Question

Overview – Subnetting Class C AddressesIn this assignment, first we explain how to find the subnet address

for each subnet, the range of host addresses, and the direct broadcast address.

Assume that you have been assigned the 192.168.1.0/24 network. You need to create 8 subnets.

1. How many bits do you need to borrow from the host field for the subnet field? We need to borrow 3 bits: 1 bit would give us 2 subnets, 2 would give us 4, and 3 would give us 8.

2. What is the maximum number of subnets that can be created with this number of bits? We can create 2^3 = 8 subnets.

3. How many bits can be used to create the host space? Each byte has 8 bits—we are using 3 bits to define the subnets, and this leaves us with 5 bits for the host space.

4. What is the maximum number of host addresses available per subnet? 2^5 – 2 = 32 – 2 = 30. We have 5 bits for the host space, and each bit can assume a value of 1 or 0 (2⁵). However, two addresses on each subnet are reserved—the first one (all zeros) for the subnet address and last address (all ones) for the broadcast address.

5. What prefix would you use? What is the subnet mask, in binary and decimal format? Recall that the prefix indicates the number of bits used to identify the network. Given that this is a Class C address and we use 3 bytes plus 3 additional bits for subnets (8+8+8+3=27)8+8+8+3=27, our prefix will be /27.

Recall that the subnet mask is a continuous stream of 1s—in our case, 27 of them. Therefore, the subnet mask in binary is 11111111.11111111.11111111.11100000.

Subnet Number

Subnet Address (first address on the subnet)

Range of Host Addresses

Direct Broadcast Address (last address on the subnet)

0

192.168.1.0

192.168.1.1-192.168.1.30

192.168.1.31

1

192.168.1.32

192.168.1.33-192.168.1.62

192.168.1.63

2

192.168.1.64

192.168.1.65- 192.168.1.94

192.168.1.95

3

192.168.1.96

192.168.1.97-192.168.1.126

192.168.1.127

4

192.168.1.128

192.168.1.129- 192.168.1.158

192.168.1.159

5

192.168.1.160

192.168.1.161-192.168.1.190

192.168.1.191

6

192.168.1.192

192.168.1.193-192.168.1.222

192.168.1.223

7

192.168.1.224

192.168.1.225-192.168.1.254

192.168.1.255

We need to convert this binary number into a decimal to get the subnet mask. Hands-On Activity 5C might come in handy here. The subnet mask in decimal is 255.255.255.224.

6. What is the increment value?

The increment value is the amount by which the subnet address increases from one subnet to the next and is given by the placeholder value of the last 1 in the subnet mask. Because the last byte in the subnet mask has three 1s, the third number 1 represents 32 (see Hands-On Activity 5C in your book for more detail). So, the increment value is 32.

7. Complete the following table; define each of the subnets, the range of host addresses on the subnet, and the directed broadcast address on the subnet. Explanation of this table:

In part b, we indicated that there were eight subnets. The best way to fill out the table is to identify the subnet addresses for all subnets. The very first subnet’s IP address is when all the bits in the last byte are 0, giving us the following decimal value: 192.168.1.0. Recall from part f that the incremental value is 32, which means that the second subnet’s IP address will have the third placeholder equal to 1, giving us the following address: 192.168.1.32. To find the third subnet’s IP address, we need to multiply the increment value (32) by 2, resulting in 192.168.1.64. You would continue until the eighth subnet, in which all the first 3 bits in the last byte equal 1, giving us 192.168.1.224.

The direct broadcast address’s value is one less than the next subnet’s IP address. Also, this address will have all the host bits in the last byte equal to 1. For simplicity, I will only convert the last byte of several broadcast addresses to binary to illustrate this:

Broadcast Address

Last Byte Converted to Binary (network bits | host bits)

192.168.1.31

0 0 0 | 1 1 1 1 1

192.168.1.63

0 0 1 | 1 1 1 1 1

192.168.1.95

0 1 0 | 1 1 1 1 1

192.168.1.127

0 1 1 | 1 1 1 1 1

The addresses between the subnet address and the broadcast address can be assigned to any hosts on the network.

Deliverables

Assume that you have been assigned 202.145.22.0/29

1. How many bits are borrowed to create the subnet field? ________________
2. What is the maximum number of subnets that can be created with this number of bits? ________________
3. How many bits can be used to create the host space? ________________
4. What is the maximum number of host addresses available per subnet? ________________
5. What is the subnet mask, in binary and decimal format? ________________
6. Complete the following table and calculate the subnet that this address is on, and define all the other subnets (the range of host addresses on the subnet and the directed broadcast address on the subnet). Make sure you list all of the subnets. You may need to increase the size of the table shown below.

Subnet Number

Subnet Address

Range of Host Addresses

Direct Broadcast Address

0

1

2

3

4

5

6

7

…

Last subnet number

 

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