hmmm broadcast address is the last address in that subnet...
but i think somethin is wrong with ur subnettin..coz the netwk is a class c...from your no of network bits...there is only 22...u have not borrowed any netwk bits....
so i dont think u have subnetted coz the number of subnets is equal to 2^n - 2
n being the number of bits u borrowed from host side,,,...hmmm tts what i ermember
Lets take a simple example..
192.168.10.0 with a subnet mask of 255.255.255.224
Now how do we find our ranges ?
We do 256 - 224 (the octet we are subnetting) this gives us
32 as our block size.. in other words, all subnets will be in multiples of 32.. so the subnets will be
192.168.10.0 -- this is subnet 0 and is not valid in Cisco exams
Now the broadcast addresses for each subnet will be one less than the next subnet address.. so for 192.168.10.32 the broadcast will be 192.168.10.63 (since 10.64 is the next subnet address)...
your valid hosts are all the ones between the subnet address and the broadcast address.. so for the 192.168.10.32 subnet, valid hosts will be between 1 - 62 (63 is broadcast).
Remember, broadcast address is one less than the next subnet address.
I've got the answer for your question right here:) You happen to be in luck because I only just recently answered a member's question similar to yours for searchnetworking.com, so here's the actual answer ... read through it as I'm sure you will find it insightful!
Variable Length Subnet Masks (VLSM) can require some serious number crunching to figure out when they start getting complicated and there are many ways to help make the process more simple.
In my experience of teaching the CCNA course, I have found that every person essentially will find their own way of figuring the fastest method to calculate them as long as they work on it hard enough.
What I can do, hoping to help you, is show you the method I use, so you can then take it and see how well it suites you!
Before we dive into the deep waters, let me show you how I calculate network ranges from given subnet masks:
Assume network 192.168.0.0 and subnet mask 255.255.255.240
I first calculate each subnet's range: 256-240=16 IP's per subnet
This means we have the following subnets:
As we said, for each subnet, we have 16 IP's this includes network and broadcast addresses. Looking at the above table, I already have the network address for each subnet, so let's reveal the broadcast address:
0 - 15
16 - 31
32 - 47
48 - 63
The logic behind this is quite simple: take the next network and subtracted one from it to get the broadcast address
Now that we have the network and broadcast address, its easy to calculate the range of VALID IP addresses:
0 (1 to 14) 15
16 (17 to 30) 31
32 (33 to 46) 47
48 (49 to 62) 63
You can clearly see that everything between the network and broadcast address of each subnet, is a VALID IP address.
This method is perhaps the best and fastest way to calculate networks and IP ranges from different subnet masks.
One step up, is VLSM, where we are given any possible subnet mask and we need to calculate the subnets and IP ranges.
For this, I have created a nice table which is available on my site, under the 'supernetting/CIDR' topic (
). The chart shows all possible subnet mask combinations and the corresponding number of networks and hosts calculated for each one.
A closer look at the table will reveal a nice pattern that is repeated:
Notice the subnet mask for /17 and /25, /18 and /26.
Both /17 and /25 have the 128 decimal number in their subnet mask, while /18 and /26 have decimal 192. The only part that changes in the pattern is the location of the last 'non-zero' decimal number.
Considering that /8, /16 and /24 have only 255 and 0 decimals, so whenever I see a number between these ranges, I use the closest of the three as a reference point.
For example, /19 is a standard 16 bit subnet mask, with an additional 3 bits. If you memorise the following, then it becomes an easy task translating VLSM numbers to real decimal subnet masks:
To figure out the subnet masks here, simply do the following calculations:
256-248=8 That's 8 IP's per subnet. We need to reserve 2 IP's (network and broadcast addresses), leaving us with 6 valid IP's per subnet.
From all the above calculations, the hardest part is memorizing the last table, which is only 7 entries. From there on, it's a matter of using the quick method I just showed, to calculate your subnets and their ranges!