VLSM Design Drill #2

 In 200-301 V1 Appendices, CCENT-OLD, IPv4 VLSM

The boss of our #CCNA is at it again today with another VLSM project. The scenario helps you exercise your brain in a VLSM world. As with all the VLSM design drills, you are given an existing internetwork with some pre-existing subnets, and you need to add some new subnet(s) to the design. What subnet would you pick so you don’t cause an overlap? Today’s post lists the question; next time, the answer. Details below the fold.

Your boss wants you to add a subnet to an existing design. The existing design already has these five subnets:

  1. 192.168.1.168/29
  2. 192.168.1.240/28
  3. 192.168.1.128/29
  4. 192.168.1.96/30
  5. 192.168.1.192/27

The boss has made these choices:

  • He will add 3 new subnets
  • All subnets will be part of class C network 192.168.1.0
  • He will use one mask only for all the new subnets
  • He has not decided which mask yet
  • He wants to use the numerically highest possible subnet numbers

While the boss works to make his final choice for a mask, he wants you to do the planning. He has narrowed down the choices to three masks, as listed here. Your job: for each mask, find the three subnet IDs that would meet all the criteria, without overlapping the existing subnets.

1) /29

2) /28

3) /27

Ask questions if you have them. Also, check out the ICND1 100-101 Cert Guide’s chapter 20 for some tips on finding these overlaps, or check the other VLSM drills here in the blog.

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Franklin

Existing Subnets

30/ 192.168.1.96 – 192.168.1.99
29/ 192.168.1.128 – 192.168.1.135
29/ 192.168.1.168 – 192.168.1.175
27/ 192.168.1.192 – 192.168.1.223
28/ 192.168.1.240 – 192.168.1.255

Answer

29/ 192.168.1.144 – 192.168.1.151
29/ 192.168.1.152 – 192.168.1.159
29/ 192.168.1.160 – 192.168.1.167

28/ 192.168.1.48 – 192.168.1.63
28/ 192.168.1.64 – 192.168.1.79
28/ 192.168.1.80 – 192.168.1.95

29/ 192.168.1.0 – 192.168.1.31
29/ 192.168.1.32 – 192.168.1.63
29/ 192.168.1.64 – 192.168.1.95

Franklin

correction on the last one (instead of /29 is /27)

27/ 192.168.1.0 – 192.168.1.31
27/ 192.168.1.32 – 192.168.1.63
27/ 192.168.1.64 – 192.168.1.95

[…] 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 […]

[…] 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 […]

Punya Atma

prefix /29 – the three subnet IDs meet all the criteria for this prefix, and their respective subnet broadcast addresses:-

192.168.1.184 – 192.168.1.191
192.168.1.224 – 192.168.1.231
192.168.1.232 – 192.168.1.239

Punya Atma

Prefix /28 – the three subnet IDs meet all the criteria, and, their respective subnet broadcast addresses:-

192.168.1.144 – 192.168.1.159
192.168.1.176 – 192.168.1.191
192.168.1.224 – 192.168.1 239

Punya Atma

The main criteria I focused much is the requirement of the numerically highest subnet IDs.

Punya Atma

Prefix: /27 – here only the first lowest three subnet IDs meet all the required criteria. Below the list of those subnet IDs, and, their respective subnet broadcast addresses:-

192.168.1.0 – 192.168.1.31
192.168.1.32 – 192.168.1.63
192.168.1.64 – 192.168.1.95

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