Chapter 5 Addressing

1. Chapter 5 Addressing Lecture

2. Example 4 A company is granted the site address 181.56.0.0 (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 (16 + 10). The total number of 0s is 6 (32 - 26). The mask is 11111111 11111111 1111111111000000 or 255.255.255.192. The number of subnets is 1024. The number of addresses in each subnet is 26 (6 is the number of 0s) or 64. Lecture

3. Example 4 Subtract 63 from 255 to get 192 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 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) 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? 198.47.32.0 198.47.33.0 198.47.34.0 198.47.32.0 198.47.42.0 198.47.52.0 198.47.62.0 198.47.31.0 198.47.32.0 198.47.33.0 198.47.34.0 198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0 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) Lecture

7. Example 8 A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0. How many blocks are in this supernet and what is the range of addresses? Solution • The default mask has 24 1s because 205.16.32.0 is a class C. • Because the supernet mask is 255.255.248.0, 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 205.16.32.0 and must be contiguous, the blocks are 205.16.32.0, 205.16.33.0, 205.16.34.0………. 205.16.39.0. • The first address is 205.16.32.0. The last address is 205.16.39.255. • The total number of addresses is 8 x 256 = 2048 Lecture

8. Explain Supernetting Conceptually Back out this bit from netid into host id Causes these 2 blocks to combine as a single block 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 ? 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 255.255.255.11000000 (192) • We could then further divide one of the 64-host subnets into two 32-host subnets by applying this mask 255.255.255.11100000 (224) after this mask of 255.255.255.11000000 (192) is used Lecture

11. Ch 5 Classless Addressing Lecture

12. 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) 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. email, 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) 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. Lecture

15. Classless Subnet Illustration Netid subnetid Lecture

16. Example 9 Which of the following can be the beginning address of a block that contains 16 addresses? 123.45.24.52 205.16.37.32190.16.42.4417.17.33.80 Solution The address 205.16.37.32 is eligible because 32 is divisible by 16. The address 17.17.33.80 is eligible because 80 is divisible by 16. Lecture

17. Example 10 Which of the following can be the beginning address of a block that contains 1024 addresses? 205.16.37.32190.16.42.017.17.32.0 123.45.24.52 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 17.17.32.0 meets this condition. 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) Lecture

19. A block in classes A, B, and C can easily be represented in slash notation as A.B.C.D/ nwhere n is either 8 (class A), 16 (class B), or 24 (class C). Lecture

20. Example 11 A small organization is given a block with the beginning address and the prefix length 205.16.37.24/29 (in slash notation). What is the range of the block? Solution The beginning address is 205.16.37.24. To find the last address we keep the first 29 bits and change the last 3 bits to 1s. Beginning:11001111 00010000 00100101 00011000 Ending : 11001111 00010000 00100101 00011111 There are only 8 addresses in this block. Lecture

21. Example 13 What is the network address if one of the addresses is 167.199.170.82/27? 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 01010010. Changing the last 5 bits to 0s, we get 01000000 or 64. The network address is 167.199.170.64/27. Lecture

22. Example 14 An organization is granted the block 130.34.12.64/26. 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. Lecture