Explain me how to do this step by step.

gouki2005

Please i am figthing with the subnetting i can find host valid range ,broadcast address thx to the subnetting made easy..but i cant do exercises like this i have some problems

I have this ip address 172.20.0.0 255.255.0 we need 30 subnet with 500 user in each subnet

how i do this one??

which is the fastest,easiest way to do exercises like this one..

explain me just this one i need just one example..and i can practice and learn with this one please i really need help here!!! step by step to understand thank you..

bermovick

Let me see if I can manage this, using what I've picked up.

To find out how many host bits you need for the number given, just find the highest bit that it fits into. In this case, 256 < 500 < 512, so you need 512 worth of 'host bits', so 11111111.11111111.11111100.00000000 (255.255.252.0).

Similarly, when searching a requested number of subnets, the same applies, you just add these to the network+subnet side. 16 < 30 < 32. 32 worth of subnet bits results in 11111111.11111111(class B network) + 11100000.00000000 (255.255.224.0)

OK, I've apparently screwed something up there, or else this question was one with multiple answers. Someone else can hopefully either correct me or verify this is one with multiple answers (111XXX00 as 3rd octet).

Edit: wow. I feel dumb now. grabbing wrong number of bits, etc.

notgoing2fail

If you want 30 subnets. You'll need to borrow 5 bits. That gives you 32. 2 to the power of 5 equals 32.


172. 20 . 11111000 . 00000000

Now when you convert to mask you're looking at: 255.255.248.0

172.20.0.0/21

With 11 zero's for your host bits, that gives you well over the 500 users that you need per subnet...

DPG

I don't really have a concrete method of subnetting but this is what I would do:

Find the smallest subnet that can hold the required amount of hosts.

500 hosts are needed per subnet.

1 2 4 8 16 32 64 128 256 512 1024 etc...

512 (510 available hosts) is the key number.

You then find the subnet mask needed for this number (use your preferred method) which is 255.255.254.0.

The first subnet would be:

172.20.0.0 255.255.254.0 (valid range is 172.20.0.1 - 172.20.1.254)

Second subnet:

172.20.2.0 255.255.254.0 (valid range is 172.20.2.1 - 172.20.3.254)

Third subnet:

172.20.4.0 255.255.254.0 (valid range is 172.20.4.1 - 172.20.5.254)

30th subnet would be:

172.20.58.0 255.255.254.0 (valid range is 172.20.58.1 - 172.20.59.254)


This probably makes no sense at all.

thehourman

I love that subnetting made easy. I use it all the time when I go to subnettingquestions.
What I do if the question is asking for x many subnet and x number of host:

Since we need 30 subnets, I kind of like convert the decimal# into binary.
00011110 or 30 in decimal. Well, you don't really have to convert the decimal# into binary, you just need to find where the 1st bit located(starts from the left) then count the bits from there. That is why I said kind of converting the decimal into binary.
e.g. 0001xxxx, you see what I mean? Once you get the first bit just count from the 1st bit to the right, and that is how you get the x number of bits.

Then I will count the x number of bits from left to right(the bold bits)(0001xxxx). In this case, 5 bits. Then add the bits to your subnet mask. Since the subnet mask is 255.255.0.0(11111111.11111111.00000000.00000000), add 5 bits to the mask then it will be like this. 255.255.248.0(11111111.11111111.11111000.00000000).

To answer the question, 172.20.0.0 255.255.0.0
2^5 = 32 Subnets(the # 5 is the extra bits we made). Since we need 30 subnets, with the help of extra 5 bits, we made 32 subnets.

2^11 - 2 = 2046 Hosts (the # 11 is the # of 0s(in italics))
So the subnet mask is 255.255.248.0 or /21
172.20.0.0/21 (255.255.248.0)

Thanks Jeremy

gouki2005

i still dont understan very well but i understand little more look this one

How many subnets and hosts per subnet can you get from the network 172.26.0.0 255.255.254.0?

thx to you i know how many host can have this subnet

255.255.254.0 = 11111111.11111111.11111110.00000000 we have 9 0 so 2^9 = 512 -2 = 510 host

Answer: 128 subnets and 510 hosts

but i dont know how to find the subnets number.

Monkerz

Ok, so is it more important that you have exactly 30 (32) subnets or subnets that can hold only 500 (510) hosts? Are we suppose to utilize the entire range or do they care?

I mean, to me this could go either way. Is this poor question wording? Do the tests have questions like this or is it pretty straight forward.

172.20.0.0\21 = 30 (32) Subnets within range

172.20.0.0\23 = 500 (510) Hosts per subnet

notgoing2fail

Monkerz wrote: »

Ok, so is it more important that you have exactly 30 (32) subnets or subnets that can hold only 500 (510) hosts? Are we suppose to utilize the entire range or do they care?

I mean, to me this could go either way. Is this poor question wording? Do the tests have questions like this or is it pretty straight forward.

172.20.0.0\21 = 30 (32) Subnets within range

172.20.0.0\23 = 500 (510) Hosts per subnet

From what I remember, you "care" about the network first, than hosts....but don't quote me on it...

