VLSM Design Drill #2

certskills
By certskills January 27, 2014 09:05

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.

Answers to Tshoot Drill – WAN Interface Down
#CCNA Tshoot Drill: OSPF Hello Timer Changed
certskills
By certskills January 27, 2014 09:05
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8 Comments

  1. Franklin January 27, 10:55

    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

    Reply to this comment
    • Franklin January 27, 10:58

      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

      Reply to this comment
  2. Punya Atma September 10, 06:02

    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

    Reply to this comment
  3. Punya Atma September 10, 07:08

    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

    Reply to this comment
  4. Punya Atma September 10, 07:11

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

    Reply to this comment
  5. Punya Atma September 10, 08:13

    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

    Reply to this comment
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