MULTIDISCIPLINRY NATURE OF THE ENVIRONMENTAL STUDIES.pptx
A seminar on subnetting by sanjay
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. 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. 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. 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. 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. 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
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. 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. 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. 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. 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. 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
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. 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