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IP range

Posted on 2011-02-19
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Last Modified: 2012-05-11
Hi all,

It's about that time of the year when I get round to trying to complete my 291 exam (swear i'm going to do it this time lol)...and no it's not home work it's me trying to learn myself  (hence why i keep giving up at 291 and starting over)

Anyway I'm stuck on the IP range section. I've been given a No of hosts (3800) and a no of servers and printers which will use reservations (100)

From this i've got the mask 255.255.240.0 - This is no problem I know how to do this

But they've also asked for the start and end IP. From the info they've provided there is no range to work with. It's a choice, and I need to work out how they arrived at their solution

They've said the solution is; start IP = 10.0.0.101 and end IP = 10.0.15.160

Other available answers were 10.0.24.160 and 10.0.10.101

Can someone please explain how they arrived at their answer. Why 10.0.0.101 and not 10.0.10.101. And how is it 10.0.15.160 the end IP and not 10.0.24.160?
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Question by:LFC1980
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by:Jon Snyderman
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So your subnet is 8 bits, 8 bits, 4 bits, 0 bits.  
So the 10 and the 0 are hard and fast part of your network portion of the address.  The last octet of 0 bytes with 20 bit subnet makes it part of the host address.

The question is the 3rd octet with 4 bits.  So your third octet options are
0-15 = 00000000 - 00001111  This works because the first four bits are all zeros. and the last four are all zeros and all once.

10 would not be logical because it doesnt fall on a even break of bits.  HOST ranges always go from all 0's (the network address) to all 1's (the broadcast address.  10 = 00001010.  With a subnet of 20 bits (255.255.240.0), this falls right in the middle of a range.  

24 would not be logical for the same reason, but in the next range up.  The 3rd octet in the next range would be .16-.31 putting 24 right in the middle.  

Make sense or did I confuse you more?  
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kdearing earned 250 total points
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OK, a subnet of 255.255.240.0 will give you 4096 addresses (4094 usable).

Now for a valid IP range, it's just math.

Assuming you're using the 10.x.x.x range:
You need 100 reservations, these are normally at the beginning.

So we'll start at 10.0.0.101; that's the start IP

One way to do it is to convert to binary

00001010 . 00000000 . 00000000 . 01100101  =  start IP
00001010 . 00000000 . 00001110 . 11011000      add 3800
00001010 . 00000000 . 00001111 . 00111101  =  end IP  10.0.15.61

I assume this was a multiple choice question and you had to pick one that met requirements.
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Expert Comment

by:pwindell
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Fast way to look at it is this:
No "binary stuff" required

Mask is 255.255.240.0
Subtract 240 from 256 and you get 16 for the 3rd Octect
The zero counts as a number, so the Range is:
x.x.0.x thru x.x.15.x
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