The document discusses subnetting a Class C network using the "moving stick" and "magic number" methods. It shows how to borrow bits from the last octet to create subnets and calculates the subnet, range, and broadcast addresses. Using the IP address 192.168.12.0 with requirements for at least 3 subnets with more than 50 hosts each, it determines the subnet mask would be 255.255.255.192, creating 4 subnets of 64 addresses each.

1. Subnetting Made Easy?Subnetting Made Easy?
The “moving stick” and the “magic number”The “moving stick” and the “magic number”
Jim BlancoJim Blanco
Aparicio-Levy Technical CenterAparicio-Levy Technical Center

2. Subnetting Made EasySubnetting Made Easy
•First let’s look at the overall requirement.
•A class C network consists of 4 octets totaling 32 bits.
•If we use a Class C network such as 192.168.12.0, we can only
make use of the last octet or 8 bits.
•There are 256 possible combinations of bits “on” or “off” in one octet.

3. Subnetting Made EasySubnetting Made Easy
•256 addresses would result in a very large collision domain.
•256 hosts using the “wire” one at a time would render the LAN
unusable.
•In business environments, host addresses are usually divided
into groups or subnets for management and security reasons.
•In addition the first address is reserved for the subnet address
and the last for a broadcast address.
•So we really have 254 available host addresses.

4. Subnetting Made EasySubnetting Made Easy
 We could just divide the addresses in the last octet into more manageable
blocks or “subnets”:
256/4 = 64 or 4 subnets each with 64 addresses
256/8 = 32 or 8 subnets each with 32 addresses
256/16 = 16 or 16 subnets each with 16 addresses
 But this is too simple. We must also keep track of subnet and broadcast
addresses.

5. Subnetting: Class CSubnetting: Class C hosthost addressaddress
192192 168168 1212 00
192192 168168 1212 0000000000000000
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
•First convert the last octet, represented by the decimal number “0”,
into 8 binary “0”s to represent 8 bits.

6. Subnetting: Class CSubnetting: Class C hosthost addressaddress
192192 168168 1212 00
192192 168168 1212 000000000000000
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
• By just utilizing the last bit, we have two possible IP addresses.

7. Subnetting: Class CSubnetting: Class C hosthost addressaddress
192192 168168 1212 00
192192 168168 1212 000000000000000 bit offbit off
192192 168168 1212 00 IP addressIP address
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
•First, with the bit remaining at “0” or off, the IP address is 192.168.12.0

8. Subnetting: Class CSubnetting: Class C hosthost addressaddress
192192 168168 1212 00
192192 168168 1212 0000000000000000 bit offbit off
192192 168168 1212 00 IP addressIP address
192192 168168 1212 000000000000001 bit onbit on
192192 168168 1212 11 IP addressIP address
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
•Second, when the bit is “1” or turned on, the IP address is 192.168.12.1
•Thus we have 2 possible IP addresses just utilizing the last bit

9. Subnetting: Class CSubnetting: Class C hosthost addressaddress
192192 168168 1212 00
192192 168168 1212 0000000000000000 bit offbit off
192192 168168 1212 00 .0.0
192192 168168 1212 0000000100000001 bit onbit on
192192 168168 1212 11 .1.1
192192 168168 1212 00000000000000 .0.0
192192 168168 1212 00000000000001 .1.1
192192 168168 1212 00000000000010 .2.2
192192 168168 1212 00000000000011 .3.3
•If we use the two last bits, in the on an off positions, we have four possible IP
addresses.
•We could continue with combinations of 3, 4 and more bits up to 8 which
would result in 256 combinations of 1 and 0 or potential IP addresses.
•Remember “0” is a number.

10. Subnetting: Class CSubnetting: Class C hosthost addressaddress
192192 168168 1212 00
192192 168168 1212 0000000000000000 bit offbit off
192192 168168 1212 00 .0.0
192192 168168 1212 0000000100000001 bit onbit on
192192 168168 1212 11 .1.1
192192 168168 1212 00000000000000 .0.0
192192 168168 1212 00000000000001 .1.1
192192 168168 1212 00000000000010 .2.2
192192 168168 1212 00000000000011 .3.3
•Since we cannot use the first address (subnet), 0 or the last address 255
(broadcast) we have 256-2=254 usable addresses.
•That’s one big collision domain.
•We need to divide it up into smaller blocks or “subnets”.

11. Subnetting: Class CSubnetting: Class C hosthost addressaddress
 Hold on. Thought we had 256 addresses?Hold on. Thought we had 256 addresses?
Or is it 254?Or is it 254?
 There are 256 combinations of 1 and 0.There are 256 combinations of 1 and 0.
 Possible addresses run from .0 to .255.Possible addresses run from .0 to .255.
 ““0” is a number.0” is a number.
 0-255 yields 256 addresses.0-255 yields 256 addresses.
 The first “0” is reserved for the subnet andThe first “0” is reserved for the subnet and
the last “255” is the broadcast address.the last “255” is the broadcast address.

12. Subnetting: Class CSubnetting: Class C hosthost addressaddress
 So that’s 256 – 2 or 254 usable addressesSo that’s 256 – 2 or 254 usable addresses
in our one big subnet.in our one big subnet.
 Next we need to decide how manyNext we need to decide how many
subnets will meet our networkingsubnets will meet our networking
requirement.requirement.

