VLSM Subnetting

xkaijinxxkaijinx Posts: 90Member ■■□□□□□□□□
Working through some VLSM overlapping problems and I am going about it totally wrong. I cannot understand how I am so off on what the actual answer is..

Problem 1:
10.1.34.9/22
Subnet ID = 10.1.32.0
Broadcast Address = 10.1.35.255

My answer:
Subnet ID = 10.1.34.0
Broadcast Address = 10.1.37.255


Problem 2:
10.1.17.1/21
Subnet ID = 10.1.16.0
Broadcast Address = 10.1.23.255

My answer:
Subnet ID = 10.1.17.0
Broadcast Address = 10.1.24.255

Comments

  • dsgmdsgm Posts: 228Member
    Im trying to understand why your answer is off
    Have you gotten your /22 /21 down properly a /22 is increment of 4, So your subnets would be 0,4,8,12...,28,32 so thats why the answer is Subnet ID 10.1.32.0 - 10.1.36.0 so the broadcast address would be 10.1.35.255 last valid address would be 10.1.35.254.


    I hope this helps
  • theodoxatheodoxa Posts: 1,340Member
    The address they're giving you is a random IP Address within that subnet. It is possible that some or [if they use a mask smaller than /8 which is unlikely] all of the octets may not be the same as those in the Subnet ID or Broadcast Address.
    R&S: CCENT CCNA CCNP CCIE [ ]
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  • shabeermshabeerm Posts: 29Member ■□□□□□□□□□
    Problem 1:

    /22 means subnet mask will be 255.255.252.0 .So block size will be 4 and changes will be happening in 3rd octet.

    10.1.0.0
    10.1.4.0
    10.1.8.0
    10.1.12.0
    10.1.16.0
    .
    .
    .
    .
    .
    10.1.28.0
    10.1.32.0
    10.1.36.0
    .
    .
    .
    Ip 10.1.34.9 will be coming in the network range 10.1.32.0 - 10.1.35.255

    Problem 2

    /21
    means subnet mask will be 255.255.248.0 and block size will be 8.changes will be happening in the 3 rd octet

    10.1.0.0
    10.1.8.0
    10.1.16.0
    10.1.24.0
    10.1.32.0
    .
    .
    .

    Here Ip 10.1.17.1 comes in the network range 10.1.16.0 - 10.1.23.255

    Hope you get an idea now :)
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