Wow this concept is so much easier then any video or book I have read. I actually like this.
Wow this concept is so much easier then any video or book I have read. I actually like this.
Hi folks,
I have two questions :
1) How can I count the next boundary?
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?Quote:
Originally Posted by kriscamaro68
Ok so I got this question on subnettingquestions.com and didn't see a way to answer it with your method:
Question: You are designing a subnet mask for the 10.0.0.0 network. You want 3800 subnets with up to 3800 hosts on each subnet. What subnet mask should you use?
Answer: 255.255.240.0
For your first question, how many bits do you need to borrow to accommodate 3800 subnets? The answer would be 12 bits as 2 ^ 12 = 4096. Your network address is a /8 by default as it is a Class A address so 8 + 12 = /20 mask.
2) may you explain It more by detail?
My regards
Thanks everyone for the great post. It helps alot.
Thanks for the post, it helped out a lot. Though I still find myself struggling on class a & b. Guess more practice will help.
thanks for the write up I hate to even think about sub-netting when I think about study. I would rather learn how routing protocols work and how routers talk to each other than subnet. Back to the books
bookmarked this thread for future reference going to get heavy on subnetting!
is there an easy to work this one out
class a network 10.0.0.0 create a subnet mask for the 600 subnets. THen Identify the 100th subnet
I can work out doing the subnet easy but is there an easier way to finding out the 100th subnet rather than writing each subnet down like this
10.0.0.0
10.0.64.0
10.0.128.0
10.0.192.0
It would take forever
I'm kinda confused. How do you know which octet to subtract from?
Like this:
"What subnet does 192.168.12.78/29 belong to?
You may wonder where to begin. Well to start with let's find the next boundary of this address.
Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8."
How do you know which boundary to use, 8, 16, 24, 32 to subtract?
Last edited by cleanwithit; 10-11-2009 at 11:06 PM.
Look at it this way. You have a /29 mask. This means that there are 29 "on" bits in the subnet mask.
11111111.11111111.11111111.11111 | 000
So, since you have a /29 you see the split in the fourth octect. That's where you know how to determine the subnet blocked and the next available block.
Subnet block: 256 - 248 = 8, this means that each subnet block size is 8
192.168.12.0 /29
192.168.12.8 /29
192.168.12.16 / 29
192.168.12.24 /29
192.168.12.32 /29
192.168.12.40 /29
192.168.12.48 /29
192.168.12.56 /29
192.168.12.64 /29
192.168.12.72 /29
192.168.12.80 /29
192.168.12.88 /29
192.168.12.96 /29
192.168.12.104 /29
and so on.
you are almost there
since the mask is 29 = 3 the block size is 8 as you have stated.
the first subnet is 192.168.12.0 hosts 192.168.12.1 - 192.168.12.6 broacast 192.168.12.7
2nd subnet 192.168.12.8 hosts 192.168.12.9 - 192.168.12.14 broadcast 192.168.12.15
etc....
the address you were given is 198.168.12.78 /29
multiplying by 8 you would realize 80 is subnet, and one less is 72....
192.168.12.72 subnet
hosts 192.168.12.73-192.168.12.78
broadcast 192.168.12.79
hope that helps
Hi guys. New to subnetting, and having a few problems. I found out how to get the network number through this great method, but I'm still having trouble.
I need to figure out :
If the IP (Host) valid?
What is the Network (Subnet) Number? ( I think this is covered by using the method above)
What is the Broadcast Number?
What is the valid IP (Host) Range?
What is the subnet mask? If given dotted decimal convert to the prefix length and vice versa?
The first one is :
192.168.10.23 / 25
The second one is :
192.168.1.253
255.255.255.240
I'm just looking for a little help.
I've gotten the basic concept, but that's about it.
Thanks guys.
practice.......you can figure out all of the answers by working through the methods listed in the thread
192.168.10.23 / 25 = 255.255.255.128 1 bit is borrowed...
128
64
32
16
8
4
2
1
that determines the subnets possible
256/128 = 2 subnets .0 and .128
256/64 = 4 subnets .0 and .64 and .128 and .192
256/32 = 8 subnets .0 and .32 and .64 and .96 and .128 etc
256/16 = 16 subnets .0 and .16 and .32 and .48 and .64 etc
256 /8 = 32 subnets .0 and .8 and .16 and .24 and .32 etc
256/4 - 64 subnets .0 and .4 and .8 and .12 and .16 etc
192.168.10.0 is the subnet
valid host range is 192.168.10.1 - 192.168.10.126
broadcast address 192.168.10.127
so your address falls into the range so yes it is valid
the second subnet starts at 192.168.10.128
this allowed them to take a network and divide it... subnetting
192.168.1.253 255.255.255.240 = /28 = 16 bit address blocks
subnet 192.168.1.240
valid host range 192.168.1.241 -192.168.1.254
broadcast 192.168.1.255
so yes it is valid host because it is in the valid range
in this case 192.168.1.0 was subnetted as follows
1 subnet 192.168.1.0 /28
2 subnet 192.168.1.16 /28
3 subnet 192.168.1.32 /28
4 subnet 192.168.1.48 /28
5 subnet 192.168.1.64 /28
6 subnet 192.168.1.80 /28
etc
your example was the last subnet available with the /28
find the subnet size by examining the mask
one ip above the subnet it the first available host address
one ip address below the next subnet is the broadcast
one ip address below the broadcast is the last host address
well this was a great tut to learn subnetting.
i have a problem
Question :-
What is the broadcast address of the network 10.4.64.0 255.255.240.0?
