First of all get the subnet mask.
Since 4 bits are being used, which translates into 128+64+32+16 = 240
So our mask is 255.255.240.0
Since we want to use the other four bits to create 14 sub-subnets => Thus
resulting in the subnet mask of 255.255.255.0 for the new sub-subnets(just for
the clarification).
Because the given subnet is 172.16.176.0 the host range is 172.16.176.0 -
172.16.191.255
Since we sub-subnetted again the new subnets will be
172.16.176.0 ignoring this one
172.16.177.0
172.16.178.0
172.16.179.0
172.16.180.0
172.16.181.0
172.16.182.0
172.16.183.0
172.16.184.0
172.16.185.0
172.16.186.0
172.16.187.0
172.16.188.0
172.16.189.0
172.16.190.0
172.16.191.0 ignore this one
(2n-2)=#of subnets thus giving us the 14 subnets
____________________Reply Separator____________________
Subject: Subnet question
Author: Hunt Lee <[EMAIL PROTECTED]>
Date: 1/28/2001 2:09 PM
Can anyone please explain to me how to derive the answer of this
question?
A company has been assigned a subnet of 172.16.176.0/20, and wants the
next four available bits to create 14 subents, each containing an equal
number of hosts. Which of the following could represent one of these
subnets?
A) 172.16.255.0/24
B) 172.16.193.0/24
C) 172.16.183.0/24
D) 172.16.16.0/24
E) 172.16.0.0/24
F) 172.16.190.0/24
Answer is C and F
Regards,
Hunt Lee
IP Solution Analyst
Cable and Wireless (Sydney)
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