The block size you're using is for subnets and you could find the last valid subnet that way. The last valid host on your last valid subnet would be 10.57.255.254.
No longer work in IT. Play around with stuff sometimes still and fix stuff for friends and relatives.
OK I think I have got it...
To calculate the block size would it be
(A)
1. The above IP address is a class A ( i.e. 255.0.0.0)
2. 240 is equal to four ones ( 128+64+32+16)
3. So 255(8 ones) +240 (4 ones) = 12 ones
4. The next boundry is 16, so 16-12 = 4
5. block size is 4
or (B)
1. The above IP address is a call A (i.e.255.0.0.0)
2 but the subnet mask which has been given is a 255.255.0.0
3. So 255 (8 ones) +255(8 ones) +240 (4 onnes) = 20 ones
4. The next boundry is 24, so 24-20 = 4
5. Block size is 4
6. 2 to the power of 4 = 16
Based on the block size of 16
172.57.0.0
172.57.16.
172.57.32
172.57.48
172.57.64
miss a few....
172.57.224
172.57.240
172.57.256
Host range betweeen 172.57.240 and 172.57.256 is:
241 -255 ?
Though I thought you cannot use the last broadcast address which is 255 or does this only apply to the last octect ?
Am I meant to subreact 4 from 16 or 24 in terms of the next boundry ? If the subnet mask is provided do I follow the subnet mask or the default subnet mask for the IP Address?
Thanks
Hi all, I'm currently studying for CCNA. This post is in regards to the "broadcast subnet."
For example, take the following:
130.4.0.0 / 24 (255.255.255.0)
130.4.0.0 --> Classful network/subnet zero
130.4.1.0 --> First Nonzero subnet
.
.
.
130.4.255.0 --> which is the "broadcast subnet." Thus as are all addresses from 130.4.255.0 - 130.4.255.255 all "broadcast addresses"? or just 130.4.255.255 and 130.4.255.1 to 130.4.255.254 can be used as valid addresses? (This example follows the guidelines that the subnet number address and broadcast address are reserved.)
Please let me know if any of the above info or my questions need clarification. Thank you.
Last edited by lstudent; 09-13-2010 at 06:10 PM.
Either I missed something in my CCNA classes or you are barking up the wrong tree.
There are really three types of addresses. Network, host, and broadcast. The first address in a subnet is the network address, and the last is the broadcast address. All inbetween is the host addreses.
Using your example:
130.4.0.0/24 is the network address, with 130.4.0.255 as the broadcast address.
130.4.1.0/24 is the network address, with 130.4.1.255 as the broadcast address
.
.
.
130.4.255.0 is the network address, with 130.4.255.255 as the broadcast address.
The broadcast address storms to all host addresses in the subnet, where as the network address is a unique identifier for the subnet.
I have never heard of a broadcast subnet.
To answer your question, 130.4.255/24 is a usable subnet for hosts, excluding of course the network and broadcast address.
Last edited by chmorin; 09-13-2010 at 06:13 PM.
That answers my question, its probably just a general term for the last subnet in a list of subnets for a given address/mask. p.389 of CCENT/CCNA ICND1 uses the term "Broadcast Subnet" in chart 12-37 as the last subnet when listing all the subnets for a given address/mask.
Thanks!
Hi all
I have a little problem with cbtnuggets method to calculate...for example
10.0.0.0 255.0.0.0
and I have to divide in 4 networks
can someone try to calculate and give me a feedback how to do it (in cbtnugets method)...
I need 4 networks:
00000100 is 4 (So it takes 3 bits for number 4)
so next step is to add those bits in subnet mask to get increment and new sub-net mask:
11111111.11100000.00000000.00000000
what makes
255.224.0.0 and increment of 32(last bit in mask)
but subnet calculator gives me this:
Am I making a obvious mistake somewhere...please help..Im gona explode
Sorry guys, found it at the end of lecture, I noticed before that 4 and 32 ect. dont give wright result...so
2, 4, 8 , 16, 32 , 64, 128 give off calculations!!!! Those are exceptions!!!
Last edited by Gargamel; 09-17-2010 at 04:54 PM. Reason: found the answer to my question :)
?
That is the binary number 4 -- but 3 bits gives you 8 possible bit choices/options/networks -- 000 001 010 011 100 101 110 111 (0 through 7)
1 bit gives you 2 choices (or 2 networks) -- 0 or 1 (0 through 1)
2 bits gives you 4 choices (or 4 networks) -- 00 01 10 11 (0 through 3)
Just like to say thanks to lordflasheart
could any help to me undetstand 23 bit mask
I have been using lordflasheart process but still cant get 23bit mask
the bit which is confusing is where you draw the boundry 24 - 23 =1 this is very a get confused
please help
Thanks storma
Hosts howto:
max = /32
Determining number of host’s:
2^(number of zeroes in the subnet mask)-2
if /26, 32-26=6
2^6-2=62 total hosts
----------------------------------------
8, 16, 24, 32 boundries
Subnet howto:
1) determine octet position from netmask
2) Take netmask subtract from boundry position
3) do 2^boundry difference = x, block size
4) count up by block size value
5) stop when block size value is close to ip address.
