Network guys, help with test question.
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
/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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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.
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.
1 network bit
1 broadcast
__________________
no sig
no sig
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.
The second part of my test for networking was take-home so I wanted to make sure I got everything right.
__________________
no sig
no sig
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.
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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