Answer Part 1: VLSM Design Drill #1

Time for some VLSM work for #CCENT and #CCNA today! Today’s post begins the analysis for the VLSM design drill posted earlier. The original post posed a scenario, with an existing subnet design that uses VLSM. Your job: plan a new subnet with mask /26. Can you do it? Enjoy!

The general idea with the process as outlined in the books 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 Answers (Before the Explanation)

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:

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 (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

Problem 1: a New /26 Subnet

At this point, when thinking about adding a new subnet with mask /26, first treat it as a nice clean subnet design problem. What’s the classful network? 192.168.1.0. What mask does the question ask you to use? /26. What are the possible subnets of class C network 192.168.1.0, when using mask /26? Now you have a question that you should know how to answer if you’ve already mastered in the ICND1 half of CCNA.

In this case, the possible /26 subnets of class C network 192.168.1.0 will follow a patter of each being 64 more than the previous subnet ID in the 4^th octet. They are:

1. 192.168.1.0
2. 192.168.1.64
3. 192.168.1.128
4. 192.168.1.192

Then 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:

1. 192.168.1.0 – 192.168.1.63
2. 192.168.1.64 – 192.168.1.127
3. 192.168.1.128 – 192.168.1.191
4. 192.168.1.192 – 192.168.1.255

Finally, when you compare this list to the original five subnets’ address ranges, back in Table 1, you’ll find that there’s overlap for all four potential new /26 subnets. So in this fictitious scenario, you can politely say something like “boss, I know you were just testing me with that first one, because there aren’t any /26 subnets that work!?” That way he can save face. 😉

I’ll discuss the other two problems in this drill when I wrap up this exercise in an upcoming post.