-------------------------------------------------------
What subnet does 192.168.12.78/29 belong to?
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.
192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
Our address is 192.168.12.78 so it must sit on the
192.168.12.72 subnet.
Glad I found this post, it really helped me understand.
I've just gone through the first page of this post now and already understand subnetting a lot better than from going through a lot of other books and tutorials. Thanks a lot!
Ok, I'm still having problems with these type of questions, where they give an ip and subnet mask, and ask for the host etc. It's been answered before, but I still don't get it.
How does 240 mean that you're borrowing the first 4 bits of the third octet???
Third octect would look like this
128|64|32|16|8|4|2|1
1|1|1|1|0|0|0|0
11110000 = 240
the easiest way to figure those out is to do the following
determine the interesting octet. (the non 255. one)
in this case it is the third octet, which is 240
each octet totals 256 (0-255)
so 256-240 = 16
your block size is 16.
so with the address 10.5.178.10 you would start with the non interesting octets (255.255.0.0)
10.5.x.x
since your block size is 16 you would apply it to the base
10.5.0.0
10.5.16.0
10.5.32.0
10.5.48.0
10.5.64.0
10.5.80.0
10.5.96.0
10.5.112.0
10.5.128.0
10.5.144.0
10.5.160.0
10.5.176.0 (winner winner chicken dinner)
10.5.192.0
now you have determined the subnet it belongs in. the next ip address 10.5.176.1 is the first available host
the broadcast address is easily determined by taking the next subnet (10.5.192.0) and subtracting 1 = 10.5.191.255
the last available host would be one address below the broadcast
10.5.191.255 - 1 = 10.5.191.254
hope that helps
Could anyone seem to help me with these type of problems? i have no clue how to start. every other question takes me 4 seconds.
heres an example.
thanks
Question: How many subnets and hosts per subnet can you get from the network 172.19.0.0/28?
Answer: 4096 subnets and 14 hosts
this is fairly easy to do. It is all based on the powers of 2....
see the chart below.
1 - 2
2 - 4
3 - 8
4 - 16
5 - 32
6 - 64
7 - 128
8 - 256
9 - 512
10 - 1024
11 - 2048
12 - 4096
13 - 8192
14 - 16384
15 - 32768
16 - 65536
So in your example the address given is a class B address. So under normal circumstances it is using a 16 bit mask. 255.255.0.0
In this example it is borrowing 12 of the possible bits for the subnets, and 4 bits for hosts.
so 2 to the 12 = 4096 = to the number of subnets
and then 4 bits are used for the number of hosts 2 to the 4 or 16... but you need to subtract 2 bits, one for the network address and one for the broadcast on each of the 4096 subnets......
Hope that helps.
Thanks for this post. I agree with the poster above, out of all the ways I have learned sub netting this week, this way is by far the fastest.
I am horrible at math well not horrible, I just do not trust my self doing it in my head, so the less processes I have to do for sub netting the better, for I always write everything out.
Hi
I joined to say thanks for this method - it has started to come together for me after struggling with subnetting using other methods for ages - (even Jeremy's CBT Nuggets method)
However, I am still struggling with these types of questions:
Question: You are designing a subnet mask for the 172.27.0.0 network. You want 30 subnets with up to 2000 hosts on each subnet. What subnet mask should you use?
Answer: 255.255.248.0
Is there an easy way of quickly knowing what mask(s) give you number of networks and hosts?
I also have problems subnetting in the 3rd octet when it's a class B address - this gets me every time - class C I have no problems with! For example:
Question: What is the last valid host on the subnetwork 10.194.176.0 255.255.240.0?
Answer: 10.194.191.254
Thanks again - great method and resource - Jason
EDIT - Actually I worked both these out from post#6 on this thread! I am buzzing!!
Last edited by Jas21; 01-16-2011 at 04:34 PM.
hi Jas. This guy helped me alot with how he does it. he uses a system called the magic number.
basically all you have to do is find the magic number and its all easy from there.
you should watch a have of his subnetting vids
YouTube - danscourses's Channel
What is the last valid host on the subnetwork 10.194.176.0 255.255.240.0?
the "important octet is the third octet as it is something other than 255.
what number bit is 240 ? it is the number 16 bit thats your magic number!
so youre only interested in counting up in 16's in the THIRD octet.
in this case you can just say to yourself. 10 x 16 is 160. 160 + 16 is 176 + another 16 = 192
the subnet range is therefor 176 - 192
host range 177 - 191.254 with the broadcast being 191.255
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