Transcript Subnetting: Class C address

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.
Subnetting: Class C host address
192
168
12
0
192
168
12
00000000
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
•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
168
12
0
192
168
12
00000000
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
• By just utilizing the last bit, we have two possible IP addresses.
Subnetting: Class C host address
192
168
12
0
192
168
12
00000000
bit off
192
168
12
0
IP address
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
192
168
12
•First, with the bit remaining at “0” or off, the IP address is 192.168.12.0
Subnetting: Class C host address
192
168
12
0
192
168
12
00000000
bit off
192
168
12
0
IP address
192
168
12
00000001
bit on
192
168
12
1
IP address
192
168
12
192
168
12
192
168
12
192
168
12
•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
168
12
0
192
168
12
00000000
bit off
192
168
12
0
.0
192
168
12
00000001
bit on
192
168
12
1
.1
192
168
12
00000000
.0
192
168
12
00000001
.1
192
168
12
00000010
.2
192
168
12
00000011
.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
168
12
0
192
168
12
00000000
bit off
192
168
12
0
.0
192
168
12
00000001
bit on
192
168
12
1
.1
192
168
12
00000000
.0
192
168
12
00000001
.1
192
168
12
00000010
.2
192
168
12
00000011
.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
168
12
0
192
168
12
0000000
bit off
192
168
12
0
1st subnet
192
168
12
10000000
bit on
192
168
12
1
2ed subnet
192
168
12
192
168
12
192
168
12
192
168
12
•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
168
12
0
192
168
12
0000000
192
168
12
0
192
168
12
10000000
192
168
12
1
192
168
12
00000000
0
192
168
12
10000000
1
192
168
12
01000000
2
192
168
12
11000000
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
•We move the stick back to the left.
•64 is our “magic” number”.
128 256 possible number of subnets
Calculate the subnets
Subnet
Range
Broadcast
address
192.168.12.0
192.168.12.64
192.168.12.128
192.168.12.192
•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
Range
192.168.12.1 - 192.168.12.62
Broadcast
address
192.168.12.63
192.168.12.64
192.168.12.128
192.168.12.192
•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”
Calculate the subnets
Subnet
Range
Broadcast
address
192.168.12.0
192.168.12.1 - 192.168.12.62
192.168.12.63
192.168.12.64
192.168.12.65 - 192.168.12.126
192.168.12.127
192.168.12.128
192.168.12.129 - 192.168.12.191
192.168.12.192
192.168.12.192
192.168.12.193 - 192.168.12.254
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