Classless-Addressing.pdf.. network lecture
- 2. Classless Addressing To overcome address depletion and give more organizations access to the Internet, classless addressing was designed and implemented. In this scheme. Address Blocks In classless addressing, when an entity, needs to be connected to the Internet, it is granted a block (range) of addresses. The size of the block (the number of addresses) varies based on the nature and size of the entity. An ISP, as the Internet service provider, may be given thousands or hundreds of thousands based on the number of customers it may serve.
- 3. Restriction To simplify the handling of addresses, the Internet authorities impose three restrictions on classless address blocks: 1. The addresses in a block must be contiguous, one after another. 2. The number of addresses in a block must be a power of 2 (1, 2,4,8,.etc) 3. The first address must be evenly divisible by the number of addresses. Example Figure shows a block of addresses, in both binary and dotted-decimal notation, granted to a small business that needs 16 addresses.
- 4. We can see that the restrictions are applied to this block: The addresses are contiguous. The number of addresses is a power of 2 (16 = 24), and the first address is divisible by 16.
- 5. The address and the /n notation completely define the whole block (the first address, the last address, and the number of addresses). First Address The first address in the block can be found by setting the 32 - n rightmost bits in the binary notation of the address to 0s. Example A block of addresses is granted to a small organization. We know that one of the addresses is 205.16.37.39/28. What is the first address in the block? Solution The binary representation of the given address is 11001101 00010000 00100101 00100 111 If we set 32−28 rightmost bits to 0, we get 11001101 00010000 00100101 00100000 or 205.16.37.32
- 6. Last Address The last address in the block can be found by setting the 32 - n rightmost bits in the binary notation of the address to 1s. The last address in the block can be found by setting the rightmost : 32 − n bits to 1s. Example Find the last address for the block. 205.16.37.39/28 Solution The binary representation of the given address is 11001101 00010000 00100101 00100111 If we set 32 − 28 rightmost bits to 1, we get 11001101 00010000 00100101 00101111 or 205.16.37.47
- 7. Number of Addresses The number of addresses in the block can be found by using the formula : 232−n Example Find the number of addresses in Example 6.6. Solution The value of n is 28, which means that number of addresses is 2 32−28 or 16. 205.16.37.32 205.16.37.47 → 205.16.37.39/28
- 8. Example Another way to find the first address, the last address, and the number of addresses is to represent the mask as a 32bit binary (or 8-digit hexadecimal) number. This is particularly useful when we are writing a program to find these pieces of information. Ex: The 205.16.37.39/28 , /28 can be represented as (Mask Definition) 11111111 11111111 11111111 11110000 (twenty-eight 1s and four 0s). Find a. The first address b. The last address c. The number of addresses.
- 9. Solution a. The first address can be found by ANDing the given addresses with the mask. ANDing here is done bit by bit. The result of ANDing 2 bits is 1 if both bits are 1s; the result is 0 otherwise. 205.16.37.32
- 10. b. The last address can be found by ORing the given addresses with the complement of the mask. ORing here is done bit by bit. The result of ORing 2 bits is 0 if both bits are 0s; the result is 1 otherwise. The complement of a number is found by changing each 1 to 0 and each 0 to 1. 205.16.37.47
- 11. c. The number of addresses can be found by complementing the mask, interpreting it as a decimal number, and adding 1 to it.
- 12. A block of addresses is granted to a small organization. We know that one of the addresses is 190.100.0.136/26. using Classless and Mask Definition methods What is the:- first address in the block: 190.100.0.128 Last address in the block: 190.100.0.191 Number of Addresses in the block: 64 Example: