Basic Subnetting Algorithm

April 20, 2018 | Author: Hasib Hamidul Haque | Category: Ip Address, Communications Protocols, Computer Network, Computer Architecture, Computer Networking

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Basic Subnetting Algorithm...

Description

Computer Networks CSE 323

Handout # 1

Subnetting

IP Addressing and Subnetting How to identify a student in a university? UAP

UIU

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How to identify a PC in a network?

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Example – IP Address

Example – Network Address

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Example – Broadcast Address

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Example: �� . �. 20�.34.�4.12 �. 23�.34.2.1 �. 114.34.2.� �. 12�.14.�.�

Decimal-to-Binary and Binary-to-Decimal

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Example: �� 201, �11, 1��.

Example: �� 11011101, 11110111, 10001100.

Example: �� . �. 114.34.2.� �. 12�.14.�.� �. 20�.34.�4.12 �. 23�.34.2.1 Example: �� . 114.34.2.� �. 132.��.�.� �. 20�.34.�4.12

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Public and Private IP addresses

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Subnetting:

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Subnetting Example

3. Steps

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June 21, 2014

Computer Networks CSE 323

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Handout # 1

Page # 8

Subnetting

June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Example: �� (/�). �. 2��.2��.2��.0 �. 2��.0.0.0 �. 2��.2��.224.0 �. 2��.2��.240.0

A Simple Algorithm for Subnetting

�� N (Network Bits)

S (Subnetting Bits)

H (Host Bits)

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

A Step by Step Guide to Subnetting Address 172.16.0.0/16

=

default subnet 255.255.0.0

ie Class B address

16 network bits

IP Address Subnet Mask

Network Host 10101100.00010000.00000000.00000000 11111111.11111111.00000000.00000000 n

You may be asked to supply 500 subnets; using the formula Subnets = 2 , Where n = no. of bits to borrow

Step 1

Step 2 n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

From the table on the right; to obtain 500 subnets. We need to borrow 9 host bits as subnet bits. This will give us; 29 = 512 subnets (Remember you may end up with unused subnets)

n

2 1 2 4 8 16 32 64 128

To determine the number of usable hosts per subnet we subtract the borrowed bits (9) from the total number of host bits (16) which gives us 7 remaining host bits. Using the formulae Hosts = 2n – 2 = 27 – 2 . From the table 27 = 128. Thus usable hosts = 128 – 2 =126 (Remember the first and last addresses are for network and broadcast, hence the minus 2 in the formula)

256 512 1024 2048 4096 8192 16384 32768 65536

Step 3 Mark the borrowed bits as subnet bits IP Address Subnet Mask

Network Subnet Host 10101100.00010000.00000000.00000000 11111111.11111111.11111111.10000000 Remember the borrowed bits in the mask change from 0 (zero) to 1 (one)

Step 4 Define the subnetted network; Address 172.16.0.0/25

=

subnet mask 255.255.255.128 (custom Subnet Mask) The first 3 octets (24 bits) are network bits, hence each octet = 255. The final octet has only the MSB (Most Significant Bit) as a network bit, hence 128.

There are now 25 network bits. (16 original + 9 borrowed)

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Step 5: Calculate the subnet addresses and usable hosts for each subnet IP Address Mask

Network Subnet Host 10101100.00010000.00000000.00000000 11111111.11111111.11111111.10000000

We are only interested in this part of the address when calculating the subnets

00000000.00000000 For each subnet, the binary value of the subnet bits will increment by 1, so the first 5 subnets and last 2 are; *See appendix 00000000.0 00000000.1 00000001.0 00000001.1 00000010.0 …etc. 11111111.0 11111111.1

First usable network

Last usable network

(zero subnet, may not be used)

(one subnet, may not be used)