ANSWER : 10.4.79.255
but the problem is how to find out broadcast address when there are 4096 subnets and 4094 host. because in exam you have to do it quickly without wasting time. is there a trick to find out quickly first valid host, last valid host and broadcast address.
1. find the subnet size by examining the mask
2. one ip address above your network address is the first available host address
3. one ip address below the next subnet is the broadcast (use the subnet size to find the boundary)
4. one ip address below the broadcast is the last host address
Try my whitepaper on subnetting, I tried to make it as easy and understandable as possible
http://www.linkedin.com/in/firemarshalbill
and if you are on linkedin send me a link request
You've almost answered it yourself...
For every 4 increments in your third octet you 2nd octet will increase by 1.
so....100 / 4 = 25 increments of your second octet will get you to 100
10.0.0.0 - 10.24.255.255 is your first 100 subnets (2nd octet 0 thru 24)
The last subnet in that range (the 100th) should be 10.24.192.0
It's very late and I'm completely out of it right now...so if I'm completely wrong and off-base with this, forgive me
I hope this helps...
George
Everytime I think I finally think I get subnetting I try a few questions and get them wrong. My blood pressure cant take much more of this. Here are a few of my main concerns:
1. What if my mask matches the next boundary? For instance, 172.21.197.0/24 well 24-24 = 0 and 2 to the power of zero .... well I think you can see that this is not going to work.
2. I am still baffled by which octet you are supposed to be working with and when. I followed the tutorial here to a tee when doing questions on subnettingquestions.com and ALWAYS turned out wrong. For instance it would seem I would use the 3rd octect since that is where the 24th bit is but I get the answer wrong all the time with this logic.
3. I am still 100% confusedon how to answer questions when given the mask in decimal form.
I very much appreciate all of your great information but I cant help but get frustrated when I constantly get this wrong over and over and over again.
OK as you can see everyone on the posting does it a little different, I think it depends on how your brain is wired.
I assume when you say decimal format you mean something like 255.255.255.240. Remember that when we are talking mask we are starting at the left with 1's. each group between the periods represents 8 bits hence 255 = 11111111 because all 8 bits are 1's. When you see that value change as in my example here 240 that is the field we are going ot work in. the way to add up the value to determine the 1's is like this
on a scrap piece of paper at the top right this (these are the 2 raised to the powers)
128 64 32 16 8 4 2 1
then you are going to subtract these values from the 240 until you reach 0 and put a 1 in the column if you can subtract.
so you get
128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0
because
240 - 128 = 112
112 - 64 = 48
48 - 31 = 16
16 - 16 = 0
so we now see that the first 4 bits are ones and the last 4 are 0's.
to give us the / (cidr) value we add up the one's. so for 255.255.255.240 we get 8+8+8+4 = 28 or as we normally write it /28
Good luck I hope this helps
If im right this is one of the most easyiest subnet. 172.21.0.0 is the network address. 255.255.255.0 is the subnet mask.
Subnet = 256
Host = 254
Valid subnet - 1.0 all the way to 255.
Remember that this is a class B address by looking at the first octect. therefore the first two octect dont change at all. the third is the subnet which in this case is all 1's = 256 the host is all 0's = 256 - 2 = 254
lol well I couldve swore Cisco standard practice did not allow subnet zero but I guess I was wrong, no wonder I was off most of the time.
Dude, I signed up for the forum just to THANK YOU for posting this!!!!
I was able to understand subnetting from the long route used in the Bryant method...but I knew that it took too long to answer the questions. It took me a bit to streamline things but now I'm able to answer EVERY question at subnettingquestions.com with ease.
I write out the powers of 2: 2 4 8 16 32 64 128 256 etc...up to 65,536 and I write out the possible masks i.e. 128 192 224 240 248 252 254 255. This allows me to answer most every question within just a few seconds (many I can even do in my head now!!)
My only regret is not finding this easy method before drudgging through all the other garbage out there.
I cannot thank you enough for this! I thought i would never understand sub netting and then found this! I must admit it took me a whole day before it started to click. As time went by it started making a little bit more sense and then a little more and then BAM!!!
I can now go onto subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online and answer all the questions in my head. Phew!!
Now that part is over i guess its time to get WAN's down pat
But thank you again for introducing me to this concept. I LOVE IT!! I even just showed my manager a few questions and answered them all in about 8 seconds and he was utterly impressed
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