![In decimal notation
• In decimal notation subnet mask value 1 to 255
represent network address and value 0 [Zero]
represent host address.
IP address 192.168.1.10
Subnet mask 255.255.255.0
• Network address is 192.168.1 and host address
is 10.](https://image.slidesharecdn.com/subenttingsupernettingandvlsmpresentation-170809122514/85/Subentting-Supernetting-and-VLSM-presentation-5-320.jpg)
![In binary notation
• In binary notation subnet mask ON bit [ 1] represent
network address while OFF bit[0] represent host
address.
IP address
11000000.10101000.00000001.00001010
Subnet mask
11111111.11111111.11111111.00000000
• Network address is 11000000.10101000.00000001
and host address is 00001010](https://image.slidesharecdn.com/subenttingsupernettingandvlsmpresentation-170809122514/85/Subentting-Supernetting-and-VLSM-presentation-6-320.jpg)
Subentting, Supernetting and VLSM presentation
- 1. Welcome to presentation on Subnetting, Supernetting, VLSM
- 2. Classfull Addressing • IP addresses, when started a few decades ago, used the concept of classes. • This architecture is called classful addressing.
- 3. Type of Classes and ranges • Class A = 0 - 127 • Class B = 128 - 191 • Class C = 192 - 223 • Class D = 224 - 239 • Class E = 240 - ∞
- 4. Subnet Mask • Subnet mask is a 32 bits long address used to distinguish between network address and host address in IP address. • For example how will we figure out network partition and host partition from IP address 192.168.1.10 ? Here we need subnet mask to get details about network address and host address.
- 5. In decimal notation • In decimal notation subnet mask value 1 to 255 represent network address and value 0 [Zero] represent host address. IP address 192.168.1.10 Subnet mask 255.255.255.0 • Network address is 192.168.1 and host address is 10.
- 6. In binary notation • In binary notation subnet mask ON bit [ 1] represent network address while OFF bit[0] represent host address. IP address 11000000.10101000.00000001.00001010 Subnet mask 11111111.11111111.11111111.00000000 • Network address is 11000000.10101000.00000001 and host address is 00001010
- 7. Supernetting • Supernetting is the opposite of subnetting • In subnetting you borrow bits from the host part • Supernetting is done by borrowing bits from the network side. • And combine a group of networks into one large supernetwork.
- 8. Example • A company needs 600 addresses. Which of the following set of class C blocks can be used to form a supernet for this company? • 1. 198.47.32.0 198.47.33.0198.47.34.0 • 2. 198.47.32.0198.47.42.0 198.47.52.0 198.47.62.0 • 3. 198.47.31.0198.47.32.0 198.47.33.0 198.47.34.0 • 4. 198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0
- 9. Solution • 1: No, there are only three blocks. • 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.
- 10. CIDR • Classless Inter Domain Routing • CIDR is a way to allow more flexible allocation of Internet Protocol (IP) addresses than was possible with the original system of IP address classes.
- 12. Rules of CIDR blocks • All IP address are should be contiguous • Block size always 2n • First IP addrees in the block should be evenly divisable by size of the block
- 13. VLSM • VLSM: Variable Length Subnet Masking. • Any prefix length is allowed in the network at any point. • Q: You have 4 subnets with the following numbers of hosts: 58, 29, 14, 3 and the address is 192.168.10.0/24 • Determine what subnets you could use to fit these hosts into it and the address. 10.1.1.0/24 10.1.2.0/25 10.1.2.128/26 10.1.2.192/27 All of these are valid in the same network!
- 14. Working with VLSM • We have 58, 29, 14, 3 Hosts. Start with 58. • Smallest number larger than (58+2) is 64. 64 is 2 bits. • 24 bits of prefix length in the address space given, add 2 for 26. • First network is 192.168.10.0/26. • The next network is 192.168.10.0+64, so we start the next “round” at 192.168.10.64 8 7 6 5 4 3 2 1 1 2 4 8 16 32 64 128 32 < (58+2) < 64 So, 192.168.10.0/26 takes care of the first 58 hosts 24 + 2 = 26 Start the next block at 192.168.10.64
- 15. Working with VLSM • Now this block has 29 hosts. • Smallest number larger than (29+2) is 32. 32 is 3 bits. • 24 bits of prefix length in the address space given, add 3 for 27. • We start this block at 192.168.10.64, so network is 10.1.1.64/27 • The next network is 192.168.10.64+32, so we start the next “round” at 192.168.10.96 8 7 6 5 4 3 2 1 1 2 4 8 16 32 64 128 16 < (29+2) < 32 So, 10.1.1.64/27 takes care of the first 29 hosts 24 + 3 = 27 Start the next block at 10.1.1.96
- 16. Working with VLSM So, the subnets for - 58 host: 192.168.10.0/26 29 host: 192.168.10.64/27 14 host: 192.168.10.96/28 3 host: 192.168.10.112/29
- 17. Any Question...
- 18. Zakaria Hossain