VLSM moving onto a larger subnet

Good Evening,

I have a question about VLSM and I am really confused on what it is I need to do for this

Subnet 192.168.0.0/26
Mask 255.255.255.192
Range 192.168.0.1 to 192.168.0.63
Broadcast 192.168.0.64

I would like to now create a new subnet on top of this as follows so that it can have 253 IP addresses

Subnet 192.168.0.65/?
Mask ???
Range 192.168.0.66 to 192.168.1.?
Broadcast 192.168.1.?

What do I need to do in my subnet block tables to work this out?

Thanks you in advance for any feedback or support!

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DPG

Perhaps you mean supernet?
Subnet 192.168.0.0/26
Mask 255.255.255.192
Range 192.168.0.1 to 192.168.0.63
Broadcast 192.168.0.64

becomes

Subnet 192.168.0.0/24
Mask 255.255.255.0
Range 192.168.0.1 to 192.168.0.253
Broadcast 192.168.0.255

If you need a /24 in addition to the /26 your next subnet would need to be 192.168.1.0/24.

GDaines

redgunner wrote: »

Range 192.168.0.1 to 192.168.0.63
Broadcast 192.168.0.64

Same mistake as the other thread - range ends at 62, broadcast address is 63, 64 is the network address of the next subnet.

I've read yours and DPG's answers in two ways, so the answer depends on which interpretation is correct?

1. I found an answer here when I googled supernet which suggests the answer to your question is to use a /23 mask if you need an additional 254 addresses on top of those you already have.

Network: 192.168.0.0
Subnet mask: 255.255.254.0
Host range: 192.168.0.1 to 192.168.1.254
Broadcast: 192.168.1.255

To be honest you should be using class B addresses instead.

2. If you simply need a total of 254 addresses then use a /24 mask.

Network: 192.168.0.0
Subnet mask: 255.255.255.0
Host range: 192.168.0.1 to 192.168.0.254
Broadcast: 192.168.0.255

pinkiaiii

simple rule ive remember that /24 is always 254 addresses thus whenever vlsm subnet or suppernet question comes up asking for say network needs 310 addresses to fit i know my mid point is /24 thus if going down one its 512-2 hosts ,if need 1000 then /22 1024 ,so whenever get question where it might ask for say 200 hosts,30,400 i always go back to /24 since i know its 256-2

and for example numbers i gave you would go like this on class c

192.168.0.0 /23 -fits 400 range 192.168.0.1-192.168.1.254 br address 255
192.168.2.0 /24 -fits 200 range 192.168.2.1-192.168.2.254 br 255
192.168.3.0 /27 fits 30 range 192.168.3.1-192.168.3.30 br 31

thus to make it easier to subnet i always use /24 as my mid point to start either vlsm or subneting,suppernetting since its easiest number to remember.

volfkhat

redgunner wrote: »

I would like to now create a new subnet on top of this as follows so that it can have 253 IP addresses

What do I need to do in my subnet block tables to work this out?

Thanks you in advance for any feedback or support!

Easily Solved.
First, figure out how many "host bits" you need in order to fit 253 addresses.
Answer:
2 ^ 8(bits) = 256.

So there you have it: 8 bits.
Now, subtract that from 32 total bits.... leaves you with 24; also known as /24.
Also known as: 255.255.255.0

So, the answer that you seek is:
Subnet 192.168.1.0
Mask 255.255.255.0
Range 192.168.1.1 to 192.168.1.254
Broadcast 192.168.1.255

THE END.

Now, if you find yourself arguing that "It's Not contiguous. The subnet needs to start with 192.168.0.65
blah blah blah."

Well, You are Wrong.

When you are dealing with VLSM subnets... you start with the LARGEST subnet First; then work your way down to the smaller ones.
This is NOT an official rule.. but it's best practice if you want to get contiguous addresses.

So, Actually, your first subnet should be:
Subnet 192.168.0.0
Mask 255.255.255.0
Range 192.168.0.1 to 192.168.0.254
Broadcast 192.168.0.255

and your second subnet would be:
Subnet 192.168.1.0
Mask 255.255.255.192
Range 192.168.1.1 to 192.168.1.62
Broadcast 192.168.1.63

And now you have two contiguous subnets.

hope that helps.