Subnetting..again - Sorry :-)

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

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Fotro

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

miller811

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?

control

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...

miller811

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 (.12 = 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

control

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?

miller811

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

billyr

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...

control

Thanks to all of you.

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