# Answer Part 1: VLSM Design Drill #2

#CCENT and #CCNA candidates need to be comfortable with VLSM. The previous post gave you a chance to exercise;  today’s post lets you check your answers to the first part of VLSM design drill 2. Of course, reading the original problem helps to understand the answer. Ask questions if you have them. Note that you can mentally use any process that finds the right answer, but the explanation shown here follows the logic spelled out in my ICND1 Official Cert Guide. Enjoy!

## The Process

The general idea with the process as outlined in the books is as follows:

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.

But first… for those of you who want a quick answer check, here is the answer to the first of the three parts of the drill. I’ll get to the other two parts next time. Then I’ll get on to the explanation:

• 192.168.1.232 /29
• 192.168.1.224 /29
• 192.168.1.184 /29

## The Explanation for the First Mask (/29)

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 like to list the subnets in the subnet order either low to high, or high to low, depending on the problem. Because this problem asks us to find the numerically-highest subnet IDs to add to the design, this explanation lists them high to low.

If you have questions on the math, ask, but I will leave that as an exercise. (That math is really an ICND1 topic, and mostly I keep that kind of discussion in the CCENT Skills blog.)

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

 Mask Subnet ID Subnet Broadcast /28 192.168.1.240 192.168.1.255 /27 192.168.1.192 192.168.1.223 /29 192.168.1.168 192.168.1.175 /29 192.168.1.128 192.168.1.135 /30 192.168.1.96 192.168.1.99

#### Step 2: List All Subnets of 192.168.1.0 with Mask /29

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, /29. That is, the boss may choose to add all three subnets, each using mask /29, per the problem statement. Table 2 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 8 in the last octet.

#### Table 2: Potential New Subnets of 192.168.1.0, /29 Mask

 Subnet ID Subnet Broadcast 192.168.1.248 192.168.1.255 192.168.1.240 192.168.1.247 192.168.1.232 192.168.1.239 192.168.1.224 192.168.1.231 192.168.1.216 192.168.1.223 192.168.1.208 192.168.1.215 192.168.1.200 192.168.1.207 192.168.1.192 192.168.1.199 192.168.1.184 192.168.1.191

#### 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 2, but do not overlap with the existing address ranges listed in Table 1:

• 192.168.1.232 /29
• 192.168.1.224 /29
• 192.168.1.184 /29
Write a comment

1. Franklin February 1, 16:11

so, the new mask does not need to be contiguous. good to know!

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