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VLSM Problem Help

mag1gurumag1guru Member Posts: 4 ■□□□□□□□□□
in CCNA & CCENT
Given 172.16.0.0/23

Need: 2 Hosts Per Network (WAN LINKS)

We only need two host IP addresses for each WAN link. 2 to power of 2 – 2 = 2 (Need 2 bits for WAN hosts)
This leaves us with 7 bits for additional subnets 2 to power of 7 = 128 additional networks and our new mask is now a 30 bit mask 172.16.0.0/30
The last network bit place used is 4 so that will be our IP block size

128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
(Network Portion Cannot Change)
(Host Bits)
(Extra Bits Used For Additional Subnets)

Ranges would start off as:
172.16.0.0 172.16.0.1 – 172.16.0.2 172.16.0.3
172.16.0.4 172.16.0.5 – 172.16.0.6 172.16.0.7
172.16.0.8 172.16.0.9 – 172.16.0.10 172.16.0.11
172.16.0.12 172.16.0.13 – 172.16.0.14 172.16.0.15
Etc.
172.16.0.252/30 172.16.0.253 - 172.16.0.254 172.16.0.255 (Last Network)

My issue with this is that it appears we stop at 172.16.0.252 however according to the math we should have 128 Networks not 64?
Can someone help me with this? I know I am missing something?
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    fourmyskullfourmyskull Member Posts: 6 ■□□□□□□□□□
    mag1guru wrote: »
    Given 172.16.0.0/23

    Need: 2 Hosts Per Network (WAN LINKS)

    We only need two host IP addresses for each WAN link. 2 to power of 2 – 2 = 2 (Need 2 bits for WAN hosts)
    This leaves us with 7 bits for additional subnets 2 to power of 7 = 128 additional networks and our new mask is now a 30 bit mask 172.16.0.0/30
    The last network bit place used is 4 so that will be our IP block size

    128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
    (Network Portion Cannot Change)
    (Host Bits)
    (Extra Bits Used For Additional Subnets)

    Ranges would start off as:
    172.16.0.0 172.16.0.1 – 172.16.0.2 172.16.0.3
    172.16.0.4 172.16.0.5 – 172.16.0.6 172.16.0.7
    172.16.0.8 172.16.0.9 – 172.16.0.10 172.16.0.11
    172.16.0.12 172.16.0.13 – 172.16.0.14 172.16.0.15
    Etc.
    172.16.0.252/30 172.16.0.253 - 172.16.0.254 172.16.0.255 (Last Network)

    My issue with this is that it appears we stop at 172.16.0.252 however according to the math we should have 128 Networks not 64?
    Can someone help me with this? I know I am missing something?

    Sounds like you are working on a simular problem as I am, I have to set mine up using 20 on 192.168.0.0 network and I am haveing issues too. I will follow this posting to see if it will help me with my issue.
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    arsalankhan007arsalankhan007 Member Posts: 11 ■□□□□□□□□□
    /30 has a block size of 4 and the no. of valid host per network is 2 and no. of networks are 64 you can check by dividing 256 by 4
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    mag1gurumag1guru Member Posts: 4 ■□□□□□□□□□
    That would mean you are disregarding the last available bit in the 3rd octet in this VLSM scenario and are only using the bits in the last octet for networks which would give you 64 indeed.

    I understand that, but want to know why that bit is disregarded?
    You have 7 bits availble for aditional segments because you only need 2 bits for hosts and you started with a 23 bit mask.

    Had 23 + 7 taken from host portion for aditional networks + 2 that are left for hosts = 32Bit

    To get 64 networks you are only using 6 bits: 2 to the power of 6 is 64

    There is no doubt you cant go past the 172.16.0.252/30 172.16.0.253 - 172.16.0.254 172.16.0.255 (Last Network)

    I just want to know about that darn bit we are disregarding?:D
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    PhaintPhaint Member Posts: 4 ■□□□□□□□□□
    Hmm, since you are using a class B address 172.16.0.0/30, it should have 16384 Networks with 2 usable hosts on each network from my calculations. The next one after 172.16.0.252 should be 172.16.1.0 I believe... then again I've just learned subnetting a month ago so I could be wrong :)
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    mag1gurumag1guru Member Posts: 4 ■□□□□□□□□□
    "Well what you have missed here my friend is that originally given netmask is /23. You have missed that last bit in the third octect. So the last network would be,
    172.16.1.252/30 172.16.1.253 - 172.16.1.254

    So this gives you another 64 addresses, giving you the original 128 addresses"

    Man that was driving me nuts! I knew there where 7 bits avaiable for subnets but could not figure out how to form the networks past the 64th network.

    The correct answer was given to me by LooseEnd @ ******** IT Certification forums icon_cheers.gif
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    EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    mag1guru wrote: »
    The correct answer was given to me by LooseEnd @ ******** IT Certification forums icon_cheers.gif

    The correct answer was given by Phaint!
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
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    mag1gurumag1guru Member Posts: 4 ■□□□□□□□□□
    Sorry Phaint missed your post man!

    Thanks to everyone who contributed.:D

    I actualy had it written that way at one point but thought I was wrong.
    Anyway I got it now!
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