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# A seminar on subnetting by sanjay

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This topic is based on subneting. I interoduce about how to devide network into subnetwork so that we can manage the wastege of IP address

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• See page 10 in the workbook and show students what happens with 255 octets in the mask when converted to binary for ANDing.
• ### A seminar on subnetting by sanjay

1. 1. A seminar on Subnetting The Institution of electronics and telecommunication engineers, Delhi centre. Presented by: Sanjay Stream: CS Guided by: Mr. Nibhesh kr. Das
3. 3.  To break the network down into pieces, each of which can be addressed separately.  Controls network traffic  Reduces broadcasts  Organization of IP address space
4. 4.  Default Class C address is divided into network and host portions as follows: N . N . N . H  To subnet we “borrow” bits from the host portion of the address (8 bits for Class C) N . N . N . x x x x x x x x  Borrowing n bits yields 2n – 2 subnets.  Leaving n bits yields 2n – 2 hosts.  For a class C, we can borrow from 2 to 6 bits.  Why not 1 bit? (How many usable subnets?)  Why not 7 bits? (How many usable hosts?)
5. 5.  Suppose we need 14 usable subnets, how many bits do we borrow?  Remember, borrowing n bits give us:  2n – 2 subnets  Try borrowing 3 bits (n = 3):  23 – 2 = 8 – 2 = 6 usable subnets (not enough)  Try borrowing 4 bits  24 – 2 = 16 – 2 = 14 usable subnets (enough)
6. 6.  Write it with the network octet in binary: 200.129.41.0000 0000 break here  Borrowing 4 bits yields 14 usable subnets  How many usable hosts per subnet?  Same formula as subnets (2n – 2)  4 host bits (n = 4)  24 – 2 = 16 – 2 = 14 usable hosts per subnet subnet bits host bits
7. 7.  Examples:  First usable 200.129.41.0001 ^ 0000 subnet address: 200.129.41.16  First usable host 200.129.41.0001 ^ 0001 on the first subnet: 200.129.41.17  Second usable host 200.129.41.0001 ^ 0010 on the first subnet: 200.129.41.18 . . .  Last usable host 200.129.41.0001 ^ 1110 on the first subnet: 200.129.41.30  Broadcast address 200.129.41.0001 ^ 1111 for the first subnet: 200.129.41.31
8. 8.  Examples:  Second usable 200.129.41.0010 ^ 0000 subnet address: 200.129.41.32  Third usable 200.129.41.0011 ^ 0000 subnet address: 200.129.41.48  Fourth usable 200.129.41.0100 ^ 0000 subnet address: 200.129.41.64 . . .  Last usable 200.129.41.1110 ^ 0000 subnet address: 200.129.41.224
9. 9.  The subnet mask (in binary) has:  all ones in the network and subnet portion of the address  all zeros in the host potion of the address  The subnet mask for the previous example is: 255.255.255. 240 255.255.255. 1111^ 0000 (128 + 64 + 32 + 16 =240)  ANDing this mask with any valid host address on the network will always yield the subnet address for that host.
10. 10.  Example (our subnet mask is 255.255.255.240) IP host address: 200.129. 41.23 Last octet to binary: 200.129. 41.0001 0111 AND subnet mask: 255.255.255.1111 0000 200.129. 41.0001 0000 Subnet Address: 200.129. 41.16 So the host address 200.129. 41.23 is on the 200.129.41.16 subnet.
11. 11.  Default Class B address is divided into network and host portions as follows: N . N . H . H  To subnet we “borrow” bits from the host portion of the address (16 bits for Class B) N . N . x x x x x x x x . x x x x x x x x  For a class B, we can borrow from 2 to 14 bits.
12. 12.  Suppose we need 80 usable subnets, how many bits do we borrow?  Remember, borrowing n bits give us:  2n – 2 subnets  Try borrowing 6 bits (n = 6):  26 – 2 = 64 – 2 = 62 usable subnets (not enough)  Try borrowing 7 bits  27 – 2 = 128 – 2 = 126 usable subnets (enough)
13. 13.  Write it with the network octets in binary: 132.178.0000000 0.00000000 break here  Borrowing 7 bits yields 126 usable subnets  How many usable hosts per subnet?  Same formula as subnets (2n – 2)  9 host bits (n = 9)  29 – 2 = 512 – 2 = 510 usable hosts per subnet subnet bits host bits
14. 14.  Examples:  First usable 132.178.0000001 ^ 0.00000000 subnet address: 132.178.2.0  First usable host 132.178.0000001 ^ 0.00000001 on the first subnet: 132.178.2.1  Second usable host 132.178.0000001 ^ 0.00000010 on the first subnet: 132.178.2.2 . . .  Last usable host 132.178.0000001 ^ 1.11111110 on the first subnet: 132.178.3.254  Broadcast address 132.178.0000001 ^ 1.11111111 for the first subnet: 132.178.3.255
15. 15.  Examples:  Second usable 132.178.0000010 ^ 0.00000000 subnet address: 132.178.4.0  Third usable 132.178.0000011 ^ 0.00000000 subnet address: 132.178.6.0 . . .  Ninety-first usable 132.178.1011011 ^ 0.00000000 subnet address: 132.178.182.0 . . .  Last usable 132.178.1111110 ^ 0.00000000 subnet address: 132.178.252.0
16. 16.  The subnet mask for this example is: 255.255.254.0 255.255.1111111 ^ 0.00000000  ANDing this mask with any valid host address on this network will always yield the subnet address.
17. 17.  Example: IP host address: 132.178.119.112 Last octets to binary: 132.178.0111011 ^ 1.01110000 AND subnet mask: 255.255.1111111 ^ 0.00000000 132.178.0111011 ^ 0.00000000 Subnet Address: 132.178.118.0
18. 18. Thank You….