1. Chapter 5 Addressing
Dr. Clincy
Lecture

2. Example 4 Solution The mask is 11111111 11111111 11111111 11000000 or
A company is granted the site address (class B). The company needs 1000 subnets. Design the subnets.
Solution
The number of 1s in the default mask is 16 (class B).
The company needs 1000 subnets. This number is not a power of 2. The next number that is a power of 2 is 1024 (210). We need 10 more 1s in the subnet mask.
The total number of 1s in the subnet mask is 26 ( ).
The total number of 0s is 6 ( ).
The mask is or
The number of subnets is 1024.
The number of addresses in each subnet is 26 (6 is the number of 0s) or 64.
Dr. Clincy
Lecture

3. Example 4
Subtract 63 from 255 to get 192
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Lecture

4. SUPERNETTING
Although class A and B addresses are dwindling – there are plenty of class C addresses
The problem with C addresses is, they only have 256 hostids – not enough for any midsize to large size organization – especially if you plan to give every computer, printer, scanner, etc. multiple IP addresses
Supernetting allows an organization the ability to combine several class C blocks in creating a larger range of addresses
Note: breaking up a network = subnetting
Note: combining Class-C networks = supernetting
Dr. Clincy
Lecture

5. Assigning or Choosing Class C Blocks
When assigning class C blocks, there are two approaches: (1) random and (2) superblock
Random Approach: the routers will see each block as a separate network and therefore, for each block there would be an entry in the routing table – a router contains an entry for each destination network
Superblock Approach: instead of multiple routing table entries, there would be a single entry. However, the choices of blocks need to follow a set of rules:
#1 – the # of blocks must be a power of 2 (ie. 1, 2, 4, 8 …)
#2 – blocks must be contiguous (no gaps between blocks)
#3 – the 3rd byte of the first address in the superblock must be evenly divisible by the number of blocks – ie. if the # of blocks is N, the 3rd byte must be divisible by N
Number of 1s removed from Default mask is dictated by the number of C blocks combined (ie 1 for 2, 2 for 4, 3 for 8, etc)
Dr. Clincy
Lecture

6. Example 5
A company needs 600 addresses. Which of the following set of class C blocks can be used to form a supernet for this company?
Solution
1: No, there are only three blocks. Must be a power of 2
2: No, the blocks are not contiguous.
3: No, 31 in the first block is not divisible by 4.
4: Yes, all three requirements are fulfilled. (1. Power of 2, 2. Contiguous and 3. 3rd byte of 1st address is divisible by 4: 32/4=8)
Dr. Clincy
Lecture

7. Example 8
A supernet has a first address of and a supernet mask of How many blocks are in this supernet and what is the range of addresses?
Solution
The default mask has 24 1s because is a class C.
Because the supernet mask is , the supernet has 21 1s.
Since the difference between the default and supernet masks is 3, there are 23 or 8 blocks in this supernet.
Because the blocks start with and must be contiguous, the blocks are , , ………
The first address is The last address is
The total number of addresses is 8 x 256 = 2048
Dr. Clincy
Lecture

8. Explain Supernetting Conceptually
Back out this bit from netid into host id
Causes these 2 blocks to combine as a single block
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Lecture

9. Variable-length subnetting
Suppose you were granted a Class C address – this mean you would have 8 bits to play with
Also, suppose you needed 5 subnets consisting of the following # of hosts: 60, 60, 60, 30 and 30
If you used a 2 bit subnet mask – can get 4 subnets with 64 stations each (too big)
If you used a 3 bit subnet mask – can get 8 subnets with 32 stations each (too small)
What’s the solution ?
Dr. Clincy
Lecture

10. Variable-length Subnetting
Solution: used 2 subnet masks – one applied after the other
Could use a 2 bit subnet mask and get 4 subnets with 64 stations each - this would satisfy the three 60-host subnet requirement – therefore the subnet mask would be (192)
We could then further divide one of the 64-host subnets into two 32-host subnets by applying this mask (224) after this mask of (192) is used
Dr. Clincy
Lecture

