Answer Part 2: VLSM Design Drill #2

This post wraps up the answers for VLSM design drill 2. Read the original problem if you didn’t start there, and look at answer, part 1, for the first part of the answer. This post gives the answers if the boss chooses to use mask /28 or /27. Note that you can mentally use any process that finds the right answer, and read for more background info in the ICND1 Official Cert Guide. Enjoy!

The Process

Just as a reminder, here’s a big picture look at the process shown here.

Analyze existing: Look at the existing addresses or subnets, and find the range of addresses in each subnet, including the subnet ID and subnet broadcast addresses.

Find potential new subnets: Take one of the mask for the new subnet(s), and find all possible subnets of that network

Find Subnets with no overlap: Compare the lists, starting at either the low or high end (depending on the problem statement). Find the first N subnets (again per the problem statement) that do not overlap with the list from the first step.

The Answers for the Second Mask (/28)

But first… for those of you who want a quick answer check, here are the literal answers:

For /28:

• 192.168.1.224 /28
• 192.168.1.176 /28
• 192.168.1.144 /28

And for /27:

• 192.168.1.64 /27
• 192.168.1.32 /27
• 192.168.1.0 /27

The Explanation for the Second Mask (/28)

The beginning of this post listed three steps, so this explanation mirrors those steps.

Step 1: List Existing Subnets

The first step lists the exiting subnets and their address ranges; I’ve listed them numerically highest to lowest, to match the problem statement.

Table 1: Pre-Existing Subnet IDs and Address Ranges

Step 2: List All Subnets of 192.168.1.0 with Mask /28

At this step, list all the subnets of the class C network, 192.168.1.0, with the presumed mask for this part of the problem, /28. That is, the boss may choose to add all three subnets, each using mask /28, per the problem statement. Table 3 lists some of those potential subnets, going from the highest subnet ID towards the numerically lowest subnet ID, just to make the comparisons to Table 1 easier. Note that all the subnet IDs are multiples of 16 in the last octet.

Table 3: Potential New Subnets of 192.168.1.0, /28 Mask

Step 3: Compare the Lists, and Choose Non-overlapped Subnets from Table 2

At this step, just use the old mark-1 eyeball and look at the lists. In this case, the following subnets are in the list of potential subnets in Table 3, but do not overlap with the existing address ranges listed in Table 1:

• 192.168.1.224 /28
• 192.168.1.176 /28
• 192.168.1.144 /28

The Explanation for the Third Mask (/27)

The beginning of this post listed three steps, so this explanation mirrors those steps.

Step 1: List Existing Subnets

The first step lists the exiting subnets and their address ranges; Just look back to Table 1 for the list.

Step 2: List All Subnets of 192.168.1.0 with Mask /27

At this step, list all the subnets of the class C network, 192.168.1.0, with the presumed mask for this part of the problem, /27. That is, the boss may choose to add all three subnets, each using mask /27, per the problem statement. Table 4 lists those potential subnets, going from the highest subnet ID towards the numerically lowest subnet ID, just to make the comparisons to Table 1 easier. Note that all the subnet IDs are multiples of 32 in the last octet.

Table 3: Potential New Subnets of 192.168.1.0, /28 Mask

Step 3: Compare the Lists, and Choose Non-overlapped Subnets from Table 2

At this step, just use the old mark-1 eyeball and look at the lists. In this case, the following subnets are in the list of potential subnets in Table 4, but do not overlap with the existing address ranges listed in Table 1:

• 192.168.1.64 /27
• 192.168.1.32 /27
• 192.168.1.0 /27