I have been looking all over at sub netting tutorials. From this I have managed to figure out how to subnet based on given number of hosts and subnet numbers and work out the ranges based on the subnet mask given and answer questions here http://www.subnettingquestions.com relating to it. The issue I am having is that this site asks me to find out "How many subnets and hosts per subnet can you get from the network 172.30.0.0/26?".

Whats the process for working this out?

Merci

✔ Best Answer

That's a good tutorial mikelinusI'm a big fan of learning the binary of subnetting when learning it. For me it helped make sense out of it all.

When I have need of a quick check and I have a computer and the internet handy I use the following online subnet calculator.

http://www.subnet-calculator.com/

Knowing your CIDR notation is a huge help in subnetting. When first learning how to subnet I made up a cheat sheet I've been using ever since when I don't have a computer handy. I tend to not draw the sheet out anymore as I pretty much have it memorized now after so many years of using it.

It consists of a single octet. I make eight vertical lines (spaced) and on top of each line going from right to left I show the 2^ exponent. Right to left it's 2^0, 2^1, 2^2 etc on to 2^7 Then, above that I write the result of the exponent. Again, right to left, 1 (2^0 = 1), 2 (2^1 = 2), 4 (2^2 = 4), 8 (2^3 = 8) etc, 16, 32, 64, 128

Then on the bottom, I sum the results of the exponents on top going this time from left to right, 128, 192 (128 + 64 = 192), 224 (192 + 32 = 224), 240 (224 + 16 = 240) etc, 248, 252, 254, 255

If you write it out on a sheet of paper like that, you can calculate your subnet mask from it easily.

So with regard to the question above:

"How many subnets and hosts per subnet can you get from the network 172.30.0.0/26?"You know at a glance that:

/26means you're subnetting in the 4th octet and the172.30.0.0tells you it's a Class B address. Your first 24 bits are the first 3 octets. Knowing that, I now know I have 2 bits remaining. If you use the cheat sheet, count 2 bits into the 4th octet (left to right always) and get a mask of (second place from left)192(128 + 64 = 192)You have 10 network bits and 2^10 = 1024 networks (Class B subnet, masked in the 4th octet = 8 bits for third octet plus 2 bits in 4th octet giving you the 10 "network" bits)

You have 6 host bits and 2^6 = 64 hosts per subnet (apply the n-2 rule and you get 62 actual usable addresses per subnet)

Double check my math on the online subnet calculator if you'd like but I'm pretty sure I've got it correct. NOTE: I did this without drawing up my cheat sheet. Like I said, I have it memorized I've done it so many times.

I hope this makes sense.

If you can't get the cheat sheet right, let me know I'll draw it up, take a pic and post the pic.

It matters not how straight the gate,

How charged with punishments the scroll,

I am the master of my fate;

I am the captain of my soul.***William Henley***

message edited by Curt R

Thanks Mike. Eventually I figured it out. Thanks a lot

For those of us who can't be bothered, or haven't the time just now... what is the answer? Off to play polo and then tea 'n cake with my Aunt (she's 90 this year...)

That's a good tutorial mikelinusI'm a big fan of learning the binary of subnetting when learning it. For me it helped make sense out of it all.

When I have need of a quick check and I have a computer and the internet handy I use the following online subnet calculator.

http://www.subnet-calculator.com/

Knowing your CIDR notation is a huge help in subnetting. When first learning how to subnet I made up a cheat sheet I've been using ever since when I don't have a computer handy. I tend to not draw the sheet out anymore as I pretty much have it memorized now after so many years of using it.

It consists of a single octet. I make eight vertical lines (spaced) and on top of each line going from right to left I show the 2^ exponent. Right to left it's 2^0, 2^1, 2^2 etc on to 2^7 Then, above that I write the result of the exponent. Again, right to left, 1 (2^0 = 1), 2 (2^1 = 2), 4 (2^2 = 4), 8 (2^3 = 8) etc, 16, 32, 64, 128

Then on the bottom, I sum the results of the exponents on top going this time from left to right, 128, 192 (128 + 64 = 192), 224 (192 + 32 = 224), 240 (224 + 16 = 240) etc, 248, 252, 254, 255

If you write it out on a sheet of paper like that, you can calculate your subnet mask from it easily.

So with regard to the question above:

"How many subnets and hosts per subnet can you get from the network 172.30.0.0/26?"You know at a glance that:

/26means you're subnetting in the 4th octet and the172.30.0.0tells you it's a Class B address. Your first 24 bits are the first 3 octets. Knowing that, I now know I have 2 bits remaining. If you use the cheat sheet, count 2 bits into the 4th octet (left to right always) and get a mask of (second place from left)192(128 + 64 = 192)You have 10 network bits and 2^10 = 1024 networks (Class B subnet, masked in the 4th octet = 8 bits for third octet plus 2 bits in 4th octet giving you the 10 "network" bits)

You have 6 host bits and 2^6 = 64 hosts per subnet (apply the n-2 rule and you get 62 actual usable addresses per subnet)

Double check my math on the online subnet calculator if you'd like but I'm pretty sure I've got it correct. NOTE: I did this without drawing up my cheat sheet. Like I said, I have it memorized I've done it so many times.

I hope this makes sense.

If you can't get the cheat sheet right, let me know I'll draw it up, take a pic and post the pic.

It matters not how straight the gate,

How charged with punishments the scroll,

I am the master of my fate;

I am the captain of my soul.***William Henley***

message edited by Curt R

Oh, the linked question! I used the online calculator (much faster than pencil and paper......lol)

Select Class C

Plug in the IP: 192.168.37.128

Pick the subnet mask from the pulldown menu: 255.255.255.224

and voila! The "broadcast address" field gives you: 192.168.37.159

It matters not how straight the gate,

How charged with punishments the scroll,

I am the master of my fate;

I am the captain of my soul.***William Henley***

I totally missed the linked question, apologies to Rider3. ::mike

Thanks very much for this response. Its perfect.

Ask Your Question

Weekly Poll

Do you think Monopoly should update its pieces?

Discuss in The Lounge

Poll History