Equipment Support > Hardware By Brand > Cisco > [Other] Help Subnetting? CISCO! CCNA

reply to connellyg

Re: [Other] Help Subnetting? CISCO! CCNA

sharrahl... it wasnt 80 hosts, it was 80 subnets. Which cannot be done.

I think my large post above kinda showed that. :/

Regards,

Lee

reply to connellyg
Here's a nice link:

»www.silverdragon.com/punkie/howt···ing.html

reply to connellyg
For future reference here is a another way to VLSM. I posted on over at the networking forum. Maybe something we can add to the FAQ here.

»[Other] Vlsm

DP
--
"Knowledge is contagious, infect"

Thanks for the suggestion :

»Cisco Forum FAQ »Quick and Easy Subnetting on Routing, Switching and Network Design Relationship

reply to connellyg
That is right .15 would be a brodcast 9 is the first host on the .8 network and .14 is the last .15 is the brodacast and .16 is the next avaiable network then .24 and .32 and so on. The magic number is 8

Jimbo694 please consider the benefits and join our community? Register, if you have not done so already, and the site will remember which forums are your favorites, and which threads you have posted in. Our site instant message feature will allow you to contact other members more easily, and more features open up.

Registration only takes a second and you can start using the site immediately, no need to wait for verification email before logging in. If you are not already logged in, just scroll the page way up and look to the right, at the top, to see the registration link.
--
Even if only one cure is to be found, wouldn't you like to help?
Join us won't you?
Each moment is a chance to turn it all around.

reply to Jimbo694
Looks like this thread will never die...

Only joking! I might as well give my input since it seems that the whole world and its dad have.

First of all, you get the mask for the particular IP address you are allocated, e.g. the mask for network address 172.16.1.0 would be 255.255.0.0.

With me so far, good!

Now, you are given a task where you have to subnet that network address into 50 subnets and a minimum number of 100 hosts.

As you all should know, there are 4 octets (4 groups of 8 bits) within a mask.

From the example above, the mask would be:

255.255.0.0 = 11111111.11111111.00000000.00000000

The combined total of 11111111 is 255, hence the 255 mask.

Now, since we need 50 subnets and 100 hosts minimum, we would look at the 2 last octets and use the two times table to derive the subnet, i.e. each bit corresponds to 2. Using the remaining last 2 octets which were part of the host address, we would multiply the bits (2x) by each other until we reach a number that is equal to or greater than the required subnets. Using the example:

each bit "0" = 2.

00000000 = third octet

2x2x2x2x2x2 = 64

Minus 64 from 2 = 62 subnets present.

We had to multiply 2 into it self 6 times to get a number equal to or greater than the subnets required, therefore, 6 bits would be changed to 1s:

11111100

Which when converted to decimal is: 252.

Now you know that subnet mask 255.255.252.0 gives you over 50 subnets but you are meant to have 100 hosts minimum too. Well, since an octet field of all 0s is equal to 254 hosts (2x2x2x2x2x2x2x2=256- 2), you also satisfy those requirements. In actual fact, since you have 10 host bits (last 2 bits from the third octet and 8 bits from the fourth), you have 1022 hosts.

Hope this clarifies things.
--
If only my employers can see how much effort I put into this forum. They would then understand why I sleep at my desk.