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

2. 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.

3. 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.

4. 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.

5. Subnetting: Class C host address 192168120 1921681200000000 19216812 19216812 19216812 19216812 19216812 19216812 19216812 First convert the last octet, represented by the decimal number “0”, into 8 binary “0”s to represent 8 bits.

6. Subnetting: Class C host address 192168120 19216812 0000000 00000000 19216812 19216812 19216812 19216812 19216812 19216812 19216812 By just utilizing the last bit, we have two possible IP addresses.

7. Subnetting: Class C host address 192168120 19216812 0000000 00000000 bit off 192168120 IP address 19216812 19216812 19216812 19216812 19216812 19216812 First, with the bit remaining at “0” or off, the IP address is 192.168.12.0

8. Subnetting: Class C host address 192168120 1921681200000000 bit off 192168120 IP address 19216812 0000000 00000001 bit on 192168121 IP address 19216812 19216812 19216812 19216812 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 C host address 192168120 1921681200000000 bit off 192168120.0 1921681200000001 bit on 192168121.1 19216812 000000 00000000.0 19216812 000000 00000001.1 19216812 000000 00000010.2 19216812 000000 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.

10. Subnetting: Class C host address 192168120 1921681200000000 bit off 192168120.0 1921681200000001 bit on 192168121.1 19216812 000000 00000000.0 19216812 000000 00000001.1 19216812 000000 00000010.2 19216812 000000 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”.

11. Subnetting: Class C host address Hold on. Thought we had 256 addresses? Or is it 254? Hold on. Thought we had 256 addresses? 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 last “255” is the broadcast address. The first “0” is reserved for the subnet and the last “255” is the broadcast address.

12. Subnetting: Class C host address So that’s 256 – 2 or 254 usable addresses in our one big subnet. 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. Next we need to decide how many subnets will meet our networking requirement.

13. Subnetting: Class C subnet address 192168120 19216812 000000 0000000 bit off 192168120 1 st subnet 19216812 0000000 10000000 bit on 192168121 2 ed subnet 19216812 19216812 19216812 19216812 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 C subnet address 192168120 192168120000000 192168120 1921681210000000 192168121 19216812 000000 000000000 19216812 000000 100000001 19216812 000000 010000002 19216812 000000 110000003 Borrowing two bits yields four combinations of bits on and off, or four different combinations and 4 possible subnets

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

17. The moving stick 256 128 64 32 16 8 4 2 possible number of host addresses 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 stick 256 128 64 32 16 8 4 2 possible number of host addresses 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 stick 256 128 64 32 16 8 4 2 possible number of host addresses 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 stick 256 128 64 32 16 8 4 2 possible number of host addresses 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 subnets Use this IP address 192.168.12.0 Our company requires at least 3 subnets with more than 50 hosts per subnet.

22. The moving stick 256 128 64 32 16 8 4 2 possible number of host addresses 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 stick 256 128 64 32 16 8 4 2 possible number of host addresses 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 stick add the “magic number” 256 128 64 32 16 8 4 2 possible number of host addresses 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 subnets SubnetRange Broadcast address 192.168.12.0 192.168.12. 192.168.12.64 192.168.12. 192.168.12.128 192.168.12.1 192.168.12.192 Add to the “0” subnet by increments of 64, our magic number. We find our 4 subnet addresses.

26. Calculate the subnets SubnetRange Broadcast address 192.168.12.0192.168.12.1 - 192.168.12.62192.168.12.63 192.168.12.64 192.168.12.128 192.168.12.192 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”

27. Calculate the subnets SubnetRange 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

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

30. Just because you graph subnets on a piece of paper doesn’t mean your router orJust 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 propertiesWe need a subnet mask to enter into the router CLI or your PC’s local area connection properties

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

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

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