# Answer Part 2: VLSM Design Drill #1

More VLSM work for #CCENT and #CCNA today! Today’s post is simple enough: this post wraps up the answer to the VLSM design drill posted earlier. Dive right in! Here’s the list of earlier posts in this exercise:

The original problem statement

The general idea with the process begins with an analysis of the existing subnets, defining the range of addresses in each subnet. Then you take each mask for a new subnet, one at a time, and on paper, find all the possible subnets of that classful network when using that mask. Then you just have to compare the lists to find a slot for the new subnet – a new subnet that doesn’t overlap with the preexisting subnets.

## The Rest of the Answers (Before the Explanation)

But first… the answers. Then I’ll get on to the explanation:

1)     (Mask/26): No subnets meet the requirements!

## The Pre-Existing Subnets

All three questions require that you think about the pre-existing set of subnets. So, the original five subnet IDs, and the ending number in the range for each subnet, are listed in Table 1. I’m not going through the math on how to get these unless someone asks, but it’s a great exercise.

#### Table 1: Pre-Existing Subnet IDs and Address Ranges

 Mask Subnet ID Subnet Broadcast /29 192.168.1.32 192.168.1.39 /28 192.168.1.0 192.168.1.15 /29 192.168.1.128 192.168.1.135 /30 192.168.1.96 192.168.1.99 /27 192.168.1.192 192.168.1.223

## Problem 2: a New /27 Subnet

• What’s the classful network? (192.168.1.0)
• What’s the mask? (/27, or 255.255.255.224)
• What are the possible subnets of this classful network, with this mask, if you used only that mask?

The first two of these questions are a given, and the last one is actually an ICND1-level task (although a lot of people need practice before taking the ICND2 or CCNA exam.) In this case, the possible /27 subnets of class C network 192.168.1.0 are:

1. 192.168.1.0
2. 192.168.1.32
3. 192.168.1.64
4. 192.168.1.96
5. 192.168.1.128
6. And so on, adding 32

(Note: those of you using my books: take mask 255.255.255.224, find the magic number as 256 – mask 224 = 32, and the subnets are all multiples of the magic number.)

Next, you need to compare the range of addresses in each of these potential subnets to the list of ranges for the pre-existing subnets. At this point, you only have the subnet IDs listed, so go ahead and calculate the range of addresses in each subnet. It’s actually easy at this point: the broadcast address of one subnet is one less than the subnet ID of the next subnet. Here’s the list, stopping with the first five:

1. 192.168.1.0 – 192.168.1.31
2. 192.168.1.32 – 192.168.1.63
3. 192.168.1.64 – 192.168.1.95
4. 192.168.1.96 – 192.168.1.127
5. 192.168.1.128 – 192.168.1.159

Finally, when you compare this list to the original five subnets’ address ranges, back in Table 1, you’ll find that there’s overlap with the first two potential new subnets (192.168.1.0/27 and 192.168.1.32/27). The numerically lowest with no overlap is 192.168.1.64/27.

## Problem 3: a New /28 Subnet

• What’s the classful network? (192.168.1.0)
• What’s the mask? (/28, or 255.255.255.240)
• What are the possible subnets of this classful network, with this mask, if you used only that mask?

In this case, the first five possible /28 subnets of class C network 192.168.1.0 are:

1. 192.168.1.0
2. 192.168.1.16
3. 192.168.1.32
4. 192.168.1.48
5. 192.168.1.64

(Note: those of you using my books: take mask 255.255.255.240, find the magic number as 256 – mask 240 = 16, and the subnets are all multiples of the magic number. So the rest of the subnet IDs end in 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240.)

Next, you need to compare the range of addresses in each of these potential subnets to the list of ranges for the pre-existing subnets. At this point, you only have the subnet IDs listed, so go ahead and calculate the range of addresses in each subnet. It’s actually easy at this point: the broadcast address of one subnet is one less than the subnet ID of the next subnet. Here’s the list, stopping with the first five:

1. 192.168.1.0 – 192.168.1.15
2. 192.168.1.16 – 192.168.1.31
3. 192.168.1.32 – 192.168.1.47
4. 192.168.1.48 – 192.168.1.63
5. 192.168.1.64 – 192.168.1.79

Finally, when you compare this list to the original five subnets’ address ranges, back in Table 1, you’ll find that there’s overlap with only the first potential new subnet (192.168.1.0/28). The numerically lowest with no overlap is 192.168.1.16/28.

Write a comment

1. ruissalo March 19, 17:13

HI Wendell,

i think you made a typo here:

The Rest of the Answers (Before the Explanation)

(…)

should be /28 at the end?

and thanks for all this exercises btw 😉

2. CCENTSkills March 19, 17:32

Ruissalo,
You are right. Just fixed it. Thanks for the heads up! And you’re welcome for the exercises…
Wendell

• David Peel August 3, 02:45

Man iiiii!!!! Just figured this thing out I’m so happy that it’s clear to me now thank GOD!!!!! For HIS WISDOM!!

Ya that Chapter 20 is small but powerful!!!

3. Ccent soon? November 15, 17:11

Just wondering, how quick should we be able to answer questions like this?

• CCENTSkills November 27, 10:33

Hi,
I typically avoid putting a number of the time for this type of exercise, but if I had to, I’d say 3 minutes for each. If you break it down, you first need a list of possible new subnets (call that 30 seconds), and then to calculate the range of addresses for the existing subnets (30 seconds each)… call it 3 minutes.

## Write a comment

#### Subscribe

Subscribe to our mailing list and get interesting stuff and updates to your email inbox.

Thank you for subscribing.

Something went wrong.