Subnetting questions Broadcast Address

CiscoNewb

Hi,

Please could someone break this down for me?

What is the broadcast address of the network 192.168.63.0/27?

Answer: 192.168.63.31

Thanks in Advance.

amp2030

CiscoNewb wrote: »

Hi,

Please could someone break this down for me?

What is the broadcast address of the network 192.168.63.0/27?

Answer: 192.168.63.31

Thanks in Advance.

CCNA - Subnetting Tips

How I personally deal with it?

27 > 24 (so last octet is the interesting one)
32 - 27 = 5 (so last 5 bits used for host)
2^5 = 32 (length of an interval)
0 32 64 ... (0 is in the first interval)
32 - 1 = 31 (broadcast address is last in the interval)
192.168.63.31 (final answer)

The above process also tells you 192.168.63.1 is first valid host, 192.168.63.30 is last valid host.

LBC90805

OK 192.168.63.0 is a Class C network. Which by itself will be a /24, 255.255.255.0, network. This will have 256 Addresses, which 192.168.63.0 is the Network address, not usable by a host, and 192.168.63.255 is the broadcast address, again not usable by a host.

Now if we need more networks out of that 192.168.63.0 address we can break it up. A /25 network will produce 2 total networks out of the 192.168.63.0 network with 128 Addresses, of which only 126 addresses are usable by hosts do to the Network and Broadcast addresses. You always subtract two from the total number of addresses due to the network and broadcast addresses. For a /25 on the 192.168.63.0 network you will have Network 192.168.63.0 with broadcast address of 192.168.63.127, and 192.168.63.128 with a broadcast of 192.168.63.255!

Now 192.168.63.0 with a /26 will give us four usable networks with 64 IPs but subtract two due to the network and broadcast addresses, that will produce 62 usable IPs for the hosts.

For 192.168.63.0 with /27 you have eight total network with 30 usable by the host. With the 192.168.63.0 network you will have a broadcast address of 192.168.63.31. 0 - 31 IP addresses adds up to 32 addresses. That is 256 divided by 8, gives you 32, but you need the first address to be the network address and the last address, of .31, is the broadcast address. Everything in the middle of those two IPs can be used for end host. Now the next network up from 192.168.63.0 would be 192.168.63.32 with a broadcast address of 192.168.63.63, after that 192.168.63.64 with a broadcast address of 192.168.63.95 and so. What would be the next Network addresses and broadcast addresses for the remaining five networks for 192.168.63.0 /27 subnetted?

Might be too wordy, I can simplfy.

itdaddy

CiscoNewb

pay attention to key words like network or subnet or host id or host ip.
it all depends on what they give you so you can find something:


1. They give you network 192.168.63.0 (which is the 1st non-usable ip in this range)
2. They give you the mask in bit notation.
3. Find the increment to find the range to find the broadcast address.
4. 32 bit total subtract 27 = 5 bits remainin for hosts.
5. Since it is class C (24 bits), 24 bits for class, then 3 bits for subnet, 5 for hosts =32
6. 128 64 32 (then the 5 bits for hosts) so the increment is 32.
7. 32-1 (the 1 is 192.168.63.0) so then 192.168.63.0 + 31 = 192.168.63.0 - 192.168.63.31)
8. you then have your range 0 and 31 and the .31 is the broadcast address.

done.The key is to understand how to find the increment and to know the play on words.
In other words, what is given and what you need to find. That is it



192.168.63.0 192.168.63.31 (see how you add 1 and then add 31 again)
192.168.63.32 192.168.63.63

http://subnettingquestions.com

try this sight...and use a subnet calc from BOSON it is free. use them both
because in the Boson subnet calc you see bit notation which is key to understanding
the subnets and increment values.

aordal

The way I do it is I subtract the amount of bits from / for the class of the address (A,B,C). So a class C is /24. So I do: 27-24 = 3. And since 3 is equal to (in binary) 11100000. I just add those 3 bits together, which is 224(128+64+32). Then I take that # and subtract from 256 and I get 32. And since it was a class C address we are working with the "intersting octet" is the last octet.

And that 32 we got are basically your subnets.
0
32
64
96 etc..

The broadcast is -1 ip address below the subnet so for the subnets i put above:
0 (network) 31(broadcast)
32 (network) 63(broadcast)
64(network) 95(broadcast)
96(network) 127(broadcast)



Hopefully that's not confusing.

itdaddy

What Amp2030 said.

Just read his post. I do it exactly like him and it is fast.

LBC90805

I know I'm long winded, but that can help some folks out.

Truthfully everyone will have a way to work out subnetting in their own little way. Personally I just internalized the numbers. Coming from a math background and tutoring algebra for way over a decade it wasn't that big of a deal.

Personally the way I do it is this.

I know that 255.255.255.0 is a /24 network. Has 256 addresses minus 2 total that are Usable end host ip, total of 254 addresses.

Now all I ever do for my math for subnetting is either Double 256 or half it depending on in which direction the subnet mask is going. When I was younger I memorized all the numbers doubled from 1 on up. That is I knew these numbers where doubled from the number before it:
1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 etc

Always start from /24, 255.255.255.0 because I know it to be 256.
If I need to figure out /23, 255.255.254.0 I just double 256, which is 512, minus 2 for 510 usable IP Host. Simple, remember if we are going to take subnet bits from the left we double how many usable IPs we have minus two!

If I need to figure out /25, 255.255.255.128 I just half 256, which is 128, minus 2 for 126 usable IP Host. Simple, remember if we go right to take subnet bits then we half how many usable IPs we have minues two.

