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Introduction to Networking (Yarnfield)

Learn about Variable Length Subnet Masking (VLSM), the advantages of VLSM over classful subnetting, and how to perform VLSM operations on given IP addresses. Includes a practical example.

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Introduction to Networking (Yarnfield)

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  1. Introduction to Networking (Yarnfield) Variable Length Subnet Masking (VLSM)

  2. Objectives • Define VLSM • Describe the difference between classful subnetting • Describe the advantages of VLSM • Be able to perform VLSM operations on give IP addresses

  3. Classful subnetting exercise • 172.80.0.0 255.255.248.0 • Find • The first five subnet addresses • First host, last host and broadcast of each subnet • Default gateway • How many subnets can be made? • How many hosts per subnet?

  4. VLSM defined • More than one subnet mask • Using classful subnetting wastes IP addresses • Why? 172.80.40.0 172.80.40.1 – 47.254 172.80.32.0 172.80.32.1 – 39.254 172.80.24.0 172.80.24.1 – 31.254 172.80.8.0 172.80.8.1 – 15.254 172.80.16.0 172.80.16.1 – 23.254

  5. We need • An IP address to perform VLSM on • The number of hosts involved in each part of the network

  6. We will... • Create a number of subnet masks that suit our needs more efficiently than a classful subnetting scheme could

  7. Example using a Class C network address 192.168.1.0 60 hosts 120 hosts 30 hosts

  8. Process • Find the segment with the largest number of hosts connected to it • Find an appropriate subnet mask for the largest segment • Write down the subnet addresses to fit the subnet mask • Take one of the newly created subnet addresses and apply a new subnet mask to it that is more appropriate • Write down the subnet addresses to fit the new subnet mask • Repeat from step 4 for smaller segments

  9. Example continued • Find the segment with the largest number of hosts connected to it • In the example the largest segment has 120 hosts connected so we must start with this segment • To accomodate120 hosts we need to use 7 bits from the host portion of the address (27 - 2 = 126)

  10. Example continued • Find an appropriate subnet mask for the largest segment • If we have borrowed 7 bits for our hosts the subnet mask (in binary) will be 11111111.1111111.1111111.1000000 • Convert this to decimal and we get 255.255.255.128

  11. Example continued • Write down the subnet addresses to fit the subnet mask • Now we need to find the subnet addresses that this subnet mask will create • 256 – 128 = 128 • Therefore the subnets would be 192.168.1.0 and 192.168.1.128 (remember we can now use subnet zero!) • We can now assign 192.168.1.0/25 to accommodate the 120 segment and have 192.168.1.128 to use for the other two segments

  12. 60 hosts (62 in total) 120 hosts (126 in total) 192.168.1.0/25 30 hosts (30 in total)

  13. Example continued • Take one of the newly created subnet addresses and apply a new subnet mask to it that is more appropriate • We still have two segments to deal with and we have a new subnet address to work with of 192.168.1.128 • We must start with the larger segment, which has 60 hosts • To accommodate 60 hosts we need to borrow 6 bits from the host portion of the given IP address • 26 – 2 = 62 hosts • This will give us a subnet mask of 1111111.1111111.1111111.11000000 which is the same as 255.255.255.192

  14. Example continued • Write down the subnet addresses to fit the new subnet mask • Now we need to find the subnet addresses that this subnet mask will create • 256 – 192 = 64 • Therefore the new subnet addresses would be 192.168.1.128 and 192.168.1.192 • We can now use 192.168.1.128/26 for the segment with 60 hosts

  15. 60 hosts (62 in total) 192.168.1.128/26 120 hosts (126 in total) 192.168.1.0/25 30 hosts (30 in total)

  16. Example continued • Take one of the newly created subnet addresses and apply a new subnet mask to it that is more appropriate • We still have the segment with 30 hosts to deal with • We work this out in the same way as before • To accommodate 30 hosts we need to borrow 5 bits from the host portion of the IP address • 25 – 2 = 30 hosts • This will give us a subnet mask of 1111111.1111111.1111111.11100000 which is 255.255.255.224

  17. Example continued • Write down the subnet addresses to fit the new subnet mask • Now we need to find the subnet addresses that this subnet mask will create • 256 – 224 = 32 • Therefore the new subnet addresses would be 192.168.1.192 and 192.168.1.224 • We can now use 192.168.1.192/27 for the segment with 30 hosts • We still have the new 192.168.1.224 subnet which could be used for future growth

  18. Result 192.168.1.0 60 hosts (62 in total) 192.168.1.128/26 120 hosts (126 in total) 192.168.1.0/25 30 hosts (30 in total) 192.168.1.192/27

  19. Summary • To determine the number of hosts a subnet can support use the formula 2n – 2 • Always start the process with the segment with the largest amount of hosts to accommodate • Classless subnetting deals with the hosts as opposed to classful subnetting which deals more with subnets

  20. Exercise • 192.168.2.0/24 • 7 remote sites, 30 hosts each • P to P links between routers Remote A 30 hosts Remote B 30 hosts Remote C 30 hosts Remote D 30 hosts Central Remote E 30 hosts Remote F 30 hosts Remote G 30 hosts

  21. Exercise • 192.168.3.0 30 hosts 6 hosts Backbone 126 hosts 6 hosts 30 hosts 6 hosts 30 hosts

  22. Questions... • ...are there any?

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