Creating a Logical Network Topology.

Given the IP Address 172.20.0.0 /24, create a logical network that incorporates the following:

1. Subnet D - 125 Hosts
2. Subnet C - 47 Hosts
3. Subnet B - 6 Hosts
4. Subnet A - 2 Hosts

Step1: Begin with SubnetD as it requires the most number for hosts.

We are given the Network Address in the question and the First host is always Network +1 (N+1).

To find the Subnet mask we need to look at how many bits need to be used to create a network that will have at least 125 hosts. To find this we use 2^n where n= the number of bits used in the subnet mask.

From the table you can see we need to borrow 7 bits from the host portion of the subnet mask to create a subnet that will use at least 125 hosts. In this case borrowing 7 bits will give us 128 addresses but we lose 2. One for the network and one for broadcast leaving us with 126 addresses for Hosts. 

If the Subnet Mask has 32 bits and we use 7 of them for the host portion, this means 25 bits remain as subnet bits in the subnet mask. This subnet mask can be represented by /25, where this means the first 25 bits off the subnet mask are all 1’s.

Broadcast address is your network address plus 128 addresses and the last use-able host in one less than that (BC-1). From here you can work out the following table:

Use this method for Subnet A,B and C, going from the subnet with the most number of hosts to least number of hosts until you come to the following:

Sub-netting a C Class IP address 101.

You have been given the network address 192.168.26.0/24 to subnet and provide the
IP addressing for the networks shown in the Topology Diagrams. You must determine the number of networks needed then design an appropriate addressing scheme. Place the correct address and mask in the Addressing Table. In this example, the number of hosts is not important. You are only required to determine the number of subnets per topology example.
Topology Diagram:





Task 1: Determine the Number of Subnets in the Topology Diagram.
Step 1: How many networks are there? 6

Step 2: How many bits should you borrow to create the required number of subnets?
How many usable host addresses and usable subnets did this give you?
What is the new subnet mask in decimal form?
Now for a class C network, we only need to look at the last Octet. Now on a /24 network, the network has 256 addresses with 254 useable hosts. But we want to limit the amount of addresses to create more subnets. To do this we “borrow” Host Bits (H) from the Subnet Mask. When we borrow host bits, they become Subnet Bits (SN).
Since there are six networks, we need to figure out how many bits to borrow. This is done with the formula 2^n where n = the number of bits borrowed.

As you can see, borrowing one or two host bits won’t give us the number of networks we need. But borrowing 3 Host bits for Subnet Bits will give us 8 networks, which will leave us 2 networks for possible expansion. After borrowing 3 H bits for SN bits, the last octet of the Subnet Mask 255.255.255.0 will now read 255.255.255.224 and will look like this:

To find out how many hosts per subnet we look at the last octet of the new Subnet Mask.
There are a few ways to work this out, see which is easiest for you:

1. Using the 2^n formula to find out how many addresses we are given by 5 Host bits. 2^5=32-2=30 hosts per subnet.
2. All Class C addresses (/24 or above) have 256 addresses so we simply subtract the decimal value of the last octet (224). 256 addresses – 224 addresses leaves us with 32 addresses, -2 for network and broadcast addresses leaving us with 30 Host addresses per network.
3. Or we know we have 8 sub-networks so 256 addresses \ 8 networks = 32 addresses -2 = 30 Hosts per subnet.

Step 3: How many subnets are available for future use?
Simple! We borrowed 3 bits which gave us 8 sub networks. We only need 6 so 8-6= 2 spare subnets are available for future use.

Step 4: First Useable Host Address.
This is easy enough. All we need to do here is take the network address and add one. (NA+1)
Network Address: 192.168.26.0
First Useable Address: 192.168.26.1 (NA+1)

Step 5: Last Useable Host address.
The Last Useable Host Address is the last address before the Broadcast Address (BC-1)
If we have 8 subnets with 32 addresses each starting with a network address of 192.168.26.0, the broadcast address for the first subnet will be the 32nd address and last address in that subnet. 192.168.26.31 (BC). The Last useable host address being BC-1 will be 192.168.26.31.

Task 2: Record Subnet Information.
Step 1: Fill in the following chart with the subnet information.

Task 3: What If?

What if you were given a host address and not the network address? Well you would have to find the network address yourself. Here is how you do it.

Step 1: Find the Network Address using the ANDing Method.
If you weren’t supplied with the network address but rather a host address (192.168.26.54), this would throw a spanner in the works. So it’s important you know how to use the ANDing process to determine the network address.
Firstly you will need to translate the IP address and Subnet mask down to their binary form.
IP: 192.168.26.54
11000000.10101000.00011010.00110110
SM: 255.255.255.224 (/27)
11111111.11111111.11111111.11100000
Then put the two values side by side. By using the ANDing process, you compare the two values. If a 1 in the IP column and the SM column line up it results in a 1 below. A 1 and 0 or 0 and 0 results in a zero.
       11000000101010000001101000110110
+     11111111111111111111111111100000
=     11000000101010000001101000100000
The 32 bit address resulting from the comparison on the IP and Subnet Mask Address is the network address for the host address 192.168.26.54. When you translate this address from its binary form to its decimal form you get 192.168.26.32.

Step 2: Find the Broadcast Address, number of subnets and number of hosts per subnet using the Subnet Mask.
Reverse Engineering the Subnet Mask will give you the number of subnets, number of hosts per subnet and from this we can calculate the broadcast address for the first subnet.
For this we look at the last octet of the Subnet Mask 255.255.255.224.
Next we need to know how many subnets there are. First we need to find out how many bits were borrowed to create the subnets. So convert the last octet of the subnet mask from decimal to binary.



In a normal network you would expect to see the last octet read all 0’s (8 host bits). But in a subnet you start to see some Subnet bits creep in which result in some 1’s creeping in. Anytime the last octet of the Subnet Mask reads higher than .0, bits have been borrow for a subnet.

As you can see from the table, once 224 is translated into binary, 3 host bits (H) have been borrowed and become Subnet Bits (SN). Following the formula 2^n where n= the number of bits (SN) borrowed, you can find out the number for subnets created by borrowing n number of bits.
Since in this example n=3, the formula will read 2^3 which results in 8 Subnets are created by borrowing 3 bits for subnetting.

To find out the number of hosts per subnet we use the same formula but n now equals the number of Host bits. So the formula becomes 2^5 which results in 32 addresses per subnet. But don’t forget 2 addresses are used per subnet for the network address and broadcast address. So it’s as simple as 32 addresses – 2 for network and broadcast leaves us with 30 useable host addresses per subnet.

Finally we need to find the broadcast address for the subnet 192.168.26.32. Earlier we figured out there are 32 addresses per subnet. So the last address or 32nd address in the subnet is the broadcast address. So 192.168.26.32 + 32 addresses (includes 192.168.26.32) gives us 192.168.26.63.
This is the broadcast address for the subnet 192.168.26.32.