how to design this network?

October 23, 2012 at 07:00:30
Specs: Windows 7
i have 1 router, the default gateway is 172.16.1.1
and this router will connect to 18 access point.
and each access point need 30 usable host...
how to design this netwotk, what subnet should i use...
there is only 1 router, so just have only 1 default gateway,

it is if the network have too many host, the speed will slow down, because they need wait others host to broadcast?


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#1
October 23, 2012 at 07:13:54
We make it a habit to not help people do their homework here at Computing.net. We feel you'll learn more if you do it yourself.

I would like to point something out to you. If you plan on working in the IT industry you're going to be expected to show better use of capialization that you did above. You're not texting, you're in a venue where you should at leaste attempt to seem literate. It's very unprofessional to type without caplitalization, proper punctuation, spelling, and use of grammar. You wouldn't type like that on your resume, so don't do it here either. It's hard to take someone serious who appears to be barely literate.

Now, having said that, if you attended classes regularly, paid attention in class, took adequate notes and did homework every night, you'd know this isn't a tough question and should only take about 5 minutes (tops) to figure out.

It matters not how straight the gate,
How charged with punishments the scroll,
I am the master of my fate;
I am the captain of my soul.

***William Henley***


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#2
October 23, 2012 at 07:39:00
Since you need 540 hosts you will need to supernet. You will need a router that supports supernetting.

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#3
October 23, 2012 at 08:05:31
Why would you need to supernet with a Class B address range?

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#4
October 23, 2012 at 09:27:28
I hate to give the answer away, but I suspect the OP won't understand anyhow so here goes:

Because you're working in the third octet, not the second.

http://www.firewall.cx/networking-t...


It matters not how straight the gate,
How charged with punishments the scroll,
I am the master of my fate;
I am the captain of my soul.

***William Henley***


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#5
October 23, 2012 at 10:30:38
But 172.16.x.x is a Class B network. That the default gateway is at 172.16.1.1 is irrelevant. That works with a subnet mask of 255.255.0.0 (or several other masks - you wouldn't have to set it that wide for such a small number of computers).

The gateway often occupies one of the extreme ends of the available address range, but there's nothing that says it has to. So why the need to supernet? Subnetting sounds more like it (although even that is not necessary).


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#6
October 23, 2012 at 13:14:19
255.255.0.0 = /16 = 65,534 addresses

That's a bit excessive for the required 540. Since the closest you can get, wasting the least number of addresses possible is /22 that means you are masking in the third octet.

So, is it still a Class B network given you're now masking in the third octet?

:)

It matters not how straight the gate,
How charged with punishments the scroll,
I am the master of my fate;
I am the captain of my soul.

***William Henley***


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#7
October 23, 2012 at 13:37:19
Yes. It's a subnetted Class B network. Why would a router have a problem with that?

Anyway, it's a private network range - there are plenty more where that came from.


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#8
October 23, 2012 at 15:40:20
lots of soho routers don't support super or subnetting
I suggested the supernet of a class c because it fits the criteria of the amount of ips needed.

It is also the common test answer when presented with this scenerio :-)

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#9
October 23, 2012 at 23:06:58
But there are no Class C networks to supernet here. 172.16.1.0-172.16.1.255, for example, isn't a Class C network.

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#10
October 24, 2012 at 11:33:18
What lan ip range you use is not set in stone.

As Curt R pointed out you can still do the /22 for the class b address which results in 1022 ip addresses with a range of 172.16.0.0 - 172.16.3.255.

172.16.1.0-172.16.1.255 range doesn't yield 540 ip addresses.

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#11
October 24, 2012 at 11:35:43
Exactly. No supernetting involved, just straightforward subnetting.

I didn't say that the example I gave yielded more than 253 addresses; I just said it isn't a Class C network. It isn't.


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