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

2. 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

3. 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

4. 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!

5. # 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

6. 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.

7. Class C Sample Problem
Subnet the 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.

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

9. 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

10. Class C Problem Binary subnets
For example subnet 11 is

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

12. Class C Subnets Dotted Decimal notation#
Bits
Dotted
Decimal 8 1 9 2 10 3 11 4 12 5 13 6 14 7 15

13. # 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

14. Possible Hosts on Subnet 11#
Bits
Dotted
Deceimal 8 1 9 2 10 3 11 4 12 5 13 6 14 7 15
Subnet and host bits are combined to create the dotted decimal

15. Special host addresses
A host address of all 0 bits is the network identifier or network address.
/24
/28
Subnet 11 from our sample
A host address of all 1 bits is the layer 3 broadcast address for that network.
/24
/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

16. 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.
/24
/28
A broadcast to the all 1’s subnet could not be distinguished from a broadcast to all hosts of the original network
/24
/28
Broadcast to subnet 15

17. 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

18. Class B Sample Problem
Subnet the 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.

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

20. Class B Problem Step 3.
Identify your subnets: The formula determines that there are 8 possible Subnets. #
Bits
Dotted
Decimal 4 1 5 2 6 3 7
Which subnets can not be used?

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

22. Host Addresses The 8,192 host addresses would range from to to

23. Host Addresses Dotted Decimal
Using Dotted Decimal notation with Subnet 1, they would range from to
Network address
Broadcast address
Subnet and host bits are combined to create the dotted decimal

24. 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 ( ) to one less than the next network address ( ).
The last subnet address is the same as the mask #
Subnets 4 1 5 2 6 3 7

25. Subnet Table # Subnet 1st Host Last Host Broadcast Mask 172.123.0.0 1 2 3 4 5 6 7
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!

26. “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

27. 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.

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

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

30. Subnet Table, first subnets#
Subnet
Mask
1st Host
Last Host
Broadcast 1 2 3 4 5 6 7 8
+64
+64

31. Subnet Table, last subnets#
Subnet
Mask
1st Host
Last Host
Broadcast
1017
1018
1019
1020
1021
1022
1023
- 64
Last subnet has the same value in the mixed octet as does the mask

32. Subnetting across octet boundaries
A closer look at the transition from subnet 3 to subnet 4
Subnet 3:
Subnet 4:
A closer look at the transition from subnet 1019 to subnet 1020
Subnet 1019:
Subnet 1020:

33. 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

34. HAPPY SUBNETTING!