Subnetting

Question 1

Calculate subnet address for host:

Answer 1

10.10.9.9/23

/23 = 255.255.254.0

11111111.11111111.11111110.00000000

Block size = 2 in 3rd octet (2^1 = 2 — 1 “0” in 3rd octet)

Count by 2 until host address of 9 is passed (0, 2, 4, 6, 8, 10)

10.10.8.0 is last subnet before passing 9

Question 2

Calculate broadcast address:

Answer 2

192.168.1.199/28

/28 = 240 = block size of 16 in 4th octet (2^4 = 16)

Subnets = 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208

192 is last subnet before passing host of 199

207 is last address in 192 subnet = broadcast address

Question 3

You need to subnet your office. You need six subnets, each with 20 hosts. What Class C subnet mask should you use?

Answer 3

/27

255.255.255.224 / 11111111.11111111.11111111.11100000

Block size of 32 in 4th octet

Class C network = 24 network bits

Leaves 8 subnets

2^8 = 32 -2 = 30 host per subnet

Question 4

Calculate network ID

Answer 4

192.168.0.123/29

Block size = 3 in 4th octet

2^3 = 8 address per subet

0, 8, 16, 24 … 112, 120, 128

123 falls between 120 + 128

Lowest value in range = 120 = network ID

Question 5

Calculate number of subnets and hosts

Answer 5

172.16.0.0/19

Determine the network class + associated number of network bits

A = 8, B = 16, C = 24

11111111.11111111.11100000.00000000

Network = 16, Host = 13, Subnet = 3

2^3 = 8 subnets

2^13 = 8,192 - 2 = 8,190 hosts

Question 6

How many hosts provided by subnet mask 255.255.255.192

Answer 6

62

Subnet mask with 255 in first 3 octets found by:

(256 - 4th octet value) - 2

256 - 192 = 64

64 - 2 = 62

Question 7

Subnet block size

Answer 7

Size of subnet blocks

Use block size to count up from 0 to determine which subnet block IP address belongs to

2 ^ host bits (2 ^ 8)

Question 8

Subnet mask to use in WAN point-to-point links to reduce waste of IP address

Answer 8

/30

255. 255.255.252
256. 11111111.11111111.11111100

2^2 = 4 - 2 = 2 hosts per subnet