CCENT 5 S01 E20 – Adv TCP: IP Subnetting, Part 2

  CCENT, Networking, Subnets, TCP/IP

This article outlines the steps required to break a Class A, B or C network down into multiple Subnets as defined by the number of Hosts required for each subnet.

Several examples are provided below, along with practice exercises (actual answers not provided.)

Use these examples to work on your speed!  Goal: 2 – 3 Minutes per problem.

Steps Required to Break a Network into Subnets by the Number of Hosts

  1. Determine the Number of Hosts and Convert to Binary
  2. Reserve Bits in Subnet Mask and find the Increment
  3. Use Increment to find the Network Ranges

Network Scenario #1: 216.21.5.0 with 30 Hosts Each

1. Determine Bits required for Hosts

  • 30 = 0001xxxx = 5 Bits (32 Increment)

2. Reserve Bits in Subnet Mask and Find Increment

  • 11111111.11111111.1111111.11100000 = 255.255.255.224 (aka /27)
  • Increment = -.-.-.11100000 = 32
  • Max Networks = 8
  • Max Hosts = 2^5 – 2 = 32 – 2 = 30

3. Calculate the ranges

  • 216.21.5.0 – 216.21.5.31
  • 216.21.5.32 – 216.21.5.63
  • 216.21.5.64 – 216.21.5.95
  • 216.21.5.96 – 216.21.5.127
  • 216.21.5.128 – 216.21.5.159
  • 216.21.5.160 – 216.21.5.191
  • 216.21.5.192 – 216.21.5.223
  • 216.21.5.224 – 216.21.5.255

Network Scenario #2: 195.5.20.0 with 50 Hosts Each

1. Convert Hosts to Bits

  • 50 = 001xxxxx = 6 Bits
  • Increment = 2^6 = 64

2. Reserve Bits and find Increment

  • Save the Hosts -.-.-.11000000
  • Find the Increment -.-.-.11000000 = 255.255.255.192 = /26
  • Increment = 64
  • Max Networks = 2^2 = 4
  • Hosts / Subnet = 2^6 – 2 = 64 – 2 = 62

3. Calculate Ranges

  • 195.5.20.0 – 195.5.20.63
  • 195.5.20.64 – 195.5.20.127
  • 195.5.20.128 – 195.5.20.191
  • 195.5.20.192 – 195.5.20.255

Network Scenario #3: 150.5.0.0 with 500 Hosts Each

  1. Convert Hosts to Binary
  • 500 = 00000001.xxxxxxxx = 9 Bits
  • Increment = 2^9 = 512

2. Reserve Bits and determine Mask

  • -.-.11111110.00000000 = 255.255.254.0 = /23
  • Increment = -.-.11111110.00000000 = 2 in 3rd Octet
  • Max Networks = 2^7 = 128
  • Host / Subnet = 2^9 – 2 = 512 – 2 = 510 Hosts

3. Calculate Ranges

  • 150.5.0.0 – 150.5.1.255
  • 150.5.2.0 – 150.5.3.255
  • 150.5.4.0 – 150.5.5.255
  • 150.5.6.0 – 150.5.7.255

Network Scenario #4: 12.0.0.0 with 100 Hosts Each

  1. Convert Hosts to Binary
  • 100 = 01xxxxxx =  7 Bits
  • Increment = 128

2. Reserve Bits

  • -.11111111.11111111.10000000 = 255.255.255.128  /25
  • -.11111111.11111111.10000000 = 128 Increment
  • Max Networks = 2^17 = 131,072
  • Hosts / Subnet = 2^7 – 2 = 128 – 2 = 126

3. Calculate Ranges

  • 12.0.0.0 – 12.0.0.127
  • 12.0.0.128 – 12.0.0.255
  • 12.0.1.0 – 12.0.1.127
  • 12.0.1.128 – 12.0.1.255
  • 12.0.2.0 – 12.0.2.127

On My Own

200.1.1.0 with 40 Hosts / Network

Class C

40 = 001xxxxx = 6 Bits = 64 Increment

Networks = -.-.-.11000000 = 4 Networks (2 ^ 2bits)

  • 200.1.1.0 – .63
  • 200.1.1.64 – .127
  • 200.1.1.128 – .191
  • 200.1.1.192 – .255

199.9.10.0 with 12 Hosts / Network

Class C

12 = 00001xxx = 4 bits = 16 Increment

Mask = -.-.-.11111000 = 255.255.255.248  / 29

  • 199.9.10.0 – 199.9.10.15
  • 199.9.10.16 – 199.9.10.31
  • 199.9.10.32 – 199.9.10.47
  • 199.9.10.48 – 199.9.10.63
  • 199.9.10.64 – 199.9.10.79

170.50.0.0 with 1000 Hosts / Network

Class B

1000 = 0000001x.xxxxxxxx = 10 Bits = 1024 Increment = 1022 Hosts/Network

Mask -.-.11111100.00000000 = 255.255.252.0 = /14

  • 170.50.0.0 – 170.50.3.255
  • 170.50.4.0 – 170.50.7.255
  • 170.50.8.0 – 170.50.11.255
  • 170.50.12.0 – 170.50.15.255

12.0.0.0 with 100 Hosts / Network

Class A

100 = 01xxxxxx = 7 Bits = 128 Increment

Mask = 255.255.255.128

Networks = 2^17

  • 12.0.0.0 – .127
  • 12.0.0.128 – 255
  • 12.0.1.0 – 127
  • 12.0.1.128 – 255

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