Subnetting Made Easy?

The “moving stick” and the “magic number” Jim Blanco Aparicio-Levy Technical Center

Subnetting 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.

Subnetting 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.

Subnetting 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.

192 192 192 192 192 192 192 192 192 Subnetting: Class C host address 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 00000000 •First convert the last octet, represented by the decimal number “0”, into 8 binary “0”s to represent 8 bits.

Subnetting: Class C host address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 0000000 0 • By just utilizing the last bit, we have two possible IP addresses.

Subnetting: Class C host address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 0000000 0 0 bit off IP address •First, with the bit remaining at “0” or off, the IP address is 192.168.12.0

Subnetting: Class C host address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 00000000 0 0000000 1 1 bit off IP address bit on IP address •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

Subnetting: Class C host address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 00000000 0 00000001 1 000000 00 000000 01 000000 10 000000 11 bit off .0

bit on .1

.0

.1

.2

.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.

Subnetting: Class C host address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 00000000 0 00000001 1 000000 00 000000 01 000000 10 000000 11 bit off .0

bit on .1

.0

.1

.2

.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”.

Subnetting: Class C host address  Hold on. Thought we had 256 addresses? Or is it 254?

 There are 256 combinations of 1 and 0.

 Possible addresses run from .0 to .255.

 “0” is a number.

 0-255 yields 256 addresses.

 The first “0” is reserved for the subnet and the last “255” is the broadcast address.

Subnetting: Class C host address  So that’s 256 – 2 or 254 usable addresses in our one big subnet.

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

Subnetting: Class C subnet address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 0 000000 0 1 0000000 1 bit off 1 st subnet bit on 2 ed subnet •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 subnet address 192 192 192 192 192 192 192 192 192 168 168 168 168 168 168 168 168 168 12 12 12 12 12 12 12 12 12 0 0000000 0 10000000 1 00 000000 10 000000 01 000000 11 000000 0 1 2 3 •Borrowing two bits yields four combinations of bits on and off, or four different combinations and 4 possible subnets

The moving stick 192.168.12.0

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

The 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.

The 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.

The 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.

The 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.

The 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.

Calculate the subnets Use this IP address 192.168.12.0

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

The 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.

The 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.

The moving stick add the “ magic 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”.

Subnet

192.168.12.0

192.168.12.

64 192.168.12.

128 192.168.12.

192 Calculate the subnets

Range Broadcast address

•Add to the “0” subnet by increments of 64, our magic number.

•We find our 4 subnet addresses.

Calculate the subnets

Subnet

192.168.12.0

192.168.12.64

192.168.12.128

192.168.12.192

Range

192.168.12.1 192.168.12.62

Broadcast address

192.168.12.63

•The first usable address is 192.168.12.1

•The last usable address is 192.168.12.62

in our first subnet •The broadcast address is 192.168.12.63

•.0 through .63 totals 64 addresses, our “ magic number ”

Calculate the subnets

Subnet Range

192.168.12.0

192.168.12.64

192.168.12.1 - 192.168.12.62

192.168.12.65 - 192.168.12.126

192.168.12.128

192.168.12.129 - 192.168.12.191

192.168.12.192

192.168.12.193 - 192.168.12.254

Broadcast address

192.168.12.63

192.168.12.127

192.168.12.192

192.168.12.255

•Fill in the remaining columns

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

•Just because you graph subnets on a piece of paper doesn’t mean your router or PC has any idea what you did.

•We need a subnet mask to enter into the router CLI or your PC’s local area connection properties

Subnet 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.

Subnet 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.

Subnet 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.

Problem completed  Our company required us to borrow 2 bits so our IP address and subnet mask is: 192.168. 12. 0 255.255.255.192