Network Range confusion

Hey guys,
I am practicing subnetting and looking at two separate network ranges. Really confused on what IP addresses to assign in the middle of these ranges, if you could please enlighten me. Also if any tips to lessen my confusion?

Example 1: 150.5.0.0 - 150.5.1.255
Example 2: 150.5.2.0 - 150.5.3.255

Ok Class B address. /23 s/n mask.

For Example 1 Would this be...
150.5.1.1
150.5.1.2 .... etc

or
150.5.0.1
150.5.0.2

Same for Example 2....
150.5.2.0
150.5.2.1
150.5.2.2

Or
150.5.3.1
150.5.3.2

It seems to be for the Class B Address when you reach the last number in the third octet which is 255 than than the next number would be the broadcast address?

Could you please provide an example of how you would assign these, as well as what is the IP address right BEFORE the broadcast address for each example?

Sorry for the long post, just very confused.

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Comments

jdballinger

None of those are correct. When doing subnetting, remember that you always increment the last octet you borrowed from.

150.5.0.0 /23 means a mask of 255.255.254.0, right? The third octet was the last one you borrowed from. The mask is 254, which makes your 'magic number' 2 (256-254) so you would increment accordingly, 150.5.0.0, 150.5.2.0, 150.5.4.0, etc.

I'm including a link to the explanation my Cisco Academy instructor used when he taught us VLSM, and it really helped me to understand it. Hopefully it helps you as well.

http://dl.dropbox.com/u/3412229/DCampbell_Subnetting_Magic_Number.doc

xkaijinx

jd, thanks for the reply but I am not confused about the actual range. Going from the Subnet ID - Broadcast address

I am confused about the IP Addresses that go in between the subnet ID and the Broad address

For instance, one of the ranges are 150.5.2.0 - 150.5.3.255

What IP addresses go in between this range?

Jason R

xkaijinx wrote: »

jd, thanks for the reply but I am not confused about the actual range. Going from the Subnet ID - Broadcast address

I am confused about the IP Addresses that go in between the subnet ID and the Broad address

For instance, one of the ranges are 150.5.2.0 - 150.5.3.255

What IP addresses go in between this range?

Hello Xka,

The addresses that would go in the 150.5.2.0 subnet given that you are borrowing 23 bits are: 150.5.2.1 - 150.5.3.254 with 150.5.3.255 as its subnet broadcast address. Keep in mind that when you borrow 23 bits you have 510 available addresses.

MickQ

With any subnet, the very first IP address is the network address, and the last IP is the subnet's broadcast address.
With this case of the /23 mask the IPs 150.5.0.255 and 150.5.1.0 are actually useable host IPs.

For /23 subnets you have 512 IPs in total, first one for network, last for broadcast. This leaves 510 for the hosts.

Ltat42a

xkaijinx wrote: »

jd, thanks for the reply but I am not confused about the actual range. Going from the Subnet ID - Broadcast address

I am confused about the IP Addresses that go in between the subnet ID and the Broad address

For instance, one of the ranges are 150.5.2.0 - 150.5.3.255

What IP addresses go in between this range?

Add 1 to the network address, subtract 1 from the broadcast address. (150.5.2.1 - 150.5.3.254)

Node Man

This thread makes me think about the Cisco "Area 51" Arcade Game. That may be good practice regarding this thread. I like that game.

gbdavidx

jdballinger wrote: »

None of those are correct. When doing subnetting, remember that you always increment the last octet you borrowed from.

150.5.0.0 /23 means a mask of 255.255.254.0, right? The third octet was the last one you borrowed from. The mask is 254, which makes your 'magic number' 2 (256-254) so you would increment accordingly, 150.5.0.0, 150.5.2.0, 150.5.4.0, etc.

I'm including a link to the explanation my Cisco Academy instructor used when he taught us VLSM, and it really helped me to understand it. Hopefully it helps you as well.

http://dl.dropbox.com/u/3412229/DCampbell_Subnetting_Magic_Number.doc

forget 150.5.1.0?

MickQ

150.5.1.0 would be the next subnet if it were a /24 mask.

For those looking for the full details of why, check the (many) other related posts in this forum.

Edit: Don't forget that the value (x) after the slash (/x) indicates how far the '1's in binary digits go.

E.g. 11111111.11111111.11111111.00000000
1..........8............16............24............32

a /23 gives (32-23=9) us 2^9, which is 512.
Take the first IP as the network address, and the last one as the subnet broadcast address.
This leaves us with 512-2 = 510 host addresses.

Because it's a /23, we'll be incrementing in the equivalent of twice as many /24 subnets - that's where 150.5.1.0 is actually a host IP.

goldenlight

The Magic number Trick and the Ip Calculator helped me understand sub netting. Once it clicked it was smooth sailing throughout the course.