13. Subnetting: Class CSubnetting: Class C subnetsubnet addressaddress
192192 168168 1212 00
192192 168168 1212 0000000000000 bit offbit off
192192 168168 1212 00 11stst
subnetsubnet
192192 168168 1212 100000000000000 bit onbit on
192192 168168 1212 11 22eded
subnetsubnet
192192 168168 1212
192192 168168 1212
192192 168168 1212
192192 168168 1212
•The same rule applies to borrowing bits for subnet addresses.
•Start at the left side.
•The first bit can be borrowed and turned on or off resulting in 2 subnets.

14. Subnetting: Class CSubnetting: Class C subnetsubnet addressaddress
192192 168168 1212 00
192192 168168 1212 00000000000000
192192 168168 1212 00
192192 168168 1212 1000000010000000
192192 168168 1212 11
192192 168168 1212 00000000000000 00
192192 168168 1212 10000000000000 11
192192 168168 1212 01000000000000 22
192192 168168 1212 11000000000000 33
•Borrowing two bits yields four combinations of bits on and off, or four
different combinations and 4 possible subnets

15. The moving stickThe moving stick
0 0 0 0 0 0 0 0
192.168.12.0
•Now let’s put it all together with our “moving stick” method
•Write the last octet in binary

16. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible host addresses
0 0 0 0 0 0 0 0
•Start on the right.
•Number to the left to show possible numbers of host addresses.

17. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
Start on the left.
Number to the right to show possible numbers of subnets.

18. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
•Draw the “moving stick.”
•You could have a combination of 4 subnets with 64 addresses each.

19. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
Move the “stick” to the right.
You could have a combination of 8 subnets with 32 addresses each.

20. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
Move it again.
You could have a combination of 16 subnets with 16 addresses each.

21. Calculate the subnetsCalculate the subnets
Use this IP addressUse this IP address
192.168.12.0192.168.12.0
Our company requires at least 3 subnets withOur company requires at least 3 subnets with
more than 50 hosts per subnet.more than 50 hosts per subnet.

22. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
•Look back to our first example.
•We borrowed two bits.
•This fits the requirement of our company – 4 subnets each with up to 64 addresses.

23. The moving stickThe moving stick
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
•We could move the “stick” to the right.
•But a combination of 8 subnets with 32 addresses each
does not meet our company’s requirement.

24. The moving stickThe moving stick
add the “add the “magic numbermagic number””
256 128 64 32 16 8 4 2 possible number of host addresses
0 0 0 0 0 0 0 0
2 4 8 16 32 64 128 256 possible number of subnets
•We move the stick back to the left.
•64 is our “magic” number”.

25. Calculate the subnetsCalculate the subnets
SubnetSubnet RangeRange BroadcastBroadcast
addressaddress
192.168.12.0192.168.12.0
192.168.12.192.168.12.64
192.168.12.192.168.12.128
192.168.12.192.168.12.1192
•Add to the “0” subnet by increments of 64, our magic number.
•We find our 4 subnet addresses.

26. Calculate the subnetsCalculate the subnets
SubnetSubnet RangeRange BroadcastBroadcast
addressaddress
192.168.12.0192.168.12.0 192.168.12.1 - 192.168.12.62 192.168.12.63
192.168.12.64192.168.12.64
192.168.12.128192.168.12.128
192.168.12.192192.168.12.192
•The first usable address is 192.168.12.1 in our first subnet
•The last usable address is 192.168.12.62
•The broadcast address is 192.168.12.63
•.0 through .63 totals 64 addresses, our “magic number”

27. Calculate the subnetsCalculate the subnets
SubnetSubnet RangeRange BroadcastBroadcast
addressaddress
192.168.12.0192.168.12.0 192.168.12.1 - 192.168.12.62192.168.12.1 - 192.168.12.62 192.168.12.63192.168.12.63
192.168.12.64192.168.12.64 192.168.12.65 - 192.168.12.126192.168.12.65 - 192.168.12.126 192.168.12.127192.168.12.127
192.168.12.128192.168.12.128 192.168.12.129 - 192.168.12.191192.168.12.129 - 192.168.12.191 192.168.12.192192.168.12.192
192.168.12.192192.168.12.192 192.168.12.193 - 192.168.12.254192.168.12.193 - 192.168.12.254 192.168.12.255192.168.12.255
•Fill in the remaining columns

29. Ok, I lied. You still haveOk, I lied. You still have
to figure out that peskyto figure out that pesky
subnet mask.subnet mask.

30. •Just because you graphJust because you graph
subnets on a piece of papersubnets on a piece of paper
doesn’t mean your router ordoesn’t mean your router or
PC has any idea what you did.PC has any idea what you did.
•We need a subnet mask toWe need a subnet mask to
enter into the router CLI or yourenter into the router CLI or your
PC’s local area connectionPC’s local area connection
propertiesproperties

31. Subnet MaskSubnet Mask
128 64 32 16 8 4 2 1 binary numbers
0 0 0 0 0 0 0 0
•Renumber your last 8 bits to show the binary equivalent.
•Draw your stick to show the two borrowed bits.
•Your subnet mask is 128 + 64 = 192.

32. Subnet MaskSubnet Mask
128 64 32 16 8 4 2 1 binary numbers
0 0 0 0 0 0 0 0
•If you had borrowed 3 bits.
•Your subnet mask would be 128 + 64 + 32 = 224.

33. Subnet MaskSubnet Mask
128 64 32 16 8 4 2 1 binary numbers
0 0 0 0 0 0 0 0
•Move the stick to borrow 4 bits.
•Your subnet mask would be 128 + 64 + 16 + = 240.

34. Problem completedProblem completed
 Our company required us to borrow 2 bitsOur company required us to borrow 2 bits
so our IP address and subnet mask is:so our IP address and subnet mask is:
192.168. 12. 0192.168. 12. 0
255.255.255.192255.255.255.192