Now of course I don't ever do this math any longer because I have everthing memorized, but that is how I use to do it when I was learning.

CiscoNewb

Hi All,

Thank you everyone for all the help. I think I am starting to get it now.

I have been doing the questions on subnettingquestions.com and I really see what itdaddy means regarding the play on words. I figured out the answer to one of the questions and when I revealed the answer mine was wrong. It was only when I looked more closely at the question that I noticed it wanted a valid host range and not just the host range which was my answer.

I'll have to watch out for things like that.

Thanks again all.

kevin31

Guys

This is great people posting there different techniques it does really help. Be good if we couuld get more people posting there methods?????

Im still trying to grasp the concept so not doing much icnd 1 studying till I grasp the subnetting seems to be a fundamental part of the exam.

Method I have tried is taking the last part of the subnet away from 256 below:

192.168.63.0/27 = 255.255.255.224

256 -224 = 32

so 0
32
64

64 is the next host so 32 - 1 is the broadcast address (i think that right) lol

Please correct if that not right?

Havent got my head roound working out amount of hosts yet although LBC90805 method looks very good.

Thanks

Kevin

LBC90805

64 would not be the next host. It is the Next Network Address, 192.168.63.64!

Within the 192.168.63.64 network, the next host is 192.168.63.65! The broadcast address for this network would be 192.168.63.95!!!
0
32
64
96
128
160
192
224
are all the network addresses in the fourth Octet, they would be unusable to assign to host with a /27 subnet!

First address in a range is alwaysthe Network address, last address in the range is always he Broadcast address.
31
63
95
127
159
191
223
255

kevin31 wrote: »

Guys

This is great people posting there different techniques it does really help. Be good if we couuld get more people posting there methods?????

Im still trying to grasp the concept so not doing much icnd 1 studying till I grasp the subnetting seems to be a fundamental part of the exam.

Method I have tried is taking the last part of the subnet away from 256 below:

192.168.63.0/27 = 255.255.255.224

256 -224 = 32

so 0
32
64

64 is the next host so 32 - 1 is the broadcast address (i think that right) lol

Please correct if that not right?

Havent got my head roound working out amount of hosts yet although LBC90805 method looks very good.

Thanks

Kevin

itdaddy

CiscoNewb

keep watching what they give you and then you can work from there.

Kevin31

dude email me and I can find you some great pdfs on subnetting.my email is
robert@itdaddy.net

Stuff I used and the webiste subnettingquestions.com daily 30 minutes after I got good.
60 minutes daily when I sucked


Aordal
weird how you get the increment that way all the time freaking numbers are weid.
i always take the last bit in the subnets 128 64 32 and bam i got the increment/interval.
freaking weird..but if it works cool.

Kaminsky

aordal wrote: »

The way I do it is I subtract the amount of bits from / for the class of the address (A,B,C). So a class C is /24. So I do: 27-24 = 3. And since 3 is equal to (in binary) 11100000. I just add those 3 bits together, which is 224(128+64+32). Then I take that # and subtract from 256 and I get 32. And since it was a class C address we are working with the "intersting octet" is the last octet.

And that 32 we got are basically your subnets.
0
32
64
96 etc..

The broadcast is -1 ip address below the subnet so for the subnets i put above:
0 (network) 31(broadcast)
32 (network) 63(broadcast)
64(network) 95(broadcast)
96(network) 127(broadcast)



Hopefully that's not confusing.

Exactly the way I do it and it is very fast in mental arithmetic. The more you practice this, the quicker and more used to the bits you get.

itdaddy

Kaminsky/aordal

to get the increment/interval just do this vs adding and subtracting

27-24 = 3 bits for subnets
32-27=5 bits for host
since you know the subnet bits are in the last octet and use up the first 3 bits
128 64 32 just take the 32 as interval. this method I got from jeremy ciora and
it is fast and easy to find the inetrval/increment..;)I always look at the last bit
of the subnet to the right and take it as the increment..

CiscoNewb

Not sure if I am asking too much here and tell me if I am, but are the questions on subnettingquestions.com typical of those subnetting questions found on the exam?

Thanks,

kevin31

itdaddy wrote: »

CiscoNewb

keep watching what they give you and then you can work from there.

Kevin31

dude email me and I can find you some great pdfs on subnetting.my email is
robert@itdaddy.net

Stuff I used and the webiste subnettingquestions.com daily 30 minutes after I got good.
60 minutes daily when I sucked


Aordal
weird how you get the increment that way all the time freaking numbers are weid.
i always take the last bit in the subnets 128 64 32 and bam i got the increment/interval.
freaking weird..but if it works cool.

Hi Itdaddy

thanks i have mailed you! your help would be great thanks!

Anyone here use a site called ************? (sorry if not allowed to mention other sites please remove if the case?)

thanks
kev

Neeko

To work out the increment I just count down the subnet bits from the start of the last octet being used. So if it’s /28 I know 4 bits stops at 16 and therefore the increment is 16. When you’re working in the last octet its simple to use the above method to work out the number of hosts, but if you’re working lower than /24 it isn’t so simple.

Lets say we have /18, I know 2 bits stops at 64 so the increment is 64, but we have 14 bits left for the host range. Although the first and last valid hosts are easy to identify and stating whether a host is valid with this mask is straight forward, the thing that isn’t simple is the number of hosts unless you memorize high multiples of 2.

As said above you can double 256 for any bits below the last octet. So for the above example we have 256 hosts in the 4th octet, and working backwards in the third we have 512, 1024, 2048, 4096, 8192 and finally 16384 which is 2^14.