Some help with subnetting class A

Hi Everyone,

I've been trying out questions at subnettingquestions.com

and i came across this question that i cannot work out the correct answer.

What is the broadcast address of the network 10.104.128.0/20?

Answer: 10.104.143.255

A detailed explanation of how this is the answer would be very helpful, please bear with me as i am a complete newb and still need some hand-holding.

I just started studying for the exam that i have scheduled in 2 months any and all help is appreciated.

And please accept my apologies if i have posted in the wrong place.

Sincerely
bundleboy

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Comments

millworx

Pretty easy. So each you have a /20 mask. Which is using four bits in the 3rd octect as host bits. The network section is 20 bits long out of 32 total.

In binary a /20 mask would be represented as
11111111.11111111.11110000.00000000
1 = Network Section 20 bits long.
0 = Host Section.
In the third octet we need to calculate the host bits.

So the last four bits in the 3rd section are 8 + 4 + 2 + 1. That gives you 15.

Now 128 + 15 = 143. 143 = Last network in your range.

So you are using all networks from 10.104.128.0 - 10.104.143.255

The last address is your broadcast. 10.104.143.255. 255 cannot be assigned as a host since it is the last address in the range.

bundleboy wrote: »

Hi Everyone,

I've been trying out questions at subnettingquestions.com

and i came across this question that i cannot work out the correct answer.

What is the broadcast address of the network 10.104.128.0/20?

Answer: 10.104.143.255

A detailed explanation of how this is the answer would be very helpful, please bear with me as i am a complete newb and still need some hand-holding.

I just started studying for the exam that i have scheduled in 2 months any and all help is appreciated.

And please accept my apologies if i have posted in the wrong place.

Sincerely
bundleboy

gosh1976

I'll try and explain it as best I can. It's a slash /20 which puts us in the third octet so we don't worry about the first two octets.

/20 borrows 4 bits form the third octet so we know out increment or chunk size is 16:

increments    128  64 32 16   8   4  2   1 
bits           1   1   1  1    0  0  0   0

I don't even think about binary there is no need to...
so the network we are working with is 10.104.128.0 if you simply add 16 (the increment size) to the third octet you get the next network ID 10.104.144.0 but we want the broadcast of the same network and that's the last IP address in that subnet so just take one away from that third octet and the fourth octet is the highest possible number 255. so you have 10.104.143.255..

That is actually how I think about it in my head but I guess it might be easier just to add 15 to the third octet in 10.104.128.0 and then use the highest possible number in the 4th octet giving us 10.104.143.255

I don't think my explanation is very good but that is more or less the thought process I use and I'm able to solve a subnetting question like that in seconds.:)

bundleboy

millworx wrote: »

Pretty easy. So each you have a /20 mask. Which is using four bits in the 3rd octect as host bits. The network section is 20 bits long out of 32 total.

In binary a /20 mask would be represented as
11111111.11111111.11110000.00000000
1 = Network Section 20 bits long.
0 = Host Section.
In the third octet we need to calculate the host bits.

So the last four bits in the 3rd section are 8 + 4 + 2 + 1. That gives you 15.

Now 128 + 15 = 143. 143 = Last network in your range.

So you are using all networks from 10.104.128.0 - 10.104.143.255

The last address is your broadcast. 10.104.143.255. 255 cannot be assigned as a host since it is the last address in the range.

Awesome that was such quick response!

Wow I've never understood it better! and i can see how that works but...
My understanding was that since it is a class A address the mask by right should be

N H H H right? shouldn't we be taking the 2nd octet rather than the 3rd octet? I know i am understanding something wrongly, can you help me point out where im going wrong?

another question.. would the method be the same if it were a class B or class C address?
Please pardon my ignorance.

miller811

bundleboy wrote: »

Awesome that was such quick response!

Wow I've never understood it better! another question.. would the method be the same if it were a class B or class C address? and why do we have to choose the 3rd octet? Pardon my ignorance.

you just need to determine the interesting octet...
the one that is not 255....

millworx

bundleboy wrote: »

Awesome that was such quick response!

Wow I've never understood it better! another question.. would the method be the same if it were a class B or class C address? and why do we have to choose the 3rd octet? Pardon my ignorance.

Yes it works the same with class A B and C

Doesnt matter if its a /18 /19 /20 etc

That notation (CIDR) just tells you how many network bits are set. You can ignore any octects that are set to all 1 in your calculations

so for example if you had a /8
It would be represented as:
11111111.00000000.00000000.00000000
So you would ignore that first octect.

