For this assignment you will explore subnet addresses, host addresses and direct broadcast addresses. You will be provided an IP addresses and then asked to answer questions related to the IP.

In this assignment, first we explain how to find the subnet address for each subnet, the range of host addresses, and the direct broadcast address. Then you will be asked to do a similar exercise by yourself. Assume that you have been assigned the 192.168.1.0/24 network. You need to create 8 subnets. 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. What is the maximum number of subnets that can be created with this number of bits? We can create 2^3 = 8 subnets. How many bits can be used to create the host space? Each byte has 8 bitswe are using 3 bits to define the subnets, and this leaves us with 5 bits for the host space. 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 reservedthe first one (all zeros) for the subnet address and last address (all ones) for the broadcast address. 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 1sin 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)

0 192.168.1.0 192.168.1.1192.168.1.30 192.168.1.31 1 192.168.1.32 192.168.1.33192.168.1.62 192.168.1.63 2 192.168.1.64 192.168.1.65 192.168.1.94 192.168.1.95 3 192.168.1.96 192.168.1.97192.168.1.126 192.168.1.127 4 192.168.1.128 192.168.1.129 192.168.1.158 192.168.1.159 5 192.168.1.160 192.168.1.161192.168.1.190 192.168.1.191 6 192.168.1.192 192.168.1.193192.168.1.222 192.168.1.223 7 192.168.1.224 192.168.1.225192.168.1.254 192.168.1.255

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.

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.

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 subnets 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 subnets IP address will have the third placeholder equal to 1, giving us the following address: 192.168.1.32. To find the third subnets 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 addresss value is one less than the next subnets 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 205.235.48.0/28 How many bits are borrowed to create the subnet field? ________________ What is the maximum number of subnets that can be created with this number of bits? ________________ How many bits can be used to create the host space? ________________ What is the maximum number of host addresses available per subnet? ________________ What is the subnet mask, in binary and decimal format? ________________ 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 Note this table may need to be expanded to show all IP Subnets Answer the following: What subnet number is the IP address of 205.235.48.129 on? A junior network administrator is trying to assign 192.168.111.96 as a static IP address for a computer on the network but is getting an error message. Why? Can 205.235.48.254 be assigned as a Host IP address? Why?