Valid IP addresses range and Broadcast address

Dear Fellows,

I am having trouble understanding “How to determine valid IP addresses ranges” by given IP address as well finding broadcast address. Any help on this will greatly appreciate it.

Many Thanks

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Comments

dublin_101

i would recommend doing a subnetting workbook.........if you do a google search you should find some.......

broadcast in i.p is the last address in the range...

so an example would be the example of class c address:

network address: 192.168.10.0

subnet mask: 255.255.255.0

on this network address, the network address is as stated, 192.168.10.0, the first usable host of this would be 192.168.10.1 up until the last usable being 192.168.10.254....note that the .254 in binary is 11111110

the broadcast address would be 192.168.10.255, where the 255 is represented in binary as 11111111.........so as you have noticed, the broadcast address is all 1s.

so for this network address, the usable range would be:

192.168.10.1 to 192.168.10.254

i hope you have understood this!.

bmauro

I really think it all comes down to what you find is the easiest for you to figure out.

I'll go over what I find is the easiest for me.

255.0.0.0 - 255.255.0.0 - 255.255.255.0 - are all easy masks, not too much thought needs to be put in to figure out the valid hosts and broadcast addresses.

For me the hard ones were 255.255.248.0 or 255.255.255.240 or 255.255.224 - you get the idea. I would focus on the first octet that was not equal to 255 - thats were you need to look.

In the first "hard" mask I mentioned 255.255.248.0 - the interesting octet is the third one which is at 248. All I would do is:

256 - 248 = 8 <---- This 8 represents how large your subnets are

So if you would take 172.16.0.0 for example using 255.255.248.0 mask all you do is count by 8's

172.16.8.0
172.16.16.0
172.16.24.0
172.16.32.0 ....... and so on

So for the valid host everything between 172.16.8.1 and 172.16.15.254 is a valid host. 172.16.16.0 is the next subnet - so all you do is subtract one from the last octet and you would have 172.16.15.255 .... and that is your broadcast address. Subtract one more and you have your last valid ip address.

Thats how I do subnetting - it works for me

dublin_101

bmauro wrote:

So if you would take 172.16.0.0 for example using 255.255.248.0 mask all you do is count by 8's

172.16.8.0
172.16.16.0
172.16.24.0
172.16.32.0 ....... and so on

Thats how I do subnetting - it works for me

very well said........even though you said more difficult, the more you practise the more you remember the number.....and seeing there is only a few numbers to remember, it only takes little practice to be able to remember the subet number ..

subnet of 128 ...means 1 bit is borrowed and means the subnets will jump by 128

subnet of 192 ...means 2 bits are borrowed and means the subnets will jump by 64

subnet of 224...means 3 bits are borrowed and means the subnets will jump by 32

subnet of 240...means 4 bits are borrowed and means the subnets will jump by 16

etc , etc

practise this for a couple hours and you'll start to know them automatically!

remember, practise makes perfect!

imran1

Thank you all for quick response, I been studying for CCNA exam using “Ultimate CCNA study package” from thebrayantadvantage.com, I was able to practice and understood binary/decimal conversions as well determining number of subnet and number of hosts. Going through the text I seem to get halted on the part for determining valid IP addresses ranges as well broadcast address. DarbyWeaver did a good job by breaking into a part with explanation. (Off course all you were helpful). I still can’t figure out why my answers are not coming right when comparing from the text book I been studying.

Example (1) Valid IP range for 140.10.10.0/23
Answer: 140.10.10.1 – 140.10.11.254 with broadcast address 14.10.11.255. Now how the number 11 does comes in third octet?

Example (2) Valid IP range for 222.23.48.64/26
Answer: 222.23.48.65 – 222.23.48.126 with broadcast address 222.23.48.127 this one I couldn’t figure out where 127 came from.

I am going to try and review this, maybe this time I may pick up what I miss, by any chance anyone of you has any helpful link or care to chat on yahoo/msn (my id imran11755) I would be happy to go over with my problem.

Thanks again for a

bmauro

Those are the types of subnetting questions that got me in the beginning too.

I can't recommend www.subnettingquestions.com enough. Great site that just hits you time and time again with hard subnetting questions. I would do this anytime I had a free second. Soon you memorize the different subnet masks and can quickly do the math in your head.(Just remember that zero subnet is allowed over there).

For your first example: 140.10.10.0/23

/23 is the same as 255.255.254.0. All I do is quickly take 256 - 254 which gives me 2. That is your subnet number. So your subnets would be (if subnet zero is disabled)

140.10.2.0
140.10.4.0
140.10.6.0..... and it just increases by 2.

So for your example the subnet is 140.10.10.0 and the next subnet is 140.10.12.0 - again just another multiple of 2.

That means that from 140.10.10.1 (first valid host) all the way to 140.10.11.254 (last valid host) are the valid host ip addresses. The way I look at it is once you map out the valid subnets (or at least the subnet above the one you're working on) it's easy to see what belongs to the subnet you're working on.

Now to your second Q. 222.23.48.64/26

/26 is the same as 255.255.255.192

Again I take 256 - 192 which give us 64.

So 222.23.48.64 is the subnet. Add another 64 to find the next subnet which would be 222.23.48.128. To find the broadcast - just subtract one - which would give you 222.23.48.127.

Hope it helps.

dublin_101

ok, you seem to be getting close..........

remember this:

the broadcast address is all 1s of the host address....we are talking binary, so on the host side, all 1s will give you the broadcast address.

140.10.10.0

divnth

During prep, an exam dais this was not a Class A address. I don't understand why.
Are they wrong or am I thick?

00001010.10101100.11001100.00000111

LOkrasa

00001010.10101100.11001100.00000111 = 10.172.204.7
Class A address but its a reserved address bc it falls in the 10.0.0.0 - 10.255.255.255 range.

datchcha

Binary to decimal and decimal to binary - is pretty easy when you get the basics down. Let me show a little trick that I use, which might help you out.

1.0 Binary-to-decimal conversion

10011010
Remember the order: 128, 64, 32, 16, 8, 4, 2, 1
128 + 16 + 8 + 2 = 154

2.0 Decimal-to-Binary conversion

185
I always start with the left most value in the binary string, which just so happens to be the largest value. Start at the left, and work your way right, subtracting the value from you’re the number you want to convert. Example:

Step 1: 185 – 128 = 57 -->10000000
Note: you can not subtract 64 from 57, so the next most number is 32
Step 2: 57-32 = 25 -->> 10100000
Step 3: 25 – 16 = 9 -->> 10110000
Step 4: 9 – 8 = 1 -->> 10111000
Step 5: 1 -1 = 0 - -> 10111001

hope this helps a little... good luck