Subnetting..again - Sorry :-)

controlcontrol Member Posts: 309
Didn't want to hijack someone elses thread so here goes....very simple I'm sure.

I thought I had this subnetting sorted, and then tried the subnettingquestions website and seem to be failing miserably.

A simple case e.g here the is the question and answer.

What is the broadcast address of the network 172.23.87.128/25?
Answer: 172.23.87.255

Could someone run through how they go this as I really don't know where I went wrong.

Thank You

Comments

  • FotroFotro Member Posts: 5 ■□□□□□□□□□
    Well /25 is 255.255.255.128

    so the networks go up in 128's

    so first network is 172.32.87.0
    and the broadcast of that is 172.32.87.127
    next network is 172.32.87.128
    and the broadcast is 172.32.87.255

    so i basically just kept going up in 128's till i couldn't any more :P

    check out the subnet sheet on this site **** Sheets - Packet Life
    Makes things alot easier.

    Sorry if i didn't explain it correctly i'm not the best either lol
  • miller811miller811 Member Posts: 897
    control wrote: »
    Didn't want to hijack someone elses thread so here goes....very simple I'm sure.

    I thought I had this subnetting sorted, and then tried the subnettingquestions website and seem to be failing miserably.

    A simple case e.g here the is the question and answer.

    What is the broadcast address of the network 172.23.87.128/25?
    Answer: 172.23.87.255

    Could someone run through how they go this as I really don't know where I went wrong.

    Thank You

    172.23.87.0 with a 255.255.255.0 0r /24 mask can be broken into two distinct networks by subnetting.

    by moving into a 255.255.255.128 mask you create two distinct networks

    Network 1st valid host last vaild broadcast address
    172.23.87.0 172.23.87.1 172.23.87.126 172.23.87.127
    172.23.87.128 172.23.87.129 172.23.87.254 172.23.87.255

    the same network could have been broken into 4 distinct networks with a /26
    172.23.87.0
    172.23.87.64
    172.23.87.128
    172.23.87.192

    or it could have been split with multiple masks

    Network 1st valid host last vaild broadcast address
    172.23.87.0 /24 172.23.87.0 172.23.87.254 172.23.87.255
    172.23.87.0 /25 172.23.87.1 172.23.87.126 172.23.87.127
    172.23.87.128 /26 172.23.87.129 172.23.87.190 172.23.87.191
    172.23.87.192 /27 172.23.87.193 172.23.87.222 172.23.87.223
    172.23.87.224 /28 172.23.87.225 172.23.87.238 172.23.87.239

    etc

    does that help?
    I don't claim to be an expert, but I sure would like to become one someday.

    Quest for 11K pages read in 2011
    Page Count total to date - 1283
  • controlcontrol Member Posts: 309
    miller811 wrote: »
    172.23.87.0 with a 255.255.255.0 0r /24 mask can be broken into two distinct networks by subnetting.

    by moving into a 255.255.255.128 mask you create two distinct networks

    QUOTE]


    Hi,

    Thanks for that. I have a query regarding your first statement above, you say the /24 mask can create 2 distinct networks, and also the /25 mask creates 2 distinct networks? it just confused me as you said both give 2 networks...
  • miller811miller811 Member Posts: 897
    control wrote: »
    miller811 wrote: »
    172.23.87.0 with a 255.255.255.0 0r /24 mask can be broken into two distinct networks by subnetting.

    by moving into a 255.255.255.128 mask you create two distinct networks

    QUOTE]


    Hi,

    Thanks for that. I have a query regarding your first statement above, you say the /24 mask can create 2 distinct networks, and also the /25 mask creates 2 distinct networks? it just confused me as you said both give 2 networks...

    sorry, any network can be broken down (subnetted) multiple ways.
    you have to go back to the base line in this case a /24 since it is a class C address to start

    so start with X.X.X.0
    then figure out your block size.
    256- interesting octet (.12icon_cool.gif = 128 block size

    /25 can break it into 2 networks

    x.x.x.0 - x.x.x.127
    x.x.x.128 - x.x.x.255
    I don't claim to be an expert, but I sure would like to become one someday.

    Quest for 11K pages read in 2011
    Page Count total to date - 1283
  • controlcontrol Member Posts: 309
    miller811 wrote: »
    control wrote: »

    sorry, any network can be broken down (subnetted) multiple ways.
    you have to go back to the base line in this case a /24 since it is a class C address to start

    Hi Miller,

    You are helping and I know how to get my original answer now so thanks.

    Just think I'm confusing myself a bit. You say the original address is Class C, but is it not a class B address?
  • miller811miller811 Member Posts: 897
    control wrote: »
    miller811 wrote: »

    Hi Miller,

    You are helping and I know how to get my original answer now so thanks.

    Just think I'm confusing myself a bit. You say the original address is Class C, but is it not a class B address?
    you are correct,
    I was actually posting on two different questions at the same time.
    sorry for the confusion
    I don't claim to be an expert, but I sure would like to become one someday.

    Quest for 11K pages read in 2011
    Page Count total to date - 1283
  • billyrbillyr Member Posts: 186
    Yes originally a class B, I think he was trying to keep it simple for you so as not to confuse.

    The easiest way to deal with these is simply to ignore the address class. All you are interested in is the actual slash notation i.e /25.

    /25 tells you that 25 bits have been reserved for use by the network, anything else thats left can be used to address your hosts.

    If you take a look at the placement value of the last bit that is turned on, i.e the 25th bit you will see it has a value of 128. That is your increment size.

    11111111.11111111.11111111.10000000
    1st subnet = 0
    2nd subnet = 128

    e.g a /27 subnet
    11111111.11111111.11111111.11100000
    27 bits reserved for the networks, placement value of the last bit is 32, that again is your increment size.

    1st subnet =0
    2nd subnet =32
    3rd subnet =64
    etc...
  • controlcontrol Member Posts: 309
    Thanks to all of you.

    Actually a lot more clearer and I have learned a quicker way to increment now too icon_smile.gif
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