Modules 11 – 13: IP Addressing

Question 1

What is the prefix length notation for the subnet mask 255.255.255.224?

1. /25
2. /26
3. /27
4. /28

Answer 1

3. /27

Hint

The binary format for 255.255.255.224 is 11111111.11111111.11111111.11100000. The prefix length is the number of consecutive 1s in the subnet mask. Therefore, the prefix length is /27.

Question 2

How many valid host addresses are available on an IPv4 subnet that is configured with a /26 mask?

1. 254
2. 190
3. 192
4. 62
5. 64

Answer 2

4. 62

Hint

63 addresses - 1 for network range address

= 62 valid host addresses

11111111.11111111.11111111.11000000

Question 3

Which subnet mask would be used if 5 host bits are available?

1. 255.255.255.0
2. 255.255.255.128
3. 255.255.255.224​
4. 255.255.255.240

Answer 3

3. 255.255.255.224​

Hint

11111111.11111111.11111111.11100000

Counting from the right = 5 host bits

Question 4

A network administrator subnets the 192.168.10.0/24 network into subnets with /26 masks. How many equal-sized subnets are created?

1. 1
2. 2
3. 4
4. 8
5. 16
6. 64

Answer 4

3. 4

Hint

11111111. 11111111.11111111.00000000 = /24 CIDR
11111112. 11111111.11111111.11000000 = /26 CIDR

Therefore we can create 4 equal size subnets of /26 with 8 host bits spare.

Question 5

Match the subnetwork to a host address that would be included within the subnetwork. (Not all options are used.)

192. 168.1.32/27 = ?
193. 168.1.64/27 = ?
194. 168.1.96/27 = ?

1. 192.168.1.48

2. 192.168.1.63

3. 192.168.1.128

4. 192.168.1.68

5. 192.168.1.121

Answer 5

192. 168.1.32/27 = 1. 192.168.1.48
193. 168.1.64/27 = 4. 192.168.1.68
194. 168.1.96/27 = 5. 192.168.1.121

Hint

/27 CIDR = 31 addresses

Therefore /27 subnets are divided at:

Network 0 | Valid Hosts 1-30 | Broadcast 31

Network 32 | Valid Hosts 33-62 | Broadcast 63

Network 64 | Valid Hosts 65-94 | Broadcast 95

Network 96 | Valid Hosts 97-126 | Broadcast 127

Question 6

An administrator wants to create four subnetworks from the network address 192.168.1.0/24. What is the network address and subnet mask of the second useable subnet?

1. subnetwork 192.168.1.64 subnet mask 255.255.255.192
2. subnetwork 192.168.1.32 subnet mask 255.255.255.240
3. subnetwork 192.168.1.64 subnet mask 255.255.255.240
4. subnetwork 192.168.1.128 subnet mask 255.255.255.192
5. subnetwork 192.168.1.8 subnet mask 255.255.255.224

Answer 6

1. subnetwork 192.168.1.64 subnet mask 255.255.255.192

Hint

/24 CIDR = 4x /26 with 64 addresses per subnet

Subnet mask = 11111111.11111111.11111111.11000000

Subnet mask = 128 + 64 = 192

Subnet mask = 255.255.255.192

Therefore /26 subnets are divided at:

Network 0 | Valid Hosts 1-62 | Broadcast 63 = 64 addresses

Network 64 | Valid Hosts 65-127 | Broadcast 128 = 64 addresses

Question 7

How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?

1. two
2. three
3. four
4. five

Answer 7

2. three

Hint

Each network that is directly connected to an interface on a router requires its own subnet. The formula 2n, where n is the number of bits borrowed, is used to calculate the available number of subnets when borrowing a specific number of bits.

Question 8

How many host addresses are available on the 192.168.10.128/26 network?

1. 30
2. 32
3. 60
4. 62
5. 64

Answer 8

4. 62

Hint

A /26 prefix gives 6 host bits, which provides a total of 64 addresses, because /26 = 64.

Subtracting the network and broadcast addresses leaves 62 usable host addresses.

Question 9

How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?

1. 510
2. 512
3. 1022
4. 1024
5. 2046
6. 2048

Answer 9

3. 1022

Hint

A mask of 255.255.252.0 is equal to a prefix of /22.

A /22 prefix provides 22 bits for the network portion and leaves 10 bits for the host portion.

The 10 bits in the host portion will provide 1022 usable IP addresses.

11111111.11111111.11111100.00000000

If we count from the right = 1024 - 2 for network and broadcast addresses.