/16 would be
11111111.11111111.00000000.00000000

So you can ignore the first 2 octects since the network bits are all 1's

You are correct that a class A address is N H H H buts thats ONLY if your using subnet masks that are classful such as 255.0.0.0 in this case you are not! a /20 address mask would be 255.255.240.0, so you actually using the first two octects as N N and part of the 3rd octect as N.

Hope that helps

bundleboy

thanks alot guys understood it alot more clearer now!! I'll be forever grateful!

gosh1976

bundleboy wrote: »

Awesome that was such quick response!

Wow I've never understood it better! another question.. would the method be the same if it were a class B or class C address? and why do we have to choose the 3rd octet? Pardon my ignorance.

You aren't choosing the third octect a /20 subnet mask means you are borrowing 20 bits there are 8 bits to an octet so you would be in the third octet. 11111111.11111111.11110000.00000000

the method would be the same to figure out the boradcast of 129.104.128.0 /20 you would still be borrowing 4 bits from the third octet.

If you are unsure of why you would be working in the third octet with a /20 subnet mask then you may want to take a look at some of the videos floating around: professor messer has a binary math and 2 subnetting videos Professor Messer's Free CompTIA Network+ Certification Training Course | Professor Messer - CompTIA A+, CompTIA Network+, Certification Training

I've never watched these but I've seen many people recommend them so I started to Learn To Subnet: A Free, Lecture-Based Presentation on IP Addressing and Subnetting

then once you have the concepts down and you can do ANDing and all that I really like this guys method which is basically what I use though I changed it a bit and If I write down a chart it's a little different: https://learningnetwork.cisco.com/docs/DOC-3849

also you can use the binary game and 3 subnetting games at cisco
Cisco Binary Game
https://learningnetwork.cisco.com/docs/DOC-7561
https://learningnetwork.cisco.com/docs/DOC-1802
https://learningnetwork.cisco.com/docs/DOC-7558

I haven't played those games in a while i think the last one might be pretty hard. You may have to sign up for a free cisco account to play as well.

bundleboy

gosh1976 wrote: »

You aren't choosing the third octect a /20 subnet mask means you are borrowing 20 bits there are 8 bits to an octet so you would be in the third octet. 11111111.11111111.11110000.00000000

the method would be the same to figure out the boradcast of 129.104.128.0 /20 you would still be borrowing 4 bits from the third octet.

If you are unsure of why you would be working in the third octet with a /20 subnet mask then you may want to take a look at some of the videos floating around: professor messer has a binary math and 2 subnetting videos Professor Messer's Free CompTIA Network+ Certification Training Course | Professor Messer - CompTIA A+, CompTIA Network+, Certification Training

I've never watched these but I've seen many people recommend them so I started to Learn To Subnet:* A Free, Lecture-Based Presentation on IP Addressing and Subnetting

then once you have the concepts down and you can do ANDing and all that I really like this guys method which is basically what I use though I changed it a bit and If I write down a chart it's a little different: https://learningnetwork.cisco.com/docs/DOC-3849

also you can use the binary game and 3 subnetting games at cisco
Cisco Binary Game
https://learningnetwork.cisco.com/docs/DOC-7561
https://learningnetwork.cisco.com/docs/DOC-1802
https://learningnetwork.cisco.com/docs/DOC-7558

I haven't played those games in a while i think the last one might be pretty hard. You may have to sign up for a free cisco account to play as well.

Cool thanks for the resources!

Technology1

First thing to identify is the 20 bit mask. /20 mask is 255.255.240.0

So you immediately know that the IP Address is 10.104.128.0 and the subnet mask is 255.255.240.0

You take the 240 and subtract it from 256, which gives you the network increment number of 16. The subnets are in increments of 16, in the 3rd octet.

The Network ID is 10.104.128.0

The Broadcast Address is 1 less than the next subnet, which is 10.104.143.255

Anything in between the network address and the broadcast address is the ip host range, which would be 10.104.128.1 to 10.104.143.254

Always remember that network addresses are even numbers and broadcast addresses are odd numbers.

The 8 numbers to always remember for the subnet mask are 128, 192, 224, 240, 248, 252, 254, 255.

bundleboy wrote: »

Hi Everyone,

I've been trying out questions at subnettingquestions.com

and i came across this question that i cannot work out the correct answer.

What is the broadcast address of the network 10.104.128.0/20?

Answer: 10.104.143.255

A detailed explanation of how this is the answer would be very helpful, please bear with me as i am a complete newb and still need some hand-holding.

I just started studying for the exam that i have scheduled in 2 months any and all help is appreciated.

And please accept my apologies if i have posted in the wrong place.

Sincerely
bundleboy