00000000 172.123.64.0 6 11000000.00000000 172.123.196.0 3 01100000.00000000 172.123.96.0 7 11100000.00000000 172.123.224.0 Which subnets can not be used?

# of Host Addresses = 2Number of host bits 11111111.11111111.11100000.00000000 Number of host bits = 32 – mask bits(19) = 13. 8192 Possible host addresses The Powers of 2 table again!

Host Addresses The 8,192 host addresses would range from to 10101100 01111011 00100000 00000000 10101100 01111011 00100000 00000001 10101100 01111011 00100000 00000010 to 10101100 01111011 00100000 11111111 10101100 01111011 00100001 00000000 10101100 01111011 00100001 00000001 10101100 01111011 00111111 11111110 10101100 01111011 00111111 11111111

Host Addresses Dotted Decimal Using Dotted Decimal notation with Subnet 1, they would range from 10101100.01111011.00100000.00000000 172.123.32.0 10101100.01111011.00100000.00000001 172.123.32.1 10101100.01111011.00100000.00000010 172.123.32.2 to 10101100.01111011.00111111.11111110 172.123.63.254 10101100.01111011.00111111.11111111 172.123.63.255 Network address Broadcast address Subnet and host bits are combined to create the dotted decimal

Address Patterns In this example each subnet address is 32 more than the previous subnet address. The host addresses on subnet 1 range from the subnet address (172.123.32.0) to one less than the next network address (172.123.63.255). The last subnet address is the same as the mask # Subnets 10101100.01111011 00000000.00000000 172.123.0.0 4 10000000.00000000 172.123.128.0 1 00100000.00000000 172.123.32.0 5 10100000.00000000 172.123.160.0 2 01000000.00000000 172.123.64.0 6 11000000.00000000 172.123.192.0 3 01100000.00000000 172.123.96.0 7 11100000.00000000 172.123.224.0

Subnet Table # Subnet 1st Host Last Host Broadcast Mask 172.123.0.0 172.123.0.0 225.255.224.0 172.123.0.1 172.123.31.254 172.123.31.255 1 172.123.32.0 172.123.32.1 172.123.63.254 172.123.63.255 2 172.123.64.0 172.123.64.1 172.123.95.254 172.123.95.255 3 172.123.96.0 172.123.96.1 172.123.127.254 172.123.127.255 4 172.123.128.0 172.123.128.1 172.123.159.254 172.123.159.255 5 172.123.160.0 172.123.160.1 172.123.191.254 172.123.191.255 6 172.123.192.0 172.123.192.1 172.123.223.254 172.123.223.255 7 172.123.224.0 172.123.224.1 172.123.255.254 172.123.255.255 Shortcut! Notice that every item (except the mask) is 32 more than the previous value in the octet containing both subnet and host bits. Only true for 3 bit subnets!

“Magic Number” shortcuts The number 32 in the previous example is sometimes called the “magic number.” It is a result of using dotted decimal notation. It only applies in the octet that contains both subnet & host bits. There are two ways to determine this number It is the value of the right most network bit position It also equals 256 – subnet mask

Last Example! Class A Subnet 121.0.0.0 /8 to create 1000 subnets. Step 1. Determine # of bits to borrow The formula 2 N ≥ 1000 Look at your Powers of 2 table to find a value ≥ 1000 You will need to borrow 10 bits.

Class A Problem Step 2 Step 2: Determine your mask. Start with the Class A default mask: 11111111.00000000.00000000.00000000 Change the 10 leftmost host bits into subnet bits 11111111.11111111.11000000.00000000 Write the mask: 255.255.192.0 or /18

Determine Magic Number Network: 121.0.0.0 Mask:255.255.192.0 Subnet 0: 01111001.00000000.00000000.00000000 Subnet 1: 01111001.00000000.01000000.00000000 Subnet 1: 121.0.64.0 or 256 – 192 = 64 Magic Number! Only use in the octet with both subnet and host bits.

Subnet Table, first subnets # Subnet Mask 1st Host Last Host Broadcast 121.0.0.0 225.255.192.0 121.0.0.1 121.0.63.254 121.0.63.255 1 121.0.64.0 121.0.64.1 121.0.127.254 121.0.127.255 2 121.0.128.0 121.0.128.1 121.0.191.254 121.0.191.255 3 121.0.192.0 121.0.192.1 121.0.255.254 121.0.255.255 4 121.1.0.0 121.1.0.1 121.1.63.254 121.1.63.255 5 121.1.64.0 121.1.64.1 121.1.127.254 121.1.127.255 6 121.1.128.0 121.1.128.1 121.1.191.254 121.1.191.255 7 121.1.192.0 121.1.192.1 121.1.255.254 121.1.255.255 8 121.2.0.0 121.2.0.1 121.2.63.254 121.2.63.255 +64 +64

Subnet Table, last subnets # Subnet Mask 1st Host Last Host Broadcast 1017 121.254.0.0 225.255.192.0 121.254.0.1 121.254.63.254 121.254.63.255 121.254.64.0 121.254.64.1 121.254.127.254 121.254.127.255 1018 121.254.128.0 121.254.128.1 121.254.191.254 121.254.191.255 1019 121.254.192.0 121.254.192.1 121.254.255.254 121.254.255.255 1020 121.255.0.0 121.255.0.1 121.255.63.254 121.255.63.255 1021 121.255.64.0 121.255.64.1 121.255.127.254 121.255.127.255 1022 121.255.128.0 121.255.128.1 121.255.191.254 121.255.191.255 1023 121.255.192.0 121.255.192.1 121.255.255.254 121.255.255.255 - 64 Last subnet has the same value in the mixed octet as does the mask

Subnetting across octet boundaries A closer look at the transition from subnet 3 to subnet 4 Subnet 3: 01111001.00000000.11000000.00000000 Subnet 4: 01111001.00000001.00000000.00000000 A closer look at the transition from subnet 1019 to subnet 1020 Subnet 1019: 01111001.11111110.11000000.00000000 Subnet 1020: 01111001.11111111.00000000.00000000

Summary # of Possible Subnets = 2Number of subnet bits # of Possible Hosts = 2Number of host bits # of Usable Hosts = (2Number of host bits) – 2 # of RIP v1 Usable Subnets = (2Number of subnet bits) – 2 Magic Number – used in octet with both subnet and host bits. Right most subnet bit value 256 – subnet mask Last possible subnet = mask of mixed octet

HAPPY SUBNETTING!