Monkerz

gouki2005 wrote: »

i still dont understan very well but i understand little more look this one

How many subnets and hosts per subnet can you get from the network 172.26.0.0 255.255.254.0?

thx to you i know how many host can have this subnet

255.255.254.0 = 11111111.11111111.11111110.00000000 we have 9 0 so 2^9 = 512 -2 = 510 host

Answer: 128 subnets and 510 hosts

but i dont know how to find the subnets number.

Convert the mask to binary:
11111111.11111111.11111110.00000000

To Find Subnets: Since this is a Class B IP, you know the first two octets are reserved. So count the 1s in the last two octets. You have 7. 2^7=128 so you have 128 available subnets.

thehourman

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html
Thanks to lordflashheart technique.

172.26.0.0 255.255.254.0
To find the subnets
The IP is in class B, so the default subnet is 255.255.0.0 or 11111111.11111111.00000000.00000000.
As you can see you already added 7 bits to the class B mask, and made it 255.255.254.0 11111111.11111111.11111110.00000000.
Since you already know how to get the host, do the same thing finding the subnet. Instead of counting 0s, count the extra 1s. The question you have provided, there are 7 extra 1s or bits
2^7 is 128. 7 is the number of bits you added to the class B mask.

For example, 192.168.1.0 255.255.255.252
The 192.168 is in class C, so the default mask is 255.255.255.0; but the mask in the example is 255.255.255.252. So in the example, they added 6 bits to the class C mask and that makes it 252.
Class C: 11111111.11111111.11111111.00000000 <-- 255.255.255.0
Classless: 11111111.11111111.11111111.11111100 <-- 255.255.255.252

Therefore, the formula is 2^x where x is the number of bits you added to the mask. In this case is 6 bits.
2^6 = 64. So there are 64 subnets in 192.168.1.0 255.255.255.252
About the host, we have two 0s left, so 2^2 - 2 = 2 hosts per subnet

256-252 = 4 (where 4 is the block size)

[B][U]What if they give me the subnet mask in dotted decimal?[/U][/B]
 
If you're lucky and they give you a mask in dotted decimal format then  you should have an even easier time. All you need again is your block  size.
 
Let's say they have given a mask of 255.255.255.248 and you wish to know  the block size. Here's the technique:
 
1. Starting from the left of the mask find which is the first octet to  NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size  e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step  1 as you have learned above (the example above would be in the last  octet).
 
Another example is a mask of 255.255.192.0 - you would simply count up  in 256 - 192 = 64 in the third octet.
 
One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the  second octet.

It would be like this:
192.168.1.0 (1st subnet)
192.168.1.1 (1st host)
192.168.1.2 (last host)
192.168.1.3 (broadcast)
192.168.1.4 (next subnet)
192.168.1.5 (1st host)
192.168.1.6 (last host)
192.168.1.7 (broadcast)
192.168.1.8 (next subnet)
so on and so on.

Monkerz

I thought he wanted to know how to find the 128 subnets...

gouki2005

another one guys even with the subnetting made easy questions like this kill me is a class A subnet so what advice can you give me to "beat" questions like this.

What is the last valid host on the subnetwork 10.31.16.0 255.255.240.0?

Monkerz

If you break the mask down to binary, you get:
11111111.11111111.11110000.00000000
What bit does the rightmost 1 land on? It lands on 16 so your increment is 16. I like to state this as i=16. Now find the subnets…
10.31.0.0 -- 10.31.15.255
10.31.16.0 -- 10.31.31.255
10.31.32.0 -- 10.31.47.255
10.31.48.0 -- 10.31.63.355
When you look at this table your subnet’s last IP is 10.31.31.255, this is not a usable IP because it is the Broadcast IP of that subnet. So your last usable(host) IP address for that subnet would be 10.31.31.254.

gouki2005

i understand i can do subnettingquestions.com withou problems using 20-30 seconds for each one thx to subnetting made easy and your explains thx

Stuppored

Can anyone link me a site that deals strictly with questions like these... i like them cause they're more of a challenge... but subnettingquestions.com doesnt have enough of em to satisfy me: I have this ip address 172.20.0.0 255.255.0 we need 30 subnet with 500 user in each subnet... please post a link!

wbosher

Stuppored wrote: »

Can anyone link me a site that deals strictly with questions like these... i like them cause they're more of a challenge... but subnettingquestions.com doesnt have enough of em to satisfy me: I have this ip address 172.20.0.0 255.255.0 we need 30 subnet with 500 user in each subnet... please post a link!

Google.com

But seriously, if you've got the Cisco Press books they come with the Boson practice exams. There's heaps of subnetting questions there.

Project2501

Stuppored wrote: »

Can anyone link me a site that deals strictly with questions like these... i like them cause they're more of a challenge... but subnettingquestions.com doesnt have enough of em to satisfy me: I have this ip address 172.20.0.0 255.255.0 we need 30 subnet with 500 user in each subnet... please post a link!

I'd rather not post a link here as I feel like I'm spamming. I sent you a PM instead.

Monkerz

How can something that tastes so good, get a bad rep from something it had nothing to do with? Why call it spamming? Why not Brussel-Spouting....or in-laws-ing....you know what I mean?