This Subnetting Project Warmup works through several common subnetting challenges and presents a step-by-step process for solving them, along with interim and final solutions. It is suggested that you work through these problems, and perhaps print out this Warmup for reference when completing the Subnetting Project lab.
IP addresses have a network portion followed by a host portion. Using classless inter-domain routing (CIDR) notation, the number of initial bits of the network portion of the address is the number following the slash. Another way to express this is with an IP address and subnet mask, which is the decimal equivalent of the leading binary ones (the mask) followed by binary zeros.
Example: The IP address 192.168.155.27/16 would have a subnet mask of 255.255.0.0
From the subnet mask, we can derive the network and broadcast addresses and the number of addressable hosts for a network.
Example: Given a random host with address 173.42.153.163/28, find:
The network address
The broadcast address
The number of addressable hosts
- From the /28, we know the subnet mask has 28 leading ones. The first 3 octets are /24, so the mask starts 255.255.255.x. The final octet has 4 leading ones: 11110000, equal to decimal 240. The subnet mask is then 255.255.255.240. The first three octets are passed unchanged to the network address. The fourth octet is converted to binary and masked with 11110000 = 240:
163 = 1010 0011
240 = 1111 0000
1010 0000 = 160 So, the network address is 173.42.153.160/28.
- To find the broadcast address, take the network portion (first 28 bits) and make the remaining bits all ones. The first 3 octets will be unchanged. The network portion of the final octet is 1010 (from part 1), so we append all ones, 1111 for 10101111
10101111 = 175, the broadcast address is 173.42.153.175
- Since the network portion of the address uses the first 28 bits, only the final 4 (32-28 ) bits are available for addressable hosts, 24=16
But, 2 of these are reserved for the network and broadcast addresses, so there are
24-2=16-2=14 possible host addresses (173.42.153.161-174)
To create subnets to an existing network, figure out how much of the host portion must be used and borrow the corresponding number of leading bits to create the network (subnet) addresses.
Ex: Create 6 subnets of the network 192.168.53.0/24
We need 3 bits to uniquely identify 6 subnets: 23=8 (Note: 22=4<> )
The subnet mask for the current network is 255.255.255.0 (because of the /24)
The new subnets’ network addresses are /27 (24 + 3), so the first 3 octets stay the same. The subnet mask’s final octet has binary representation 11100000 (leading 3 bits)
11100000 = 224, so the subnet mask for the new subnets is 255.255.255.224
The 6 subnets must be assigned addresses using combinations of the first 3 bits of the final octet. We use the first 6 available binary combinations: 000, 001, 010, 011, 100, 101
Now we fill out the octet with zeros to obtain the subnet network addresses: 00000000=0, 00100000=32, 01000000=64, 01100000=96, 10000000=128, 10100000=160
Converting to decimal, we get the final answer. Our subnets are:
192.168.53.0/27 192.168.53.32/27 192.168.53.64/27
192.168.53.96/27 192.168.53.128/27 192.168.53.160/27
I have to use the info above to answer these questions below. Could someone answer them
you are the network administrator for the new startup MMORPG developer, Rowdy Gamerz. The organization has obtained the block of public IPv4 network addresses represented by 224.151. .0/24, where the third octet is equal to the last two digits of your myUTSA ID (abc123). For example, if your myUTSA ID is xyz574, then the third octet for your assignment is 74, so you have the block of network addresses 224.151.74.0/24.
Write down your assigned network address: 224.151. .0/24
- What is your assigned network address?
- What is the subnet mask (dotted decimal form) for this network?
- What is the broadcast address for this network?
- How many addressable hosts are available for this network?
Following good network design practices, you decide to segment this network into 4 equal subnets, to be used or shared among the different Rowdy Gamerz departments: Software Development, Technical Support, Marketing/Sales, and Management & Human Resources.
- To create 4 subnets, you must borrow how many bits from the host portion of the network? (Hint: solve 4 = 2n)
- What is the subnet mask (dotted decimal form) for these new subnets?
The Correct Answer and Explanation is :
Here are the answers to the subnetting questions, along with a detailed explanation for each:
1. What is your assigned network address?
The block of network addresses you have is 224.151.[last two digits of your myUTSA ID].0/24. Since the last two digits of your myUTSA ID are abc123 (with 23 as the last two digits), the assigned network address is:
- 224.151.23.0/24
2. What is the subnet mask (dotted decimal form) for this network?
The subnet mask for a /24 network means that the first 24 bits of the IP address are reserved for the network portion. In dotted decimal form, the subnet mask for a /24 network is:
- 255.255.255.0
This is because the first three octets (24 bits) are set to 1 (for the network portion), and the last octet (8 bits) is set to 0 (for the host portion).
3. What is the broadcast address for this network?
The broadcast address is the last address in the network, where all the host bits are set to 1. Since the subnet mask is /24, the first three octets of the network address are fixed, and the last octet will have all bits set to 1. This gives us:
- 224.151.23.255
This is the broadcast address, as the host portion of the address is filled with ones.
4. How many addressable hosts are available for this network?
A /24 subnet has 8 bits for the host portion of the address (since there are 32 bits in an IPv4 address). The number of addressable hosts is determined by the formula 2^n – 2, where n is the number of bits available for hosts. Here, n = 8 (for the last octet), so:
- 2^8 – 2 = 256 – 2 = 254 addressable hosts
The subtraction of 2 accounts for the network address and the broadcast address, which cannot be assigned to hosts.
5. To create 4 subnets, you must borrow how many bits from the host portion of the network?
To create 4 subnets, we need to find how many bits we need to borrow from the host portion. The formula for determining the number of subnets is 2^n, where n is the number of bits borrowed. You need enough bits to create at least 4 subnets, so:
- 2^2 = 4 subnets (this is the minimum number of bits needed to create 4 subnets)
Thus, you must borrow 2 bits from the host portion of the network.
6. What is the subnet mask (dotted decimal form) for these new subnets?
After borrowing 2 bits, the new subnet mask will have 24 + 2 = 26 bits for the network portion. The new subnet mask in dotted decimal form is:
- 255.255.255.192
This is because the first 26 bits are set to 1 (network portion), and the last 6 bits are set to 0 (host portion). To convert this to dotted decimal, the binary representation is:
- 11111111.11111111.11111111.11000000 Which is equivalent to 255.255.255.192 in decimal.
Summary
- Assigned network address: 224.151.23.0/24
- Subnet mask for /24: 255.255.255.0
- Broadcast address: 224.151.23.255
- Number of addressable hosts: 254
- Bits to borrow for 4 subnets: 2
- Subnet mask for new subnets: 255.255.255.192
Explanation:
Subnetting helps efficiently manage IP addresses by dividing larger networks into smaller, more manageable subnets. By borrowing bits from the host portion, we create additional subnets, which is essential when dividing a single network for different departments (e.g., Software Development, Technical Support, etc.). By using the 2^n formula, where n is the number of bits borrowed, we determine how many subnets we can create and how to allocate the remaining bits for hosts. The subnet mask defines the size of the network and host portions of the address, and adjusting the subnet mask allows for greater flexibility in managing different network segments.