11. Ch 5 Classless Addressing
Dr. Clincy
Lecture

12. Classful Addressing is Obsolete
Guess What ?
Classful Addressing is Obsolete
However, understanding the classful approach will help you easily understand the classless approach
Quickly explain classless vs classful (leave address aggregation for the routing topics)
Dr. Clincy
Lecture

13. CLASSLESS ADDRESSING
Recall the problems with Classful addressing – you have to get a predefined block of addresses – in most cases, the block is either too large or too small
In the 1990’s, ISP came into prominence – they provide Internet access for individuals to midsize organizations that don’t want sponsor their own Internet service (ie. , etc).
The ISP’s are granted several B and C blocks of addresses and they subdivide their address space into groups of 2, 4, 8, 16, etc.. – blocks can be variable length
Because of the up rise of ISP’s, in 1996, the Internet Authorities announced a new architecture called Classless Addressing (making classful addressing obsolete)
Dr. Clincy
Lecture

14. Number of Addresses in a Classless Block
There are two conditions
Condition 1: the number of addresses in a block; it must be a power of 2 (2, 4, 8, . . .). A household may be given a block of 2 addresses. A small business may be given 16 addresses. A large organization may be given 1024 addresses.
Another Condition:
The beginning address must be evenly divisible by the number of addresses.
For example, if a block contains 4 addresses, the beginning address must be divisible by 4. If the block has less than 256 addresses, we need to check only the rightmost byte. If it has less than 65,536 addresses, we need to check only the two rightmost bytes, and so on.
Dr. Clincy
Lecture

15. Classless Subnet Illustration
Netid subnetid 1
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Lecture

16. Example 9
Which of the following can be the beginning address of a block that contains 16 addresses?
Solution
The address is eligible because 32 is divisible by 16. The address is eligible because 80 is divisible by 16.
Dr. Clincy
Lecture

17. Example 10
Which of the following can be the beginning address of a block that contains 1024 addresses?
Solution
To be divisible by 1024, the rightmost byte of an address should be 0 because any value in that first byte will be a fraction of 1024 (ie. 0 to 255).
To be divisible by 1024, the rightmost byte should be 0 and the second rightmost byte must be divisible by 4 because for every unique number in the second byte position, there exist 256 addresses in the first byte position that maps to it. To get 1024 addresses overall, you will need an increment of 4 in the 2nd byte position.
Therefore, the 2nd byte needs to be divisible by 4.
Only the address meets this condition.
Dr. Clincy
Lecture

18. Mask
Recall the Classful approach, only given an IP – the user defined their mask
For the Classless approach, when an org is given a block, it’s given both the starting address and the mask – these two pieces of info defines the entire block
For classless case, instead of writing out the full mask, we just specify the number of 1’s in the mask and append it to the address – this is called slash notation or CIDR (classless interdomain routing) notation
For classless addressing, the prefix refers to the common part of the address (ie. network portion)
For classless addressing, the suffix refers to the varying part of the address (ie. host portion)
Dr. Clincy
Lecture

19. A block in classes A, B, and C can easily be represented in slash notation as A.B.C.D/ n where n is either 8 (class A), 16 (class B), or (class C).
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Lecture

20. There are only 8 addresses in this block.
Example 11
A small organization is given a block with the beginning address and the prefix length /29 (in slash notation). What is the range of the block?
Solution
The beginning address is To find the last address we keep the first 29 bits and change the last 3 bits to 1s.
Beginning:
Ending :
There are only 8 addresses in this block.
Dr. Clincy
Lecture

21. Example 13
What is the network address if one of the addresses is /27?
Solution
The prefix length is 27, which means that we must keep the first 27 bits as is and change the remaining bits (5) to 0s. The 5 bits affect only the last byte. The last byte is Changing the last 5 bits to 0s, we get or 64. The network address is /27.
Dr. Clincy
Lecture

22. Example 14
An organization is granted the block /26. The organization needs to have four subnets. What are the subnet addresses and the range of addresses for each subnet?
Solution
The suffix length is 6. This means the total number of addresses in the block is 64 (26). If we create four subnets, each subnet will have 16 addresses.
Dr. Clincy
Lecture