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Network guys, help with test question.

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Old Mar 6, 2006 | 01:59 PM
  #11  
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Originally Posted by Driverman5777
Why do you have to do this?
my guess is he is taking a test
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Old Mar 6, 2006 | 02:19 PM
  #12  
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D because....

/20 = 255.255.240.0
256 - 240 = 16

so you know your networks are in subnets of 16
16, 32, 48, 64........208, 224

172.16.208.0 < 172.16.213.16 < 172.16.224.0
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Old Mar 6, 2006 | 02:20 PM
  #13  
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Originally Posted by Pete
The IP address 172.16.213.16/20 is a host address in which of the following subnets:

A) 172.16.213.224
B) 172.16.213.240
C) 172.16.0.0
D) 172.16.208.0
E) 172.16.176.0

All I can figure out is that the subnet mask is 255.255.240.0 and that there are 12 bits reserved for hosts (I think). Can someone help me out? Thanks.
14 bits reserved for hosts
1 network bit
1 broadcast
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Old Mar 6, 2006 | 02:57 PM
  #14  
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Makes sense now. Thanks guys.

The second part of my test for networking was take-home so I wanted to make sure I got everything right.
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Old Mar 6, 2006 | 09:31 PM
  #15  
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Originally Posted by Pete
Makes sense now. Thanks guys.

The second part of my test for networking was take-home so I wanted to make sure I got everything right.
np
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Old Mar 6, 2006 | 09:36 PM
  #16  
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Its D.

Because

/20 = 255.255.240.0
256 - 240 = 16 x 2 = 32 / 2 = 16

so you know your networks are in subnets of 16
16, 32, 48, 64........208, 224 and so on

172.16.208.0 < 172.16.213.16 < 172.16.224.0

Let me know if you want me to go into more detail.
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Old Mar 6, 2006 | 09:38 PM
  #17  
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fuggin will :rofl:
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Old Mar 6, 2006 | 09:41 PM
  #18  
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that stuff was easy...but I forgot how to do it. h:
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