VLSM Subnetting Task

Computer ScienceComputer NetworkingMedium

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Question 4

You are employed as the network administrator of the above network, and you are asked to create subnets using VLSM for the IP address range 19.0.0.0/16 for the above network. Subnet accordingly and complete the following table:

| | Network ID | Subnet Mask | CIDR Netmask | Broadcast ID | Wildcard Mask |

|---|---|---|---|---|---|

| Network 1 | | | | | |

| Network 2 | | | | | |

| Network 3 | | | | | |

| Network 4 | | | | | |

[10 Marks]

This question includes visual content: A table for networking calculations. The columns are 'Network ID', 'Subnet Mask', 'CIDR Netmask', 'Broadcast ID', and 'Wildcard Mask'. The rows correspond to 'Network 1', 'Network 2', 'Network 3', and 'Network 4'.

Animated Video Solution

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Step by Step Written Solution

1
Step 1

Hi baxter, let's look at how to solve this networking problem using Variable Length Subnet Masking, or VLSM.

Subnetting with VLSM

Base Network: 19.0.0.0/16

2
Step 2

The goal of VLSM is to allocate IP addresses efficiently by starting with the largest host requirements first. Since the exact host counts are in the diagram above our crop, we will use a common scenario of four networks with decreasing sizes to demonstrate the methodology.

VLSM Principles

1. Rank networks by size (Largest to Smallest).

2. Find the smallest power of 2 that accommodates hosts + 2.

3. Assign the subnet mask and move to the next block.

3
Step 3

Let's assume our host requirements are thirty thousand, fifteen thousand, seven thousand, and three thousand. We start with our base block of nineteen point zero point zero point zero slash sixteen.

Step-by-Step Allocation

$$19.0.0.0/16 = 65,536 \text{ total addresses}$$
4
Step 4

For Network one with thirty thousand hosts, we need two to the power of fifteen, which is thirty-two thousand seven hundred and sixty-eight. This gives us a slash seventeen mask.

$$\text{Network 1: } 19.0.0.0/17$$
5
Step 5

The subnet mask is two fifty-five point two fifty-five point one twenty-eight point zero. The broadcast ID is the last address in this block, which is nineteen point zero point one twenty-seven point two fifty-five.

$$\text{Broadcast: } 19.0.127.255$$
6
Step 6

Next, Network two starts at nineteen point zero point one twenty-eight point zero. For fifteen thousand hosts, we use a slash eighteen mask, which covers sixteen thousand three hundred and eighty-four addresses.

$$\text{Network 2: } 19.0.128.0/18$$
7
Step 7

The mask is two fifty-five point two fifty-five point one ninety-two point zero, and the broadcast ID is nineteen point zero point one ninety-one point two fifty-five.

$$\text{Broadcast: } 19.0.191.255$$

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About This Question

Subject
Computer Science
Topic
Computer Networking
Difficulty
Medium

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