Subnetting Basics

Internetworking Basics

Chapter 9

Subnetting Basics

Answers to the questions

Activity A
Activity B
Activity C
Activity D

Activity A

Specify the subnet mask if each network below is divided into 4 subnets with 2 of the subnets useable.
1. 120.0.0.0
2. 140.64.0.0
Specify the subnet mask if each network below is divided into into 8 subnets with 6 of the subnets useable.
1. 120.0.0.0
2. 140.64.0.0

Answers

For 4 subnets we need to borrow 2 bits from the host part - 11000000 (decimal 192). So the subnet mask for each network is...
1. 120.0.0.0 is a Class A address so the standard subnet mask is 255.0.0.0. Borrowing bits gives a new subnet mask of 255.192.0.0
2. 140.64.0.0 is a Class B address so the standard subnet mask is 255.255.0.0. Borrowing bits gives a new subnet mask of 255.255.192.0
For 8 subnets we need to borrow 3 bits from the host part - 11100000 (decimal 224). So the subnet mask for each network is...
1. 120.0.0.0 is a Class A address so the standard subnet mask is 255.0.0.0. Borrowing bits gives a new subnet mask of 255.224.0.0
2. 140.64.0.0 is a Class B address so the standard subnet mask is 255.255.0.0. Borrowing bits gives a new subnet mask of 255.255.224.0

Activity B

For each subnet mask calculate the number of useable subnets possible
1. Subnet Mask: 255.255.255.224
2. Subnet Mask: 255.255.255.240
3. Subnet Mask: 255.255.248.0

Answers

The subnets are defined by the last octet 224, which is 11100000 in binary, showing that 3 bits have been borrowed. Since 2³ - 2 = 6 then we can have 6 useable subnets.
The subnets are defined by the last octet 240, which is 11110000 in binary, showing that 4 bits have been borrowed. Since 2⁴ - 2 = 14 then we can have 14 useable subnets.
The subnets are defined by the last octet 248, which is 11111000 in binary, showing that 5 bits have been borrowed. Since 2⁵ - 2 = 30 then we can have 30 useable subnets.

Activity C

Calculate the subnet address for each host given below
1. IP Address: 200.100.2.68
  Subnet Mask: 255.255.255.224
2. IP Address: 130.40.85.3
  Subnet Mask: 255.255.240.0
3. IP Address: 113.203.2.3
  Subnet Mask: 255.248.0.0

Answers

This is a Class C address so a Boolean AND should be carried out on the last octet of the IP address and subnet mask. 68 (01000100) AND 224 (11100000) = 64 (01000000)
Subnet Address is: 200.200.2.64
This is a Class B address so a Boolean AND should be carried out on the third octet of the IP address and subnet mask. 85 (01010101) AND 240 (11110000) = 80 (01010000)
Subnet Address is: 130.40.65.0
This is a Class A address so a Boolean AND should be carried out on the second octet of the IP address and subnet mask.
203 (11001011) AND 248 (11111000) = 200 (11001000)
Subnet Address is: 113.200.0.0

Activity D

A network needs to be divided into subnets.Two separate subnets are required to contain up to 50 hosts each. Two other subnets will contain up to 25 hosts each. To allow for expansion, another subnet may be needed containing 25 to 50 hosts.
1. You are required to choose a suitable IP Address Class for this network and specify how you will subnet the network. A private IP network address range can be used.
2. Now choose a suitable IP Address Class for this network and specify how you will subnet the network using a public IP network address. Hint: Class A addresses are the most expensive, Class C are the least.

Answers

The total number of hosts may reach 150. At least 5 useable subnets are needed. Three of those subnets must be able to accommodate up to 50 hosts each. Two of those subnets must be able to accommodate up to 25 hosts each.

A Class C private network address would allow up to 254 hosts. However, to create at least 5 subnets, 3 bits would need to be borrowed, creating 6 useable subnets. This is enough, but each subnet can only take up to 30 hosts, so this is no good.

A Class B private network address would allow up to 8,190 hosts in 6 subnets. This is sufficient for the job.
From above, we already know that a Class B network address, divided into 6 subnets is sufficient for this network. However, Class B public addresses are expensive. An alternative would be to purchase 2 Class C network addresses. One Class C network could be divided into 2 subnets, containing up to 62 hosts each. The second Class C network could be divided into 6 subnets, with up to 30 hosts each.


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