subnetting a class_c_address

Subnetting Made Easy?Subnetting Made Easy?The “moving stick” and the “magic number”The “moving stick” and the “magic number”

Jim BlancoJim BlancoAparicio-Levy Technical CenterAparicio-Levy Technical Center

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.


•256 addresses would result in a very large collision domain.

•256 hosts using the “wire” one at a time would render the LANunusable.

•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.


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



But this is too simple. We must also keep track of subnet and broadcast addresses.

Subnetting: Class C Subnetting: Class C hosthost address address

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.


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.


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


192192 168168 1212 00

192192 168168 1212 0000000000000000 bit offbit off


192192 168168 1212 000000000000001 bit onbit on


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


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 IPaddresses.•We could continue with combinations of 3, 4 and more bits up to 8 whichwould result in 256 combinations of 1 and 0 or potential IP addresses.•Remember “0” is a number.


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”.


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 and The first “0” is reserved for the subnet and

the last “255” is the broadcast address. the last “255” is the broadcast address.


So that’s 256 – 2 or 254 usable addresses So that’s 256 – 2 or 254 usable addresses in our one big subnet.in our one big subnet.

Next we need to decide how many Next we need to decide how many subnets will meet our networking subnets will meet our networking requirement.requirement.

Subnetting: Class C Subnetting: Class C subnetsubnet address address

192192 168168 1212 00

192192 168168 1212 0000000000000 bit offbit off

192192 168168 1212 00 11stst subnet subnet

192192 168168 1212 100000000000000 bit onbit on

192192 168168 1212 11 22eded subnet subnet

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.

Subnetting: Class C Subnetting: Class C subnetsubnet address address

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

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


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.


256 128 64 32 16 8 4 2 possible number of host addresses

0 0 0 0 0 0 0 02 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.




•Draw the “moving stick.”•You could have a combination of 4 subnets with 64 addresses each.




Move the “stick” to the right.You could have a combination of 8 subnets with 32 addresses each.




Move it again.You could have a combination of 16 subnets with 16 addresses each.

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 with Our company requires at least 3 subnets with more than 50 hosts per subnet.more than 50 hosts per subnet.




•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.




•We could move the “stick” to the right.•But a combination of 8 subnets with 32 addresses eachdoes not meet our company’s requirement.

The moving stickThe moving stickadd the “add the “magic numbermagic number””



•We move the stick back to the left.•64 is our “magic” number”.


SubnetSubnet RangeRange Broadcast Broadcast 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.



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”



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

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

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

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.


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.


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.

Problem completedProblem completed

Our company required us to borrow 2 bits Our 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

subnetting a class_c_address

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