Thus first usable subnet (Subnet 1) = 10101100.00010000.00000000.10000000 = 172 . 16 . 0 . 128 the first usable host (Subnet 1) = 10101100.00010000.00000000.10000001 172 . 16 . 0 . 129 the last usable host (Subnet 1) = 10101100.00010000.00000000.11111110 172 . 16 . 0 . 254 the broadcast address (Subnet 1) = 10101100.00010000.00000000.11111111 172 . 16 . 0 . 255 Thus host range (Subnet 1) = 10101100.00010000.00000000.10000001 to 10101100.00010000.00000000.11111110 172 . 16 . 0 . 129 172 . 16 . 0 . 254 Thus 2nd usable subnet (Subnet 2) = 10101100.00010000.00000001.00000000 = 172 . 16 . 1 . 0 the first usable host (Subnet 2) = 10101100.00010000.00000001.00000001 172 . 16 . 1 . 1 the last usable host (Subnet 2) = 10101100.00010000.00000001.01111110 172 . 16 . 1 . 126 the broadcast address (Subnet 2) = 10101100.00010000.00000001.01111111 172 . 16 . 1 . 127 Thus host range (Subnet 2) = 10101100.00010000.00000001.00000001 to 10101100.00010000.00000001.01111110 172 . 16 . 1 . 1 172 . 16 . 1 . 126 the host range (Subnet 3) = 10101100.00010000.00000001.10000001 172 . 16 . 1 . 129

to 10101100.00010000.00000001.11111110 172 . 16 . 1 . 254

and the host range (Subnet 4) = 10101100.00010000.00000010.00000001 172 . 16 . 1 . 129

to 10101100.00010000.00000010.01111110 172 . 16 . 1 . 126 and so on………

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June 21, 2014

Computer Networks CSE 323

Handout # 1

Subnetting

Appendix IP Value Subnet Number Subnet 0( Not Used) Subnet 1 Address Subnet 1 Host 1 Subnet 1 Host 126 Subnet 1 Broadcast Subnet 2 Address Subnet 2 Host 1 Subnet 2 Host 126 Subnet 2 Broadcast Subnet 3 Address Subnet 3 Host 1 Subnet 3 Host 126 Subnet 3 Broadcast

128 256 0 0 0 0 0 0 0 0 0 0 0 0 0

64 128 0 0 0 0 0 0 0 0 0 0 0 0 0

32 64 0 0 0 0 0 0 0 0 0 0 0 0 0

16 32 0 0 0 0 0 0 0 0 0 0 0 0 0

8 16 0 0 0 0 0 0 0 0 0 0 0 0 0

4 8 0 0 0 0 0 0 0 0 0 0 0 0 0

2 4 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

Subnet 510 Address Subnet 510 Host 1 Subnet 510 Host 126 Subnet 510 Broadcast Subnet 511(Not Used)

1 128 2 1 0 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 etc. 1 0 1 0 1 0 1 0 1 1

64

32

16

8

4

2

1

0 0 0 1 1 0 0 1 1 0 0 1 1

0 0 0 1 1 0 0 1 1 0 0 1 1

0 0 0 1 1 0 0 1 1 0 0 1 1

0 0 0 1 1 0 0 1 1 0 0 1 1

0 0 0 1 1 0 0 1 1 0 0 1 1

0 0 0 1 1 0 0 1 1 0 0 1 1

0 0 1 0 1 0 1 0 1 0 1 0 1

0 0 1 1 0

0 0 1 1 0

0 0 1 1 0

0 0 1 1 0

0 0 1 1 0

0 0 1 1 0

0 1 0 1 0

Note: Only the final two octets are shown for clarity.

Example: �� .��.�.��. �� . �. �� . �. �� . �. �� 1. �. �� 12. Example: �� , �� .��.��.��. ��

Example: �� 1�0.�0.0.0, �� 2��.2��.2��.12�. �� 170.60.25.1 170.60.26.127 170.60.40.13� 170.60.212.212 170.60.255.1�0 170.60.252.16 170.60.100.63

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Page # 12

June 21, 2014

Basic Subnetting Algorithm

Short Description

Description

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